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http://physics.stackexchange.com/questions/19237/why-is-it-concluded-that-the-cosmos-is-expanding-when-in-fact-the-observations-a
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# Why is it concluded that the cosmos is expanding when in fact the observations are for events further back in time?
Why is it concluded that the cosmos is expanding at an ever increasing rate when looking further out and therefore further back in time?
Surely, if the time were reversed to play forwards, the conclusion would be that the expansion of the universe is actually deaccelerating, and the need to postulate 'Dark Energy' would be avoided.
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## 2 Answers
What we directly observe is that the Universe was expanding and the expansion was accelerating during the recent five billions of years or so (it was actually not accelerating before that because the dark energy wasn't dominating). The prediction that it will continue to expand and accelerate results from a "clever scientific extrapolation" – from writing the most convincing equations that describe the past and from solving them in the future.
The past observations imply that the Universe is expanding because we see that distant galaxies were moving away from us in recent billions of years (because the light is redder – has a lower frequency – than it should have according to spectroscopy, a fact we attribute to the Doppler effect). If things are moving away from each other, their distance is increasing as you go from a smaller $t$ to a larger $t$ and this increase is called expansion, by definition. Your argument by which you want to confuse expansion and contraction looks utterly bizarre to me and I am pretty sure that I won't be the only one. Something that looks and behaves like a bomb after it detonates is called "expansion" and something that behaves like the rubber after we create a hole to a rubber balloon with a needle is called contraction or implosion or shrinking and most people know how to distinguish these two processes without asking questions on Physics Stack Exchange.
There's a clear difference between expansion and contraction, a difference that even most of the laymen and kids from the kindergarten probably appreciate. These two opposite processes lead to different observations and one may clearly say that the Universe has been expanding for quite some time.
In an analogous way, one may determine that the expansion was accelerating. This is found by reconstructing the acceleration rate that was valid "very recently" and the acceleration rate that was valid "billions of years ago": one needs to look at objects at very different distances to get the rates at at least two moments in this case. When the two rates are found, we see that the rate (=speed) of the expansion in recent years was faster than the rate (=speed) that applied billions of years ago. This increase of the rate (=speed) is what is called acceleration, by definition.
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Perhaps I could appeal for somebody else who understands my question better to reply by throwing more light on it – DDD Jan 8 '12 at 12:54
I think DDD's point was that if we see more distant objects moving faster but more distant objects are also older then that is saying that the universe is slowing - older=faster, younger=slower! In fact the expansion of the universe doesn't quite work likethat – Martin Beckett Jan 8 '12 at 18:02
How does it work then? – DDD Jan 8 '12 at 18:27
Of course I should have said that my conclusion is that the expansion of the universe is decelerating not contracting (which may come later) – DDD Jan 9 '12 at 9:22
@DDD: if that's what you meant, edit your question to say so. It'll be easier to give you a helpful answer that way. – David Zaslavsky♦ Jan 10 '12 at 0:05
show 3 more comments
You should try to think of the Universe expanding in terms of the space itself expanding. Not objects moving away from each other in a fixed space. The accelerated expansion refers to the way space in between objects is growing. This is where quantities like proper and comoving distances become very important. If you want more details on that I would refer you to Friedman equations and the scale factor therein. The bottom line is, the space itself between objects is growing therefore the Universe is expanding. The rate at which it is growing is also a positive quantity therefore it is accelerating as well.
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But if it is say expanding x times faster y lightyears away than we see here, why isn't it thought that we are looking at the speed it was y years ago and we are just seeing the faster rate of expansion it used to be. But in actual fact if you could see y light years away in the present time the rate of expansion would be the same as here – Jonathan. Jan 10 '12 at 0:41
@Jonathan. When we observe a large distance, high redshift, away we are observing the rate of expansion in the past. However, I don't know what you mean by "... if you could see y light years away in the present time ...". When you see y light years away, you are observing the past. The expansion rate of the Universe, or more precisely the scale factor, is a function of time and therefore has changed over the history of the Universe. – Omar Jan 10 '12 at 1:49
Exactly so the rate of expansion is not what we are seeing. We are seeing the rate of expansion 10 years ago if looking 10 light years away. In that 10 years the rate of expansion could hava slowed. – Jonathan. Jan 10 '12 at 15:38
– Omar Jan 10 '12 at 16:44
surely the rate of expansion is the same thing as the acceleration? – Jonathan. Jan 10 '12 at 16:56
show 1 more comment
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http://nrich.maths.org/2670/solution
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Where Can We Visit?
Charlie and Lynne put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
Gr8 Coach
Can you coach your rowing eight to win?
Babylon Numbers
Can you make a hypothesis to explain these ancient numbers?
Arithmagons
Stage: 3 Challenge Level:
Esther did some wonderful "detective work" and made a number of very useful observations:
1) The totals in the squares are double the totals in the circles. I think this is because each circle number is used twice to solve the puzzle.
2) If the numbers in the squares are all even, the numbers in the circles are either all odd or all even.
If the numbers in the squares are all odd, I cannot solve the puzzle with whole numbers. This is because an even number minus an odd number equals an odd number.
If you have two odd and one even in the squares then the numbers in the circles are either two odd and one even or two even and one odd.
However, if you have two even and one odd number in the squares you cannot solve the puzzle using whole numbers.
3) If you draw the lines of symmetry through a circle and the opposite square, the difference in the numbers in the squares on either side equals the difference in the numbers in the circles on either side.
Also the sum of the numbers on each line of symmetry is the same. This also works for decimals and negative numbers.
4) I found a rule for working out the numbers in the circles. You add the numbers in the squares next to it, subtract the number in the opposite square, and divide the answer by two. This is why you sometimes get decimals and negative numbers. I have only ever found one solution to each puzzle using my rule.
I think you can always get answers as long as you can use decimals and negative numbers using my rule as zero can always be made of +2 and -2 for example.
Harriet and Laura from The Mount School offered the following explanation:
We found that when you added together the numbers in the squares, (the ones already given) it totalled twice the accumulated amount of the numbers in the circles. The reason for this can be shown using algebra and is illustrated in the diagram below.
You can find the numbers being represented by $a$, $b$ or $c$ if you have numbers in the squares.
We will show this by finding b using an example found on the website:
Using the equation $(b + a) + (b + c) = 2b + a + c$
you can take away $a$ and $c$ to find $2b$:
$(2b + a + c) - (a + c) = 2b$
So $(b + a) + (b + c) - (a + c) = 2b$
$2b = 12 + 4 - 10 = 6$
$b = 3$
This also works with negative numbers.
Robert from Leventhorpe School used a similar analysis:
To work out the formula for working out these arithmagons, we must first substitute the numbers as such:
If we know $b, c$ and $e$, then we can say that
$a + d = b$,
$d + f = e$ and
$a + f = c$
In the first one, $b = 12, c =10$ and $e = 4$,
so $a + d = 12$, $d + f = 4$ and $a + f = 10$
If we then add all of these together,
we get $(a+d) + (d+f) + (a+f) = 12 + 4 + 10$
We do not need the brackets,
so we can make it $a + d + d + f + a + f = 26$
This can be cancelled down to $2a + 2d + 2f = 26$
If we then divide everything by $2$ we get $a + d + f = 13$.
Because we know that $a + d = 12$, $f$ must be $1$.
Since $d + f = 4$, $d$ must be $3$.
Finally, we can say that a must be $9$.
Double checking this, it works. Take a look:
Tom from Colyton Grammar School also used some algebraic thinking to analyse the problem:
If the circles were $z, y$ and $x$ from the top clockwise and the squares are $q, r$ and $p$ from the left clockwise then
$x + z = q$,
$z + y = r$
and $x + y = p$.
$q - r = x - y$
so $q + p - r = 2x$
so $x = (q + p - r) / 2$.
This means that $y = ( p + r - q ) / 2$
and $z = (q + r - p) / 2$, by symmetry.
Shaun from Nottingham High School came to the same conclusion:
It struck me this problem would be most easily investigated using simultaneous equations as a means of deriving a common rule for all arithmagons.
I will call the three circles of the arithmagon $a, d$ and $f$.
Hence, each squares can be expressed as the sum of two of the variables.
Since we have three variables, and three equations involving two of them each, we know it can be solved this way.
Some preliminary doodlings proved this to be true.
Let the squares be defined as follows:
(1) $a + d = b$
(2) $d + f = e$
(3) $f + a = c$
Therefore:
(4) $b + f = c + d$
(5) $b + f = e + a$
(6) $e + a = c + d$
From (4): $f = d + c - b$
Combining this result and (2):
$e - d = d + c - b$
$2d = e - c + b$
$d = (b + e - c)/2$
And so, using this formula, the value in a circle, when the three squares in the arithmagon are known, can be found.
Of course, the variable $b$ can be used to denote any of the three circles, and $x$ and $y$ the adjacent squares (does not matter which), and $z$ the opposite.
Following these rules, the formula can also be written as:
$a = (b + c - e) / 2$
$c = (c + e - b) / 2$
I think the process I have gone through above explains why all arithmagons can be completed - they all exhibit these attributes, and so can all be solved in the same way.
Lindsay's solution is here.
It explains why Tom from Cottenham Village College found that:
If you add up all of the numbers in the squares and halve it, then subtract the number on the opposite side to a circle (the number in the square) you will get the number that goes in the circle.
Aurora, from the British School in Manila, explained her strategy clearly:
Firstly, let's call the bottom left vertex z, the top one x, and the bottom right y.
Assume:
x+y= 17
y+z= 20
z+x= 15
If x+y is 17, and z+x is 15, the difference between z and y must be 2:
z+2=y
y+z= 20
z+z+2 =20
z=9
If z=9, y is 11 and x is 6
Here is how John and Karl, from King's School Grantham, explained how they derived a formula for finding the values of the vertices.
Here is how Charlotte, from Llandovery College, summarised two of the most popular strategies, and here is how Krystof, from Uhelny Trh in Prague, applied his strategy.
We also received good solutions to this problem from Leighton, Sia Jia Rui from Raffles Institution in Singapore, Bhavik, Brenna from Bream Bay College, Heer, Sarah & Heledd from St Stephen's Carramar in Perth, Alex and Luke from Llandovery College in Wales, Sam from Shrewsbury House School, Robert from West Hoathly Primary School, Amy from Hanham High School, Ryan and Alisha from Lacon Childe School, Lauren from St.Peters Primary School and Michael, Alexander, Jake, Hussein and Charlie from Wilson's School.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://mathhelpforum.com/algebra/9887-anybody-good-logs.html
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# Thread:
1. ## anybody good with logs??
i have log_L (1-r)
with L = 2 and r = 0.02
does any1 know how to put log to the base into a calculator or any other way of calculating it?
2. You appear to have $log_{2}(1-0.02)=log_{2}(0.98)$
When entering into a calculator, use the change of base formula.
$log_{b}(u)=\frac{log_{a}(u)}{log_{a}(b)}$
In this case, $\frac{log(0.98)}{log(2)}\approx{-0.02914}$
Do not confuse the change of base formula with the quotient rule for logs.
3. Originally Posted by galactus
You appear to have $log_{2}(1-0.02)=log_{2}(0.98)$
When entering into a calculator, use the change of base formula.
$log_{b}(u)=\frac{log_{a}(u)}{log_{a}(b)}$
In this case, $\frac{log(0.98)}{log(2)}\approx{-0.02914}$
Do not confuse the change of base formula with the quotient rule for logs.
having Tr = 1/1-log_L(1-r)
with L = 2 and r = 0.02
i get this to be equal to 0.97168 and i know the answer is approx 100 so should my answer be multiplied by 100?? is this the rite procedure?
4. No. Where does the 100 come from?.
$log_{2}(1267650600228229401496703205376)=100$
5. the answer has to be 100 years its a return period for a flood
6. Originally Posted by question
the answer has to be 100 years its a return period for a flood
Perhaps you should post the actual question you are trying to answer.
RonL
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http://mathoverflow.net/questions/18817/does-2m-3n-r-have-finitely-many-solutions-for-every-r
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## Does 2^m = 3^n + r have finitely many solutions for every r?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is it true that for every integer $r$, the equation $2^m = 3^n + r$ has at most a finite number of integer solutions? I understand that this is a special case of Pillai's conjecture, which is unsolved.
If the statement is true, then can we verify the finiteness of the solution set using modular arithmetic? To be precise, is the following proposition true?
$$\forall r,\ \exists M,\ \exists N,\ \forall m,n \ge N,\ \ 2^m \not\equiv 3^n + r \pmod{M}$$
I have verified the proposition for $0 \le r \le 12$, and found the least possible modulus $M(r)$ for each $r$ in this interval. Note that $M(r) = 2$ if $r$ is even.
$M(1) = 8$, $M(3) = 3$, $M(5) = 1088$, $M(7) = 1632$, $M(9) = 3$, $M(11) = 8$.
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2
I have extended the calculation to $0 \le r \le 100$. M(r) = 3 for r = 3 (mod 6); M(r) = 8 for r = 17, 19, 25, 35, 41, 43, 49, 59, 65, 67, 73, 83, 89, 91, 97; M(r) = 60 for r = 31, 53, 79, 85, 95; M(13) = 131584; M(23) = 1088; M(29) = 117; M(37) = 21951; M(47) = 65972; M(55) = 999; M(61) = 252; M(71) = 63; M(77) = 28. – Dave R Mar 21 2010 at 5:46
## 2 Answers
Yes, it is true that this kind of equation ax+by=c, where a,b,c are non-zero and fixed and x,y are allowed to only have prime factors in a finite set, has only finitely many solutions. This is a special case of Siegel's theorem on integral points on curves.
Your second question may be unknown in the generality you pose. It is interesting that it holds. A remark: if there is a solution to $2^m = 3^n + r$, then $2^{m+k\phi(M)} \equiv 3^{n+k\phi(M)} + r (\mod M)$ for all $k,M$ if $(M,6)=1$, so if $M$ exists in this case, then $(M,6)>1$. If there is no solution to the equation $2^m = 3^n + r$, then the existence of $M$ (with $N=0$) is a special case of a conjecture of Skolem.
T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad. Oslo, 12 (1937), 1–16.
Another comment. There are no solutions when $r=11$ but $M=8$ doesn't work since $2^2 \equiv 3^2 + 11 \mod 8$. $M(11)=205$. (Edit: $M(11)=8$ is OK. I misunderstood the definition, see comments)
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Thanks you for your very helpful comments. I am intrigued by the result concerning ax+by=c . I should really learn about elliptic curves! But $M = 8$ does work when $r = 11$, since we may assume that $m \ge 3$. – Dave R Mar 20 2010 at 3:34
What you say about M=8,r=11 doesn't make sense. You can also take m=6, if you don't like m=2. – Felipe Voloch Mar 20 2010 at 3:50
2^6 = 0 (mod 8). 3^n + 11 is never divisible by 8. – Dave R Mar 20 2010 at 3:53
Felipe, there seems to be some confusion over the definition of $M(r)$. You only need $2^m\not\equiv3^n+11 (\mod 8)$ for sufficiently large $m$ and $n$ to conclude that $M(11)\leq8$. – Jonas Meyer Mar 20 2010 at 7:30
OK, I understand now, M=8 shows that the set of solutions is finite and M=205 that is empty. What I said about m=6 doesn't make sense. – Felipe Voloch Mar 20 2010 at 12:18
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I have no comment on your methods, and I know very little about this, but that case of Pillai's conjecture appears to have been solved in the 80's by Stroeker and Tijdeman [Edit: see below]. Here's a paper by Bennett from 2001 that shows more: http://www.math.ubc.ca/~bennett/B-CJM-Pillai.pdf. In particular, the number of solutions is at most 2 for each fixed $r$. More generally, Bennett shows that for fixed integers $a\geq2$, $b\geq2$, and $r\neq0$, there are at most 2 solutions $(m,n)$ to the equation $a^m=b^n+r$. The more general form of Pillai's conjecture allows $a$ and $b$ to vary and appears to still be unsolved.
Edit: What Stroeker and Tijdeman actually did was sharpen the result by showing that except when $r$ is in `$\{-1,5,13\}$`, your equation has at most one solution, and that in the exceptional cases it has two. The finiteness of the set of solutions $(m,n)$ to the equation $a^m=b^n+r$ had long been known, and Pallai himself gave some quantitative results on this using Siegel's Theorem. For finiteness alone without quantification, Bennett cites this 1918 Polya paper. My source for all of this is Bennett's paper.
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What I said about Stroeker and Tijdeman is misleading (I will edit to clarify) and Felipe Voloch is right on with Siegel's theorem. Felipe also addressed the second question, and I'm not sure why this answer was accepted. – Jonas Meyer Mar 20 2010 at 6:27
I selected your answer because I was intrigued by Bennett's result, and because I could not accept both answers. I am grateful for both of your contributions. – Dave R Mar 20 2010 at 7:17
Thanks for explaining. I guess it seemed strange because I was merely interested and did some internet searching, whereas Felipe wrote based on his expertise. Also, I have nothing useful to say about your second question. – Jonas Meyer Mar 20 2010 at 8:07
I don't want this to be a point of contention, so I have accepted Felipe's answer instead. Thanks again for your help. – Dave R Mar 20 2010 at 9:18
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http://math.stackexchange.com/questions/180653/subsets-of-not-connected-topology/180657
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# Subsets of not connected topology
I was taking a look at this book. It says (p 102) that
If a topological space (X, T) is not connected, than X contains a subset U such that U isn't X, U isn't empty and U is clopen.
The book brings no proof of the fact. How does this imply?
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## 2 Answers
This follows immediately from the fact that if $X$ is not connected, we can write
$$X = U \cup V$$
with $U$ and $V$ disjoint, not empty and both open in $X$. Since they are disjoint and their union is the whole space, we have $U = X- V$ and so $U$ is closed at the same time.
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The answer depends on which definition of connectedness is being used. In this case it’s trivial, because the definition used in this book is as follows:
3.3.4 Definition. Let $(X,\tau)$ be a topological space. Then it is said to be $\color{red}{\text{connected}}$ if the only clopen subsets of $X$ are $X$ and $\varnothing$.
In other words, $X$ is not connected if and only if it contains a clopen set other than $X$ and $\varnothing$.
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http://mathhelpforum.com/calculus/112272-need-help-some-question.html
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# Thread:
1. ## Need help with some question
A company has found that if it produces x thousand units/day of a product, The total cost would be T(x) thousand \$, where T(x) = x^3 - 6x^3 + 13x + 15, when 0<x<9. They sell it for 28 \$/unit.
a) State the profit V(x) thousand \$ as a function of x.
b) Find the maximum profit per day.
2. I'm assuming you meant to type $T(x)=x^3-6x^2+13x+15$
So profit is sales-cost. Therefore, $V(x)=28000x-(x^3-6x^2+13x+15)=-x^3+6x^2+27987x-15$
To find the maximum, you must first differentiate giving you $V'(x)=-3x^2+12x+27987$.
Then find the critical numbers by setting V'(x) to 0.
$0=x^2-4x-9329$
Quadratic formula gives you the critical values of x=98 and x=-94 but those aren't in the interval (0,9). So the maximum is when x is either 0 or 9. You can eliminate 0 because they would obviously make no profit if they didn't make a product. So plug x=9 into the profit equation.
$V(9)=-9^3+6*9^2+27987*9-15=-729+486+251883-15=251625$
The maximum profit is \$251625
3. The answer to a) was -x^3 -6x^2 +15x -15. Thats because they used 28x, instead of 28000x.
After that i couldn't get the answer to b), they get 85 000. I dont really understand how.
4. So $V(x)=-x^3-6x^2+15x-15$
First you need to find the critical numbers and see if they are in your interval (0,9). To find the critical numbers, derive the equation and set it equal to 0.
$V'(x)=-3x^2-12x+15$
$0=x^2+4x-5=(x+5)(x-1)$
$x=1$ $x=-5$
So you have a critical value at x=1. To find the maximum, you must plug in the endpoints of your interval (i.e. x=0 and x=9) and the critical number within the interval (i.e. x=1) into the original equation and find which one is the highest. That is your maximum.
$V(0)=-15$
$V(1)=-1-6+15-15=-7$
$V(9)=-9^3+6*9^2+15*9-15=-729+486+135-15=-123$
Well, that's what I got and it doesn't make sense. Can anyone else help? I am pretty sure that I have the right method of getting the maximum. I just don't understand how the answer could be negative.
5. I have figured out where i was wrong! Thanks anyways!
6. ## Thank you very much!
Thank you very much!
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http://mathoverflow.net/revisions/114051/list
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## Return to Answer
2 significantly expanded
Some more references and quotes, and some commentary.
Gower and Wagstaff (Square form factorization, Math of Computation, 2008) in a paper on SQUFOF:
On a 32-bit computer SQUFOF is the clear champion factoring algorithm for numbers between $10^{10}$ and $10^{18}$, and will likely remain so.
More recently William Hart proposed a Fermat-variant he called A One Line Factoring Algorithm, which he describes as competitive with SQUFOF in practise (while only being $O(n^{1/3})$ ). In the respective paper, to be precise a preprint thereof I do not have the actual paper, he writes (J. Aust. Math. Soc. 2012)
Most modern computer algebra packages implement numerous algorithms for factoring. For number that fit into a single machine word Shanks' SQUFOF is popular as it has run time $O(n^{1/4})$ with a very small implied constant.
So, for the range asked for in the question I am quite confident that SQUFOF would be a good choice. It should however be noted, this is also discussed in Cohen's book, that as the numbers get larger (beyond the mentioned threshold) SQUFOF becomes unattractive while, eg, Pollard rho stays interesting. The rough reason for this seems to be that SQUFOF does not profit from 'small' factors, as opposed to, e.g. Pollard (cf Laurent Berger's answer). However, for numbers in that range and after trial divison (Cohen then suggested trial division by primes up to 500000) there are not that 'small' factors anyway.
As already pointed out a big plus for SQUFOF is that the involved numbers are only size $2\sqrt{N}$, in contrast to other methods requiring often $N$ or even $N^2$. This affects only the constant in the running time, but this is also important, and in addition in practise allows to get by with simple datatypes for longer.
1
Shanks's SQUFOF (Square FOrm Factorisation) might well be worth a detailed look.
Henri Cohen comments:
This method is very simple to implement and has the big advantage of working exclusively with numbers which are at most $2\sqrt{N}$ [...] Therefore it is eminently fast and practical when one wants to factor numbers less than $10^{19}$, even on a pocket calculator.
And $2^{60}$ is just below $10^{19}$; and indeed the $10^{19}$ should arise as the number such that $2\sqrt{N}$ still fits in an "int". For reference, the complexity is $O(N^{1/4+\varepsilon})$; but this is not the main point.
Typically, this would have to be preceeded with trial division for small factors and (possibly) some primality test.
In general, I recommend the relevant chapters of the book "A course in computational algebraic number theory" by Henri Cohen (from which the quote is taken); he typically in addition to theory and algorithms, also discusses practical aspects of implementation (the book being not recent some of them might have to be modified, but still the practical aspect and discussion of ranges is present).
Also, various other references of SQUFOF are easy to find via a web-search.
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http://mathoverflow.net/questions/92696?sort=votes
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Excellent uses of induction and recursion
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Can you make an example of a great proof by induction or construction by recursion?
Given that you already have your own idea of what "great" means, here it can also be taken to mean that the chosen technique :
• is vital to the argument;
• sheds new light on the result itself;
• yields an elegant way to fulfill the task;
• conveys a powerful and simple view of an intricate matter;
• is just the only natural way to deal with the problem.
Here induction and recursion are meant in the broadest sense of the words, they can span from induction on natural numbers to well-founded recursion to transfinite induction, and so on...
Elementary examples are especially appreciated, but non-elementary ones are welcome too!
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Structure Theorem for finitely generated abelian groups. – i707107 Mar 30 2012 at 17:02
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My favorite induction proof is the joke proof that all natural numbers are interesting. Clearly 0 is interesting. Suppose m is interesting for all $0\leq m\leq n$. If n+1 were not interesting, then it would be the smallest non-interesting number, which is pretty interesting, a contradiction. So n+1 is interesting. – Benjamin Steinberg Apr 2 2012 at 20:32
@Benjamin: Your proof has earned my first laughter of the day! Nice proof! – Lorenzo Lami Apr 3 2012 at 10:04
12 Answers
A Classic: Fix a positive integer $n$. Show that it is possible to tile any $2^n \times 2^n$ grid with exactly one square removed using 'L'-shaped tiles of three squares.
It serves as a wonderful introductory example to proof by induction. Indeed, the proof can almost be represented with two appropriate figures. Yet, for those just learning induction, it is a significant problem where the application of the inductive hypothesis is far from obvious.
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It's a great nontrivial exercise for those learning induction; it's also a nice, entertaining fact for those who already know induction well. Great answer! – Lorenzo Lami Apr 8 2012 at 22:16
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There's no reason to insist $n > 0$; works for $n = 0$ as well! – I. J. Kennedy Oct 28 at 16:54
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Tsirelson's space (1974) is a good example from Banach space theory. His space is the completion of a $c_{00}$ (all finitely supported scalar sequences) under an inductively defined norm. The base norm is the sup-norm $\|\cdot \|_0$.
For $n \in \mathbb{N}$ the norm $\|x\|_{n+1}$ norm is defined by
$\|x\|_{n+1}= \sup{\frac{1}{2} \sum^k_i \|E_ix\|_{n} }$
where the supremum is taken over all sets $(E_i)_{i=1}^k$, where $E_i$ is a finite interval in $\mathbb{N}$, $\max E_i < \min E_{i+1}$ and $k \leq E_1$ (here $Ex$ denotes the restriction of $x$ to the coordinates of $E$). The Tsirelson norm is $\|x\|_T = \sup_n \|x\|_n$ and satisfies the following implicit equation
$\|x\|_T= \max ( \|x\|_0 , \sup \frac{1}{2} \sum^k_i \|E_ix\|_T ).$
The space $T$ does not contain a copy of any $\ell_p$ or $c_0$. This solved a major open problem at the time (I should point out that Tsirelson actually defined the dual of $T$ which also has the property).
The, inductive, method he devised for producing this space eventually lead to the solutions of numerous problems in Banach space theory (way to numerous to mention). Moreover, the `necessity' of the inductive construction to produce spaces not containing any $\ell_p$ of $c_0$ is a problem that has been considered by Gowers as a polymath project (unfortunately not much progress here): http://gowers.wordpress.com/2009/02/17/must-an-explicitly-defined-banach-space-contain-c_0-or-ell_p/
Check out Boris Tsirelson's website for more info on his space: http://www.math.tau.ac.il/~tsirel/Research/myspace/main.html
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Definitely a great advanced example! – Lorenzo Lami Apr 8 2012 at 22:21
1)Proof of Euler's formula, V-E+F=2, with induction on F (number of faces).
2)Backward induction proof of generalized AM-GM inequality.
3)Proof of Heine-Borel theorem using Transfinite Topological induction.
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2
I first saw "topological induction" in one of the early drafts of Thurston's notes, around 1980. A more recent study of the idea is Iraj Kalantari's "Induction over the Continuum", which can be partly viewed at springerlink.com/content/uj314q217n7ln2n3 – John Stillwell Mar 31 2012 at 11:09
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Sorry, I might be dense, but as far as I understand, an induction that is "not based on an underlying well ordering" is just a bogus mathematical argument. – André Henriques Apr 1 2012 at 14:36
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@John I learnt this method of "topological induction" in Dieudonne's "Calcul infinitesimal", where he uses it to prove for example the uniform continuity of continuous functions on closed intervals. I actually used this in class, and students seemed to like it, visualizing the proof as a zipper that you can zip from a to b without being blocked in the middle... – Sylvain Bonnot Apr 1 2012 at 16:28
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@Andre: you are not correct, induction can be performed on any well-founded relation, it need not be linear. For example, proof theorists and computer scientists often use induction on well-founded trees. – Andrej Bauer Apr 2 2012 at 6:30
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Following up on "topological induction" or "induction over the continuum," there is a recent paper on it by Pete L. Clark at math.uga.edu/~pete/… He traces versions of the idea back as far as papers by Khinchin and Perron in the 1920s. – John Stillwell Apr 2 2012 at 7:09
show 6 more comments
The following famous puzzle is a great example:
http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/
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The $n$-level Tower of Hanoi can be solved in $2^n - 1$ moves.
Not only does induction prove this, it actually shows you the solution!
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Goodstein's theorem hasn't yet been mentioned. A straightforward-looking arithmetic theorem with a surprise proof using transfinite induction. Also (the main interesting characteristic of the theorem), there is NO proof from ordinary first-order Peano arithmetic. It's actually equivalent to the formalized $\Sigma^0_1$-soundness (aka 1-consistency) of PA.
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Nice pick, I should have thought about that (and about other statements not provable in PA)! – Lorenzo Lami Apr 3 2012 at 10:09
3
Goodstein's theorem, which is a $\Pi^0_2$ statement whose existential quantifier is witnessed by a function growing faster than any provably total recursive function of PA, is not equivalent to the consistency of PA, which is merely a $\Pi^0_1$ statement. I'll fix that. – Emil Jeřábek Apr 3 2012 at 10:35
A problem I enjoyed in my undergraduate algorithms course is as follows:
Suppose you have a computing machine with the following architecture. There are $k$ stacks (for some $k$), input can be pushed onto the first stack, output is popped off of the last, and intermediate operations pop from one stack and push to the next in a line. The top of the stack may also be inspected and compared. Given a permutation of ${1,\ldots,n}$ in order as input, how many stacks $k$ do you require to sort the permutation? Describe an algorithm that achieves this bound.
One can prove the bound ($\log_2 n$) by induction, and then just state that this gives a natural recursive algorithm. The same technique was useful for a couple of other problems in a similar vein.
I think this certainly fits the bill of an elegant way to fulfill the task (prove a bound and give an achieving algorithm) in a nice class of cases.
The problem is originally from Knuth Vol. 1, and stack sorting is further elaborated on in this survey.
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Simultaneous induction as used in combinatorial group theory, for example in the proof of the Adyan-Novikov theorem providing the counterexample to the General Burnside Problem:
Some nice references about the nuts and bolts of this were supplied by Mark Sapir in an answer to one of my questions about this proof:
http://mathoverflow.net/questions/48184/a-synopsis-of-adyans-solution-to-the-general-burnside-problem
Certainly the simultaneous induction technique is an important idea in constructing such monster groups.
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Let $P(p)$ = "there is no natural $q$ such that $(p/q)^2=2$". A simple induction argument shows that P holds for all naturals $p$ and hence that $\sqrt 2$ is irrational. All descent arguments are basically induction.
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Among infinite descents, why did you choose this one? – Lorenzo Lami Apr 1 2012 at 20:14
Is this not the canonical example that sets the stage for the others? – Dan Piponi Apr 2 2012 at 1:43
The "Ercules and the Hydra" problem, as found in "L. Kirby and J. Paris. Accessible independence results for peano arithmetic. London Mathematical Society, 4:285 293, 1982.".
Using transfinite induction, it is possible to show that Hercules will always kill the hydra (with a finite number of blows) regardless of the strategy chosen to chop off hydra's heads. Moreover, this fact is not provable within Peano Arithmetic.
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The original proof of Van der Waerden's theorem on monochromatic arithmetic progressions comes to mind. Well, the more recent ones too by the way.
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Could you please provide a reference? – Lorenzo Lami Mar 31 2012 at 9:36
See the wikipedia article en.wikipedia.org/wiki/Van_der_Waerden's_theorem – Johan Wästlund Mar 31 2012 at 10:40
In pro-algebraic geometry you get to see some nice arguments by induction. For example, M. Kim proves that the continuous cohomology `$$H^1(G_{\mathbf{Q}_p},\pi_{1,et}^{uni}(X))$$` is representable by induction on the terms in the lower central series of the $\mathbf{Q}_p$ pro-unipotent algebraic group associated to the etale fundamental group of a curve $X$. Not very surprising, but still crucial for the argument.
For a reference, see page 639 in http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf
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http://mathhelpforum.com/advanced-algebra/192433-groups-category-theory-print.html
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# Groups in Category Theory
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• November 21st 2011, 02:50 PM
albi
Groups in Category Theory
There are some ways to introduce groups in Category Theory.
1. Given object $A$ and morphisms $\mu: A \times A \rightarrow A$, $\tau: A \rightarrow A$. We can introduce group on $\mathrm{Hom}(X, A)$ for any object $X$.
Morphisms $\mu$ and $\tau$ have to be such that the respective functions:
$\mathrm{Hom}(X, A) \times \mathrm{Hom}(X, A) \simeq \mathrm{Hom}(X, A \times A) \rightarrow \mathrm{Hom}(X, A)$, $\mathrm{Hom}(X, A) \rightarrow \mathrm{Hom}(X, A)$ satisfy group axioms. The arrows are defined in the natural way.
I think, that 'in the natural way', means that for $f, g \in \mathrm{Hom}(X, A)$ we have $f \cdot g = \mu \circ f \times g$, where $\times$ denotes product of morphisms (which exists due to the definition of product in category). And $f^{-1} = \tau \circ f$.
One problem is, what properties has to satisfy $\mu$ and $\tau$ in order to induce group structure.
But i don't want to think on this right now.
2. The second way, is to suppose we have group structure on each $\mathrm{Hom}(X, A)$ for given $A$ and any $X$. Of course the groups have to be related somehow. And the relation is, that for any morphism $\mathrm{Hom}(Y, X)$ the respective arrow $\mathrm{Hom}(X, A) \rightarrow \mathrm{Hom}(Y, A)$ is a group homomorphism.
Finally the question is. We have group structure given in the way 2, and we want to define group structure in the way 1. It is, we want to define morphisms $\mu$ and $\tau$ in some way.
• November 21st 2011, 03:01 PM
albi
Re: Groups in Category Theory
My idea was to start with a group on $\mathrm{Hom}(A, A)$. Then we know, that the multiplication should be: $f \cdot g = \mu \circ f \times g$ for some $\mu$.
On the other hand we have to use the fact that the groups on $\mathrm{Hom}(X, A)$ are related... Hmm, I think I've just got one idea (looks like writing this thing on the forum is helpful..., but I will think on this tomorrow)
• November 21st 2011, 09:11 PM
Drexel28
Re: Groups in Category Theory
I'm sorry this is hard to read. what precisely are you trying to come up with? Are you trying to define, in general, a group operation on the hom-sets? Or, are you looking for some notion that, when satisfied, is useful, and has to do with the hom-sets being $\mathbb{Z}$-modules? If the former is true you can't always, in the sense that you can't make them into precisely the natural sort of group that distributes over composition. In other words, you can't always make them into a group in such a way that the category becomes preadditive.
• November 22nd 2011, 11:05 AM
albi
Re: Groups in Category Theory
No, no. I think it has nothing to do with preaddtive categories or abelian groups. But writing this whole thing down, helped me to understand a bit of it, and I know to solve it.
So we have a group structures (not necessarly abelian) on $\mathrm{Hom}(X, A)$ where $A$ is fixed. In particular we have a group structure for $\mathrm{Hom}(A, A)$. Let us denote its group operations by $\cdot$ and ${}^{-1}$.
I am looking for a morhpism $\mu$ such that the multiplication is $f \cdot g = \mu \circ f \times g$. I want to express it by the group operations, neutral element, etc.
What is important to notice, that $\pi_1 \times \pi_2 = \mathrm{id}_{A \times A}$ where $\pi_i: A \times A \rightarrow A$ are projections. So substituting them for $f$ and $g$ we get $\mu = \pi_1 \cdot \pi_2$.
Which is not true. But it was the thing I came up yesterday. The problem is that the projections are not good morphisms. So I still don't know.
• November 22nd 2011, 03:00 PM
Deveno
Re: Groups in Category Theory
i think what you are trying to do is define "point-wise" multiplication. for example, if our category is Top, then Hom(X,A) is continuous functions from X to A.
so we can make this into a group by taking:
$(f*g)(x) = f(x)g(x)$
$f^{-1}(x) = (f(x))^{-1}$
$e_{\mathrm{Hom}(X,A)}(x) = e_A$.
• November 22nd 2011, 03:42 PM
albi
Re: Groups in Category Theory
Yes. In Top we get topological groups, for smooth manifolds we get Lie groups (and the operations on hom-sets are poinwise). But does it help me in the general case?
• November 22nd 2011, 07:34 PM
Drexel28
Re: Groups in Category Theory
Quote:
Originally Posted by albi
Yes. In Top we get topological groups, for smooth manifolds we get Lie groups (and the operations on hom-sets are poinwise). But does it help me in the general case?
Ah, perhaps what you mean then is a group object in a category. This is slightly more general than you are considering, but should cover your case nicely, no? Sorry if I have double misunderstood.
• November 23rd 2011, 05:06 AM
Deveno
Re: Groups in Category Theory
i think you're talking about the same thing. if $X$ lies in a category $\mathcal{C}$, and we have a functor $F:\mathcal{C} \to \bold{Grp}$ given by: $F(X) = \mathrm{Hom}(X,A)$, then $A$ is a group object in $\mathcal{C}$.
clearly, since $F$ is a functor, we also have a group morphism $F(f)$ for every $f \in \mathrm{Hom}(X,Y)$, in $\mathrm{Hom}(\mathrm{Hom}(X,A),\mathrm{Hom}(Y,A))$.
what does this mean?
we have a morphism $\mu:\mathrm{Hom}(X,A) \times \mathrm{Hom}(X,A) \to \mathrm{Hom}(X,A)$, a morhpism $\eta:1 \to \mathrm{Hom}(X,A)$, and a morphism $\iota:\mathrm{Hom}(X,A) \to \mathrm{Hom}(X,A)$ such that:
$\mu \circ (\mu \times \mathrm{id}_{\mathrm{Hom}(X,A)}) = \mu \circ (\mathrm{id}_{\mathrm{Hom}(X,A)} \times \mu)$
$\mu \circ (\eta \times \mathrm{id}_{\mathrm{Hom}(X,A)}) = \pi_1,\ \mu \circ (\mathrm{id}_{\mathrm{Hom}(X,A)} \times \eta) = \pi_2$
$\mu \circ (\mathrm{id}_{\mathrm{Hom}(X,A)} \times \iota) \circ \Delta = \mu \circ (\iota \times \mathrm{id}_{\mathrm{Hom}(X,A)}) \circ \Delta = \eta \circ 1$
where:
$\pi_1,\ \pi_2$ are the canonical projections on the product, $\Delta$ is the diagonal map, and $1$ is the unique map: $1: \mathrm{Hom}(X,A) \to 1$
now, i think albi's question is this:
given only $F$, how do we construct $\mu,\ \eta,$ and $\iota$? and the answer is: "point-wise in A".
• November 26th 2011, 11:28 AM
albi
Re: Groups in Category Theory
I don't understand anything after "what does it mean?".
In particular, I don't understand how
$\mathrm{id}_A \times \mu$ makes sense. Because the property of of product is such that, when we have $f_i : X \rightarrow A$ there exists unique $f_1 \times f_2: X \rightarrow A \times A$ such that $f_i$ factorizes on f and $\pi_i$. But $\mathrm{id}_A$ and $\mu$ have diffrent "domains" (or whatever is the name in category theory).
• November 26th 2011, 12:57 PM
Deveno
Re: Groups in Category Theory
a categorical product does not have involve a single object. that is, you can have A x B and not just A x A. in which case, if you have f: X→Y and g:Z→W you can have f x g:X x Z → Y x W. for example, if we are talking about sets, (f x g)(x,y) = (f(x),g(z)). the "coordinates" of f x g are simply $\pi_1 \circ (f \times g) = f$ and $\pi_2 \circ (f \times g) = g$, and we can use this formulation in an arbitrary category that has products.
so, explicitly, $(\mathrm{id}_A \times \mu)(a,b,c) = (a,\mu(b,c))$ for those categories where morphisms are maps of some sort.
• November 26th 2011, 03:01 PM
albi
Re: Groups in Category Theory
I still don't understand how it should work (unless products are isomorphic to coproducts, or something :P). In your example cannot be $\pi_1 \circ (f \times g) = f$ because LHS is $X \times Z \rightarrow Y$ and RHS is $X \rightarrow Y$.
• November 26th 2011, 06:07 PM
Deveno
Re: Groups in Category Theory
well, that's true...the RHS should more properly be something like $f \circ \pi_1$ to match up the domains (i was thinking of just the values, d'oh).
but i think you're missing the point: given something in Hom(X,Y) and something in Hom(Z,W), there is a natural way to create something in Hom(X x Y, Z x W).
• November 28th 2011, 07:37 AM
albi
Re: Groups in Category Theory
I get it now. The trick is to compose our morhphisms with respective $\pi$'s. Then we get morhphisms which have the same "domain". So there exists unique (and hence natural) $\times$.
I will have some more questions on this, when I get home and have time to think about it...
PS But i have to admit, that it is the reason why I could not derive properties of $\mu$. But since I can $\times$ morphisms like that, things come easier for me.
All times are GMT -8. The time now is 05:16 PM.
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http://cstheory.stackexchange.com/questions/14361/difference-between-weak-duality-and-strong-duality
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# Difference between weak duality and strong duality?
For an optimization problem $(P)$ and its dual $(D)$, I have read about two concepts: Weak Duality, and strong Duality. What I don't understand is how they are different:
Weak duality: If $\bar{x}$ is a feasible solution to $(P)$ and $\bar{y}$ is a feasible solution to $(D)$, then:
1. $c^T \bar{x} \le b^T\bar{y}$
2. if equality holds in the above inequality, then $\bar{x}$ is an optimal solution to $(P)$ and $\bar{b}$ is an optimal solution to $(D)$.
Strong duality: If there exists an optimal solution $x'$ for $(P)$, then there exists an optimal solution $y'$ for $(D)$ and the value of $x'$ in $(P)$ equals the value of $y'$ in $(D)$.
Are these two statements not saying the same thing? In other words, isn't the second statement (2.) in definition of Weak duality saying the same thing as strong duality?
Let's say we are given an LP $(P)$ and we find a dual $(D)$. Then can the same dual be used to deduce either strong duality or weak duality?
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## 1 Answer
Weak duality is a property stating that any feasible solution to the dual problem corresponds to an upper bound on any solution to the primal problem. In contrast, strong duality states that the values of the optimal solutions to the primal problem and dual problem are always equal. Was this helpful enough?
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Oh so its strong in the sense that it gives a stronger statement: one just gives an inequality while the other gives a equality? – mtahmed Nov 18 '12 at 9:43
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Yes, exactly. The names are good hints for that. – Ilan Kom Nov 18 '12 at 10:00
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http://mathoverflow.net/questions/69703/examples-of-non-simply-connected-manifolds-with-trivial-h1/69707
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## Examples of non-simply connected manifolds with trivial H^1
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It is known that, if a topological space is simply connected,its first homology group vanishes. The converse is not true, since for every presentation of a (say, finite) perfect group G we can construct a CW-complex, via generators and relations, having G as a fundamental group. Are there such examples in the class of topological or differentiable manifolds? In other words, does there exist a non-simply connected manifolds with trivial first homology group?
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5
You've almost answered your own question. You can realise any finitely presented group as the fundamental group of a closed 4-manifold. You take your 2-dimensional CW complex, embed it in R^5 (dim 5 is necessary so you can separate all the 2-cells) and then take the boundary of a tubular neighbourhood (and perhaps smooth off any corners). The result is a closed 4-manifold with the same fundamental group as the CW complex, unless I'm mistaken. – Joel Fine Jul 7 2011 at 14:27
@Joel: why is the fundamental group of the boundary of a regular neighbourhood the same as that of the original complex? – Neil Strickland Jul 7 2011 at 19:19
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To explain Joel Fine's answer. By transversality you can make a loop in the regular neighborhood miss the 2-complex and hence a smaller regular neighborhood. Then the loop is homotopic into the boundary. If a loop in the boundary is null homotopic in the regular neighborhood the disk it bounds in the regular neighbor can similarly be made to miss the 2-complex and hence is homotopic into the boundary. Thus the two complex and the boundary of the regular neighborhood have the same fundamental group. – Tom Mrowka 11 hours ago – Tom Mrowka Jul 8 2011 at 9:48
## 5 Answers
The classical examples are homology spheres.
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1
The "easiest" family of examples in this class is provided by ±1-surgery on any nontrivial knot (as pointed out in the page you linked): the first homology group is trivial by Mayer-Vietoris, and it's easy to write down a presentation for the fundamental group of the manifold using Seifert-Van Kampen. – Marco Golla Jul 7 2011 at 18:39
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Fake Projective planes
These are smooth complex projective surfaces with the same betti numbers as $\mathbb{CP}^2$, but with infinite fundamental group $\pi_1(X)$ (in fact it is isomorphic to a torsion-free cocompact arithmetic subgroup of $PU(2,1)$).
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Enriques surfaces.
These are complex projective surfaces (hence, real $4$-manifolds) with $p_g(X)=q(X)=0$, obtained by taking the quotient of a $K3$ surface (which is simply connected) by a fixed-point free involution.
So, if $X$ is such a surface we have $\pi_1(X)=\mathbb{Z}/ 2 \mathbb{Z}$.
On the other hand, for any compact complex surface $X$, the first cohomology group $H^1(X, \mathbb{Z})$ injects into $H^1(X, \mathcal{O}_X)= \mathbb{C}^{b^1(X)}$ by the standard exponential sequence of sheaves
$$0 \to \mathbb{Z} \to \mathcal{O}_X \stackrel{\textrm{exp}}{\longrightarrow} \mathcal{O}_X^* \to 0$$
(in particular, it follows that $H^1(X, \mathbb{Z})$ has no torsion).
Since for an Enriques surface $X$ we have $b^1(X)=\frac{q(X)}{2}=0$, we have $H^1(X, \mathbb{Z})=0$.
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2
This is a little misleading; the vanishing of $H^1$ here is an artifact of the universal coefficient theorem. If $\pi_1 = \mathbb{Z} / 2 \mathbb{Z}$ then $H_1$ is also $\mathbb{Z} / 2 \mathbb{Z}$. But $H^1 = Hom(H_1, Z)$ is 0 because $H_1$ is torsion; a fact which makes its presence felt in $H^2$ through an Ext term. The same trick will hold for any space with torsion $H_1$; e.g., projective spaces, lens spaces, products of these... – Craig Westerland Jul 7 2011 at 23:24
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Why misleading? For any complex projective surface the rank of $H^1(X, Z)$ is equal to $q/2$, where $q$ is the irregularity, by the Hodge decomposition and the Dolbeault isomorphism. Moreover $H^1(X, Z)$ has no torsion by the exponential sequence. Since $q=0$ for an Enriques surface, we are done. This is a complex-analytic argument, and no universal coefficient theorem is needed. Of course you can see the vanishing of $H^1$ in the way you said, but it is a matter of taste. – Francesco Polizzi Jul 8 2011 at 6:52
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And, strictly speaking, the argument about the torsion is no really necessary. The vanishing of $q(X)=H^1(X,\mathcal{O}_X)$ and the inclusion given by the exponential sequence are sufficient to conclude. These arguments of course apply only for complex projective varieties, in the case of other spaces with torsion $\pi_1$ one must use the universal coefficient theorem. – Francesco Polizzi Jul 8 2011 at 7:09
I agree that it is a matter of taste! Your machinery undoubtedly makes you as happy as mine does me. I'm just saying that the way that the question is phrased (although, admittedly, not the title of the question), Mari is asking for manifolds with trivial first homology (not cohomology). – Craig Westerland Jul 8 2011 at 9:59
Needless to say, I appreciated your comment. And yes, I answered the question in the title. The question about homology seems to me also quite easy. For any finitely presented group $G$ it is easy to construct a (smooth) compact $4$-manifold $X$ with $G$ as its fundamental group. Now take as $G$ a finite perfect group, for instance $A_5$. Then the abelianization of $G$ is trivial, hence $H_1(X, Z)=0$ (and consequently, also $H^1(X, Z)=0$). – Francesco Polizzi Jul 8 2011 at 10:22
If you are willing to use manifolds with boundary then the question is easy. For any finitely presented group $G$ you can build a finite simplicial complex $X$ with $\pi_1(X)=G$, then embed $X$ in a simplex and let $M$ be a regular neighbourhood of $X$ in the second barycentric subdivision; this will be a manifold with boundary homotopy equivalent to $X$.
If you want to restrict to smooth closed manifolds then the problem is harder, but I think that the answer is the same. Fix a sufficiently large number $n$ (I think $5$ will do) and let $P_k$ be the connected sum of $k$ copies of the $n$-torus. By a small exercise with the van Kampen theorem, $\pi_1(P_k)$ is the free product of $k$ copies of $\mathbb{Z}^n$. Thus, for any finitely presented $G$ there is an epimorphism $\pi_1(P_k)\to G$ for some $k$, with finitely generated kernel. Each generator of the kernel can be represented by a map $u:S^1\to P_k$, which we can assume to be an embedding by a transversality argument. If the normal bundle to $u$ is trivial then we can thicken it to an embedding $S^1\times B^{n-1}\to P_k$, remove the interior, and replace it with $B^2\times S^{n-3}$. (In other words, we perform surgery on $u$). This gives a new manifold, and using van Kampen again we see that the new $\pi_1$ is obtained from the old one by killing $u$. After repeating this process for each generator we get a smooth closed manifold with $\pi_1=G$.
I am not sure what to do if the normal bundle of $u$ is nontrivial, but I doubt that this is a serious problem. I also think that I have seen a more efficient construction in the literature, but I do not remember it at the moment.
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I thought you can achieve this for a manifold without boundary by just doubling what you get above. But I have to admit, the details of why this works eludes me right now. – Donu Arapura Jul 7 2011 at 12:43
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@Donu. Indeed you can double the manifold and preserve the fundamental group if its built from handles of small index. From the presentation build handle one handles corresponding to the generators and 2-handles corresponding to the relations. Doubling will add n-2 and n-1 handles to arrive at a closed manifold. If n>5 adding these doesn't change the fundamental group. – Tom Mrowka Jul 8 2011 at 11:36
Tom, thanks for the explanation. – Donu Arapura Jul 8 2011 at 12:37
Ops, I have just seen the mistake in the title. Sorry, in fact I mean H_1 and not H^1... . I would like to thank you all for your quick answers and useful examples.
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Luciano: you should have either added a comment to your question or, even better, edited it. In any case, welcome :) – Mariano Suárez-Alvarez Jul 9 2011 at 20:45
Some of these answers have $H_1=0$ also. – J.C. Ottem Jul 9 2011 at 20:47
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http://mathhelpforum.com/differential-geometry/142913-taylor-expansion-ln-1-x.html
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# Thread:
1. ## Taylor expansion of ln(1+x)
Okay, so the question is regarding the relation between the interval of convergence and the interval for which $f(x) = \ln (1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^{n}$. I know they are the same interval, but I am trying to show this by showing that the remainder function, $R_{n}$ tends to $0$ only when $x\in (-1,1]$. So I have to show that $\frac{f^{(n)}(y)}{n!}x^{n}$ tends to $0$ where $y$ lies between $0$ and $x$ (if this approach is wrong, please correct me). It can be observed that $f^{(n)}(x)=(-1)^{n+1}(n-1)!(1+x)^{-n}$ so plugging that in I have to show that $\frac{1}{n}|\frac{x}{1+y}|^{n}$ tends to $0$ (because if $|R_{n}|$ tends to $0$, then so does $R_{n}$). So the condition for that to tend to zero is that $|\frac{x}{1+y}| \le 1$. And now this is where I get stuck showing that $x$ must lie in $(-1,1]$.
A secondary question is why does this still converge on some interval even though on that interval, not all the derivatives $f^{(n)}(x)$ are not bounded by some single constant $C$ on that interval?
2. Do you know that an alternating series converges if its terms go to zero in magnitude?
3. Yes, that part I have about the radius of convergence for the series. I am trying to show that the same interval of convergence is the interval for which the Taylor series represents the function, and for that I believe I have to show that $R_{n}$ tends to $0$.
Another follow up question would be why is there this relationship between the interval of convergence of the series and the interval for which the function is equal to the Taylor expansion. Is there a generalized proof that shows that the two intervals must always be equal?
4. Oh, I see. The two intervals don't have to be equal actually (smooth bump functions, or http://www.math.niu.edu/~rusin/known.../nowhere_analy).
edit: oh sorry, I'd suggest using Cauchy's form of the remainder rather than Lagranges'.
edit2: arg, I meant integral remainder.
5. So intervals of convergence do not imply interval for which the function is analytic. Is there a proof for why in this case of $\ln (1+x)$ the two intervals are the same? Right now I'm just trying to show that they are the same (which I still need help with proving if $x\notin (-1,1]$ that there is always a case where there exists $y$ between $0$ and $x$ where $|y+1| > |x|$) but is there a proof where one interval implies the other for this particular function?
Okay, even if I use Cauchy's, I think I'll still run into the same problem that I need to show there's always a case where if $x > 1$ there will be a $y$ such that $x - y > 1$.
6. Maybe try Bernoulli's inequality: for nonzero y>-1 and integers n>1, $(1+y)^n>1+ny$.
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Distributivity_(order_theory)
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# All Science Fair Projects
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# Distributivity (order theory)
In the mathematical area of order theory, there are various notions of the common concept of distributivity, applied to the formation of suprema and infima. Most of these apply to partially ordered sets that are at least lattices, but the concept can in fact reasonably be generalized to semilattices as well.
Contents
## Distributive lattices
Probably the most common type of distributivity is the one defined for lattices, where the formation of binary suprema and infima provide the total operations of join ($\vee$) and meet ($\wedge$). Distributivity of these two operations is then expressed by requiring that the identity
$x \wedge (y \vee z) = (x \wedge y) \vee (x \wedge z)$
holds for all elements x, y, and z. This distributivity law defines the class of distributive lattices. Note that this requirement can be rephrased by saying that binary meets preserve binary joins. The above statement is known to be equivalent to its order dual
$x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z)$
such that one of these properties suffices to define distributivity for lattices. Typical examples of distributive lattice are totally ordered sets, Boolean algebras, and Heyting algebras.
## Semilattices - distributivity with one operation
Semilattices are partially ordered sets for which only one of the lattice operations is available, leading to the concepts of meet-semilattices and join-semilattices, respectively. Obviously, in the presence of only one operation, distributivity cannot be defined in a classical way. Yet, due to the interaction of the meet (resp. join) operation with the given order, one can still define a reasonable notion of distributivity. A meet-semilattice is distributive, if
if a $\wedge$ b ≤ x then there are elements a' and b' such that a ≤ a' , b ≤ b' and x = a' $\wedge$ b'
holds for all elements a, b, and x. Distributive join-semilattices are defined dually. This definition is justified by the fact that the following statements are equivalent for any lattice L:
• L is distributive as a meet-semilattice
• L is distributive as a join-semilattice
• L is a distributive lattice.
Thus any distributive meet-semilattice in which binary joins exist is a distributive lattice. Using this insight, some statements that are usually shown for distributive lattices can be generalized to distributive semilattices.
## Distributivity laws for complete lattices
For a complete lattice, arbitrary subsets have both infima and suprema and thus infinitary meet and join operations are available. Several extended notions of distributivity can thus be described. For example, finite meets may distribute over arbitrary joins, i.e.
$x \wedge \bigvee S = \bigvee \{ x \wedge s \mid s \in S \}$
may hold for all elements x and all subsets S of the lattice. Complete lattices with this property are called frames, locales or complete Heyting algebras. They arise in connection with pointless topology and Stone duality. This distributive law is not equivalent to its dual statement
$x \vee \bigwedge S = \bigwedge \{ x \vee s \mid s \in S \}$
which defines the class of dual frames.
Now one can go even further and define orders where arbitrary joins distribute over arbitrary meets. However, expressing this requires formulations that are a little more technical. Consider a doubly indexed family {xj,k | j in J, k in K(j)} of elements of a complete lattice, and let F be the set of choice functions f choosing for each index j of J some index f(j) in K(j). A complete lattice is completely distributive if for all such data the following statement holds:
$\bigwedge_{j\in J}\bigvee_{k\in K(j)} x_{j,k} = \bigvee_{f\in F}\bigwedge_{j\in J} x_{j,f(j)}$
Complete distributivity is again a self-dual property, i.e. dualizing the above statement yields the same class of complete lattices. Completely distributive complete lattices (also called completely distributive lattices for short) are indeed highly special structures. Various different characterizations exist. For example, the following is an equivalent law that avoids the use of choice functions. For any set S of sets, we define a the set S# to be the set of all subsets X of the complete lattice that have non-empty intersection with all members of S. We then can define complete distributivity via the statement
$\bigwedge \{ \bigvee Y \mid Y\in S\} = \bigvee\{ \bigwedge Z \mid Z\in S^\# \}$
The operator ( )# might be called the crosscut operator. The latter version of complete distributivity only implies the original notion when admitting the Axiom of Choice. However, the latter version is always equivalent to the statement:
$\bigwedge \{ \bigvee Y \mid Y\in S\} = \bigvee\bigcap S$
for all sets S of subsets of a complete lattice.
In addition, it is known that the following statements are equivalent for any complete lattice L
• L is completely distributive (in the original sense).
• L can be embedded into a direct product of chains [0,1] by an order embedding that preserves arbitrary meets and joins.
• Both L and its dual order Lop are continuous posets .
Direct products of [0,1], i.e. sets of all functions from some set X to [0,1] ordered pointwise , are also called cubes. The final theorem also explains why completely distributive lattice are so special. There is still more to say about complete distributivity and its intuitionistic variants: see the article on completely distributive lattices.
## Literature
Distributivity is a basic concept that is treated in any textbook on lattice and order theory. See the literature given for the articles on order theory and lattice theory. More specific literature includes:
• G. N. Raney, Completely distributive complete lattices, Proceedings of the American Mathematical Society, 3: 677 - 680, 1952.
03-10-2013 05:06:04
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http://gilkalai.wordpress.com/2010/11/25/emmanuel-abbe-polar-codes/?like=1&source=post_flair&_wpnonce=eb9f8a7f00
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Gil Kalai’s blog
## Emmanuel Abbe: Erdal Arıkan’s Polar Codes
Posted on November 25, 2010 by
Click here for the most recent polymath3 research thread. A new thread is comming soon.
Emmanuel Abbe and Erdal Arıkan
This post is authored by Emmanuel Abbe
A new class of codes, called polar codes, recently made a breakthrough in coding theory.
In his seminal work of 1948, Shannon had characterized the highest rate (speed of transmission) at which one could reliably communicate over a discrete memoryless channel (a noise model); he called this limit the capacity of the channel. However, he used a probabilistic method in his proof and left open the problem of reaching this capacity with coding schemes of manageable complexities. In the 90′s, codes were found (turbo codes and LDPC rediscovered) with promising results in that direction. However, mathematical proofs could only be provided for few specific channel cases (pretty much only for the so-called binary erasure channel). In 2008, Erdal Arıkan at Bilkent University invented polar codes, providing a new mathematical framework to solve this problem.
Besides allowing rigorous proofs for coding theorems, an important attribute of polar codes is, in my opinion, that they bring a new perspective on how to handle randomness (beyond the channel coding problem). Indeed, after a couple of years of digestion of Arıkan’s work, it appears that there is a rather general phenomenon underneath the polar coding idea. The technique consist in applying a specific linear transform, constructed from many Kronecker products of a well-chosen small matrix, to a high-dimensional random vector (some assumptions are required on the vector distribution but let’s keep a general framework for now). The polarization phenomenon, if it occurs, then says that the transformed vector can be split into two parts (two groups of components): one of maximal randomness and one of minimal one (nearly deterministic). The polarization terminology comes from this antagonism. We will see below a specific example. But a remarkable point is that the separation procedure as well as the algorithm that reconstructs the original vector from the purely random components have low complexities (nearly linear). On the other hand, it is still an open problem to characterize mathematically if a given component belongs to the random or deterministic part. But there exist tractable algorithms to figure this out accurately.
Let us consider the simplest setting. Let $X_1,...,X_n$ be i.i.d. Bernoulli($p$) and assume that \$n\$ is a power of 2. Define $G_n$ to be the matrix obtained by taking $\log_2(n)$ Kronecker products of $G_2=\begin{pmatrix}1&0\\1&1\\\end{pmatrix}$, and multiply $X=(X_1,...,X_n)$ with $G_n$ over $GF(2)$ to get $U=(U_1,...,U_n)$. Note that $U$ has same entropy as $X$, since $G_n$ is invertible (its inverse is itself). However, if the entropy of $X$ is uniformly spread out over its components, i.e., $H(X)=nH(X_1)$, the entropy of $U$ may not be, since its components are correlated. In any case, we can write $H(U)= \sum_i H(U_i|U^{i-1})$ where $U^{i-1}=(U_1,...,U_{i-1})$ are the `past components’. The polarization phenomenon then says that, except for a vanishing fraction of indices $i$ (w.r. to $n$), each term $H(U_i|U^{i-1})$ tends to either 0 or 1.
Here $H(\cdot | \cdot)$ denotes the Shannon conditional entropy (it is not a random variable like the conditional expectation but a scalar); if you don’t know the definition, what matters is to know that `$H(Y|Z)$ close to 0′ means that $Y$ is nearly a deterministic function of $Z$, and `$H(Y|Z)$ close to 1′ means that $Y$ is nearly independent of $Z$ and uniformly distributed.
Also note that defining $S$ to be the set of indices where $H(U_i|U^{i-1})$ is close to 1, we have that $U$ restricted on $S$ is roughly i.i.d. Bernoulli(1/2). Thus, randomness has been extracted, and what is remarkable is that the compression procedure (the matrix multiplication) and the reconstruction algorithm (recovering the deterministic entries from the pure bits) can be done in $O(n \log n)$. This is a consequence of the Kronecker structure of $G_n$, which allows the use of `divide and conquer’ algorithms (a la FFT). The error probability for the reconstruction is roughly $O(2^{-\sqrt{n}})$.
Over the past two years, this result has been extended to random variables valued in $GF(q)$, to some mixing processes, to some universal settings and to some multi-dimensional objects. For example, if the $X_i$‘s are i.i.d. random vectors of dimension $d$ rather than random variables, we can use a (component-wise) polarization transform that produces again special $U_i$‘s, which are, conditioned on the past, equivalent to linear forms of independent pure bits. The number of extremal structures is hence no longer 2 as for $d=1$ but is equal to the number of linear forms over $GF(2)^d$. This also raised interesting connections with combinatorics topics (in particular I found matroid theory to be useful in this problem).
The original paper is “Channel polarization: a method for constructing capacity-achieving codes for symmetric binary-input memoryless channels“, published in July 2009 in the IEEE Transactions on Information Theory. Arıkan received the IEEE Information Theory best paper award this year. This paper deals with the polarization of `channels’, whereas here I discussed the polarization of `sources’ (less general but same technique).
Emmanuel Abbe, emmanuelabbe@gmail.com
Additional links: historical take on polar codes;
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### 17 Responses to Emmanuel Abbe: Erdal Arıkan’s Polar Codes
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2. Arya says:
If it is not already obvious from the context: the construction of polar codes is nothing but identifying the coordinates where the entropy polarizes to 1. For the trivial binary erasure channel (for which capacity achieving efficiently encodable-decodable codes were already known) it is a relatively easy task. However, for the other channels, like binary symmetric, a systematic construction is still an open question. Recently Tal and Vardy claimed that they have a construction; no manuscript is available for scrutiny though.
The big advantages of polar codes are: being linear, and formed with the similar structure of Reed-Muller code, they can be easily encoded and decoded, once constructed. Also to be noted, these codes are “asymptotically bad,” their distance is a vanishing proportion of length.
In any way, this is the only true breakthrough in coding theory in at least the past 10 years IMHO. A great paper.
3. Sam says:
I don’t understand, what was the breakthrough theorem? (or was it a breakthrough approach ?)
A new proof of Shannon’s coding theorem?
A coding scheme which has a near-linear time algorithm for decoding, and which meets Shannon capacity? Wasn’t this achieved by Forney in the 1970s (by concatenation), that too by an explicit code?
Is it that polar codes are “simple” (even though they are not explicit)?
Or is it that once we do make them explicit, they might have better dependence on the parameter epsilon, when you use these codes at rate R = C – epsilon, where C is the capacity?
This confusion might be a result of the difference between theoretical computer science approach to coding theory, and the electrical engineering approach to coding theory.
• Emmanuel says:
Sam: It’s not a proof of Shannon’s coding theorem.
The concatenation code of Forney did not reach the capacity with a near linear-time algorithm as far as I remember (when you approach capacity the outer code part start mattering). For any epsilon>0 (as you wrote), polar codes allow an error probability of O(2^{-\sqrt{n}}) (with an exponent in epsilon that is strictly positive) and a complexity of O(n log n) (independent of epsilon and taking into account the procedure that figures out the classification of the indexes). Even if you ask for weaker decays in the error probability, I am not sure that there exist other codes providing such mathematical guarantees. Do you know any? What do you mean by “better dependence”?
In any case, when Gil asked to write on his blog, I thought it would be more interesting/general to put emphasis on the probabilistic phenomenon of polarization (the 0-1 result using the Kronecker structure) rather than the channel coding problem. This is however only one step (although the key one I think) used in the coding scheme of Arikan (see paper). Hope this helps.
• Arya says:
This is not a new proof of Shannon’s coding theorem. The breakthrough result is the following “We have efficiently encodable/decodable class of codes that achieves the capacity of symmetric channels, with the given decoding.”
A coding scheme which has a near-linear time algorithm for decoding, and which meets Shannon capacity? Wasn’t this achieved by Forney in the 1970s (by concatenation), that too by an explicit code?
No. Forney’s `explicit’ concatenated codes do not achieve capacity. When you say explicit construction what do you mean? Outer Reed-Solomon codes, I assume; what are the inner codes ? GMD decoding?
Is it that polar codes are “simple” (even though they are not explicit)?
They are simple enough. So far, there is no computational shortcut in construction though, for general symmetric channels.
Or is it that once we do make them explicit, they might have better dependence on the parameter epsilon, when you use these codes at rate R = C – epsilon, where C is the capacity?
The decoding complexity here is independent of such epsilon. I think that is what you have in mind. The decoding probability error is already discussed above.
This confusion might be a result of the difference between theoretical computer science approach to coding theory, and the electrical engineering approach to coding theory.
I think finding capacity achieving codes is one central topic in coding theory, and historically most important; the other problems being bounds on the size of codes (closing the gap), explicit codes that achieves GV bound for all rate etc.
4. Gil Kalai says:
I find the polar codes fascinating from several reasons.
First, given n and p it is a nice mystery which are the bits which carries maximal entropy and which are the bits that carries almost no entropy. In his lecture at Yale, Emmanuel mentioned that this is an open problem and that there are some indication that the indices of the “useful” bits has some fractal nature.
Second, this looks like a very general phenomena and will probably work for other matrices. (Although it will be less useful).
Third, there are various possible analogs. Suppose, for example you replace bits with qubits the Bernouli (p) distribution by some quantum operation (or just a simple depolarizing channel) and the matrix G by a hadamard matrix. I suppose that the polarization phenomena will prevail and this may be useful too.
Forth, the (naked) polarization phenomenon reminds me of the phenomenon of noise synchronization that I studied. (See also this mathoverflow question.)
Actually, it would be useful to indicate how the proof of the polarization property goes. from Emmanuel’s lecture I remember it is quite simple.
• Arya says:
Second, this looks like a very general phenomena and will probably work for other matrices. (Although it will be less useful).
This is already done here:
http://www.stanford.edu/~korada/papers/journal/polar_exponent_journal.pdf
Also discusses the effect of Matrix size on error exponent.
• Emmanuel says:
I think that what Gil has in mind regarding
“this looks like a very general phenomena and will probably work for other matrices. (Although it will be less useful)”
is even more general than what has been looked at in the paper you refer to Arya.
The paper you mention considers other choices for the small matrix (the [1 1, 0 1]) which can be of dimension k x k where k is fixed. It characterizes the matrices for which polarization happens (indeed it happens for most matrices). Interestingly, the paper concludes that it is not worth considering anything else than the initial 2×2 matrix: to have any gain in the convergence rate (and error decay), one has to consider k to be at least 16, with a very small improvement with respect to a big cost in complexity and dimensionality growth (n is now a power of 16 now).
However:
1. one may consider matrices which are not of this kronecker form; of course it’s kind of the crux of polarization to have the kronecker structure, but still, the problem of having radical changes between two sets of components may happen in more general settings. Gil may be referring to this?
2. one may keep the kronecker structure and consider non linear mappings, or linear in GF(q). That I think can give interesting results to be analyzed with our current tools.
• Boaz Barak says:
The definition of conditional entropy depends on the order of the particular index, with a stricter condition on each index as it grows. Does this mean that for simple cases such as p-biased coins, the indices where conditional entropy is one will be the first pn indices? (Intuitively this seems to me to be the case if you take a random matrix instead of Kronecker product.)
• Emmanuel says:
Dear Boaz Barak,
Yes I would agree with your intuition for some random matrix ensembles (provided that there is a sharp transition, and I guess you meant H(p)n instead of pn) . Indeed, it would be good to check this.
Gil: have you tried it?
I remember you asking a similar question.
Regarding the Kronecker case, it s a bit different. That s where the fractal-like shape comes in. It is true that at the beginning you see many 1′s (O(n) i guess but no proofs) and at the end you see many 0′s. However in between we don’t quite understand what happens.
For an arbitrary integer between 1 and n, it s hard to predict what happens. It can jump back and forth from o to 1…
• Yuval says:
First, we want _low_ conditional entropy, since this means _high_ mutual information. So you should reverse your directions.
When computing the transformation (x,y) -> (x+y,y), the second coordinate is indeed much better. That means that for each “1″ bit in your index you’re winning, and for each “0″ bit you’re losing. But the order matters!
The case of binary erasure channels is worked out in the original paper (it’s pretty easy). If the probability of non-erasure is e, then the original mutual information is e, a 0-bit corresponds to applying x->x^2, and a 1-bit corresponds to applying x->1-(1-x)^2. Figure 4 on page 3 of the paper shows the messy results (though perhaps bit-reversing the indices would have resulted in a nicer graph).
In general, define V_t as the subspace generated by all rows (or columns) but the first t. Let H_t = H(C(v_t)), where C is your channel, and v_t is a random vector from V_t. The mutual information of bit t is H_t – H_{t+1} (unless I made a silly mistake), so the good bits are the ones for which the transition from V_t to V_{t+1} reduces the entropy significantly.
• Yuval says:
Actually the graph in Arikan’s paper uses the correct bit-order, the other bit-order looks even more erratic.
5. Anthony says:
Nice post!
I have one question: is there any intuitive way to understand this polarization phenomenon?
6. Emmanuel says:
Gil, Anthony: regarding the proof of the 0/1 result (and maybe some intuition), here is the idea.
Check first what happens when n=2, i.e., we take [X,X'] iid and we map them to [X+X',X']. Since G_2 is invertible:
2H[X]=H[X,X']=H[X+X',X']=H[X+X']+H[X'|X+X'] (*)
(the last equality results from factorizing the distribution of [X+X',X'].)
We must have H[X'|X+X']=H[X]. So if you compare what happens before and after the G_2 map, we initially had a random vector with the randomness spread out uniformly, and we have created a new random vector which has a first component that is more random and a second one that is less given the first one.
But the total amount of randomness is the same, it’s just been shuffled.
In a game where you had to guess one component and could query the other one, and you have to choose with which of the 2 random vectors you want to play, you should chose the second one.
Now, the rest of the proof relies on the Kronecker structure. Multiplying an i.i.d. vector with G_n=G_2^{\otimes log(n)} can be done recursively. A good way to represent the output of G_n is with a tree of depth log_2(n). You start with X at the root, and at each level you are using G_2 (if you go up you add an iid copy of yourself and if you go down you don’t do anything). You can check that at the leaves you exactly get the output of G_n. Then you need to track the CONDITIONAL entropies (not the entropies directly). For this, you introduce a random process to simplify the analysis. You start at the root and go up and down with probability 1/2. The probability on the path set is uniform, so that asking that the conditional entropy of a path ends in a given interval is precisely counting the proportions of conditional entropies leaving in this interval. The key step it that the process tracking the conditional entropies in this tree is a martingale, because of (*) . It’s clearly bounded, so it must converge almost surely. Finally, notice that when it converges, you are in a situation where the action of G_2 does not shuffle much the entropy anymore, and it’s easy to check that this happens only if the random variable is deterministic or purely random (given the past), that is with an entropy of 0 or 1!
I will take a look at your noise sync pb.
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9. Emmanuel says:
By the conservation of entropy: the number of indices which are purely random must be nH(p) and all other indices must be deterministic (up to a vanishing fraction). You can also see this from the post below.
Data of the indexes you meant? you can find some info in Arikan’s paper (or email me)
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http://mathoverflow.net/questions/107101?sort=votes
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## Is the category of vector bundles over a topological space abelian?
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Consider the trivial bundle $V=\mathbb{R}\times\mathbb{R}$ and the map $f:V\rightarrow V$ given by $(t,x)\mapsto(t,tx)$. This has fibrewise kernels and cokernels, but the ranks jump at 0, so the kernel and cokernel of $f$ (as sheaves) are not vector bundles. This example (or similar) is often given to show that Vect($X$) is not in general abelian (or even preabelian).
But this doesn't strike me as correct. Just because $f$ has a kernel (say) $K$ in Sh($X$) which is not an object of Vect($X$) does not mean that there is not an object of Vect($X$) which is a kernel for $f$ in Vect($X$). In the example above, the zero bundle seems to do the job? Indeed I think you can always fix this rank-jumping behaviour by extending smoothly over the bad points (or am I wrong?).
The situation is further confused by Serge Lang claiming (in Algebra, p 134) "the category of vector bundles over a topological space is an abelian category."
As a counter-appeal to authority Ravi Vakil has (Foundations of AG notes) "locally free sheaves (i.e. vector bundles), along with reasonably natural maps between them (those that arise as maps of $\mathcal{O}_X$-modules), don’t form an abelian category."
So is the category of vector bundles over a topological space abelian or not?
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I think Lang's claim is mistaken. – Damian Rössler Sep 14 at 5:30
Intuitively, the problem is that taking fiberwise kernels or cokernels will not, in general, give you vector bundles because the rank can jump. – Rasmus Sep 14 at 8:04
## 3 Answers
There are two candidates for a category whose objects are vector bundles on $X$. Candidate (1) is the one you and Ravi Vakil's notes describe: the morphisms are all maps of $\mathcal{O}_X$-modules. As you observe, this fails to have the obvious kernels and cokernels, and since every coherent sheaf is locally a quotient of free $\mathcal{O}_X$-modules, the actual ones have to coincide with the obvious ones (which are the co/kernels in the category of coherent sheaves).
Candidate (2) has the morphisms being only those of constant rank; equivalently, those for which the sheaf quotient is a vector bundle. This makes the sheaf kernel a vector bundle as well, but this candidate has several problems identified in the comments and is as a result neither abelian nor a category.
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Unfortunately, the second category is not additive. – Laurent Moret-Bailly Sep 13 at 16:11
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View $\bf R$ (resp. ${\bf R}^2$) as the trivial (real) vector bundle of rank $1$ (resp. rank $2$) over the real line $\bf R$, viewed as a topological space with the ordinary topology. Let $f:{\bf R}\to {\bf R}^2$ be the map of vector bundles st $f(t,v)=(t,v\cdot(t,1))$. Let $g:{\bf R}^2\to{\bf R}$ be the map of vector bundles $g(t,v,w)=(t,v)$. Then $g$ and $f$ have constant rank but $g\circ f$ does not (because $g\circ f$ restricted to $0$ as rank $0$ and not otherwise). So the class you describe is not a category. – Damian Rössler Sep 13 at 16:15
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You shouldn't expect vector bundles to form an abelian category because they are projective module objects over local ring objects in a category of sheaves over a space. In other words, there is really no more reason to expect this of vector bundles than you would of a category of finitely generated projective modules over a ring.
It might help to add that taking stalks at a point is an exact functor (it preserves all colimits and all finite limits), so kernels and cokernels would be preserved under taking stalks at points. Moreover, if the space is sober, then taking stalks collectively reflects kernels and cokernels as well. This should be a reassurance as to whether one is computing kernels and cokernels correctly.
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Re: the first paragraph, when the space $X$ is compact Hausdorff, the Serre-Swan theorem asserts that the category of vector bundles over $X$ is precisely the category of finitely-generated projective modules over $C(X)$ (no sheaves required). – Qiaochu Yuan Sep 13 at 17:07
Here is an argument to show that this category doesn't have kernels. Let's take $X$ to be the interval $[-1,1]$, and consider the self-map of the trivial bundle $X \times \mathbb{R}$ given by $$(x,t) \mapsto (x, (x + |x|)t).$$ Suppose this had a kernel $K$, fitting into a sequence $K \to X \times \mathbb{R} \to X \times \mathbb{R}$. We get a sequence of modules of global sections $$\Gamma(K) \to C(X) \stackrel{x + |x|}{\longrightarrow} C(X)$$ over the ring $C(X)$ of continuous functions on $X$. (The Serre-Swan theorem has been mentioned already; this is equivalent data.)
However, the global section functor is another name for $Hom(X \times \mathbb{R}, -)$ in the category of vector bundles, and so the "kernel" property implies that the module $\Gamma(K)$ would actually be the kernel of multiplication by $x + |x|$.
This is impossible. Every element in $\Gamma(K)$ is annihilated by $x + |x|$ by definition, and $\Gamma(K)$ is nonempty, but every nonzero vector bundle on $X$ is some power of the trivial bundle (by the homotopy invariance property) and has many sections which do not vanish on $(0,1]$.
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http://physics.stackexchange.com/questions/8156/eigenvalues-of-an-operator-correspond-to-energy-states-in-quantum-mechanics-why?answertab=oldest
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Eigenvalues of an operator correspond to energy states in quantum mechanics, why?
When finding the discrete energy states of a operator I have been taught to use the time-independent Schrodinger equation which restates the definition of eigenvalues and eigenvectors. What I don’t understand is why the eigenvalues are the energy states, is there firstly a mathematical reason and secondly a physical reason?
Does this arise from Hamiltonian or Lagrangian mechanics which I am not familiar with?
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Sorry I mean eigenvalues of the operator not wave funciton – Josh Apr 6 '11 at 17:00
Keep in mind that we are only dealing with Hermetian operators, because their eigenvalues are real, and hence correspond to positive definite probabilities. – Matt Apr 8 '11 at 16:44
7 Answers
As has been remarked by others and explained clearly, and mathematically, the eigenvalues are important because a) they allow you to solve the time-dependent equation, i.e., solve for the evolution of the system and b) a state which belongs to the eigenvalue $E$, i.e., as we say, a state which is an eigenstate with eigenvalue $E$, has an expectation value of the energy operator which is easy to see has to be $E$ itself. But those explanations are advanced and rely on the maths. And they do not explain why $E$ should be considered 'an energy level'. At some risk, I will try to answer your question more physically.
What is the physical reason why the energy states of a system, e.g., an atom, are the eigenvalues of the operator $H$ that appears in the time-independent Schroedinger equation? Well, first, note that it's absolutely the same $H$ that appears in the time-dependent Schrodinger equation, $$H\cdot \psi = -i{\partial \psi \over \partial t}$$ which controls the rate of change of $\psi$.
The answer doesn't come from the classical Hamiltonian or Lagrangian mechanics, but from the then-new quantum properties of Nature. A non-classical feature of QM is that some states are stationary, which means they do not change in time. E.g., the electron in a Bohr orbit is actually not moving, not orbiting at all, and this solves the classical paradoxes about the atom (why the rotating charge doesn't radiate its energy away and fall into the centre).
The first key point is that an eigenstate is a stationary state: what is the explanation for this? well, Schroedinger's time dependent equation clearly says that, up to a constant of proportionality, the time-rate of change of any state $\psi$ is found by applying the operator $H$ (the Hamiltonian: we do not yet know it is also the energy operator) to it: the new vector or function $H\cdot\psi$ is the change in $\psi$ per unit time. Obviously if this is zero, $\psi$ does not change (this was the only classical possibility). But also if $H\cdot\psi$ is even a non-zero multiple of $\psi$, call it $E\psi$, then $\psi$ plus this rate of change is still a multiple of $\psi$, so as time goes on, $\psi$ changes in a trivial fashion: just to another multiple of itself. In QM, a multiple of the wave function represents the same quantum state, so we see the quantum state does not change.
Now the next key point is that a state with a definite energy value must be stationary. Why? In QM, it is not automatic that a system has a definite value of a physical quantity, but if it does, that means its measurement always leads to the same answer, so there is no uncertainty. So if there is no uncertainty in the energy, by Heisenberg's uncertainty principle there must be infinite uncertainty in something else, whatever is 'conjugate' to energy. And that is time. You cannot tell the time using this system, which implies it is not changing. So it is stationary. (remember, we are not assuming that $H$ is also the energy operator and we are not assuming the formula for expectations).
Thus being an eigenstate of $H$ implies $\psi$ is stationary. And having a definite energy value implies it is stationary. Being physicists, we now conclude that being an eigenstate implies it has a definite energy value, which answers your question, and these are the 'energy levels' of a system such as an atom: a system, even an atom, might not possess a definite energy, but if it doesn't, it won't be stationary, and being microscopic, the time-scale in which it will evolve will be so rapid we are unlikely to be able to observe its energy, or even care (since it won't be relevant to molecules or chemistry). So, 'most' atoms for which we can actually measure their energy must be stationary: this is 'why' the definite values of energy which a stationary state can possess are called the 'energy levels' of the system, and historically were discovered first, before Schroedingers equation. From a human perspective, most atoms that we care about spend most of their time that matters to us in an approximately stationary state.
In case you are wondering why time is the conjugate to energy, whereas Heisenberg's original analysis of his uncertainty principle showed that position was conjugate to momentum, we rely on relativity: time is just another coordinate of space-time, and so is analogous to position. And in relativistic mechanics, momentum in a spatial direction is analogous to energy (or mass, same thing). In the standard relativistic equation $$p^2-m^2=E^2,$$ we see that momentum ($p$) and mass $m$ are symmetric (except for the negative sign) with each other. So since momentum is conjugate to position, $m$ or energy must be conjugate to time. For this reason, Bohr was able to extend Heisenberg's analysis, of the uncertainty relations between measurements of position and measurements of momentum, to show the same relations between energy and time.
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Both eigenvalues and eigenstates belong to some operator. In your case, this is the Hamiltonian operator $\hat H$. It's fundamental because of many reasons. First is that it is indeed an operator that represents energy in a sense that possible energy levels are encoded in its spectrum (i.e. a set of eigenvalues). The second important reason is that it is the operator that can be found in Schrodinger equation $i \hbar \partial_t \left | \psi(t) \right > = \hat H \left | \psi(t) \right >$. This equation can then be solved by writing $\left | \psi(t) \right >$ as superposition of eigenstates of $\hat H$: $\left | \psi(t) \right > = \sum_n c_n(t) \left | \psi_n \right >$. If we can find these states, we are done as $c_n(t) = \exp({-iE_n t \over \hbar}) c_n(t=0)$ solves the equation (and it also shows the importance of these eigenstates because they are preserved by time evolution).
So this means the problem of time evolution in quantum mechanics can be reduced to the problem of finding the eigenvalues and eigenstates of $\hat H$, the equation for that being $\hat H \left | \psi_n \right> = E_n \left| \psi_n \right>$.
Note: the above assumes that $\hat H$ is time-independent. If it's not (as is the case in basically all practical applications, using perturbation theory) then we use different techniques, e.g. of path integration, or various scattering formulas.
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A transformation from one set to another can be regarded as a matrix if we define a particular basis. Likewise, an operator can be thought of as a matrix. What is the matrix equation that relates eigenvalues and eigenvectors? You are solving an eigenvalue problem when you are solving the time independent Schrodinger equation.
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The reason why it is the eigenvalues of the Hamiltonian and not some other operator that will give you the energy states is that in classical Mechanics, the Hamiltonian function is just the energy of your system, expressed as a function of position $x$ and momentum $p$. As a simple example, the Hamiltonian for a harmonic oscillator is $$H(x,p) = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ Note that this really is just the sum of kinetic and potential energy, so we could write $$H(x,p) = E$$.
To get to quantum mechanics, one now performs what is called canonical quantization. There is no mathematically rigorous reason why this will give you a correct quantum mechanics. Since quantum isn't classical, we cannot really expect to find a seamless and watertight derivation of the former from the latter. To my knowledge, this approach has, however, always given correct results.
So, in canonical quantization, what one does is to replace the variables of the Hamiltonian, i.e., $x$ and $p$, with their operator versions, $\hat x$ and $\hat p$. Now we cannot simply write $H(x,p) = E$ anymore, since the energy is a scalar, but the Hamiltonian $H$ is now an operator. Operators are functions that take a wavefunction and modify it in some way and give you a new wavefunction. Now, another postulate of quantum mechanics is that you get the expectation value of an operator $\hat A$ in a given state $\Psi$ by calculation the integral $$\int dx \Psi^*(x) \hat A \Psi(x)$$ Hence, we get the expectation value of the energy by calculating $$\int dx \Psi^*(x) H \Psi(x)$$ Obviously, if $H\Psi(x) = E\Psi(x)$, then the expectation value yields $E$, and it is not hard to show that for such an eigenstate, the variance of $E$ will be $0$, i.e. every measurement of the energy in state $\Psi$ will yield the same value $E$.
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You seem to be confusing two things, namely the eigenstates of an operator and Schrödinger's equation. A priori, these two have nothing to do with each other.
In Quantum Mechanics, measurable quantities are represented by (hermitian) operators on a Hilbert space. For instance there is an operator $P$ corresponding to the momentum. In general, when measuring the momentum of a state $|\psi \rangle$, the result will not be deterministic. However, the average over several measurements will be equal to the expection value
$$\langle \psi | P | \psi \rangle$$
However, when $|\psi\rangle$ is an eigenvector of the operator, $P|\psi\rangle = \lambda |\psi\rangle$, then the measurement will always be the same value $\lambda$.
In particular, there is an operator corresponding to the total energy, the Hamiltionian $H$. The form of this operator can be obtained from classical physics if you replace momentum and location by their corresponding operators. For instance, the Hamiltonian of an electron in an electric potential $V$ is
$$H = \frac1{2m} P^2 + eV(X) .$$
Thus, the expectation value for the energy of a state $|\psi\rangle$ is $\langle \psi|H|\psi\rangle$.
Now, the Hamiltionian is a very interesting operator because it features prominently in the equation of motion, the Schrödinger equation.
$$i\hbar \partial_t |\psi(t)\rangle = H |\psi(t)\rangle .$$
What does this have to do with the eigenvalues of the Hamiltionian? A priori nothing, but the point is that knowing the eigenvectors and -values of $H$ allow you to solve this equation. Namely, if you have an eigenvator $|\psi_n\rangle$, then you have
$$i\hbar \partial_t |\psi_n(t)\rangle = H |\psi_n(t)\rangle = E_n|\psi(t)\rangle$$
which can be solved to
$$|\psi_n(t)\rangle = e^{-\frac{i}{\hbar} E_nt} |\psi(0)\rangle$$
To summarize, the eigenvalues of an operator tells you something about what happens when you perform measurements, but in addition, the eigenvalues of the energy operator help you solve the equations of motion.
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The basic experimental fact the inventors of QM had to deal with was the uncertainty principle. The mathematics behind this principle has two major parts, one involving linear algebra and another involving Fourier analysis.
In other words, the operator algebra of QM is necessary in order to have a theory which obeys the uncertainty principle, and if you want to know why this is true, you have to study the mathematics.
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The physics of this is the DeBroglie relation for particles, which relates the energy to the frequency of some wave. The energy of a photon is the frequency of the emitted electromagnetic wave.
When a quantum mechanical atom is weakly interacting with the photon field, and goes from a state with frequency f to a state with frequency f', it can only emit photons with frequency f-f'. The reason is that the transition process is only resonant with waves of frequency equal to the beat frequency \Delta f= f-f'. The atomic relative phases during the transition process recurs with time $1\over \Delta f$, and for any outgoing wave whose frequency does not match this, the process will be cancelling at long times, and no wave will be emitted.
This means that atomic transitions from f to f' are accompanied by a loss of energy of $h\Delta f$, so that one must identify the frequency with the energy in general quantum systems. The Schrodinger waves of definite frequency are the solutions of the time independent problem, since when
$$i{d\over dt} \psi = H \psi$$
and $H\psi = E\psi$, that is, if $\psi$ is an eigenvector of H, then $\psi(t) = e^{-iEt} \psi(0)$, so the time dependence of the wave has a definite frequency. I am giving a physical argument here, because the notion that energy is frequency is engrained into the foundation of quantum mechanics, and it is hard to argue that it is true using a formalism built upon this as a foundation.
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http://mathhelpforum.com/pre-calculus/110426-incredibly-hard-malthus-problem-help-needed.html
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# Thread:
1. ## Incredibly Hard MALTHUS Problem Help Needed
I am here once again because I have no idea what I am doing. I am forced into taking this math class to graduate and we have this assignment which I can't figure out. I would be forever grateful for any help possible. The question is stated:
Using Malthus's sequence for unchecked population increase, what would the world's population be in 1900 and in 2000? First write the terms of the sequence. The T(zero) term, which is 1, represents the pop. in the year 1800. So under 1, write 1800 and the worlds population in 1800. Under the rest of the terms in the sequence write the appropriate years and the corresponding populations according to Malthus's sequence. Find the terms in the sequence corresponding to the years 1900 and 2000. You should have the unchecked population numbers for those years.
Then it says "If there was sufficient food for all the people in 1800, how many people would this sequence indicate enough food for in 1900 and in 2000? Use the sequence from the above problem and procedure . The T(zero) term which is 1, represents both the population in 1800 and the food population, since we are assuming enough food for the entire world population at that time.
the last part: "Does this sequence indicate enough food for the number you found above for the year 2000 World Population?"
Oh lord....shoot me now.
2. The population would grow according to a geometric series. The exact numbers would depend on the rate of growth (some percent per year, or common ratio).
The resources would grow according to an arithmetic series - linear growth. Again, the exact numbers would depend on the rate of growth (some number per year, or common difference).
The formulas are fairly straight forward. But without knowing the rates of growth, this is an impossible question. Did your text or teacher give you any rates you can use for these?
For the geometric (population) series, the $n^{th}$ term (n=1 fpr 1800, n=101 for the year 1900, and n=201 for the year 2000) is $u_n$. This can be found with the following formula: $u_n = u_1 * r^{n-1}$ where $u_1$ is the first term (1 in this case) and n is the number of the term you want to find. In 1900, for example, the population $u_{101}$ is: $1*r^{100}$ or just $r^{100}$. If the population is growing at a rate of 2% per year, for example, then r=1.02. If the population is growing at a rate of 5% per year, then r=1.05. If the population is doubling each year (growing at a rate of 100% per year), then r=2.
For the arithmatic (resource) series, the $n^{th}$ term (n=1 for 1800, n=101 for the year 1900, and n=201 for the year 2000) is $u_n$. This can be found with the following formula: $u_n = u_1 +d(n-1)$ where $u_1$ is the first term (1 in this case) and n is the number of the term you want to find. In 1900, for example, the population that could be fed is $u_{101}$ is: $1*d(100)$ or just $100d$. If the amount of resources is increasing by 4 per year, then d=4. If the amount of resources is increasing by 6 per year, then d=6. And so on.
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# Tagged Questions
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### How do I get the solutions of a cubic equation in trigonometric form?
I know $x^3-3 x+1=0$ has three roots that can be expressed in trigonometric form: $\{2\sin(10^\circ),\,-2\cos(20^\circ),\,2\cos(40^\circ)\}$. How can I get this result with Mathematica?
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### Both Sin[x]==0 && Cos[x]==0 as a solution
If I do FullSimplify[Reduce[Sin[p1] == 0 && Cos[p1] == 0, Reals]] I get Cos[p1] == 0 && Sin[p1] == 0 ...
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### How do I work with Root objects?
I want to solve the trigonometric equation : $$(3-\cos 4x )\cdot (\sin x - \cos x ) = 2.$$ I tried Solve[(3 - Cos[4*x])*(Sin[x] - Cos[x]) == 2, x] It returns the ...
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### Solving a trig equation
I am trying to find an explicit solution to a trigonometric equation. It is running for a long time and I didn't get any result, but Maple solves it quickly. Why ? ...
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# Talk:Law of total probability
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## Statement
I like the statement of the definition of total probability. It is readable and specialist terms are well referenced which enable the lay reader to gain some understanding of what is meant. Is there a good example of this proposition working in practice? Blueawr (talk) 19:04, 31 August 2012 (UTC)
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## Notations
I read somewhere that $P(A|B):=\frac{P(A\cap B)}{P(B)}$
in which case the two "different" notions of total probability would be aequivalent, since obvisously
$P(A\cap B_n)=P(A\mid B_n)P(B_n)$ for $n=1,2,3,\ldots$
Note however that i have very little knowledge of the whole domain, so maybe i just missed the point...
--lu 10:42, 26 Apr 2005 (UTC)
This is good. The article really should related to this. Jackzhp (talk) 15:55, 12 March 2011 (UTC)
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## conditional probability
The conditional probablity of A given B is only defined if the probability of B is non-zero. This is really unsatisfactory in this context, since you end up having to partition the space, avoiding zero-measure sets.
There must be a way around this tecnicallity. Or is it necessary? --67.103.110.175 02:55, 27 May 2006 (UTC)
OK, self, the statement should run more like, given Bi, mutually exclusive, whose probabilities sum to one ... Mathworld[1] So you can ignore that part of the space with zero probability, or include it, according to the situation. All that is important is that the part of the space with positive probability be covered. --67.103.110.175 02:55, 27 May 2006 (UTC)
It is absolutely necessary to be able to condition on continuous random variables. That entails conditioning on events of probability zero and getting nonzero conditional probabilities. Michael Hardy 21:52, 27 May 2006 (UTC)
Thanks for reverting and keeping the page accurate. However, in this case, it is a vanity and disservice not to mention that P(A|B) will not be defined for P(B)=0, in the discrete case. As for the continuous case, these are going to be a countable collection of events covering the space, so zero probability events remain insubstantial. Ozga 00:02, 28 May 2006 (UTC)
They are not at all insubstantial in the continuous case. When one finds E(Pr(A | X)). where X is a continuous random variable, the event on which one conditions always has probability 0. Michael Hardy 01:04, 30 May 2006 (UTC)
As in E(Pr(A | X))=Pr(A), as one reintegrates up the slices of A? --Ozga 17:28, 31 May 2006 (UTC)
I find this sentence confusing: "In the discrete case, the statements above are equivalent to the following statement, which also holds in the continuous case and is not the same as the law of alternatives." I think it should be reworded. Also, it is rather tricky to define the conditional probability of an event given the value of a continuous random variable. I think it is more or less
$\mathrm{Pr}(A|X=x) = \lim_{\epsilon\rightarrow 0} \frac{\mathrm{Pr}(A\cap \{x-\epsilon < X < x+\epsilon\})}{\mathrm{Pr}(\{x-\epsilon < X < x+\epsilon\})},$
provided that the density function of X is continuous at x, but take that with a grain of salt. --130.94.162.64 23:48, 15 June 2006 (UTC)
Indeed, it can be considered "tricky", since probably most people who grasp the idea intuitively do not know the Radon-Nikodym theorem, on which one of the usual definitions relies. But nonetheless perfectly doable. Michael Hardy 00:56, 16 June 2006 (UTC)
I still find that sentence confusing. From reading the entire article, I get the impression that the "law of total probability" has a perfectly unambiguous meaning that can be applied in a consistent way to both the discrete and continuous cases. The disclaimer that "nomenclature is not wholly standard" seems both a little unnecessary and rather foreboding to me. Perhaps instead, one could mention (in the beginning of that section) that in the discrete case it is also called the "law of alternatives". Also, the article should have references. --130.94.162.64 18:46, 17 June 2006 (UTC)
I think that's exactly what we should do. MisterSheik 22:34, 16 February 2007 (UTC)
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## conditional probability no2
The text states: "The law of total probability can also be stated for conditional probabilities. Taking the $B_n$ as above, and assuming $X$ is not mutually exclusive with $A$ or any of the $B_n$". Note that is assumes that $X$ is not mutually exclusive with any of the $B_n$. Doesn't this imply that $X$ is the sample space if each $B_i$ has a single event? — Preceding unsigned comment added by 134.58.253.57 (talk) 08:06, 11 May 2012 (UTC)
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## Definition
From my user talk page 83.67.217.254 14:39, 26 August 2007 (UTC)
Your edit was horribly wrong. We DO NOT want to condition on the event's actually occurring; we DO want to allow general random variables and not just indicator variables of events; and the fact that we sometimes condition on events of probability 0 and therefore need to talk about densiety functions in order to apply the result to those cases in no way means the identity is wrong. Michael Hardy 14:27, 26 August 2007 (UTC)
Hey, thanks for the flowers Michael! :-) You were referring to this edit, which you reverted with a charming and welcoming comment. I'll be back later. 83.67.217.254 14:39, 26 August 2007 (UTC)
In the meantime you may want to reflect on the fact that the definition of conditional probability does not involve random variables, but events. In other words, in the definition of P(A|B), both A and B are events, not random variables. 83.67.217.254 14:44, 26 August 2007 (UTC)
You are writing nonsense. The definition there is the definition of conditional probability given an event. There is also such a thing as conditional probability given a random variable. That is the concept needed here. I have a Ph.D. in statistics and I know this material. Michael Hardy 17:02, 28 September 2007 (UTC)]
And I repeat: the edit I reverted was horribly horribly wrong. Michael Hardy 17:04, 28 September 2007 (UTC)
I agree the definition is inconsistent —Preceding unsigned comment added by 221.47.185.2 (talk) 01:54, 28 September 2007 (UTC)
OK, I've got a minute or two and I will elaborate (barely):
• The conditional probability of an event given another event is just a number, not a random variable, so there would be no sense in evaluating its expected value, as this law does.
• To say that this is true of any event is clearly nonsense because almost any specific concrete example you pick would be a counterexample proving that the proposed law is in fact false. And I would claim that that is trivial and obvious.
• This law is found in many many many textbooks. Look it up. Don't insist on making a complete crackpot of yourself.
Michael Hardy 18:01, 28 September 2007 (UTC)
PS: The word definition that you've chosen as a heading is wrong. Concepts have definitions; propositions have statements. This is a proposition, not a concept. If you don't even know such elementary things, you shouldn't be acting that way you are. Michael Hardy 18:02, 28 September 2007 (UTC)
I have now added a new section to conditional probability that I hope will clear up the confusion on this point. Michael Hardy 19:47, 25 October 2007 (UTC)
I find it very interesting that your new definition almost exactly replicates my original edit. While I'm glad you are starting to understand, conditional probability given a continuous random variable is still not covered (unsurprisingly) by this novel definition of yours. Therefore, as I was suggesting in my edit comment, the definition of the law of total probability for continuous variables is still ill-defined. Horribly horribly yours, 83.67.217.254 13:01, 27 October 2007 (UTC)
It comes nowhere near replicating your edit, since you wrote of the conditional probability given an event, not about the conditional probability given a random variable, and your statement was clearly erroneous: you didn't need to understand what the law of total probability says in order to see that the statement as you wrote it was wrong, since virtually any example you picked would be a counterexample. I have not given any new definition in this article. I added a new section to a different article. There is nothing "novel" in the definition I gave. It is standard. I did not give the case of continuous random variables. The case I did give should make the idea clear.
Do not speak of "the definition of the law". One defines concepts; one states propositions. This law is a proposition, not a concept.
It is becoming apparent that you are here only to have fun arguing and you're dishonest. Michael Hardy 20:47, 27 October 2007 (UTC)
Other editors can judge for themselves whether or not your definition of conditional probability given a (discrete) random variable is equivalent to my edit on this article. You are entitled to your opinion and I am no longer interested in this aspect.
Since this definition of probability given a random variable is standard, you should have no problems referencing it (with page numbers if possible) and extending it (with references) to continuous variables. Until then, the current definition of the law of probability will not cover continuous variables.
By the way, as a minor point, I inform you that, despite your insistence to the contrary, I will keep writing about the "definition of the law", as opposed to, I suppose, the "statement of the law", since I find no ambiguity and you seem to understand what I mean perfectly well.
As for your accusation of "fun arguing"; I don't know what your concept of "fun" is, but I can assure you that your groundless and repeated personal attacks are not. In particular, I'd be interested to know when exactly I have been dishonest.
Also, in my user page you accuse me of "cowardice" for editing Wikipedia "anonymously". Firstly, I am not editing any more anonymously than you are. Wikipedia does not require the user names to correspond to real names, and provides no way to anybody to verify whether that is the case. If by "anonymous" you mean "unregistered", I'll have you know that I have been a happy unregistered user for quite some time, that Wikipedia welcomes unregistered users (and so should its editors) and that edits by unregistered users should not be discriminated against. This means by the way that you are not assuming good faith. Anyway, if you think I am a "coward" for not registering under my real name, I beg to differ and put to you that I find it foolish for anybody to register under their real name and run the risk (for some users apparently higher than for others) to have a permanent public record of being incompetent and behaving like a dick.
Finally, can I ask you to kindly rein in your use of bold and italics? It makes you look hysterical and desperate, and it's not conductive to a civil discussion. Thanks. 83.67.217.254 12:32, 28 October 2007 (UTC)
Definition of the law? The definition of a law, proposition or theorem makes as much sense as the definition of a sentence. Would you say, "The first sentence of the Gettysburg Address has 40 words, one main clause, one adverbial phrase, two adjectival phrases, . . ."? No, you would say, "The first sentence of the Gettysburg Address is, 'Four score and seven years ago . . .'" What you are calling the definition of the law is the law. What you see is what you get. Robert O'Rourke (talk) 04:37, 8 September 2010 (UTC)
It's starting to become apparent that you are a retired lawyer with a lot of time on his hands. Now you're trying to make it appear that I said it's cowardly to edit Wikipedia anonymously. You know that I said I have nothing against anonymous editing and that I sometimes edit anonymously myself. I never said you're a coward simply for editing anonymously. I said that your behavior is that of a coward and that in your case it is out of cowardice that you edit anonymously. Michael Hardy 16:19, 28 October 2007 (UTC)
Can you prove this accusation? Or are you just failing to assume good faith? Actually, don't answer that, please let's stick to the more important issue at hand. 83.67.217.254 16:28, 28 October 2007 (UTC)
I think I can prove that in some of your comments you acted as if the accusation is true. Michael Hardy 23:34, 31 October 2007 (UTC)
So you can't prove your accusation(s). 83.67.217.254 00:26, 1 November 2007 (UTC)
As I said, it's becoming apparent that you're a retired lawyer with time on his hands. Michael Hardy 02:40, 1 November 2007 (UTC)
I agree with Michael Hardy. The edit was clearly wrong and the reversion was necessary. You can't change the definition of a term in a proposition and expect it to remain true. In this case it is untrue. If N is an event, Pr(A|N) is a number. The expectation of a number is the number, so E[Pr(A|N)]=Pr(A|N). The law now states that Pr(A|N)=Pr(A), which isn't always true. Robert O'Rourke (talk) 22:27, 22 August 2010 (UTC)
Another problem with the edit [2] is its recommendation for bringing in the conditional density function. An authority in probability theory recommends an alternative approach: "We have now defined a conditional expectation E(Y|X) in terms of a conditional distribution, and this is quite satisfactory as long as one deals only with one fixed pair of random variables X, Y. However, when one deals with whole families of random variables the non-uniqueness of the individual conditional probabilities leads to serious difficulties, and it is therefore fortunate that it is in practice possible to dispense with this unwieldy theory. Indeed, it turns out that a surprisingly simple and flexible theory of conditional expectation can be developed without any reference to conditional distributions." [1]. He goes on to define condtional expectation given a sigma-algebra and conditional expectation given one or more random variables as the conditional expectation given the sigma algebra they generate. With condtional expectation defined, condtional probability can be defined as the conditional expectation of an indicator function. Robert O'Rourke (talk) 03:29, 25 August 2010 (UTC)
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## Law of total probability, conditional probability and conditional expectation
As per discussion above, the description currently given does not cover continuous random variables. I am no expert in this subject, but I suspect that the definition of the law currently suggested by this article cannot be extended to continuous variables without introducing the concepts of Borel sets, sigma field generated by a variable and the conditional expectation (as opposed to conditional probability) of a variable given that sigma field (see e.g. Mikosch 1998, sections 1.4.2--1.4.3).
I also suspect that the unreferenced definition of conditional probability given a random variable is original research. I believe that the "law of alternatives" reported in a separate section of the article is a better description covering discrete random variables. Incidentally, I think that contrary to what the article currently states, the partition defined by a discrete random variable is indeed a special case of a generic partition of the probability space.
Thanks for any insights. 83.67.217.254 23:09, 31 October 2007 (UTC)
It is not original research; it is found in many textbooks. I'll dig one out and cite it. For the most general definition, some measure-theoretic stuff is needed. Maybe I'll add them when I'm feeling more ambitious. But I think the very simplest example, which is the only one I gave, fully conveys what the concept is supposed to accomplish. 23:33, 31 October 2007 (UTC)
Yes, you do need to introduce Borel sets and sigma-algebras. To paraphrase Harry Truman, if you can't stand the integral, get away from the law of total probability.
It is a problem that Conditional expectation#Definition of conditional probability only goes as far as defining the conditional probability given a Borel sigma-algebra. Here [3] is a definition of conditional probabability given a random variable as the condtional probability given the Borel sigma-algebra generated by the random variable. With this definition, the proof is simple. Robert O'Rourke (talk) 22:52, 22 August 2010 (UTC)
Please do not be overly ambitious if this is not your area of expertise. Hopefully this request will attract an expert third opinion. 83.67.217.254 00:21, 1 November 2007 (UTC)
The discrete version doesn't require sigma-algebras; the most general versions do. Michael Hardy (talk) 23:24, 22 August 2010 (UTC)
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## References
1. Feller, William (1971). An Introduction to Probability Theory and Its Applications, Vol. 2, 2nd ed. New York: John Wiley & Sons. p. 162. ISBN 0471257095.
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Physics Forums
## What is the difference between electric field and magnetic field
What is the difference between electric field and magnetic field
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
From a special relativistic perspective, nothing--they're just different aspects of the same combined EM field, components that arise from different kinds of currents. I suspect this isn't the answer that would help you most, though. Perhaps you have a more specific question in mind?
Quote by Lizwi What is the difference between electric field and magnetic field
$$\nabla \circ \vec{E} = \frac{\rho}{\epsilon_{0}}$$
$$\nabla \times \vec{E} = -\frac{{\partial}{\vec{B}}}{{\partial}{t}}$$
vs.
$$\nabla \circ \vec{B} = 0$$
$$\nabla \times \vec{B} = \mu_{0} \vec{J} + \mu_{0} \epsilon_{0} \frac{{\partial}{E}}{{\partial}{t}}$$
heh, sorry, I couldn't resist..
## What is the difference between electric field and magnetic field
classically electric fields are caused by charge and magnetic fields are caused by moving charges in the static case. and if I have a time-varying E or B field they can induce the other field. And my relative motion with some stationary charge I could see a B field in my frame or an E field depending on the configuration.
Originally Posted by Lizwi View Post What is the difference between electric field and magnetic field ................... vs. heh, sorry, I couldn't resist..
This is not the complete story, please include Lorentz Force equation. To make difference
between E and B, it is very important to remember how do they interact with some particle, that is the only way to know about them( E or B).
as sanjib ghosh is getting at. The E field will affect a charged particle at rest by a B field will not cause a charged particle at rest to move.
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# Tagged Questions
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### Robust-Bayesian optimization in Markowitz framework
Suppose we are in the mean-variance optimization setting with a vector of returns $\alpha$ and a vector of portfolio weights $\omega$. In a robust setting, the returns are assumed to lie in some ...
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## Return to Answer
3 Answer GB's comment
If I understood your question correctly, the answer is yes. More precisely, the following statement should hold:
If $X$ is a closed manifold of positive sectional curvature and $Y\subset X$ is a codimension one totally geodesic submanifold that disconnects $X$, then $X$ is homeomorphic to a sphere.
This follows, as the OP suggests, from the Soul Argument of Cheeger-Gromoll, extended to Alexandrov spaces by Perelman (see, e.g., Section 6 of Perelman's notes). As mentioned in the comments, Cheeger-Gromoll's version of the argument actually suffices to get the conclusion.
A few details: denote by $C_1$ and $C_2$ the closure of the two connected components of $X\setminus Y$. These are positively curved compact Alexandrov spaces with boundary $Y$. On each of them, since the curvature is positive, the distance function to the boundary is concave. Therefore, the set of points at maximal distance (the soul) consists of a unique point. This implies that each $C_i$ is homeomorphic to a disk, hence $X=C_1\cup_{Y} C_2$ is a twisted sphere.
edit (to answer GB's comment): As discussed above, if $C$ a compact Alexandrov space with curvatures $\geq k>0$, then the soul $S=\{p\}$ is a point. Moreover, according to Perelman, the pairs $(C,\partial C)$ and $(\overline K(\Sigma_S),\Sigma_S)$ are homeomorphic (see 6.2 for proof), where $\Sigma_S$ is the space of directions at the soul and $\overline K(\Sigma_S)$ is the closure of the topological cone over $\Sigma_S$, i.e., the join of $\Sigma_S$ and a point. If $C$ is a manifold, the space of directions are spheres, so $(\overline K(\Sigma_S),\Sigma_S)$ is simply a pair $(D,\partial D)$, where $D$ is a disk.
2 Incorporated that Cheeger-Gromoll's version is enough
If I understood your question correctly, the answer is yes. More precisely, the following statement should hold:
If $X$ is a closed manifold of positive sectional curvature and $Y\subset X$ is a codimension one totally geodesic submanifold that disconnects $X$, then $X$ is homeomorphic to a sphere.
This follows, as you saythe OP suggests, from the Soul Argument of Cheeger-Gromoll, extended to Alexandrov spaces by Perelman (see, e.g., Section 6 of Perelman's notes).
More preciselyAs mentioned in the comments, Cheeger-Gromoll's version of the argument actually suffices to get the conclusion.
A few details: denote by $C_1$ and $C_2$ the closure of the two connected components of $X\setminus Y$. These are positively curved compact Alexandrov spaces with boundary $Y$. On each of them, since the curvature is positive, the distance function to the boundary is concave. Therefore, the set of points at maximal distance (the soul) consists of a unique point. This implies that each $C_i$ is homeomorphic to a disk, hence $X=C_1\cup_{Y} C_2$ is a twisted sphere.
1
If I understood your question correctly, the answer is yes. More precisely, the following statement should hold:
If $X$ is a closed manifold of positive sectional curvature and $Y\subset X$ is a codimension one totally geodesic submanifold that disconnects $X$, then $X$ is homeomorphic to a sphere.
This follows, as you say, from the Soul Argument of Perelman (see, e.g., Section 6 of Perelman's notes).
More precisely, denote by $C_1$ and $C_2$ the closure of the two connected components of $X\setminus Y$. These are positively curved compact Alexandrov spaces with boundary $Y$. On each of them, since the curvature is positive, the distance function to the boundary is concave. Therefore, the set of points at maximal distance (the soul) consists of a unique point. This implies that each $C_i$ is homeomorphic to a disk, hence $X=C_1\cup_{Y} C_2$ is a twisted sphere.
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http://mathematica.stackexchange.com/questions/tagged/function-construction+variable-definitions
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# Tagged Questions
2answers
73 views
### Creating functions from output of other calculations
Apologies in advance if the title is vague, I'm not really sure what to call this. I have a function (call it 'foo') that generates a largeish polynomial, and it is natural to make the variables be ...
1answer
109 views
### Module with “local functions”
I am trying to use a Module having "local functions", i.e., those which I need to define only inside this module. So I tried this: ...
1answer
87 views
### Is it safe to assign a variable and function of the same name for different things?
I'm writing out a notebook that goes through the van der Waals Equation of State for gases, and I run into a situation where I want to assign (simplified) Tc[b_]:=5b, use that to solve for b in terms ...
0answers
76 views
### Evaluating a function on permutations of its arguments
Say I have a function "temp" of $n+1$ variables, $y,z1,z2,z3,...,zn$. I want to test if my function has certain symmetries like swapping $y$ with square of any $z$, swapping any two of the zs, ...
3answers
137 views
### Better solution than returning a list of 3 values?
I have a function (using SetDelayed) that currently returns 3 values in a list. Later on I use the result of this list along with ...
1answer
143 views
### Is it possible to Clear all variables (but not functions)?
I have written a Mathematica script in which I define functions and variables. Here is a vastly simplified example: ...
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http://agtb.wordpress.com/2009/04/28/a-new-randomized-mechanism-for-multi-unit-auctions/
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# Turing's Invisible Hand
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## A new randomized mechanism for multi-unit auctions
April 28, 2009 by algorithmicgametheory
Shahar Dobzinski and Shaddin Dughmi just posted a new paper to the arXiv: On the Power of Randomization in Algorithmic Mechanism Design.
The mechanism design problem that they study in this paper is the basic one of a multi-unit auction: There are $M$ identical indivisible units of some good for sale among $n$ bidders, and each bidder $i$ has a valuation function $v_i : \{1 ... M\} \rightarrow \Re^+$, where $v_i(k)$ specifies his value from acquiring $k$ units. The normal assumptions are free disposal i.e. that the valuation functions $v_i$ are monotone non-decreasing. The goal is to find an allocation that gives $k_i$ units to each each bidder $i$ (and thus $\sum_i k_i \le M$) optimizing the social welfare $\sum_i v_i(k_i)$.
This problem is perhaps the simplest one exhibiting the key clash between computational complexity and incentive compatibility: since this is a social welfare maximization problem, the VCG mechanism that finds the optimal allocation is incentive compatible. However, if we view the number of units $M$ as large, and require algorithms that work in time polynomial in $n$ and $\log M$, (assuming that the valuation functions are either concisely represented or are given as “black boxes”) then finding the optimal allocation is a knapsack problem and is thus computationally hard. On the other hand, it is well known that the knapsack problem can be well-approximated efficiently (there is an FPTAS). The clash is that we don’t know how to turn such an approximation algorithm into an incentive-compatible approximation mechanism.
This paper manages to find a randomized incentive-compatible approximation mechanism. It also highlights the role and necessity of randomization in this setting.
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1. [...] On the Power of Randomization in Algorithmic Mechanism Design by Shahar Dobzinski and Shaddin Dughmi. (A link to the paper and some discussion in this blog post.) [...]
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http://physics.stackexchange.com/questions/5705/most-elegant-fundamental-formulations-of-the-laws-of-classical-mechanics/5710
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# Most elegant/fundamental formulations of the laws of classical mechanics?
Newton tried to do it with three laws/statements. While the first can be derived from the second, the three form a pretty nice framework.
Later on, I've encountered Lagrangian Mechanics, which involves, from what I gather, one statement:
• Objects seek a path that minimizes total action.
Which, while sort of simple, doesn't sound too natural or fundamental.
I've heard some of these rather elegant statements:
• The universe is translation-invariant
• The universe is invariant under different inertial frames of reference
And from either of these, one may derive Newton's laws and F=ma. However, I'm not completely sure how this can be done. Were these claims true? How is this possible?
Has anyone ever come across a surprisingly/particularly stunningly elegant/fundamental formulation of classical mechanics, and a proof that they are equivalent to Newtonian Mechanics?
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@Justin: could you give a reference to statements you post? I would not know how you could construct some sort of formulation of mechanics from these assumptions on symmetries/conserved quantities. I think that the principle of stationary action might be already quite fundamental as it seems to explain most of the physics we know of today, from quantum field theory to general relativity. – Robert Filter Feb 23 '11 at 9:01
"Which, while sort of simple, doesn't sound too natural or fundamental." -> it's not fundamental because it's derived from quantum mechanics but on the classical level you can't get anything better than this. It's the single most unifying principle in classical physics and you can use it for everything, not just mechanics. If you can somehow guess the right Lagrangian, the rest of the physics falls out. – Marek Feb 23 '11 at 9:01
@Robert: quantum field theory? In quantum theory the statement definitely doesn't hold, even non-classical trajectories contribute to the final amplitude. You are perhaps thinking of free fields (where the only contribution to path integral is classical thanks to their being gaussian) and instantons? – Marek Feb 23 '11 at 9:04
@Robert - I can't, really; I think I picked it up somewhere as an offhand statement. But from the Lagrangian, it can be derived that a translation-invariant system is one that conserves momentum, so I figured they weren't too off. Perhaps they are completely wrong; I just appreciated the elegance of the statements as a formulation of the natural laws. – Justin L. Feb 23 '11 at 9:05
@Marek - Maybe it's just my own personal bias, but there seems to be just something less-natural about the principle of stationary action, than something like "objects resist changes to their velocity proportional to their mass". – Justin L. Feb 23 '11 at 9:07
show 8 more comments
## 3 Answers
The principle of least action is much more elegant and fundamental than the explicit formulation of the laws of mechanics using three laws of Newton which disproves your intuition that the principle of least action is not elegant or fundamental.
The two symmetries you mention are symmetries that constrain the evolution of physical systems but they don't totally determine what the evolution looks like. It is not true that the symmetries you mention are enough to derive the forces among a set of objects. $F=ma$, without specifying how $F$ depends on the positions (e.g. gravitational force), is just a definition of $F$ and it is a vacuous statement so you may derive it from anything - even from no assumptions at all.
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I feel that there is something about the principle of least action's elegance that I am not understanding. Can you explain the naturalness in it? I've so far taken no courses with it; I've only worked with the mathematics and found out that it worked. – Justin L. Feb 23 '11 at 9:15
š Not that I disagree with the elegance of least action, but your opening paragraph is (unusually for your answers) rather empty. You just state that least action is elegant and fundamental, then claim that this statement disproves the idea that it is not. You might want to consider editing out the second half of that sentence, at least. – Mitchell Feb 23 '11 at 14:35
Dear Mitchell, well, the elegance or non-elegance is obviously an aesthetic issue, and I didn't start with the evaluation of it. I just gave an answer to a question which contained a proposition that it was not elegant. The more properly one understands classical physics, the more he knows that the principle of least action is more elegant than Newton's laws themselves. It generalizes to a universal (Feynman) rule in quantum mechanics, and so on. The point of my paragraph was just that aesthetic prejudices may be showed wrong, and this was the case here, too. – Luboš Motl Feb 24 '11 at 6:36
Dear Justin, I don't understand what you mean by "naturalness of the principle" and how such a naturalness may be "showed". The principle of least action is natural in the colloquial sense because experiments show that its predictions are obeyed in Nature; it is natural in the physics sense because it doesn't introduce new free parameters with values very different from one; and for a similar reason, it is a robust and universal principle. I am afraid that every other interpretation of "naturalness" you may have adopted is just totally unscientific. Maybe you just meant you're not used to it? – Luboš Motl Feb 24 '11 at 6:39
The most geometric formulation of classical mechanics is in terms of symplectic geometry. However in terms of the question asked about the principle of least action, the issue is that this formulation
Objects seek a path that minimizes total action.
is mathematically vague, and that the phrase (and definition of action) can have different variational forms. Some of these are more intuitive and some are more general than others.
The most general formulation is Hamilton's Principle:
The motion of the system from time $t_1$ to time $t_2$ is such that the line integral
$I = \int_{t1}^{t2} L dt$
where $L= T - V$ is a stationary point for the path of motion.
The elegance of this is that it identifies one path from amongst all those between initial position at time $t_1$ and final position at time $t_2$ which extremizes the path. The points and paths here belong to the multidimensional configuration space of the system. Not only that but the remaining equations of motion can now be derived from this for all classical systems, with an extension into quantum mechanics as well.
However there are other (older) formulations of similar variational principles which can also refer to "action" and which have use in classical mechanics too. The classic text Goldstein introduces a "Principle of Least Action" which is defined a little differently. If we introduce the Hamiltonian of a system H and assume that it is conserved:
$\Delta \int_{t1}^{t2} \Sigma p_i q_{i}^{dot} dt = 0$
The term inside the integral is also known as the Action and the variation involved (which is minimized) is denoted by $\Delta$.
To understand this formulation let us consider only the simplest case of it: assume non-relativistic mechanics with no time dependence and no external forces. Then this expression reduces to:
$\Delta(t_1 - t_2) =0$
This says that of all paths possible between two points, consistent with conservation of energy, the system moves along that particular path for which the time of transit is the least.
Other variations of this formulation can have the particle moving along geodesics within configuration space. Some of these formulations are similar to Fermat's principle of Geometric Optics.
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the most elegant form of classical mechanics comes from newton's laws:
1. f =ma
2. action = reaction
3. bodys remain at rest unless acted on by other forces.
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http://mathoverflow.net/questions/102374/extensions-of-topological-groups/102424
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## Extensions of topological groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose I have a central extension $1 \to U(1) \to \hat{G} \to G \to 1$ of a topological group $G$ by the circle group $U(1)$ in such a way that $\hat{G} \to G$ is a principal $U(1)$-bundle. Moreover, suppose that $G$ is $3$-connected. In particular, $\hat{G} \cong G \times U(1)$ as topological spaces.
Is the $3$-connectedness of $G$ enough to deduce that $\hat{G} \cong G \times U(1)$ as topological groups?
Such a central extension is described by a class in continuous group cohomology, but I can't see whether the connectedness of $G$ is any help in proving that this class is actually trivial.
In my case, $\hat{G}$ and $G$ are Banach Lie groups. So, feel free to assume that as well.
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Sounds dodgy! Why cannot you just do a quotient of the Heisenberg group? I mean $R\times R \times S^1$ with the product $(u,v,a)\cdot (x,y,b)=(u+x,v+y,ab(uy-vx+Z))$... – Bugs Bunny Jul 16 at 20:58
@Bugs Bunny: The quotient you describe is not $3$-connected. – Mark Grant Jul 17 at 14:23
You can induce from the exponential sequence a long exact sequence in continuous group cohomology. If $G$ is compact, it is known that $H^n(BG,\mathbb{R})=0$, and you get $H^2(BG,U(1))=H^3(BG,\mathbb{Z})$. The latter vanishes by your connectedness assumptions. But I guess your $G$ is not compact... – Konrad Waldorf Jul 17 at 21:26
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@Mark Grant, so? The quotient is $\widehat{G}$ and it cannot be 3-connected at all. My $G$ is 3-connected: it was even contractible this morning... – Bugs Bunny Jul 18 at 7:16
I guess it is even an answer then.... – Bugs Bunny Jul 18 at 7:16
show 1 more comment
## 2 Answers
No, $G=({\mathbb R}^2 ,+)$ and a certain quotient $\widehat{G}$ of the Heiseberg group is an example. Namely, $\widehat{G}={\mathbb R}^2\times S^1$ with the product $(x,a)⋅(y,b)=(x+y,ab(\omega(x,y)+{\mathbb Z}))$ where $\omega$ is a nonzero bilinear form.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Bugs is right. If $G$ is locally compact abelian and multiplication by $2$ (or squaring) is an automorphism of $G$, then $H^2(G,U(1))$ is isomorphic to the group of alternating bi-characters $G\times G\to U(1)$. This is discussed, for example, in Mumford's Tata Lectures on Theta III, and in my paper Locally Compact Abelian Groups with Symplectic Self-duality.
This points to a nice criterion for the vanishing of $H^2$ in this case, namely, $G$ should not admit non-trivial alternating bicharacters. $\mathbf R$ satisfies this condition. So do all groups which have dense locally cyclic subgroups (see Section 5 of my article).
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http://www.physicsforums.com/showthread.php?t=14475
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Physics Forums
## Simple Motion Problem I Cannot Figure Out
Hi everyone, I am new to these forums and thought this would be a good place to get some help.
I picked up a three dollar russian made math problem book at a local book sale. I am working on the questions it has in it and I came across the first problem on motion it has and I cant quite figure it out. This is high-school level stuff by the way so I would hope its easy for many here.
Problem:
The train left station A for station B. Having travelled 450 km, which constitues 75 percent of the distance between A and B, the train was stopped by a snow-drift. Half an hour later the track was cleared and the engine-driver, having increased the speed by 15 km per hour arrived at station B on time. Find the initial speed of the train.
I have tried to visualize the problem. This is what I have come up with.
The question has no total time and you have to find speed. So are there two unknowns to solve? And how do I do that in this problem? Or is there a different way to solve this? Thanks for any help.
Let the required time be T If there would have been no problem due to snow then the train wud have reached the station B in time T Now find sepearate times for two part of motion arising dua to a break of 0.5 hr + the time for break equate it will T
Mathematical Form $$\frac{D}{v}=\frac{3D}{4v}+\frac{D}{4(v+15)}+ 0.5$$
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## Simple Motion Problem I Cannot Figure Out
Note that this problem does require an assumption that is not explicitly stated in the problem discription (but which could however be quite reasonably argued as being "common sense"). The assumption of course is that the inital speed of the train was precisely the correct speed to make the train arrive at the destination exactly on time.
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http://www.reference.com/browse/wiki/Tensor_product_of_quadratic_forms
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Definitions
# Tensor product of quadratic forms
The is most easily understood when one views the quadratic forms as quadratic spaces. So, if (V, q_1) and (W, q_2) are quadratic spaces, which V,W vector spaces, then the tensor product is a quadratic form q on the tensor product of vector spaces $V otimes W$.
It is defined in such a way that for $v otimes w in V otimes W$ we have $q\left(v otimes w\right) = q_1\left(v\right)q_2\left(w\right)$. In particular, if we have diagonalizations of our quadratic forms (which is always possible when the characteristic is not 2) such that
$q_1 cong langle a_1, ... , a_n rangle$
$q_2 cong langle b_1, ... , b_m rangle$
then the tensor product has diagonalization
$q_1 otimes q_2 = q cong langle a_1b_1, a_1b_2, ... a_1b_m, a_2b_1, ... , a_2b_m , ... , a_nb_1, ... a_nb_m rangle.$
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http://mathoverflow.net/questions/99638?sort=newest
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## Quotient space of algebraic group
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $H \subset G$ closed subgroup of an algebraic group. We want to prove the existence of the quotient $G/H$ which is a quasi-projective variety and homogeneous G-space.
We can find a vector $0 \ne y_0 = [v] \in \mathbb{P}^n$ such that $\forall h \in H: h\cdot y_0 = y_0$ and thus $H = \mathrm{Stab}_G(y_0) =G_{y_0}$ (stabilizer) and $G/H =$ the orbit of $y_0$ under $G$.
We define a group action G-morphism $\varphi: G \to Y$ by $g \mapsto g\cdot y_0$. We want to prove that the differential $(d \varphi)_e$ is separable and surjective.
The tangent space of an alebraic group is a derivative module $\mathrm{Der}_k(k[G],k)$ where $D \in \mathrm{Der}_k(k[G],k)$ is a k-linear derivation (i.e. $\forall x \in k: D(x)=Dx = 0$). We also define a derivation $\delta \in \mathrm{Der}_k(k[G],k[G])$ which returns a derivative function. We call $$\mathcal{L}(G) = \left( { \delta \in \mathrm{Der}_k(k[G],k[G]) | \delta \mbox{ is left invariant}, \delta \lambda_g(f(x)) = \lambda_g \delta f(x) . } \right)$$ There is isomorphism between these spaces, $D \mapsto \delta_D$.
According to [Springer, "Linear Algebraic Group", 4.4.7, p.72] we deine $J$ to be the ideal in $k[G]$ of functions that vanish on $H$. We want to prove $$T_eH = ( { D \in T_eG | \delta_D I \subseteq I } )$$ (this is a set and a tagent space).
After that we want to show that $T_eH = \ker (d \varphi)_e$ and by a theorem it is equivalent that $\varphi$ is separable.
I tried but didn't managed to prove the last two assertions (the two equalities about $T_eH$), so I need some help with this.
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I would change the title to "Quotient space of algebraic group," since you are asking about taking the quotient by an arbitrary (not necessarily normal) closed subgroup. And why do you say $G/H$ is projective? This is often false: e.g. take $G$ to be the Heisenberg group and $H$ its center. – Justin Campbell Jun 14 at 19:32
We build $Y=G/H$ as a projective space in order to have a H-invariant vector $0\ne v \in k^n$ such that $\forall h \in H : h \cdot v = kv \in \mathrm{Span}_k(v)$. However, this is not the important thing in my question. – Zachi Evenor Jun 14 at 19:43
1
There is the problem that $H$ had better be the scheme theoretic stabilizer of the vector $v$, not just the algebraic group stabilizer. In other words, things may go wrong if you do not require them to go right. It is a condition on the construction of $v$ that the Lie algebra of $H$ is the full stabilizer of the point $v$ in the Lie algebra of $G$. It can fail. – Wilberd van der Kallen Jun 15 at 10:15
@Wilberd van der Kallen We proved a Lemma that said that the ring of regular functions on the relevant open sets is stabilized by $H$. Let $U \subset Y$ be an open set, then one can show (it is a long proof, though), that $$\mathcal{O}_Y(U) \circ \varphi = \left( \mathcal{O}_G(\varphi^{-1}(U)) \right)^H$$ so I guess this satisfies the demands regarding the scheme. Note that in class we havn't used the "schemes" and "sheaves" concepts and use more basic tools (Alg. group, Lie algebras, ringed spaces, differential etc). – Zachi Evenor Jun 15 at 11:18
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@ Zachi Evenor. I am not worried if $H$ stabilizes, I am worried if there may be more elements in the Lie algebra of $G$ that also stabilize $v$. In that case we say that the scheme theoretic stabilizer is bigger. So my problem is that what you try to prove is not always true. For example, suppose $G$ is the general linear group and we are in positive characteristic $p$. If you compose the action of $G$ with the Frobenius map from $G$ to $G$ (that raises all entries to the $p$-th power), then suddenly the full Lie algebra of $G$ acts trivially and there is no way to prove the desired formula. – Wilberd van der Kallen Jun 15 at 12:00
## 2 Answers
@Wilberd van der Kallen : The comment option doesn't work for me right now (probably because I am logged from two different computers) so I write here.
I think I understand that by left invariance a function vanishing in $H$ can be checked for $e$ only, since $$\delta \lambda_h f(x) = \lambda_h \delta f(x)$$ and we may factor $x = he$ for $x \in H$. Tell me if that is correct.
However, I don't understand the $m / m^2$ construction, and why it represents the module (I know that $m/m^2$ is a representive element with $\delta f(x) = f - f(e)$ but don't understand the implementation to a given derivative module, how to represent the general derivative by $\delta$).
(in the general case: let $B = A \oplus M$ direct sum of mudules where $A \subset B$ and $M$ is a maximal ideal in the ring. Then we defined $$\pi_A : B \to A \ ; \ \pi_A(a+m)=a$$ and $\delta (f) = (f - \pi_A(f))$ and then $M/M^2$ is a representative for a derivative module, since $\delta$ is a derivative.).
I also don't understand what do you mean by $m$-adic completion.
Thanks and sorry for the late answer.
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@Zachi Evenor The definition of a tangent space in terms of $m/m^2$ is standard and the notion of $m$-adic completion is also standard. The latter just means that one computes with power series expansions in terms of local coordinates. Ask anyone knowing enough algebraic geometry. I am not using the description of the tangent space in terms of left invariant derivations because for me it obscures what is happening. – Wilberd van der Kallen Jun 30 at 12:36
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You want to show that $T_eH=\{D\in T_eG\mid\delta_DI\subset I\}$. This is something general about smooth subvarieties of a variety. The left hand side maps to the right hand side by functoriality of tangent spaces. Now $H$ is a smooth subvariety and the dimension of its tangent space is the dimension of $H$. So we must understand that the right hand side is no bigger. But for this one may look in local coordinates at $e$. Say $m$ is the ideal of functions vanishing at $e$. Then one may compute with $m/m^2$ to check that preserving $I$ imposes enough linear restrictions to bound the dimension. As an algebraist I would first pass to the $m$-adic completion which is a ring of power series. In that ring the ideal $I$ looks very simple.
The other equality cannot be proved without further data on the construction of $v$. But basically it is again a statement that things are OK in local coordinates, now around $v$. The condition on an element of $T_eG$ that it does not move $v$ must now impose enough linear restrictions to bound the dimension again.
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I have to think about it. – Zachi Evenor Jun 17 at 19:57
@Wilberd van der Kallen I replied your answer in a different answer (see below) due to technical reasons. I still have questions as written there. Thanks, – Zachi Evenor Jun 24 at 12:53
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http://www.physicsforums.com/showthread.php?p=4078268
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## probability of overlapping random pulses
I have a problem calculating the following probability.
There are two signals A and B each consisting of a series of "pulses" at times
{tA0, tA0+Δt, tA1, tA1+Δt, tA2,tA2+Δt, ...} and
{tB0, tB0+Δt, tB1, tB1+Δt, tB2, tB2+Δt,...}
The signal A is "on" in the time intervals [tAn, tAn+Δt], and it's off in the time intervals [tBn+Δt, tBn+1].
There is a given probability for the signal A to be "on" depending on the (random) times between the pulses; the same applies for the signal B.
How can one calculate the probability that for a certain time T both signals are "on"
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Recognitions: Science Advisor You haven't stated a precise question. What does "between the pulses mean?". What are the "the pulses"? Is signal A the only signal that is pulsing. What's the difference between being "on" and being in the state of emitting a pulse? Do the times $t^B$ have anything to do with signal B? What are the given probabilities and what events do they describe? Does the situation involving independent events of some kind? Are you trying to find the probability that A and B are both "on" for an instant of time? For at least some interval of time? Do you want the calculate that this ever happens once ( between time= 0 and "infinity"?) or do you want to calculate the mean number of time this events happens in an hour - or something like that?
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Quote by Stephen Tashi You haven't stated a precise question.
Really? OK, not very precise, I agree.
Quote by Stephen Tashi What does "between the pulses mean?".
Between means
Quote by tom.stoer The signal A is ... off in the time intervals [tBn+Δt, tBn+1].
That's between the pulses [where it's on]
Quote by Stephen Tashi What are the "the pulses"?
A pulse is when a signal is "on"
Quote by Stephen Tashi Is signal A the only signal that is pulsing.
No; there are two sequences of pulses:
Quote by tom.stoer There are two signals A and B each consisting of a series of "pulses" at times {tA0, tA0+Δt, tA1, tA1+Δt, tA2,tA2+Δt, ...} and {tB0, tB0+Δt, tB1, tB1+Δt, tB2, tB2+Δt,...}
Quote by Stephen Tashi What's the difference between being "on" and being in the state of emitting a pulse?
Nothing; during a pulse the signal is “on”; between the pulses it’s "off".
Quote by Stephen Tashi Do the times $t^B$ have anything to do with signal B?
Of course, what else should the superscript B indicate??
Quote by Stephen Tashi What are the given probabilities and what events do they describe? Does the situation involving independent events of some kind?
I haven’t specified the probability for the times between the pulses. I hope one can find some kind of general ansatz. If not one may assume some kind of normal distribution. The different times where the signals are off are independent. So both two subsequent off-times of the same signal are independent as well as the two signals A and B.
Quote by Stephen Tashi Are you trying to find the probability that A and B are both "on" for an instant of time? For at least some interval of time?
I try to find the probability that that for some time t (picked randomly) both signals are “on”.
Quote by Stephen Tashi … do you want to calculate the mean number of time this events happens in an hour - or something like that?
yes.
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## probability of overlapping random pulses
Quote by tom.stoer Of course, what else should the superscript B indicate??
You said:
The signal A is "on" in the time intervals [tAn, tAn+Δt], and it's off in the time intervals [tBn+Δt, tBn+1].
Did you mean to say A is off in the time intervals $[t_{A_n} + \triangle t, t_{A_{n+1}}]$?
I haven’t specified the probability for the times between the pulses. I hope one can find some kind of general ansatz. If not one may assume some kind of normal distribution.
You'd have to define the random variables that are going to have the distribution. What would they represent? Duration of pluse? Time between pulses? Time between change of state between pulse and non-pulse? Without knowing the physics of the the problem, it isn't clear what random variables to represent.
Normal distributions can produce values arbitrarily smaller than their mean so it isn't clear how to interpret these as durations of time, which are bounded below by 0.
I think this general type of problem has well-known solutions but the details are going to depend on how the phenomena of the pulses is modeled. For example, two computers trying to communicate on the same ethernet wire is one scenario. The overlap of two claps of thunder is another.
Hey tom.stoer. For your problem, can we safely say that you want to find a distribution where you have two independent processes (corresponding to A and B) where you want to find P(A in a = X, B in b = Y) where a is an interval region in A (corresponding to a time interval) and b is the same for B where X corresponds to the number of pulses in that interval and B corresponds to the number of pulses in that respective interval?
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Quote by Stephen Tashi Did you mean to say A is off in the time intervals $[t_{A_n} + \triangle t, t_{A_{n+1}}]$?
Oh sh..; of course I mean that!
Quote by Stephen Tashi You'd have to define the random variables that are going to have the distribution. What would they represent? ... Time between pulses?
Yes, time between the pulses. The duration of the pulses is Δt = const.
Quote by Stephen Tashi I think this general type of problem has well-known solutions but the details are going to depend on how the phenomena of the pulses is modeled. For example, two computers trying to communicate on the same ethernet wire is one scenario. The overlap of two claps of thunder is another.
Yes, something like that. Do you know any reference?
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Quote by chiro For your problem, can we safely say that you want to find a distribution where you have two independent processes (corresponding to A and B) where you want to find P(A in a = X, B in b = Y) where a is an interval region in A (corresponding to a time interval) and b is the same for B where X corresponds to the number of pulses in that interval and B corresponds to the number of pulses in that respective interval?
As far as I understand - no.
The distribution of "off-times" between the pulses shall be some reasonable distribution (not a normal distribution - that's nonsense - I agree, but e.g. uniform distribution or Poisson distribution, ...).
Then I would like know a general ansatz how to calculate the probability that for some arbitrary time t both signals are "on".
Let's look at one signal. The probability for one signal to be "on" is just
$$P^A_\text{on} = \lim_{T\to\infty} \frac{T_\text{on}}{T}$$
Is it allowed to use a different limit, namely
$$T^A(n) = t^A_n$$
$$T^A_\text{on}(n) = n\,\Delta t$$
$$P^A_\text{on} = \lim_{n\to\infty} \frac{T^A_\text{on}(n)}{T^A(n)} = \lim_{n\to\infty} \frac{n\,\Delta t}{t^A_n}$$
That would mean that all one has to do is to calculate the times, i.e. to write down and evaluate a sum over random variables. Then my guess would be that the probability for both signals being "on" is just the product
$$P^{A\,\text{and}\,B}_\text{on} = P^A_\text{on} \cdot P^B_\text{on}$$
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Quote by tom.stoer or Poisson distribution
A Poission distribution is a distribution of "a number of counts", so you'd being using integer multiples of time if you did that.
$$P^A_\text{on} = \lim_{n\to\infty} \frac{T^A_\text{on}(n)}{T^A(n)} = \lim_{n\to\infty} \frac{n\,\Delta t}{t^A_n}$$
It isn't clear what expression means because ${t_A}_n$ is a random variable, not an ordinary variable. ${t_A}_n$ it isn't a deterministic function of $n$. If we think of $\{ {t_A}_n\}$ as a sequence of things, the things aren't single numbers. It's a sequence of random variables and the usual definition for such a sequence (if it has one) is that the limit is another random variable.
If you want a "ansatz" in the sense of a mere guess, you might try taking the limit, as time T, approaches infinity of a fraction involving the products: (the expected number of pulses of the signal that happen in time T)( delta T).
You haven't described any practical goal of you analysis. If you are trying to settle a academic controversy or a bet, then calculating the probability that both signal are on "at a randomly selected time" may answer that purpose. I don't see that this probabiliy has practical use otherwise. For example, you can't take that answer and formulate a distribution for the length of time intervals during which the signals overlap.
A guess about the general mathematics of your problem is that it is a "Markov renewal process". The four states of the process are: (A off, B off), (A on, B off), (A on, B off), (A on, B on). However, we'd have to think about it carefully to verify that.
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Quote by Stephen Tashi A Poission distribution is a distribution of "a number of counts", so you'd being using integer multiples of time if you did that.
Not really. The question regarding pulses in a certain time intervall can be asked regarding arbitrary real values [t, t+Δt]; the Δt which is the duration of the pulse "on" has nothing to do with the Poisson distribution (or any other distributation).
Quote by Stephen Tashi It isn't clear what expression means ... It's a sequence of random variables and the usual definition for such a sequence ... is that the limit is another random variable.
I think you know what I have in mind; what would be the correct ansatz?
Quote by Stephen Tashi You haven't described any practical goal of you analysis.
Think about two signals A and B with fixed Δt and some distribution of the times {tA0, tA1, ...} and {tB0, tB1, ...}. What is the probability that for any time t two pulses from the signal A and B are overlapping?
Quote by Stephen Tashi A guess about the general mathematics of your problem is that it is a "Markov renewal process". The four states of the process are: (A off, B off), (A on, B off), (A on, B off), (A on, B on).
Thanks for the hint. I checked the properties of the Marnkov renewal process. It seems that what I have in mind is not one renewal process (with above mentioned states) but a pair of such processes.
One question for the OP has to do with these pulses. If these are pulses in the form of a Dirac-delta type pulse, then the only way they can over-lap is if they occupy the exact same position. You have mentioned that you want to consider an interval with some delta t value, but if these delta's go to infinity and your pulse is basically a pure theoretical Dirac-delta t kind of event, then if you are considering a continuum for the domain, then the probability of getting a pulse in this Dirac-delta style event will be 0 anyway. If you want to consider a non-zero length finite interval, then this is a little different because you can use something based on a Poisson process, or if you want to stick to the continuum, you use a distribution for the "waiting time" till an event takes place as opposed to using a rate parameter (like the Poisson does). So if you consider modelling your pulse process as a "waiting time" model for both signals and then look at the relevant probabilities, that might serve your interest.
The problem is non-trivial but significantly easier if you can assume that the pulses follow a Poisson process - so that, conditional on the number of pulses, the pulse times are uniformly distributed within the interval (with a slight modification if the inter-pulse times follow a Poisson process, rather than the pulse start times). I'm not sure if this will lead to an analytic solution but it should at least be approximated with Mone Carlo methods.
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Quote by chiro If these are pulses in the form of a Dirac-delta type pulse, then the only way they can over-lap is if they occupy the exact same position.
It's about rectangular pulses which are "on" / "off".
Quote by chiro You have mentioned that you want to consider an interval with some delta t value, but if these delta's go to infinity and your pulse is basically a pure theoretical Dirac-delta t kind of event, then if you are considering a continuum for the domain, then the probability of getting a pulse in this Dirac-delta style event will be 0 anyway.
Δt is finite, i.e. 0 < Δt < ∞, and const.
Quote by chiro If you want to consider a non-zero length finite interval, then this is a little different because you can use something based on a Poisson process, or if you want to stick to the continuum, you use a distribution for the "waiting time" till an event takes place as opposed to using a rate parameter (like the Poisson does).
A Poisson process is fine.
Quote by chiro So if you consider modelling your pulse process as a "waiting time" model for both signals and then look at the relevant probabilities, that might serve your interest.
Thanks, nice.
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Quote by bpet The problem is non-trivial but significantly easier if you can assume that the pulses follow a Poisson process - so that, conditional on the number of pulses, the pulse times are uniformly distributed within the interval (with a slight modification if the inter-pulse times follow a Poisson process, rather than the pulse start times). I'm not sure if this will lead to an analytic solution but it should at least be approximated with Mone Carlo methods.
Thanks, I will try to find a solution; I was thinking about Monte Carlo as well.
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http://math.stackexchange.com/questions/65543/subtraction-of-elements-in-sets/65544
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# Subtraction of elements in sets
Let A be a set of finite elements.
$A=\{1,2,3,4,5\}$
If I want to remove one element and show I removed one element, how should I do?
Pseudo mathematical notation:
$A - \{2\} = \{1,3,4,5\}$
Thank you very much!
n
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People often prefer to write $A\setminus\{2\}$ or $A \smallsetminus \{2\}$ to simply $A -\{2\}$ but what you wrote is fine. See here for example where `\setminus` $\setminus$ is used. – t.b. Sep 18 '11 at 17:45
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Why is your notation psuedo-mathematical? – Srivatsan Sep 18 '11 at 17:45
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i thought it wasn't mathematical enough to be mathematical – graphtheory92 Sep 18 '11 at 18:21
Somehow, the previous comments sound like a Zen koan. – Ilmari Karonen Sep 18 '11 at 18:48
## 1 Answer
Your notation above is actually used in set theory.
In general if you have two sets $A$ and $B$, the difference $A - B$ is the set
$A - B = \{x \in A : x \notin B\}$
Also, note that $\{1, 2, 3, 4, 5\} - \{2, 6\} = \{1,3,4,5\}$. $B$ need not be a subset of $A$.
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One reason to prefer $A \setminus B$ is that I've also seen $A - B$ occasionally used for the set $\{a-b: a \in A, b \in B\}$. That is, by that definition $\{1,2,3,4,5\} - \{2,6\} = \{-5,-4,-3,-2,-1,0,1,2,3\}$. Of course, there's no problem as long as you explain which definition you're using, but $A \setminus B$ avoids that ambiguity completely. – Ilmari Karonen Sep 18 '11 at 18:55
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http://mathforum.org/mathimages/index.php?title=Fractal_Dimension&oldid=20633
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Fractal Dimension
From Math Images
Revision as of 10:12, 16 June 2011 by Kderosier (Talk | contribs)
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
Contents
What is Fractal Dimension?
The fractal dimension, $D$, of a particular fractal is a measure of how the complexity of the figure increases as it scales. The dimension is the exponent that relates the scaling factor to the measure of the figure. That is, scale$^D$ = number of copies of the figure.
Or, $D = \frac{log(N)}{log(e)}$, where N is the number of copies of the figure and e is scale.
Example: Sierpinski's Triangle
For example, consider Sierpinski's triangle. If we double its size (as in the picture to the right), we create three copies of it. So its dimension can be calculated by $2^D=3\,$. So $D = \frac{log(3)}{log(2)} = 1.58$.
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http://physics.stackexchange.com/questions/22776/in-electrostatics-why-the-conductor-is-an-equipotential-surface/22777
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# In electrostatics, why the conductor is an equipotential surface?
Since the electric field inside a conductor is zero that means the potential is constant inside a conductor, which means the "inside" of a conductor is an equipotential region.
Why books conclude also that, the surface is at the same potential as well?
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## 4 Answers
The "first level" answer was given by nibot in a comment.
The entire conductor must be equipotential. If there were a potential difference from one part of a conductor to another, free electrons would move under the influence of that potential difference to cancel it out.
However, since I have similar curiosity myself I'm going to try to answer in greater depth.
Imagine a conducting sphere with a negative charge on it. There exist a certain number of surplus electrons. A conductor has electrons that are bound in their orbit to a given nucleus and electrons in the conduction band. The entire volume of the sphere beyond a certain distance from the surface is almost perfectly balanced on the scale of a few atoms. That is, the number of electrons balances perfectly with the number of nuclei, and in reality, the "orbit" of the conduction band electrons span many nuclei.
The critical discussion is what happens to the "surplus" electrons. According to the electrostatic potential equations they must all exist exactly on the surface of the sphere. This, of course, is physically absurd. There is, however, nothing absurd about the mathematics of a surface charge. A surface charge gives a well behaved:
• field - which is simply normal to the surface and points toward the surface in the case of electrons. It has the mathematical form of $-|x|/x$ if the surface is at $x=0$.
• potential - which is piecewise linear, having the mathematical form of $|x|$
In the case of the electron-rich conductor I am speaking of, the field and potential can be revised due to the existence of a field from the curved geometry of the sphere. The field is $0$ within the sphere and a negative value just above the surface. The potential is constant within the sphere and linearly increasing (due to negative charge) just above the surface.
The density of surplus charge mathematically follows a Dirac-delta function around the surface. Let's give the surplus charge density the notation of $\rho(x)=-\delta(x)$ (still using negative because these are electrons). How do we resolve this absurdity? Surely it is absurd, because it would imply that the electrons pile up, saturate the conduction band, and would inhabit even higher energy levels. Here is a basic illustration for the conduction band:
I want to call attention to the y-axis, energy. Energy of what? This is the energy of the orbital (for a single electron I believe), which comes from physics that I do not understand myself, including the Pauli exclusion principle and quantum physics. For answering this question, however, I think we need a simple take-away and I will propose such a thing here. Since electrons are not bound to a single nucleus, we will do best to speak of the total charge density at some point. At a given point, the surplus charge has two potentials associated with it, these being the electrostatic potential (I'll call $V_{C}$) and what I will call the conduction band potential (I'll call $V_{B}$) (although I would appreciate someone proposing better terminology).
$$V_C(x) = \int_x^\infty k \rho(x') dx'$$
$$V_B(x) = V_B( \rho(x) )$$
I really don't know what this function for conduction band energy is like. I would imagine that for a conductor in a differential sense it would be linear (see image), and since $\rho$ is a measure of surplus charge it could probably be formalized as $V_B(x) = C \rho(x)$ where $C$ is a constant.
My point is that you can use the combinations of these two potentials to resolve the absurdity of the infinite charge pileup. What fundamental equation should we seek to satisfy? I suggest that every free charge in the conductor will assume the lowest energy state it can. Charges will always move to a lower potential, unless something is holding them back, that "something" is $V_B$, or the conduction band potential, quantum pileup pressure, or whatever we should call it. You can use some math to determine:
$$V_B(x) + V_C(x) = constant = V_C$$
Let $V_C$ by itself represent the potential in the center of the sphere
It's useful to note that $\rho(x)=0$ in the majority of the sphere, so for the majority of the sphere $V_B=0$, and this is why we don't often entertain this concept in basic physics classes (and why it's so hard to you and I to get straight answers about the question). This has an interesting implication that only a small thickness within close proximity to the surface has these interesting dynamics associated with it.
Now, if you combine all of the equations I have presented so far you can obtain a differential equation for the surplus charge density within the vicinity of the surface. I believe this results in a simple 1st order equation that yields a decreasing exponential. I find the boundary conditions a little difficult, because the surplus charge density beyond the surface is $0$, but the function is allowed to be discontinuous there, so I think the needed condition in place of a boundary condition would be $V(-\infty)=V_C$, which is easy to implement.
So my answer is that a conductor is not an equipotential surface if you consider the orbital/quantum effects. In the vicinity of the surface the potential will have the following general form if the surface is at $x=0$ and the conductor is on the -x side.
$$V(x) = \begin{cases} V_C + c e^{\lambda x}, & \mbox{if} x<0 \\ V_C + c + d x, & \mbox{if} x \ge 0 \end{cases}$$
c and d are constants I don't know the value of
While writing this answer I stumbled upon the concept of the Stern layer. This may be what I'm describing, but I'm not entirely sure.
http://en.wikipedia.org/wiki/Double_layer_%28interfacial%29
This looks an awful lot like what I was describing. I think the Debye length might be useful as well, which seems to be the general scale over which these effects (of charge pileup) matter. So the sound-byte refinement of nibot's answer may just be that the potential of most conductors is very nearly constant because the Debye length is small relative to their total dimensions.
This isn't my area, but the question always drove me nuts personally, and I hope my mega-overkill answer is helpful to someone someday.
## Another reference
If one was only interested in a quick qualitative picture (which is our entire readership), they might to better to disregard everything written in this answer and instead stare at this image from a professor at the University of Kiel, Dr. Helmut Föll.
They introduce the helpful terminology of the electrochemical potential, which I've already written about thoroughly up to this point. This potential is comprised on the electrostatic potential added to something else that I still can't find a name for! Nonetheless, the website gives the following formula for the nameless potential.
$$V_B(\rho) = \frac{kT}{e} ln \rho(c)$$
This reference also makes arguments for the linearization of the above function. Why? And what are the limits of this?
We may thus assume within a very good approximation that the carrier density at any point is given by the constant volume density c0 of the field free material, plus a rather small space dependent addition c1(x);
So yes, they did what I thought they did. What about the justification? Here:
As you might know, the Debye length is a crucial material parameter not only in all questions concerning ionic conducitvity (the field of "Ionics"), but whenever the carrier concentration is not extremely large (i.e. comparable to the concenetration of atoms, i.e in metals).
Bam. This is a very strong and well-reasoned argument. When we deal with electrostatics, the electron surplus numbers are extraordinarily small compared to total atom numbers. Even if the surplus is spread among an microscopic Debye length on the surface, it will still be vanishingly small compared to proton density (number per volume). Obviously, these statements break down in special cases, like semiconductors, which are specifically engineered to violate the assumptions laid out here. At that point, you'll head back to the first principles. However, I do have some hesitation with the $ln(\rho)$ form. My intuition is that it would apply specific to a single conduction band. Of course, all energy levels are quantized in the first place, but this is getting far more complicated than what we ever needed.
As a bit of a personal note, I always struggled with chemistry in my early studies. If 10 years ago someone had come up to me and said "hey Alan, chemistry is really about the balancing of electrochemical potential which includes electrostatic potential and other potentials that can be empirically quantified", I think I might be a chemical engineer right now. As it happened, however, I had a fantastic physics teacher and an okay chemistry teacher. It just goes to show the impact of teachers can have and the importance of using non-subjective arguments in the classroom!
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This result can be understood mathematically. Suppose the system has reached equilibrium and all charges have stopped moving so that electrostatics applies. Then the potential is a harmonic function $\Delta \varphi = 0$ in $\mathbb{R}^3$ and the conductor is a closed and bounded region in $\mathbb{R}^3$.
A general property of harmonic functions is the maximum principle. In short, if $\varphi$ is harmonic on an open set $\Omega$ and $B \subset \Omega$ is a closed and bounded region, then $\varphi$ has no local minimum or maximum in the interior of $B$ and the absolute maximum and minimum of $\varphi$ occur on the boundary of $B$. In particular, if $\varphi$ is constant on the interior of $B$ it must be (the same) constant on the boundary of $B$.
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The change in potential between two points is
$$\Delta V = \int_a^b \mathrm{d}\vec{\ell} \cdot \vec{E}$$
but inside the conductor $\vec{E} = 0$ so that integral between any two interior points is also zero, accordingly the interior is all at the same potential.
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Yes I know that, that is what I have said already, I am ok with that part. The inside of the conductor is an equipotential region. I am asking about why the surface of the conductor is at the same potential as the interior ? – Revo Mar 24 '12 at 22:14
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The entire conductor must be equipotential. If there were a potential difference from one part of a conductor to another, free electrons would move under the influence of that potential difference to cancel it out. – nibot Mar 24 '12 at 22:19
The surface charge is still free to move about the conductor, so approaching the surface from the inside the above argument still applies. If you want to be pedantic let $a$ lie in the interior and let $b$ approach the surface from the inside and take the limit. – dmckee♦ Mar 24 '12 at 22:22
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@dmckee Don't take this personally, but I don't understand your comment, and I think there's just not enough space here to clearly lay out the specifics of what you mean by your limit approach suggestion. – AlanSE Mar 25 '12 at 4:24
@AlanSE: This answer just lives in PhysicsLand (tm) where the surface layer is infinitesimal. That is it ignores all the detail that you've written so nicely about. – dmckee♦ Mar 25 '12 at 17:48
Because if the surface is not equipotential then it would mean that there is a tangential component of electric field along the surface.
This component will result in motion of electrons, but since we have static fields this is not possible. Thus by contradiction we can say that surface must be equipotential.
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Hello again! Usually it's better to not answer if you don't have anything different to add-- the other answers cover this in depth.. The entire content of your answer is contained in the quoted section of AlanSE's post, so there's nothing new here. Instead, you can upvote the posts which you like. – Manishearth♦ Apr 5 '12 at 15:04
Sorry for repetition, I should have read other answer too. I sure will keep that in mind next time. – anuragsn7 Apr 5 '12 at 15:08
I'm more curious how the charges create a field tangential to the surface O_O – AlanSE Apr 5 '12 at 15:09
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http://mathoverflow.net/questions/66240?sort=votes
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Topological spaces, uncountable subsets and separability
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Hi, the following is a well known theorem
Let $M$ be a metric space. If every uncountable subset of $M$ has a limit point, then $M$ is separable.
Question: Is there a similar result for topological spaces?
I have almost no knowledge of topology so I can only hope that this is not trivial.
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4 Answers
The statement, "Let $M$ be a topological space. If every uncountable subset of $M$ has a limit point, then $M$ is separable," is false. Consider the first uncountable ordinal $\omega_1$, under the order topology (see http://en.wikipedia.org/wiki/First_uncountable_ordinal). $\omega_1$ is countably compact, hence also weakly countably compact (that is, every infinite subset has a limit point), but $\omega_1$ is not separable.
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Consider the topological space $[0, \omega_1]$ where $\omega_1$ is the first uncountable ordinal. Every uncountable subset has a limit point, namely $\omega_1$, because the complement of any neighbourhood of $\omega_1$ is countable. However, it is not separable, since the supremum of a countable subset of countable ordinals is countable.
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The property "Every uncountable set has a limit point" is related to the Lindelöf property (every open cover has a countable subcover). For metrisable spaces these notions are equivalent, and in general Lindelöf implies the limit point property. And there are many Lindelöf spaces that are not separable, as the other examples show. For metrisable spaces, being Lindelöf, separable, second countable, etc. are all equivalent.
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Every compact space satisfies "every infinite set has a limit point". (I am assuming here that by limit point you mean what I would call an accumulation point, i.e., a point $x$ such that evey neighborhood contains infinitely many elements of the set.) So in particular, in a compact space every uncountable set has a limit point. It follows that every compact space that is not separable (for instance a sufficiently high power of the closed unit interval or Robert Israel's example) shows that the theorem you mention does not hold for topological spaces in general.
The theorem you mention of course implies that compact metric spaces are separable.
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http://mathoverflow.net/questions/41036?sort=oldest
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## How to find which subset of bitfields xor to another bitfield?
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I have a somewhat coding-oriented problem. I have a bunch of bitfields and would like to calculate what subset of them to xor together to achieve a certain other bitfield, or if there isn't a way to do it discover that no such subset exists.
I'd like to do this using a free library, rather than original code, and I'd strongly prefer something with Python bindings (using Python's built-in math libraries would be acceptable as well, but I want to port this to multiple languages eventually). Also it would be good to not take the memory hit of having to expand each bit to its own byte.
Some further clarification: I only need a single solution. My matrices are the opposite of sparse. I'm very interested in keeping the runtime to an absolute minimum, so using algorithmically fancy methods for inverting matrices is strongly preferred. Also, it's very important that the specific given bitfield be the one outputted, so a technique which just finds a subset which xor to 0 doesn't quite cut it.
And I'm generally aware of gaussian elimination. I'm trying to avoid doing this from scratch!
cross-posted to stackoverflow, because it's unclear where the right place for this question is - http://stackoverflow.com/questions/3855479/how-to-find-which-subset-of-bitfields-xor-to-another-bitfield
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I am inclined to think this question is a bit outside the scope of MO. But I may be wrong. Discussion at meta.mathoverflow.net/discussion/695/… – Willie Wong Oct 4 2010 at 16:49
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SO gave a right answer - your problem is equivalent to solving linear equation Ax=b in GF(2), given matrix A and b. – sdcvvc Oct 4 2010 at 16:51
Hey, aren't you the guy who invented bittorrent? – Harry Gindi Oct 4 2010 at 16:52
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@Harry: Yes. See his user profile. – Willie Wong Oct 4 2010 at 17:04
I don't know if this is suitable for MathOverflow - it seemed a bit over peoples's heads on StackOverflow, but they seem to have pulled through with some answers, so perhaps I should have just been a bit more patient. A reasonably mathy question is whether it's possible to go faster than gaussian elimination when A is non-square, although I have to admit that I'm mostly interested in using a library to hide all this stuff. – Bram Cohen Oct 4 2010 at 23:49
show 3 more comments
## 1 Answer
The ways to solve it may vary based on which varies more frequently: the dictionary $A$ used to generate the wanted bitstring $B$, or the bitstring $B$.
First, fix the length of the words we are talking about to $n$ bits.
The dictionary $A$ is a set of $n$-bit long words $A=${$a_1,a_2,...a_m$}, $a_i \in${$0,1$}${}^n$.
Given an insufficient dictionary, it may not be possible to generate all possible bit patterns of length $n$. For example, if $C$={ 1000, 0011, 0001}, then it is impossible for the dictionary $C$ to generate the bit-pattern $x_4 1 x_2 x_1$, a 4-bit string with the $2^2$ value set to $1$.
It may make sense given a dictionary $A$ of $n$-bit length words to test the dictionary as a viable signal generator by seeing if it is possible to create the "single-bit-on" patterns in the dictionary $C$ defined as
• $C=${$c_1, c_2, ..., c_{n}$}
• such that {$c_m = d_n d_{n-1}...d_2 d_1$} where
• $d_j=1$ if $j=m$, and
• $d_j=0$ if $j\ne m$
If it is not possible for the dictionary $A$ to generate the dictionary $C$, then there will be certain bit patterns which are not reachable by using words in the dictionary $A$ and the binary-operation XOR.
Once a mapping is generated from $A$ to the single-bit-on dictionary $C$, it is a simple task to create the mapping from $A$ to an arbitrary bit pattern, $B$. Take the bits which are on in $B$, and take the mappings which generate those single bits on in $C$, and concatenate them together.
An even number of XOR's for any particular bit pattern in $C$ cancel each other out, leaving a single count of whichever elements in $A$ would generate bit-pattern $B$.
One quick observation: the dictionary $A$ of $n$-bit long words must contain at least $n$ words for it to be able to generate all possible $n$-bit long strings, and none of them should be linear combinations of the other.
For example, the alphabet X={0001, 1000} is too small to be able to generate all possible 4-bit long words, simply from the observation that it only contains two words of 4-bit length.
The alphabet Y={0001, 0011, 0010, 1000, 1001} has enough words to possibly span all possible 4-bit length words, however $Y_2 = Y_1$ XOR $Y_3$, and $Y_5 = Y_1$ XOR $Y_4$. It is not possible to generate the bit patterns $a1cd$, where $a,c,d \in${0,1} using alphabet $Y$. Even though $Y$ is defined as $5$ elements, it really only contains 3 degrees of freedom, as two of the elements can be defined as linear combinations of the others.
In other words, using gaussian elimination on your dictionary using XOR as the operation on the right may be the best way to test or assess your dictionary, with the caveat that if your dictionary does not contain at least as many words as there are bits in each word then your dictionary will not suffice to generate all possible bit patterns.
It's also possible to think of this as operations of a message being passed along the nodes of an $n$-dimensional hypercube. But that's just a different way of thinking of it.
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http://physics.stackexchange.com/questions/52126/spring-extensions
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# Spring extensions
A certain spring has attached to it a mass of 25 units: on increasing the load by 6 units it extends 2.5 cm.
a) What is the time of oscillation under the original load?
b) What will be the velocity and acceleration when it is midway between its lowest and mean positions if it is loaded as at first, pulled down 5 cm and let go?
I am guessing the spring is hung vertically. Do I work out the modulus of elasticity first? I guess the equations $\omega=\sqrt{k/m}$ will be helpful.
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## 1 Answer
Let's use SI units so mass is in kilograms and length is in meters. Let the original mass be $m_1 = 25 \,\mathrm{kg}$ and the final mass be $m_2 = m_1 +\Delta m$ where $\Delta m = 6 \,\mathrm{kg}$. Let the original extension of the spring be $x_1$ and the final extension be $x_1 + \Delta x_1$ where $\Delta x = 2.5\, \mathrm{cm}$. We can determine the spring constant by noting that in each case, the gravitational force of the mass on the spring must balance the spring force pulling up; this gives two equations:
$m_1g = k(x_1 - x_\mathrm{eq})$
$m_2g = k(x_2 - x_\mathrm{eq})$
so that subtracting the first equation from the second gives
$(m_2 - m_1)g = k(x_2-x_1)$
which given the notation above gives
$\Delta m g = k \Delta x$
so the spring constant is
$k = \frac{\Delta m}{\Delta x} g = \frac{6\,\mathrm{kg}}{2.5\,\mathrm{cm}} g$
With this in hand, you can indeed compute the angular frequency for the original load $m_1$ using the formula you wrote. This in turn will give you the period, which is what I assume part (a) is asking for. I'll let you think about part (b) since this sounds like a homework question to me.
Let me know of any typos.
Cheers!
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Just remember not to give away too much when you're answering a homework question ;-) – David Zaslavsky♦ Jan 25 at 0:05
@DavidZaslavsky Yea; my apologies. I'll make my comments more general next time. – joshphysics Jan 25 at 0:21
So the time period is $\omega = \sqrt{k/m_1}=\sqrt{12/(5gm_1)} 2\pi f=\sqrt{12/5gm_1} T=2\pi \sqrt{5gm_1/12}$ – bbr4in Jan 26 at 13:43
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http://mathoverflow.net/revisions/14245/list
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## Return to Question
3 deleted 2 characters in body
Consider the optimization problem
$$\min_x ||Ax||_1 + \lambda||x-b||^2,$$
where $A \in \mathbb{R}^{n \times n}$, $x,b \in \mathbb{R}^n$ and $\lambda$ is strictly greater than 0. (This problem is closely related to the "lasso" problem in basis pursuit.) Can anything be said about the value of $\lambda$ for which $Ax^*$ is sparsest? Clearly some values are bad: for instance, if $\lambda$ is huge and $b$ is dense then it is unlikely that $Ax^\star$ will be very sparse.
In other words: among all $\lambda > 0$ there is at least one value $\lambda^\star$ such that $||Ax^\star(\lambda)||_0$ is minimized. Are there, say, bounds on $\lambda^\star$ in terms of $A$ and $b$? I'd also be interested in results pertaining to basis pursuit or other similar problems.
Edit: I'm primarily interested in problems where optimal ideal sparsity cannot be achieved, i.e., $||Ax^\star(\lambda^\star)||_0 > 0.$ (Assume that $A$ is square w/ full rank and $b \ne 0$.)
2 getting rid of the null space; added 9 characters in body
Consider the optimization problem
$$\min_x ||Ax||_1 + \lambda||x-b||^2,$$
where $A \in \mathbb{R}^{m mathbb{R}^{n \times n}$, $x,b \in \mathbb{R}^n$ and $\lambda$ is strictly greater than 0. (This problem is closely related to the "lasso" problem in basis pursuit.) Can anything be said about the value of $\lambda$ for which $Ax^*$ is sparsest? Clearly some values are bad: for instance, if $\lambda$ is huge and $b$ is dense then it is unlikely that $Ax^\star$ will be very sparse.
In other words: among all $\lambda > 0$ there is at least one value $\lambda^\star$ such that $||Ax^\star(\lambda)||_0$ is minimized. Are there, say, bounds on $\lambda^\star$ in terms of $A$ and $b$? I'd also be interested in results pertaining to basis pursuit or other similar problems.
Edit: I'm primarily interested in problems where optimal sparsity cannot be achieved, i.e., $||Ax^\star(\lambda^\star)||_0 > 0.$ (Assume that $A$ is square w/ full rank and $b \ne 0$.)
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# Maximizing Sparsity in l1 Minimization?
Consider the optimization problem
$$\min_x ||Ax||_1 + \lambda||x-b||^2,$$
where $A \in \mathbb{R}^{m \times n}$, $x,b \in \mathbb{R}^n$ and $\lambda$ is strictly greater than 0. (This problem is closely related to the "lasso" problem in basis pursuit.) Can anything be said about the value of $\lambda$ for which $Ax^*$ is sparsest? Clearly some values are bad: for instance, if $\lambda$ is huge and $b$ is dense then it is unlikely that $Ax^\star$ will be very sparse.
In other words: among all $\lambda > 0$ there is at least one value $\lambda^\star$ such that $||Ax^\star(\lambda)||_0$ is minimized. Are there, say, bounds on $\lambda^\star$ in terms of $A$ and $b$? I'd also be interested in results pertaining to basis pursuit or other similar problems.
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http://mathoverflow.net/questions/8648/easiest-way-to-determine-the-singular-locus-of-projective-variety-resolution-of/8650
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## Easiest way to determine the singular locus of projective variety & resolution of singularities
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For an affine variety, I know how to compute the set of singular points by simply looking at the points where the Jacobian matrix for the set of defining equations has too small a rank.
But what is the corresponding method for a variety that is a projective variety,and also a variety is a subset of a product of some projective space and affine space? The way I can think of is covering it by sets that are affine, and doing it for each affine set in this open cover - but that seems tedious for practical purposes (but fine for theoretical definitions & theoretical properties).
Also for resolution of singularities, what is a simple method that is guaranteed to work? The way suggested in the definitions in Hartshorne and other books, is to blow up along the singular locus, then look at the singular locus of the blow-up, and blow up again, and so on - is that guaranteed to terminate? What are some more efficient methods? I have looked at the reference "Resolution of Singularities", a book by someone - that's what he also seems to suggest (though his proof is very general, and I didn't read all of it).
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## 4 Answers
The Jacobian condition for smoothness is valid also for projective varieties as well as affine varieties: you just take a homogeneous defining ideal and compute the rank of the Jacobian matrix at the point p, see e.g. p. 4 of
http://www.ma.utexas.edu/users/gfarkas/teaching/alggeom/march4.pdf
For a general variety: yes, I think the most computationally effective way to do it is to cover it by affine opens and apply the Jacobian condition on each one separately.
About your question on resolution of singularities: this is tricky in high dimension! As I understand it, you can indeed resolve singularities just by a combination of blowups and normalizations (at least in characteristic 0), but starting in dimension 3 you have to be somewhat clever about where in and what order you perform your blowups. It is not the case that if you just keep picking a closed subvariety and blowing it up (and then normalizing) that you will necessarily terminate with a smooth variety.
For more details presented in a user-friendly way, I recommend Herwig Hauser's article
http://homepage.univie.ac.at/herwig.hauser/Publications/The%20Hironaka%20Theorem%20on%20resolution%20of%20singularities/The%20Hironaka%20Theorem%20on%20resolution%20of%20singularities.pdf
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In the scheme-theoretic setting, the Jacobian criterion only works for closed points, right? How to do it for non-closed points, can we still find the singular non-closed points using the Jacobian criterion? – Wanderer Mar 2 2010 at 1:46
The set of non-closed points that are in a set is determined by the set of closed points that are in it. If the ideal generated by the appropriately-sized minors is contained in a prime ideal $p$, then $p$ is singular. – Will Sawin Sep 30 2011 at 4:53
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The other answers are already very good. A few additional notes:
(1) As others have explained, the Jacobian ideal method works for projective space. It also works for other toric varieties. Any smooth toric variety can be written as $(\mathbb{C}^n \setminus \Sigma)/(\mathbb{C}^{*})^k$ where $\Sigma$ is an arrangement of linear spaces and $(\mathbb{C}^{*})^k$ acts on $\mathbb{C}^n$ by some linear representation. For example, $\mathbb{P}^n = (\mathbb{C}^{n+1} \setminus \{ 0 \} )/\mathbb{C}^*$. So you can use Greg's trick of unquotienting, using the ordinary Jacobi criterion, and ignoring singularities on $\Sigma$. See David Cox's notes for how to write a toric variety in this manner.
(2) For varieties of dimension greater than $1$, resolution of singularities is very computationally intensive. It has been implemented in Macaulay 2, but it tends to tax the memory resources of the system.
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Thanks much for this answer. Even though I had seen all pieces of this, I somehow never grokked the unquotienting trick for a general toric variety. – Greg Kuperberg Dec 13 2009 at 1:48
You're welcome! Note that I added the word "smooth" to the above. I think there is some sense in which unquotienting works for non-smooth toric varieties, but I realized I don't understand it well enough to say how the Jacobi criterion will be transformed by it. – David Speyer Dec 13 2009 at 14:19
For many questions, the easiest way to see the nuts and bolts of a projective variety $V \subseteq P^n$ is to look at its cone $CV \subseteq A^{n+1}$. After all, the graded ring whose Proj is $V$ is the same as the ungraded ring whose Spec is $CV$. Obviously, there is almost always a singularity at the origin; but if you ignore that point, the other singular points all correspond between $V$ and $CV$. You can also think of the grading as geometrically represented by multiplication by $k^*$, if you are working over an algebraically closed field $k$. (Because the homogeneous polynomials are then eigenvectors of that group action.) You can think of $V$ as obtained from $CV$ by and then dividing by scalar multiplication.
The atlas-of-charts analysis of a projective variety is certainly important, but to some extent it is meant as an introduction to intrinsic algebraic geometry rather than as the best computational tool.
Your second question is reviewed in Wikipedia. As Wikipedia explains, Hironaka's big theorem was that it is possible to resolve all singularities of a variety by iterated blowups along subvarieties. I do not know a lot about this theory, but if so many capable mathematicians went to so much trouble to find a method, then surely there is no simple method.
On the other hand for curves, there is a stunning method that I learned about (or maybe relearned) just recently. Again according to Wikipedia, taking the integral closure of the coordinate ring of an affine curve, or the graded coordinate ring of a projective curve, solves everything. The claim is that it always removes the singularities of codimension 1, which are the only kind that a curve has.
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The statement for curves comes down to the usual Galois antiequivalence between smooth curves over a field and their fields of functions. – S. Carnahan♦ Dec 12 2009 at 5:45
I should have inserted the word "projective" between "smooth" and "curves". – S. Carnahan♦ Dec 12 2009 at 5:47
I'd say it rather comes down to the fact that an integrally closed local ring of dimension one is regular, so that the normalization of the curve is smooth at each of its points. – Mariano Suárez-Alvarez Dec 12 2009 at 6:02
There are several ways to argue why it works for curves. In any case, it works. – Greg Kuperberg Dec 12 2009 at 7:11
About resolutions: yes, that is essentially the algorithm, and it does terminate, but you have to be careful about what you blow up (you have to blow up smooth subvarieties, for example, to be sure this converges, &c) and in what order.
Two very nice textbooks on resolution with plenty of both examples and details are Kollár's book and Cutkosky's book. There is also a (more recent?) nice one by O. Villamayor.
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Tagged Questions
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determinant of an $n\times n$ matrix type [duplicate]
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1answer
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Is it true that in $Mat(n,n)$ the set of singular matrices forms a hyperplane?
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Calculating determinant of a matrix whose non-zero elements are two sub-matrices on its diagonal
Given a $m \times m$ square matrix $M$: $$M = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$$ $A$ is an $a \times a$ and $B$ is a $b \times b$ square matrix; and of course $a+b=m$. All the ...
2answers
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Determinant of a block matrix with $\mathrm{Id}$ and $0$ in the diagonal
How to compute the determinant $\det A$ depending on $B$ and $C$, where $$A = \left(\begin{matrix}\mathrm{Id} & B \\ C & 0 \end{matrix} \right),$$ a) when $C$ is square, b) $C$ has more ...
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http://mathoverflow.net/questions/103688/is-any-simple-abelian-variety-covered-by-a-non-simple-abelian-variety
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## Is any simple abelian variety covered by a non-simple abelian variety
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $A/k$ be a simple abelian variety.
Does there exist a non-simple abelian variety $B/k$ and a finite homomorphism $f:B\to A$ over $k$?
I don't need $f:B\to A$ to be etale.
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I believe the question should truly be "Is any simple abelian variety covered by a non-simple abelian variety?" – David Corwin Aug 1 at 14:48
I think I see what you mean. Thanks for correcting my English. – Harry Aug 1 at 14:56
It isn't necessarily a question of English, more of mathematics. – David Corwin Aug 1 at 15:43
## 1 Answer
No. If $End_A$ and $End_B$ are the endomorphism rings then an isogeny $B\to A$ will give an isomorphism $End_A\otimes\mathbb Q\to End_B\otimes\mathbb Q$. But the first of these algebras is a division algebra and the second is not.
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2
In case this wasn't clear: a finite homomorphism is necessarily etale since if it etale over one point it is etale over every point, and finite morphisms are etale on a nonempty open subset, thus it is an isogeny. – Will Sawin Aug 1 at 14:10
2
To elaborate: If $f:B\to A$ is finite and separable (e.g. if $char\, k=0$) then Riemann-Hurwitz implies that the ramification divisor $R= K_A-f^*K_B=0$. Therefore $f$ is etale. – Donu Arapura Aug 1 at 14:53
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# Thread:
1. ## Continuity & inverse image of closed sets being closed
Question: Let $f$ be a function defined on a closed domain $D$. Show that $f$ is continuous if and only if the inverse image of every closed set is a closed set.
One approach is to solve it using the fact that the complement of open sets are closed. However, I am wondering how I can prove it using the definition of a closed set using limit points?
I know that a set is closed if it contains all of its limit points. How could I use this definition to prove the above question? In other words, how do I show that if a set in the range holds all of its limit points, mapping it backwards to the domain will create a set also containing all of its limit points?
Thanks a lot!
2. Hello,
Here is something I did : http://www.mathhelpforum.com/math-he...790-post2.html , using neighbourhoods (continuous => ...)
The key for that is : a set is open if it is a neighbourhood of any of its point.
I'm not used to limit points... so I hope it will sort of help you !
3. Originally Posted by Last_Singularity
Question: Let $f$ be a function defined on a closed domain $D$. Show that $f$ is continuous if and only if the inverse image of every closed set is a closed set.
I know that a set is closed if it contains all of its limit points.
What sort of space are you working with? Is a metric space?
If it is a general top-space, if so what properties do it have?
To use the limit point approach you need to consider that.
4. Originally Posted by Plato
What sort of space are you working with? Is a metric space?
If it is a general top-space, if so what properties do it have?
To use the limit point approach you need to consider that.
This is real analysis, so $f: X \rightarrow Y$ where $X,Y \subseteq \mathbb{R}$
5. Originally Posted by Last_Singularity
This is real analysis, so $f: X \rightarrow Y$ where $X,Y \subseteq \mathbb{R}$
Point number 1. Real analysis is about more than $\mathbb{R}$ so why would you expect us to know in what setting your problem is casted?
Now to your question, for continuous functions we have the following.
The inverse image of an open set is an open set.
The complement of the inverse image is the inverse of the complement of the image.
The complement of a closed set is open.
If $M$ is closed then $M^c$ is open.
$f^{ - 1} \left( {M^c } \right) = \left( {f^{ - 1} (M)} \right)^c$
Can you finish?
6. Yes, thanks a lot!
And sorry about not specifying that it was in a real analysis context; I was not aware that such concepts can be generalized to metric spaces and other areas of mathematics...
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Elementary Algebra, v. 1.0
by John Redden
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4.1 Solving Linear Systems by Graphing
Learning Objectives
1. Check solutions to systems of linear equations.
2. Solve linear systems using the graphing method.
3. Identify dependent and inconsistent systems.
Definition of a Linear System
Real-world applications are often modeled using more than one variable and more than one equation. A system of equationsA set of two or more equations with the same variables. consists of a set of two or more equations with the same variables. In this section, we will study linear systemsIn this section, we restrict our study to systems of two linear equations with two variables. consisting of two linear equations each with two variables. For example,
A solution to a linear systemAn ordered pair that satisfies both equations and corresponds to a point of intersection., or simultaneous solutionUsed when referring to a solution of a system of equations., to a linear system is an ordered pair (x, y) that solves both of the equations. In this case, (3, 2) is the only solution. To check that an ordered pair is a solution, substitute the corresponding x- and y-values into each equation and then simplify to see if you obtain a true statement for both equations.
Example 1: Determine whether (1, 0) is a solution to the system ${ x−y=1 −2x+3y=5$.
Solution: Substitute the appropriate values into both equations.
Answer: Since (1, 0) does not satisfy both equations, it is not a solution.
Try this! Is (−2, 4) a solution to the system ?
Answer: Yes
Solve by Graphing
Geometrically, a linear system consists of two lines, where a solution is a point of intersection. To illustrate this, we will graph the following linear system with a solution of (3, 2):
First, rewrite the equations in slope-intercept form so that we may easily graph them.
Next, replace these forms of the original equations in the system to obtain what is called an equivalent systemA system consisting of equivalent equations that share the same solution set.. Equivalent systems share the same solution set.
If we graph both of the lines on the same set of axes, then we can see that the point of intersection is indeed (3, 2), the solution to the system.
To summarize, linear systems described in this section consist of two linear equations each with two variables. A solution is an ordered pair that corresponds to a point where the two lines in the rectangular coordinate plane intersect. Therefore, we can solve linear systems by graphing both lines on the same set of axes and determining the point where they cross. When graphing the lines, take care to choose a good scale and use a straightedge to draw the line through the points; accuracy is very important here. The steps for solving linear systems using the graphing methodA means of solving a system by graphing the equations on the same set of axes and determining where they intersect. are outlined in the following example.
Example 2: Solve by graphing: ${x−y=−42x+y=1$.
Solution:
Step 1: Rewrite the linear equations in slope-intercept form.
Step 2: Write the equivalent system and graph the lines on the same set of axes.
Step 3: Use the graph to estimate the point where the lines intersect and check to see if it solves the original system. In the above graph, the point of intersection appears to be (−1, 3).
Answer: (−1, 3)
Example 3: Solve by graphing: ${2x+y=2−2x+3y=−18$.
Solution: We first solve each equation for y to obtain an equivalent system where the lines are in slope-intercept form.
Graph the lines and determine the point of intersection.
Answer: (3, −4)
Example 4: Solve by graphing: ${3x+y=6y=−3$.
Solution:
Answer: (3, −3)
The graphing method for solving linear systems is not ideal when the solution consists of coordinates that are not integers. There will be more accurate algebraic methods in sections to come, but for now, the goal is to understand the geometry involved when solving systems. It is important to remember that the solutions to a system correspond to the point, or points, where the graphs of the equations intersect.
Try this! Solve by graphing: .
Answer: (−2, 4)
Dependent and Inconsistent Systems
Systems with at least one solution are called consistent systemsA system with at least one solution.. Up to this point, all of the examples have been of consistent systems with exactly one ordered pair solution. It turns out that this is not always the case. Sometimes systems consist of two linear equations that are equivalent. If this is the case, the two lines are the same and when graphed will coincide. Hence the solution set consists of all the points on the line. This is a dependent systemA system that consists of equivalent equations with infinitely many ordered pair solutions, denoted by (x, mx + b).. Given a consistent linear system with two variables, there are two possible results:
The solutions to independent systemsA system of equations with one ordered pair solution (x, y). are ordered pairs (x, y). We need some way to express the solution sets to dependent systems, since these systems have infinitely many solutions, or points of intersection. Recall that any line can be written in slope-intercept form, $y=mx+b$. Here, y depends on x. So we may express all the ordered pair solutions $(x, y)$ in the form $(x, mx+b)$, where x is any real number.
Example 5: Solve by graphing: .
Solution: Determine slope-intercept form for each linear equation in the system.
In slope-intercept form, we can easily see that the system consists of two lines with the same slope and same y-intercept. They are, in fact, the same line. And the system is dependent.
Answer: $(x, 23x−3)$
In this example, it is important to notice that the two lines have the same slope and same y-intercept. This tells us that the two equations are equivalent and that the simultaneous solutions are all the points on the line $y=23x−3$. This is a dependent system, and the infinitely many solutions are expressed using the form $(x, mx+b)$. Other resources may express this set using set notation, {(x, y) | $y=23x−3$}, which reads “the set of all ordered pairs (x, y) such that y equals two-thirds x minus 3.”
Sometimes the lines do not cross and there is no point of intersection. Such systems have no solution, Ø, and are called inconsistent systemsA system with no simultaneous solution..
Example 6: Solve by graphing: ${−2x+5y=−15−4x+10y=10$.
Solution: Determine slope-intercept form for each linear equation.
In slope-intercept form, we can easily see that the system consists of two lines with the same slope and different y-intercepts. Therefore, they are parallel and will never intersect.
Answer: There is no simultaneous solution, Ø.
Try this! Solve by graphing: ${x+y=−1−2x−2y=2$.
Answer: $(x, −x−1)$
Key Takeaways
• In this section, we limit our study to systems of two linear equations with two variables. Solutions to such systems, if they exist, consist of ordered pairs that satisfy both equations. Geometrically, solutions are the points where the graphs intersect.
• The graphing method for solving linear systems requires us to graph both of the lines on the same set of axes as a means to determine where they intersect.
• The graphing method is not the most accurate method for determining solutions, particularly when the solutions have coordinates that are not integers. It is a good practice to always check your solutions.
• Some linear systems have no simultaneous solution. These systems consist of equations that represent parallel lines with different y-intercepts and do not intersect in the plane. They are called inconsistent systems and the solution set is the empty set, Ø.
• Some linear systems have infinitely many simultaneous solutions. These systems consist of equations that are equivalent and represent the same line. They are called dependent systems and their solutions are expressed using the notation $(x, mx+b)$, where x is any real number.
Topic Exercises
Part A: Solutions to Linear Systems
Determine whether the given ordered pair is a solution to the given system.
1. (3, −2); ${x+y=−1−2x−2y=2$
2. (−5, 0); ${x+y=−1−2x−2y=2$
3. (−2, −6); ${−x+y=−43x−y=−12$
4. (2, −7); ${3x+2y=−8−5x−3y=11$
5. (0, −3); ${5x−5y=15−13x+2y=−6$
6. $(−12, 14)$; ${x+y=−14−2x−4y=0$
7. $(34, 14)$; ${−x−y=−1−4x−8y=5$
8. (−3, 4); ${13x+12y=123x−32y=−8$
9. (−5, −3); ${y=−35x−10y=5$
10. (4, 2); ${x=4−7x+4y=8$
Given the graph, determine the simultaneous solution.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Part B: Solving Linear Systems
Solve by graphing.
21. ${y=32x+6y=−x+1$
22. ${y=34x+2y=−14x−2$
23. ${y=x−4y=−x+2$
24. ${y=−5x+4y=4x−5$
25. ${y=25x+1y=35x$
26. ${y=−25x+6y=25x+10$
27. ${y=−2y=x+1$
28. ${y=3x=−3$
29. ${y=0y=25x−4$
30. ${x=2y=3x$
31. ${y=35x−6y=35x−3$
32. ${y=−12x+1y=−12x+1$
33. ${2x+3y=18−6x+3y=−6$
34. ${−3x+4y=202x+8y=8$
35. ${−2x+y=12x−3y=9$
36. ${x+2y=−85x+4y=−4$
37. ${4x+6y=362x−3y=6$
38. ${2x−3y=186x−3y=−6$
39. ${3x+5y=30−6x−10y=−10$
40. ${−x+3y=35x−15y=−15$
41. ${x−y=0−x+y=0$
42. ${y=xy−x=1$
43. ${3x+2y=0x=2$
44. ${2x+13y=23−3x+12y=−2$
45. ${110x+15y=2−15x+15y=−1$
46. ${13x−12y=113x+15y=1$
47. ${19x+16y=019x+14y=12$
48. ${516x−12y=5−516x+12y=52$
49. ${16x−12y=92−118x+16y=−32$
50. ${12x−14y=−1213x−12y=3$
51. ${y=4x=−5$
52. ${y=−3x=2$
53. ${y=0x=0$
54. ${y=−2y=3$
55. ${y=5y=−5$
56. ${y=2y−2=0$
57. ${x=−5x=1$
58. ${y=xx=0$
59. ${4x+6y=3−x+y=−2$
60. ${−2x+20y=203x+10y=−10$
Set up a linear system of two equations and two variables and solve it using the graphing method.
61. The sum of two numbers is 20. The larger number is 10 less than five times the smaller.
62. The difference between two numbers is 12 and their sum is 4.
63. Where on the graph of $3x−2y=6$ does the x-coordinate equal the y-coordinate?
64. Where on the graph of $−5x+2y=30$ does the x-coordinate equal the y-coordinate?
A regional bottled water company produces and sells bottled water. The following graph depicts the supply and demand curves of bottled water in the region. The horizontal axis represents the weekly tonnage of product produced, Q. The vertical axis represents the price per bottle in dollars, P.
Use the graph to answer the following questions.
65. Determine the price at which the quantity demanded is equal to the quantity supplied.
66. If production of bottled water slips to 20 tons, then what price does the demand curve predict for a bottle of water?
67. If production of bottled water increases to 40 tons, then what price does the demand curve predict for a bottle of water?
68. If the price of bottled water is set at \$2.50 dollars per bottle, what quantity does the demand curve predict?
Part C: Discussion Board Topics
69. Discuss the weaknesses of the graphing method for solving systems.
70. Explain why the solution set to a dependent linear system is denoted by (x, mx + b).
Answers
1: No
3: No
5: Yes
7: No
9: Yes
11: (5, 0)
13: (2, 1)
15: (0, 0)
17: $(x, 2x−2)$
19: $∅$
21: (−2, 3)
23: (3, −1)
25: (5, 3)
27: (−3, −2)
29: (10, 0)
31: $∅$
33: (3, 4)
35: (−3, −5)
37: (6, 2)
39: $∅$
41: $(x, x)$
43: (2, −3)
45: (10, 5)
47: (−9, 6)
49: $(x, 13x−9)$
51: (−5, 4)
53: (0, 0)
55: $∅$
57: $∅$
59: (3/2, −1/2)
61: The two numbers are 5 and 15.
63: (6, 6)
65: \$1.25
67: \$1.00
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http://mathhelpforum.com/advanced-math-topics/8755-another-nasty-integral.html
|
# Thread:
1. ## Another nasty integral
I've got another one for you.
I ran across this theorem in my Field Theory text. The theorem is this:
Given a sufficiently smooth function f(x) we have that:
$\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}$
The term "sufficiently smooth" does not require that f(x) is a $C^{\infty}$ function, but as most wavefunctions in Physics are $C^{\infty}$ I would be satisfied with a proof of this case. (Note: Obviously we are concerned only with f(x) that have limits that existat $\pm \infty$.)
The only f(x) I can figure out how to integrate (as a test of the theorem) is a Gaussian. All the other functions I know of with the required properties at infinity is discontinous at at least one point on the real line, so it doesn't qualify.
-Dan
2. Originally Posted by topsquark
I've got another one for you.
I ran across this theorem in my Field Theory text. The theorem is this:
What does this have to do with Field theory?
3. Originally Posted by topsquark
I've got another one for you.
I ran across this theorem in my Field Theory text. The theorem is this:
Given a sufficiently smooth function f(x) we have that:
limit(x -> + infinity)f(x) + limit(x -> - infinity) = limit(e -> 0+) e Integral( dx f(x) exp(-e*abs(x)) ) where the integral is over the whole real line.
The term "sufficiently smooth" does not require that f(x) is a C(infinity) function, but as most wavefunctions in Physics are C(infinity) I would be satisfied with a proof of this case. (Note: Obviously we are concerned only with f(x) that have limits that exist at (+/-) infinity.)
The only f(x) I can figure out how to integrate (as a test of the theorem) is a Gaussian. All the other functions I know of with the required properties at infinity is discontinous at at least one point on the real line, so it doesn't qualify.
-Dan
This is difficult to read without LaTeX, but this might be a consequence
of the Laplace transform derivative properties.
RonL
4. Originally Posted by ThePerfectHacker
What does this have to do with Field theory?
It comes in handy when deriving the interaction matrix (S matrix) using the path integral method. In the long run we are going to take the natural log of a quantity that is the integral of this, so it is convenient to change all factors into exponential terms. That's as much as I can give you without quoting some seriously nasty integrals. (Which I'm not about to do without LaTeX!)
-Dan
5. Originally Posted by CaptainBlack
This is difficult to read without LaTeX, but this might be a consequence
of the Laplace transform derivative properties.
RonL
Interesting thought. I'll look into it. Thanks!
-Dan
6. Still working on the clue, but since LaTeX is back I thought I'd fix up the original post in case it gives any thoughts.
-Dan
7. Pffl. I'm running into a problem almost immediately.
$\int_{-\infty}^{\infty}dx F(x) e^{-\epsilon |x|} = \int_0^{\infty}dx F(-x) e^{-\epsilon x} + \int_0^{\infty}dx F(x) e^{-\epsilon x}$
Although many wavefunctions are parity eigenstates I cannot assume this, so I have no way to simplify the first integral in general. My Laplace transforms abilities are introductory. Is there a formula in terms of $L \{F(x) \}$ for the first integral?
If I DO assume that F(x) is even/odd then I get:
$\int_{-\infty}^{\infty}dx F(x) e^{-\epsilon |x|} = \pm f(\epsilon) + f(\epsilon)$
So when I take $\lim_{\epsilon \to 0^+} \epsilon (\pm f(\epsilon) + f(\epsilon))$ I get nothing that resembles $F(\infty) + F(-\infty)$, it's in terms of the transform not the original function?
-Dan
8. I was thinking about this approach also. I tend to make things more complicated than they need to be so I thought I'd simply try an integration by parts:
$\int_{-\infty}^{\infty}dx f(x)e^{-\epsilon |x|} = - \lim_{x \to \infty} f(x) \frac{e^{-\epsilon x}}{\epsilon} - \lim_{x \to -\infty} f(x) \frac{e^{\epsilon x}}{\epsilon} -$ $\int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}$
$= - \lim_{x \to \infty} \left [ (f(x) + f(-x)) \frac{e^{-\epsilon x}}{\epsilon} \right ] - \int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}$
To finish this I need to operate $\lim_{\epsilon \to 0^+} \epsilon$ on this, so I get:
$= - \lim_{\epsilon \to 0^+} \epsilon \lim_{x \to \infty} \left [ (f(x) + f(-x)) \frac{e^{-\epsilon x}}{\epsilon} \right ] -$ $\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f'(x) e^{-\epsilon |x|}$
$= - \lim_{\epsilon \to 0^+} \lim_{x \to \infty} \left [ (f(x) + f(-x)) e^{-\epsilon x} \right ] - 0?$
But how does one take this limit?
(I'm also having a problem arguing the limit of the integral, but I'll worry about that later.)
-Dan
9. Originally Posted by topsquark
Pffl. I'm running into a problem almost immediately.
$\int_{-\infty}^{\infty}dx F(x) e^{-\epsilon |x|} = \int_0^{\infty}dx F(-x) e^{-\epsilon x} + \int_0^{\infty}dx F(x) e^{-\epsilon x}$
I am sure you know this, you cannot take a Laplace transform of any given function. If a function is "too large" then it will not work. For example $e^{x^2}$.
I think the condition is (peicewise smooth),
$|f(x)|<K\cdot e^{-\epsilon x}$.
Futhermore, should it not be,
$\int_{-\infty}^0$
You have it the other way around.
10. Originally Posted by ThePerfectHacker
I am sure you know this, you cannot take a Laplace transform of any given function. If a function is "too large" then it will not work. For example $e^{x^2}$.
I think the condition is (peicewise smooth),
$|f(x)|<K\cdot e^{-\epsilon x}$.
Futhermore, should it not be,
$\int_{-\infty}^0$
You have it the other way around.
Actually, no. I subbed $x' = -x$ into the $\int_{-\infty}^0$ integral to get it into the form I posted.
So
$\int_{-\infty}^0 dx F(x) e^{\epsilon x} = \int_{\infty}^0 (-dx') F(-x') e^{-\epsilon x'}$ $= \int_0^{\infty} dx' F(-x') e^{-\epsilon x'}$
(I did this to put it into the "correct" form for a Laplace transform.)
As to the "largeness" argument, I am assuming a function that approaches a constant value at either infinity (ie I am assuming $\lim_{x \to \pm \infty}f(x)$ exists) and since I'm assuming f(x) is a $C^{\infty}$ function I am also assuming no discontinuities. (Perhaps I wasn't clear enough about that in my first post.) I believe (please correct me if I'm wrong) that guarentees the existence of the integral. If not then I am simply forced to add the assumption that the integral does exist and I'll work out later what restrictions that puts on the Physics.
-Dan
11. Originally Posted by topsquark
Actually, no. I subbed $x' = -x$ into the $\int_{-\infty}^0$ integral to get it into the form I posted.
As to the "largeness" argument, I am assuming a function that approaches a constant value at either infinity (ie I am assuming $\lim_{x \to \pm \infty}f(x)$ exists) and since I'm assuming f(x) is a $C^{\infty}$ function I am also assuming no discontinuities. (Perhaps I wasn't clear enough about that in my first post.) I believe (please correct me if I'm wrong) that guarentees the existence of the integral. If not then I am simply forced to add the assumption that the integral does exist and I'll work out later what restrictions that puts on the Physics.
-Dan
I would have a look at this, but its approaching Christmas, so I am run
off my feet here with getting stuff and attending end of term shows etc.
It may be a week before I get time in anything more than 10 minute slots.
RonL
12. Originally Posted by CaptainBlack
I would have a look at this, but its approaching Christmas, so I am run
off my feet here with getting stuff and attending end of term shows etc.
It may be a week before I get time in anything more than 10 minute slots.
RonL
Understood. I am going on a two week stint in Myrtle Beach with my parents (they're staying there until April) so I might not be around much myself.
(Chuckles) Don't worry. I suspect the problem will still be around when you get back. (Unless someone has a flash of inspiration anyway. But as I'll be away from my personal "library" it won't likely be me! )
-Dan
13. I got a tip from another forum on this. I want to bounce it off you guys.
For starters this is NOT rigorous, however it does explain why the assumption of a "sufficiently smooth" function was made as opposed to a more precise statement.
$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|}$
Split the integral into two parts, x < 0 and x > 0. I am going to look at the latter integral.
$\lim_{\epsilon \to 0^+} \epsilon \int_0^{\infty}dx f(x) e^{-\epsilon x}$
(*) Now make the substitution $y = \epsilon x$:
$\lim_{\epsilon \to 0^+} \epsilon \int_0^{\infty}dx f(x) e^{-\epsilon x} = \lim_{\epsilon \to 0^+} \int_0^{\infty}dy f \left ( \frac{y}{\epsilon} \right ) e^{-y}$
If the function is "sufficiently smooth" we may interchange the limit and integration operations:
$\lim_{\epsilon \to 0^+} \int_0^{\infty}dy f \left ( \frac{y}{\epsilon} \right ) e^{-y} = \int_0^{\infty}dy e^{-y} \lim_{\epsilon \to 0^+} f \left ( \frac{y}{\epsilon} \right )$
Now, $\lim_{\epsilon \to 0^+} f \left ( \frac{y}{\epsilon} \right )$ if it exists is just the number $\lim_{x \to \infty}f(x)$ and is a constant, so we may take it outside of the integration.
$\int_0^{\infty}dy e^{-y} \lim_{\epsilon \to 0^+} f \left ( \frac{y}{\epsilon} \right ) = f(\infty) \int_0^{\infty}dy e^{-y} = f(\infty) \cdot 1 = f(\infty)$
where I have used an obvious notation.
The x < 0 integration may be done in the same way. Thus we have:
$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty}dx f(x) e^{-\epsilon |x|} = f(\infty) + f(-\infty)$
One comment. Go back to the (*) line. Being more careful, this integral should be written as:
$\lim_{\epsilon \to 0^+} \epsilon \lim_{N \to \infty} \int_0^{N}dx f(x) e^{-\epsilon x}$
So when I make the y substitution:
$\lim_{\epsilon \to 0^+} \epsilon \lim_{N \to \infty} \int_0^{N}dx f(x) e^{-\epsilon x} = \lim_{\epsilon \to 0^+} \lim_{N \to \infty} \int_0^{\epsilon N}dy f \left ( \frac{y}{\epsilon} \right ) e^{-y}$
In order to get the expression I wrote I implicitly took $\lim_{\epsilon \to 0^+} \lim_{N \to \infty} \epsilon N \to \infty$. The argument is that $\epsilon$, however small, is always a positive number so upon taking the N limit $\epsilon N$ should still go to infinity. This, aside from switching the limit and the integration, is the weakest point in the argument. Any comments?
Thanks!
-Dan
14. Originally Posted by topsquark
I've got another one for you.
I ran across this theorem in my Field Theory text. The theorem is this:
Given a sufficiently smooth function f(x) we have that:
$\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}$
Rough outline:
1. Every function can be written as the sum of an odd function and an even function zero at the origin and a constant.
2. The result is trivially true for a sufficiently well behaved odd function.
3. The result is a result of the fundamental theorem of calculus, and the
Laplace transform of a derivative for an even function (sufficiently well behaved) zero at the origin.
3a. The result is true for a constant (we need to use the Laplace transform of a constant result here)
4. Hence the result is true for a sufficiently well behaved function.
I can amplify the details if you need (though I doubt that I will feel like
going for full rigour)
RonL
15. Originally Posted by CaptainBlack
Rough outline:
1. Every function can be written as the sum of an odd function and an even function.
2. The result is trivially true for a sufficiently well behaved odd function.
3. The result is a result of the fundamental theorem of calculus, and the
Laplace transform of a derivative for an even function (sufficiently well behaved).
4. Hence the result is true for a sufficiently well behaved function.
I can amplify the details if you need (though I doubt that I will feel like
going for full rigour)
RonL
No, I don't need "full rigor." (Sounds dangerous anyway! ) I just wanted to check that there were no glaring logical arguments in the proof. For some reason when I integrate over arbitrary functions my intuition and confidence goes right out the window. Thanks for the review!
-Dan
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|
http://math.stackexchange.com/questions/93898/trying-to-figure-out-how-an-approximation-of-a-logarithmic-equation-works
|
Trying to figure out how an approximation of a logarithmic equation works
The physics books I'm reading gives $$\triangle\tau=\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\left(r_{1}-r_{2}+2m\ln\frac{r_{1}-2m}{r_{2}-2m}\right).$$ We are then told $2m/r$ is small for $r_{2}<r<r_{1}$ which gives the approximation$$\triangle\tau\approx\frac{2}{c}\left(r_{1}-r_{2}-\frac{m\left(r_{1}-r_{2}\right)}{r_{1}}+2m\ln\left(\frac{r_{1}}{r_{2}}\right)\right).$$ I can see how $$\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\approx\frac{2}{c}$$ but can't see how the rest of it appears. It seems to be saying that$$2\ln\frac{r_{1}-2m}{r_{2}-2m}\approx\left(-\frac{\left(r_{1}-r_{2}\right)}{r_{1}}+2\ln\left(\frac{r_{1}}{r_{2}}\right)\right)$$
I've tried getting all the lns on one side, and also expanding $\ln\frac{r_{1}-2m}{r_{2}-2m}$ to $\ln\left(r_{1}-2m\right)-\ln\left(r_{2}-2m\right)$ and generally juggling it all about but with no luck. Any suggestions or hints from anyone?
It's to do with the gravitational time delay effect. It seems a bit more maths than physics which is why I'm asking it here.
Many thanks
-
1
Is $m$ a small mass? I.e. is $m^2/r_1$ considered negligible? – Raskolnikov Dec 24 '11 at 17:02
@Raskolnikov - Thanks. m equals Gm/c^2. It's part of the Schwarzschild metric. – Peter4075 Dec 24 '11 at 17:17
@Raskolnikov - sorry, that's nonsense. m equals GM/c^2 – Peter4075 Dec 24 '11 at 17:28
OK, for Earth, this would mean that $GM/c^2\approx (GM/R^2)\cdot (R^2/c^2)\approx 10\cdot(R/c)^2$ where $R$ is the Earth's radius. This is indeed a small number and is therefore negligible when squared. I presume the time delay is related to a GPS application or something? – Raskolnikov Dec 24 '11 at 17:32
@Raskolnikov. It's to do with bouncing a radar signal sent from Earth off Venus or Mercury and measuring the gravitational time delay effect due to the Sun curving spacetime. All planets are on the same side of the Sun. M is the mass of the Sun. r1 and r2 are the Schwarzschild radial coordinates of Earth and Venus/Mercury. The book is A short course in general relativity by Foster and Nightingale, page 130. It's on Google Books. Thanks – Peter4075 Dec 24 '11 at 17:47
2 Answers
As Raskolnikov says, the first approximation is actually
$$\frac{2}{c}\left(1-\frac{2m}{r_1}\right)^{1/2}\approx\frac{2}{c}\left(1-\frac{m}{r_1}\right)$$
This is a valid approximation because the power series for $(1-x)^{1/2}$ is
$$1 -\frac{1}{2}x+\cdots$$
So as long as $x=\frac{2m}{r_1}$ is close to zero, the above approximation is a valid first-degree approximation. Expanding this substitution,
$$\begin{align} \Delta\tau&\approx\frac{2}{c}\left(1-\frac{m}{r_1}\right)\left(r_{1}-r_{2}+2m\ln\frac{r_{1}-2m}{r_{2}-2m}\right)\\ & = \frac{2}{c}\left(r_{1}-r_{2}-\frac{m(r_1-r_2)}{r_1}+2m\left(1-\frac{m}{r_1}\right)\ln\frac{r_{1}-2m}{r_{2}-2m}\right) \end{align}$$
So the second approximation that has been made is
$$\begin{align} \left(1-\frac{m}{r_1}\right)\ln\frac{r_{1}-2m}{r_{2}-2m}&\approx\ln\left(\frac{r_1}{r_2}\right)\\ \end{align}$$
This is equivalent to the following approximation using logarithm rules
$$\begin{align} \left(1-\frac{m}{r_1}\right)\left(\ln(r_1)+\ln(1-2m/r_1)-\ln(r_2)-\ln(1-2m/r_2)\right)&\approx\ln\left(\frac{r_1}{r_2}\right)\\ \end{align}$$
Now you just drop all the terms that have $\frac{m}{r_i}$, and your approximation is another logarithm rule. It is valid to drop these terms, because presumably $\ln(r_1)$, $\ln(r_2)$, and $1$ are relatively much larger.
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Yes! I would never have seen that in a month of Sundays. Thanks. – Peter4075 Dec 24 '11 at 18:28
It actually seems to me they use
$$\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\approx\frac{2}{c}\left(1-\frac{m}{r_{1}}\right)$$
and
$$2m\ln\frac{r_{1}-2m}{r_{2}-2m}\approx 2m\ln\left(\frac{r_{1}}{r_{2}}\right) \; .$$
EDIT: Just realized the following:
$$2m\ln\frac{r_{1}-2m}{r_{2}-2m}\approx 2m\ln\left(\frac{r_{1}}{r_{2}}\right) + 2m\left(\frac{2m}{r_2}-\frac{2m}{r_1}\right) \; .$$
Now the last term can be rewritten as
$$2m\left(\frac{2m}{r_2}-\frac{2m}{r_1}\right) = \frac{(2m)^2}{r_1 r_2}(r_1-r_2) = \left(\frac{2m}{r_1}\right)\left(\frac{2m}{r_2}\right)(r_1-r_2)$$
which is negligible.
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http://mathematica.stackexchange.com/questions/tagged/computational-geometry+graphs-and-networks
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http://telescoper.wordpress.com/tag/anthropic-principle/
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# In the Dark
A blog about the Universe, and all that surrounds it
## Insignificance
Posted in The Universe and Stuff with tags anthropic principle, astronomy, Cosmology, life, Moon, partial eclipse, Sun, total eclipse on January 4, 2011 by telescoper
I’m told that there was a partial eclipse of the Sun visible from the UK this morning, although it was so cloudy here in Cardiff that I wouldn’t have seen anything even if I had bothered to get up in time to observe it. For more details of the event and pictures from people who managed to see it, see here. There’s also a nice article on the BBC website. The BBC are coordinating three days of programmes alongside a host of other events called Stargazing Live presumably timed to coincide with this morning’s eclipse. It’s taking a chance to do live broadcasts about astronomy given the British weather, but I hope they are successful in generating interest especially among the young.
As a spectacle a partial solar eclipse is pretty exciting – as long as it’s not cloudy – but even a full view of one can’t really be compared with the awesome event that is a total eclipse. I’m lucky enough to have observed one and I can tell you it was truly awe-inspiring.
If you think about it, though, it’s a very strange thing that such a thing is possible at all. In a total eclipse, the Moon passes between the Earth and the Sun in such a way that it exactly covers the Solar disk. In order for this to happen the apparent angular size of the Moon (as seen from Earth) has to be almost exactly the same as that of the Sun (as seen from Earth). This involves a strange coincidence: the Moon is small (about 1740 km in radius) but very close to the Earth in astronomical terms (about 400,000 km away). The Sun, on the other hand, is both enormously large (radius 700,000 km) and enormously distant (approx. 150,000,000 km). The ratio of radius to distance from Earth of these objects is almost identical at the point of a a total eclipse, so the apparent disk of the Moon almost exactly fits over that of the Sun. Why is this so?
The simple answer is that it is just a coincidence. There seems no particular physical reason why the geometry of the Earth-Moon-Sun system should have turned out this way. Moreover, the system is not static. The tides raised by the Moon on the Earth lead to frictional heating and a loss of orbital energy. The Moon’s orbit is therefore moving slowly outwards from the Earth. I’m not going to tell you exactly how quickly this happens, as it is one of the questions I set my students in the module Astrophysical Concepts I’ll be starting in a few weeks, but eventually the Earth-Moon distance will be too large for total eclipses of the Sun by the Moon to be possible on Earth, although partial and annular eclipses may still be possible.
It seems therefore that we just happen to be living at the right place at the right time to see total eclipses. Perhaps there are other inhabited moonless planets whose inhabitants will never see one. Future inhabitants of Earth will have to content themselves with watching eclipse clips on Youtube.
Things may be more complicated than this though. I’ve heard it argued that the existence of a moon reasonably close to the Earth may have helped the evolution of terrestrial life. The argument – as far as I understand it – is that life presumably began in the oceans, then amphibious forms evolved in tidal margins of some sort wherein conditions favoured both aquatic and land-dwelling creatures. Only then did life fully emerge from the seas and begin to live on land. If it is the case that the existence of significant tides is necessary for life to complete the transition from oceans to solid ground, then maybe the Moon played a key role in the evolution of dinosaurs, mammals, and even ourselves.
I’m not sure I’m convinced of this argument because, although the Moon is the dominant source of the Earth’s tides, it is not overwhelmingly so. The effect of the Sun is also considerable, only a factor of three smaller than the Moon. So maybe the Sun could have done the job on its own. I don’t know.
That’s not really the point of this post, however. What I wanted to comment on is that astronomers basically don’t question the interpretation of the occurence of total eclipses as simply a coincidence. Eclipses just are. There are no doubt many other planets where they aren’t. We’re special in that we live somewhere where something apparently unlikely happens. But this isn’t important because eclipses aren’t really all that significant in cosmic terms, other than that the law of physics allow them.
On the other hand astronomers (and many other people) do make a big deal of the fact that life exists in the Universe. Given what we know about fundamental physics and biology – which admittedly isn’t very much – this also seems unlikely. Perhaps there are many other worlds without life, so the Earth is special once again. Others argue that the existence of life is so unlikely that special provision must have been made to make it possible.
Before I find myself falling into the black hole marked “Anthropic Principle” let me just say that I don’t see the existence of life (including human life) as being of any greater significance than that of a total eclipse. Both phenomena are (subjectively) interesting to humans, both are contingent on particular circumstances, and both will no doubt cease to occur at some point in perhaps not-too-distant the future. Neither tells us much about the true nature of the Universe.
Let’s face it. We’re just not significant.
61 Comments »
## Ergodic Means…
Posted in The Universe and Stuff with tags anthropic principle, Cosmology, ergodic hypothesis, multiverse, Physics, probability, statistical physics on October 19, 2009 by telescoper
The topic of this post is something I’ve been wondering about for quite a while. This afternoon I had half an hour spare after a quick lunch so I thought I’d look it up and see what I could find.
The word ergodic is one you will come across very frequently in the literature of statistical physics, and in cosmology it also appears in discussions of the analysis of the large-scale structure of the Universe. I’ve long been puzzled as to where it comes from and what it actually means. Turning to the excellent Oxford English Dictionary Online, I found the answer to the first of these questions. Well, sort of. Under etymology we have
ad. G. ergoden (L. Boltzmann 1887, in Jrnl. f. d. reine und angewandte Math. C. 208), f. Gr.
I say “sort of” because it does attribute the origin of the word to Ludwig Boltzmann, but the greek roots (εργον and οδοσ) appear to suggest it means “workway” or something like that. I don’t think I follow an ergodic path on my way to work so it remains a little mysterious.
The actual definitions of ergodic given by the OED are
Of a trajectory in a confined portion of space: having the property that in the limit all points of the space will be included in the trajectory with equal frequency. Of a stochastic process: having the property that the probability of any state can be estimated from a single sufficiently extensive realization, independently of initial conditions; statistically stationary.
As I had expected, it has two meanings which are related, but which apply in different contexts. The first is to do with paths or orbits, although in physics this is usually taken to meantrajectories in phase space (including both positions and velocities) rather than just three-dimensional position space. However, I don’t think the OED has got it right in saying that the system visits all positions with equal frequency. I think an ergodic path is one that must visit all positions within a given volume of phase space rather than being confined to a lower-dimensional piece of that space. For example, the path of a planet under the inverse-square law of gravity around the Sun is confined to a one-dimensional ellipse. If the force law is modified by external perturbations then the path need not be as regular as this, in extreme cases wandering around in such a way that it never joins back on itself but eventually visits all accessible locations. As far as my understanding goes, however, it doesn’t have to visit them all with equal frequency. The ergodic property of orbits is intimately associated with the presence of chaotic dynamical behaviour.
The other definition relates to stochastic processes, i.e processes involving some sort of random component. These could either consist of a discrete collection of random variables {X1…Xn} (which may or may not be correlated with each other) or a continuously fluctuating function of some parameter such as time t, i.e. X(t) or spatial position (or perhaps both).
Stochastic processes are quite complicated measure-valued mathematical entities because they are specified by probability distributions. What the ergodic hypothesis means in the second sense is that measurements extracted from a single realization of such a process have a definition relationship to analagous quantities defined by the probability distribution.
I always think of a stochastic process being like a kind of algorithm (whose workings we don’t know). Put it on a computer, press “go” and it spits out a sequence of numbers. The ergodic hypothesis means that by examining a sufficiently long run of the output we could learn something about the properties of the algorithm.
An alternative way of thinking about this for those of you of a frequentist disposition is that the probability average is taken over some sort of statistical ensemble of possible realizations produced by the algorithm, and this must match the appropriate long-term average taken over one realization.
This is actually quite a deep concept and it can apply (or not) in various degrees. A simple example is to do with properties of the mean value. Given a single run of the program over some long time T we can compute the sample average
$\bar{X}_T\equiv \frac{1}{T} \int_0^Tx(t) dt$
the probability average is defined differently over the probability distribution, which we can call p(x)
$\langle X \rangle \equiv \int x p(x) dx$
If these two are equal for sufficiently long runs, i.e. as T goes to infinity, then the process is said to be ergodic in the mean. A process could, however, be ergodic in the mean but not ergodic with respect to some other property of the distribution, such as the variance. Strict ergodicity would require that the entire frequency distribution defined from a long run should match the probability distribution to some accuracy.
Now we have a problem with the OED again. According to the defining quotation given above, ergodic can be taken to mean statistically stationary. Actually that’s not true. ..
In the one-parameter case, “statistically stationary” means that the probability distribution controlling the process is independent of time, i.e. that p(x,t)=p(x,t+Δt) . It’s fairly straightforward to see that the ergodic property requires that a process X(t) be stationary, but the converse is not the case. Not every stationary process is necessarily ergodic. Ned Wright gives an example here. For a higher-dimensional process, such as a spatially-fluctuating random field the analogous property is statistical homogeneity, rather than stationarity, but otherwise everything carries over.
Ergodic theorems are very tricky to prove in general, but there are well-known results that rigorously establish the ergodic properties of Gaussian processes (which is another reason why theorists like myself like them so much). However, it should be mentioned that even if the ergodic assumption applies its usefulness depends critically on the rate of convergence. In the time-dependent example I gave above, it’s no good if the averaging period required is much longer than the age of the Universe; in that case even ergodicity makes it difficult to make inferences from your sample. Likewise the ergodic hypothesis doesn’t help you analyse your galaxy redshift survey if the averaging scale needed is larger than the depth of the sample.
Moreover, it seems to me that many physicists resort to ergodicity when there isn’t any compelling mathematical grounds reason to think that it is true. In some versions of the multiverse scenario, it is hypothesized that the fundamental constants of nature describing our low-energy turn out “randomly” to take on different values in different domains owing to some sort of spontaneous symmetry breaking perhaps associated a phase transition generating cosmic inflation. We happen to live in a patch within this structure where the constants are such as to make human life possible. There’s no need to assert that the laws of physics have been designed to make us possible if this is the case, as most of the multiverse doesn’t have the fine tuning that appears to be required to allow our existence.
As an application of the Weak Anthropic Principle, I have no objection to this argument. However, behind this idea lies the assertion that all possible vacuum configurations (and all related physical constants) do arise ergodically. I’ve never seen anything resembling a proof that this is the case. Moreover, there are many examples of physical phase transitions for which the ergodic hypothesis is known not to apply. If there is a rigorous proof that this works out, I’d love to hear about it. In the meantime, I remain sceptical.
5 Comments »
## Multiversalism
Posted in The Universe and Stuff with tags anthropic principle, Cosmology, multiverse on June 17, 2009 by telescoper
The word “cosmology” is derived from the Greek κόσμος (“cosmos”) which means, roughly speaking, ”the world as considered as an orderly system”. The other side of the coin to “cosmos” is Χάος (“chaos”). In one world-view the Universe comprised two competing aspects: the orderly part that was governed by laws and which could (at least in principle) be predicted, and the “random” part which was disordered and unpredictable. To make progress in scientific cosmology we do need to assume that the Universe obeys laws. We also assume that these laws apply everywhere and for all time or, if they vary, then they vary in accordance with another law. This is the cosmos that makes cosmology possible. However, with the rise of quantum theory, and its applications to the theory of subatomic particles and their interactions, the field of cosmology has gradually ceded some of its territory to chaos.
In the early twentieth century, the first mathematical world models were constructed based on Einstein’s general theory of relativity. This is a classical theory, meaning that it describes a system that evolves smoothly with time. It is also entirely deterministic. Given sufficient information to specify the state of the Universe at a particular epoch, it is possible to calculate with certainty what its state will be at some point in the future. In a sense the entire evolutionary history described by these models is not a succession of events laid out in time, but an entity in itself. Every point along the space-time path of a particle is connected to past and future in an unbreakable chain. If ever the word cosmos applied to anything, this is it.
But as the field of relativistic cosmology matured it was realised that these simple classical models could not be regarded as complete, and consequently that the Universe was unlikely to be as predictable as was first thought. The Big Bang model gradually emerged as the favoured cosmological theory during the middle of the last century, between the 1940s and the 1960s. It was not until the 1960s, with the work of Hawking and Penrose, that it was realised that expanding world models based on general relativity inevitably involve a break-down of known physics at their very beginning. The so-called singularity theorems demonstrate that in any plausible version of the Big Bang model, all physical parameters describing the Universe (such as its density, pressure and temperature) all become infinite at the instant of the Big Bang. The existence of this “singularity” means that we do not know what laws if any apply at that instant. The Big Bang contains the seeds of its own destruction as a complete theory of the Universe. Although we might be able to explain how the Universe subsequently evolves, we have no idea how to describe the instant of its birth. This is a major embarrassment. Lacking any knowledge of the laws we don’t even have any rational basis to assign probabilities. We are marooned with a theory that lets in water.
The second important development was the rise of quantum theory and its incorporation into the description of the matter and energy contained within the Universe. Quantum mechanics (and its development into quantum field theory) entails elements of unpredictability. Although we do not know how to interpret this feature of the theory, it seems that any cosmological theory based on quantum theory must include things that can’t be predicted with certainty.
As particle physicists built ever more complete descriptions of the microscopic world using quantum field theory, they also realised that the approaches they had been using for other interactions just wouldn’t work for gravity. Mathematically speaking, general relativity and quantum field theory just don’t fit together. It might have been hoped that quantum gravity theory would help us plug the gap at the very beginning of the Universe, but that has not happened yet because there isn’t such a theory. What we can say about the origin of the Universe is correspondingly extremely limited and mostly speculative, but some of these speculations have had a powerful impact on the subject.
One thing that has changed radically since the early twentieth century is the possibility that our Universe may actually be part of a much larger “collection” of Universes. The potential for semantic confusion here is enormous. The Universe is, by definition, everything that exists. Obviously, therefore, there can only be one Universe. The name given to a Universe that consists of bits and pieces like this is the multiverse.
There are various ways a multiverse can be realised. In the “Many Worlds” interpretation of quantum mechanics there is supposed to be a plurality of versions of our Universe, but their ontological status is far from clear (at least to me). Do we really have to accept that each of the many worlds is “out there”, or can we get away with using them as inventions to help our calculations?
On the other hand, some plausible models based on quantum field theory do admit the possibility that our observable Universe is part of collection of mini-universes, each of which “really” exists. It’s hard to explain precisely what I mean by that, but I hope you get my drift. These mini-universes form a classical ensemble in different domains of a single-space time, which is not what happens in quantum multiverses.
According to the Big Bang model, the Universe (or at least the part of it we know about) began about fourteen billion years ago. We do not know whether the Universe is finite or infinite, but we do know that if it has only existed for a finite time we can only observe a finite part of it. We can’t possibly see light from further away than fourteen billion light years because any light signal travelling further than this distance would have to have set out before the Universe began. Roughly speaking, this defines our “horizon”: the maximum distance we are in principle able to see. But the fact that we can’t observe anything beyond our horizon does not mean that such remote things do not exist at all. Our observable “patch” of the Universe might be a tiny part of a colossal structure that extends much further than we can ever hope to see. And this structure might be not at all homogeneous: distant parts of the Universe might be very different from ours, even if our local piece is well described by the Cosmological Principle.
Some astronomers regard this idea as pure metaphysics, but it is motivated by plausible physical theories. The key idea was provided by the theory of cosmic inflation, which I have blogged about already. In the simplest versions of inflation the Universe expands by an enormous factor, perhaps 1060, in a tiny fraction of a second. This may seem ridiculous, but the energy available to drive this expansion is inconceivably large. Given this phenomenal energy reservoir, it is straightforward to show that such a boost is not at all unreasonable. With inflation, our entire observable Universe could thus have grown from a truly microscopic pre-inflationary region. It is sobering to think that everything galaxy, star, and planet we can see might from a seed that was smaller than an atom. But the point I am trying to make is that the idea of inflation opens up ones mind to the idea that the Universe as a whole may be a landscape of unimaginably immense proportions within which our little world may be little more than a pebble. If this is the case then we might plausibly imagine that this landscape varies haphazardly from place to place, producing what may amount to an ensemble of mini-universes. I say “may” because there is yet no theory that tells us precisely what determines the properties of each hill and valley or the relative probabilities of the different types of terrain.
Many theorists believe that such an ensemble is required if we are to understand how to deal probabilistically with the fundamentally uncertain aspects of modern cosmology. I don’t think this is the case. It is, at least in principle, perfectly possible to apply probabilistic arguments to unique events like the Big Bang using Bayesian inference. If there is an ensemble, of course, then we can discuss proportions within it, and relate these to probabilities too. Bayesians can use frequencies if they are available but do not require them. It is one of the greatest fallacies in science that probabilities need to be interpreted as frequencies.
At the crux of many related arguments is the question of why the Universe appears to be so well suited to our existence within it. This fine-tuning appears surprising based on what (little) we know about the origin of the Universe and the many other ways it might apparently have turned out. Does this suggest that it was designed to be so or do we just happen to live in a bit of the multiverse nice enough for us to have evolved and survived in?
Views on this issue are often boiled down into a choice between a theistic argument and some form of anthropic selection. A while ago I gave a talk at a meeting in Cambridge called God or Multiverse? that was an attempt to construct a dialogue between theologians and cosmologists. I found it interesting, but it didn’t alter my view that science and religion don’t really overlap very much at all on this, in the sense that if you believe in God it doesn’t mean you have to reject the multiverse, or vice-versa. If God can create a Universe, he could create a multiverse to0. As it happens, I’m agnostic about both.
So having, I hope, opened up your mind to the possibility that the Universe may be amenable to a frequentist interpretation, I should confess that I think one can actually get along quite nicely without it. In any case, you will probably have worked out that I don’t really like the multiverse. One reason I don’t like it is that it accepts that some things have no fundamental explanation. We just happen to live in a domain where that’s the way things are. Of course, the Universe may turn out to be like that - there definitely will be some point at which our puny monkey brains can’t learn anything more – but if we accept that then we certainly won’t find out if there is really a better answer, i.e. an explanation that isn’t accompanied by an infinite amount of untestable metaphysical baggage. My other objection is that I think it’s cheating to introduce an infinite thing to provide an explanation of fine tuning. Infinity is bad.
3 Comments »
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http://crypto.stackexchange.com/questions/1296/security-of-simple-xor-and-s-box-cipher/1307
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# Security of simple xor and s-box cipher?
What weaknesses (or strengths) do block ciphers based on only key xor and s-box have when operating in CBC mode?
A cipher's internal primitive might be a simple as this:
$C = S[M \oplus k]$, where $C$ is ciphertext, $M$ is the plaintext message, $k$ is the key and $S$ is an S-box.
Assume the follwoing:
• The key $k$ is sufficiently large and is random.
• The initialization vector used by CBC is random.
• The block size is reasonable, e.g. 128 or 256 bits.
• The S-box is chosen such that correlation bias between $S[x]$ and $x$ is low.
• The S-box is chosen such that $S[a]$ and $S[b]$ are independant for any random $a$ and $b$.
EDIT: Completely re-written to be more general, removing all specific implementation details.
Here's the original custom scheme, which was considered too broad to be a valid question:
$k$ is a 256-bit key, $S$ is a 16-bit S-box that has some set properties.
The cipher operates in CBC mode (as you'll see in the second step), has a block size of 256 bits, and operates as follows for a total of 16 rounds:
• $k_r = rot(k, r)$, where $k$ is the key, $r$ is the round number and $rot()$ is a circular shift operation.
• $C_n = S[M_n \oplus k_r] \oplus C_{n-1}$ where $n$ is the block number (assume $C_{-1}$ is the IV)
• Select $a = 3(r+n) \space mod\space 256$
• Select $b = (k_r \space\& \space 255) \space mod\space 256$
• Swap the bits $C_n[a]$ and $C_n[b]$
• $C_n = S[C_n]$
-
1
I don't readily see how you decrypt ... do you have an inverse S-box, too? Also, you seem to have mixed the mode of operation (CBC) with your block cipher design itself, which seems to be not a good idea. – Paŭlo Ebermann♦ Nov 23 '11 at 13:44
1
"In cryptography, a block cipher is a symmetric key cipher operating on fixed-length groups of bits, called blocks, with an unvarying transformation" - I'd say that my cipher fits that definition. Regardless, it's semantics, and its name doesn't affect its security (or lack thereof). I'm really just trying to learn here. – Polynomial Nov 23 '11 at 14:06
1
You state that the cipher has 16 rounds; which steps are repeated in a round? Step 2 refers to $M_n$ (which is presumably the plaintext message); is that final value of $C_n$ from the previous round? – poncho Nov 23 '11 at 14:26
1
Just a note to anyone reading this and being confused (e.g. with poncho's mentioning of "Step 2"), there was a cleanup and the question was changed. You can view the original question in the edit history. – Polynomial Nov 23 '11 at 14:48
1
@EthanHeilman - Done. – Polynomial Nov 23 '11 at 15:43
show 10 more comments
## 3 Answers
Let a "block cipher" be defined with a fixed S-box $S$ (i.e. a permutation of some space) and a key $K$ (same size than a block), such that the encryption of a block $M$ is $C = S[P\oplus K]$. Everybody knows $S$ and can apply and invert it (that's a "S-box", not a "key" -- if the S-box is "key dependent" then the S-box is itself a block cipher in its own right, and that's another question). Since the key has the size of a block, blocks are supposed to be large enough to defeat exhaustive search on the key (say, we use 128-bit blocks).
Hence, given a single known plaintext block and the corresponding ciphertext, an attacker can trivially recover $K$: $K = S^{-1}[C] \oplus M$. This is weak unless you ever encrypt a single block with a given key, in which case this is just One-Time Pad. The S-box here is a red herring: as a generic rule, any fixed public permutation for which either the input or output is known can be "removed" at will, so it adds nothing to security. For instance, in DES, the "Initial Permutation" and the "Final Permutation" (which swap bits around in a 64-bit word) can be abstracted away from the security analysis, because anybody can apply and unapply them. Without the S-box, we get $C = M \oplus K$, the well-known equation for OTP.
Since a fixed permutation is as good as gone when it operates on known values, and since we usually consider that the attacker knows some plaintext/ciphertext pairs, this means that a S-box, to be useful, must be "isolated" from both plaintext and ciphertext. So we can imagine the following design:
$$C = K \oplus S[M \oplus K]$$
This is what @Ethan suggests to study as an exercise (except the he talks about 16-bit blocks, hence a 16-bit key, which is trivially undone through exhaustive search; here, let's assume that $C$, $M$ and $K$ are sequences of 128 bits). If you look at the slides that Ethan points to, you will learn that this is a special case of a more generic construction from Even and Mansour, which goes like this:
$$C = K_2 \oplus S[M \oplus K_1]$$
with two $n$-bit keys $K_1$ and $K_2$. John Daemen has then shown that although the total key material length is $2n$ bits, you get much less security, actually no more than $2^{n/2}$. With $n = 128$, that's "64-bit security", a rather low value against today's technology (64-bit exhaustive key search is expensive but has been done at least once with thousands of machine and 5 years of computation). With $n = 256$, we could hope for 128-bit security, albeit with 512 bits of key material.
However, there are two important points about that:
• The subkeys $K_1$ and $K_2$ should be unrelated. Using the same key for both, or simply keys which admit simple algebraic relations, may allow for many additional attacks.
• The $2^{n/2}$ bound is for a generic S-box, i.e. a permutation taken at random over all the permutations of $n$-bit blocks. But the S-box is computable. There exists a rather compact piece of code which can run it in both directions. Yet, there are $2^n!$ permutations of $n$-bit blocks, which implies that the minimal representation of a random permutation will need, on average, at least $2^{264}$ bits, if $n = 256$. This is ludicrously high (it will not fit in the known Universe). Conclusion: an actual implementation of that cipher will use a specific S-box out of the very limited set of S-boxes than can be computed by some code which fits in a few kilobytes of opcodes and constant arrays. The S-box necessarily has some structure. Structure might be exploitable, and often is. The $2^{n/2}$ theoretical strength may then be hard to reach in practice.
So the Even-Mansour scheme is not "secure enough". What do cryptographers do in such a situation ? They add more rounds ! So we are talking about something like:
$$C = K_5 \oplus S[K_4 \oplus S[K_3 \oplus S[K_2 \oplus S[K_1 \oplus S[K_0 \oplus M]]]]]$$
for, say, 5 rounds. How many rounds do we need ? This depends on the specific unavoidable structure of $S$, and also on how "different" the $K_i$ are (we would, in practice, generate the $K_i$, dubbed "subkeys", from a master key $K$ through a deterministic process -- the "key schedule" -- which itself will have some potentially exploitable structure). We want the whole thing to be decently efficient, so we cannot add thousands of rounds, and we must cope with a relatively simple $S$.
Down that road lies... the AES itself. With 128-bit blocks and a 128-bit master key, the AES uses 10 rounds, hence 11 subkeys. Each round $S$ is a combination of a few operations, some linear (in a given vector space over $\mathbb{F}_{256}$), and an internal fixed permutation over 8-bit blocks (what the AES specification calls "the S-box", but not what we call $S$ in this text !). The AES is believed secure for what it is meant to be. It has raised a few concerns about "related keys": pairs of keys which induce the corresponding AES instances to behave in a related way, which can be detected; nothing serious for encryption, but something to watch for if you want to use the AES as part of, say, a hash function. This is due to the mathematical structure in the key schedule. The Whirlpool hash function is a hash function derived from the AES, and the first thing the Whirlpool designers do was to replace the AES key schedule with something with less exploitable structure (but also slower).
Summary: designing a block cipher as a sequence of XOR with (sub-)keys mixed with a fixed permutation of the whole block space is a valid block cipher design. But with a single "perfect" S-box and two sub-keys, you do not get your money worth of security (you get at most $2^{n/2}$ of a $n$-bit block). And since both the S-box and the key schedule cannot be "perfect", you have to add more rounds, and be very careful about the mathematical structures that you must necessarily employ. The current best-in-class design with that kind of structure is the AES, and it took many smart cryptographers and a lot of time to actually build some trust in its security.
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+1, this is an excellent answer, thanks Thomas. – Ethan Heilman Nov 23 '11 at 19:32
Wow. This answer is precisely what I was looking for. The only part I don't quite get is the S-box section - are you implying that a $S[x]$ is meant to exist for every $n$-bit $x$, where $n$ is the block size? Wouldn't this result in ludicrously sized S-boxes? I know you mentioned something about huge S-boxes that can't fit in the universe (yay for the $10^{89}$ particle limit!) but I'm not sure how this works practically. – Polynomial Nov 24 '11 at 7:07
Hmm, do we compute each $S[x]$ for $x$ on the fly, using some function $f$, rather than "caching" the entire $S$ field in an array somewhere? – Polynomial Nov 24 '11 at 7:14
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$S$ "exists" as code which can compute a value ($S[x]$) for every $x$. One possible implementation is as a big array. Or you could have a sequence of instructions. That's equivalent here: if you have, e.g., 2 kB of ROM (code + arrays) then you have 16384 bits, and you can possibly compute only $2^{16384}$ distinct functions with that. But there are far many more possible functions with 256-bit inputs than that. Whether you use your ROM budget for code or for data is irrelevant in this computation. – Thomas Pornin Nov 24 '11 at 13:26
I'm going to ignore the "CBC" part of the question and focus on "What are the strengths and weaknesses of a s-box and xor cipher. I'm going to assume that the s-box size is smaller than the message block size since any cipher that has a block size that is equal to it's s-box size is going to have a block size small enough to be brute forced.
Using xor and s-boxes we need some way to "mix" or diffuse the information in the message block to all the bits in ciphertext block. Typically this is down using a permutation as is done in a permutation-substitution network.
To reframe this within the scope of the question, a cipher that did not use permutations to mix the bits across substitution boundaries will be easy to brute force since the ciphertext block can we broken into chunks and each chunk can be attacked independently.
EDIT: These attacks on the scheme apply to the original scheme (which is posted at the bottom of the question) and may not apply to the new scheme (although they are general methods of attack).
I like this question a lot because it gets at the heart of what makes a strong block cipher and what makes a weak block cipher. If I have made mistakes in understanding your scheme, please don't hesitate to correct me. Here are the weaknesses in the design that a strong block cipher would protect against:
Known plaintext attack - Given that an inverse s-box $S^{-1}$ exists the final step $C_n = S[C_n]$ is always invertible by an attacker. The second to the last step is only 256 possibilities since we know $S^{-1}[C_n]$ and we can check all possibilities. We now know several bits of the round key (which is bad as we will discuss below) we also know the result of $S[M_n \oplus k_r] \oplus C_{n-1}$ which we will call $R_n =S[M_n \oplus k_r] \oplus C_{n-1}$. This is really bad because the ciphertext $C_{n-1}$ is public knowledge and $S[M_n \oplus k_r]$ is invertible. Given these two facts we can find $M_n \oplus K_r$ by
$$S^{-1}[R_n \oplus C_n] = M_n \oplus K_r$$
Given a known plaintext $M_n$, and $M_n \oplus K_r$ we learn the round key $K_r$. As I'll talk about below, the round key gives us the full key $K$ which allows us to decrypt the rest of the message.
Weak Keys Schedule - Rotations don't mix up the round keys ($k_r$) enough. Consider these three attacks:
1. Weak Keys - Someone chooses a key that is the same key under some or all of the rotations, for example consider the key $00000...000$ or the key $010101010...01010$. Under these conditions each round or some percentage of rounds is using the same key making attacks far easier.
2. Round keys are invertible to full keys - Given a round key $K_r$ we can easily perform the inverse rotation to get the full key $K$. Therefore breaking one round (the last round) breaks all the rounds.
3. Related Keys Attacks - Related keys have related round keys only different by a fixed rotation and therefore they have the same number of 1's, the same distance between these 1's, etc... This is bad.
The number of round keys is related to the number of message blocks - If I read your scheme correctly the each round key is used for a different message block. That is $K_0$ is used for $M_0$, $K_1$ is used for $M_1$, and so on. If you are only encrypting one block, you are only performing one round. This is extremely dangerous because you aren't performing 16 cipher rounds per ciphertext block, but 1 round per ciphertext block meaning that the equations to determine the input bits from the output bits are going to be extremely short and easy to solve. I also doubt that you would get sufficient active S-boxes in that many rounds.
You might enjoy these slides on minimalist cryptography.
For the purposes of inquiry consider the simpler and most secure scheme, though still not actually secure.
$$C_n = S[K \oplus M_n] \oplus K$$
Where $S$ is a 16-bit s-box and the block size is also 16-bits.
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Does the known plaintext not get prevented by the use of multiple rounds? Since it's essentially $C_n = S[S[S[K_n \oplus M_n] \oplus M_{n+1}] \oplus M_{n+2} ... ]]]$ – Polynomial Nov 23 '11 at 15:15
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The above scheme does not use multiple rounds but rather uses only one round with a different round key per message block. A multiple round scheme would have the same the same message block be enciphered by more than one round key. You may have conflated CBC with rounds. I think you should probably abandon the CBC stuff until you have a secure block cipher and then figure out how to use that secure block cipher in a CBC mode. – Ethan Heilman Nov 23 '11 at 15:32
Sorry, must've missed it off my re-write. If you take a look at the edit history, you'll see my original design. It specifies 16 rounds of $C_n = S[K_r \oplus M_n]$ on each block. Would this result in any significant changes in the feasibility of the attacks you described? – Polynomial Nov 23 '11 at 15:36
@Polynomial - Consider the case in which you only have one message block, $C_n = S[K_0 \oplus M_0]$. $S$ is invertible and so we are back to the known plaintext attack. You need the second xor of the key to prevent an attacker from inverting S ($C_n = S[K_0 \oplus M_0] \oplus K_0$) . The approach you should take is, to build a secure block cipher that only processes one message block. If it isn't secure with only one block it is unlikely to be secure with more than one block. – Ethan Heilman Nov 23 '11 at 15:48
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– Ethan Heilman Nov 23 '11 at 16:17
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As I already said in my comment, what you have here is not what one usually calls a block cipher.
A (standard) block cipher is a pair of functions $$E : K \times M \to M \text{ and } D : K \times M \to M$$ This means $E$ takes an element of $K$ and an element of $M$ as input, and gives an element of $M$ as output – same for $D$.
(where $K$ is called the key space and $M$ the "block space") such that for each $y \in K$ and $x \in M$, we have $$D(y, E(y, x)) = x.$$
Normally $K$ and $M$ are finite sets of the form $K = \{0,1\}^k$ and $M = \{0,1\}^m$, where $k$ is called the key size, $m$ is called the block size. (But block ciphers for other block types can be constructed from these, see "format-preserving encryption").
We then evaluate the security of the primitive $(E, D)$ (considering the key secret), before going on to fit the block cipher into a mode of operation (which itself can then have its security evaluation independent of the cipher).
What you are using, is a function pair of the signature $$E, D : K \times I \times M \times M \to M$$ (where I is some index space – in your case $I = \mathbb Z_m = \{0, ..., m-1\}$ is enough, with $m = 256$ being the block size, but we could also use $I = \mathbb N$, the set of all natural numbers).
I.e. both your encryption function $E$ and your decryption function $D$ each take a key, an index, and two message blocks as input, and output a message block.
These functions should have this property: $$D(y, n, c, E(y,n,c, x)) = x \qquad \forall y \in K, n \in I, c \in M, x \in M.$$
Giving the same key, index and "additional block" to encryption and decryption function, the decryption function just gives back the original plaintext provided to the encryption function.
Function pairs such as these are known as tweakable block ciphers, though usually the tweak is smaller than the key and block. (The tweak in your case would be the pair $(n, c)$.) An example for a known tweakable block cipher is the Threefish cipher which is used as a primitive inside of the Skein hash function (one of the SHA-3 candidates).
This is then used in a special mode (not CBC, though it looks similar) where the block number is used as the $n$ input, and the previous ciphertext block as the $c$ input (to both encryption and decryption), in addition to the key $y$ – we could call this "ciphertext-to-tweak chaining" mode.
Now, you have multiple things to do:
• define a suitable security notion for your cipher type.
• make it credible that your concrete cipher has these security properties
• make it credible that your chaining mode, provided the cipher itself is secure, has the necessary properties to use it for message encryption.
This is more complicated than defining a block cipher (in the ordinary sense) and using a standard mode of operation (which already comes with a security proof), and thus less likely to succeed. Also, you will likely get less cryptanalysis this way, as it is a non-standard model.
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Buh? Any chance you could clear up your explanation to avoid use of stuff like $\mathbb{Z}_m$, or at least explain the meaning of them? I'm not versed in such formal notation. I didn't understand a single part of anything you typed in LaTeX. – Polynomial Nov 23 '11 at 17:04
$\mathbb{Z}_m$ is simply the set of integer numbers from $1$ to $m$. In your case, this is the same $m$ as the block size, since after each 256 blocks the effect of the block number $n$ to the cipher is the same as before. I'll try to reword it, though. – Paŭlo Ebermann♦ Nov 23 '11 at 17:09
Same issues for all the stuff like $E : K \times M \to M$. It's completely opaque to me without a description. – Polynomial Nov 23 '11 at 17:16
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Probably a question on notation would be in order but let me try to provide a concise explanation here. $E : K \times M \rightarrow M$ means $E$ is a function that uses $K$ to map values of $M$ to other values of $M$. That is the encryption function $E$ uses a key $K$ to map a message to another message with message-space $M$ where messsage-space is the space of all possible messages. It is a notational definition of encryption. – Ethan Heilman Nov 23 '11 at 17:29
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@Polynomial: I added some (small-font) explanation for the notation to my article. Though in general, if you don't even know this basic mathematical notation, you are nowhere near the knowledge level necessary to design a secure cipher, sorry. – Paŭlo Ebermann♦ Nov 23 '11 at 17:47
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http://unapologetic.wordpress.com/2008/07/17/
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# The Unapologetic Mathematician
## Set — the card game
I’d like to talk about the card game Set. Why? Because the goal is to find affine lines! Huh?
Well, first you have to understand that we’re not working over the usual fields we draw our intuition from. We’re using the field $\mathbb{Z}_3$ of integers modulo $3$. A bit more explicitly, we start with the ring $\mathbb{Z}$ of integers and quotient out by the ideal $3\mathbb{Z}$ of multiples of $3$. This is a maximal ideal, because if we add in any other integer we can subtract off the closest multiple of $3$. Then we’re left with either ${1}$ or $-1$, and this gives us all the integers in our expanded ideal. But we know that the quotient of a ring by a maximal ideal is a field! Thus adding and multiplying integers modulo $3$ gives us a field. For the three elements of $\mathbb{Z}_3$ we’ll use $\{1,2,3\}$ (using $3$ instead of the usual ${0}$).
So what does this have to do with the card game? Well, each one has four features, each picked from three choices:
• Colors: red, green, or purple
• Symbols: squiggles, diamonds, or ovals
• Patterns: solid, striped, or outlined
• Numbers: one, two, or three
Let’s consider the color alone for now. We choose (somewhat arbitrarily) to correlate the colors red, green, and blue with the elements ${1}$, $2$, and $3$ of $\mathbb{Z}_3$, respectively. We don’t want to think of them as “being” the field elements, nor even as the elements of a one-dimensional vector space over $\mathbb{Z}_3$. The arbitrariness of our correspondence points to the fact that the colors constitute an affine line over $\mathbb{Z}_3$!
Similarly, we have lines for symbols, patterns, and numbers. And it’s pretty straightforward to see that we can take the product these affine lines to get a four-dimensional affine space. That is, the cards in Set form an affine four-space over $\mathbb{Z}_3$, and this is the stage on which we play our game.
Now what’s the goal of the game? You have to identify a collection of three cards so that in each of the four characteristics they’re all the same or all different. What I assert is that these conditions identify affine lines in the affine four-space. So what’s an affine line in this context?
Well, first let’s move from the affine four-space of cards to the affine four-space of four-tuples of field elements. We’ll just use the lists of choices above and identify them with ${1}$, $2$, and $3$, respectively, and read them in the same order as above. That is, the four-tuple $\left(3,1,2,2\right)$ would mean the card with two purple, striped squiggles. We can do this because (since they’re torsors) all affine spaces of the same dimension over the same field are isomorphic (but not canonically so!).
Now we can say that an affine line is a map from the standard affine line (given by “one-tuples” of field elements) to this affine four-space. But it can’t be just any function. It has to preserve relative differences! Since we have only three points to consider, we can reduce this question somewhat: an affine line is a list of three four-tuples of field elements, and the difference from the first to the second must equal the difference from the second to the third.
So we can take any two points, and then we can work out what the third one has to be from there. And we can tell that each of the four components works independently from all the others. So let’s look at colors alone again. Let’s say that the first two cards have colors $c_1$ and $c_2$. Then the affine condition says that $c_3-c_2=c_2-c_1$. That is, $c_3=2c_2-c_1$. It’s straightforward from here to see that if $c_1=c_2$, then $c_3$ is again the same as those two. On the other hand, if $c_1$ and $c_2$ are different, then $c_3$ must be the remaining choice.
What have we seen here? If we start with two cards with the same color, the third card on the affine line will have the same color as well. If the first two cards have different colors, the third one will be the remaining color. And the same analysis applies to symbols, patterns, and numbers. Thus affine lines are sets of three cards which are all the same or all different in each of the four characteristics.
Now that we know that, what can we do with it? Well, look at the gameplay. We don’t have all of the cards spread out at once: we deal out a certain number at a time. So the question is: how many points in the affine space $\mathbb{Z}_3^4$ can we take without them containing an affine line? This is analogous to the problem of placing eight queens on a chessboard. A more advanced problem is to give the expected number of affine lines in a subset of size $n$.
Posted by John Armstrong | Algebra, Linear Algebra | 18 Comments
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathoverflow.net/revisions/44988/list
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## Return to Answer
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The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1)$$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is nothing more than just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended for by $0$ to $t\leq 0$) which belongs belonging to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)\cap L^2(\mathbb R)$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\ \\ 0, & t\leq 0 \end{cases}$$
One can show also that the Parseval identity holds
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt,$$ so there is a complete analogy with the standard Fourier transform.
Executive summary. A function $f$ is a the Laplace transform of some function $\phi$ satisfying condition (1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space$H^2(\sigma>c)$ on a (right) half-plane.
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The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1)$$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is nothing more than the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended for $t\leq 0$) which belongs to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\ \\ 0, & t\leq 0 \end{cases}$$
Summary
One can show also that the Parseval identity holds
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt,$$ so there is a complete analogy with the standard Fourier transform.
Executive summary. A function $f$ is a Laplace transform of some function $\phi$ satisfying condition (1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space $H^2(\sigma>c)$ on a (right) half-plane.
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The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1)$$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is nothing more than the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended for $t\leq 0$) which belongs to $L^2$ L^2(\mathbb R)$for$\sigma=c$and to$L^1$L^1(\mathbb R)$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\ \\ 0, & t\leq 0 \end{cases}$$
Summary. A function $f$ is a Laplace transform of some function $\phi$ satisfying condition (1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space $H^2(\sigma>c)$ on a (right) half-plane.
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The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1)$$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is nothing more than the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ which belongs to $L^2$ for $\sigma=c$ and to $L^1$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\ \\ 0, & t\leq 0 \end{cases}$$
Summary. A function $f$ is a Laplace transform of some function $\phi$ which satisfies satisfying condition (1) 1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space $H^2(\sigma>c)$ on a (right) half-plane.
1
The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1)$$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is nothing more than the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ which belongs to $L^2$ for $\sigma=c$ and to $L^1$ for $\sigma>c$. Taking the inverse Fourier transform, we get that $$e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$ and $$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$ or, equivalently, $$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\ \\ 0, & t\leq 0 \end{cases}$$
Summary. A function $f$ is a Laplace transform of some function $\phi$ which satisfies (1) if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space $H^2(\sigma>c)$ on a (right) half-plane.
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http://physics.stackexchange.com/questions/23668/how-do-i-calculate-output-voltage-from-a-given-supply-voltage-and-resistance
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# How do I calculate output voltage from a given supply voltage and resistance?
I have searched everywhere but I can't seem to find out how I can calculate the output voltage (or voltage over a component) from the resistance of a resistor (or other component) and the supply voltage. Everything seemed to be about Ohm's Law, which looks like it could help but I don't know how.
In the case I need to solve, there is a supply of 5v and I need to get the resistance needed to lower the voltage for an LED (the current and resistance at optimum voltage I can get if needed) to 3v. I will definitely need to do this operation again so if you could give something like an equation, rather than the necessary resistance.
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Something to do with the internal resistance of the power supply? – leongz Apr 12 '12 at 17:20
## 2 Answers
You need to know the resistance of the LED: call this $R_{LED}$ and call the resistor $R$. The current flowing out of the power supply is just the voltage (5 volts) divided by total resistance, $I = V/R$, so:
$$I = \frac {5}{R + R_{LED}}$$
The voltage drop across your resistor, $R$, is just $V = IR$ and you want this to be two volts to leave a three volt drop across the LED, so:
$$2 = IR = \frac {5R}{R + R_{LED}}$$
and a quick rearrangement gives:
$$R = \frac{2}{3}R_{LED}$$
More generally, suppose the power supply voltage is $V$ and the voltage drop across the resistor is $V_R$, then you get:
$$V_R = \frac {VR}{R + R_{LED}}$$
so:
$$R = \frac{V_R}{V - V_R} R_{LED}$$
But as user1631 says, this assumes the LED can be treated as a simple resistor, and in practice this isn't true. A quick Google found http://www.oksolar.com/led/led_color_chart.htm, which gives some graphs of current against voltage for LEDs. In fact the same Google found http://en.wikipedia.org/wiki/LED_circuit, which describes exactly the problem you're asking about.
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So basically, resistance = Vtarget/Vdrop x Rled? Or am I getting this the wrong way? – Prehistoricman Apr 12 '12 at 22:23
I've updated my answer with a few more details – John Rennie Apr 13 '12 at 5:47
Thanks, I realised that I got the Vtarget and Vdrop the wrong way around. – Prehistoricman Apr 13 '12 at 10:46
LED's are not like your standard circuit elements. Voltage current relation is highly nonlinear, so you don't generally speak the LED having a unique resistance.
The voltage vs current relation is almost vertical at the operating voltage, i.e. you will have a huge range of currents with voltage very close to say 3V. The resistance of the LED can be almost anything, depending on current. Choosing the resistor to make the voltage across the LED is getting it backwards. The LED will adjust its own internal resistance to keep its voltage drop near 3V (within some range), the purpose of the external resistor is to limit the current through the LED so that it does not damage itself. You start with the current Imax you want to limit to, assume 3V accross the LED, solve 5V = 3V + R*Imax.
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The current for the LED is 0.2A@3v. With both of your methods, I get ~10. That sounds a bit low. Is it right though? – Prehistoricman Apr 12 '12 at 22:18
10 ohms sounds good to me. Why do you think it's low? – user1631 Apr 12 '12 at 22:54
Not sure. I've done a few more calculations and they all sound right. I just hope they work in practice. I am trying to replace some green/red LEDs with some blue/red LEDs but when I got the new ones, I found out their required voltage was higher than the power supplied so I am using the 5v of the system to get a bit more power to the lights. I need a 68 ohm resistor for the blue light and 165 for the red. Thanks guys, this was really useful. – Prehistoricman Apr 13 '12 at 10:42
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# Tagged Questions
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### Laplace Transform Help
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http://physics.stackexchange.com/questions/tagged/thermodynamics?page=2&sort=votes&pagesize=30
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http://mathoverflow.net/questions/75749/virtual-dimension-of-moduli-of-stable-maps
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Virtual dimension of moduli of stable maps
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I apologize if this question was already asked somwhere else on this website. Let us consider $f:C \to X$ a stable map. This is a point in the moduli space of stable maps. It seems intuitively to me that any holomorphic global section of $f^*(T_X)$ gives a deformation of the stable map. It is well-known, but not clear to me, why the moduli space of stable maps in a neighbourood of this point looks like a closed subvariety in $H^0(C, f^*T_X)$ cut out by $h^1(C, f^*T_X)$ equations. Where do this equations come from? Is there a concrete way to see them? (e.g. some sections of $f^*(T_X)$ give forbidden deformations, or the same deformation,...).
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This is a consequence of the deformation theory of morphisms with smooth target: deformations lie in $H^0(f^{\ast}T_X)$ and obstructions lie in $H^1(f^{\ast}T_X)$. This is explicitly covered, for example, in Hartshorne's notes on deformation theory (go to the section on deformations of a morphism). – Mike Skirvin Sep 18 2011 at 14:44
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-1 for the choice of nickname! (just kidding. But I think with such a nickname you'll be perceived as an administrator or something by the new users..) – Qfwfq Sep 18 2011 at 15:09
we have a map from deformation space $H^0$ to obstruction space $H^1$ such that its zero locus corresponds to actual deformations. In symplectic geometry, this map is given by Cauchy-Riemann equation. – Mohammad F.Tehrani Sep 18 2011 at 15:14
1 Answer
If we are looking at the space of stable maps, we need to allow $C$ to deform as well. Let $C$ be a curve with the set of marked points $P$. There is a following exact sequence: \begin{eqnarray*} 0 &\to& \text{Aut}(C,P) \to H^0(C, f^*(T_X)) \\ \to \text{Def}(C,P,f) &\to& \text{Def}(C,P) \to H^1(C,f^*T_X)) \\ \to \text{Ob}(C,P,f) &\to& 0 \end{eqnarray*} Thus the virtual dimension is \begin{eqnarray*} \text{virdim} &=& \dim \text{Def} (C,P,f) - \dim \text{Ob}(C,P,f) \\ &=& \dim \text{Def}(C,P) - \dim \text{Aut}(C,P) +h^0(f^*T_X) - h^1(f^*T_X) \end{eqnarray*}
As to why we need to substract $h^1$ from the dimension. Let $(O,m)$ be the local ring of the map $f$ at the moduli space. If the space is smooth at $f$ then the dimension would be $h^0$. But this is not often the case. Generally $h^0$ would be the dimension of Zariski tangent space ( $= \dim_k m/m^2$). If $O$ is not regular, one way to measure its irregularity is by using infinitesimal lifting property: given a map $\phi: O \to A$, can we lift it to an infinitesimal extension $A':$ $0\to J \to A' \to A \to 0$. This is the same as extending a deformation of $f$ over $A$ to a higher order. Such a lift is obstructed by an element $\alpha \in H^1(f^*T_X)\otimes J$ (local to global).
Let $O = (R,m_P)/I$ such that $R$ is regular having same Zariski tangent space. Then $\dim O$ is roughly $\dim R - \text{rank}(I) = \dim R - \dim I/m_pI$. Now $I/mP_I$ is the canonical obstruction for $O$ and it can be embedded into $H^1(C,f^*T_X)$, thus the dimension of $O$ is roughly $\dim R - \dim I/m_pI \approx h^0 -h^1$
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http://en.wikibooks.org/wiki/A-level_Chemistry/AQA/Module_5/Thermodynamics/Thermodynamic_Temperature
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# A-level Chemistry/AQA/Module 5/Thermodynamics/Thermodynamic Temperature
## What is Temperature?
There are a few ways we can describe temperature. But a large amount of them are surprisingly poor. Sadi Carnot for instance imagined heat as a liquid, and it was this view that allowed him to develop heat engine theory.
Here we will develop a way of viewing temperature that makes it useful in a Thermodynamic point of View.
## Two systems in Thermal Contact
Imagine two systems, 1 and 2, with total energy E = E1 + E2 and Entropy S = S1 + S2.
Conservation of energy of the system means that E2 = E - E1 always.
Differentiating by E1 we see that:
$\begin{matrix} \frac{d E_2}{d E_1} &=& \frac{d(E - E_1)}{dE_1} \\ &=& -\frac{d E_1}{dE_1} \\ &=& -1 \end{matrix}$
Now if we look at Entropy changing with Energy.
$\frac{\partial S}{\partial E_1} = \frac{d S_1}{d E_1} + \frac{d S_2}{d E_1} = \frac{d S_1}{d E_1} + \frac{d S_2}{d E_2}\frac{d E_2}{d E_1} = \frac{d S_1}{d E_1} - \frac{d S_2}{d E_2}$
At equilibrium there is no change in entropy, so when the systems are at the same temperature
$\frac{\partial S}{\partial E_1} = \frac{d S_1}{d E_1} - \frac{d S_2}{d E_2} = 0$
We use this to define Temperature.
$T^{-1} = \frac{\partial S}{\partial E}_{N,V}$
N and V means the Volume and Number of Particles is fixed.
Or for a specific system
$T_i^{-1} = \frac{\partial S_i}{\partial E_i}_{N_i,V_i}$
## Usage
If we were to look at how entropy changed with temperature, it is easy to show that as entropy always increases, energy always flows from the hotter body to the cooler one. Try it.
### Authors
--Frontier 11:51, 23 Apr 2005 (UTC)
#### Note
One of the equations wasn't parsing correctly at the time of writing. I've left it as is because the error message was 'Unknown Function \begin' which should be recognised. I'm not really sure what the problem is.
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http://mathoverflow.net/revisions/30234/list
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## Return to Answer
2 added 215 characters in body
If your compact Riemann surfaces $M$ and $N$ have a hyperbolic metric in which the boundary curves are totally geodesic of the same length, then this follows from the Fenchel-Nielsen coordinate parametrization of Teichmuller space. Any compact Riemann surface with boundary has a uniformization to a hyperbolic surface with totally geodesic boundary (one may see this by doubling, and applying the uniformization theorem). If you perform this uniformization, and then compare the two sides, your identification between boundary components may differ from the parametrization given by Fenchel-Nielsen coordinates (for example, the lengths of the boundary components may differ), and so it's more complicated to see how the conformal structure changes. In fact, if $M$ or $N$ is a disk or annulus, then the conformal structure may not be changing at all. If $M$ and $N$ both have Euler characteristic less than zero, then large twists should change the conformal structure. For example, a twist by $2\pi$ will change the Riemann surface by Dehn twist, and therefore gives a different conformal structure (up to isotopy).
There's also a slight issue of moduli, in that a rotation by $2\pi$ changes the point in Teichmuller space, but not in moduli space. I'm implicitly assuming you're asking for the Teichmuller parameter. If you're asking for the parameter in moduli space, then the twist will take you along a non-trivial closed loop in moduli space, so at least there will be uncountably many different conformal moduli as you twist.
1
If your compact Riemann surfaces $M$ and $N$ have a hyperbolic metric in which the boundary curves are totally geodesic of the same length, then this follows from the Fenchel-Nielsen coordinate parametrization of Teichmuller space. Any compact Riemann surface with boundary has a uniformization to a hyperbolic surface with totally geodesic boundary (one may see this by doubling, and applying the uniformization theorem). If you perform this uniformization, and then compare the two sides, your identification between boundary components may differ from the parametrization given by Fenchel-Nielsen coordinates (for example, the lengths of the boundary components may differ), and so it's more complicated to see how the conformal structure changes. In fact, if $M$ or $N$ is a disk or annulus, then the conformal structure may not be changing at all. If $M$ and $N$ both have Euler characteristic less than zero, then large twists should change the conformal structure. For example, a twist by $2\pi$ will change the Riemann surface by Dehn twist, and therefore gives a different conformal structure (up to isotopy).
There's also a slight issue of moduli, in that a rotation by $2\pi$ changes the point in Teichmuller space, but not in moduli space. I'm implicitly assuming you're asking for the Teichmuller parameter.
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http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/102653
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Not especially famous, long-open problems which anyone can understand
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Question: I'm asking for a big list of not especially famous, long open problems that anyone can understand. Community wiki, so one problem per answer, please.
Motivation: I plan to use this list in my teaching, to motivate general education undergraduates, and early year majors, suggesting to them an idea of what research mathematicians do.
Meaning of "not too famous" Examples of problems that are too famous might be the Goldbach conjecture, the $3x+1$-problem, the twin-prime conjecture, or the chromatic number of the unit-distance graph on ${\Bbb R}^2$. Roughly, if there exists a whole monograph already dedicated to the problem (or narrow circle of problems), no need to mention it again here. I'm looking for problems that, with high probability, a mathematician working outside the particular area has never encountered.
Meaning of: anyone can understand The statement (in some appropriate, but reasonably terse formulation) shouldn't involve concepts beyond (American) K-12 mathematics. For example, if it weren't already too famous, I would say that the conjecture that "finite projective planes have prime power order" does have barely acceptable articulations.
Meaning of: long open The problem should occur in the literature or have a solid history as folklore. So I do not mean to call here for the invention of new problems or to collect everybody's laundry list of private-research-impeding unproved elementary technical lemmas. There should already exist at least of small community of mathematicians who will care if one of these problems gets solved.
I hope I have reduced subjectivity to a minimum, but I can't eliminate all fuzziness -- so if in doubt please don't hesitate to post!
To get started, here's a problem that I only learned of recently and that I've actually enjoyed describing to general education students.
http://en.wikipedia.org/wiki/Union-closed_sets_conjecture
Edit: I'm primarily interested in conjectures - yes-no questions, rather than classification problems, quests for algorithms, etc.
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You might get more success if you sampled certain open problem lists and indicated which ones fit your list and which ones did not. I could mention various combinatorial problems such as integer complexity, determinant spectrum, covering design optimization, but I can't tell from your description if they would be suitable for you. Gerhard "They Are Suitable For Me" Paseman, 2012.06.21 – Gerhard Paseman Jun 21 at 19:11
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Here is some collection of some other "collect open problems" quests. on MO: mathoverflow.net/questions/96202/… PS Nice question ! PSPS may be add tag "open-problems" – Alexander Chervov Jun 21 at 20:53
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Nice question!! – S. Sra Jun 22 at 3:25
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To save the search for explanation of cryptic acronyms for those of us outside US, K-12 means high school. @Mahmud: You are using a wrong meaning of the word “problem”. The TSP is not an unproved mathematical statement, it is a computational task. – Emil Jeřábek Jun 22 at 12:05
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More precisely, K-12 means anything up to high school (K = Kindergarten, 12 = 12th grade, and K-12 covers this range). – Henry Cohn Jun 22 at 13:05
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73 Answers
The following problem is very well-known among algebraic geometers:
````Does there exist a cubic 4-fold that is not rational?
````
It's probably not well-known outside of algebraic geometry, even though it can easily be explained in every elementary terms:
Does there exist a polynomial equation $F$ of degree three in five variables with the following property: Let $X \subset \mathbb C^5$ be the solution set of $F = 0$. Then there exists no chart $U \subset \mathbb C^4, \phi \colon U \to X$ such that $\phi$ is defined by rational functions (i.e., quotients of polynomials).
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Cool ! what are the references for current state of art ? – Alexander Chervov Jul 24 at 9:57
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You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
What is the least $S$ (if any) such that any subset of a plane of area $S$ contains $3$ vertices of a triangle of unit area?
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How many trees are there?
Let $T(n)$ be the number of trees on $n$ vertices up to graph isomorphism. There is no known closed formula for $T(n)$.
In 1947 Richard Otter proved[Source] the asymptotic result $$T(n) \sim A \cdot B^n \cdot n^{-\frac{5}{2}}$$ where $A \approx 0.535$ & $B \approx 2.996$.
By way of contrast, let $L(n)$ be the number of labelled trees, i.e. trees formed from vertices labelled $1,...,n$ where isomorphism additionally preseves the label. In 1889, Arthur Cayley showed[Source] that $$L(n)=n^{n-2}$$
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• Is the Ring of Periods actually a field? (most likely, no)
• Is the equality of periods decidable? (hopefully, yes)
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problems which anyone can understand ? Uhhh – Denis Serre Sep 25 at 7:50
Let $R(x)=P(x)/Q(x)$ where $P(x)$ and $Q(x)$ are polynomials with integer coefficients and $Q(0)\neq 0$. Is there an algorithm that given $P(x)$ and $Q(x)$ as an input always halts and decides if the Taylor series of $R(x)$ at $x=0$ has a coefficient $0$?
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Waring's problem inequality
One of the oldest (Since 1770) and famous open problem in number theory is Waring's problem. It has been conjectured that if
$$Frac\bigg[\bigg(\frac{3}{2}\bigg)^n\bigg] \le 1 - \bigg[\bigg(\frac{3}{4}\bigg)^n\bigg].$$
(where $Frac$ denotes the fractional part) true then, the general solution of Waring's problem is
$$g(n) = 2^n + Int\bigg[\bigg(\frac{3}{2}\bigg)^n\bigg] - 2.$$
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I think you could give an accessible K-12 formulation of the definition of a group (as a group of permutations, for instance) and of an integral group ring. The Zero Divisor Conjecture (Kaplansky, 1940) then states, in one version, that if $G$ is a torsion-free group then the group ring $\mathbb{Z}[G]$ has no zero divisors besides the number $0$.
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>I think you could give an accessible K-12 formulation of the definition of a group ... In any case this is perfect for my follow-up question which the net gods decided to close for the time being. – David Feldman Jul 13 at 5:36
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@David just wanted to write the same :) mathoverflow.net/questions/101169/… – Alexander Chervov Jul 13 at 6:18
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http://en.wikipedia.org/wiki/Keller%27s_conjecture
From Wikipedia:
Keller's conjecture is the conjecture introduced by Ott-Heinrich Keller (1930) that in any tiling of Euclidean space by identical hypercubes there are two cubes that meet face to face.
Keller's original cube-tiling conjecture remains open in dimension 7.
Conjecture was shown to be true in dimensions at most 6 by Perron (1940a, 1940b). However, for higher dimensions it is false, as was shown in dimensions at least 10 by Lagarias and Shor (1992) and in dimensions at least 8 by Mackey (2002), using a reformulation of the problem in terms of the clique number of certain graphs now known as Keller graphs. Although this graph-theoretic version of the conjecture is now resolved for all dimensions, Keller's original cube-tiling conjecture remains open in dimension 7.
The related Minkowski lattice cube-tiling conjecture states that, whenever a tiling of space by identical cubes has the additional property that the cube centers form a lattice, some cubes must meet face to face. It was proved by György Hajós in 1942.
Szabó (1993), Shor (2004), and Zong (2005) give surveys of work on Keller's conjecture and related problems.
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The following conjecture by Carsten Thomassen:
If $G$ is a 3-connected graph, every longest cycle in $G$ has a chord.
Thomassen has proven the conjecture true for 3-connected cubic graphs.
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Is there a triangle that can be cut into $7$ congruent triangles? (no)
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Nice problems but what are the sources? – Alexander Chervov Jul 24 at 20:23
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I heard this in a personal communication. But it turns out this is already settled negatively in 2008: michaelbeeson.com/research/papers/… – Vladimir Reshetnikov Jul 28 at 20:34
More than ten years ago I posed the following problem in a couple of math-related mailing lists:
Let $G_n$ be the graph with vertex set `$\{1, 2, \dots, 2n\}$` such that `$\{i,j\}$` is an edge if and only if $i+j$ is a prime number. Is it true that $G_n$ is eulerian for every $n \geq 2$?
It is a simple consequence of Bertrand's Postulate (there is always a prime between $k$ and $2k$) that $G_n$ is connected and has a perfect matching for every $n$.
The problem turned out to be an old one. I believe that some variation of it appears in Richard K. Guy's "Unsolved Problems in Number Theory" and according to this article, it was originally posed in the Journal of Recreational Mathematics in 1982.
Michael A. Jones and Leslie Cheteyan, "Two observation on unsolved problem #1046 on prime circles of `$\{1, 2, . . . , 2m\}$`", J. Recreational Mathematics Vol.35(1) (2006), 15--19.
The whole issue can be downloaded here:
http://www.baywood.com/comppdf/0022-412x.pdf
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N. M. Katz: "Simple Things we don't know": https://web.math.princeton.edu/~nmk/pisa16.pdf
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Ore's odd Harmonic number conjecture
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This is a notorious problem. The OP asked for “not too famous”. – Emil Jeřábek Jun 27 at 14:18
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http://mathoverflow.net/revisions/96232/list
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## Return to Answer
5 Correct typo
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse.
The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
More details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$". It is possible to "eliminate" this these bigons one by one, leaving a sphere with only 3 faces as desired.
Remark.
Alternatively to combinatorial complexes, you can think in terms of pictures in the sense of Rourke, as in
1. Colin P. Rourke, Presentations and the trivial group, Topology of low-dimensional manifolds (Berlin), Lecture Notes in Math., vol. 722, Springer, 1979, Proc. Second Sussex Conf., Chelwood Gate, 1977, pp. 134–143.
2. Johannes Huebschmann, Aspherical 2-complexes and an unsettled problem of J. H. C. Whitehead, Math. Ann. 258 (1981), 17–37.
4 Format
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse.
The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
More details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$". It is possible to "eliminate" this bigons one by one, leaving a sphere with only 3 faces as desired.
Remark.
Alternatively to combinatorial complexes, you can think in terms of pictures in the sense of Rourke, as in
1. Colin P. Rourke, Presentations and the trivial group, Topology of low-dimensional manifolds (Berlin), Lecture Notes in Math., vol. 722, Springer, 1979, Proc. Second Sussex Conf., Chelwood Gate, 1977, pp. 134–143.
2. Johannes Huebschmann, Aspherical 2-complexes and an unsettled problem of J. H. C. Whitehead, Math. Ann. 258 (1981), 17–37.
3 Format
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse.
The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
UPDATE: more
More details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$".
It is possible to "eliminate" this bigons one by one, leaving a sphere with only 3 faces as desired.
Remark.
Alternatively to combinatorial complexes, you can think in terms of pictures in the sense of Rourke, as in
1. Colin P. Rourke, Presentations and the trivial group, Topology of low-dimensional manifolds (Berlin), Lecture Notes in Math., vol. 722, Springer, 1979, Proc. Second Sussex Conf., Chelwood Gate, 1977, pp. 134–143.
2. Johannes Huebschmann, Aspherical 2-complexes and an unsettled problem of J. H. C. Whitehead, Math. Ann. 258 (1981), 17–37.
2 add a more detailed explanation
UPDATE: more details.
If such a "sphere" (more properly it would be called a "combinatorial sphere") glued of 3 faces exists, it is more or less clear that $x$ is the product of a conjugate of $y$ and a conjugate of $z$.
Suppose now $x=gyg^{-1}hzh^{-1}$. Then it is possible to use 3 faces with contour labels $x$, $y$, and $z$, and additional "degenerate" faces with contour labels "$a1a^{-1}1$", "$b1b^{-1}1$", "$a1a^{-1}$", "$b1b^{-1}$", "$a^{-1}1a$", "$b^{-1}1b$", "$111$" to glue a "big" sphere as above, except with additional "degenerate" faces.
Then it is possible to "collapse" all edges labelled with "1". The sphere will "fall apart" into several spheres. Assuming that none of the words $x$, $y$, $z$ is trivial, none of the 3 "essential" faces can be left on its own, so all 3 will be in the same smaller sphere. Other faces in this smaller sphere will be digons with contour labels "$aa^{-1}$" and "$bb^{-1}$".
It is possible to "eliminate" this bigons one by one, leaving a sphere with only 3 faces as desired.
1
Andreas, I believe so, and for free groups with more than 2 generators too. As well as the analogous question for more than 3 elements: for example, if $w=fxf^{-1}gyg^{-1}hzh^{-1}$.
To see if $x=gyg^{-1}hzh^{-1}$, it is enough to study first the cases $x=1$, $y=1$, $z=1$, and if none of these elements is trivial, to see if it is possible to "glue" a sphere out of 3 discs labelled on the boundary with $x$, $y$, and $z$, respecting the labels, and so that the orientations of the discs labelled with $y$ and $z$ were the same, and the orientation of the disc labelled with $x$ was the inverse. The length of the $1$-skeleton of such a sphere will be $(|x|+|y|+|z|)/2$, so there are finitely many cases to consider.
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http://math.stackexchange.com/questions/66677/heine-borel-finite-intersection-equivalence-in-subspaces
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# Heine-Borel - Finite Intersection equivalence in subspaces
I'm doing a course in Calculus/Real Analysis, and we're covering really basic topological ideas (metric spaces, open/closed sets etc.). From my notes it looks like we proved the following in class (Heine-Borel):
(1) A set K is compact iff (2) for every open cover of K there is a finite subcover.
And then we claimed without proof that the right hand side is clearly equivalent to the following property: (3) for every set of closed subsets from K that have the finite intersection property, the intersection of said sets is non-empty.
It looks pretty easy to prove this equivalence when K is an entire space. Like this proof here. But from my notes it appears to be a theorem that holds true with any subset of the metric space.
My question is how that proof holds up when K (as mentioned in class) is not the whole space but a subset of the space. The proof relies on the complement of the sets, which means that given an open cover, the complement of the sets are closed but are still in the space (obviously, since the complement is with respect to the space). However when we take the complement w.r.t. the space and K is no longer the whole space, the sets we have are no longer from K necessarily. It also appears of no help to intersect them with K, since we cannot assume K is closed (that's part of what we're trying to prove in the sense that compactness is closed+bounded).
The difference seems petty but it's bugging me to no end.
Apologies about the wordiness, my latex skills are about as poor as my topology skills. Thanks.
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1
Note that (2) is the usual definition of compactness. I don’t know what definition your course is using, but it may be something that’s actually not equivalent to compactness in all topological spaces. – Brian M. Scott Sep 22 '11 at 18:44
@BrianM.Scott, we defined compactness of K as: every sequence in K has a subsequence that converges to an element in K, which I think is pretty general. – davin Sep 23 '11 at 0:27
@davin: It is actually a property called sequential compactness, which in metric spaces is equivalent to compactness. – Asaf Karagila Sep 23 '11 at 6:38
## 1 Answer
Every set with a topology is a space.
In particular, if $X$ is any space (metric, in this case) and $K\subseteq X$ is a compact subset of $X$ then $K$ is a space equipped with the relative topology.
$U\subseteq K$ is open if and only if there exists some $V\subseteq X$ which is open in the original topology and $K\cap V=U$.
Now if $K$ is a compact subset of $X$ then $K$ forms a compact space with the relative topology. This is since every cover of $K$ by open sets can be extended to an open cover in $X$, where we know $K$ is compact.
From here we can just assume that $K$ is now the entire space, and prove as in the link.
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I didn't really get your answer at first, because it sounds like you're saying "if we know K is compact then we can consider K a space", which only works to prove one direction of the theorem (still better than what I had). And after some thought I realised that that direction is the only problematic one. Thanks. – davin Sep 23 '11 at 0:33
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http://unapologetic.wordpress.com/2010/03/19/measures/?like=1&source=post_flair&_wpnonce=c1ce40f6e1
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# The Unapologetic Mathematician
## Measures
From this point in, I will define a “set function” as a function $\mu$ whose domain is some collection of subsets $\mathcal{E}\subseteq P(X)$. It’s important to note here that $\mu$ is not defined on points of the set $X$, but on subsets of $X$. For some reason, a lot of people find that confusing at first.
We’re primarily concerned with set functions which take their values in the “extended real numbers” $\overline{\mathbb{R}}$. That is, the value of $\mu(E)$ is either a real number, or $+\infty$, or $-\infty$, with the latter two being greater than all real numbers and less than all real numbers, respectively.
We say that such a set function $\mu:\mathcal{E}\to\overline{\mathbb{R}}$ is “additive” if whenever we have disjoint sets $E_1$ and $E_2$ in $\mathcal{E}$ with disjoint union $E_1\uplus E_2$ also in $\mathcal{E}$, then we have
$\displaystyle\mu(E_1\uplus E_2)=\mu(E_1)+\mu(E_2)$
Similarly, we say that $\mu$ is finitely additive if for every finite, pairwise disjoint collection $\{E_1,\dots,E_2\}\subseteq\mathcal{E}$ whose union is also in $\mathcal{E}$ we have
$\displaystyle\mu\left(\biguplus\limits_{i=1}^nE_i\right)=\sum\limits_{i=1}^n\mu(E_i)$
And we say that $\mu$ is countably additive of for every pairwise-disjoint sequence $\{E_i\}_{i=1}^\infty$ of sets in $\mathcal{E}$ whose union is also in $\mathcal{E}$, we have
$\displaystyle\mu\left(\biguplus\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu(E_i)$
Now we can define a “measure” as an extended real-valued, non-negative, countably additive set function $\mu$ defined on an algebra $\mathcal{A}$, and satisfying $\mu(\emptyset)=0$. With this last assumption, we can show that a measure is also finitely additive. Indeed, given a collection $\{E_1,\dots,E_n\}$, just define $E_i=\emptyset$ for $i>n$ to get a sequence. Then we find
$\displaystyle\mu\left(\biguplus\limits_{i=1}^nE_i\right)=\mu\left(\biguplus\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu(E_i)=\sum\limits_{i=1}^n\mu(E_i)$
If $\mu$ is a measure on $\mathcal{A}$, we say a set $A\in\mathcal{A}$ has finite measure if $\mu(A)<\infty$. We say that $A$ has “$\sigma$-finite” measure if there is a sequence of sets $\{A_i\}_{i=1}^\infty$ of finite measure ($\mu(A_i)<\infty$) so that $A\subseteq\bigcup_{i=1}^\infty A_i$. If every set in $A$ has finite (or $\sigma$-finite) measure, we say that $\mu$ is finite (or $\sigma$-finite) on $A$.
Finally, we say that a measure is “complete” if for every set $A$ of measure zero, $\mathcal{A}$ also contains all subsets of $A$. That is, if $A\in\mathcal{A}$, $\mu(A)=0$, and $B\subseteq A$, then $B\in\mathcal{A}$. At first, this might seem to be more a condition on the algebra $\mathcal{A}$ than on the measure $\mu$, but it really isn’t. It says that to be complete, a measure can only assign ${0}$ to a set if all of its subsets are also in $\mathcal{A}$.
### Like this:
Posted by John Armstrong | Analysis, Measure Theory
## 17 Comments »
1. This stuff is SO important, and so rarely done well. I’m confident that you can make it delightful.
Comment by | March 20, 2010 | Reply
2. thanks for these excellent posts, i been trying to teach my self measure theory i’m finding these posts very useful.
looking forward to the rest
Comment by learningToMeasure | March 20, 2010 | Reply
3. [...] I say that any measure on an algebra is both monotone and subtractive. Indeed, since is an algebra, then is guaranteed [...]
Pingback by | March 22, 2010 | Reply
4. [...] I say that a measure is continuous from above and [...]
Pingback by | March 23, 2010 | Reply
5. [...] as Metric It turns out that a measure turns its domain into a sort of metric space, measuring the “distance” between two [...]
Pingback by | March 24, 2010 | Reply
6. [...] We’re going to want a modification of the notion of a measure. But before we introduce it, we have (of course) a few [...]
Pingback by | March 25, 2010 | Reply
7. [...] a Measure to an Outer Measure Let be a measure in a ring (not necessarily an algebra) , and let be the hereditary -ring generated by . For every [...]
Pingback by | March 26, 2010 | Reply
8. [...] measure on a hereditary -ring is nice and all, but it’s not what we really want, which is a measure. In particular, it’s subadditive rather than additive. We want to fix this by restricting to [...]
Pingback by | March 29, 2010 | Reply
9. [...] additive. That is, if we define for , then is actually a measure. Even better, it’s a complete [...]
Pingback by | March 30, 2010 | Reply
10. [...] and Induced Measures If we start with a measure on a ring , we can extend it to an outer measure on the hereditary -ring . And then we can [...]
Pingback by | March 31, 2010 | Reply
11. [...] of a Measure At last we can show that the set function we defined on semiclosed intervals is a measure. It’s clearly real-valued and non-negative. We already showed that it’s monotonic, and [...]
Pingback by | April 16, 2010 | Reply
12. [...] We’ve spent a fair amount of time discussing rings and -rings of sets, and measures as functions on such collections. Now we start considering how these sorts of constructions relate [...]
Pingback by | April 26, 2010 | Reply
13. [...] we don’t particularly care if the set where is false is itself measurable, although if is complete then all -negligible sets will be measurable. This sort of language is so common in measure theory [...]
Pingback by | May 13, 2010 | Reply
14. [...] If is a.e. non-negative, then will also be non-negative, and so the indefinite integral is a measure. Since is integrable we see [...]
Pingback by | June 21, 2010 | Reply
15. [...] We continue what we started yesterday by extending the notion of a measure. We want something that captures the indefinite integrals of every function for which it’s [...]
Pingback by | June 22, 2010 | Reply
16. [...] I say that each of these set functions — , , and — is a measure, and that . If is (totally) finite or -finite, then so are and , and at least one of them will [...]
Pingback by | June 25, 2010 | Reply
17. [...] real-valued functions on them. Given a function on a Boolean ring , we say that is additive, or a measure, -finite (on -rings), and so on analogously to the same concepts for set functions. We also say [...]
Pingback by | August 5, 2010 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://en.m.wikipedia.org/wiki/Character_group
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# Character group
In mathematics, a character group is the group of representations of a group by complex-valued functions. These functions can be thought of as one-dimensional matrix representations and so are special cases of the group characters which arises in the related context of character theory. Whenever a group is represented by matrices, the function defined by the trace of the matrices is called a character; however, these traces do not in general form a group. Some important properties of these one-dimensional characters apply to characters in general:
• Characters are invariant on conjugacy classes.
• The characters of irreducible representations are orthogonal.
The primary importance of the character group for finite abelian groups is in number theory, where it is used to construct Dirichlet characters. The character group of the cyclic group also appears in the theory of the discrete Fourier transform. For locally compact abelian groups, the character group (with an assumption of continuity) is central to Fourier analysis.
## Preliminaries
Let G be an abelian group. A function $f:G\rightarrow \mathbb{C}\backslash\{0\}$ mapping the group to the non-zero complex numbers is called a character of G if it is a group homomorphism—that is, if $\forall g_1,g_2 \in G\;\; f(g_1 g_2)=f(g_1)f(g_2)$.
If f is a character of a finite group G, then each function value f(g) is a root of unity (since $\forall g \in G \;\; \exists k \in \mathbb{N}$ such that $g^{k}=e$, $f(g)^{k}=f(g^{k})=f(e)=1$).
Each character f is a constant on conjugacy classes of G, that is, f(h g h−1) = f(g). For this reason, the character is sometimes called the class function.
A finite abelian group of order n has exactly n distinct characters. These are denoted by f1, ..., fn. The function f1 is the trivial representation; that is, $\forall g \in G\;\; f_1(g)=1$. It is called the principal character of G; the others are called the non-principal characters. The non-principal characters have the property that $f_i(g)\neq 1$ for some $g \in G$.
↑Jump back a section
## Definition
If G is an abelian group of order n, then the set of characters fk forms an abelian group under multiplication $(f_j f_k)(g)= f_j(g) f_k(g)$ for each element $g \in G$. This group is the character group of G and is sometimes denoted as $\hat {G}$. It is of order n. The identity element of $\hat {G}$ is the principal character f1. The inverse of fk is the reciprocal 1/fk. Note that since $\forall g \in G\;\; |f_k(g)|=1$, the inverse is equal to the complex conjugate.
↑Jump back a section
## Orthogonality of characters
Consider the $n \times n$ matrix A=A(G) whose matrix elements are $A_{jk}=f_j(g_k)$ where $g_k$ is the kth element of G.
The sum of the entries in the jth row of A is given by
$\sum_{k=1}^n A_{jk} = \sum_{k=1}^n f_j(g_k) = 0$ if $j \neq 1$, and
$\sum_{k=1}^n A_{1k} = n$.
The sum of the entries in the kth column of A is given by
$\sum_{j=1}^n A_{jk} = \sum_{j=1}^n f_j(g_k) = 0$ if $k \neq 1$, and
$\sum_{j=1}^n A_{j1} = \sum_{j=1}^n f_j(e) = n$.
Let $A^\ast$ denote the conjugate transpose of A. Then
$AA^\ast = A^\ast A = nI$.
This implies the desired orthogonality relationship for the characters: i.e.,
$\sum_{k=1}^n {f_k}^* (g_i) f_k (g_j) = n \delta_{ij}$ ,
where $\delta_{ij}$ is the Kronecker delta and $f^*_k (g_i)$ is the complex conjugate of $f_k (g_i)$.
↑Jump back a section
## See also
↑Jump back a section
## References
• See chapter 6 of Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3, MR 0434929, Zbl 0335.10001
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http://wiki.ece.cmu.edu/ddl/index.php/Excel_Solver
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# Excel Solver
## What is Solver?
Solver is an add in tool found in Microsoft Excel. It is primarily used to determine the maximum or minimum value of a problem, which is assigned to a cell in an Excel spreadsheet. In optimization, Excel Solver is a very powerful tool. The Solver is a combination of a numerous amount of programs. These programs functions are composed of among which are the Graphical User Interface (GUI), an algebraic modeling language (for example, GAMS), and optimizers for linear, non-linear, and integer programs. With these powerful tools put together, Solver is able to find an optimal value for a target cell in a worksheet.
## How Solver Works
Excel Solver is a tool that falls under the category of what-if analysis. In optimization, this means that Solver is the type of tool that will determine what would happen in a problem's outcome if one parameter was changed. Within one spreadsheet, there are essentially three basic types of cell.
Target Cell: This is typically one cell in a spreadsheet. It is usually a function that inputs other values that are found within the same spreadsheet. These cells fall within the next two categories.
Adjustable Cells: These cells must be given an initial value. When Solver is run, it will change the values of these cells in order to reach the optimum solution in the target cell.
Constraint Cells: These are set values that will restrict values that Solver will use. They can refer to other cells in the spreadsheet.
## Example
The following provides an example of how Solver could be used. Suppose the goal is to minimize f(x) = πr2Lρ. The following is an illustration of how this problem could be executed in Excel Solver (where cell B5 is the location of the objective function):
Cell B% inputs values for the radius, length, and rho which are specified in cell D12, D15, and D17 respectively under the design parameter category. This set up also incorporates some constraints, designated in cells D8 and D9. The two constraints represented by those cells are $\sigma(yield) = \frac{My}{I} \leq 0$ and $\delta(maximum) = \frac{F L^3}{3 E I} \leq 0$.
The Solver tool is an Add-In of Microsoft Excel located under "Tools" of the main menu. The three things that must be specified are the target cell, the adjustable cells, and the constraint cells. The target cell should simply reference the cell in which the objective function was inserted. This is the cell that solver wants to optimize. In order for it to optimize well, Solver must know which variables of the target function it is allowed to change. In this particular example, Solver is allowed to change the radius, r. However, in this example, Solver has also been specified to comply to some constraints, of which the cells have been specified. Once the set up is complete, click Solve, and the worksheet will be update the cells with an optimum solution in the target cell as well as update the values that are allowed to change, which in this case is the radius. With Solver, more constraints can be added as well as more changeable parameters to find different optimal values.
## Links
http://office.microsoft.com/en-us/help/HP051983681033.aspx
http://en.wikipedia.org/wiki/What-if_analysis
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http://mathoverflow.net/revisions/78688/list
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## Return to Answer
2 added 292 characters in body; added 39 characters in body; edited body
The problem can be solved by running some Integer Relation algorithm (e.g., PSLQ) on the numbers $1, r, r^2, \dots, r^N$ where $r$ is a given root.
See http://en.wikipedia.org/wiki/Integer_relation_algorithm
For example, here is computation in PARI/GP which gives a better result than the polynomial shown in question:
? r = 28.552622898861801; algdep(r,10)
%1 = 3*x^10 + 38*x^9 - 3695*x^8 + 4582*x^7 + 3016*x^6 + 1435*x^5 + 4552*x^4 - 1219*x^3 - 9920*x^2 - 2402*x + 3087
? subst(%1,x,r)
%2 = -2.7334689816478450022 E-24
1
The problem can be solved by running some Integer Relation algorithm (e.g., PSLQ) on the numbers $1, r, r^2, \dots, r^N$ where $r$ is a given root.
See http://en.wikipedia.org/wiki/Integer_relation_algorithm
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http://mathoverflow.net/questions/107298/realizable-order-sequences-for-finite-groups/108037
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## Realizable Order Sequences for Finite Groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
My post is motivated at least in part by this MO question.
Has there been any work done on realizable order sequences for finite groups? By an "order sequence" I mean a non-decreasing list of the orders of the elements of the group. By "realizable" I mean there is some finite group that has that particular order sequence.
For example, $(1, 2, 4, 4)$ is a realizable order sequence; it is the order sequence of $\mathbb{Z}/4\mathbb{Z}$.
Is this something that has been studied in any depth? (Perhaps under a different name?) Are there any non-trivial theorems about realizable order sequences? (Examples of trivial theorems: ones that fall immediately out of Lagrange's Theorem; the degree sequence of $\mathbb{Z}/n\mathbb{Z}$)
I know that there are some nice theorems about degree sequences in graph theory (e.g., Erdős-Gallai theorem; Havel-Hakimi theorem), and though I have seen "order sequence" defined in a few Abstract Algebra texts, I have yet to come across any results of much interest.
I also wonder whether results such as this problem solution (Problem 6636, F. Schmidt, Amer. Math. Monthly, Vol. 98, No. 10 (Dec., 1991), pp. 970-972) or this paper (Isaacs et al (2009). Sums of element orders in finite groups. Commun. Alg. 37(9):2978-2980) contain ideas that would be applicable to such a topic.
Finally, is there an easily accessible (and organized) database that lists order sequences for all finite groups up to a certain not-too-small size?
Edit 1: Is anyone up to computing such a list and posting it somewhere accessible?
Edit 2: Now that Alexander Gruber has kindly posted computations for a fair number of order sequences, I wonder: (a) when does the same order sequence correspond to more than one group? (b) given an order sequence that corresponds to precisely one group, how difficult is it to recover the corresponding group's structure?
Edit 3: Mr. Gruber has most recently pointed me toward a related area of research on "OD-characterizability." One mathematician who has done a fair bit of work in this area is AR Moghaddamfar. See, for example, Recognizing Finite Groups Through Order and Degree Pattern.
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You could compute such a list using, say, GAP and the smallgroups library. – Arturo Magidin Sep 16 at 4:52
## 4 Answers
Here's a quick list for groups up to order 512. Format is group order, followed by a list of possible order sequences for groups of that order, then an empty space, then a list of the number of groups with each respective order sequence (in the order they were listed). You can see it starts to get pretty long. Let me know if there is another specific range of orders you'd like me to run this code on.
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Since the motivating question has a purported proof for solvable groups, it would be great to see the code run on orders of non-solvable groups, which can be found at oeis.org/A056866 – Benjamin Dickman Sep 27 at 9:19
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Here it is for the nonsolvable groups of those orders. docs.google.com/… – Alexander Gruber Sep 27 at 14:53
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I don't know if this constitutes an answer but... You might be interested in the paper by Mazurov called The set of orders of elements in a finite group.
Given a group $G$, Mazurov defines $\omega(G)$ to be the set of element of orders of a group $G$. (So $\omega(G)$ isn't quite what you're looking for. But, still, information about the behaviour of $\omega(G)$ as $G$ varies over all finite groups, will tell you a lot about the behaviour of order sequences as you define them in your question.)
Apparently there was a long-standing conjecture to the effect that if two distinct groups $G$ and $H$ satisfied the identity $\omega(G)=\omega(H)$, then there were infinitely many such groups. In the paper above Mazurov proves a counter-example to this but he also remarks that "we believe that almost all groups $G$ are characterized by $\omega(G)$ uniquely." Now what he means by "almost all groups" I couldn't possibly say but, still, if this is true then it has significant implications for your order sequences.
In a different direction you might also be interested in various work on $GK(G)$, the Gruenberg-Kegel graph, or prime graph, of a group $G$. This is a graph whose vertices are labelled by primes dividing $|G|$, with two such vertices $p$ and $q$ being connected if an element of order $pq$ exists in $G$. Lots of information about $GK(G)$ has been determined for different groups $G$, including recognition theorems of the form "Suppose the underlying graph of $GK(G)$ is isomorphic to the graph $X$, then $G$ is the group $Y$".
If $G$ is simple, there is also a series of papers by Vasil'ev and Vdovin (and probably others) which give information about different properties of $GK(G)$, including maximum clique size etc. Again information about order sequences can be derived from these results.
P.S. I have e-copies of all the papers I mention above and can email them if you need me to.
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No nilpotent group $\neq 1$ is characterized by $\omega(G)$, and I would be surprised to see a solvable group that is, so maybe Mazurov wanted to say "almost all simple groups"? – F. Ladisch Sep 25 at 13:26
I should have also mentioned the notion of the spectrum' of a group, which is the set of maximal elements in the poset of element orders of the group (ordered by divisibility). I don't know the state of the literature on the spectrum' but I know the term crops up in various places. – Nick Gill Sep 25 at 16:40
Just to develop Ladish's comment. Every $p$-group is nilpotent. Yet it is not determined by its order sequence. For instance, both ${\mathbb F}_3^3$ and the unit-triangular group $UT_3({\mathbb F}_3)$ have $27$ elements, all of them, but the unit, being of order $3$.
This is related to the question http://mathoverflow.net/questions/39848 .
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Yes, indeed there are many non-isomorphic examples of $p$-groups of exponent $p,$ each pair providing a counterexample to the assertion that order sequence determines the isomorphism type of the group – Geoff Robinson Sep 28 at 6:18
Thanks for the note, example, and link. It's interesting that of the five groups of order $27$, there is of course the cyclic group $C_{27}$, and then not only do $UT(3,3)$ and $C_{3}^{3}$ have the same order sequence, but so do the other two groups of this order: namely, the direct product of $C_9$ and $C_3$ as well as the semidirect product of $C_9$ and $C_3$. From these examples alone, a whole host of possible conjectures on order sequences can be dismissed outright... – Benjamin Dickman Sep 28 at 15:22
At least for finite abelian groups the problem has been solved. See
Isomorphism of Finite Abelian Groups, by Ronald McHaffey, The American Mathematical Monthly, Vol. 72, No. 1 (Jan., 1965), pp. 48-50, Stable URL: http://www.jstor.org/stable/2313001
The author essentially proves that if the sequence of orders of two finite abelian groups are the same then the groups are isomorphic.
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http://quant.stackexchange.com/questions/4593/how-to-simulate-stock-prices-using-variance-gamma-process?answertab=oldest
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# How to simulate stock prices using variance gamma process?
I want to simulate stock prices with the variance gamma process. The model is given by:
$S_T=S_0 e^{ {[}(r-1)T + \omega + z{]}}$
where
$S_0=$ starting value
$T=$ Time
$\omega=\frac{T}{\nu}ln(1-\theta \nu - \sigma^2 \frac{\nu }{2})$
$r=$ interest rate
$z=$ normally distributed variable with mean $\theta g$ and standard deviation $\sigma \sqrt{g}$
I know, that I have to simulate first the g values by a random generator (using gamma function with parameters), then generate random numbers z using the g's. But my problem is, how does I specify the three parameters $\nu$ and $\theta$ and r? The T means years, so if I have e.g. 10 trading days, this would be 10 divided by 365. I had a another simulation with the geometric brownian motion before, there I used the sample mean, sample standard deviation, 22 trading days, and starting value 20. So I thought to make it comparable:
$T=22/365$
$S_0=20$
Nut what about $\theta$, $\nu$ and r? Is r just the sample mean?
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Hi, welcome to Quant.SE. Please don't hesitate to register in order to help the site grow and make it out of beta. – SRKX♦ Nov 22 '12 at 14:42
## 2 Answers
If you say stock prices are following GBM then you can say
$dS_t = \mu S_tdt + \sigma S_t dW_t$
solving which it brings
where $\sigma$ is volatility and $r$ is risk free rate .
**EDITED
For a Variance Gamma process theta is the deterministic drift in subordinated Brownian motion and sigma standard deviation in subordinated Brownian motion. I choose mu in (0.1,0.3) and volatility estimate by GARCH or around 15% lower to 30% upper for a typical simulation
HTH
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@ Ashwani Roy no, this is not an appropriate answer to my question, I did a GBM simulation already, now I want to do it with VG – user1690846 Nov 22 '12 at 9:54
Sorry , totally misunderstood . Edited based on what i know about Levy's process models – Ash Nov 22 '12 at 10:32
Why are these equations presented as images? You can enter $\LaTeX$ code and it will be converted into text appropriate for a browser. – chrisaycock♦ Nov 22 '12 at 13:44
@chrisaycock Sorry I am new to answering here. Will try this in future. Apologies if it caused any problems – Ash Nov 22 '12 at 13:55
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@AshwaniRoy I edited the answer and convereted the first equation to $\LaTeX$ for you so you can use it as an example to convert the other and future. Thanks for taking the time to answer! – Louis Marascio Nov 22 '12 at 15:54
show 2 more comments
This paper seems to outline what you are looking for. You want to be careful about mean/variance/kurtosis to make sure you are working in the correct measure.
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Any chance you could summarize the contents of that paper here? – chrisaycock♦ Dec 26 '12 at 17:12
It's a basic survey/study of a variance gamma model vs black-scholes, using calibrations to both historical as well as implied data. – experquisite Dec 27 '12 at 4:12
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http://mathhelpforum.com/advanced-statistics/93497-test-statistic.html
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# Thread:
1. ## test statistic?
X and Y stand for the size of the male and femal of a specific species of moth. Assume the distributions of X and Y are $N(\mu_{X},\sigma^{2}_{X})$ and $N(\mu_{Y},\sigma^{2}_{Y})$. $\sigma^{2}_{Y}>\sigma^{2}_{X}$. Use the modification of Z to test the hypothesis $H_{0}$: $\mu_{X}-\mu_{Y}=0$ against the alternative hypothesis $H_{1}$: $\mu_{X}-\mu_{Y}<0$.
How do you define the test statistic and a critical region that has a significance level of $\alpha=0.025$?
thank you very much
2. 1 Do you know the population variances?
If so the rejection region is $(-\infty ,-1.96)$
and the test statistic is ${\bar X-\bar Y\over \sqrt{ {\sigma^{2}_{Y}\over n_Y}+{\sigma^{2}_{X}\over n_X}}}$.
2 If you don't know $\sigma^{2}_{Y}$ and $\sigma^{2}_{X}$, then it's a t test.
BUT now I need to know if you're allowed to assume that $\sigma^{2}_{Y}=\sigma^{2}_{X}$ or not.
3 HOWEVER, if the sample sizes are large then you can approximate this via the Central Limit Theorem and use 1.
3. That is all the info I have (and part of why I am so stuck)
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http://thetoiletdoor.net/
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# The Toilet Door
uninteresting reading whilst you're on the can
## Efficiency
May 7th, 2013 by Fireslide
Ok, so I installed a math plugin based on LaTeX so I can type equations. The added bonus is I'll get to teach myself how to use them.
The purpose of this is so I can make things look a bit neater. So onto efficiency.
I've been thinking about efficiency my entire life, always applying it to various tasks, like driving, working, walking, eating, exercise. The problem is, that there's so many ways to define it. The most common and general definition is
Where maximum output is defined as some maximum achievable quantity. The energy efficiency of an engine say could be
So naturally, I strive to be efficient in most things I do. When I spend money, I calculate the dollar per hour entertainment value for purchases, I evaluate the efficiency of my work output and spending.
So relating this to the main point, is that governments and oppositions always claim to be able to increase the efficiency of the public service and cut costs. Now I'm sure everyone has their own horror stories about how inefficient their visit to Centrelink or Medicare was, but on the whole, these organisations have been streamlined quite a lot. It's not usually the end of line workers that are terribly inefficient anyway. So how do we define efficiency for government spending?
Obviously, everyone wants their tax dollars to be spent well and not wasted, but where it gets interesting is that these people themselves are probably no more efficient than the government when it comes to spending. Take for example, a bottle of shampoo. Suppose it costs \$10 and is a generous 1 litre volume. That gives a cost of 1 cent per mL. Now, most people have trouble getting the last few bits out of the shampoo bottle. There might be 10 or 20 mLs leftover that are just too difficult to extract, it's possible to do it but the time it takes is not worth it, so we accept some waste. The efficiency in this case is about 99.8% which is a high number. Food wastage is probably much lower, on every plate there might be 5% left, or if you get too full you can often leave about 10 to 50% of a meal. Sometimes it might be saved and eaten again later, but it could be thrown out. A mobile phone contract that includes some dollar value of free calls and messages, many people would not extract full value or efficiency out of that as well.
The point I'm trying to illustrate, is that people are inefficient, there's nothing wrong with that. It's just there's a lack of realisation that the government is made up from people too, so all these inefficiencies add up. So next time someone complains about government wastage, point out that they probably equally waste on the same scale food, shampoo, consumables etc.
Tags: math, politics
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## Habit updates
May 1st, 2013 by Fireslide
So when I started my attempt to blog regularly a little over a month and a half ago, I had an overly ambitious goal of adding one meaningful entry per day. Obviously that requires a significant time investment and whilst I managed it for a week, I've fallen off. This wasn't unexpected however, as with all habit building/breaking it's unlikely to change the first time you try. I reaslied it follows the trend of dieting or the gym that most people have. They are pretty vigilant for the first week or two, then they miss a day because of a work function, or they are sick. Then they miss another for a different reason. The routine is then broken and they stop going, they might have spurts of activity where they regain or rediscover their original motivation but the routine always gets interrupted or broken.
What I've realised is that isn't a bad thing, it'd be ideal if we could robotically acquire and discard habits at will. I'd program myself to wake at 6, go running, work efficiently, not eat crap and so on. It's not a bad thing to have routines broken because then the feeling of achievement and progress towards self improvement is much greater for all the obstacles we face. My goal of not having any soft drink has failed fairly miserably, but that's always going to be one of the hardest things to break, but at least I'm noting every day I do or don't have it. There's a few days a month where I don't have any, if I can slowly build on that, that's progress.
So back to this blog, I wanted to update every day, but that hasn't happened, but I have been updating about once every 9 days or so, which is a lot more than I used to do. So I may have failed at my original stated goal, but I set that knowing that I'd likely not be able to keep it up, but if it gets me in the habit of thinking about making blog posts more often, then it's working. So all up, it's been relatively successful. In 90 days or so I've transformed from making one blog post a year maybe to probably an average of 30 or 40 a year at this current rate.
Tags: habit, improvement, self
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## Numbers
April 23rd, 2013 by Fireslide
So there's an election due in September and this means that both major political parties are starting to have some 'serious' debate about the state of the economy, debt, policies and so forth. One thing that I've noticed is the way all parties use numbers as talking points as metrics for success or failure.
A hypothetical example would be, "We've created 150,000 jobs and unemployment has dropped to 5.2%". At face value this seems like a good statement to make, but the more I think about it, the more it seems meaningless. Creating jobs is obviously a good thing but it's the numbers they throw around in a debate that concerns me. What does creating 150,000 jobs mean? Did the public service grow by 150,000? Are the jobs in industries or fields we should be investing in or pulling out of? Are they full time jobs or is this casual work? Similarly with the unemployment rate dropping to 5.2%, what was it before? What is the underemployment rate?
These are all questions I'm thinking of when I hear someone mention something with numbers, what do those numbers really mean in the larger context of everything else. Now obviously numbers are important, having any metrics, even flawed ones are better than having no metrics, part of the issue I realise now is that people don't really understand numbers.
A study was performed testing if people would behave differently for a reward if it was \$3 or 300 cents. The results, surprisingly indicated that some people preferred 300 cents. Even if cents and dollars were switched around, people were more easily swayed by the larger number.
This knowledge has some interesting consequences for looking at political discourse. Are the politicians aware of this effect and use it to mislead or confuse citizens about the state of things? It's possible, there's limited time in media segments to accurately and adequately describe what a number truly represents, it's probably more important that the reader or viewer simply remembers that it was 150,000 jobs created or that a policy will cost \$94 billion.
Speaking of policy costs, it's interesting to observe that the cost of everything is often put in vacuum. $94 billion sounds like a lot and it rightly is for an individual to own, but in the context of an entire country that has a yearly GDP in the order of$1.5 trillion ($1500 billion), it doesn't seem as large, it'll seem even smaller if instead of stating the total cost over 10 years and comparing to a yearly GDP, we state the yearly cost$9.4 billion.
I'm going to keep an eye on how it progresses and see if there's a correlation between the way the numbers are presented and how they are meant to be viewed. Obviously positive achievements would be promoted and negative achievements downplayed.
Tags: communication, math, politics
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## Monopoly. Gambling or Trading game?
April 13th, 2013 by Fireslide
So one thing I enjoy is games and exploring the core concepts and skills they test when taken to high levels. One thing that nearly all games have in common is the concept of trading.
Now trading in a traditional sense could be two parties coming to a mutually agreeable set of terms to net benefit of both parties. In most competitive games however, trading isn't mutually agreeable. In Chess say, both players start with the set of pieces. Each turn a player is trading positional advantage for material advantage, or vice versa. Sometimes a player may initiate a trade by taking a piece. The opposing player may even the trade up by taking a piece back of equal value. The most common example of this is in Chess, where it's rare to not trade Queens. You don't want to give up your queen without getting a significant material or positional advantage out of it.
So onto Monopoly, it has a trading element core to the rules of the game. Players can make trades for property, cash at any time. The main reason to do this, is to gain a cash advantage or material advantage over your opponent. Now, a lot of people when they play monopoly use a flow of logic when it comes to trades along this line. "I don't want to trade with you, because you're making this trade to get an advantage over me, no matter how much you are seemingly offering me". So if there's no trading, then monopoly is really a game of chance, you're hoping that your dice rolls are favourable on average, and the opponents are unfavourable on average.
Where it gets interesting, is that monopoly is what I'd call a 'solved' game. A number of people have calculated the probabilities of landing on any given square, the average number of rolls for a certain investment in a set of properties to pay off. The limitations on human players would to be to remember and evaluate all the tables of data to know the likelihood of winning from a given board position. So taking it to the next level, let's suppose that we have perfect players that know all the probabilities for any given game state to calculate a winner. Trading properties then becomes simply betting. Investing heavily into mayfair and boardwalk has a low probability of paying off, so most people wouldn't make it, but it essentially becomes an agreement between players that in the next x dice rolls Player A thinks they wont land on it, and Player B thinks they will.
http://www.amnesta.net/other/monopoly/
http://www.tkcs-collins.com/truman/monopoly/monopoly.shtml
I think it's interesting that many games and sports rely on limitations in various skills or abilities to make them fun. Monopoly can be fun because most people are incapable of correctly evaluating the probability of any given player winning from a certain position, but even taken to the extreme, it's betting on dice rolls.
Tags: games, monopoly
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## Update
April 8th, 2013 by Fireslide
So I've been playing around a bit with LaTeX because I like the idea of what it can do. It sorts out all the tricky formatting depending on how you want to present your work. It's not ideal for collaborative editing though. Microsoft Word wins in that department with comments and tracking changes.
One of the ways I'm going to investigate using LaTeX is with some custom tags. This xkcd comic illustrates which characters interact with each other as a function of time over a story. Now that was produced by reading the books, and manually going through and making note of which characters are where and with who at each time. Ideally though, if that information is put into custom tags at various chapters or at points in the story it can be invisible to the reader, but a simple program or script could extract the information and produce a plot like that. Similarly, if we could generate one for characters, we can also do it for action, or emotional content.
Having the ability to produce graphs to demonstrate the action during a story or comedy say gives an author new tools with which to view the overall story. Is the front action heavy? Is it too dry? Is there enough story progression? Humans are visual creatures, so I'm hoping I can make something to produce these graphs, so that when I begin writing, I have a large number of tools available to guide my story.
Tags: improvement, phd, science, self, writing
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## Self referential references
March 29th, 2013 by Fireslide
So I need to come up with a name for a phenomena I've observed.
Basically I tell my students at the start of the practical the same things. Check the marking scheme, don't forget to talk about errors, work quickly.
Inevitably they ignore me and forgot to talk about errors, so I started telling the students. Check the marking scheme, don't forget to talk about errors, work quickly. I tell all the other students this and they usually fail, so maybe you'll be different.
Inevitably they think I'm joking or something, and they do poorly. So I started telling the students. Check the marking scheme, don't forget to talk about errors, work quickly. I tell all the other students this and they usually fail, and I tell them what I've just told you now and they usually fail, so maybe you'll be different.
So we can really just state it's
Message Content. + self referencing warning
I need to print something like that out for my students and just show them what happens
Tags: funny, teaching
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## Musings on privacy
March 27th, 2013 by Fireslide
So I just read a substantial paper that debunks the argument many people use to support legislation or technology that would infringe on privacy. "If you've got nothing to hide, what's there to worry about?"
https://papers.ssrn.com/sol3/papers.cfm?abstract_id=998565&
Basically he states that privacy is a fairly ill defined term, and it has a range of classifications. He goes on to define the problems that the term privacy can encompass. The two main ones are Orwellian surveillance and Kafkaesque problems. Kafkaesque being the kind of situation where another party has so much information about you that before you can take any actions, you're already limited in what you can do.
Anyway, it had me thinking about trying to break down privacy to what it really means. All I've come up with so far is that privacy is like a currency denominated by information. The majority of interactions we have in the world are essentially trades in information. With friends, we share secrets, which is an information trade to build up trust. Similarly, when we go for job interviews, or are looking for romantic partners, we have the ability to release and share information in a way that makes us more desirable.
Where privacy is important is that it allows us options in how we reveal information and when. If you remove privacy, it allows other parties you may want to interact with to assess you on either a) incomplete or incorrect information or b) correct information presented in an unfavorable order. For most cases it's probably a), which is why privacy is important.
I guess an appropriate analogy would be suggesting that life is a bit like a card game. You can spend some time trying to trade for better cards, you can play cards in a beneficial order, you can make intelligent guesses and inferences about other people based on what actions they take. The privacy component is that your hand is hidden information, there should be no desire to give up the advantages that privacy affords even if you're doing nothing wrong, because it can in no way help you in the future. What is happening with big data and data mining is that we're getting so much information from everyone about what they are discarding, what they play and what hands they have that we can start building fairly accurate models about what moves you'll make next.
They've already built a computer program that will consistently beat most humans at rock paper scissors based on big data http://www.nytimes.com/interactive/science/rock-paper-scissors.html?_r=0
It's an interesting thing to consider, but for the time being, fight to hold onto your privacy, otherwise you're giving up your social currency too easily and gaining nothing for it. As I'd say in most games, that's a bad trade.
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## Game Theory
March 26th, 2013 by Fireslide
So I haven't updated for a while. I think I was perhaps a bit ambitious to think I could make a meaningful entry every day. I either need to rapidly increase the speed at which I can produce an entry, or I need to make them less frequently. Part of the issue is unbalanced work weeks, but anyway. On with the content.
So one thing I find interesting in Game Theory. I'm talking about things like the Prisoner's Dilemma, Pascal's Wager. They can be applied to all sorts of situations wherein you are uncertain about the state of one particular function. For example, you could apply it to the weather raining or not raining. Then your choices are wear rain appropriate clothes, or not. Ideally you want to only wear rain appropriate clothes when it's raining, but it comes down to what will bother you more, wearing rain clothes when you don't need them, or getting soaked in your normal clothes.
The other situation where they are interesting is when you're relatively sure of one of the states, but not certain, but the payoff for being correct is small or insignifcant compared to being wrong. For example, funding some unlikely to work research but for a defense purpose. Funding the research and having it pay off is huge, Funding the research for no pay off is not ideal, but on the other side, not funding the research and not having it pay off achieves nothing, but not funding and having it succeed is disasterous. Other situations turn up in social contexts as well.
I'll expand on this more later when I can actually draw some in here, it's too awkward to talk about them without
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## Paradox of existence
March 21st, 2013 by Fireslide
We spend a good deal of time learning how to perform certain tasks, so our brain can pretty much complete them on autopilot, freeing up our brain to do other tasks. The problem is that in doing so, our brain stops forming memories of that task unless they are significant, eg a car crash. So we repeat tasks and form habits and routines, so our brain can be lazy.
The paradox is this. Do we spend our time in the same routines, and remembering on highlights of our existence, but live efficiently. Or do we force ourselves to not use the easy pathways formed in our brain, so we're constantly learning and relearning and remember more in detail, but live less efficiently.
It's a tough concept for me to resolve.
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## The future of TV
March 20th, 2013 by Fireslide
http://www.wired.com/underwire/2013/03/nielsen-family-is-dead/
So I read this article tonight and it covers a number of highly relevant points. I'm not going to cover everything, but I do want to talk about the first point they bring up, that is, how network executives measure the popularity of a show is shifting. Rather than surveying the habits of random households, some companies are now just sampling the available data on twitter. It means going from a sample size of 25,000 with a good deal of selection bias to an entire, worldwide demographic of millions of people, also with some selection bias.
It speaks to the complacency of companies when they aren't constantly trying to innovate and improve, only maintain share price that they didn't start doing this sooner. What is really interesting though, is whilst it's starting with TV shows, there's no reason the traffic on twitter can't be used measure a whole range of social phenomena. Journalists are already tweeting details of events, court cases live and generating interest and discussion. As more people become involved, the information that can be gleamed from twitter analysis will become more reliable cross sections of public opinion.
There is definitely self selection bias though, the demographic of older people or those without reliable access to the internet are unable to make full use of the power of social media. It shouldn't be solely relied upon just yet, but for now it's large enough that institutions can no longer ignore it. I'd like to think that in 30 years time, when the older generation has passed away, that the entirety of the population could be reliably sampled through the internet, there'd be no need to rely on phone polls, face to face polls. Polling companies themselves would have shifted to big data analysis on trends from twitter or facebook. Public opinion on legislation could be reliably sampled by simple search query, or a tweet asking for opinions from a popular politician or celebrity.
I'm excited about the efficiency of the future, I can't wait for it to get here.
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http://quant.stackexchange.com/questions/3012/creating-an-n-factor-certainty-equivalent-discounting-formula/3055
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# Creating an n-factor Certainty Equivalent Discounting Formula
Brealey & Myers provide a certainty-equivalent version of the present value rule, using CAPM, as follows:
$$PV_0=\frac{C_1 - \lambda_m *cov(C_1, r_m)}{1 + r_f}$$
$PV_0$ - Present Value of cash flow 1 at time 0.
$\lambda_m$ - Market price of risk = $\frac{r_m-r_f}{\sigma_m^2}$
$cov(C_1, r_m)$ - Covariance of the cash flow at time 1 with the return on the market.
I want to create an n-factor version of this same model. However, using Fama French 3-factor model as an example, the following doesn't seem to work on a toy example I've set up:
$$PV_0=\frac{C_1 - \lambda_m *cov(C_1, r_m)- \lambda_{smb} *cov(C_1, r_{smb})- \lambda_{hml} *cov(C_1, r_{hml})}{1 + r_f}$$
$\lambda_m$ = $\frac{r_m-r_f}{\sigma_m^2}$
$\lambda_{smb}$ = $\frac{r_s-r_b}{\sigma_{smb}^2}$
$\lambda_{hml}$ = $\frac{r_h-r_l}{\sigma_{hml}^2}$
Question: what am I doing wrong? Is there some way I need to adjust for the covariance amongst the factors?
===Update===
In checking my toy example again, I realized that I might in fact have the right formula above. So points/checkmarks to anyone who can either prove the above right or wrong or provide a citation to the more general form:
$$PV_0=\frac{C_1 - \displaystyle\sum_{i=1}^n\lambda_i *cov(C_1, r_i)}{1 + r_f}$$
for orthogonal risk factors $i_1,i_2,\dotsc,i_n$.
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## 1 Answer
Alright, here's the proof (I think):
Statement of APT:
$$E(r_a)=r_f + \displaystyle\sum_{i=1}^n\lambda_i * cov(E(r_a), r_i)$$
Expand $E(r_a)$:
$$\frac{E(C_1)}{PV_0} - 1 =r_f + \displaystyle\sum_{i=1}^n\lambda_i * cov(\frac{E(C_1)}{PV_0} - 1, r_i)$$
Since $PV_0$ doesn't have any covariance with $r_i$, we can reduce the above to the following:
$$\frac{E(C_1)}{PV_0} - 1 =r_f + \displaystyle\sum_{i=1}^n\frac{\lambda_i * cov(E(C_1), r_i)}{PV_0}$$
Rearrange:
$$\frac{E(C_1) -\displaystyle\sum_{i=1}^n\lambda_i * cov(E(C_1), r_i)}{PV_0} = 1 + r_f$$
And finally:
$$\frac{E(C_1) -\displaystyle\sum_{i=1}^n\lambda_i * cov(E(C_1), r_i)}{1 + r_f} = PV_0$$
QED (until somebody points out a dumb error I've made).
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I'm not qualified to judge this work, which is why I personally couldn't help on the question anyway. But since I want to award the bounty, and since you've taken the time to answer your own question, I've award the bounty to you. Hopefully someone more qualified can assess what's going on. – chrisaycock♦ Mar 11 '12 at 18:04
Appreciate it. I'd be happy to give it back since I was too lazy to figure out it before I posted the original question. – MikeRand Mar 11 '12 at 22:07
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http://mathoverflow.net/questions/105208?sort=oldest
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## Are there better upper bounds on the rank of the commutant of a fusion module than the global dimension?
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Suppose I have a fusion category $\mathcal{C}$ and an indecomposable module category $\mathcal{M}$ over it. The commutant $\mathcal{C}_\mathcal{M}^*$ is the category of module endofunctors, and gives another fusion category, Morita equivalent to the original $\mathcal{C}$.
Can I bound the rank of $\mathcal{C}_\mathcal{M}^*$?
Recall that the rank is the number of isomorphism classes of irreducible objects. Certainly $\mathcal{C}$ and $\mathcal{C}_\mathcal{M}^*$ have the same global dimension, so easily $\operatorname{rank}(\mathcal{C}_\mathcal{M}^*) \leq \operatorname{dim}(\mathcal{C})$. Are there better upper bounds available?
Update: I'm happy to consider all the 'decategorified' data of $\mathcal{C}$ and $\mathcal{M}$, that is, the Grothendieck groups of both, along with the ring and module structures thereon, when trying to come up with an estimate, not just the rank of $\mathcal{C}$.
As examples:
• the Haagerup subfactor gives a Morita equivalence between two fusion categories with ranks 4 and 6, and global dimension $\approx 35.725$
• $\operatorname{Rep}(G)$ and $\text{Vec}_G$ are Morita equivalent, with global dimension $|G|$. Here $\operatorname{rank}(\text{Vec}_G) = |G|$, while when $G$ is non-commutative $\operatorname{rank}(\operatorname{Rep}(G))$ may be much smaller.
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## 2 Answers
In the special case of Vec(G) and Rep(G) there's a lot of results in the literature. Usually the phrasing is in terms of conjugacy classes (e.g. the strongest proved bound as far as I know is Keller's "Finite groups have even more conjugacy classes"). I learned about this from Pavel Etingof when Eric Rowell asked him about a similar question for ranks of centers.
Do I understand correctly though that you're happy to have conditions that involve more than just the rank of C? E.g. if I say wanted to have the full list of dimensions of objects in C as input to the bound would that be a problem?
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Thanks Noah for the link. I clarified that I'm happy to use all the information in the fusion ring and its module. – Scott Morrison♦ Aug 22 at 22:01
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If $\mathcal{C}$ and $\mathcal{D}$ are Morita equivalent by a pair `$({_\mathcal{C}}\mathcal{M}_{\mathcal{D}}, {_\mathcal{D}}\mathcal{N}_{\mathcal{C}})$` of bimodules, then there is a natural map of fusion bimodules `${_D}N \otimes_{C} M_{D} \to {_D}D_D$` that preserves Frobenius-Perron dimension (see section 5.1 of Noah's latest preprint with Pinhas Grossman). So the Frobenius-Perron dimensions of elements of $N \otimes_C M$ (which do not depend on knowing $D$) will give Frobenius-Perron dimensions of elements of $D$. Then I think you may be able to use the known possible small Frobenius-Perron dimensions of objects (from your paper with Noah and Frank Calegari) to determine some nontrivial lower bounds on Frobenius-Perron dimensions of simple objects in $\mathcal{D}$, and thus improve on the global dimension bound.
(Or do arbitrarily small numbers of the form $2 \cos (\pi / n)$ already generate the full ring of real cyclotomic integers? Even if so, this method could at least reduce the bound by 1 for weakly integral categories, although that's not a great improvement.)
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Thanks Evans, this does seem promising at least for some small cases. For example for Haagerup the module has an object of dimension $\alpha = \sqrt{(5+\sqrt{13})/2}$, so $D$ has a (not-necessarily simple) object of dimension $\alpha^2 = (5+\sqrt{13})/2 \simeq 4.30278$. It's easy to see that this can only be decomposed as $1+1+(1+\sqrt{13})/2$, $2+(1+\sqrt{13})/2$ or $2 \cos(\pi/n) + x$, where $x > 76/33$. There's a finite list of such $n$, and it turns out $x$ is never maximal amongst its Galois conjugates except when $n = 2$ or $3$. – Scott Morrison♦ Aug 24 at 2:31
Thus we have to split into four cases already, but in each we've already identified the dimensions of some simple objects in $D$, and hence improved the bound on the rank. – Scott Morrison♦ Aug 24 at 2:32
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http://mathoverflow.net/questions/111193/monoidal-structure-on-a-category-with-products-and-with-terminal-object/111223
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## Monoidal structure on a category with products and with terminal object
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Let $K$ be a category with products $(X,Y)\mapsto X\sqcap Y$ and with a terminal object $T$. It seems obvious to me that $\sqcap$ and $T$ define a structure of a monoidal category on $K$, but I can't find a reference. When I try to prove this myself I come to amazingly bulky constructions. Is there a text where this is accurately proved, or at least formulated?
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3
I don't know a reference, but I do know this has a name: ncatlab.org/nlab/show/cartesian+monoidal+category . – Eric Peterson Nov 1 at 19:41
Eric, thank you, that's interesting. So this means that the accurate proof exists... It would be nice to look at it... – Sergei Akbarov Nov 1 at 20:00
Is easy (elements check) that $Set$ is cartesian (i.e. for finite products) monoidal, then for a general cartesian category you apply the (general) representable $(X, -)$ to the axioms diagrams (and use the result in $Set$), then the commutativity of each diagrams follow from Yoneda Lemma (need only the faithful part). – Buschi Sergio Nov 1 at 20:03
Sergio, I don't understand this trick. Is it possible, for example, to prove the diagram of associativity (the pentagon) in this way? – Sergei Akbarov Nov 1 at 20:20
1
@Sergei, using the Yoneda lemma it is, because that tells you the functor $X\mapsto Hom(-,X)$ is fully faithful. – David Roberts Nov 1 at 23:15
## 3 Answers
Is very easy prove that $(Set, \times, 1)$ is monoidal (by elements checking). Now let $\mathcal{C}$ a category by finite product $\times$ and (then) with a final object $1$. Consider the axioms of monoidal category for $(\mathcal{C}, \times , 1)$ stated by diagrams (see for example p.462 of "Closed Categories" by Eilenberg & Kelly, LA Jolla 1967), now it remains to prove that these diagrams are commutative. COnsider a such diagram $\textbf{D}$ and a (general) object $X\in \mathcal{C}$ and the representable $(X, -): \mathcal{C}\to Set: A \mapsto (X, A)$, acting by $(X, -)$ on this diagram, we get a similar diagram in $Set$, say $X(\textbf{D})$, and $(X, -)$ preserve the product $\times$ and the final object $1$, now we just know that $(Set, \times, 1)$ is monoidal, then $X(\textbf{D})$ is commutative. Because this is true for each object $X$, by Yoneda lemma follow that $\textbf{D}$ is commutative (more easily observe that given $f, g: A \to B$, if $(X, f)=(X, g): (X, A)\to (X, B)$ for each $X$ then $f=g$ (consider $X=A$ and $1_A$)).
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Interesting... OK, I need a time to verify the details. – Sergei Akbarov Nov 2 at 7:54
It would amount to the same thing if we thought about this in terms of generalized elements. – Spice the Bird Nov 2 at 17:03
Generalized elements? What's this? Anyway, I accept Sergio's answer. – Sergei Akbarov Nov 2 at 17:36
Let $\mathcal{C}$ be a category and $X$ be an object. Then a generalized element of $X$ of shape $Y$ is a morphism, $f:Y\rightarrow X$. We may also write this as $f\in_{Y} X$, for $f$ is a $Y$ shaped element of $X$. See ncatlab.org/nlab/show/generalized+element and also see the book at the website patryshev.com/books/Sets%20for%20Mathematics.pdf. – Spice the Bird Nov 3 at 2:30
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You need to have chosen products for every pair of objects using the axiom of choice, otherwise you don't get a product functor, just a product anafunctor and I recommend you don't try to use those just yet. Then you can prove - and this is the key step - that any two bracketings of an iterated product are isomorphic in a unique way when you demand the isomorphism respects the all the projections. The unique such isomorphism for a triple product is then the associator, and the uniqueness of the isomorphism for the 4-fold product means that the pentagon commutes. Ditto with the other coherence conditions.
Note that you can choose the product of any object $X$ with the terminal object to be $X$. This makes the unit conditions automatic.
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David, yes, I agree that the operation $(X,Y)\mapsto X\sqcap Y$ must be a mapping. Suppose this is so, then you say that the Yoneda lemma allows to simplify the proof? Can you recommend a text where this trick is used (not necessarily for proving what I am asking about, but just for something...), I would like to look how this trick works. – Sergei Akbarov Nov 2 at 6:25
You can find it as an example of a monoidal category in Tom Leinster's "Higher Operads, Higher Categories", which contains loads of coherence proofs for higher categories.
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I didn't understand, is this proved there, or just formulated? – Sergei Akbarov Nov 1 at 20:06
Just formulated, but the book gives a lot of information that will allow you to prove it yourself. – Wouter Stekelenburg Nov 1 at 23:17
Thank you, I'll try to find this book. – Sergei Akbarov Nov 2 at 7:55
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http://mathhelpforum.com/calculus/185778-anyone-want-verify-answers-lagrange-multiplier-question.html
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# Thread:
1. ## Anyone want to verify answers for lagrange multiplier question?
Would all be helping me out here alot
Find the local maxima and minima of the following function:
$f(x_1,x_2,x_3)=x_1 + 2x_2 + x_3$
Subject to the constraints:
$x_1^2+x_2^2+x_3^2=2$ and, $x_1^2+(x_2-1)^2+x_3^2=3$
The answers I got were:
$(x_1,x_2,x_3)=(1,0,1)$ with $\lambda_1=-\frac{3}{2}, \lambda_2=1$ ... This gives a local maximum.
Also...
$(x_1,x_2,x_3)=(-1,0,-1)$ with $\lambda_1=-\frac{1}{2}, \lambda_2=1$ ... This gives a local minimum.
I hope these are right. If I have made an error, and you want my working just reply... thanks!!
2. ## Re: Anyone want to verify answers for lagrange multiplier question?
Did you notice that your second constraint can be nicely simplified with your first constraint? Rather simplifies the whole problem, in fact.
3. ## Re: Anyone want to verify answers for lagrange multiplier question?
Yes, I did notice this. And this was my first step used in the problem - that is $x_2=0$
I was only asking to see if anyone can verify my answers....
4. ## Re: Anyone want to verify answers for lagrange multiplier question?
Why do you doubt? Did you make any mistakes on purpose?
I'm just questioning why you can't verify your own answers. How did you determine the two points were local max and min?
5. ## Re: Anyone want to verify answers for lagrange multiplier question?
Haha, I doubt myself all the time - although the fact that they are nice answers gives me some confidence. I determined local max/min by doing the whole Bordered hessian matrix technique, confirming they weren't non-degenerate critical points, and then checking that $\textbf{h}^TH\textbf{h}$, where $\textbf{h}$ is the tangent vector at that point, was always positive to give a local min, negative to give a local max....
??
6. ## Re: Anyone want to verify answers for lagrange multiplier question?
Well, there you go. If you can even SAY "Bordered Hessian" I have to think that you ahve at least a little clue. As long as you used the right criteria for various minors, you have it! Did you? :-)
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http://mathhelpforum.com/geometry/41526-area-quadrilateral.html
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# Thread:
1. ## area of quadrilateral
bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$
area:
$\frac 1 2 \begin{vmatrix}<br /> 0 & 0 \\<br /> -t & 0 \\ <br /> 0 & -\frac 1 t \\<br /> -\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\<br /> 0 & 0<br /> <br /> \end{vmatrix}$
$= \frac 1 2 (\frac {t^4-1}{1+t^4})$
cannot prove that the area is a constant
thanks
2. Originally Posted by afeasfaerw23231233
bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$
Yes the co-ordinates are right.
But the area is wrong. The area is a constant and it is 1.
I did it the long way. First let $O_1O_2$ be the line joining the centers of $C_1$ and $C_2$.
Then equation of the line $O_1O_2$ is $\frac{x}{t} + yt + 1 = 0$.
Now the base of the triangle is $|O_1O_2| = \sqrt{t^2 + \frac1{t^2}}$.
The perpendicular distance(height) from $(0,0)$ to the line is $\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}}$. So the area of the triangle is $\frac12 |O_1O_2|\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |\frac{0}{t} + 0t + 1|= \frac12$.
The perpendicular distance(height) from $(-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4})$ to the line is $\frac{|\frac{-\frac {2t}{1+t^4}}{t} -\frac {2t^3 }{1+t^4}t + 1|}{\sqrt{t^2 + \frac1{t^2}}}$. So the area of the triangle is $\frac12 |O_1O_2|\frac{|-\frac {2}{1+t^4} -\frac {2t^4 }{1+t^4} + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |-\frac {2+2t^4 }{1+t^4} + 1| =\frac12 |-2 + 1| = \frac12$.
So the area of the quadrilateral is the sum of the area of triangles.
And that is 1, a constant.
3. Originally Posted by afeasfaerw23231233
bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$
area:
$\frac 1 2 \begin{vmatrix}<br /> 0 & 0 \\<br /> -t & 0 \\ <br /> 0 & -\frac 1 t \\<br /> -\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\<br /> 0 & 0<br /> <br /> \end{vmatrix}$
$= \frac 1 2 (\frac {t^4-1}{1+t^4})$
cannot prove that the area is a constant
thanks
When t = 1, A = 1. But your answer gives A = 0 ......
4. Originally Posted by mr fantastic
When t = 1, A = 1. But your answer gives A = 0 ......
$A = \frac{1}{2} \, |(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + (x_3 y_4 - x_4 y_3) + (x_4 y_1 - x_1 y_4)|$
where the order of the points is such that 1 connects to 2 connects to 3 connects to 4 connects to 1.
So your answer is wrong because you have the order of the points wrong in the formula you used ......
You should get $A = \frac{1}{2} \left( \frac{2 t^4}{1 + t^4} + \frac{2}{1 + t^4}\right) = 1$, as expected from the special case I considered in my earlier reply.
5. Originally Posted by afeasfaerw23231233
bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) ...
b) find that the area of the quadrilateral is a constant.
...
Here is a completely different attempt:
1. Calculate the coordinates of the centers of the 2 circles:
$x^2+2tx+{\color{red}t^2}+y^2 = {\color{red}t^2} ~\iff~ (x+t)^2+y^2 = t^2$
$x^2+y^2 +\frac {2y}t +{\color{red} \left(\frac1t \right)^2}={\color{red} \left(\frac1t \right)^2}~\iff~ x^2+\left(y+\frac1t\right)^2= \left(\frac1t \right)^2$
Then the area of the quadrilateral consists of 2 congruent right triangles. The legs of one of these right triangles have the lengthes(?) $R = t$ and $r = \frac1t$.
Therefore the area of the quadrilateral is:
$A = 2 \cdot \frac12 \cdot t \cdot \frac1t = 1$
Attached Thumbnails
6. i joined the points in a wrong order.
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http://mathoverflow.net/questions/5553/which-graphs-have-incidence-matrices-of-full-rank/6650
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## Which graphs have incidence matrices of full rank?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This is a follow-up to a previous question. What graphs have incidence matrices of full rank?
Obvious members of the class: complete graphs.
Obvious counterexamples: Graph with more than two vertices but only one edge.
I'm tempted to guess that the answer is graphs that contain spanning trees as subgraphs. However, I haven't put much thought into this.
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"I haven't put much thought into this". +1 for honestly stating what some large proportion of posters on this website fail to admit. – Lavender Honey Nov 14 2009 at 19:11
## 5 Answers
The first answer identifies "incidence matrix" with "adjacency matrix". The latter is the vertices-by-vertices matrix that Sciriha writes about. But the original question appears to concern the incidence matrix, which is vertices-by-edges. The precise answer is as follows.
Theorem: The rows of the incidence matrix of a graph are linearly independent over the reals if and only if no connected component is bipartite.
Proof. Some steps are left for the reader :-)
Note first that the sum of rows indexed by the vertices in one color class of a bipartite component is equal to the sum of the rows indexed by the other color class. Hence if some component is bipartite, the rows of the incidence matrix are linearly dependent.
For the converse, we have to show that the incidence matrix of a connected non-bipartite graph has full rank. Select a spanning tree $T$ of our graph $G$. Since $G$ is not bipartite, there is an edge $e$ of $G$ such that the subgraph $H$ formed by $T$ and $e$ is not bipartite.
The trick is to show that the columns of the incidence matrix indexed by the edges of H form an invertible matrix. We see that $H$ is built by "planting trees on an odd cycle". We complete the proof by induction on the number of edges not in the cycle.
The base case is when $H$ is an odd cycle. It is easy to show that its incidence matrix is invertible. Otherwise there is a vertex of valency one, $x$ say, such that $H \setminus x$ is connected and not bipartite. Then the incidence matrix of $H \setminus x$ is invertible and again it is easy to see this implies that the incidence matrix of $H$ is invertible.
Remark: I do not know who first wrote this result down. It is old, and is rediscovered at regular intervals.
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Awesome, thanks! Do you have a reference for this? A graph theory textbook would be just fine. – Jiahao Chen Nov 17 2009 at 20:15
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The earliest reference I could find in 30 minutes was: @article {MR0441791, AUTHOR = {Van Nuffelen, Cyriel}, TITLE = {On the incidence matrix of a graph}, JOURNAL = {IEEE Trans. Circuits and Systems}, YEAR = {1976}, NUMBER = {9}, PAGES = {572} It might be in Scheinermann's "Fractional Graph Theory", but I am travelling and cannot check this. It's surely in some of the texts on combinatorial optimization, but I am travelling... – Chris Godsil Nov 19 2009 at 0:50
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An interesting example of a class of graphs for which it not known when the adjacency matrix $A$ has full rank is the Hasse diagrams of the lattices $L(k,j)$. See pages 21-22 of arXiv:math/0501230. What makes this interesting is that explicit trigonometric formulas are known for the eigenvalues of $A$ and their multiplicities, but it is unclear from these formulas whether there is a zero eigenvalue of positive multiplicity.
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As Chris pointed out, this answer is off the mark. The graphs with adjacency matrices (not incidence matrices) of non-full rank are called singular graphs. It seems that it is a known and difficult open problem to characterize them. See this paper by Sciriha.
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I think I can find a graph with a spanning tree that does not have an adjacency matrix of full rank. Start with a path of length five join the first point to the fourth and the fifth two the secod. Then the first and the fifth will have the same adjacency row and that will be a dependency and that will be a graph with a spanning tree that does not have full rank. However the question is not about adjacency matrices but incidence matrices. However since any graph with a connected bipartite component does not have a incidence matrix of full rank as noted in another post we can take any connected tree and that graph will have a spanning tree and it will have bipartite component so it will not have a singular matrix of full rank.
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This refers to the signed incidence matrix, the question asks about the unsigned incidence matrix. – Chris Godsil May 25 2011 at 11:46
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http://mathoverflow.net/questions/17006/linear-algebra-proofs-in-combinatorics/33505
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## Linear Algebra Proofs in Combinatorics?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Simple linear algebra methods are a surprisingly powerful tool to prove combinatorial results. Some examples of combinatorial theorems with linear algebra proofs are the (weak) perfect graph theorem, the Frankl-Wilson theorem, and Fisher's inequality.
Are there other good examples?
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@Francois- thanks for fixing the link. – Tony Huynh Mar 3 2010 at 20:50
As long as this remains a link list, I will put it in comments: mathlinks.ro/viewtopic.php?t=5976 mathlinks.ro/viewtopic.php?t=167503 mathlinks.ro/viewtopic.php?t=290708 perso.univ-rennes1.fr/xavier.caruso/articles/… ima.umn.edu/preprints/Feb92Series/918.pdf – darij grinberg Mar 3 2010 at 20:50
@Tony: No problem. URL encoding gets tricky. en.wikipedia.org/wiki/Percent-encoding – François G. Dorais♦ Mar 3 2010 at 20:56
You could probably find plenty by searching MathSciNet for papers with a combinatorics MSC code in the journal "Linear Algebra and its Applications". – Douglas S. Stones Mar 3 2010 at 22:53
@Douglas - Thanks for the tip. Unfortunately, I am MathSciNetless for the next little while... – Tony Huynh Mar 4 2010 at 2:25
## 14 Answers
Some other examples are the Erdos-Moser conjecture (see R. Proctor, Solution of two difficult problems with linear algebra, Amer. Math. Monthly 89 (1992), 721-734), a few results at http://math.mit.edu/~rstan/312/linalg.pdf, and Lovasz's famous result on the Shannon capacity of a 5-cycle and other graphs (IEEE Trans. Inform. Theory 25 (1979), 1-7). For a preliminary manuscript of Babai and Frankl on this subject, see http://www.cs.uchicago.edu/research/publications/combinatorics.
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Right, I thought immediately of the lovely manuscript by Babai and Frankl. Too bad it is not more easily available: I was lucky enough to take a combinatorics class from Babai in 1998 and thereby acquire a copy. I still have it! (Do you know why they have apparently lost interest in publishing it?) – Pete L. Clark Mar 4 2010 at 0:12
I don't know why Babai and Frankl have lost interest in publishing it, but the preliminary manuscript is still readily available from the link above. – Richard Stanley Mar 4 2010 at 0:49
@Richard - Much thanks! There is a lot of good information in your post. – Tony Huynh Mar 4 2010 at 2:24
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I know someone tried to contact the UChicago department several times about getting that manuscript without luck, so here is a scanned copy: ifile.it/6wl3uv2 – Steven Sam Mar 4 2010 at 17:05
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I've received a few emails about the link being dead, so here is a new one: ifile.it/wpolrs – Steven Sam Jan 29 2012 at 5:11
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The Lindstrom-Gessel-Viennot Lemma uses the reflection principle on $S_n$ to say that the number of nonintersecting families of lattice paths in the plane equals the determinant of a matrix so that the $i,j$-th entry is the number of paths from the $i$th source to the $j$th sink.
This was not a linear algebra proof. However, this determinant can be used to enumerate plane partitions inside an $a\times b \times c~$ box, to $q$-enumerate plane partitions by weight, and to count domino tilings of an Aztec diamond. The resulting determinants can be manipulated and evaluated in ways which are natural in linear algebra, but not as clear on the objects, such as factoring the matrices. These enumerations can be viewed as applications of simple results in linear algebra.
Notes:
Lattice paths are defined and the sources and sinks are restricted so that any nonintersecting family must be an even permutation from source indices to sink indices, usually the identity.
Others independently discovered this result, e.g., Karlin and McGregor.
The same idea applies to Brownian motion.
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The AMS has a new book out, Jiri Matousek, Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra. Info at http://www.ams.org/bookstore-getitem/item=STML-53
"This volume contains a collection of clever mathematical applications of linear algebra, mainly in combinatorics, geometry, and algorithms."
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Thanks Gerry. The book looks good, I'll probably pick it up. I really enjoyed Using the Borsuk-Ulam Theorem by the same author. – Tony Huynh Mar 10 2010 at 21:28
Hoffman and Singleton proved that a regular graph with girth 5 and diameter 2 has to have degree 2, 3, 7, or 57. If I recall correctly, the proof used spectral properties of the adjacency matrix to produce some non-polynomial equation for which these were the integer solutions.
There are unique examples of the first three cases: degree 2 is a pentagon, degree 3 is the Petersen graph, and degree 7 is the Hoffman-Singleton graph. The existence of the degree 57 graph is still open (as far as I know).
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The proof that every $n\times n$ semi-magic square can be written as an integer linear combination of $n^2-2n+2$ permutation matrices.
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There is nice linear-algebra proof of the following result in discreet geometry:
Any $n$ lines in general position cut from the plane at least $n-2$ triangles.
See А. Я. Белов Об одной задаче комбинаторной геометрии (thanks to Arseny and Garry). You will find there more examples of such problems.
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Alexey Kanel-Belov – akopyan Mar 4 2010 at 3:29
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Perhaps the paper in question is A problem in combinatorial geometry, Uspekhi Mat Nauk 47 (1992), no. 3 (285), 151-152, translation in Russian Math Surveys 47 (1992), no. 3, 167-168, MR 93h:52016. Maybe it's also in A Kanel'-Belov and A Kovaldzhi, Taking on triangles: in search of answers between the lines, Quantum 11 (2001), no. 4, 10-16. – Gerry Myerson Mar 4 2010 at 4:02
Here is a link to a Tricki article that has some further examples.
http://www.tricki.org/article/Dimension_arguments_in_combinatorics
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This article is very nice, for the trick and also the exposition. Thanks!. A curiosity, who is its author? Sorry, but the author isn't listed at the site. Maybe you? :-) – Anonymous Mar 4 2010 at 22:14
Actually yes ... It's possible to see who Tricki articles are by if you look at their revision history. – gowers Mar 5 2010 at 11:39
It's not quite what you have asked for, but very close:
Some facts - and proofs! - in combinatorics can be interpreted as linear algebra over the "field with one element". In this very nicely written article Henry Cohn gives a concrete meaning to this and shows how to make a proof from linear algebra into a proof about a combinatorical statement by rephrasing it into axiomatic projective geometry.
(by the way: Lior's answer is an instance of linear algebra over field with one element)
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These references may be more shallow than you desired, but they are both fun and lucid.
1) Noga Alon's Tools From Higher Algebra contains many things (or at least references to those things) that only require linear algebra at heart, such as Rayleigh's Principle.
2) A Course in Combinatorics by van Lint and Wilson is laced with gems in self-contained sections, such that each page is an adventure. You'll find Lots of techniques here that only require linear algebra, including the awkward-looking "interlacing property" of eigenvalues that have popped up way too much for me to ignore by now.
My favorite is actually the aforementioned Babai/Frankl manuscript, which is still very readable and useful. In theory you can still get it; in practice more difficult. First time I tried to order it I didn't get a reply at all.
-Yan
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Thanks. The book by van Lint and Wilson is indeed one of my favorites. You're quite right that it's hard to ignore (no matter how hard I try) interlacing eigenvalues. – Tony Huynh Mar 10 2010 at 16:17
The following is a good illustration:
Let $P$ be a finite set ("points"), and let `$L\subset 2^P$` ("lines") be such that distinct lines intersect in at most one point and any two distinct points are contained in a line. Let $V$ be the real vector space with basis $P$, $W$ the vector space with basis $L$. There are natural linear maps $T\colon V\to W$ and $S\colon W\to V$ mapping every point to the sum of the lines containing it, and every line to the sum of the points in it. Then $ST = J+D-I$ where $J$ is the all-ones matrix (through every two distinct points there is a unique line), $I$ the identity matrix and $D$ is diagonal with entries counting the lines through each point.
Assume that not all points are collinear. Then all the diagonal entries of $D-I$ are at least one; it is then easy to verify that the determinant of $ST$ is positive, and conclude that `$|L| \geq |P|$`.
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Here is an example I learned about this month: The edges of the complete graph cannot be partitioned into fewer than $n-1$ complete bipartite graphs. Apparently the only known proofs involve linear algebra.
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I should add that Richard Brualdi, from whom I learned this, will be writing a book over the course of the next year largely on the connections between linear algebra and graph theory. – Tracy Hall Jul 27 2010 at 13:27
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Recently Sundar Vishwanathan gave a combinatorial proof: arxiv.org/abs/1007.1553 – Konrad Swanepoel Sep 7 2010 at 13:57
Nice proof in arxiv.org/abs/1007.1553 ! (Typo: $\sum_{i=1}^n$ should be $\sum_{i=1}^{n-2}$ in the last formula of the proof; also, $k > n^n$ can just as well be replaced by $k > n^{n-1}$.) Can something similar be done with this problem: artofproblemsolving.com/Forum/… ? – darij grinberg Mar 27 2012 at 23:05
The polynomial method in combinatorial incidence geometry relies crucially on linear algebra to locate a non-trivial polynomial of controlled degree that vanishes at a specified set of points. A good example of the method in action is Dvir's proof of the finite field Kakeya conjecture, see e.g. http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/ .
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This is also in Matousek's book mentioned by Gerry Myerson. – Richard Stanley Mar 28 2012 at 0:26
This is a crosspost from http://mathoverflow.net/questions/33911/, suggested by Kevin O'Bryant. I think it's relevant here. Everything below is verbatim from the earlier post.
My favorite application of linear algebra, as introduced to me by Fan Chung, is Oddtown (which I learned about from a manuscript of Lovasz, but may not be due to him).
The $n$ residents of Oddtown love to form clubs; call the family of these $\mathcal{F}$. If $F_1$ and $F_2$ are in $\mathcal{F}$, then $|F_1|$ must be odd (this is Oddtown!) and $|F_1 \cap F_2|$ must be even unless $F_1 = F_2$ ($\scriptsize{go\;Oddtown?}$). The question is, how many clubs may these $n$ people form?
The answer (taken from Tibor Szabó's lecture notes) is this:
Let $\mathcal{F} = {F_1,\ldots,F_m} \subseteq 2^{[n]}$ be a set of clubs in Oddtown. Let $\mathbf{v}_i \in \{0,1\}^n$ be the characteristic vector of $F_i$; the $j$th coordinate is 1 iff $j \in F_i$.
Note that $\mathbf{v}_i^T \mathbf{v}_j = |F_i \cap F_j|$.
Now, $\mathbf{v}_1,\ldots,\mathbf{v}_m$ is independent over $\mathbb{F}^n_2$: if $\lambda_1\mathbf{v}_1 + \cdots + \lambda_m\mathbf{v}_m = 0$, then for each $i$ we have $$0 \;=\; (\lambda_1\mathbf{v}_1 + \cdots + \lambda_m\mathbf{v}_m)^T\mathbf{v}_i \;=\; \lambda_1\mathbf{v}_1^T\mathbf{v}_i + \cdots + \lambda_i\mathbf{v}_i^T\mathbf{v}_i + \ldots + \lambda_m\mathbf{v}_m^T\mathbf{v}_i \;=\; \lambda_i$$
Since $\mathbf{v}_1,\ldots,\mathbf{v}_m$ are linearly independent vectors over $\mathbb{F}^n_2$, $m \leq n$, and Oddtown can have at most $n$ clubs.
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There is also a book in Russian, Линейно-алгебраический метод в комбинаторике by Raygorodsky, that deals with this.
http://www.ozon.ru/context/detail/id/3625051/
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http://www.cfd-online.com/W/index.php?title=Introduction_to_turbulence/Statistical_analysis/Estimation_from_a_finite_number_of_realizations&diff=7839&oldid=7838
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# Introduction to turbulence/Statistical analysis/Estimation from a finite number of realizations
### From CFD-Wiki
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| | A procedure for answering these questions will be illustrated by considerind a simple '''estimator''' for the mean, the arithmetic mean considered above, <math>X_{N}</math>. For <math>N</math> independent realizations <math>x_{n}, n=1,2,...,N</math> where <math>N</math> is finite, <math>X_{N}</math> is given by: | | A procedure for answering these questions will be illustrated by considerind a simple '''estimator''' for the mean, the arithmetic mean considered above, <math>X_{N}</math>. For <math>N</math> independent realizations <math>x_{n}, n=1,2,...,N</math> where <math>N</math> is finite, <math>X_{N}</math> is given by: |
| - | | + | |
| - | <table width="100%"><tr><td> | + | :<math>X_{N}=\frac{1}{N}\sum^{N}_{n=1} x_{n}</math> |
| - | :<math> | + | |
| - | X_{N}=\frac{1}{N}\sum^{N}_{n=1} x_{n} | + | |
| - | </math> | + | |
| - | </td><td width="5%">(2)</td></tr></table> | + | |
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| | <font color="orange" size="3">Figure 2.9 not uploaded yet</font> | | <font color="orange" size="3">Figure 2.9 not uploaded yet</font> |
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| | Now, as we observed in our simple coin-flipping experiment, since the <math>x_{n}</math> are random, so must be the value of the estimator <math>X_{N}</math>. For the estimator to be ''unbiased'', the mean value of <math>X_{N}</math> must be true ensemble mean, <math>X</math>, i.e. | | Now, as we observed in our simple coin-flipping experiment, since the <math>x_{n}</math> are random, so must be the value of the estimator <math>X_{N}</math>. For the estimator to be ''unbiased'', the mean value of <math>X_{N}</math> must be true ensemble mean, <math>X</math>, i.e. |
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| - | <table width="100%"><tr><td> | + | :<math>\lim_{N\rightarrow\infty} X_{N} = X</math> |
| - | :<math> | + | |
| - | \lim_{N\rightarrow\infty} X_{N} = X | + | |
| - | </math> | + | |
| - | </td><td width="5%">(2)</td></tr></table> | + | |
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| | It is easy to see that since the operations of averaging adding commute, | | It is easy to see that since the operations of averaging adding commute, |
| | | | |
| - | <table width="100%"><tr><td> | | |
| | :<math> | | :<math> |
| | \begin{matrix} | | \begin{matrix} |
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| | \end{matrix} | | \end{matrix} |
| | </math> | | </math> |
| - | </td><td width="5%">(2)</td></tr></table> | | |
| | | | |
| | (Note that the expected value of each <math>x_{n}</math> is just <math>X</math> since the <math>x_{n}</math> are assumed identically distributed). Thus <math>x_{N}</math> is, in fact, an ''unbiased estimator for the mean''. | | (Note that the expected value of each <math>x_{n}</math> is just <math>X</math> since the <math>x_{n}</math> are assumed identically distributed). Thus <math>x_{N}</math> is, in fact, an ''unbiased estimator for the mean''. |
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| | The question of ''convergence'' of the estimator can be addressed by defining the square of '''variability of the estimator''', say <math>\epsilon^{2}_{X_{N}}</math>, to be: | | The question of ''convergence'' of the estimator can be addressed by defining the square of '''variability of the estimator''', say <math>\epsilon^{2}_{X_{N}}</math>, to be: |
| | | | |
| - | <table width="100%"><tr><td> | | |
| | :<math> | | :<math> |
| | \epsilon^{2}_{X_{N}}\equiv \frac{var \left\{ X_{N} \right\} }{X^{2}} = \frac{\left\langle \left( X_{N}- X \right)^{2} \right\rangle }{X^{2}} | | \epsilon^{2}_{X_{N}}\equiv \frac{var \left\{ X_{N} \right\} }{X^{2}} = \frac{\left\langle \left( X_{N}- X \right)^{2} \right\rangle }{X^{2}} |
| | </math> | | </math> |
| - | </td><td width="5%">(2)</td></tr></table> | | |
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| | Now we want to examine what happens to <math>\epsilon_{X_{N}}</math> as the number of realizations increases. For the estimator to converge it is clear that <math>\epsilon_{x}</math> should decrease as the number of sample increases. Obviously, we need to examine the variance of <math>X_{N}</math> first. It is given by: | | Now we want to examine what happens to <math>\epsilon_{X_{N}}</math> as the number of realizations increases. For the estimator to converge it is clear that <math>\epsilon_{x}</math> should decrease as the number of sample increases. Obviously, we need to examine the variance of <math>X_{N}</math> first. It is given by: |
| | | | |
| - | <table width="100%"><tr><td> | | |
| | :<math> | | :<math> |
| | \begin{matrix} | | \begin{matrix} |
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| | \end{matrix} | | \end{matrix} |
| | </math> | | </math> |
| - | </td><td width="5%">(2)</td></tr></table> | | |
| | | | |
| - | since <math>\left\langle X_{N} \right\rangle = X</math> from equation 2.46. Using the fact that operations of averaging and summation commute, the squared summation can be expanded as follows: | + | since <math>\left\langle X_{N} \right\rangle = X</math> from the equation for <math>\langle X_{N} \rangle</math> above. Using the fact that operations of averaging and summation commute, the squared summation can be expanded as follows: |
| | | | |
| - | <table width="100%"><tr><td> | | |
| | :<math> | | :<math> |
| | \begin{matrix} | | \begin{matrix} |
| Line 72: | | Line 57: | |
| | \end{matrix} | | \end{matrix} |
| | </math> | | </math> |
| - | </td><td width="5%">(2)</td></tr></table> | | |
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| - | where the next to last step follows from the fact that the <math>x_{n}</math> are assumed to be statistically independent samples (and hence uncorrelated), and the last step from the definition of the variance. It follows immediately by substitution into equation 2.49 that the square of the variability of the estimator, <math>X_{N}</math>, is given by: | + | where the next to last step follows from the fact that the <math>x_{n}</math> are assumed to be statistically independent samples (and hence uncorrelated), and the last step from the definition of the variance. It follows immediately by substitution into the equation for <math>\epsilon^{2}_{X_{N}}</math> above that the square of the variability of the estimator, <math>X_{N}</math>, is given by: |
| | | | |
| - | <table width="100%"><tr><td> | | |
| | :<math> | | :<math> |
| | \begin{matrix} | | \begin{matrix} |
| Line 83: | | Line 66: | |
| | \end{matrix} | | \end{matrix} |
| | </math> | | </math> |
| - | </td><td width="5%">(2)</td></tr></table> | | |
| | | | |
| | Thus ''the variability of the stimator depends inversely on the number of independent realizations, <math>N</math>, and linearly on the relative fluctuation level of the random variable itself <math>\sigma_{x}/ X</math>''. Obviously if the relative fluctuation level is zero (either because there the quantity being measured is constant and there are no measurement errors), then a single measurement will suffice. On the other hand, as soon as there is any fluctuation in the <math>x</math> itself, the greater the fluctuation ( relative to the mean of <math>x</math>, <math>\left\langle x \right\rangle = X</math> ), then the more independent samples it will take to achieve a specified accuracy. | | Thus ''the variability of the stimator depends inversely on the number of independent realizations, <math>N</math>, and linearly on the relative fluctuation level of the random variable itself <math>\sigma_{x}/ X</math>''. Obviously if the relative fluctuation level is zero (either because there the quantity being measured is constant and there are no measurement errors), then a single measurement will suffice. On the other hand, as soon as there is any fluctuation in the <math>x</math> itself, the greater the fluctuation ( relative to the mean of <math>x</math>, <math>\left\langle x \right\rangle = X</math> ), then the more independent samples it will take to achieve a specified accuracy. |
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| | '''Example:''' In a given ensemble the relative fluctuation level is 12% (i.e. <math>\sigma_{x}/ X = 0.12</math>). What is the fewest number of independent samples that must be acquired to measure the mean value to within 1%? | | '''Example:''' In a given ensemble the relative fluctuation level is 12% (i.e. <math>\sigma_{x}/ X = 0.12</math>). What is the fewest number of independent samples that must be acquired to measure the mean value to within 1%? |
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| - | '''Answer'''Using equation 2.52, and taking <math>\epsilon_{X_{N}}=0.01</math>, it follows that: | + | '''Answer'''Using the equation for <math>\epsilon^{2}_{X_{N}}</math> above, and taking <math>\epsilon_{X_{N}}=0.01</math>, it follows that: |
| | | | |
| - | | | |
| - | <table width="100%"><tr><td> | | |
| | :<math> | | :<math> |
| | \left(0.01 \right)^{2} = \frac{1}{N}\left(0.12 \right)^{2} | | \left(0.01 \right)^{2} = \frac{1}{N}\left(0.12 \right)^{2} |
| | </math> | | </math> |
| - | </td><td width="5%">(2)</td></tr></table> | | |
| | | | |
| | or <math>N \geq 144</math>. | | or <math>N \geq 144</math>. |
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| | | + | | [[Statistical analysis in turbulence|Up to statistical analysis]] | [[Multivariate random variables|Back to multivariate random variables]] | [[Generalization to the estimator of any quantity|Forward to generalization to the estimator of any quantity]] |
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## Estimators for averaged quantities
Since there can never an infinite number of realizations from which ensemble averages (and probability densities) can be computed, it is essential to ask: How many realizations are enough? The answer to this question must be sought by looking at the statistical properties of estimators based on a finite number of realization. There are two questions which must be answered. The first one is:
• Is the expected value (or mean value) of the estimator equal to the true ensemble mean? Or in other words, is yje estimator unbiased?
The second question is
• Does the difference between the and that of the true mean decrease as the number of realizations increases? Or in other words, does the estimator converge in a statistical sense (or converge in probability). Figure 2.9 illustrates the problems which can arise.
## Bias and convergence of estimators
A procedure for answering these questions will be illustrated by considerind a simple estimator for the mean, the arithmetic mean considered above, $X_{N}$. For $N$ independent realizations $x_{n}, n=1,2,...,N$ where $N$ is finite, $X_{N}$ is given by:
$X_{N}=\frac{1}{N}\sum^{N}_{n=1} x_{n}$
Figure 2.9 not uploaded yet
Now, as we observed in our simple coin-flipping experiment, since the $x_{n}$ are random, so must be the value of the estimator $X_{N}$. For the estimator to be unbiased, the mean value of $X_{N}$ must be true ensemble mean, $X$, i.e.
$\lim_{N\rightarrow\infty} X_{N} = X$
It is easy to see that since the operations of averaging adding commute,
$\begin{matrix} \left\langle X_{N} \right\rangle & = & \left\langle \frac{1}{N} \sum^{N}_{n=1} x_{n} \right\rangle \\ & = & \frac{1}{N} \sum^{N}_{n=1} \left\langle x_{n} \right\rangle \\ & = & \frac{1}{N} NX = X \\ \end{matrix}$
(Note that the expected value of each $x_{n}$ is just $X$ since the $x_{n}$ are assumed identically distributed). Thus $x_{N}$ is, in fact, an unbiased estimator for the mean.
The question of convergence of the estimator can be addressed by defining the square of variability of the estimator, say $\epsilon^{2}_{X_{N}}$, to be:
$\epsilon^{2}_{X_{N}}\equiv \frac{var \left\{ X_{N} \right\} }{X^{2}} = \frac{\left\langle \left( X_{N}- X \right)^{2} \right\rangle }{X^{2}}$
Now we want to examine what happens to $\epsilon_{X_{N}}$ as the number of realizations increases. For the estimator to converge it is clear that $\epsilon_{x}$ should decrease as the number of sample increases. Obviously, we need to examine the variance of $X_{N}$ first. It is given by:
$\begin{matrix} var \left\{ X_{N} \right\} & = & \left\langle X_{N} - X^{2} \right\rangle \\ & = & \left\langle \left[ \lim_{N\rightarrow\infty} \frac{1}{N} \sum^{N}_{n=1} \left( x_{n} - X \right) \right]^{2} \right\rangle - X^{2}\\ \end{matrix}$
since $\left\langle X_{N} \right\rangle = X$ from the equation for $\langle X_{N} \rangle$ above. Using the fact that operations of averaging and summation commute, the squared summation can be expanded as follows:
$\begin{matrix} \left\langle \left[ \lim_{N\rightarrow\infty} \sum^{N}_{n=1} \left( x_{n} - X \right) \right]^{2} \right\rangle & = & \lim_{N\rightarrow\infty}\frac{1}{N^{2}} \sum^{N}_{n=1} \sum^{N}_{m=1} \left\langle \left( x_{n} - X \right) \left( x_{m} - X \right) \right\rangle \\ & = & \lim_{N\rightarrow\infty}\frac{1}{N^{2}}\sum^{N}_{n=1}\left\langle \left( x_{n} - X \right)^{2} \right\rangle \\ & = & \frac{1}{N} var \left\{ x \right\} \\ \end{matrix}$
where the next to last step follows from the fact that the $x_{n}$ are assumed to be statistically independent samples (and hence uncorrelated), and the last step from the definition of the variance. It follows immediately by substitution into the equation for $\epsilon^{2}_{X_{N}}$ above that the square of the variability of the estimator, $X_{N}$, is given by:
$\begin{matrix} \epsilon^{2}_{X_{N}}& =& \frac{1}{N}\frac{var\left\{x\right\}}{X^{2}} \\ & = & \frac{1}{N} \left[ \frac{\sigma_{x}}{X} \right]^{2} \\ \end{matrix}$
Thus the variability of the stimator depends inversely on the number of independent realizations, $N$, and linearly on the relative fluctuation level of the random variable itself $\sigma_{x}/ X$. Obviously if the relative fluctuation level is zero (either because there the quantity being measured is constant and there are no measurement errors), then a single measurement will suffice. On the other hand, as soon as there is any fluctuation in the $x$ itself, the greater the fluctuation ( relative to the mean of $x$, $\left\langle x \right\rangle = X$ ), then the more independent samples it will take to achieve a specified accuracy.
Example: In a given ensemble the relative fluctuation level is 12% (i.e. $\sigma_{x}/ X = 0.12$). What is the fewest number of independent samples that must be acquired to measure the mean value to within 1%?
AnswerUsing the equation for $\epsilon^{2}_{X_{N}}$ above, and taking $\epsilon_{X_{N}}=0.01$, it follows that:
$\left(0.01 \right)^{2} = \frac{1}{N}\left(0.12 \right)^{2}$
or $N \geq 144$.
## Credits
This text was based on "Lectures in Turbulence for the 21st Century" by Professor William K. George, Professor of Turbulence, Chalmers University of Technology, Gothenburg, Sweden.
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Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Tag Archive
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## Szemeredi’s regularity lemma via random partitions
26 April, 2009 in expository, math.CO, math.PR | Tags: random neighbourhoods, szemeredi regularity lemma | by Terence Tao | 7 comments
In the theory of dense graphs on ${n}$ vertices, where ${n}$ is large, a fundamental role is played by the Szemerédi regularity lemma:
Lemma 1 (Regularity lemma, standard version) Let ${G = (V,E)}$ be a graph on ${n}$ vertices, and let ${\epsilon > 0}$ and ${k_0 \geq 0}$. Then there exists a partition of the vertices ${V = V_1 \cup \ldots \cup V_k}$, with ${k_0 \leq k \leq C(k_0,\epsilon)}$ bounded below by ${k_0}$ and above by a quantity ${C(k_0,\epsilon)}$ depending only on ${k_0, \epsilon}$, obeying the following properties:
• (Equitable partition) For any ${1 \leq i,j \leq k}$, the cardinalities ${|V_i|, |V_j|}$ of ${V_i}$ and ${V_j}$ differ by at most ${1}$.
• (Regularity) For all but at most ${\epsilon k^2}$ pairs ${1 \leq i < j \leq k}$, the portion of the graph ${G}$ between ${V_i}$ and ${V_j}$ is ${\epsilon}$-regular in the sense that one has
$\displaystyle |d( A, B ) - d( V_i, V_j )| \leq \epsilon$
for any ${A \subset V_i}$ and ${B \subset V_j}$ with ${|A| \geq \epsilon |V_i|, |B| \geq \epsilon |V_j|}$, where ${d(A,B) := |E \cap (A \times B)|/|A| |B|}$ is the density of edges between ${A}$ and ${B}$.
This lemma becomes useful in the regime when ${n}$ is very large compared to ${k_0}$ or ${1/\epsilon}$, because all the conclusions of the lemma are uniform in ${n}$. Very roughly speaking, it says that “up to errors of size ${\epsilon}$“, a large graph can be more or less described completely by a bounded number of quantities ${d(V_i, V_j)}$. This can be interpreted as saying that the space of all graphs is totally bounded (and hence precompact) in a suitable metric space, thus allowing one to take formal limits of sequences (or subsequences) of graphs; see for instance this paper of Lovasz and Szegedy for a discussion.
For various technical reasons it is easier to work with a slightly weaker version of the lemma, which allows for the cells ${V_1,\ldots,V_k}$ to have unequal sizes:
Lemma 2 (Regularity lemma, weighted version) Let ${G = (V,E)}$ be a graph on ${n}$ vertices, and let ${\epsilon > 0}$. Then there exists a partition of the vertices ${V = V_1 \cup \ldots \cup V_k}$, with ${1 \leq k \leq C(\epsilon)}$ bounded above by a quantity ${C(\epsilon)}$ depending only on ${\epsilon}$, obeying the following properties:
• (Regularity) One has
$\displaystyle \sum_{(V_i,V_j) \hbox{ not } \epsilon-\hbox{regular}} |V_i| |V_j| = O(\epsilon |V|^2) \ \ \ \ \ (1)$
where the sum is over all pairs ${1 \leq i \leq j \leq k}$ for which ${G}$ is not ${\epsilon}$-regular between ${V_i}$ and ${V_j}$.
While Lemma 2 is, strictly speaking, weaker than Lemma 1 in that it does not enforce the equitable size property between the atoms, in practice it seems that the two lemmas are roughly of equal utility; most of the combinatorial consequences of Lemma 1 can also be proven using Lemma 2. The point is that one always has to remember to weight each cell ${V_i}$ by its density ${|V_i|/|V|}$, rather than by giving each cell an equal weight as in Lemma 1. Lemma 2 also has the advantage that one can easily generalise the result from finite vertex sets ${V}$ to other probability spaces (for instance, one could weight ${V}$ with something other than the uniform distribution). For applications to hypergraph regularity, it turns out to be slightly more convenient to have two partitions (coarse and fine) rather than just one; see for instance my own paper on this topic. In any event the arguments below that we give to prove Lemma 2 can be modified to give a proof of Lemma 1 also. The proof of the regularity lemma is usually conducted by a greedy algorithm. Very roughly speaking, one starts with the trivial partition of ${V}$. If this partition already regularises the graph, we are done; if not, this means that there are some sets ${A}$ and ${B}$ in which there is a significant density fluctuation beyond what has already been detected by the original partition. One then adds these sets to the partition and iterates the argument. Every time a new density fluctuation is incorporated into the partition that models the original graph, this increases a certain “index” or “energy” of the partition. On the other hand, this energy remains bounded no matter how complex the partition, so eventually one must reach a long “energy plateau” in which no further refinement is possible, at which point one can find the regular partition.
One disadvantage of the greedy algorithm is that it is not efficient in the limit ${n \rightarrow \infty}$, as it requires one to search over all pairs of subsets ${A, B}$ of a given pair ${V_i, V_j}$ of cells, which is an exponentially long search. There are more algorithmically efficient ways to regularise, for instance a polynomial time algorithm was given by Alon, Duke, Lefmann, Rödl, and Yuster. However, one can do even better, if one is willing to (a) allow cells of unequal size, (b) allow a small probability of failure, (c) have the ability to sample vertices from ${G}$ at random, and (d) allow for the cells to be defined “implicitly” (via their relationships with a fixed set of reference vertices) rather than “explicitly” (as a list of vertices). In that case, one can regularise a graph in a number of operations bounded in ${n}$. Indeed, one has
Lemma 3 (Regularity lemma via random neighbourhoods) Let ${\epsilon > 0}$. Then there exists integers ${M_1,\ldots,M_m}$ with the following property: whenever ${G = (V,E)}$ be a graph on finitely many vertices, if one selects one of the integers ${M_r}$ at random from ${M_1,\ldots,M_m}$, then selects ${M_r}$ vertices ${v_1,\ldots,v_{M_r} \in V}$ uniformly from ${V}$ at random, then the ${2^{M_r}}$ vertex cells ${V^{M_r}_1,\ldots,V^{M_r}_{2^{M_r}}}$ (some of which can be empty) generated by the vertex neighbourhoods ${A_t := \{ v \in V: (v,v_t) \in E \}}$ for ${1 \leq t \leq M_r}$, will obey the conclusions of Lemma 2 with probability at least ${1-O(\epsilon)}$.
Thus, roughly speaking, one can regularise a graph simply by taking a large number of random vertex neighbourhoods, and using the partition (or Venn diagram) generated by these neighbourhoods as the partition. The intuition is that if there is any non-uniformity in the graph (e.g. if the graph exhibits bipartite behaviour), this will bias the random neighbourhoods to seek out the partitions that would regularise that non-uniformity (e.g. vertex neighbourhoods would begin to fill out the two vertex cells associated to the bipartite property); if one takes sufficiently many such random neighbourhoods, the probability that all detectable non-uniformity is captured by the partition should converge to ${1}$. (It is more complicated than this, because the finer one makes the partition, the finer the types of non-uniformity one can begin to detect, but this is the basic idea.)
This fact seems to be reasonably well-known folklore, discovered independently by many authors; it is for instance quite close to the graph property testing results of Alon and Shapira, and also appears implicitly in a paper of Ishigami, as well as a paper of Austin (and perhaps even more implicitly in a paper of myself). However, in none of these papers is the above lemma stated explicitly. I was asked about this lemma recently, so I decided to provide a proof here.
Read the rest of this entry »
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Thermodynamics
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Thermodynamics is a branch of natural science concerned with heat and its relation to energy and work. It defines macroscopic variables (such as temperature, internal energy, entropy, and pressure) that characterize materials and radiation, and explains how they are related and by what laws they change with time. Thermodynamics describes the average behavior of very large numbers of microscopic constituents, and its laws can be derived from statistical mechanics.
Thermodynamics applies to a wide variety of topics in science and engineering—such as engines, phase transitions, chemical reactions, transport phenomena, and even black holes. Results of thermodynamic calculations are essential for other fields of physics and for chemistry, chemical engineering, aerospace engineering, mechanical engineering, cell biology, biomedical engineering, and materials science—and useful in other fields such as economics.[1][2]
Much of the empirical content of thermodynamics is contained in the four laws. The first law asserts the existence of a quantity called the internal energy of a system, which is distinguishable from the kinetic energy of bulk movement of the system and from its potential energy with respect to its surroundings. The first law distinguishes transfers of energy between closed systems as heat and as work.[3][4][5] The second law concerns two quantities called temperature and entropy. Entropy expresses the limitations, arising from what is known as irreversibility, on the amount of thermodynamic work that can be delivered to an external system by a thermodynamic process.[6] Temperature, whose properties are also partially described by the zeroth law of thermodynamics, quantifies the direction of energy flow as heat between two systems in thermal contact and quantifies the common-sense notions of "hot" and "cold".
Historically, thermodynamics developed out of a desire to increase the efficiency of early steam engines, particularly through the work of French physicist Nicolas Léonard Sadi Carnot (1824) who believed that the efficiency of heat engines was the key that could help France win the Napoleonic Wars.[7] Irish-born British physicist Lord Kelvin was the first to formulate a concise definition of thermodynamics in 1854:[8]
Thermo-dynamics is the subject of the relation of heat to forces acting between contiguous parts of bodies, and the relation of heat to electrical agency.
Initially, the thermodynamics of heat engines concerned mainly the thermal properties of their 'working materials', such as steam. This concern was then linked to the study of energy transfers in chemical processes, for example to the investigation, published in 1840, of the heats of chemical reactions[9] by Germain Hess, which was not originally explicitly concerned with the relation between energy exchanges by heat and work. Chemical thermodynamics studies the role of entropy in chemical reactions.[10][11][12][13][14][15][16][17][18] Also, statistical thermodynamics, or statistical mechanics, gave explanations of macroscopic thermodynamics by statistical predictions of the collective motion of particles based on the mechanics of their microscopic behavior.
## Introduction
The plain term 'thermodynamics' refers to macroscopic description of bodies and processes.[19] "Any reference to atomic constitution is foreign to classical thermodynamics."[20] The qualified term 'statistical thermodynamics' refers to descriptions of bodies and processes in terms of the atomic constitution of matter, mainly described by sets of items all alike, so as to have equal probabilities.
Thermodynamics arose from the study of energy transfers that can be strictly resolved into two distinct components, heat and work, specified by macroscopic variables.[21][22]
Thermodynamic equilibrium is one of the most important concepts for thermodynamics.[23] The temperature of a system in thermodynamic equilibrium is well defined, and is perhaps the most characteristic quantity of thermodynamics. As the systems and processes of interest are taken further from thermodynamic equilibrium, their exact thermodynamical study becomes more difficult. Relatively simple approximate calculations, however, using the variables of equilibrium thermodynamics, are of much practical value in engineering. In many important practical cases, such as heat engines or refrigerators, the systems consist of many subsystems at different temperatures and pressures. In practice, thermodynamic calculations deal effectively with these complicated dynamic systems provided the equilibrium thermodynamic variables are nearly enough well-defined.
Basic for thermodynamics are the concepts of system and surroundings.[14][24] The surroundings of a thermodynamic system are other thermodynamic systems that can interact with it. An example of a thermodynamic surrounding is a heat bath, which is considered to be held at a prescribed temperature, regardless of the interactions it might have with the system.
There are two fundamental kinds of entity in thermodynamics, states of a system, and processes of a system. This allows two fundamental approaches to thermodynamic reasoning, that in terms of states of a system, and that in terms of cyclic processes of a system.
A thermodynamic system can be defined in terms of its states. In this way, a thermodynamic system is a macroscopic physical object, explicitly specified in terms of macroscopic physical and chemical variables that describe its macroscopic properties. The macroscopic state variables of thermodynamics have been recognized in the course of empirical work in physics and chemistry.[15]
A thermodynamic system can also be defined in terms of the processes it can undergo. Of particular interest are cyclic processes. This was the way of the founders of thermodynamics in the first three quarters of the nineteenth century.
For thermodynamics and statistical thermodynamics to apply to a process in a body, it is necessary that the atomic mechanisms of the process fall into just two classes:
• those so rapid that, in the time frame of the process of interest, the atomic states effectively visit all of their accessible range, bringing the system to its state of internal thermodynamic equilibrium; and
• those so slow that their progress can be neglected in the time frame of the process of interest.[25][26]
The rapid atomic mechanisms mediate the macroscopic changes that are of interest for thermodynamics and statistical thermodynamics, because they quickly bring the system near enough to thermodynamic equilibrium. "When intermediate rates are present, thermodynamics and statistical mechanics cannot be applied."[25] Such intermediate rate atomic processes do not bring the system near enough to thermodynamic equilibrium in the time frame of the macroscopic process of interest. This separation of time scales of atomic processes is a theme that recurs throughout the subject.
For example, classical thermodynamics is characterized by its study of materials that have equations of state or characteristic equations. They express relations between macroscopic mechanical variables and temperature that are reached much more rapidly than the progress of any imposed changes in the surroundings, and are in effect variables of state for thermodynamic equilibrium. They express the constitutive peculiarities of the material of the system. A classical material can usually be described by a function that makes pressure dependent on volume and temperature, the resulting pressure being established much more rapidly than any imposed change of volume or temperature.[27][28][29][30]
The present article takes a gradual approach to the subject, starting with a focus on cyclic processes and thermodynamic equilibrium, and then gradually beginning to further consider non-equilibrium systems.
Thermodynamic facts can often be explained by viewing macroscopic objects as assemblies of very many microscopic or atomic objects that obey Hamiltonian dynamics.[14][31][32] The microscopic or atomic objects exist in species, the objects of each species being all alike. Because of this likeness, statistical methods can be used to account for the macroscopic properties of the thermodynamic system in terms of the properties of the microscopic species. Such explanation is called statistical thermodynamics; also often it is also referred to by the term 'statistical mechanics', though this term can have a wider meaning, referring to 'microscopic objects', such as economic quantities, that do not obey Hamiltonian dynamics.[31]
The thermodynamicists representative of the original eight founding schools of thermodynamics. The schools with the most-lasting effect in founding the modern versions of thermodynamics are the Berlin school, particularly as established in Rudolf Clausius’s 1865 textbook The Mechanical Theory of Heat, the Vienna school, with the statistical mechanics of Ludwig Boltzmann, and the Gibbsian school at Yale University, American engineer Willard Gibbs' 1876 On the Equilibrium of Heterogeneous Substances launching chemical thermodynamics.[33]
## History
The history of thermodynamics as a scientific discipline generally begins with Otto von Guericke who, in 1650, built and designed the world's first vacuum pump and demonstrated a vacuum using his Magdeburg hemispheres. Guericke was driven to make a vacuum in order to disprove Aristotle's long-held supposition that 'nature abhors a vacuum'. Shortly after Guericke, the physicist and chemist Robert Boyle had learned of Guericke's designs and, in 1656, in coordination with scientist Robert Hooke, built an air pump.[34] Using this pump, Boyle and Hooke noticed a correlation between pressure, temperature, and volume. In time, Boyle's Law was formulated, stating that for a gas at constant temperature, its pressure and volume are inversely proportional. In 1679, based on these concepts, an associate of Boyle's named Denis Papin built a steam digester, which was a closed vessel with a tightly fitting lid that confined steam until a high pressure was generated.
Later designs implemented a steam release valve that kept the machine from exploding. By watching the valve rhythmically move up and down, Papin conceived of the idea of a piston and a cylinder engine. He did not, however, follow through with his design. Nevertheless, in 1697, based on Papin's designs, engineer Thomas Savery built the first engine, followed by Thomas Newcomen in 1712. Although these early engines were crude and inefficient, they attracted the attention of the leading scientists of the time.
The concepts of heat capacity and latent heat, which were necessary for development of thermodynamics, were developed by professor Joseph Black at the University of Glasgow, where James Watt worked as an instrument maker. Watt consulted with Black on tests of his steam engine, but it was Watt who conceived the idea of the external condenser, greatly raising the steam engine's efficiency.[35] Drawing on all the previous work led Sadi Carnot, the "father of thermodynamics", to publish Reflections on the Motive Power of Fire (1824), a discourse on heat, power, energy and engine efficiency. The paper outlined the basic energetic relations between the Carnot engine, the Carnot cycle, and motive power. It marked the start of thermodynamics as a modern science.[17]
The first thermodynamic textbook was written in 1859 by William Rankine, originally trained as a physicist and a civil and mechanical engineering professor at the University of Glasgow.[36] The first and second laws of thermodynamics emerged simultaneously in the 1850s, primarily out of the works of William Rankine, Rudolf Clausius, and William Thomson (Lord Kelvin).
The foundations of statistical thermodynamics were set out by physicists such as James Clerk Maxwell, Ludwig Boltzmann, Max Planck, Rudolf Clausius and J. Willard Gibbs.
From 1873 to '76, the American mathematical physicist Josiah Willard Gibbs published a series of three papers, the most famous being "On the equilibrium of heterogeneous substances".[10] Gibbs showed how thermodynamic processes, including chemical reactions, could be graphically analyzed. By studying the energy, entropy, volume, chemical potential, temperature and pressure of the thermodynamic system, one can determine if a process would occur spontaneously.[37] Chemical thermodynamics was further developed by Pierre Duhem,[11] Gilbert N. Lewis, Merle Randall,[12] and E. A. Guggenheim,[13][14] who applied the mathematical methods of Gibbs.
The lifetimes of some of the most important contributors to thermodynamics.
### Etymology
The etymology of thermodynamics has an intricate history.[38] It was first spelled in a hyphenated form as an adjective (thermo-dynamic) and from 1854 to 1868 as the noun thermo-dynamics to represent the science of generalized heat engines.[38]
The components of the word thermodynamics are derived from the Greek words θέρμη therme, meaning heat, and δύναμις dynamis, meaning power.[39][40][41]
Pierre Perrot claims that the term thermodynamics was coined by James Joule in 1858 to designate the science of relations between heat and power.[17] Joule, however, never used that term, but used instead the term perfect thermo-dynamic engine in reference to Thomson’s 1849[42] phraseology.[38]
By 1858, thermo-dynamics, as a functional term, was used in William Thomson's paper An Account of Carnot's Theory of the Motive Power of Heat.[42]
## Branches of description
Thermodynamic systems are theoretical constructions used to model physical systems that exchange matter and energy in terms of the laws of thermodynamics. The study of thermodynamical systems has developed into several related branches, each using a different fundamental model as a theoretical or experimental basis, or applying the principles to varying types of systems.
### Classical thermodynamics
Classical thermodynamics accounts for the adventures of thermodynamic systems in terms, either of their time-invariant equilibrium states, or else of their continually repeated cyclic processes, but, formally, not both in the same account. It uses only time-invariant, or equilibrium, macroscopic quantities measurable in the laboratory, counting as time-invariant a long-term time-average of a quantity, such as a flow, generated by a continually repetitive process.[43][44] Classical thermodynamics does not admit change over time as a fundamental factor in its account of processes. An equilibrium state stands endlessly without change over time, while a continually repeated cyclic process runs endlessly without change over time.
In the account in terms of equilibrium states of a system, a state of thermodynamic equilibrium in a simple system (as defined below in this article), with no externally imposed force field, is spatially homogeneous.
In the classical account strictly and purely in terms of cyclic processes, the spatial interior of the 'working body' of a cyclic process is not considered; the 'working body' thus does not have a defined internal thermodynamic state of its own because no assumption is made that it should be in thermodynamic equilibrium; only its inputs and outputs of energy as heat and work are considered.[45] It is of course possible, and indeed common, for the account in terms of equilibrium states of a system to describe cycles composed of indefinitely many equilibrium states.
Classical thermodynamics was originally concerned with the transformation of energy in cyclic processes, and the exchange of energy between closed systems defined only by their equilibrium states. For these, the distinction between transfers of energy as heat and as work was central.
As classical thermodynamics developed, the distinction between heat and work became less central. This was because there was more interest in open systems, for which the distinction between heat and work is not simple, and is beyond the scope of the present article. Alongside amount of heat transferred as a fundamental quantity, entropy, considered below, was gradually found to be a more generally applicable concept, especially when chemical reactions are of interest. Massieu in 1869 considered entropy as the basic dependent thermodynamic variable, with energy potentials and the reciprocal of thermodynamic temperature as fundamental independent variables. Massieu functions can be useful in present-day non-equilibrium thermodynamics. In 1875, in the work of Josiah Willard Gibbs, the basic thermodynamic quantities were energy potentials, such as internal energy, as dependent variables, and entropy, considered as a fundamental independent variable.[46]
All actual physical processes are to some degree irreversible. Classical thermodynamics can consider irreversible processes, but its account in exact terms is restricted to variables that refer only to initial and final states of thermodynamic equilibrium, or to rates of input and output that do not change with time. For example, classical thermodynamics can consider long-time-average rates of flows generated by continually repeated irreversible cyclic processes. Also it can consider irreversible changes between equilibrium states of systems consisting of several phases (as defined below in this article), or with removable or replaceable partitions. But for systems that are described in terms of equilibrium states, it considers neither flows, nor spatial inhomogeneities in simple systems with no externally imposed force field such as gravity. In the account in terms of equilibrium states of a system, descriptions of irreversible processes refer only to initial and final static equilibrium states; rates of progress are not considered.[47][48]
### Local equilibrium thermodynamics
Local equilibrium thermodynamics is concerned with the time courses and rates of progress of irreversible processes in systems that are smoothly spatially inhomogeneous. It admits time as a fundamental quantity, but only in a restricted way. Rather than considering time-invariant flows as long-term-average rates of cyclic processes, local equilibrium thermodynamics considers time-varying flows in systems that are described by states of local thermodynamic equilibrium, as follows.
For processes that involve only suitably small and smooth spatial inhomogeneities and suitably small changes with time, a good approximation can be found through the assumption of local thermodynamic equilibrium. Within the large or global region of a process, for a suitably small local region, this approximation assumes that a quantity known as the entropy of the small local region can be defined in a particular way. That particular way of definition of entropy is largely beyond the scope of the present article, but here it may be said that it is entirely derived from the concepts of classical thermodynamics; in particular, neither flow rates nor changes over time are admitted into the definition of the entropy of the small local region. It is assumed without proof that the instantaneous global entropy of a non-equilibrium system can be found by adding up the simultaneous instantaneous entropies of its constituent small local regions. Local equilibrium thermodynamics considers processes that involve the time-dependent production of entropy by dissipative processes, in which kinetic energy of bulk flow and chemical potential energy are converted into internal energy at time-rates that are explicitly accounted for. Time-varying bulk flows and specific diffusional flows are considered, but they are required to be dependent variables, derived only from material properties described only by static macroscopic equilibrium states of small local regions. The independent state variables of a small local region are only those of classical thermodynamics.
### Generalized or extended thermodynamics
Like local equilibrium thermodynamics, generalized or extended thermodynamics also is concerned with the time courses and rates of progress of irreversible processes in systems that are smoothly spatially inhomogeneous. It describes time-varying flows in terms of states of suitably small local regions within a global region that is smoothly spatially inhomogeneous, rather than considering flows as time-invariant long-term-average rates of cyclic processes. In its accounts of processes, generalized or extended thermodynamics admits time as a fundamental quantity in a more far-reaching way than does local equilibrium thermodynamics. The states of small local regions are defined by macroscopic quantities that are explicitly allowed to vary with time, including time-varying flows. Generalized thermodynamics might tackle such problems as ultrasound or shock waves, in which there are strong spatial inhomogeneities and changes in time fast enough to outpace a tendency towards local thermodynamic equilibrium. Generalized or extended thermodynamics is a diverse and developing project, rather than a more or less completed subject such as is classical thermodynamics.[49][50]
For generalized or extended thermodynamics, the definition of the quantity known as the entropy of a small local region is in terms beyond those of classical thermodynamics; in particular, flow rates are admitted into the definition of the entropy of a small local region. The independent state variables of a small local region include flow rates, which are not admitted as independent variables for the small local regions of local equilibrium thermodynamics.
Outside the range of classical thermodynamics, the definition of the entropy of a small local region is no simple matter. For a thermodynamic account of a process in terms of the entropies of small local regions, the definition of entropy should be such as to ensure that the second law of thermodynamics applies in each small local region. It is often assumed without proof that the instantaneous global entropy of a non-equilibrium system can be found by adding up the simultaneous instantaneous entropies of its constituent small local regions. For a given physical process, the selection of suitable independent local non-equilibrium macroscopic state variables for the construction of a thermodynamic description calls for qualitative physical understanding, rather than being a simply mathematical problem concerned with a uniquely determined thermodynamic description. A suitable definition of the entropy of a small local region depends on the physically insightful and judicious selection of the independent local non-equilibrium macroscopic state variables, and different selections provide different generalized or extended thermodynamical accounts of one and the same given physical process. This is one of the several good reasons for considering entropy as an epistemic physical variable, rather than as a simply material quantity. According to a respected author: "There is no compelling reason to believe that the classical thermodynamic entropy is a measurable property of nonequilibrium phenomena, ..."[51]
### Statistical thermodynamics
Statistical thermodynamics, also called statistical mechanics, emerged with the development of atomic and molecular theories in the second half of the 19th century and early 20th century. It provides an explanation of classical thermodynamics. It considers the microscopic interactions between individual particles and their collective motions, in terms of classical or of quantum mechanics. Its explanation is in terms of statistics that rest on the fact the system is composed of several species of particles or collective motions, the members of each species respectively being in some sense all alike.
## Thermodynamic equilibrium
Equilibrium thermodynamics studies transformations of matter and energy in systems at or near thermodynamic equilibrium. In thermodynamic equilibrium, a system's properties are, by definition, unchanging in time. In thermodynamic equilibrium no macroscopic change is occurring or can be triggered; within the system, every microscopic process is balanced by its opposite; this is called the principle of detailed balance. A central aim in equilibrium thermodynamics is: given a system in a well-defined initial state, subject to specified constraints, to calculate what the equilibrium state of the system is.[52]
In theoretical studies, it is often convenient to consider the simplest kind of thermodynamic system. This is defined variously by different authors.[47][53][54][55][56][57] For the present article, the following definition is convenient, as abstracted from the definitions of various authors. A region of material with all intensive properties continuous in space and time is called a phase. A simple system is for the present article defined as one that consists of a single phase of a pure chemical substance, with no interior partitions.
Within a simple isolated thermodynamic system in thermodynamic equilibrium, in the absence of externally imposed force fields, all properties of the material of the system are spatially homogeneous.[58] Much of the basic theory of thermodynamics is concerned with homogeneous systems in thermodynamic equilibrium.[10][59]
Most systems found in nature or considered in engineering are not in thermodynamic equilibrium, exactly considered. They are changing or can be triggered to change over time, and are continuously and discontinuously subject to flux of matter and energy to and from other systems.[60] For example, according to Callen, "in absolute thermodynamic equilibrium all radioactive materials would have decayed completely and nuclear reactions would have transmuted all nuclei to the most stable isotopes. Such processes, which would take cosmic times to complete, generally can be ignored.".[60] Such processes being ignored, many systems in nature are close enough to thermodynamic equilibrium that for many purposes their behaviour can be well approximated by equilibrium calculations.
### Quasi-static transfers between simple systems are nearly in thermodynamic equilibrium and are reversible
It very much eases and simplifies theoretical thermodynamical studies to imagine transfers of energy and matter between two simple systems that proceed so slowly that at all times each simple system considered separately is near enough to thermodynamic equilibrium. Such processes are sometimes called quasi-static and are near enough to being reversible.[61][62]
### Natural processes are partly described by tendency towards thermodynamic equilibrium and are irreversible
If not initially in thermodynamic equilibrium, simple isolated thermodynamic systems, as time passes, tend to evolve naturally towards thermodynamic equilibrium. In the absence of externally imposed force fields, they become homogeneous in all their local properties. Such homogeneity is an important characteristic of a system in thermodynamic equilibrium in the absence of externally imposed force fields.
Many thermodynamic processes can be modeled by compound or composite systems, consisting of several or many contiguous component simple systems, initially not in thermodynamic equilibrium, but allowed to transfer mass and energy between them. Natural thermodynamic processes are described in terms of a tendency towards thermodynamic equilibrium within simple systems and in transfers between contiguous simple systems. Such natural processes are irreversible.[63]
## Non-equilibrium thermodynamics
Non-equilibrium thermodynamics[64] is a branch of thermodynamics that deals with systems that are not in thermodynamic equilibrium; it is also called thermodynamics of irreversible processes. Non-equilibrium thermodynamics is concerned with transport processes and with the rates of chemical reactions.[65] Non-equilibrium systems can be in stationary states that are not homogeneous even when there is no externally imposed field of force; in this case, the description of the internal state of the system requires a field theory.[66][67][68] One of the methods of dealing with non-equilibrium systems is to introduce so-called 'internal variables'. These are quantities that express the local state of the system, besides the usual local thermodynamic variables; in a sense such variables might be seen as expressing the 'memory' of the materials. Hysteresis may sometimes be described in this way. In contrast to the usual thermodynamic variables, 'internal variables' cannot be controlled by external manipulations.[69] This approach is usually unnecessary for gases and liquids, but may be useful for solids.[70] Many natural systems still today remain beyond the scope of currently known macroscopic thermodynamic methods.
## Laws of thermodynamics
Main article: Laws of thermodynamics
Thermodynamics states a set of four laws that are valid for all systems that fall within the constraints implied by each. In the various theoretical descriptions of thermodynamics these laws may be expressed in seemingly differing forms, but the most prominent formulations are the following:
• Zeroth law of thermodynamics: If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other.
This statement implies that thermal equilibrium is an equivalence relation on the set of thermodynamic systems under consideration. Systems are said to be in thermal equilibrium with each other if spontaneous molecular thermal energy exchanges between them do not lead to a net exchange of energy. This law is tacitly assumed in every measurement of temperature. For two bodies known to be at the same temperature, deciding if they are in thermal equilibrium when put into thermal contact does not require actually bringing them into contact and measuring any changes of their observable properties in time.[71] In traditional statements, the law provides an empirical definition of temperature and justification for the construction of practical thermometers. In contrast to absolute thermodynamic temperatures, empirical temperatures are measured just by the mechanical properties of bodies, such as their volumes, without reliance on the concepts of energy, entropy or the first, second, or third laws of thermodynamics.[55][72] Empirical temperatures lead to calorimetry for heat transfer in terms of the mechanical properties of bodies, without reliance on mechanical concepts of energy.
The physical content of the zeroth law has long been recognized. For example, Rankine in 1853 defined temperature as follows: "Two portions of matter are said to have equal temperatures when neither tends to communicate heat to the other."[73] Maxwell in 1872 stated a "Law of Equal Temperatures".[74] He also stated: "All Heat is of the same kind."[75] Planck explicitly assumed and stated it in its customary present-day wording in his formulation of the first two laws.[76] By the time the desire arose to number it as a law, the other three had already been assigned numbers, and so it was designated the zeroth law.
• First law of thermodynamics: The increase in internal energy of a closed system is equal to the difference of the heat supplied to the system and the work done by it: ΔU = Q - W [77][78][79][80][81][82][83][84][85][86][87]
The first law of thermodynamics asserts the existence of a state variable for a system, the internal energy, and tells how it changes in thermodynamic processes. The law allows a given internal energy of a system to be reached by any combination of heat and work. It is important that internal energy is a variable of state of the system (see Thermodynamic state) whereas heat and work are variables that describe processes or changes of the state of systems.
The first law observes that the internal energy of an isolated system obeys the principle of conservation of energy, which states that energy can be transformed (changed from one form to another), but cannot be created or destroyed.[88][89][90][91][92]
• Second law of thermodynamics: Heat cannot spontaneously flow from a colder location to a hotter location.
The second law of thermodynamics is an expression of the universal principle of dissipation of kinetic and potential energy observable in nature. The second law is an observation of the fact that over time, differences in temperature, pressure, and chemical potential tend to even out in a physical system that is isolated from the outside world. Entropy is a measure of how much this process has progressed. The entropy of an isolated system that is not in equilibrium tends to increase over time, approaching a maximum value at equilibrium.
In classical thermodynamics, the second law is a basic postulate applicable to any system involving heat energy transfer; in statistical thermodynamics, the second law is a consequence of the assumed randomness of molecular chaos. There are many versions of the second law, but they all have the same effect, which is to explain the phenomenon of irreversibility in nature.
• Third law of thermodynamics: As a system approaches absolute zero the entropy of the system approaches a minimum value.
The third law of thermodynamics is a statistical law of nature regarding entropy and the impossibility of reaching absolute zero of temperature. This law provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is the absolute entropy. Alternate definitions are, "the entropy of all systems and of all states of a system is smallest at absolute zero," or equivalently "it is impossible to reach the absolute zero of temperature by any finite number of processes".
Absolute zero is −273.15 °C (degrees Celsius), or −459.67 °F (degrees Fahrenheit) or 0 K (kelvin).
## System models
A diagram of a generic thermodynamic system
An important concept in thermodynamics is the thermodynamic system, a precisely defined region of the universe under study. Everything in the universe except the system is known as the surroundings. A system is separated from the remainder of the universe by a boundary, which may be notional or not, but by convention delimits a finite volume. Exchanges of work, heat, or matter between the system and the surroundings take place across this boundary.
Types of transfers permitted in a thermodynamic process for a type of thermodynamic system
type of system type of transfer
Mass Work Heat
Open ± ±
Closed 0 ± ±
Adiabatically enclosed 0 ± 0
Adynamically enclosed 0 0 ±
Isolated 0 0 0
The boundary is simply a surface around the volume of interest. Anything that passes across the boundary that effects a change in the internal energy needs to be accounted for in the energy balance equation. The volume can be the region surrounding a single atom resonating energy, as Max Planck defined in 1900; it can be a body of steam or air in a steam engine, such as Sadi Carnot defined in 1824; it can be the body of a tropical cyclone, such as Kerry Emanuel theorized in 1986 in the field of atmospheric thermodynamics; it could also be just one nuclide (i.e. a system of quarks) as hypothesized in quantum thermodynamics.
Boundaries are of four types: fixed, moveable, real, and imaginary. For example, in an engine, a fixed boundary means the piston is locked at its position; then a constant volume process occurs, no work being permitted. In that same engine, a moveable boundary allows the piston to move in and out, permitting work. For closed systems, boundaries are real, while open system boundaries are often imaginary.
Thermodynamics sometimes distinguishes five classes of systems, defined in terms of what is allowed to cross their boundaries. There is no mechanical boundary for the whole earth including its atmosphere, and so roughly speaking, no external work is done on or by the whole earth system. Such a system is sometimes said to be diabatically heated or cooled by radiation.[93][94] A process in which no work is transferred is sometimes called adynamic.[95]
Engineering and natural processes are often described as composites of many different component simple systems, sometimes with unchanging or changing partitions between them.
## States and processes
There are two fundamental kinds of entity in thermodynamics, states of a system, and processes of a system. This allows three fundamental approaches to thermodynamic reasoning, that in terms of states of thermodynamic equilibrium of a system, and that in terms of time-invariant processes of a system, and that in terms of cyclic processes of a system.
The approach through states of thermodynamic equilibrium of a system requires a full account of the state of the system as well as a notion of process from one state to another of a system, but may require only an idealized or partial account of the state of the surroundings of the system or of other systems.
The method of description in terms of states of thermodynamic equilibrium has limitations. For example, processes in a region of turbulent flow, or in a burning gas mixture, or in a Knudsen gas may be beyond "the province of thermodynamics".[96][97][98] This problem can sometimes be circumvented through the method of description in terms of cyclic or of time-invariant flow processes. This is part of the reason why the founders of thermodynamics often preferred the cyclic process description.
Approaches through processes of time-invariant flow of a system are used for some studies. Some processes, for example Joule-Thomson expansion, are studied through steady-flow experiments, but can be accounted for by distinguishing the steady bulk flow kinetic energy from the internal energy, and thus can be regarded as within the scope of classical thermodynamics defined in terms of equilibrium states or of cyclic processes.[43][99] Other flow processes, for example thermoelectric effects, are essentially defined by the presence of differential flows or diffusion so that they cannot be adequately accounted for in terms of equilibrium states or classical cyclic processes.[100][101]
The notion of a cyclic process does not require a full account of the state of the system, but does require a full account of how the process occasions transfers of matter and energy between the principal system (which is often called the working body) and its surroundings, which must include at least two heat reservoirs at different known and fixed temperatures, one hotter than the principal system and the other colder than it, as well as a reservoir that can receive energy from the system as work and can do work on the system. The reservoirs can alternatively be regarded as auxiliary idealized component systems, alongside the principal system. Thus an account in terms of cyclic processes requires at least four contributory component systems. The independent variables of this account are the amounts of energy that enter and leave the idealized auxiliary systems. In this kind of account, the working body is often regarded as a "black box",[102] and its own state is not specified. In this approach, the notion of a properly numerical scale of empirical temperature is a presupposition of thermodynamics, not a notion constructed by or derived from it.
### Account in terms of states of thermodynamic equilibrium
When a system is at thermodynamic equilibrium under a given set of conditions of its surroundings, it is said to be in a definite thermodynamic state, which is fully described by its state variables.
If a system is simple as defined above, and is in thermodynamic equilibrium, and is not subject to an externally imposed force field, such as gravity, electricity, or magnetism, then it is homogeneous, that is say, spatially uniform in all respects.[103]
In a sense, a homogeneous system can be regarded as spatially zero-dimensional, because it has no spatial variation.
If a system in thermodynamic equilibrium is homogeneous, then its state can be described by a few physical variables, which are mostly classifiable as intensive variables and extensive variables.[14][31][68][104][105]
Examples of extensive thermodynamic variables are total mass and total volume. Examples of intensive thermodynamic variables are temperature, pressure, and chemical concentration; intensive thermodynamic variables are defined at each spatial point and each instant of time in a system. Physical macroscopic variables can be mechanical or thermal.[31] Temperature is a thermal variable; according to Guggenheim, "the most important conception in thermodynamics is temperature."[14]
Intensive variables are defined by the property that if any number of systems, each in its own separate homogeneous thermodynamic equilibrium state, all with the same respective values of all of their intensive variables, regardless of the values of their extensive variables, are laid contiguously with no partition between them, so as to form a new system, then the values of the intensive variables of the new system are the same as those of the separate constituent systems. Such a composite system is in a homogeneous thermodynamic equilibrium. Examples of intensive variables are temperature, chemical concentration, pressure, density of mass, density of internal energy, and, when it can be properly defined, density of entropy.[106]
Extensive variables are defined by the property that if any number of systems, regardless of their possible separate thermodynamic equilibrium or non-equilibrium states or intensive variables, are laid side by side with no partition between them so as to form a new system, then the values of the extensive variables of the new system are the sums of the values of the respective extensive variables of the individual separate constituent systems. Obviously, there is no reason to expect such a composite system to be in a homogeneous thermodynamic equilibrium. Examples of extensive variables are mass, volume, and internal energy. They depend on the total quantity of mass in the system.[107]
Though, when it can be properly defined, density of entropy is an intensive variable, for inhomogeneous systems, entropy itself does not fit into this classification of state variables.[108][109] The reason is that entropy is a property of a system as a whole, and not necessarily related simply to its constituents separately. It is true that for any number of systems each in its own separate homogeneous thermodynamic equilibrium, all with the same values of intensive variables, removal of the partitions between the separate systems results in a composite homogeneous system in thermodynamic equilibrium, with all the values of its intensive variables the same as those of the constituent systems, and it is reservedly or conditionally true that the entropy of such a restrictively defined composite system is the sum of the entropies of the constituent systems. But if the constituent systems do not satisfy these restrictive conditions, the entropy of a composite system cannot be expected to be the sum of the entropies of the constituent systems, because the entropy is a property of the composite system as a whole. Therefore, though under these restrictive reservations, entropy satisfies some requirements for extensivity defined just above, entropy in general does not fit the above definition of an extensive variable.
Being neither an intensive variable nor an extensive variable according to the above definition, entropy is thus a stand-out variable, because it is a state variable of a system as a whole.[108] A non-equilibrium system can have a very inhomogeneous dynamical structure. This is one reason for distinguishing the study of equilibrium thermodynamics from the study of non-equilibrium thermodynamics.
The physical reason for the existence of extensive variables is the time-invariance of volume in a given inertial reference frame, and the strictly local conservation of mass, momentum, angular momentum, and energy. As noted by Gibbs, entropy is unlike energy and mass, because it is not locally conserved.[108] The stand-out quantity entropy is never conserved in real physical processes; all real physical processes are irreversible.[110] The motion of planets seems reversible on a short time scale (millions of years), but their motion, according to Newton's laws, is mathematically an example of deterministic chaos. Eventually a planet suffers an unpredictable collision with an object from its surroundings, outer space in this case, and consequently its future course is radically unpredictable. Theoretically this can be expressed by saying that every natural process dissipates some information from the predictable part of its activity into the unpredictable part. The predictable part is expressed in the generalized mechanical variables, and the unpredictable part in heat.
Other state variables can be regarded as conditionally 'extensive' subject to reservation as above, but not extensive as defined above. Examples are the Gibbs free energy, the Helmholtz free energy, and the enthalpy. Consequently, just because for some systems under particular conditions of their surroundings such state variables are conditionally conjugate to intensive variables, such conjugacy does not make such state variables extensive as defined above. This is another reason for distinguishing the study of equilibrium thermodynamics from the study of non-equilibrium thermodynamics. In another way of thinking, this explains why heat is to be regarded as a quantity that refers to a process and not to a state of a system.
A system with no internal partitions, and in thermodynamic equilibrium, can be inhomogeneous in the following respect: it can consist of several so-called 'phases', each homogeneous in itself, in immediate contiguity with other phases of the system, but distinguishable by their having various respectively different physical characters, with discontinuity of intensive variables at the boundaries between the phases; a mixture of different chemical species is considered homogeneous for this purpose if it is physically homogeneous.[111] For example, a vessel can contain a system consisting of water vapour overlying liquid water; then there is a vapour phase and a liquid phase, each homogeneous in itself, but still in thermodynamic equilibrium with the other phase. For the immediately present account, systems with multiple phases are not considered, though for many thermodynamic questions, multiphase systems are important.
#### Equation of state
The macroscopic variables of a thermodynamic system in thermodynamic equilibrium, in which temperature is well defined, can be related to one another through equations of state or characteristic equations.[27][28][29][30] They express the constitutive peculiarities of the material of the system. The equation of state must comply with some thermodynamic constraints, but cannot be derived from the general principles of thermodynamics alone.
#### Thermodynamic processes between states of thermodynamic equilibrium
A thermodynamic process is defined by changes of state internal to the system of interest, combined with transfers of matter and energy to and from the surroundings of the system or to and from other systems. A system is demarcated from its surroundings or from other systems by partitions that more or less separate them, and may move as a piston to change the volume of the system and thus transfer work.
#### Dependent and independent variables for a process
A process is described by changes in values of state variables of systems or by quantities of exchange of matter and energy between systems and surroundings. The change must be specified in terms of prescribed variables. The choice of which variables are to be used is made in advance of consideration of the course of the process, and cannot be changed. Certain of the variables chosen in advance are called the independent variables.[112] From changes in independent variables may be derived changes in other variables called dependent variables. For example a process may occur at constant pressure with pressure prescribed as an independent variable, and temperature changed as another independent variable, and then changes in volume are considered as dependent. Careful attention to this principle is necessary in thermodynamics.[113][114]
#### Changes of state of a system
In the approach through equilibrium states of the system, a process can be described in two main ways.
In one way, the system is considered to be connected to the surroundings by some kind of more or less separating partition, and allowed to reach equilibrium with the surroundings with that partition in place. Then, while the separative character of the partition is kept unchanged, the conditions of the surroundings are changed, and exert their influence on the system again through the separating partition, or the partition is moved so as to change the volume of the system; and a new equilibrium is reached. For example, a system is allowed to reach equilibrium with a heat bath at one temperature; then the temperature of the heat bath is changed and the system is allowed to reach a new equilibrium; if the partition allows conduction of heat, the new equilibrium is different from the old equilibrium.
In the other way, several systems are connected to one another by various kinds of more or less separating partitions, and to reach equilibrium with each other, with those partitions in place. In this way, one may speak of a 'compound system'. Then one or more partitions is removed or changed in its separative properties or moved, and a new equilibrium is reached. The Joule-Thomson experiment is an example of this; a tube of gas is separated from another tube by a porous partition; the volume available in each of the tubes is determined by respective pistons; equilibrium is established with an initial set of volumes; the volumes are changed and a new equilibrium is established.[115][116][117][118][119] Another example is in separation and mixing of gases, with use of chemically semi-permeable membranes.[120]
#### Commonly considered thermodynamic processes
It is often convenient to study a thermodynamic process in which a single variable, such as temperature, pressure, or volume, etc., is held fixed. Furthermore, it is useful to group these processes into pairs, in which each variable held constant is one member of a conjugate pair.
Several commonly studied thermodynamic processes are:
• Isobaric process: occurs at constant pressure
• Isochoric process: occurs at constant volume (also called isometric/isovolumetric)
• Isothermal process: occurs at a constant temperature
• Adiabatic process: occurs without loss or gain of energy as heat
• Isentropic process: a reversible adiabatic process occurs at a constant entropy, but is a fictional idealization. Conceptually it is possible to actually physically conduct a process that keeps the entropy of the system constant, allowing systematically controlled removal of heat, by conduction to a cooler body, to compensate for entropy produced within the system by irreversible work done on the system. Such isentropic conduct of a process seems called for when the entropy of the system is considered as an independent variable, as for example when the internal energy is considered as a function of the entropy and volume of the system, the natural variables of the internal energy as studied by Gibbs.
• Isenthalpic process: occurs at a constant enthalpy
• Isolated process: no matter or energy (neither as work nor as heat) is transferred into or out of the system
It is sometimes of interest to study a process in which several variables are controlled, subject to some specified constraint. In a system in which a chemical reaction can occur, for example, in which the pressure and temperature can affect the equilibrium composition, a process might occur in which temperature is held constant but pressure is slowly altered, just so that chemical equilibrium is maintained all the way. There is a corresponding process at constant temperature in which the final pressure is the same but is reached by a rapid jump. Then it can be shown that the volume change resulting from the rapid jump process is smaller than that from the slow equilibrium process.[121] The work transferred differs between the two processes.
### Account in terms of cyclic processes
A cyclic process[122] is a process that can be repeated indefinitely often without changing the final state of the system in which the process occurs. The only traces of the effects of a cyclic process are to be found in the surroundings of the system or in other systems. This is the kind of process that concerned early thermodynamicists such as Carnot, and in terms of which Kelvin defined absolute temperature,[123][124] before the use of the quantity of entropy by Rankine[125] and its clear identification by Clausius.[126] For some systems, for example with some plastic working substances, cyclic processes are practically nearly unfeasible because the working substance undergoes practically irreversible changes.[67] This is why mechanical devices are lubricated with oil and one of the reasons why electrical devices are often useful.
A cyclic process of a system requires in its surroundings at least two heat reservoirs at different temperatures, one at a higher temperature that supplies heat to the system, the other at a lower temperature that accepts heat from the system. The early work on thermodynamics tended to use the cyclic process approach, because it was interested in machines that converted some of the heat from the surroundings into mechanical power delivered to the surroundings, without too much concern about the internal workings of the machine. Such a machine, while receiving an amount of heat from a higher temperature reservoir, always needs a lower temperature reservoir that accepts some lesser amount of heat, the difference in amounts of heat being converted to work.[90][127] Later, the internal workings of a system became of interest, and they are described by the states of the system. Nowadays, instead of arguing in terms of cyclic processes, some writers are inclined to derive the concept of absolute temperature from the concept of entropy, a variable of state.
## Instrumentation
There are two types of thermodynamic instruments, the meter and the reservoir. A thermodynamic meter is any device that measures any parameter of a thermodynamic system. In some cases, the thermodynamic parameter is actually defined in terms of an idealized measuring instrument. For example, the zeroth law states that if two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. This principle, as noted by James Maxwell in 1872, asserts that it is possible to measure temperature. An idealized thermometer is a sample of an ideal gas at constant pressure. From the ideal gas law PV=nRT, the volume of such a sample can be used as an indicator of temperature; in this manner it defines temperature. Although pressure is defined mechanically, a pressure-measuring device, called a barometer may also be constructed from a sample of an ideal gas held at a constant temperature. A calorimeter is a device that measures and define the internal energy of a system.
A thermodynamic reservoir is a system so large that it does not appreciably alter its state parameters when brought into contact with the test system. It is used to impose a particular value of a state parameter upon the system. For example, a pressure reservoir is a system at a particular pressure, which imposes that pressure upon any test system that it is mechanically connected to. The Earth's atmosphere is often used as a pressure reservoir.
## Conjugate variables
Main article: Conjugate variables
A central concept of thermodynamics is that of energy. By the First Law, the total energy of a system and its surroundings is conserved. Energy may be transferred into a system by heating, compression, or addition of matter, and extracted from a system by cooling, expansion, or extraction of matter. In mechanics, for example, energy transfer equals the product of the force applied to a body and the resulting displacement.
Conjugate variables are pairs of thermodynamic concepts, with the first being akin to a "force" applied to some thermodynamic system, the second being akin to the resulting "displacement," and the product of the two equalling the amount of energy transferred. The common conjugate variables are:
• Pressure-volume (the mechanical parameters);
• Temperature-entropy (thermal parameters);
• Chemical potential-particle number (material parameters).
## Potentials
Thermodynamic potentials are different quantitative measures of the stored energy in a system. Potentials are used to measure energy changes in systems as they evolve from an initial state to a final state. The potential used depends on the constraints of the system, such as constant temperature or pressure. For example, the Helmholtz and Gibbs energies are the energies available in a system to do useful work when the temperature and volume or the pressure and temperature are fixed, respectively.
The five most well known potentials are:
Name Symbol Formula Natural variables
Internal energy $U$ $\int ( T dS - p dV + \sum_i \mu_i dN_i )$ $S, V, \{N_i\}$
Helmholtz free energy $F$ $U-TS$ $T, V, \{N_i\}$
Enthalpy $H$ $U+pV$ $S, p, \{N_i\}$
Gibbs free energy $G$ $U+pV-TS$ $T, p, \{N_i\}$
Landau Potential (Grand potential) $\Omega$, $\Phi_{G}$ $U - T S -$$\sum_i\,$$\mu_i N_i$ $T, V, \{\mu_i\}$
where $T$ is the temperature, $S$ the entropy, $p$ the pressure, $V$ the volume, $\mu$ the chemical potential, $N$ the number of particles in the system, and $i$ is the count of particles types in the system.
Thermodynamic potentials can be derived from the energy balance equation applied to a thermodynamic system. Other thermodynamic potentials can also be obtained through Legendre transformation.
## Axiomatics
Most accounts of thermodynamics presuppose the law of conservation of mass, sometimes with,[128] and sometimes without,[24][129][130] explicit mention. Particular attention is paid to the law in accounts of non-equilibrium thermodynamics.[131][132] One statement of this law is "The total mass of a closed system remains constant."[15] Another statement of it is "In a chemical reaction, matter is neither created nor destroyed."[133] Implied in this is that matter and energy are not considered to be interconverted in such accounts. The full generality of the law of conservation of energy is thus not used in such accounts.
In 1909, Constantin Carathéodory presented[55] a purely mathematical axiomatic formulation, a description often referred to as geometrical thermodynamics, and sometimes said to take the "mechanical approach"[85] to thermodynamics. The Carathéodory formulation is restricted to equilibrium thermodynamics and does not attempt to deal with non-equilibrium thermodynamics, forces that act at a distance on the system, or surface tension effects.[134] Moreover, Carathéodory's formulation does not deal with materials like water near 4 °C, which have a density extremum as a function of temperature at constant pressure.[135][136] Carathéodory used the law of conservation of energy as an axiom from which, along with the contents of the zeroth law, and some other assumptions including his own version of the second law, he derived the first law of thermodynamics.[137] Consequently one might also describe Carathėodory's work as lying in the field of energetics,[138] which is broader than thermodynamics. Carathéodory presupposed the law of conservation of mass without explicit mention of it.
Since the time of Carathėodory, other influential axiomatic formulations of thermodynamics have appeared, which like Carathéodory's, use their own respective axioms, different from the usual statements of the four laws, to derive the four usually stated laws.[139][140][141]
Many axiomatic developments assume the existence of states of thermodynamic equilibrium and of states of thermal equilibrium. States of thermodynamic equilibrium of compound systems allow their component simple systems to exchange heat and matter and to do work on each other on their way to overall joint equilibrium. Thermal equilibrium allows them only to exchange heat. The physical properties of glass depend on its history of being heated and cooled and, strictly speaking, glass is not in thermodynamic equilibrium.[70]
According to Herbert Callen's widely cited 1985 text on thermodynamics: "An essential prerequisite for the measurability of energy is the existence of walls that do not permit transfer of energy in the form of heat.".[142] According to Werner Heisenberg's mature and careful examination of the basic concepts of physics, the theory of heat has a self-standing place.[143]
From the viewpoint of the axiomatist, there are several different ways of thinking about heat, temperature, and the second law of thermodynamics. The Clausius way rests on the empirical fact that heat is conducted always down, never up, a temperature gradient. The Kelvin way is to assert the empirical fact that conversion of heat into work by cyclic processes is never perfectly efficient. A more mathematical way is to assert the existence of a function of state called the entropy that tells whether a hypothesized process occurs spontaneously in nature. A more abstract way is that of Carathéodory that in effect asserts the irreversibility of some adiabatic processes. For these different ways, there are respective corresponding different ways of viewing heat and temperature.
The Clausius-Kelvin-Planck way This way prefers ideas close to the empirical origins of thermodynamics. It presupposes transfer of energy as heat, and empirical temperature as a scalar function of state. According to Gislason and Craig (2005): "Most thermodynamic data come from calorimetry..."[144] According to Kondepudi (2008): "Calorimetry is widely used in present day laboratories."[145] In this approach, what is often currently called the zeroth law of thermodynamics is deduced as a simple consequence of the presupposition of the nature of heat and empirical temperature, but it is not named as a numbered law of thermodynamics. Planck attributed this point of view to Clausius, Kelvin, and Maxwell. Planck wrote (on page 90 of the seventh edition, dated 1922, of his treatise) that he thought that no proof of the second law of thermodynamics could ever work that was not based on the impossibility of a perpetual motion machine of the second kind. In that treatise, Planck makes no mention of the 1909 Carathéodory way, which was well known by 1922. Planck for himself chose a version of what is just above called the Kelvin way.[146] The development by Truesdell and Bharatha (1977) is so constructed that it can deal naturally with cases like that of water near 4 °C.[140]
The way that assumes the existence of entropy as a function of state This way also presupposes transfer of energy as heat, and it presupposes the usually stated form of the zeroth law of thermodynamics, and from these two it deduces the existence of empirical temperature. Then from the existence of entropy it deduces the existence of absolute thermodynamic temperature.[14][139]
The Carathéodory way This way presupposes that the state of a simple one-phase system is fully specifiable by just one more state variable than the known exhaustive list of mechanical variables of state. It does not explicitly name empirical temperature, but speaks of the one-dimensional "non-deformation coordinate". This satisfies the definition of an empirical temperature, that lies on a one-dimensional manifold. The Carathéodory way needs to assume moreover that the one-dimensional manifold has a definite sense, which determines the direction of irreversible adiabatic process, which is effectively assuming that heat is conducted from hot to cold. This way presupposes the often currently stated version of the zeroth law, but does not actually name it as one of its axioms.[134] According to one author, Carathéodory's principle, which is his version of the second law of thermodynamics, does not imply the increase of entropy when work is done under adiabatic conditions (as was noted by Planck[147]). Thus Carathéodory's way leaves unstated a further empirical fact that is needed for a full expression of the second law of thermodynamics.[148]
## Scope of thermodynamics
Originally thermodynamics concerned material and radiative phenomena that are experimentally reproducible. For example, a state of thermodynamic equilibrium is a steady state reached after a system has aged so that it no longer changes with the passage of time. But more than that, for thermodynamics, a system, defined by its being prepared in a certain way must, consequent on every particular occasion of preparation, upon aging, reach one and the same eventual state of thermodynamic equilibrium, entirely determined by the way of preparation. Such reproducibility is because the systems consist of so many molecules that the molecular variations between particular occasions of preparation have negligible or scarcely discernable effects on the macroscopic variables that are used in thermodynamic descriptions. This led to Boltzmann's discovery that entropy had a statistical or probabilistic nature. Probabilistic and statistical explanations arise from the experimental reproducibility of the phenomena.[149]
Gradually, the laws of thermodynamics came to be used to explain phenomena that occur outside the experimental laboratory. For example, phenomena on the scale of the earth's atmosphere cannot be reproduced in a laboratory experiment. But processes in the atmosphere can be modeled by use of thermodynamic ideas, extended well beyond the scope of laboratory equilibrium thermodynamics.[150][151][152] A parcel of air can, near enough for many studies, be considered as a closed thermodynamic system, one that is allowed to move over significant distances. The pressure exerted by the surrounding air on the lower face of a parcel of air may differ from that on its upper face. If this results in rising of the parcel of air, it can be considered to have gained potential energy as a result of work being done on it by the combined surrounding air below and above it. As it rises, such a parcel usually expands because the pressure is lower at the higher altitudes that it reaches. In that way, the rising parcel also does work on the surrounding atmosphere. For many studies, such a parcel can be considered nearly to neither gain nor lose energy by heat conduction to its surrounding atmosphere, and its rise is rapid enough to leave negligible time for it to gain or lose heat by radiation; consequently the rising of the parcel is near enough adiabatic. Thus the adiabatic gas law accounts for its internal state variables, provided that there is no precipitation into water droplets, no evaporation of water droplets, and no sublimation in the process. More precisely, the rising of the parcel is likely to occasion friction and turbulence, so that some potential and some kinetic energy of bulk converts into internal energy of air considered as effectively stationary. Friction and turbulence thus oppose the rising of the parcel.[153][154]
## See also
Entropy production
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149. Mak, M. (2011). Atmospheric Dynamics, Cambridge University Press, Cambridge UK, ISBN 978-0-521-19573-7.
## Cited bibliography
• Adkins, C.J. (1968/1975). Equilibrium Thermodynamics, second edition, McGraw-Hill, London, ISBN 0-07-084057-1.
• Bailyn, M. (1994). A Survey of Thermodynamics, American Institute of Physics Press, New York, ISBN 0-88318-797-3.
• Bryan, G.H. (1907). Thermodynamics. An Introductory Treatise dealing mainly with First Principles and their Direct Applications, B.G. Teubner, Leipzig.
• Callen, H.B. (1960/1985). Thermodynamics and an Introduction to Thermostatistics, (1st edition 1960) 2nd edition 1985, Wiley, New York, ISBN 0-471-86256-8.
• Eu, B.C. (2002). Generalized Thermodynamics. The Thermodynamics of Irreversible Processes and Generalized Hydrodynamics, Kluwer Academic Publishers, Dordrecht, ISBN 1-4020-0788-4.
• Fowler, R., Guggenheim, E.A. (1939). Statistical Thermodynamics, Cambridge University Press, Canbridge UK.
• Gibbs, J.W. (1875). On the equilibrium of heterogeneous substances, Transactions of the Connecticut Academy of Arts and Sciences, 3: 108–248.
• Grandy, W.T., Jr (2008). Entropy and the Time Evolution of Macroscopic Systems, Oxford University Press, Oxford, ISBN 978-0-19-954617-6.
• Guggenheim, E.A. (1949/1967). Thermodynamics. An Advanced Treatment for Chemists and Physicists, (1st edition 1949) 5th edition 1967, North-Holland, Amsterdam.
• Haase, R. (1971). Survey of Fundamental Laws, chapter 1 of Thermodynamics, pages 1–97 of volume 1, ed. W. Jost, of Physical Chemistry. An Advanced Treatise, ed. H. Eyring, D. Henderson, W. Jost, Academic Press, New York, lcn 73–117081.
• Kondepudi, D., Prigogine, I. (1998). Modern Thermodynamics. From Heat Engines to Dissipative Structures, John Wiley & Sons, ISBN 0-471-97393-9.
• Lebon, G., Jou, D., Casas-Vázquez, J. (2008). Understanding Non-equilibrium Thermodynamics, Springer, Berlin, ISBN 978-3-540-74251-7.
• Partington, J.R. (1949). An Advanced Treatise on Physical Chemistry, volume 1, Fundamental Principles. The Properties of Gases, Longmans, Green and Co., London.
• Pippard, A.B. (1957). The Elements of Classical Thermodynamics, Cambridge University Press.
• Planck, M.(1897/1903). Treatise on Thermodynamics, translated by A. Ogg, Longmans, Green & Co., London.
• Planck, M. (1923/1926). Treatise on Thermodynamics, third English edition translated by A. Ogg from the seventh German edition, Longmans, Green & Co., London.
• Sommerfeld, A. (1952/1956). Thermodynamics and Statistical Mechanics, Academic Press, New York.
• Tisza, L. (1966). Generalized Thermodynamics, M.I.T Press, Cambridge MA.
• Truesdell, C.A. (1980). The Tragicomical History of Thermodynamics, 1822–1854, Springer, New York, ISBN 0-387-90403-4.
## Further reading
• Goldstein, Martin, and Inge F. (1993). The Refrigerator and the Universe. Harvard University Press. ISBN 0-674-75325-9. OCLC 32826343. A nontechnical introduction, good on historical and interpretive matters.
• Kazakov, Andrei (July–August 2008). "Web Thermo Tables – an On-Line Version of the TRC Thermodynamic Tables". Journal of Research of the National Institutes of Standards and Technology 113 (4): 209–220.
The following titles are more technical:
• Cengel, Yunus A., & Boles, Michael A. (2002). Thermodynamics – an Engineering Approach. McGraw Hill. ISBN 0-07-238332-1. OCLC 45791449 52263994 57548906.
• Fermi, E. (1956). Thermodynamics, Dover, New York.
• Kittel, Charles & Kroemer, Herbert (1980). Thermal Physics. W. H. Freeman Company. ISBN 0-7167-1088-9. OCLC 32932988 48236639 5171399.
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http://math.stackexchange.com/questions/118029/proof-strategy-stirling-numbers-formula?answertab=active
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Proof strategy - Stirling numbers formula
I need to prove such a formula:
$$\left\{ \begin{matrix} n \\ k-1 \end{matrix} \right\} \cdot \left\{ \begin{matrix} n \\ k+1 \end{matrix} \right\} \le \left\{ \begin{matrix} n \\ k \end{matrix} \right\}^{2} \\$$ Where {} are Stirling numbers of the second kind.
(number of ways to partition a set of n objects into k non-empty subsets).
I tried to figure out some combinatorial proof, but failed.
I'd be grateful for any help.
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What is the source of this problem? – Aryabhata Mar 8 '12 at 21:55
3
– Sasha Mar 8 '12 at 22:26
1
– Mike Spivey Mar 8 '12 at 22:34
There is a fairly short but very non-combinatorial proof in Herbert S. Wilf’s generatingfunctionology, which may be downloaded here as a hyperlinked PDF. The desired result is Corollary 4.5.3. Do you just need a proof, or do you specifically want t combinatorial proof? – Brian M. Scott Mar 9 '12 at 8:53
Combinatorial proof was rather my personal guess. I just noticed that huge amount of Stirling-number-related formulas have clear and easy combinatorial proof, that's all. – Maria McKee Mar 9 '12 at 11:35
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http://quant.stackexchange.com/questions/1565/how-do-i-graphically-represent-the-evolution-of-a-covariance-matrix-over-time
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How do I graphically represent the evolution of a covariance matrix over time?
I am working with a set of covariance matrices evaluated at various points in time over some history. Each covariance matrix is $N\times N$ for $N$ financial time-series over $T$ periods. I would like to explore some of the properties of this matrix's evolution over time, particularly whether correlation as a whole is increasing or decreasing, and whether certain series become more or less correlated with the whole. I am looking for suggestions as to the kinds of analysis to perform on this data-set, and particularly graphical/pictorial analysis. Ideally, I would like to avoid having to look in depth into each series as $N$ is rather large.
Update
The following graphs were generated based on the accepted answer from @Quant-Guy. PC = principal component = eigenvector. The analysis was done on correlations rather than covariances in order to account for vastly different variances of the $N$ series.
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1
Those charts look great - thanks for posting! This seems like an interesting tool for identifying regime change points. – Quant Guy Aug 2 '11 at 21:29
Multivariate DCC is another way. There is support for this in the `rmgarch` package. – Jase Dec 27 '12 at 10:12
– Sid Jan 6 at 16:27
5 Answers
I would consider a motion chart that plots the eigenvalues of the covariance matrix over time.
For a static view you can create a table: rows represent dates, and columns represent eigenvectors. The entries of the table represent changes in the angle of the eigenvector from the previous row. This will show how stable your covariance structure is.
You can also create a second table this time with eigenvalues as the columns sorted from high to low (and the corresponding values below for each date). This shows the variance described by each eigenvector so you can see whether correlation as whole is increasing or decreasing
Update: You can also measure the distance between the two covariance matrices via some distance measure metric such as Kullback-Leibler divergence, euclidean distance, Mahalanabois, etc.
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1
This may be a bit abstract but here goes. Create a table: rows represent dates, and columns represent eigenvectors. The entries of the table represent changes in the angle of the eigenvector from the previous row. This will show how stable your covariance structure is. You can also create a second table this time with eigenvalues as the columns sorted from high to low (and the corresponding values below for each date). – Quant Guy Aug 2 '11 at 19:31
That's brilliant! I just tried it and it actually works really well to demonstrate major shifts in the correlation structure. Can we do some cleanup and incorporate your comment into your answer? – Tal Fishman Aug 2 '11 at 20:00
Cool! I updated the answer. I'm not sure if you are able to share/upload that output but it would be very intellectually interesting to look at! – Quant Guy Aug 2 '11 at 20:09
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Be mindful of the algorithm used to produce the eigen decomposition, since some methods (thinking eig() in Matlab) are arbitrary with regards to eigenvector sign, so the angle of rotation may not always be consistent. Another option to illustrate similar features is to plot the percentage of variation explained by a fixed number of principal components (sometimes called the absorption ratio) over time. – michaelv2 Aug 3 '11 at 12:34
@michaelv2 you are absolutely right, and in this case I took min(x,180-x) for the graph above. I think percentage variation explained is what quant-guy had in mind initially, and I plotted that as well, it is also helpful. – Tal Fishman Aug 3 '11 at 23:26
Here's an interesting possibility: correlation network analysis + motion chart.
Thanks to the hot research efforts in social network analysis (SNA), network analysis and graphics libraries such as R and Gephis are now easily accessible. I am well-versed in correlation analysis, and have a feeling that SNA can be effectively adapted for it. After all, the 'linear' correlation is just a special relationship among many others. Since SNA is hot area now. Many efforts are devoted and many resources are available. Can't help imagining that we can leverage it in finance.
Then the motion chart concept can further leverage the power of network representation. I would imagine that when a regime shifts toward high correlation, we should see the dynamics of 'clustering' effects, whereas low correlation regimes would show nodes scattered around. For example, if there is a sector-level event, we would see nodes in that sector start to cluster more tightly around few benchmark nodes. In a macro event, we would see all benchmarks or eigenvectors cluster.
In any cases, I think the motion network graph will be a much richer representation than the motion matrix, where the dynamics are usually represented in numbers, colors, or angles. If one still prefers to visualize 'covariance', then the node size will be a natural place for volatility (though I still prefer to separate correlation and variance visualization). The line width / color / distance can represent something else intuitively.
Last, I guess the reason evolving network graphs are not often used for financial correlations is that a lot of the traditional financial applications are static, in which case the advantages of network graphs over matrix representation are limited.
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Interesting idea. Do you have any references of how it can be adapted to correlation analysis? – Tal Fishman Aug 4 '11 at 20:54
you may be right that SNA could one day prove useful in finance, but I'm really not sure that this particular application (my question) is such a case. Your answer strikes me somewhat as falling under the saying, "to someone with only a hammer, everything looks like a nail." Nevertheless, I encourage you to post your own question asking whether SNA is useful in finance. – Tal Fishman Aug 4 '11 at 22:00
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I believe (more precisely, I 'guess') their underlying analysis is very similar (but please feel free to correct me if I am totally wrong). It should all start the cluster analysis, en.wikipedia.org/wiki/Cluster_analysis Then, apply certain network visualization algorithm en.wikipedia.org/wiki/Network_visualization Here is an attempt (not exactly the same way I suggest) I have seen: investuotojas.eu/2011/03/22/correlation-network – 楊祝昇 Aug 4 '11 at 22:10
I take back my earlier comments, I think this approach is interesting, too, and has its merits. – Tal Fishman Aug 8 '11 at 4:23
I'd look at the evolution of a heat map based on the correlation structure (literally the lower triangle). I'd probably write a script in R or python that writes out the heat map per t to disk, then use a command line program like imagemagick to stitch images together into an animated gif, for example. I'm sure you could do it entirely in Processing too, and there you'd be able to mouse over pairs, etc. Perhaps seriation techniques could group sets ex post within the triangle.
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Minimum spanning trees are another option, with edges between nodes based on either Euclidean distances of the matrix or another distance measure of your choosing. They can be more effective at illustrating the underlying structure of the matrix than some other methods (eg heatmaps, eigenvalue ratio plots), but this may not be practical if T is large.
You didn't specify what environment you're working in, but R has several packages (see vegan:spantree, ape:mst, igraph:minimum.spanning.tree, ade4:mstree and fAssets:assetsDendrogramPlot or fAssets:assetsCorEigenPlot) that support plotting both minimum spanning trees (with various layout types) and dendrograms. Gephi also produces plots that may be more aesthetically pleasing (and it's open source), although it does require that the data processing be done elsewhere and there is a bit of a learning curve.
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I think you misunderstood my question. In my case, it is pretty clear to me what the relationships are between the variables, and in fact I have already plotted dendrograms to investigate this. However, it is not clear how I can examine changes in the inter-relationships over time. $T$ is about 600 in my case, so clearly drawing a minimum spanning tree and/or dendrogram at each point in time is not practical. – Tal Fishman Aug 2 '11 at 18:07
Agreed. The only drawback to eigenvalue plots in general is that it's impossible to determine which variables correspond to which principal components, but if you only have a few matrices to evaluate then MSTs don't suffer from this problem. – michaelv2 Aug 3 '11 at 14:19
This appoach could be use if we first make cluster analysis, and then make minimum spanning tree for cluster centres. Beyond that change in spanning tree structure could be view as evidence of some "topological" changes in market structure. – Qbik Apr 26 '12 at 12:33
On conceptual level making cluster analysis or kmeans, for arbitrarily chosen k (in cluster analysis with hierarchical methods we simple would cut dendrogram into k pieces/subtrees) and then copmuting average correlations is much simpler then PCA.
But there are some problems with cluster analysis on correlation matrix of time series. If ,for each day, we have one correlation matrix of price changes with 15 minutes time resolution and then clusterize it into K pieces, then for each day we could have different sets of stocks in each cluster, we can fight or use it.
Fight it: if data are from the period of 6 months then we find clusters by merging daily time series of each stock into one time series and then clusterize and compute averages or some different measure. For Future data we can hold stocks memberships to clusters or merge new observations with previous set or make analysis in moving window.
Use it : compute total number of changes in membership of stocks - first day we have clusters A,B `A={stock1,stock2} B={stock3,stock4,stock5}` next day : `A={stock1,stock3,stock4} B={stock2,stock5}` so 3 stocks have changed its clusters. The more stocks have changed clusters the more market behaviour is unusual that day. This could be done easy with kmeans - we take means from day `T`, and attribute memberships for stocks `T` and `T+1`day, then count differences.
and distance matrix `=(2*(1-corelationMatrix))^0.5`
Some cool visualizations of correlation matrix by networks analysis : http://www.maths.tcd.ie/~coelhor/Palermo_Presentation_v1.0.pdf
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http://en.m.wikibooks.org/wiki/Signal_Processing/Image_Editing
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Signal Processing/Image Editing
As an image is a type of 2-D signal; instead of just time-amplitude pairs that correspond to a voice transmission, consider "time in the X domain"-"time in the Y domain"-amplitude pairs. That is, an X coordinate, a Y coordinate and an amplitude. This will give you a monochromatic image. As such, signal processing tools can be used in editing images.
Singular Value Decomposition
The Singular Value Decomposition is a matrix decomposition (another way to say factorization).
$M = U\Sigma V^*$
where
• U is an m×m unitary matrix over.
• Σ is an m×n diagonal matrix with nonnegative real numbers s_{n,n} on the diagonal where
$s_{1,1}> s_{2,2} > \ldots s_{n,n}$
• V*, an n×n unitary matrix. V* is the conjugate transpose (take the complex conjugate of all entries, and then perform a transpose operation on the matrix) of V.
The diagonal entries Σi,i of Σ are known as the singular values of M.
Image Compression
The nature of the singular values Σ is such that for a certain k,
$s_{1,1}> s_{2,2} >> s_{k,k} > \ldots s_{n,n}$
In order to transmit a 10x10 monochromatic image with 2 values ("on" or "off") it would require a matrix that has 10x10 = 100 entries. Consider the following image.
This image can be represented by the following matrix.
$M = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}$
The singular value decomposition of which is
$M = U\Sigma V^*,$
Where
$\Sigma = \begin{bmatrix} 7.557 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 2.979 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1.422 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}$
Using the matrix
$\Sigma_{sm} = \begin{bmatrix} 7.557 & 0 & 0\\ 0 & 2.979 & 0\\ 0 & 0 & 1.422\end{bmatrix}$
And corresponding "truncated" versions of U and V* (Use only the first three columns of U and the first three columns of V*), we can find that
$M_{sm}= \begin{bmatrix} 0.999 & 0.999 & 0.999 & 0.999 & 1.000 & 1.000 & 0.999 & 0.999 & 0.999 & 0.999\\ 0.999 & 0.999 & 0.999 & 0.999 & 1.000 & 1.000 & 0.999 & 0.999 & 0.999 & 0.999\\ 1.000 & 1.000 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 & 1.000 & 1.000\\ 1.000 & 1.000 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 & 1.000 & 1.000\\ 1.000 & 1.000 &-0.001 &-0.001 & 1.001 & 1.001 &-0.001 &-0.001 & 1.000 & 1.000\\ 1.000 & 1.000 &-0.001 &-0.001 & 1.001 & 1.001 &-0.001 &-0.001 & 1.000 & 1.000\\ 1.000 & 1.000 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 & 1.000 & 1.000\\ 1.000 & 1.000 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 &-0.001 & 1.000 & 1.000\\ 0.999 & 0.999 & 0.999 & 0.999 & 1.000 & 1.000 & 0.999 & 0.999 & 0.999 & 0.999\\ 0.999 & 0.999 & 0.999 & 0.999 & 1.000 & 1.000 & 0.999 & 0.999 & 0.999 & 0.999\end{bmatrix}$
A cursory examination of the previous matrix will show that M_{sm} is approximately equal to M. Note that the truncated version of U and V* use 3*10 numbers each. The total number of values needed for this type of storage is 2*30+3 = 63. Which means data usage is reduced by nearly half.
Noise Removal
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http://mathoverflow.net/questions/67290/space-of-compact-operators
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## Space of compact operators
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am interested in the Banach space $\mathcal{K}=\mathcal{K}(\ell^2)$ of compact operators on $\ell^2$, however my questions can be stated for any $\mathcal{K}(E)$, where $E$ is an arbitrary Banach space. I think that everyone who tries to study "classical" operator spaces like $\mathcal{K}$, $p$-Schatten class operators etc. immediately discovers the similarity with "commutative" counterparts, i.e. $c_0$ and $\ell^p$. This phenomenon is visible when one uses (generalised) singular numbers for certain classes of operators. Again, I have got plenty of questions concerning this stuff, let me list at least two of them:
1) what are the complemented subspaces of $\mathcal{K}$? Is $\mathcal{K}$ complemented in $\mathcal{B}(\ell^2)$? Recently, Haydon and Argyros constructed an HI-space $E$ such that $\mathcal{K}(E)$ has codimension 1 $\mathcal{B}(E)$, thus complemented.
2) is every bounded operator from $p$-Schatten class to $\mathcal{K}$ compact?
What other properties $\mathcal{K}$ shares with $c_0$?
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I've removed the "operator spaces" tag, since nowadays this usually refers to operator spaces in the sense of Effros, Ruan et al. – Yemon Choi Jun 8 2011 at 19:24
What if you search MathSciNet for "operator spaces" to see what it commonly refers to in recent years? – Gerald Edgar Jun 8 2011 at 20:17
@Gerald: Well, of the top 40 hits, all but 2 (maybe 3) seem to use the term in Yemon's sense. – Matthew Daws Jun 8 2011 at 20:33
## 2 Answers
It is easy to see that whenever a space has an unconditional basis then the space of diagonal operators of the basis is equivalent to $\ell_\infty$. If $c_0$ embeds in $K(X,Y)$ then $K(X,Y)$ is not complemented in $B(X,Y)$. One reference for this is: M. FEDER. On subspaces of spaces with an unconditional basis and spaces of operators. Illinois J. Math. 34 (1980), 196-205.
It is also a direct consequence of a result from a Studia paper of Tong and Wilken from 1971. Here they prove that if $Y$ has an unconditional basis then $K(X,Y)$ is uncomplemented in $B(X,Y)$ (assuming the spaces are not equal).
As far as I know the Argyros-Haydon space is the first example of a space for which it is known that $K(X)$ is complemented in $B(X)$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Probably in (2) you meant to ask whether every bounded operator from $K$ into a Schatten $p$ class is compact, since every operator from $c_0$ into $\ell_p$, $p<\infty$, is compact. But no either way: $K$ and any Schatten $p$ class contain complemented subpspaces isometrically isomorphic to $\ell_2$ (e.g. operators whose matrix representation has zeroes except in the first column).
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3
You might look at Rosenthal's expository paper ma.utexas.edu/users/rosenthl/pdf-papers/93.pdf which discusses properties of $K$ and other `$C^*$` algebras as Banach spaces and as operator spaces. – Bill Johnson Jun 9 2011 at 2:54
That is very useful. Thank you! – Tomek Kania Jun 9 2011 at 16:41
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http://physics.stackexchange.com/questions/18527/does-the-pauli-exclusion-principle-instantaneously-affect-distant-electrons/18581
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# Does the Pauli exclusion principle instantaneously affect distant electrons?
According to Brian Cox in his A night with the Stars lecture$^1$, the Pauli exclusion principle means that no electron in the universe can have the same energy state as any other electron in the universe, and that if he does something to change the energy state of one group of electrons (rubbing a diamond to heat it up in his demo) then that must cause other electrons somewhere in the universe to change their energy states as the states of the electrons in the diamond change.
But when does this change occur? Surely if the electrons are separated by a significant gap then the change cannot be instant because information can only travel at the speed of light. Wouldn't that mean that if you changed the energy state of one electron to be the same as another electron that was some distance away, then surely the two electrons would be in the same state until the information that one other electron is in the same state reaches the other electron.
Or can information be transferred instantly from one place to another? If it can, then doesn't that mean it's not bound by the same laws as the rest of the universe?
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$^1$: The Youtube link keeps breaking, so here is a search on Youtube for Brian Cox' A Night with the Stars lecture.
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Could you give a reference? This sound rather dubious the way you state it here. – leftaroundabout Dec 20 '11 at 11:51
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Like I said, it came from Brian Cox's Evening With The Stars lecture. If you're in the UK you should still be able to see it on iPlayer. – GordonM Dec 20 '11 at 12:00
– Arnold Neumaier Mar 5 '12 at 17:04
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If two electrons are distant, then they can't occupying the same state. The state includes the position. – Peter Shor Apr 17 at 15:58
## 10 Answers
The Pauli exclusion principle can be stated as "two electrons cannot occupy the same energy state", but this is really only a rough way of stating it. It's more precise to say that the wavefunction of a system is anti-symmetric with respect to exchange of two electrons. The trouble is that now I have to explain to a non-physicist what "anti-symmetric" means and that's hard without going into the maths. I'll have a go at doing this below.
Anyhow, Brian Cox is being a bit liberal with the truth because I'm not sure it makes sense to say the electrons in his bit of diamond and electrons in far away bits of the universe can be described by a single wavefunction. If this isn't a good description then the Pauli exclusion principle doesn't have any meaning for the system.
Suppose you have two electrons in an atom or some other small system. Then that system is described by some wavefunction $\Psi(e_1, e_2)$ where I've used $e_1$ and $e_2$ to denote the two electrons. The Pauli exclusion principle states:
$$\Psi(e_1, e_2) = -\Psi(e_2, e_1)$$
that is if you swap the two electrons $\Psi$ changes to $-\Psi$. But suppose the two electrons were exactly the same. In that case swapping the electrons cannot change $\Psi$ because they're identical. So we'd have:
$$\Psi(e_1, e_2) = \Psi(e_2, e_1)$$
but the exclusion principle states:
$$\Psi(e_1, e_2) = -\Psi(e_2, e_1)$$
therefore if both are true:
$$\Psi(e_2, e_1) = -\Psi(e_2, e_1)$$ ie $$\Psi = -\Psi$$
The only way you can have $\Psi = -\Psi$ is if $\Psi$ is zero, which means $\Psi$ doesn't exist. This is why if the Pauli exclusion is true, two electrons can't be identical i.e. they can't be in the same energy state.
But this only applies because I could write down a wavefunction $\Psi$ to describe the system. When systems become large, e.g. two footballs in a swimming pool instead of two electrons in an atom, it isn't useful to try and write a wavefunction to describe the system and the exclusion principle doesn't apply. NB this doesn't mean the exclusion principle is wrong, it just means it doesn't apply to that system.
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The answer is nonsense. The wave function of two electrons is not a function of the electrons but a function of their position. Moreover, a sign change in a wave function is irrelevant as this only changes the phase, not the probability amplitude. Thus the argument is spurious and misleading. – Arnold Neumaier Mar 5 '12 at 17:04
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@ArnoldNeumaier: you'll find exactly this argument in any graduate textbook on quantum mechanics – John Rennie Mar 5 '12 at 17:18
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...but surely not phrased in this sloppy way. You conclusion that the pauli principle doesn't apply to large systems is also wrong. It still applies, though there are so many states anyway that it is already impossible for practical reasons to create identical states, so that the Pauli principle adds no new information. – Arnold Neumaier Mar 5 '12 at 17:41
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I agree that this could use an edit. e1 and e2 should be electron positions not electrons, but moreover "Suppose the two electrons were exactly the same" is not very clear... do you mean that the two-particle state is a product of two of the same single-electron state? E.g. Psi(x,y) = psi(x)psi(y)? The exclusion principle indicates that such states are forbidden for fermions, whereas psi1(x)psi2(y) - psi2(x)psi1(y) would be allowed. I assume that's what you're getting at, but it's not very clear. – Tim Goodman Aug 15 '12 at 19:49
This answer does need to be cleaned up...perhaps Arnold could edit it to properly represent the wavefunction's antisymmetry as a function of position with concise definitions about what "exchanging electrons" is. My graduate class on the subject did use position, but in my opinion didn't define very well mathematically what "exchanging electrons" was... – daaxix Jan 29 at 18:12
show 1 more comment
Cox said that when he heats the diamond every electron in the universe shifts energy levels instantly to respect the Pauli Exclusion Principle. The problem here is not that he is talking about energy levels. The set of levels is a set of eigensates of the hamiltonian and the fact that some of them are degenerate does not invalidate what he said. In fact his diagrams showed different spin states at the same energy level so he was making this evident.
There are however some other aspects of his statement that are worth objecting to. One problem is that energy states for free electrons are not at discrete levels. It is not clear that what he says makes sense in an open system.
Another problem is that the states are not ordered in such a way that everything can move up one level.
Also the picture he is basing his claims on assumes that electrons are in energy levels of fixed systems, but the other particles are moving too. How does this affect the energy levels of the electrons?
Perhaps the strangest part of his claim was that he said everything would move instantly. Was he invoking the entanglement of the electrons? This does not make sense since he is not merely observing the diamond, he is heating it up. the kind of effect he is describing could not propagate faster than light.
Even taking into account the need to simplify for the public audience I think it is a bit of a stretch to attach any real meaning to what he said or the logic behind it.
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By the way, aside from this it was a great TV lecture. There are not many people who can stand up in front of an audience and make physics that entertaining – Philip Gibbs Dec 21 '11 at 18:58
I think that Jim Parsons would still win in this respect, Phil. He has more physically accurate writers of the script, too. – Luboš Motl Mar 13 '12 at 20:17
Brian Cox is wrong. Period. It is not at all about energy states, it is about an antisymmetric wavefunction. For example, every atomic energy level can contain 2 electrons - one with spin up and one with spin down and they can have the same energy. I wouldn't waste any more time on his rubbish...
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-1, it is an antisymmetric wavefunction, but you don't explain how it is wrong very well... – daaxix Jan 29 at 18:08
Such claims as with these electrons should not be taken as strict fact, or used to draw such conclusions about physical reality.
For two identical systems, such as two neutral hydrogen atoms, separated by a great distance, the two electrons have identical energies. When considering both atoms together as one quantum system, even when far apart, we must look at the energy levels of the sum and the difference of wavefunctions of the two electrons. Typically one is slightly larger, the other slightly smaller than the original levels. These shifts in energy are greater with more overlap of wavefunctions, and zero when the atoms are infinitely far apart.
Wavefunctions describing these electrons fall off exponentially with distance away from the system. Think of this as a "spatial half-life" with a distance scale of nanometers, maybe microns. At human-scale distances, for all practical purposes the overlap is zero.
According to ideal math, which gives us ideal mathematical forms for wavefunctions satisfying perfect wave equations in an idealized physical system, the overlap and energy shifts are nonzero, but so absurdly tiny if we're talking about human-scale situations or bigger. Never mind astronomical distances.
These ideal equations are not to be ignored - Nobel Prizes have been won for making precise measurements of electrons, precise to ten digits or so (check wikipedia for the latest) and agreeing with the predictions of quantum electrodynamics. It works well.
Using QED, we can estimate the energy shifts involved in the sorts of situations presented by Dr. Cox. These are so absurdly tiny that they'd be swamped out by much larger tiny effects such as the gravitational effects of a speck of dust anywhere in the vicinity of either atom, Doppler shifts due to even the slightest motion - like trillionths of miles-per-hour.
Heisenberg says that quantum mechanical systems don't even have precise energy levels beyond any particular precision, when you measure the energy over some corresponding time interval. To have a meaningful energy level so precise as to distinguish between the sum and difference wavefunctions, the system must be stationary, untouched, for a very long time - we're talking about cosmic time scales. While the exact math lets calculate such things, the real physical world isn't obligated to be so precise at that level.
In short, Dr. Cox is making a wild and idealized extrapolation.
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I think what Dr. Cox is assuming is that two neutral hydrogen atoms, at great distance, can be viewed as being in an energy eigenstates of the system composed of the two atoms. These energy eigenstates will be have slightly different energies. But this is only true at the time scale of hbar over the difference in the two energies, which is enormous. – Peter Shor Apr 17 at 16:10
It is a bit hard to say something that captures the essence of your question without some mathematics, but nonetheless, I'll give it a go. :)
In my view, field theory puts all these so called "paradoxes" to rest. To state is simply, there is only one electron field in the universe. What we perceive as particles are excitations in the field due to coupling with other fields. If you take this view, then all you are doing is measuring attributes of the same field at different space-time points.
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IMHO this answer would be satisfying if you mentioned how Pauli exclusion principle works in QFT. – Adam Zalcman Dec 20 '11 at 17:06
The antisymmetry of the wave function is a consequence of the anticommutation rule for fermions in QFT. And the Pauli principle is expressed by the fact that applying the same creation operator twice annihilates a state. – Arnold Neumaier Mar 5 '12 at 18:16
I wrote to Professor Cox the day after his TV lecture raising basically the issues discussed here. In particular the idea of instantaneous action seems particularly hard to swallow, and it seems the answer, which seems no answer at all, is that only information cannot be transmitted faster than light velocity. But telling another electron a billion km away of its energy state sounds like information to me. As to the other points, he invited me to read his book which I am doing but am no closer to accepting his arguments yet. It all has a slightly mystical feel to it which I know he would hate to be accused of, and indeed much of quantum mechanics has been held in such belief in the past. It is only the 'well it works approach' which has justified the seemingly outrageous propositions, there being no other serious contenders. We are on a road to understanding something fantastic but nobody has any idea of what it is.
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The Pauli exclusion principle does not explicitly reference energy. It states that no two identical fermions (in this case electrons) may occupy the same quantum state. In a system where each quantum state corresponds to a unique energy level, then you could infer that no two identical fermions may occupy the same energy level, otherwise no such constrain exists. An example already given is where two electrons may occupy the same energy level in an atom. This is because the energy level corresponds to the orbital angular momentum of the particle, but there is an additional degree of freedom, the spin of the electron. Of course the Pauli exclusion principle operates here by ensuring NO MORE than 2 electrons can be at this energy level, because there are only two possible spin states.
If I understand you correctly, the way you've interpreted what Brian Cox is saying is that, given some electrons in a diamond, which are at some energy, no other electrons in the universe may also be at that energy. So if you change the energy of the diamond by heating it up, then any electrons in the universe which happen to be at this new energy will have to adjust in order to accommodate this change. I haven't seen what Brian Cox said, so I don't know whether what you've paraphrased is accurate, but if this is what he was implying, then he's dead wrong. This is simply not true.
One could extend this reasoning to consider an electron in an atom. Given this rule that no other electrons in the universe may also be at this energy, one must conclude that no other atom may contain an electron at this energy, thus making consistent chemical behaviour impossible. The point is that being in one atom or another corresponds to a different quantum state, so no violation occurs when one electron in one atom happens to be in the same quantum state as an electron in another atom, since there's the added degree of freedom of being in one atom or the other. The same kind of thing holds for crystals and all the other electrons in the universe.
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My take on what he said was that the energy he imparted to the diamond will pass as heat to the atoms and electrons of air immediately surrounding the diamond, then on to others next to them, eventually it will be radiated into space and, as energy cannot be destroyed will continue it's journey across the universe long after the planet has been consumed by our expanding sun , minutely raising the energy levels of the electrons it meets. In other words entropy increasing from the chemical energy in his food to mechanical energy in his hand to thermal energy in the diamond/air/interstellar space.
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@GordonM One has to be careful when making arguments about energy (energy leves). There is a concept which has been overlooked in this interesting discussion. Let us accept for a moment professor Cox's argument. Now, let us assume that an electron somewhere inside the diamond gets a sufficient amout of energy, 0.5eV say, to jump from energry level E_1 to another energy level or virtual state E_2. There are about 10^59 Kg of matter in the universe, and therefore it is very likely that there are an extremely large number of electrons in the universe, which occupy that particular energy level E_2, and they should move to some other energy level, and let us assume there is no domino effect by doing this. The amount of energy needed for all those electrons in the universe to do so can be immense! It is clear that we have a problem here. Where is that extraordinarily large amount of energy going to come from? One can argue that, some electrons will move up and some will move down one level so it will all balance out in the end! If that is the case then, we should be able to observe this effect happening with atoms in our laboratories, as atoms should suddenly emit light just because a piece of diamond at some other part of the universe was warmed up a bit!!??
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Any discussion on the subject of particle physics is bound to be fatuous unless it can conducted in explicit mathmatical terms and what I understand on the matter for example would be unlikely to fill the back of a postcard so I hesitate to comment. Having seen the program however, my take on what Cox was trying to explain was the inter-relatedness of all matter. IE photons exist only in the sense of 2 events that occur in 2 different places. So Somewhere in the universe an electron jumps to a lower energy state and somewhere else another electron jumps to a higher state. We interpret this as a photon passing between the 2 locations. Before the event the path of the photon (ie the location of the 2 events) can only be predicted in terms of probability and mapped as a field. Intuitively it seems reasonable to me that the Pauli exclusion principle could apply to any system one might care to define even though different parts of the system may be physically very remote from each other. I do not believe therefore that Cox was suggesting that because an electron on one atom (say the hydrogen atom at the back of my eye) was in one energy state then no electrons attached to other hydrogen atoms in the universe could have an electron in a similar state. What he did seem to be suggesting however was that the Pauli exclusion principle applies across the system defined by the 2 events and not simply constrained to a single atom. IE the principle is a universal law.
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If this is actually what Prof. Cox said (I expect that it isn't what he intended to say, but maybe what he actually said), it's misleading. The Pauli exclusion principle just says that two electrons can't occupy the same state, but the description of the state includes position as well as energy. – Peter Shor Apr 17 at 16:04
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http://mathhelpforum.com/math-topics/179898-problem-pulley.html
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# Thread:
1. ## Problem with pulley
Hey, i have the mark scheme for this question but i simply do not understand it.the question is here : maths problem , part E . the answer is here:answer .
I don't understand, what i did was use equations of motion , from the point when Q reverses and falls 3.15 m, therefore giving me the value of V from v^=u^2+2as
v^2=0+2x(9.8x3.15)
v=7.857 m/2
so i used V=U+at
giving t=0.8017
What is wrong with my method? Would appreciate help, been bugging me all day
2. Originally Posted by chartsy
Hey, i have the mark scheme for this question but i simply do not understand it.the question is here : maths problem , part E . the answer is here:answer .
I don't understand, what i did was use equations of motion , from the point when Q reverses and falls 3.15 m, therefore giving me the value of V from v^=u^2+2as
v^2=0+2x(9.8x3.15)
this is wrong. the above equation is true for a particle in free fall. P is not in a free fall
v=7.857 m/2
so i used V=U+at
giving t=0.8017
What is wrong with my method? Would appreciate help, been bugging me all day
now try it.
3. when P hits the ground, Q is still rising at speed $v = at = 4.2 \, m/s$
time for it to rise and fall to its original position is
$t = \dfrac{2 \cdot v_0}{g} = \dfrac{6}{7} \, sec$
4. so...you kind of combine the upmove and the downmove into the one equation v=u+at? i didn't know you could do that , i thought you could only use equations of motion one direction at a time
5. what i also don't get is, why should the speed stay at 4.2 m/s? i mean, if Q falls back to where it was when P hit the ground, surely ,given that -9.8/ms^2 is greater than -4.2ms^2, the speed V would be bigger than 4.2?
6. Originally Posted by chartsy
so...you kind of combine the upmove and the downmove into the one equation v=u+at? i didn't know you could do that , i thought you could only use equations of motion one direction at a time
You aren't combining them. You are using them one at a time. You need to find the speed that Q is moving upward initially (the speed it gets to after P hits the ground). That gives you the v0 for the problem after P is on the ground.
-Dan
7. Originally Posted by chartsy
what i also don't get is, why should the speed stay at 4.2 m/s? i mean, if Q falls back to where it was when P hit the ground, surely ,given that -9.8/ms^2 is greater than -4.2ms^2, the speed V would be bigger than 4.2?
I get it! it's symmetrical because it decelerates as fast as it accelrates!
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http://math.stackexchange.com/questions/260518/which-of-the-following-sets-are-dense-in-mathbbr2-with-respect-to-the-usua?answertab=votes
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# Which of the following sets are dense in $\mathbb{R}^2$ with respect to the usual topology?
Which of the following sets are dense in $\Bbb R^2$ with respect to the usual topology.
1. $\{(x,y) \in\mathbb{R}^2:x\in \mathbb{N}\}$
2. $\{(x,y) \in\mathbb{R}^2:x+y \text{ is a rational number}\}$
3. $\{(x,y) \in\mathbb{R}^2:x^2+y^2=5\}$
4. $\{(x,y) \in\mathbb{R}^2:xy\neq 0\}$
Clearly 1 is false.
3 is false as it is bounded and closed
4 is true as it is the set of all points that are not on the axes x and y.
Am I correct.
But I am not sure about 2 but my guess is true as rationals/ irrationals are dense and 2 holds iff either both are rational or conjugate irrational .
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Looks good to me. – Clayton Dec 17 '12 at 5:51
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I assume that the $\mathbb{R}$ in the title is intended to be $\mathbb{R}^2$. – André Nicolas Dec 17 '12 at 5:59
## 2 Answers
The set in 2 contains $\mathbb Q \times \mathbb Q$ and so is dense in $\mathbb R \times \mathbb R$.
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Let $(a,b)$ be a point in $\mathbb{R}^2$, and let $\epsilon\gt 0$. We show that there exist rational numbers $s,t$ such that $\sqrt{(a-s)^2+(b-t)^2}\lt \epsilon$. In particular, $s+t$ is rational.
Since the rationals are dense in $\mathbb{R}$, there is a rational $s$ such that $|a-s|\lt \dfrac{\epsilon}{\sqrt{2}}$. Similarly, there is a rational $t$ such that $|b-t|\lt \dfrac{\epsilon}{\sqrt{2}}$.
It follows that $\sqrt{(a-s)^2+(b-t)^2}\lt \epsilon$.
A similar argument but somewhat simpler argument takes care of (4).
The fact that the set is bounded is enough for (3). Or the fact that there is no point on the circle that is anywhere near $(0,0)$.
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so only 2 and 4 are correct.am i right?? – priti Dec 17 '12 at 6:07
Yes, you had the right answers in all cases. I wrote out some details only because you expressed some doubt. – André Nicolas Dec 17 '12 at 6:12
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http://mathhelpforum.com/differential-geometry/163257-branch-cuts-integrals.html
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# Thread:
1. ## Branch Cuts and Integrals
Hello again,
I've come across a new topic that involves solving integrals that involve logarithms. The examples in the text are not too clear how this is done. Can anyone help me solve these:
$\int_{0}^{\infty} \frac{logx}{1+x^4} dx$
$\int_{0}^{\infty} \frac{(logx)^2}{1+x^2} dx$
$\int_{0}^{\infty} \frac{\sqrt{x}logx}{1+x^2} dx$
Any help would be really appreciated. I just can't seem to figure out the right steps to go about solving these problems. I learn backward and need to see examples I understand to get the concept. Thanks in advance!
2. In order to illustrate the procedure we will consider the integral...
$\displaystyle \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}$ (1)
... the others are similar. The integrand function has in $z=0$ a singulatity of the type branch point and we have to select an integration path that doesn't contain this point. The best cadidate is the 'red path' in the figure...
First step is to valuate the integral...
$\displaystyle \int_{c} \frac{\ln z}{1+z^{4}}$ (2)
... where c is the path ABCDEFGHA... how can we proceed?...
Kind regards
$\chi$ $\sigma$
3. The second step is the computation of the integral...
$\displaystyle \int_{c} \frac{\ln z}{1+z^{4}}$ (1)
... along the 'red path' in the figure. At this scope we use the residue theorem...
$\displaystyle \int_{c} f(z)\ dz = 2\ \pi\ i\ \sum_{k} r_{k}$ (2)
... where...
$\displaystyle r_{k} = \lim_{z \rightarrow z_{k}} (z-z_{k})\ f(z)$ (3)
... are the residues of the poles of f(*) inside the path c. In our case the poles are at $\displaystyle z_{k} = e^{i (2 k+1) \frac{\pi}{4}}$ , $k=0,1,2,3$ so that...
$\displaystyle r_{k} = i\ \frac{\pi}{16}\ (2k+1)\ e^{-i (2k+1) \frac{3}{4} \pi}\implies \int_{c} f(z)\ dz = 2\ \pi\ i\ \sum_{k=0}^{3} r_{k} = -i\ \frac{\pi^{2}}{\sqrt{2}}$ (4)
Honestly in pure calculus I am a little poor [] , so that I propose a little break hoping that somebody verifies my computation, in particular the result (4)...
Kind regards
$\chi$ $\sigma$
4. I'm afraid that HalsofIvy's congratulations are a little premature for the reason that we will see now. In the last post we have found that...
$\displaystyle \int_{c} \frac{\ln z}{1+z^{4}}\ dz = -i\ \frac{\pi^{2}}{\sqrt{2}}$ (1)
The (1) can be write as...
$\displaystyle \int_{r}^{R} \frac{\ln x}{1+x^{4}}\ dx + i\ R\ \int_{0}^{2 \pi} \frac{\ln R + i\ \theta}{1+R^{4}\ e^{4 i \theta}}\ e^{i \theta}\ d\theta +$
$\displaystyle + \int_{R}^{r} \frac{\ln x + 2\ \pi\ i}{1+x^{4}}\ dx + i\ r\ \int_{2 \pi}^{0} \frac{\ln r + i\ \theta}{1+r^{4}\ e^{4 i \theta}}\ e^{i \theta}\ d\theta = -i\ \frac{\pi^{2}}{\sqrt{2}}$ (2)
Now if in (2) $R \rightarrow \infty$ and $r \rightarrow 0$ the second and the fourth integral vanish and we obtain...
$\displaystyle \int_{0}^{\infty} \frac{\ln x }{1+x^{4}}\ dx - \int_{0}^{\infty} \frac{\ln x }{1+x^{4}}\ dx - 2 \pi i \int_{0}^{\infty} \frac{dx}{1+x^{4}} = -i\ \frac{\pi^{2}}{\sqrt{2}} \implies \int_{0}^{\infty} \frac{dx}{1+x^{4}} = \frac{\pi}{2\ \sqrt{2}}$ (3)
... that is a very good result but... not the goal we were searching for ... may be that we have to take a little different way...
Kind regards
$\chi$ $\sigma$
5. How about doing the same analysis but use the integral:
$\displaystyle\int \frac{\log^2(z)}{1+z^4}dz$
6. From the 'stubborn old wolf' a new attempt... the function to be integrate is again $f(z) = \frac{\ln z}{1+z^{4}}$ but the new 'red path' is illustrated in figure...
First step is the computation of the residues of the two poles inside the 'red path' and that gives...
$\displaystyle \int_{c} \frac{\ln z}{1+z^{4}}\ dz = 2\ \pi\ i\ \sum_{k=0}^{1} r_{k} = - \frac{\pi^{2}}{4 \sqrt{2}} + i\ \frac{\pi^{2}}{2 \sqrt{2}}$ (1)
As in the previous post coth the contribute of the 'small half circle' for $r \rightarrow 0$ and the contribute of the 'big half circle' for $R \rightarrow \infty$ vanish so that is...
$\displaystyle \int_{c} \frac{\ln z}{1+z^{4}}\ dz = \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}\ dx + \int_{-\infty}^{0} \frac{\ln (-x) }{1+x^{4}}\ dx =$
$\displaystyle = 2\ \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}\ dx - i\ \pi\ \int_{0}^{\infty} \frac{dx}{1+x^{4}} = - \frac{\pi^{2}}{4 \sqrt{2}} + i\ \frac{\pi^{2}}{2 \sqrt{2}}$ (2)
From (2) we derive immediately...
$\displaystyle \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}\ dx = -\frac{\pi^{2}}{8\ \sqrt{2}}$ (3)
$\displaystyle \int_{0}^{\infty} \frac{dx}{1+x^{4}} = \frac{\pi}{2\ \sqrt{2}}$ (4)
In Italy we say: due piccioni con una fava! ...
Kind regards
$\chi$ $\sigma$
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http://mathhelpforum.com/algebra/182869-help-distributive-property-print.html
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# Help with distributive property
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• June 11th 2011, 08:17 PM
tcampbell011
Help with distributive property
Hey I was doing a math problem and needed help with one certain part near the end of the problem. Here is the problem and what I have done already:
8(1/x-7)=9(1/x)+3(1/5x)
8/x-7=9/x+3/5x
then I multiply each side by the least common denominator. I understand what to do on the left hand side, I just don't understand how to distribute the x-7 on the right (I assume the 5x gets canceled out:
5x(x-7)*(8/x-7) = (9/x+3/5x)*5x(x-7)
40x=(9/x+3)*(x-7)
Any help would be greatly appreciated!!!! I'm really frustrated! I know the answer is 48x-336 but I can't see how to get there!
• June 11th 2011, 08:33 PM
Prove It
PLEASE use brackets where they're needed or learn some LaTeX, it's impossible to tell if you mean $\displaystyle \frac{1}{x - 7}$ or $\displaystyle \frac{1}{x} - 7$, or if you mean $\displaystyle \frac{1}{5x}$ or $\displaystyle \frac{1}{5}x$.
From your working, it looks like you meant
$\displaystyle \begin{align*}8\left(\frac{1}{x - 7} \right) &= 9\left(\frac{1}{x}\right) + 3\left(\frac{1}{5x}\right) \\ \frac{8}{x - 7} &= \frac{9}{x} + \frac{3}{5x} \\ \frac{8}{x - 7} &= \frac{45}{5x} + \frac{3}{5x} \\ \frac{8}{x - 7} &= \frac{48}{5x} \\ 8(5x) &= 48(x - 7) \\ 40x &= 48x - 336 \\ 336 &= 8x \\ 42 &= x \end{align*}$
• June 11th 2011, 08:36 PM
tcampbell011
Thank you, I understand it now... I see how it could be confusing, but you nailed it. I will try to figure out what LaTeX is.
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http://math.stackexchange.com/questions/239839/any-partition-of-1-2-ldots-9-must-contain-a-3-term-arithmetic-progressi
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# Any partition of $\{1,2,\ldots,9\}$ must contain a $3$-Term Arithmetic Progression
Prove that for any way of dividing the set $X=\{1,2,3,\dots,9\}$ into $2$ sets, there always exist at least one arithmetic progression of length $3$ in one of the two sets.
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What have you tried? Perhaps you could expand your post, because I think I know what you mean, but can't be sure. – Mark Bennet Nov 18 '12 at 14:49
What?? After "which" I got lost... – DonAntonio Nov 18 '12 at 14:49
I cannot believe that the "dirichlet series" tag is appropriate for this, but until the question is clarified it is a lottery to chose a better one. But if this is homework, that should be added. – Mark Bennet Nov 18 '12 at 14:55
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anyone understand me? arithmetic which has length 3 is in one of 2 sets that means the arithmetic $<a;a+d;a+2d$ – LevanDokite Nov 18 '12 at 15:01
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– sperners lemma Nov 18 '12 at 15:54
## 1 Answer
This result is part of a nice area in Ramsey theory. If you want to study generalizations, the term you want to look for is Van der Waerden number; we are saying here that $w(2,3)\le 9$, where the $3$ indicates you want an arithmetic progression of length three, and the $2$ indicates that yo are splitting the set in two pieces. In fact, $w(2,3)=9$, meaning that, in addition, there is a way to split the numbers from $1$ to $8$ into two pieces, both avoiding such triples: $\{1,2,5,6\}$ and $\{3,4,7,8\}$.
To see that $9$ suffices, the easiest argument proceeds by an analysis of cases: Consider a splitting of $X$ into two sets, and let's attempt to see what restrictions these sets must satisfy in order to avoid arithmetic triples. We must conclude that it is impossible to have such a splitting. The key in my approach is to consider $4,5,6$. They cannot all be in the same piece, but two of them must be. Let's call that piece $A$, and let $B$ be the other one.
• Case 1. $4,6\in A$.
This is the easiest case to eliminate, since $2,5,8$ must then be an arithmetic triple in $B$: Consider, respectively, the triples $2,4,6$, and $4,5,6$, and $4,6,8$. If we place any of $2,5,8\in A$, one of these three arithmetic triples ends up in $A$.
• Case 2. $4,5\in A$.
Then $3,6\in B$ (consider, respectively, the triples $3,4,5$ and $4,5,6$), so $9\in A$ (consider $3,6,9$), but then $1,7\in B$ (consider $1,5,9$ and $5,7,9$), so $2,8\in A$ (consider $1,2,3$ and $6,7,8$). We now see this case cannot be either, because the triple $2,5,8$ is in $A$.
• Case 3. $5,6\in A$.
This is really the same as case 2, by symmetry. (In this case, $4,7\in B$, so $1\in A$, so $3,9\in B$, so $2,8\in A$, and we see that the triple $2,5,8$ is in $A$.)
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do you know an argument that isn't case analysis? – sperners lemma Nov 18 '12 at 16:41
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I wanted to say that case 3 should be the same as case 1 by symmetry, and eventually realized that $[1\ldots 9]$ is not symmetric about $[3,4,5]$. I wonder if an argument similar to this one, but using $[4,5,6]$ instead of $[3,4,5]$, would be simpler. – MJD Nov 18 '12 at 16:45
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I doubt there is one. "Softer" arguments that avoid case analysis do not tend to give precise bounds. The nice argument in "Ramsey theory" by Graham-Rotschild-Spencer, for example, gives $325$ as an upper bound. (The argument there is fairly intuitive, and one sees immediately how to generalize. But it is of course terrible for concrete bounds.) – Andres Caicedo Nov 18 '12 at 16:46
@MJD You are right. Using $4,5,6$ gives us a bit more symmetry, so case 3 would just be the "reflection" of case 1. I should have used that triple. Thanks! – Andres Caicedo Nov 18 '12 at 16:48
I didn't think it through myself, so I don't know whether it happens to complicate the other two cases enough that the gain from the symmetry is not actually a win. – MJD Nov 18 '12 at 16:49
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http://mathoverflow.net/questions/34232/injective-maps-mathbbrn-to-mathbbrm/34253
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Injective maps $\mathbb{R}^{n} \to \mathbb{R}^{m}$
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Let $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ be an injection for $n>m$. Can $f$ be continuous? Why?
I got this question in mind when I was trying to find a continuous map from $\mathbb{R}^{2}$ to $\mathbb{R}$.
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4 Answers
If $f$ is injective and continuous from $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n>m$ then $f$ restricts to a continuous bijection from $S^{n-1}$, the unit sphere in $R^n$, to a compact subset $K$ of $\mathbb{R}^m$. Thus you can embed $S^{n-1}$, and a foriori $S^m$ in $\mathbb{R}^m$.
But there are homological obstructions to embedding $S^m$ in $\mathbb{R}^m$. Using the arguments of this excellent paper
Albrecht Dold, A simple proof of the Jordan-Alexander complement theorem, Amer. Math. Monthly 100 (1993), 856-85.
(essentially a cunning use of the Mayer-Vietoris theorem) it would entail the homology of the space $\mathbb{R}^m-K$ being nonzero in negative dimension, which is absurd.
Added As I replied in haste I forgot the sledgehammer that cracks this little nut, namely Alexander duality.
Added later In fact this result also follows from Brouwer's Invariance of Domain. This is Theorem 2B.3 on page 172 of Hatcher's book. This implies that if one has an embedding from $\mathbb{R}^n$ to itself, then its image is open. One gets such an embedding by composing your putative embedding with the natural embedding of $\mathbb{R}^m$ in $\mathbb{R}^n$. Adapting the proof, gives a swift proof that the answer of your original question is no.
If you have a continuous injection from $\mathbb{R}^n$ to $\mathbb{R}^m$, with $n>m$ then you have an embedding of $S^{n-1}$ into a nontrivial hyperplane in $\mathbb{R}^n$. By the Jordan-Brouwer separation theorem the image $K$ of this embedding separates $\mathbb{R}^n$ but it's easy to see that since the image is in a hyperplane any two points of the complement of $K$ can be connected by a path (exercise for reader :-)).
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There's a simpler argument when m=1. Any such map would induce a continuous map $\mathbb{R}^n\setminus\{p\}\rightarrow \mathbb{R}$ whose image is disconnected, a contradiction. Here $p$ is the preimage of any interior point of the original image. – Kevin Ventullo Aug 2 2010 at 10:18
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Dear all, Chandru usually wishes elementary (or at least "ingenious") arguments; cf. mathoverflow.net/questions/34055/… . – Wadim Zudilin Aug 2 2010 at 10:52
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Set of measure zero, the answer is different. The Baire space $J$ of irrationals is homeomorphic to $J \times J$, which gives us a 1-1 map from $J \times J$ into $\mathbb{R}$. – Gerald Edgar Aug 2 2010 at 12:28
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Chandru, is it a simple question to prove that $\mathbf{R}^m$ and $\mathbf{R}^n$ are not homeomorphic for $m\ne n$? – Robin Chapman Aug 2 2010 at 15:22
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Chandru, I don't know any way of proving that $\mathbb{R}^m$ and $\mathbb{R}^n$ for $m\ne n$ save by using the methods or results of algebraic topology such as homology theory or the Jordan-Brouwer separation theorem. Your question looks just as hard, and to me at least, it looks unlikely there's a naive proof. – Robin Chapman Aug 2 2010 at 17:20
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You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Alternatively, you may use the Borsuk-Ulam antipodal theorem, in order to prove that such a map cannot be one-to-one.
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More generally, you might ask whether there is a continuous injective map from a space of dimension $n$ to a space of dimension $m$, where $n > m$. The subject of dimension theory investigates this problem. (Hurewicz & Wallman wrote a classic, though hard to find nowadays, book on the subject; perhaps an expert could recommend a more modern reference?) Here is a representative theorem:
If $f: X \to Y$ is a continuous surjection bewteen separable compact metric spaces, where $X$ has dimension $n$ and $Y$ has dimension $m$, then there is a point $y \in Y$ whose preimage contains at least $n - m + 1$ points.
This applies to your question, as Robin Chapman explained, by restricting your map to a nice compact subspace of dimension $n$, for instance the unit closed ball.
In any case, my point is that the techniques above for Euclidean space and for spheres have appropriate generalizations to separable metric spaces.
As to the "why?", I recommend tracking down a copy of Hurewicz & Wallman from a library. It's very pleasant reading and assumes only a brief encounter with point set topology.
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For a more modern reference: Engelking, Theory of Dimensions, finite and infinite. For separable metric spaces, also Jan van Mill, "infinite-dimensional topology, prerequisites and an introduction", or the newer "the infinite-dimensional topology of function spaces", both of which have a chapter on this subject. – Henno Brandsma Aug 2 2010 at 18:56
I taught an introductory topology course this last autumn where I covered this theorem from an elementary point of view. The argument just uses the Brouwer fixed point theorem (which itself has a proof via Stokes' theorem which is readily accessible to students with several variable calculus) plus elementary point set topology. In particular no homology or Jordan--Brouwer separation theorems are used. The treatment was based upon that of Hurewicz & Wallman but was also inspired by Larry Guth's ICM-2010 presentation in Hyderabad.
If you're interested the notes are available on
http://www.maths.ed.ac.uk/uploads/assets/32_section6.pdf
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http://math.stackexchange.com/questions/222132/ideals-in-localization-of-dedekind-domains
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# Ideals in localization of Dedekind domains
If $A$ is a Dedekind domain and $a,b$ are ideals, then why does $aA_p⊂bA_p$ for every prime ideal $p$ imply that $a\subset b$?
I read it in Milne's notes but it alludes to DVR's and I'm not familiar with that concept.
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## 2 Answers
Let $x$ lie in the first ideal. Consider the set $I$ of elements $a$ of $A$ such that $ax$ is in the second ideal. Check that $I$ is an ideal and that the hypothesis implies that $I$ is not contained in any nonzero prime ideal of $A$. Hence $I=A$, and so $1$ is in $I$.
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Factor $a$ and $b$ into its finite product of prime ideals. If say $$b=(p_1) \cdots (p_n),$$ $b_{p_i} = (p_i)_p$, so you have that $a$ also have $$(p_1) \cdots (p_n)$$ in its factorization. Thus, it must be contained in $b$.
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http://www.physicsforums.com/showthread.php?p=1726905
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## Mass: Matter or Inertia?
Well my trouble stems from that I learned Inertia to be a property of all masses, the property that all masses will not accelerate unless a force is applied. To me, this was always just a property, like a square having adjacent sides at right angles.
Today in my Physics class I was told : Mass is a measure how much inertia as object has, when previously I had replaced Inertia with Matter. I was somewhat confused and even right now, I think that was wrong - its like having 2 cubes with one with edges 2 units and the other 5 units, and then asking which cubes edges were "more equal" to each other. I asked my teacher and he said just because it was a property didn't mean it wasn't quantifiable, like density. But his example of density doesn't seem to do it for me, the fact that an object has a density merely states all objects that have mass take up a finite volume.
So my question is: Is inertia quantifiable? If so, what are its SI Units? I asked my teacher that as well and he seemed to ignore that question =[
Thanks for any replies guys, greatly appreciated.
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus I agree with what you were told today: mass in a measure of inertia. In other words, mass is how me quantify inertia. Mass also 'happens' to be how we quantify the amount of matter that a particular object has. I must admit that I don't follow your cube analogy, but if density doesn't do it for you, perhaps a different analogy will. Consider electric charge, now you agree that electric charge is a property of matter, yes? It is the property that two 'like' charges will repel and two 'unlike' charges attract. Hopefully you will also agree that we can definitely quantify the 'amount' of charge an object has.
Quote by Gib Z Well my trouble stems from that I learned Inertia to be a property of all masses, the property that all masses will not accelerate unless a force is applied. To me, this was always just a property, like a square having adjacent sides at right angles. Today in my Physics class I was told : Mass is a measure how much inertia as object has, when previously I had replaced Inertia with Matter. I was somewhat confused and even right now, I think that was wrong - its like having 2 cubes with one with edges 2 units and the other 5 units, and then asking which cubes edges were "more equal" to each other. I asked my teacher and he said just because it was a property didn't mean it wasn't quantifiable, like density. But his example of density doesn't seem to do it for me, the fact that an object has a density merely states all objects that have mass take up a finite volume. So my question is: Is inertia quantifiable? If so, what are its SI Units? I asked my teacher that as well and he seemed to ignore that question =[ Thanks for any replies guys, greatly appreciated.
Inertia = mass.
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## Mass: Matter or Inertia?
Ok well that definitely is a good example Hoot, I guess this resolves itself out if I change my definition of Inertia? Because saying Inertia is a property of Masses when Inertia = mass is a bit weird lol.
Quote by Gib Z Ok well that definitely is a good example Hoot, I guess this resolves itself out if I change my definition of Inertia? Because saying Inertia is a property of Masses when Inertia = mass is a bit weird lol.
How do you define mass? Just think about it.
I think your initial ideas regarding inertia are correct, although I don’t think your analogies with geometric figures help out. You mentioned SI units…No, inertia is not an SI unit or any other unit for that matter, it is the property of matter which resists acceleration as expressed in Newtons first law. But let’s look at some SI units.. Length(meter), mass(kilogram) and Time(second) are base units in the SI system, and force(newton) is a derived unit . 1 newton is the force required to accelerate a mass of 1 kilogram 1 meter per second squared). This relationship is expressed in Newtons first law as F=m a, or as Newton expressed it “Every body persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed on it” Since mass is an SI base unit defined by a cylinder of platinum, mass represents a quantity of matter. Inertia on the other hand is the property of matter which resists acceleration as expressed in Newtons first law. This is how I see it, and I can see how your instructors view could be confusing. On the other hand, I might be the confused one.
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Quote by lightarrow How do you define mass? Just think about it.
I don't know how to, thats the problem lol! Defining it as mass or inertia still doesn't do justice to my mathematically reliant brain :( If anyone has some nice mathematical definition, please show!
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Quote by Gib Z I don't know how to, thats the problem lol! Defining it as mass or inertia still doesn't do justice to my mathematically reliant brain :( If anyone has some nice mathematical definition, please show!
Personally I find the definition,
$$m = \frac{|\mathbf{a}|}{|\mathbf{F}|}\hspace{2cm}|\mathbf{F}|\neq0$$
rather intuitive
Recognitions: Homework Help I was thinking of that, but then my definition of Force must be completely non reliant on mass, and I define a newton as the force required to accelerate a mass of 1kg by 1ms^-2 :(
Mass is a base unit in the SI system of units. It is not a derived unit. Here is what a kilogram of mass is *: The kilogram (kg) is the unit of mass: It is equal to the mass of the international prototype of the kilogram. The internal prototype is made of platitum-iridium (90% platinum, 10% iridium) and is preserve in a vault at Serves France, by the International Bureau of Weights and Measures . There are no units of force or acceleration mentioned in this definition because mass is not defined that way. In particular, the equation mass=force/acceleration does not define mass. *Metric Units and Conversion Charts", Theordore Wildi
Quote by Gib Z Well my trouble stems from that I learned Inertia to be a property of all masses [...] So my question is: Is inertia quantifiable?
Gib Z: Maybe your definition is that of a purist so to speak. Perhaps the history-based definition has meaning exactly in the case where external force is not present while others use it in the complementary case where force is present.
One definition ends up using inertia as a yes-or-no proposition: does the body behave in a certain way in the absence of force or not? Period. Others end up extending the inertia question to become: what happens when a force IS present and end up using quantifiable numbers for this usage of inertia which then relates to mass.
My suspiscion is that much of the confusion in physics is due to the fact that we use old terms in new ways.
Recognitions: Homework Help Yes kwestion! Your second paragraph is exactly it =] So I'm guessing I'm meant to change from the first to the second now?
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Quote by jimvoit Mass is a base unit in the SI system of units. It is not a derived unit. Here is what a kilogram of mass is *: The kilogram (kg) is the unit of mass: It is equal to the mass of the international prototype of the kilogram. The internal prototype is made of platitum-iridium (90% platinum, 10% iridium) and is preserve in a vault at Serves France, by the International Bureau of Weights and Measures . There are no units of force or acceleration mentioned in this definition because mass is not defined that way. In particular, the equation mass=force/acceleration does not define mass. *Metric Units and Conversion Charts", Theordore Wildi
That doesn't answer the question. It is perfectly valid to choose other quantities as base units and derive, say, length and mass from them.
In what is called the "universal" system, you take universal constants, such as the speed of light c, the gravitational constant G, and Planks constant, h, as the "base" units, then derive units for length, time, mass, etc. from them.
Quote by Gib Z So I'm guessing I'm meant to change from the first [usage] to the second now?
Gib Z: I'm not completely convinced. My take is that inertia or inertness is an important building block for the terms mass and momentum, but not necessarily the other way around. After all, doesn't a photon demonstrate the principle of inertia without demonstrating mass? (Ugh, I don't intend for this to branch into a discussion about refraction--yet :-) )
It may be that we're meant to accept cross-over terminology sometimes and figure it out by context. Fortunately, I think you'll agree that people seem to usually change to the word mass when they mean the second usage (responsiveness to force).
Quote by kwestion Gib Z: After all, doesn't a photon demonstrate the principle of inertia without demonstrating mass?
Hmm, I guess I'd like to retract that as an assertion and leave it as a question because of pathway concerns. Also, if a massless photon has inertia due to its momentum (a question), then that usage of the term inertia would necessarily mean that inertia is not a synonym for mass.
I'm new here and havn't learned how to use the Quote/original post feature yet...that's why i've used the Title to direct this response. The original question posted by Gib Z was “Is inertia quantifiable? If so, what are its SI Units?” I believe I have answered this original question. This question came about as a result of information he acquired in his physics class which contradicted his understanding of inertia. I became interested in this post in part because I thought Gib Z’s understanding was accurate. To clarify this issue it is necessary to explore how mass, force, and acceleration are defined under the SI system. This has been at the heart of all my posts and I think it is the proper path to “answering” his original question. As to the choice of base units, yes, other base units could have been chosen by the General Conference of Weights and Measures.
Mass = inertia = quantity of matter They're the same thing. It's like a triangle: A triangle may be defined as a polygon with 3 sides, or it may be defined as a polygon with 3 angles. This is not an inconsistency because the definitions are equivalent. Physics can be derived from more than one axiomatic system. There is no point in arguing over which is correct. You just use whichever one is most practical. As a student, of course, it is most practical to use the system your instructor is using. :)
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Introductory Physics Homework 1
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http://physics.stackexchange.com/questions/19732/at-the-molecular-level-how-is-the-pressure-at-the-bottom-of-a-lake-higher-than
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At the molecular level, how is the pressure at the bottom of a lake higher than at the top?
Surely the temperature of the molecules is the same throughout the water. Using $p = \rho g h$ seems to assume a constant density as well. But then how is it that the force per unit area on an object placed at the bottom of the lake will be higher than that on an object near the surface? My first thought is that the incompressible assumption is the approximation at fault, but it doesn't seem that a minute increase in the density of molecules at the bottom of the lake could account for many times more bombardments per square centimeter. What am I missing?
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Actually, the temperature of water isn't normally uniform throughout a lake. The water may be heated by sunlight or geothermal vents, or cooled by evaporation or conduction into the Earth, etc., and any of these processes will have a quicker and greater effect on the part of the lake near where they are working. But this question is still meaningful even if you're working with a body of water small enough that the temperature gradient is negligible. – David Zaslavsky♦ Jan 20 '12 at 5:49
2 Answers
First of all, the temperature and pressure of a liquid are two independent intensive variables. Either of them may be lower or higher at the bottom of a lake, independently of the other. So let's focus on the pressure.
You are totally right that the incompressibility fails and this is the reason why the lake "knows" about the higher pressure: the density of atoms or molecules becomes somewhat higher. But it's still true that the liquid is "approximately incompressible" and this is exactly the reason why the changes of the pressure are so great even if the changes of the density are minuscule. In other words, $$\frac{\partial p}{\partial \rho}$$ is a very large number or, equivalently, $$\frac{\partial \rho}{\partial p}$$ is a very small number. That's what we mean by (approximate) incompressibility and that's why the changes of the density unavoidably linked to finite (or even huge) changes of the pressure are so tiny.
Liquids are nearly incompressible because of Pauli's exclusion principle; the electrons in the atoms are just not allowed to occupy the same state. One has to change the structure of the states but this, because of the repulsion of the charged nuclei and electrons from each other, leads to immense increases of energy. That's why the density of liquids (or, equivalently, the volume occupied by a single atom or molecule) is de facto calculable independently of the pressure.
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In a hard-gas model, if you make the atoms spheres, there are indeed lots more bombardments low down in the lake than high up, since hard sphere collisions are the only source of force, and the temperature (hence the mean velocity) is the same. The reason this is violating your intuition is because you are approaching the continuous collision limit, so that when the hard spheres are nearly touching, the number of collisions per-second diverges. The differences in pressure in the hard-sphere gas lead to many more collisions at the bottom then at the top.
But this model is stupid, because the force between atoms on the scale of their separation in a liquid is smoother than a hard sphere. In the real case, you just push the atoms together a tiny amount, and they push very much more against each other, but in a smooth way.
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http://stats.stackexchange.com/questions/tagged/bernoulli-distribution
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# Tagged Questions
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http://math.stackexchange.com/questions/tagged/affine-geometry+homework
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# Tagged Questions
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http://physics.stackexchange.com/questions/57157/zeta-regularization-gone-bad
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# Zeta regularization gone bad
This may sound as a mathematical question, but it should be very familiar to physicists. I am trying to perform an expansion of the function $$f(x) = \sum_{n=1}^{\infty} \frac{K_2(nx)}{n^2 x^2},$$ for $x \ll 1$. Here, $K_2(x)$ is the modified Bessel function of the second kind. This series is a result of solving the integral $$f(x) = \frac{1}{3}\int_1^\infty \frac{(t^2-1)^{3/2}}{\mathrm{e}^{xt}-1}\mathrm{d}t.$$ The result should be $$f(x) \approx \frac{\pi^4}{45 x^4} - \frac{\pi^2}{12 x^2}+\frac{\pi}{6x}-\frac{1}{32}\left( \frac{3}{2}-2\gamma+2\ln4\pi-\ln x^2\right)+\mathcal{O}(x^2),$$ where $\gamma$ is the Euler-Mascheroni constant. It agrees numerically with $f(x)$ for small $x$. However, by using the series expansion of the Bessel function $$K_2(nx) = \frac{2}{n^2x^2}-\frac{1}{2}+\frac{1}{2}\sum_{k=0}^\infty \left[\psi(k+1)+\psi(k+3)-\ln\frac{n^2x^2}{4}\right]\frac{\left(\frac{n^2 x^2}{4}\right)^{k+1}}{k!(k+2)!},$$ with $\psi(x)$ being the digamma function and using the zeta regularization for summation over $n$, I am able to reproduce all the terms except $\frac{\pi}{6x}$. I.e., my result is $$f(x) = \frac{2\zeta(4)}{x^4} - \frac{\zeta(2)}{2x^2} + \frac{1}{8}\sum_{k=0}^\infty \left[\left(\psi(k+1)+\psi(k+3)-\ln\frac{x^2}{4}\right)\zeta(-2k) + 2 \zeta'(-2k)\right]\frac{\left(\frac{x^2}{4}\right)^{k}}{k!(k+2)!}.$$ It seems very strange that the $\frac{\pi}{6x}$ term should appear in the expansion since only even powers of $x$ appear in $K_2(nx)$. But, numerically, it is certainly there. How did I miss it?
I think that the missing term indicates that the zeta regularization isn't used properly. The term $\frac{\pi}{6x}$ appears right in the middle, separating the convergent sums $\zeta(4)$ and $\zeta(2)$ from the (regularized) divergent sums $\zeta(-2k)$ and $\zeta'(-2k)$. So, the missing term may be the price to pay for using the zeta regularization. Unfortunately, I don't know enough formal math to, so to speak, "repair the damage". Any help on the subject is more than welcome.
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http://math.stackexchange.com/questions/300203/how-can-i-find-the-lenght-of-a-side-of-a-polygon-with-known-number-of-sides-that
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# How can i find the lenght of a side of a polygon with known number of sides that has a circle with known diameter inscribed in it?
How can i find the lenght of a side of a polygon with known number of sides that has a circle with known diameter inscribed in it? I'm a web-developer intereseted in this certain problem, that would be the solution to one of my aplications. And also is there a relation betwen a polygon that has a inscribed polygon with knwon distance betwen their sides? It would help even an answer for particular cases like pentagon or hexagon. I hope i've been specific enough :)
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## 2 Answers
For an $n$-gon the relations between circumscribed radius $R$, inscribed radius $\rho$ and side length $a$ are $$\frac \rho R = \cos\frac\pi n$$ $$\frac a R = 2\sin\pi n$$ $$\frac a \rho = 2\tan\pi n.$$
For the second problem: If you have a polygon with side length $a$ and inscribed radius $\rho$, then the side length $a'$ for a smaller $n$-gon at distance $d$ is given by $$\frac{a-a'}{a}=\frac d\rho$$ i.e. $$a'=a\cdot\left(1-\frac d\rho\right).$$ Of course other data such as inscribed or circumscribed radius scale by the same factor $1-\frac d\rho$.
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You don't even need the "known distance". If the "known diameter" is $d$, then let the vertices of the pentagon be $A, B, C, D, E$ in counterclockwise order. Let $O$ be the center of the circle, and let $M$ be the midpoint of $AB$. Consider the right-angled triangle $AOM$; note that $\angle AOM = 360^\circ/10 = 36^\circ$. Hence, by trigonometry, $\overline{AM}/\overline{MO} = \overline{AM}/(d/2) = \tan(36^\circ)$. Use that to solve for $2\overline{AM}$, which is the side length of the outer pentagon.
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http://mathoverflow.net/questions/92099/how-many-models-of-peano-arithmetic-are-isomorphic-to-the-standard-model-and-how/92106
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## How many models of Peano arithmetic are isomorphic to the standard model and how many models of Peano arithmetic are non-standard?
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I am currently writing a paper on non-standard models of Peano arithmetic and I am having trouble finding references or information that discuss the relative sizes of how many models of Peano arithmetic there are in the standard and the non-standard cases.
I see it quoted all over the place that, "It is familiar that there are continuum-many pairwise non-isomorphic countable models of $\mathsf{PA}$". From this I take it that there are $\mathcal{c}$-many ($\aleph$-many) non-standard models of Peano arithmetic. Where can I find a proof of this fact? How many models of Peano arithmetic are there that are isomorphic to the standard model?
Thank you!
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"How many models of Peano arithmetic are there that are isomorphic to the standard model?" One, up to isomorphism. – Guillaume Brunerie Mar 24 2012 at 19:34
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Two excellent references: 1. "Models of Peano Arithmetic" by R. Kaye. (In particular, the specific fact you ask for and several variants are discussed there.) 2. "The structure of models of Peano Arithmetic", by R. Kossak and J. Schmerl. – Andres Caicedo Mar 24 2012 at 20:14
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Samuel, as for your question about how many isomorphic models of the standard model there are, it doesn't have a very useful answer. The simple answer is that there is as many as there are sets (in say ZFC), which is to say the class of standard models of PA is a proper class. For each set $A$ just consider the model of pairs $(n,A)$ where each $n$ is a standard natural number (properly coded). With an appropriate multiplication and addition intepretation this is a model of PA. That is why we usually count models (or algebraic structures) up to isomorphism only. – Jason Rute Mar 25 2012 at 15:44
## 2 Answers
Here is another way to do it.
By the Gödel-Rosser theorem, there are continuum many distinct consistent completions of PA. One can see this by building a tree of finite extensions of PA, using the Gödel-Rosser theorem at each node to branch with the Rosser sentence or its negation relative to that theory (and also deciding the $n^{\rm th}$ sentence), so that every branch through the tree is a complete consistent extension of PA. Every such consistent completion of PA has a countable model. Since different complete theories cannot have isomorphic models, you get continuum many non-isomorphic countable model of PA.
(Meanwhile, Andreas's answer applies not just to PA, but to any fixed theory, and so in fact, the compactness argument he mentions shows that each of these continuum many extensions of PA has continuum many non-isomorphic models.)
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@JoelDavidHamkins: This answer is very helpful! Thank you! – Samuel Reid Mar 24 2012 at 23:02
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For any set $S$ of (standard) prime numbers, there is, by a compactness argument, a non-standard countable model $M(S)$ of PA containing an element divisible by exactly the primes in $S$. (Actually there are many such models but I need just one.) The same model $M$ might serve as $M(S)$ for several differnt $S$´s, but only countably many, since $M$ is countable. Since there are continuum many choices for $S$, there must be continuum many non-isomorphic $M(S)$´s.
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http://physics.stackexchange.com/questions/429/how-can-we-describe-the-polarization-of-light-coming-from-an-arbitrary-angle?answertab=oldest
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# How can we describe the polarization (of light) coming from an arbitrary angle?
In an optics lab, where all optical beams pretty much reside in a plane, it is fairly simple to describe (linear) polarizations as vertical or horizontal (or s and p).
When we start talking about light coming from any angle (say, from any sky position), we now need a basis of polarization states at every sky position.
It seems to me that this requires a vector field on the sphere; and we know that no continuous, nonzero vector field on the sphere exists (the hairy ball theorem). One might try to use parallel transport to move a basis of polarization vectors from one position to another, but of course this is path-dependent and also leads to inconsistencies.
How is this resolved?
Does it make sense to compare the polarizations of light coming from two different directions?
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– j.c. Nov 19 '10 at 2:51
## 3 Answers
Sure, you can just express the polarization vector in terms of a Cartesian coordinate system, or any coordinate system you like. This requires that the polarization vectors are constrained to be perpendicular to the light's momentum, $\vec{k}\cdot\vec{\epsilon}=0$.
I'm not sure I quite understand the basis (no pun intended) of your objection, though. Is there a particular physical situation you have in mind in which this problem would arise?
Typically what's important for light coming from a wide range of angles, e.g. with sunglasses, is how much of it is horizontally polarized and how much is vertically polarized, so you can just split the polarization vector into a vertical component and a component in the 2D horizontal plane.
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As an example, how would we compare light polarized "north-south" coming from directly overhead with vertically polarized light coming from the east? with vertically polarized light coming from the north? – nibot Nov 9 '10 at 20:19
What sort of comparison would you be doing? e.g. what kind of physical process or measuring device would be involved? – David Zaslavsky♦ Nov 9 '10 at 20:24
1
The problem I had in mind was of expressing the incoming field as aX+bY, where a and b are scalar fields over the sphere, and X and Y are vector fields over the sphere which give the basis for the two polarizations "X" and "Y". For example, 'a' and 'b' could give the acceptance of an antenna for the two polarizations X and Y. – nibot Nov 16 '10 at 19:04
You don't need a vector field on the sphere - you just need vectors. Vectors don't have any intrinsic location, just a direction and a magnitude.
The polarization of light is independent of the propagation direction of the light. Let's examine this with a simple experiment:
Consider an ideal plane-wave laser beam, beam 1, propagating in the z-direction and striking a screen some distance away. Another beam with exactly the same intensity and wavelength, beam 2, hits the screen at the same point, but from a different angle. Both beams are vertically polarized -- that is to say, their E-vectors point in the x-direction.
At the point where these beams hit the screen, you will observe interference fringes, which will go all the way down to zero intensity at their darkest points. This is because the x-polarized components of the electric field interfere with each other. Besides the x-components, there are no y or z-components to interfere.
Now consider the same situation again, but with both beams horizontally polarized. That is to say, the E-field of each beam points at right angles to the beam's propagation direction, and lies in the yz-plane.
However, the polarizations are not the same, even though they are both known as "p-polarized". (At least I hope so -- I can never remember which is s and which is p.) Beam 1's E-field points purely in the y-direction, but beam 2's E-field has both a y and a z-component, as I've drawn on the beam in the illustration.
At the point where these beams hit the screen, the y-components of the electric fields interfere, once again producing interference fringes. However, even though both beams' amplitudes are the same, the y-components are not equal, so the dark parts of the fringes don't go all the way down to zero intensity.
So you see, polarization vectors are polarization vectors, no matter which direction they're coming from. Equally valid, you could define your coordinate axes according to the propagation direction of beam 2, so that beam 1 came in at an angle instead, and still arrive at the same result. Specifically, this means that you don't need to worry about parallel transport.
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The problem I had in mind concerns the radiation pattern of an antenna. If you are concerned only about the power radiated in a given direction, then there is no trouble: this is just a scalar field over the sphere. But suppose we want to break this down into two radiation patterns, one for "polarization A" and one for "polarization B"... then we run into the problem described in my question. Is there a sensible way to talk about polarization-dependent radiation patterns? – nibot Nov 16 '10 at 19:32
I think the solution is to simply give the radiation pattern as a mapping from angle to the field vector emitted at that angle. I think the "hairy ball theorem" then simply tells us that the antenna's radiation pattern must have nulls--which is a nice result rather than a problem. The point that "polarization vectors are polarization vectors, no matter which direction they're coming from" is well taken. – nibot Nov 16 '10 at 19:38
"However, the polarizations are not the same, even though they are both known as p-polarized." This is the essence of my question: can you extend the notion of "p-polarized" and "s-polarized" to beams going in all directions? The answer is no. It works in a 2D lab, but once you add the 3rd dimension, you can arrange optics to convert "s" to "p". In particular, if you take a horizontal beam and have it hit a mirror and be redirected upwards, you can no longer call it either 's' or 'p' polarized, even though the polarization vector is well defined. – nibot Nov 16 '10 at 19:46
Actually, "p" and "s" are defined with respect to a surface of reflection. As I said, I always mix them up, but as far as I remember, "s" is polarized in the plane of the surface of reflection, and "p" is polarized in the plane defined between the incoming and outgoing beams. So, "s" and "p" are still defined when you reflect a beam upwards. – ptomato Nov 16 '10 at 21:59
I think what you are looking for are known as "Stokes Parameters". There is also a very nice description of the polarization tensor in Sec. 9.10 of Mukhanov's "Physical Foundations of Cosmology". With these tools you can compare radiation patterns from different directions and what have you. Hope that helps.
Cheers,
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http://mathhelpforum.com/calculus/101321-inverses-second-derivatives.html
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# Thread:
1. ## Inverses and Second Derivatives
Problem:
If f is a one-to-one, twice differentiable function with inverse function g, show that:
g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )
and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
___________________________
My work so far:
For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
(f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3
I'm not sure where to go from here. What can I do about the second derivatives? Thank you!
2. Originally Posted by uberbandgeek6
Problem:
If f is a one-to-one, twice differentiable function with inverse function g, show that:
g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )
and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
___________________________
My work so far:
For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
(f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3
I'm not sure where to go from here. What can I do about the second derivatives? Thank you!
Part I
Remember the identity that for inverses, we have
$f[g(x)]=x$.
Taking the derivative, we get
$f'[g(x)]g'(x)=1$.
Divide by $f'g$ to get:
$g'(x)=(f'[g(x)])^{-1}$.
Take the derivative again, yielding
$g''(x)=(-1)(f'[g(x)])^{-2}f''[g(x)]g'(x)$.
Remember that $g'(x)=(f'[g(x)])^{-1}$ and substitute:
$g''(x)=(-1)(f'[g(x)])^{-2}f''[g(x)](f'[g(x)])^{-1}$.
And then simplify to yield
$g''(x)=-\frac{f''[g(x)]}{(f'[g(x)])^{3}}$.
Part II
Now, when we say that $f$ is increasing and concave upward, we are essentially saying that
$f'(t),f''(t)>0$ for all $t\in D$. Since $g(x)\in D$, then
$f'[g(x)],f''[g(x)]>0$.
We can use that information with our previous equations to infer that $g''(x)<0$.
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http://mathoverflow.net/questions/94302/hilbert-samuel-function-and-that-of-the-irreducible-components
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## Hilbert-Samuel function and that of the irreducible components.
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How to obtain a relation between the Hilbert-Samuel function of the local ring at a point of a reduced, but not necessarily irreducible variety, and the Hilbert-Samuel functions of the corresponding local rings of its irreducible components?
More concrete. R, a regular local Noetherian ring, complete if you wish, I an ideal in R that is the intersection of some prime ideals I_k such that there are no embedded primes.
I am looking for a formula relating the HS function of R/I and those of R/I_k? The filtration is with respect to the maximal ideal of R.
Edit: Info: There is also an analogous formula for Hilbert functions (analogous to the associativity formula for multiplicity). Proposition 3.2 in Equimultiplicity and blowing up, by Herrmann, Ikeda and Orbanz. $H^{(i)}[\underline{x},a,M]=\sum_{p∈Assh(M/aM)}e(\underline{x},R/p)H^{(i)}[aR_p,M_p]$, where $M$ is finitely generated $R$-module, $a$ and ideal in $R$, and $\underline{x}$ a multiplicity system for $M/aM$. If I put $R$ as my ring $R/I$, $a$ as the maximal ideal, and $M:=R$, if I understood their definition of $Assh$ this only gives me information about those components $p$ having $dim(R/p)=dim(M)=dim(R)$.
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1
You're going to need to think about how those components are glued together. Consider the ring $R \leq k[[x,y]]/(xy)$ generated by $x+y$ and $x^n$, two lines glued together to order $n$. Once you get to multiple components there's going to be lots of schemy inclusion-exclusion to think about. – Allen Knutson Apr 17 2012 at 17:19
@Knutson. I don't understand the example. Is $R:=k[[x+y,x^n]]/(xy)$? How does one sees the two lines? – Franklin Apr 18 2012 at 19:15
A friend told me: To take $k[[s,t]]\mapsto k[[x,y]]/(xy)$ by sending $s\mapsto x+y$ and $t\mapsto x^n$. The image is $R$ and the kernel seems to be $t(s^n-t)$, which gives us that $R=k[[s,t]]/(t(s^n-t))$. But this is a hypersurface singularity. The Hilbert-Samuel function is only going to depend on the order, in this case $2$. Perhaps that is not the ring you meant. – Franklin Apr 18 2012 at 22:06
## 1 Answer
There is an associativity formula for Hilbert-Samuel multiplicity which says, the multiplicity is the sum of the multiplicities of the irreducible components in your concrete case, i.e. $e_0(m,R/I) = \sum e_0(m,R/I_k)$. You may want to take a look at Theorem 14.7 of Matsumura, commutative ring theory for the general statement. But I don't know the answer to the relationship between Hilbert-Samuel functions.
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There is also an analogous formula for Hilbert functions. Proposition 3.2 in Equimultiplicity and blowing up, by Herrmann, Ikeda and Orbanz. $H^{(i)}[\underline{x},a,M]=\sum_{p\in Assh(M/aM)} e(\underline{x},R/p)H^{(i)}[aR_p,M_p]$, where M is finitely generated R-module, a and ideal in R, and $\underline{x}$ a multiplicity system for M/aM. If I put R as my ring, a as my ideal I, and M:=R. But if I understood their definition of Assh this only gives me information about those components p having dim(R/p)=dim(M)=dim(R). – Franklin Apr 17 2012 at 20:54
Perhaps I should put the comment above as info in the question. – Franklin Apr 17 2012 at 20:55
Not $a$ as my ideal, $R$ is already my $R/I$ and a the maximal ideal. – Franklin Apr 17 2012 at 21:06
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http://mathoverflow.net/questions/77985?sort=newest
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## locally connected versus locally compact
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In the definition of a locally connected space we demand every neighbourhood of a point to satisfy certain condition whereas for a locally compact space we demand that one neighbourhood be there with the required property.
Is there some reason for this difference? Is it so that a compact space needs(?) to be locally compact?
Connectedness is a "geometric" property whereas compactness is an "analytic" property. Is that a reason behind such different definitions?
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2
It's more general than this -- "locally" is used in both senses depending on what's being talked about, it's not particular to just connectedness and compactness. – Harry Altman Oct 13 2011 at 5:00
1
Not really every neighbourhood, but a local base of them. – Henno Brandsma Oct 13 2011 at 6:07
## 4 Answers
1. One could phrase local compactness using neighbourhood bases as well (in the Hausdorff case, at least) if desired: once one has one precompact open neighbourhood, one automatically has a whole neighbourhood base of precompact sets, since any subset of a precompact set is still precompact. (And in practice, this is often how local compactness is actually used.)
2. The most important thing that a property named "locally P" should obey is that it be local rather than global. In the case of topological spaces, this means that if a topological space X is locally P at some point $x_0$, and we have another topological space Y which agrees with X at a neighbourhood of $x_0$ (e.g. Y could be the restriction of X to a neighbourhood of $x_0$, with the relative topology), then Y should be locally P at $x_0$ as well. Both local compactness and local connectedness, as defined traditionally, have this locality property. On the other hand, the property "$x_0$ has at least one connected neighbourhood" is not local (consider for instance $({\bf Q} \times {\bf R}) \cup \{\infty\}$ in the Riemann sphere ${\bf R}^2 \cup \{\infty\}$, which is a globally connected space which is locally identical near the origin to ${\bf Q} \times {\bf R}$ in ${\bf R}^2$, which has no connected open sets), and thus this property does not deserve the name of "local connectedness at $x_0$".
It may help to think of the modifier "locally" not as a rigid recipe for converting global properties to local ones, but rather as an indicator that the property being modified is a local analogue of the global, unmodified, property. In most cases there is only one obvious such analogue to select, although in some cases (such as the notion of "locally compact" in the non-Hausdorff setting) there is some freedom of choice, which ultimately means that one has make a somewhat arbitrary convention regarding terminology at some point.
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1
This is a very neat way to justify my intuitive but not so elaborate and well justified belief that any neighborhood should have the property. Note taken! :-) I think of a neighborhood as a "sufficiently large set". Large enough to be considered a neighborhood. Often, one needs to choose a "small" set, but at the same time, big enough to make the small step big enough to achieve the goal... "path connect two points", for instance. – André Caldas Oct 13 2011 at 18:57
Thank you Terry for the explanation. Now I guess I understand the modifier "locally" better. – Shripad Oct 14 2011 at 7:24
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If $X$ is a space and "P" is an adjective that can apply to spaces, then in many cases "$X$ is locally P" means "for every point $x\in X$, for every neighborhood $N$ of $x$, there is a neighborhood $N'$ of $x$ in $X$ such that $N'\subset N$ and $N$ is P", or more briefly "every point of $X$ has arbitrarily small P neighborhoods". I would agree with Terry, though, in cautioning against a strict rule.
This gives the usual notion of locally connected (which turns out to be equivalent to saying that every component of every open set is open), and likewise for locally path-connected. It also gives one of the usual notions of locally compact, generally considered better than the other usual notion, which happens to agree with it in the Hausdorff case so that the difference is often irrelevant.
Here is one way in which people confuse each other with "locally" and "local". What do I mean if I say that a map $f:X\to Y$ is locally a homeomorphism? Surely I do not mean that every $x$ has arbitrarily small neighborhoods $N$ such the the restriction to $N$ is a homeomorphism $N\to Y$. What I mean is that every $x$ has arbitrarily small neighborhoods $N$ such the the restriction to $N$ gives a homeomorphism $N\to f(N)$. OK, so this conforms to the general rule of my first paragraph [extended to the case where "P" is allowed to be "a homeomorphism" even though that's not an adjective] if I give "homeomorphism" the old-fashioned meaning of "homeomorphism to its image". (Also I could have omitted "arbitrarily small", but that's beside the point.)
But that's not the confusion that I was thinking of. Here it comes. I have known more than one student to misinterpret the expression "locally homeomorphic" in the following way: you learn what a local homeomorphism is (perhaps while studying covering spaces), and then later someone says something about $X$ being locally homeomorphic to $Y$. And you assume that this means that there is a map $f:X\to Y$ that is a local homeomorphism, whereas you should have been thinking "every $x$ has arbitrarily small neighborhoods $N$ such that $N$ is homeomorphic to $Y$".*
*or more likely such that $N$ is homeomorphic to a neighborhood of a point in $Y$. See, that's an example of what Terry is saying.
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I think the best definition of locally compact is that the lattice of open sets is a continuous lattice in the sense of Dana Scott. This is what is needed for a space X to have the property X\times() is adjoint to Hom(X, ). See Johnstone's Stone Space book for more on this.
Added I will flesh out this answer now that I have more time. In a lattice, an element $x$ is said to be way below an element $y$, written $x\ll y$, if whenever $y$ is below a join $\bigvee A$, then $x$ is below $\bigvee F$ where $F$ is a finite subset of $A$. (This kind of means $x$ is compact in $y$). If $T$ is a topological space and $\mathcal O(T)$ is the lattice of open sets, then $U\ll V$ iff there is a compact space $K$ with $U\subseteq K\subseteq V$.
A complete lattice is a continuous lattice if each element is a directed join of elements way below it. It is not hard to prove that $\mathcal O(T)$ is continuous for a Hausdorff space iff $T$ is locally compact. It has been argued that this is the right general definition of a locally compact space. The main positive evidence, if memory serves, is that $\mathcal O(T)$ is continuous iff for all spaces (maybe some minor separation condition like sober is needed??) $Y,Z$ one has $Hom(T\times Y,Z)\cong Hom(Y,Hom(T,Z))$ (i.e. $T$ is exponentiable). I think this result holds on the nose in the context of locales (pointless spaces).
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Is $X=T$? – Emil Jeřábek Oct 14 2011 at 10:29
Yes, I'll fix it. – Benjamin Steinberg Oct 14 2011 at 10:47
I, myself, am revolted with such a definition, too.
It seems, according to Wikipedia, that there is no consensus on the definition. And probably, few people care about it, because in Hausdorff spaces, all the definitions are equivalent. In addition to the definition you present, one could also say that a space is locally compact when every point has a closed compact neighbohood. In general, even if a neighborhood is compact, it does not mean its closure will be compact as well.
In the Hausdorff case, the closure of subsets of compact sets are compact, since every compact set is closed in this case. Also, in the Hausdorff case it is true that if $K$ is a compact neighborhood of $x$, then $x$ has a neighborhood filter base made out of compact sets. This is because $K$ is a normal space.
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Wow, I never knew this! I always figured the strange definition was to give nice properties, but in fact the nice properties are all for locally compact Hausdorff spaces (e.g. they are characterized by being open subspaces of compact Hausdorff spaces), so I'm guessing point-set topology courses everywhere could just use the standard "locally <blank>" definition and avoid this confusion completely. – David White Oct 13 2011 at 13:58
1
Which one is your "standard 'locally <blank>' definition"? :-) – André Caldas Oct 13 2011 at 14:52
For all points p and for all neighborhoods U there is a smaller neighborhood V satisfying the property. I think this is the standard for every locally <blank> other than locally compact – David White Oct 13 2011 at 22:05
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http://mathoverflow.net/questions/95586/probability-of-a-random-walk-crossing-an-increasing-function-of-the-standard-devi/95608
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## Probability of a Random Walk crossing an increasing function of the standard deviation
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Let $(S_n)_{n=0}^{\infty}$ be a random walk with $S_n = \sum_{i=1}^n X_i$, and let the $X_i$ be distributed according to some (bounded) distribution function $F$ with mean $0$ and variance $1$, so that $S_n$ has mean $0$ and variance $\sigma_n^2 = n$. Let $g(n)$ be a (slowly) increasing function of $n$, and let $h(n) = g(n) \sigma_n = g(n) \sqrt{n}$. I am interested in
$$\mathcal{P}_h(n) := P(\exists \ n_0 \leq n: S_{n_0} > h(n_0)),$$
or equivalently,
$$\mathcal{P}_h(n) := P(\exists \ n_0 \leq n: \frac{S_{n_0}}{\sigma_{n_0}} > g(n_0)),$$
and in particular, proving that $\mathcal{P}_h(n)$ is small for certain fixed $F$ and $g$ and large $n$. In other words, we start a random walk, and instead of asking for the probability that at some point it ends up in, say, the rightmost $5\%$ region of all values at that time (which would correspond to $g(n)$ being constant), the region gets smaller over time. So as $n$ increases, the a priori probability $P(S_n > h(n))$ decreases.
At this point I am interested in any ideas for solving this problem. If you have any ideas on how to prove that $\mathcal{P}_h(n)$ is small, I would really like to hear your thoughts. Also if the case of a standard random walk is easier, a solution to that problem may give insight to this problem as well.
Right now I have a proof for $P_h(n) < \epsilon$ based on using a piecewise constant function $\ell$ as a lower bound for $h$, i.e. $\ell(n) \leq h(n)$ for all $n$, and showing that $\mathcal{P}_{\ell}(n) < \epsilon$. But this does not seem like a sharp bound in general, and I am curious if there are better proof methods for this problem.
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## 2 Answers
Check out the Law of iterated Logarithm. Is this enough for your purpose?
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Thanks, that looks very interesting. But I'm also interested in an "efficient" way to turn the $\mathcal{P}_h(n)$ above into something nicer. Of course we could use the upper bound $P(\exists \ n_0 \leq n: S_{n_0} > h(n_0)) \leq \sum_{i=1}^n P(S_i > h(i))$ but that does not seem very sharp. – TMM Apr 30 2012 at 20:42
I suggest taking a look at how the LIL is derived, then trying to apply similar methods for your problem. In general, you want to measure the correlation between the events $S_i > h(i)$, and it is often useful to bunch all of these events for $i=2^j,..,2^{j+1}$ together. the correlation between different $j$'s here will then decay exponentially. – Ori Gurel-Gurevich Apr 30 2012 at 21:30
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think you need to look for Large Deviation Theory for help; in particular you could check out Dembo and Zetouni's book. I think a version of equation (1.2.14) might be a good place to start.
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