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http://math.stackexchange.com/questions/309205/sudokus-as-composition-tables-of-finite-groups
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# Sudokus as composition tables of finite groups
If $G$ is a finite group then the composition table of $G$ is a latin square (ie, each row and column contains each group element exactly once).
Assume now that $|G| = n^2$ for some natural number $n$. We can then split the composition table for $G$ into $n^2$ $n\times n$ squares, and we can ask whether the resulting table is a solved sudoku (of size $n^2\times n^2$ rather than the usual $9\times 9$).
When making the composition table for $G$, there are two choices to be made. One is the ordering of the elements for the rows, and the other is an ordering of the elements for the columns. By looking at the row and column corresponding to the identity, one can see that if we pick the same ordering for rows and columns, then the composition table cannot be a sudoku (if we label the entries in the composition table as $a_{i,j}$ and the row/column corresponding to the identity is $k$ then we have $a_{k-1,k} = a_{k,k-1}$ and $a_{k+1,k} = a_{k,k+1}$ and at least one of these pairs belong to the same $n\times n$ square).
If $G$ is cyclic, then we can indeed make the composition table a sudoku as follows: Write $G = \{0,1,\dots ,n^2 - 1\}$ (identified with the integers mod $n^2$). Order the columns in the natural way (ie, in increasing order with the chosen representatives), and order the rows in blocks of $n$ such that in each block, the difference between consecutive terms is $n$ (so each block consists of those elements with a given residue mod $n$ in increasing order).
For the non-cyclic group of order $4$, it is also possible (it is easy to do by trial and error). For the non-cyclic group of order $9$, it is again possible, but it takes a while to find some orderings that work. Ones that work are as follows: Let the group be generated by two elements $a$ and $b$ and order the columns as $1,ab,a^2b^2,a,b^2,a^2b,a^2,ab^2,b$ and the rows as $1,a,b,ab,a^2,b^2,a^2b,ab^2,a^2b^2$.
My question is: For what groups is it possible to find orderings of the rows and columns that turn the composition table into a sudoku?
Edit: Note that if this is possible for some group $G$ then each of those $n\times n$ squares correspond to a pair of subsets $A,B\subseteq G$, each of size $n$, such that $AB = G$.
In fact, it is possible iff there are subsets $A_i$ and $B_i$ of $G$ all of size $n$ such that $$G = \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n B_i$$ and such that for all $i,j$ we have $A_iB_j = G$.
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And one can ask the converse problem: what sudoku tables can be viewed as a composition table? – awllower Feb 21 at 3:10
## 1 Answer
One case where a sudoku arrangement is possible is when $G$ has a subgroup $H$ of order $n$. Let $Ha_1, Ha_2, \ldots Ha_n$ be the right cosets of $H$ and let $T_1, T_2, \ldots, T_n$ be a partition of $G$ into complete sets of left coset representatives. That is, each $T_i$ contains exactly one element from each left coset of $H$ and $T_i \cap T_j = \emptyset$ for $i \neq j$.
Order the columns by listing every element of $Ha_1$ first, then every element of $Ha_2$ and so on until $Ha_n$. Similarly order the rows by listing each element of $T_1$ first, then $T_2$ and so on until $T_n$. Then each $n \times n$ block is contains elements of a $Ha_i$ in the columns and elements of some $T_j$ in the rows. It turns out that this arrangement gives us a sudoku table.
We should show that every element of $G$ occurs exactly once in the block corresponding to $Ha_i$ and $T_j$. That is, we want to show that $T_jHa_i = G$, and this follows from the fact that there are no repetitions in the block. Suppose that $s_k, s_{k_0} \in T_j$ and $h, h_0 \in H$. If $s_kha_i = s_{k_0}h_0a_i$, then $s_kH = s_{k_0}H$ so $s_k = s_{k_0}$ since $T_j$ contains exactly one representative for each left coset. Thus $h = h_0$ as well.
I found the idea for the construction here, where a bit more general problem is considered. If $|G| = ab$ and $[G:H] = a$, then the same construction as in this answer gives you a sudoku table with blocks of size $a \times b$.
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Very nice observation. So that paper discusses a more general case, but only provides a partial answer (essentially the one above). So we are left with looking at those groups of order $n^2$ and with no subgroup of order $n$. – Tobias Kildetoft Feb 20 at 17:51
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http://mathoverflow.net/questions/89105?sort=newest
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## Bound on graph domination number when min degree is 7
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a graph $G$ whose minimum vertex degree is $\delta=7$. I am seeking an upper bound on the domination number $\gamma(G)$ in terms of the number of vertices $n$ of $G$. I found a paper by Edwin Clark, Boris Shekhtman, Stephen Suen, and David Fisher, "Upper bounds for the domination number of a graph," Congressus Numerantium, 132:99–124, 1998 (CiteSeer link), that implies a bound of about $0.31 n$ (more precisely, $(18286568 / 58640175)n$). I was hoping for a smaller upper bound. Perhaps there have been advances since that paper? Any pointers to relevant literature would be much appreciated. Thanks!
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## 1 Answer
I believe the complete history is as follows. For an arbitrary graph with n vertices and minimum degree k, the result has been shown to be at most
n[1+ln(k+1)]/(k+1)
by Arnautov 1974, and Payan 1975 but the articles were written in Russian and French (resp.). For k=7, this gives a bound of 0.384930n. In 1985, Caro and Roditty improved this bound slightly to 0.325283n when k=7. For this second result see Y. Caro and Y. Roditty, On the vertex-independence number and star decomposition of graphs, Ars Combin. 20 (1985), 167-180. Next comes the result you mentioned 0.311844 in 1996 (not 1998). So, unfortunately, I think the answer is that your bound is the best known at this time. See also the discussion at the beginning of On the domination number of Hamiltonian graphs with minimum degree six by Xing, Hattingh, and Plummer Applied Math. Letters 21 (2008) 1037-1040 for the most recent paper published on this topic.
A short proof of the result of Arnautov and Payan can be found in Intro to graph Theory by Douglas West 2nd Ed page 117 which uses the greedy algorithm (by Alon). Discussion are also found in papers for k=2,3,4,5,6 by various authors (I'm guessing this may not be of use to you but these cover some history). There is a new paper by Kostochka and Stocker for cubic graphs (5n/14) which indicates that some authors are still working on this problem. This is everything that has been done (as far as I know). Let me know if you learn more.
Another approach would be to use the independent domination number i(G) since gamma(G) \le i(G) fo any graph G. I think the most recent result here is 0.18329n BUT this is only for a 7-regular graph. Not sure if is this helps. The result is by Duckworth and Wormald entitled On the Independent Domination Number of Random Regular Graphs Combinatorics, Probability and Computing submitted in 2003.
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Thank you, these are useful references to me! – Joseph O'Rourke Apr 23 2012 at 14:29
There is also a nice probabilistic proof of the Amautov & Payan result in Lesniak & Chartrand. Or at East was in the 3rd edition. – Felix Goldberg Apr 25 2012 at 21:03
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http://nrich.maths.org/7244/solution
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### Football World Cup Simulation
A maths-based Football World Cup simulation for teachers and students to use.
### Probably ...
You'll need to work in a group for this problem. The idea is to decide, as a group, whether you agree or disagree with each statement.
### In the Playground
What can you say about the child who will be first on the playground tomorrow morning at breaktime in your school?
# Winning the Lottery
##### Stage: 2 Challenge Level:
We have received many good responses for this question. It is interesting to see all the different lottery rules you have come up with.
Rosalind from Ricards Lodge High School made a good start on the first question:
There are four balls and you must choose one, so it's 1 ball which you choose out of the total of 4 balls that can be chosen. So the chance is 1:4.
Kennard Brentfield Primary School also correctly answered this part of the question.
Another student from Ricards Lodge gave a nice complete solution to the questions:
My method of finding out what the probability is of winning the new lottery, the old lottery game and if it is actually harder to win than the old one is this:
The old lottery game:
In the old lottery game the probability of winning is $\frac{1}{4}$. This is because there are four balls and you pick out one. Therefore you have a chance of $1$ in $4$.
The new lottery game:
The probability of winning the new lottery game is $\frac{1}{6}$. This is because there are six options you could have:
$1$ and $2$, $1$ and $3$, $1$ and $4$, $2$ and $3$, $2$ and $4$, $3$ and $4$.
You can only pick one pair out of the six options so the probability is $1$ in $6$. Therefore it is harder to win the new lottery game than the old one.
If I were going to make a newer even harder lottery game, this would be my method:
There would be four balls and you can pick out any two. Then if your balls were picked out in the same order as you picked out yours then you win!
This is harder because the balls have to be picked out in the correct order making it a lot more difficult to win!
Daisy, also from Ricards Lodge submitted an excellent solution with similar methods. She also investigated the chances of winning if three balls are chosen from four in the bag:
The chance of you winning the lottery in Far away land is $\frac{1}{4}$ because there are only four balls to pick from.
When they made it slightly harder the chance of winning was $\frac{1}{6}$. The possible options you could choose were (without repeating yourself):
$1$ and $2$, $1$ and $3$, $1$ and $4$, $2$ and $3$, $2$ and $4$, $3$ and $4$.
There were only six possible outcomes, therefore your chance of winning was $\frac{1}{6}$.
Using four balls the probability of you winning does not get any harder then $\frac{1}{6}$ unless you need a pattern.
For example if you needed to match with three balls then there are fewer outcomes, in fact only four which brings you back to the probability of $\frac{1}{4}$.
Alistair from Charters School, Krystof from Uhelny Trh, Prague also sent us good solutions to the first parts of the question.
Katy from Ricards Lodge told us her idea to make the lottery harder, by creating more possible outcomes:
My idea is to have $150$ national lottery balls, only one of them picked. There are $150$ possible outcomes, which is greater than $6$ or $4$ possible outcomes.
Ali from Riversdale Primary also shared how he arrived at his method to make winning more difficult:
I tried to make it harder by picking three balls, but the probability of winning was $1$ in $4$ again, because there is only one ball left in the bag.
To make it harder you have to increase the number of balls. I added one more and found that the chance of winning is $1$ in $5$ if you have to pick one or four balls, or $1$ in $10$ if you have to pick two or three balls.
Alice, Oliver, Michael, Elsa and Rosie from the Extension Maths group at St Nicolas C of E Junior School, Newbury also answered the first two parts of the problem correctly. Alice and Oliver created their own, harder version of the lottery:
Our lottery uses five balls, numbered $1$ to $5$, placed in a bag and you must choose two numbers. Your numbers must match, in the same order, the balls drawn from the bag.
We listed all the possibilities:
$1$ and $2$, $1$ and $3$, $1$ and $4$, $1$ and $5$, $2$ and $1$, $2$ and $3$, $2$ and $4$, $2$ and $5$, $3$ and $1$, $3$ and $2$, $3$ and $4$, $3$ and $5$, $4$ and $1$, $4$ and $2$, $4$ and $3$, $4$ and $5$, $5$ and $1$, $5$ and $2$, $5$ and $3$, $5$ and $4$.
That makes $20$ and they are all equally likely. We know our lottery is harder to win because there are more possibilities. The chances of winning are $\frac{1}{20}$ not $\frac{1}{4}$.
Elsa and Rosie chose a simpler version:
Number of balls: nine Pick: one
Answers: $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$
Draw: one ball
Chance of winning: $\frac{1}{9}$
There is one ball picked out of the bag and there are nine balls that have an equally likely chance of being picked. So $9$ will be the denominator. There is one draw so $1$ will be the numerator, so that will equal $\frac{1}{9}$ which means it is harder.
Benjamin from Flora Stevenson Primary School also worked out the problem using the same thinking.
Well done to all of you who sent in your solutions.
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://mathoverflow.net/questions/82547?sort=newest
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## A question on the product of element orders of a finite group
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be a finite group of order $n$ and $\psi(G)$ be the sum of element orders of $G$. Then $\psi(G)\leq\psi(C_n)$, where $C_n$ is the cyclic group of order $n$ (see "Sums of element orders in finite groups", Comm. Algebra 37 (2009), 2978-2980). Is it true a similar inequality for the product of element orders of $G$?
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1
Have you checked it for small examples? I personally have no intuition otherwise for why this should/should not be true... – Igor Rivin Dec 3 2011 at 10:40
Yes, it seems to be true. – Marius Tarnauceanu Dec 3 2011 at 11:09
## 1 Answer
Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds $$\sum_{g\in G}f(o(g))\geq \sum_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$ This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical theorem of Frobenius saying that when $k$ divides the order of a group, the number of elements of order dividing $k$ is divisible by $k$, then proceed by induction. An application of this exact idea is for example problem 10775 on the American Math Monthly. For your question we just need $f(x)=-\log x$.
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It's an interesting idea. I will try to give a complete proof. Thanks! – Marius Tarnauceanu Dec 7 2011 at 8:03
4
The proof given in that Monthly solution seems to me to be at best incomplete. I agree (in the notation of that problem) that the sets S_d for divisors d of k do not consume too much of the set G_k, but what about sets S_d where d < k but d is not a divisor. Perhaps they involve some elements of G_k. Does anyone see a way to repair this problem? – Marty Isaacs Jul 19 at 19:52
The proof seems to only want to define $S_d$ for $d \le n$ with $d$ a divisor of $n$. Am I missing something? – Aaron Meyerowitz Aug 9 at 6:24
@Aaron: The problem is with the $S_d$ with $d < k$, $d$ not a divisor of $k$, but $\gcd(d,k) \neq 1$. Those $S_d$'s might have taken away too many elements of $G_k$, leaving less than $\phi(k)$ elements to build $S_k$. – Tom De Medts Aug 9 at 13:28
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http://physics.stackexchange.com/questions/51920/amount-of-thermal-energy-in-the-earth/52004
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# Amount of thermal energy in the Earth?
Does anyone know the amount of thermal energy that the Earth's mantle and core possess? I don't mean the maximum limit of electrical power we could generate with geothermal plants, but rather: if you took the Earth and magically cooled it down so that its temperature became homogeneously 0K, what would the change in the Earth's internal energy be? (ignoring the Sun)
If we don't have that data, does anyone have an idea how to perform an order of magnitude estimate?
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2
As a start, you could look up specific heats for the substances that make up the mantle, outer core, and inner core along with their temperatures and attempt to calculate the heat that would be lost if they were to be cooled to absolute zero. This neglects phase transitions and probably other subtleties...but it's a start. – joshphysics Jan 23 at 1:14
– anna v Jan 23 at 4:39
1
@joshphysics I thought about that, but I don't know how to calculate the mass of the mantle, outer core and inner core. I could use a state equation (we do know the volume, temperature and pressure) but I have zero idea what state equation to use for molten stuff under huge pressures and temperatures. – Sandra Jan 23 at 16:10
## 1 Answer
The amount of thermal energy in the Earth's core and mantle is determined primarily by the temperature. The three-dimensional temperature field inside the earth is imprecisely known. One dimensional "Onion skin" models of the earth's interior are based upon empirical evidence from seismology, geodesy, and mineral physics, but there are almost certainly lateral variations in temperature which are important to understand.
The article mentioned in the comments can provide you with estimated temperatures at important discontinuities inside the earth. Connect these with constant gradients, or find an alternative model temperature profile.
You should be able to find some specific heat capacity measurements for periodite (mantle), perovskite, and liquid iron. To an order of magnitude these should be like $10^2$ - $10^3$ $\text{J} \text{ kg}^{-1} \text{K}$.
You'll also need the density profile, and can get that from a one dimensional model of the density of the earth. You might try Preliminary Reference Earth Model (PREM) or The Reference Earth Model Website. The latter reference models include three dimensional models.
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http://mathhelpforum.com/trigonometry/115105-arcs-arc-length.html
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# Thread:
1. ## Arcs/Arc Length
Hey guys, another question for ya:
0 = theda
Suppose that 0 = 150 degrees
a. Find the length of the circular arc corresponding to 0(theda) on a circle of radius 5.
(For this question I got, 13.09, so could someone check to see if it's correct?)
b. Find the area of the slice corresponding to 0(theda) on a circle of radius 5
c. Find sin 0 and cos 0 (again 0 = theda)
d. Find the angle that the radial line segment of the unit circle ending at (sin 0, cos 0) makes with the positive horizontal axis.
I would GREATLY appreciate it if you guys could also provide the answers just for proofing and rechecking.
2. Hello goliath
Originally Posted by goliath
Hey guys, another question for ya:
0 = theda
Suppose that 0 = 150 degrees
a. Find the length of the circular arc corresponding to 0(theda) on a circle of radius 5.
(For this question I got, 13.09, so could someone check to see if it's correct?)
b. Find the area of the slice corresponding to 0(theda) on a circle of radius 5
c. Find sin 0 and cos 0 (again 0 = theda)
d. Find the angle that the radial line segment of the unit circle ending at (sin 0, cos 0) makes with the positive horizontal axis.
I would GREATLY appreciate it if you guys could also provide the answers just for proofing and rechecking.
If $\theta$ is measured in radians, then the formulae for arc length, $s$, and area of sector, $A$, are:
$s =r\theta$
$A = \tfrac12r^2\theta$
In this question $\theta = 150^o=\frac{5\pi}{6}$ radians and $r = 5$.
So:
(a) Arc length $s = r\theta=5\times \frac{5\pi}{6}\approx 13.09\text{ cm}$. So you are right!
(b) Area of sector $A = \tfrac12r^2\theta=\frac{5^2\cdot5\pi}{2\cdot6}\app rox32.7\text{ cm}^2$
(c) $\sin\left(\frac{5\pi}{6}\right)=\sin\left(\pi - \frac{5\pi}{6}\right)=\sin\left(\frac{\pi}{6}\righ t)=\frac12$
$\cos\left(\frac{5\pi}{6}\right)=-\cos\left(\pi - \frac{5\pi}{6}\right)=-\cos\left(\frac{\pi}{6}\right)=-\frac{\sqrt3}{2}$
(d) We want the point whose coordinates are $\left(\frac12,-\frac{\sqrt3}{2}\right)$. Sketch a diagram, then, and you'll see that the point is in the fourth quadrant, with the radial line making an angle $60^o$ below the $x$-axis. So the angle we want is $-60^o$ (or $300^o$), measuring angles positive anticlockwise.
Grandad
3. Thank you so much, I appreciate the help!
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http://stats.stackexchange.com/questions/27206/what-does-r-do-when-it-plots-the-residuals-of-an-ar-fit?answertab=votes
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# What does R do when it plots the residuals of an AR fit?
This is a question that's been bugging me for some time. The problem is this: I'm modelling the residuals of a model $f(t,\vec{\theta})$ with (what I think is) an AR process plus a white noise process via MCMC using a multivariate gaussian likelihood, where I model the covariance matrix elements equal to the elements of the autocovariance of an $AR(1)$ process plus i.i.d. white noise $\varepsilon(t)$, i.e., I model
$$r(t)=AR(1)+\varepsilon(t),$$ where $r(t)$ are the residuals of my model, $r(t)=d(t)-f(t,\vec{\theta})$, where $d(t)$ is the data. The thing is that I wanted a visual check to see if my fit was ok (because I'm pretty sure via some previous analysis that my residuals can be modelled efficiently by an $AR(1)$ plus white noise: I just want to show this result to other people "quickly").
What I did then was to analize the residuals via the `arima` and `arma` functions in the `tseries` library in R, and I saw that they actually fitted the residuals with the same coefficient as the MCMC result (within the error bounds and not taking in consideration the additive white noise, which has small variance anyways). However, I also saw that the `arma` function has some nice plots: it plots the "residuals" of the AR fit.
This is my question: how do you plot the residuals of an AR fit if an AR model is stochastic by nature? I think my question goes down to: how do you generate a realization of your model that matches your particular realization of the AR process (i.e. your data) in order to substract this realization from your data? The only way I could think of is to simulate various realizations of the $AR(1)$ model until one fits your data, but this seems like stacking the deck.
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## 2 Answers
First, an AR(1) plus white noise is equivalent to an ARMA(1,1) process. So unless there is a good reason to formulate your model with a latent AR(1) process observed with error, I would suggest you use the simpler and equivalent model with ARMA(1,1) errors.
Then you can write $$d_t = f(t,\theta) + r_t$$ where $$r_t = \phi r_{t-1} + \gamma e_{t-1} + e_t$$ and $e_t$ is white noise. The residuals are $$e_t = r_t - \phi r_{t-1} - \gamma e_{t-1}$$ which can easily be found recursively beginning with $r_0=e_0=0$ so that $e_1=r_1$.
If you wish to persist with the AR(1) + WN formulation, you can compute the errors via a Kalman filter.
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Yeah, I actually have some physical reasons to model an $AR(1)$ model observed with errors. I believe that the $AR(1)$ process is the underlying signal, which is measured by pixel counts in a CCD. This measurement includes the arrival of photons, which is essentially a poisson process (which can be very well approximated in my case, where we have a high number of counts and exposure times, by a gaussian process). – Néstor Apr 29 '12 at 7:57
What I'm actually doing is to model directly the covariance matrix via the autocovariance function of an $AR(1)$ process, which for the diagonal elements I sum the variance of the white noise (I derived this result some time ago and I think it makes sense). The interesting thing about this approach is that we actually know the variance of the poisson process (we have a well understanding of the CCD response), so this also is a kind of double-check for the $AR(1)+WN$ model. – Néstor Apr 29 '12 at 8:00
All this makes sense to me: does it makes sense to you? I'm really open to second opinions :-) – Néstor Apr 29 '12 at 8:02
As Rob Hyndman wrote, you can use Kalman filter. Your interest seems on estimating the unobservable AR(1) component $\alpha_t$ in your model written as $$\left\{\begin{array}{r l} \alpha_{t} &= \phi \,\alpha_{t-1} +\eta_{t}\\ y_t &= \alpha_t + \varepsilon_t \end{array}\right.$$ where $\eta_{t}$ and $\varepsilon_t$ are independent gaussian noises. This is a standard linear State Space (SS) model. I used $y_t$ in place of your $r(t)$ to stick to more conventional notations. Assuming $\alpha_t$ to be centered as here is not essential.
Kalman filtering techniques are dedicated to the recursive computations of the conditional expectations $$a_{t|u} := \mathbb{E}\left[\alpha_t \,\vert\, y_1,\,y_2,\,\dots,\,y_u\right]$$ for times $t$ and $u$. The estimation task is called prediction if $t>u$, filtering if $t=u$ and smoothing for $t < u$. The best estimation $a_{t|u}$ of $\alpha_t$ uses all observations, hence is for $u=n$, the last observation. This kind of estimation is often called signal extraction, which makes sense for a signal $+$ noise'' model as yours.
In R, you can use e.g. the `KalmanSmooth` function (from the stats package), which will require the SS model, and will give you the estimated or "de-noised" signal $a_{t|n}$. To assess the effect of the measurement noise, then simply display a time plot of the estimation, or of the estimate of the observation noise $\varepsilon_t$. The best estimate of $\varepsilon_t$ is its expectation conditional on the whole series $e_{t|n}:=y_t - a_{t|n}$. The parameters to be estimated are $\phi$, $\sigma^2_\eta$ and also $\sigma^2_\varepsilon$, at least if you do not consider it as perfectly known. You can use the `KalmanLike` and `optim` to get Maximum Likelihood estimates which will require a little program.
The problem of simulating the series $\alpha_t$ (or the noises) conditional on the observations is known as simulation smoothing''. Efficient algorithms have been designed by (among others) Durbin and Koopman and seem to be available in the R package KFAS.
````## parameters
phi <- 0.8; sig.eta <- 1; sig.epsilon <- 2
n <- 200
## simulate AR(1) + noise
y <- arima.sim(n = n, list(ar = phi, ma = numeric(0), sd = 1))
y <- y + rnorm(n, sd = sig.epsilon)
plot(y, type = "o", pch = 16, cex = 0.6)
## write the State Space model
mod <- list(T = phi, Z = 1, h = sig.epsilon^2, V = sig.eta^2)
## initialisation part
P <- sig.eta^2 / (1-phi^2)
mod <- c(mod, list(a = 0, P = P, Pn = P))
## smooth
res <- KalmanSmooth(y = y, mod = mod, nit = -1)
lines(res$smooth, col = "orangered", lwd = 2)
legend("topleft", lty = rep(1, 1), lwd = c(1, 2),
pch = c(16, NA), col = c("black", "orangered"),
legend = c("y (raw)", "a (smoothed)"))
````
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http://math.stackexchange.com/questions/298332/finding-values-such-that-a-matrix-has-a-unique-solution
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# Finding values such that a matrix has a unique solution
I have a matrix where I'm supposed to find the values of a and b so that the matrix has a unique solution. I have looked through my textbook and there aren't any examples of how to go about this.
$$\left[ \begin{array}{@{}ccc|c@{}} 3&-1&2 & 1 \\ 0& \frac{5}{3}& \frac{14}{3} & \frac{10}{3} \\ 0&0& a-6 & b-4 \\ \end{array} \right]$$
How would I go about solving this problem? Thanks.
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Hint: For a unique solution, we can look for what in the determinant? Regards – Amzoti Feb 8 at 21:31
I'n not sure, we started determinants yesterday. – user1709173 Feb 8 at 21:32
This does not depend on $b$. This is equivalent to the invertibility of the $3\times 3$ matrix on the left. Whose determinant is? – julien Feb 8 at 21:32
The determinant must be nonsingular. Can you do it now from the last problem? Regards – Amzoti Feb 8 at 21:32
1
Consider the case where a=6 and everything else would be my hint. – JB King Feb 8 at 21:38
show 3 more comments
## 1 Answer
A system of equations represented by a coefficient matrix $A$ has a unique solution if its determinant is NOT EQUAL to $0$: iff $\det A \neq 0$.
Note that you matrix, below, is an upper triangular matrix. The determinant of a triangular matrix is equal to the product of its diagonal entries:
$$\det A = \det \begin{bmatrix} 3 & -1 &2 \\ 0 & \small\frac53 & \small\frac{14}{3} \\ 0 & 0 & a-6 \\ \end{bmatrix} = 3 \cdot \frac53 \cdot (a - 6)\;=\;5(a-6)$$
$$\det A = 0 \iff a = 6$$
Hence, there exists a unique solution to the system of equations represented by the augmented coefficient matrix if and only if $\;\;\bf{a\neq 0}$:
$$A' = \left[ \begin{array}{@{}ccc|c@{}} 3&-1&2 & 1 \\ 0& \small\frac{5}{3}& \small\frac{14}{3} & \small\frac{10}{3} \\ 0&0& a-6 & b-4 \\ \end{array} \right]$$
Now we can express $x_3$ as a function of $a$ and $b$ for $a\neq 6:$ The solution, in this case, to $x_3$, is given by $x_3 = \large\frac{b-4}{a-6},\;$ provided $a \neq 6$. It turns out that the precise value of $b$ is irrelevant to the uniqueness of a solution ($b$ may take on any value):
So a unique solution exists to the system of equations given by the augmented coefficient matrix if and only if $$a\in (-\infty, 6)\cup (6, \infty)$$
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I'd say determinants are overkill here. I'd start about in the middle of your answer; you get a (unique) value of $x_3$ if and only if $a\ne6$, and then you get unique values of $x_2$ and then of $x_1$. – Gerry Myerson Feb 9 at 5:38
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http://nrich.maths.org/6025/index
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nrich enriching mathematicsSkip over navigation
### Logic
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### Logic, Truth Tables and Switching Circuits Challenge
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and fill in the blanks in truth tables to record your findings.
# Truth Tables and Electronic Circuits
##### Stage: 2, 3 and 4
Article by Toni Beardon
Before we experiment with circuits we need to decide how to record what happens when we change the switches in the circuit from 0 to 1 or from 1 to 0. We use small letters $p$ and $q$ to represent the switches. These switches also represent statements that are true or false. A true statement has truth value 1 and a false statement has truth value 0. In circuit diagrams each switch is either 'on' (representing the number 1 or a true statement) or 'off' (representing the number 0 or a false statement). Combinations of switches called logical gates represent the logical connectives .
Logic is concerned with compound statements formed from given statements by means of the connectives 'not', 'and', 'or', 'if ... then' and combinations of these connectives. For example "Today is Tuesday" is a simple statement and "The sun is shining" is a simple statement. Both statements can be true or false. Combining these two statements will make a compound statement, for example "Today is Tuesday and the sun is shining".
Logical arguments are a combination of statements, and circuits are a combination of switches that control the flow of current. Modern computers store data in the form of '0's and '1's, and perform many operations on that data, including arithmetic. Studying the relation between truth tables and circuits will help us to understand a little of the underlying principles behind the design of computers.
Fig. 1 gives a summary of the information you need to build your own circuits. Fig. 1
First, connect a switch to a lamp as in Fig. 2 and click on the switch several times to change it from 1 to 0 and back to 1. Observe that the light goes on when the switch registers 1 and the light goes off when the switch registers 0. Fig. 2
Fig. 3 Now put a NOT gate into the circuit between the switch and the lamp, as shown in Fig. 3, and observe what happens when you change the switch from 1 to 0.
Given a statement $p$, the negation not$p$, written $\neg p$, is demonstrated by this circuit and defined by the following truth table. $$\begin{array}[ p & \neg p \\ 1 & 0 \\ 0 & 1 \end{array}$$
And
The logical connective 'and' is used to make a compound statement from two simpler statements. For example if we speak the truth when we say "Today is Friday and it is raining" then "Today is Friday" must be true and " it is raining" must be true. If either or both are false then the compound statement is false.
Make and test the circuit shown in Fig. 4 and fill in the truth table replacing question marks by 1 if the light goes on and zero if the light does not go on. Fig. 4 $$\begin{array}[ p &q & p\wedge q \\ 1 &1 &?\\ 1 &0 &?\\ 0 &1 &?\\ 0 &0 &? \\ \end{array}$$
If you want to check your work then you can click here.
Or
In everyday language we often use 'or' inclusively as in "If you want to buy cereals or soft drinks go to aisle 7 in the supermarket" when we expect people who want to buy both cereals and soft drinks to go to that aisle as well as people wanting to buy just one of them. On other occasions we use 'or' exclusively to offer two alternatives expecting the listener to choose just one of them as in "Do you want steak for dinner or chicken?". The meaning is usually clear from the context but to avoid any ambiguity in mathematics the logical connective 'or' is always used inclusively and never exclusively.
Given any two statements$p$, $q$ then $p$ or $q$, (written $p \vee q$)is given in the truth table on the right. Make the circuit shown in Fig. 5 and replace the question marks by 1's and 0's according to whether the lamp lights up or not. Fig. 5 $$\begin{array}[ p &q & p\vee q \\ 1 &1 &?\\ 1 &0 &?\\ 0 &1 &?\\ 0 &0 &? \\ \end{array}$$
Again you can check your work by clicking here.
XOR, NAND, NOR and XNOR
Now experiment with the circuits for XOR, NAND, NOR and XNOR and replace the question marks in the following truth table. $$\begin{array}[ p &q &p XOR q & p NAND q &p NOR q & p XNOR q \\ 1 &1 &? &? &? &? \\ 1 &0 &? &? &? &? \\ 0 &1 &? &? &? &? \\ 0 &0 &? &? &? &? \\ \end{array}$$ Comparing these tables to what we have already discovered we see that p NOR q means 'not(p or q)'. Also p NAND q means 'not(p and q)'.
You can check the truth tables for XOR, NAND, NOR and XNOR by clicking here.
If you want to learn more about logic and circuits then read Logic, Truth Tables and Swirching Circuits which covers the content of this article and goes further to discuss tautologies, implication and the Sheffer stroke.
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://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/RLC_circuit
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# All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary
# Science Fair Project Encyclopedia
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# RLC circuit
An RLC circuit is a kind of electrical circuit composed of a resistor (R), an inductor (L), and a capacitor (C). See RC circuit for the simpler case. A voltage source is also implied. It is called a second-order circuit or second-order filter as any voltage or current in the circuit is the solution to a second-order differential equation. Since the circuit components are assumed ideal and are linear, and RLC circuit is an example of an electrical harmonic oscillator.
The resonant or center frequency of such a circuit (in hertz) is:
$f_c = {1 \over 2 \pi \sqrt{L C}}$
It is a form of bandpass or bandcut filter, and the Q factor is
$Q = {f_c \over BW} = {2 \pi f_c L \over R} = {1 \over \sqrt{R^2 C / L}}$
There are two common configurations of RLC circuits: series and parallel.
Contents
## Series RLC Circuit
In this circuit, the three components are in series with the voltage source. An RLC series circuit has a resonant frequency and is often used as a model for analysing electronic circuits, such as calculating impedance.
Where the notations in the figure above are:
• V - the voltage of the power source (measured in volts V)
• I - the current in the circuit (measured in amperes A)
• R - the resistance of the resistor (measured in ohms = V/A)
• L - the inductance of the inductor (measured in henries = Vs/A)
• C - the capacitance of the capacitor (measured in farads = C/V = As/V)
Given the parameters V, R, L, and C, the solution for the current (I) using Kirchoff's voltage law (or KVL) is:
${V_R+V_L+V_C=V} \,$
For a time-changing voltage V(t), this becomes
$RI(t) + L { {dI} \over {dt}} + {1 \over C} \int_{-\infty}^{t} I(\tau)\, d\tau = V(t)$
Rearranging the equation will result in the following second order differential equation:
${{d^2 I} \over {dt^2}} +{R \over L} {{dI} \over {dt}} + {1 \over {LC}} I(t) = {1 \over L} {{dV} \over {dt}}$
### The ZIR (Zero Input Response) solution
Nullifying the input (voltage sources) we get the equation:
${{d^2 I} \over {dt^2}} +{R \over L} {{dI} \over {dt}} + {1 \over {LC}} I(t) = 0$
with the initial conditions for the inductor current, IL(0), and the capacitor voltage VC(0). However, in order to solve the equation properly, the initial conditions needed are I(0) and I'(0).
The first one we already have since the current in the main branch is also the current in the inductor, therefore
$I(0)=I_L(0) \,$
The second one is obtained employing KVL again:
$V_R(0)+V_L(0)+V_C(0)=0 \,$
$\Rightarrow I(0)R+I'(0)L+V_C(0)=0 \,$
$\Rightarrow I'(0)={1 \over L}\left[-V_C(0)-I(0)R \right]$
We have now a homogeneous second order differential equation with two initial conditions. Usually second order differential equations are written as:
$I''+2\xi \omega_k I' + \omega_k^2 I = 0$
In case of an electrical circuit ωk > 0 and therefore, there are three possible cases:
#### Over damping
$\alpha>1 \Rightarrow R^2 C>4L \,$
In this case, the characteristic polynomial's solutions are both negative real numbers. This is called "over damping":
Two negative real roots, the solutions are:
$I(t)=a e^{\lambda_1 t} + b e^{\lambda_2 t}$
#### Critical damping
$\alpha=1 \Rightarrow R^2 C=4L \,$
In this case, the characteristic polynomial's solutions are identical negative real numbers. This is called "critical damping":
The two roots are identical (λ1 = λ2 = λ), the solutions are:
I(t) = (a + bt)eλt
#### Under damping
$\alpha<1 \Rightarrow R^2 C<4L \,$
In this case, the characteristic polynomial's solutions are complex conjugate and have negative real part. This is called "under damping":
Two conjugate roots ($\lambda_1 = \bar \lambda_2 = \alpha + i\beta$), the solutions are:
$I(t)=e^{\alpha t} \left[ a \sin(\beta t) + b \cos(\beta t) \right]$
### The ZSR (Zero State Response) solution
This time we nullify the initials conditions and stay with the following equation:
$\left\{\begin{matrix} {{d^2 I} \over {dt^2}} +{R \over L} {{dI} \over {dt}} + {1 \over {LC}} I(t) = {1 \over L}{{dV} \over {dt}} \\ I(0^{-})=I'(0^{-})=0 \end{matrix}\right.$
Separate solution for every possible function for V(t) is impossible, however, there is a way to find a formula for I(t) using convolution. In order to do that, we need a solution for a basic input - the Dirac delta function.
In order to find the solution more easily we will start solving for the Heaviside step function and then using the fact our circuit is a linear system, its derivative will be the solution for the delta function.
The equation will be therefore, for t>0:
$\left\{\begin{matrix} {{d^2 I_u} \over {dt^2}} +{R \over L} {{dI_u} \over {dt}} + {1 \over {LC}} I_u(t) = 0 \\ I(0^{+})=0 \qquad I'(0^{+})={1 \over L} \end{matrix}\right.$
Assuming λ1 and λ2 are the roots of
$P(R)= R^2+2\xi \omega_k R + \omega_k^2$
then as in the ZIR solution, we have 3 cases here:
#### Over Damping
Two negative real roots, the solution is:
$I_u(t)={1 \over {L(\lambda_1-\lambda_2)}} \left[ e^{\lambda_1 t}-e^{\lambda_2 t} \right]$
$\Rightarrow I_{\delta}(t)={1 \over {L(\lambda_1-\lambda_2)}} \left[ \lambda_1 e^{\lambda_1 t}-\lambda_2 e^{\lambda_2 t} \right]$
#### Critical Damping
The two roots are identical (λ1 = λ2 = λ), the solution is:
$I_u(t)={1 \over L} t e^{\lambda t}$
$\Rightarrow I_{\delta}(t)={1 \over L} (\lambda t+1) e^{\lambda t}$
#### Under Damping
Two conjugate roots ($\lambda_1 = \bar \lambda_2 = \alpha + i\beta$), the solution is:
$I_u(t)={1 \over {\beta L}} e^{\alpha t} \sin(\beta t)$
$\Rightarrow I_{\delta}(t)={1 \over {\beta L}} e^{\alpha t} \left[ \alpha \sin(\beta t) + \beta \cos(\beta t) \right]$
(to be continued...)
### Sinusoidal steady state analysis
The series RLC can be analyzed in the frequency domain using complex impedance relations. If the voltage source above produces a pure sine wave with amplitude V and angular frequency ω, KVL can be applied:
$V = I \left ( R + j \omega L + \frac{1}{j \omega C} \right )$
Where I is the complex current through all components. Solving for I:
$I = \frac{V}{ R + j \omega L + \frac{1}{j \omega C} }$
Taking the magnitude of the above equation:
$I_{mag} = \frac{V}{\sqrt{ R^2 + \left ( \omega L - \frac{1}{\omega C} \right )^2 }}$
If we choose trivial values where R = 1, C = 1, L = 1, and V = 1, then the graph of magnitude of current as a function of ω is:
Note that there is a peak at ω = 1. This is known as the resonant frequency. Solving for this value, we find:
$\omega_o = \frac{1}{\sqrt{L C}}$
## Parallel RLC Circuit
A much more elegant way of recovering the circuit properties of an RLC circuit is through the use of nondimensionalization.
For a parallel configuration of the same components, where Φ is the magnetic flux in the system
$C \frac{d^2 \Phi}{dt^2} + \frac{1}{R} \frac{d \Phi}{dt} + \frac{1}{L} \Phi = I_0 \cos(\omega t) \Rightarrow \frac{d^2 \chi}{d \tau^2} + 2 \zeta \frac{d \chi}{d\tau} + \chi = \cos(\Omega \tau)$
with substitutions
$\Phi = \chi x_c, \ t = \tau t_c, \ x_c = L I_0, \ t_c = \sqrt{LC}, \ 2 \zeta = \frac{1}{R} \sqrt{\frac{L}{C}}, \ \Omega = \omega t_c .$
The first variable corresponds to the maximum magnetic flux stored in the circuit. The second corresponds to the period of resonant oscillations in the circuit.
The expressions for the linewidth in the series and parallel configuration are inverses of each other. This is particularily useful for determining whether a series or parallel configuration is to be used for a particular circuit design. However, in circuit analysis, usually the reciprocal of the latter two variables are used to characterize the system instead. They are known as the resonant frequency and the Q factor respectively.
## See also
03-10-2013 05:06:04
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http://math.stackexchange.com/questions/177208/prove-sin2a-sin2b-sin2c-2-sina-sinb-cosc-if-abc-180-deg/177493
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# Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees
I most humbly beseech help for this question.
If $A+B+C=180$ degrees, then prove $$\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$$ I am not sure what trig identity I should use to begin this problem.
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try substitution C = 180-A-B to get formulae for $\cos(C)$ and $\sin^2(C)$ in terms of trig functions of $A$ and $B$ – gt6989b Jul 31 '12 at 15:20
thanks for your post I will study them and use them to solve the problem. – Fernando Martinez Jul 31 '12 at 15:30
Fernando Martinez appreciates all the post and the work that went into answering his question. – Fernando Martinez Aug 1 '12 at 15:34
## 5 Answers
Let's take the Right Hand Side. $$2 \sin B \cos C = \sin(B+C) + \sin(B-C) = \sin A + \sin(B-C)$$ Therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin A \sin(B-C)$ Now, $$2 \sin A \sin (B-C) = \cos (A - B +C) - \cos (A + B -C) = \cos 2C - \cos 2B = 2 \sin^2 B - 2 \sin ^2 C$$ Cancelling the $2$ gives us therefore, that $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$.
To finish, we already established $2 \sin A \sin B \cos C = \sin^2 A = \sin A \sin (B-C)$. Now, since $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$, therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin^2 B - \sin^2 C$.
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Hmm which two would I cancel out. – Fernando Martinez Jul 31 '12 at 15:45
No, I meant to cancel out the $2$ in $2 \sin A \sin (B-C) = 2 ( \sin^2 B - \sin^2 C)$, giving us $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$ – Rijul Saini Jul 31 '12 at 16:02
I see my now so technically would I have now sin^2B-sin^2C=cos(a-b+c)-cos(a+b-c)=cos2c-cos2b=2sin^2B-2sin^2C – Fernando Martinez Jul 31 '12 at 16:17
What would I do next I am not sure what to do..... – Fernando Martinez Jul 31 '12 at 16:20
I edited my post. – Rijul Saini Jul 31 '12 at 17:10
One can deduce this identity from the conjunction of the law of sines and the law of cosines.
The law of sines says that for the three angles $A$, $B$, $C$ of a triangle, with opposite sides $a$, $b$, $c$, we have $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d.$$ The last equality merely defines $d$, and one can omit it and still have a statement of the law of sines. The common value $d$ is actually the diameter of the circumscribed circle.
If things are scaled so that $d=1$, then we have
$$a=\sin A,\quad b=\sin B,\quad c=\sin C\tag{1}$$
The law of cosines says $$a^2+b^2-2ab\cos C = c^2.\tag{2}$$ Substitute the expressions in $(1)$ into the appropriate places in $(2)$ and you get $$\sin^2 A+\sin^2 B - 2\sin A\sin B\cos C = \sin^2 C$$ and there's your identity.
This still leaves the problem of how to prove the law of sines and the law of cosines. And if you want to use your identity to prove the law of cosines, then this reasoning would be circular. But if you can take the two laws to be established already, then this does it.
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This may not be the shortest way, but is pretty systematic and doesn't involve any real trickery. First, we eliminate the $C$ angles from the equation, using that $C = 180 - A - B$. We write $\sin(C) = \sin(180 - A - B) = \sin(A + B)$, and $\cos(C) = \cos(180 - A - B) = -\cos(A + B)$ and what you need to prove is $$\sin^2A + \sin^2B - \sin^2(A + B) = -2\sin A\sin B \cos(A+B)$$ Inserting the sine and cosine addition formulas in this, what you need to prove becomes $$\sin^2A + \sin^2B - (\sin A\cos B + \cos A \sin B)^2 = -2\sin A \sin B(\cos A \cos B - \sin A \sin B)$$ Writing this out, it becomes $$\sin^2A + \sin^2B - \sin^2 A\cos^2 B - 2\sin A \cos A \sin B \cos B -\cos^2 A \sin^2B$$ $$= -2\sin A\sin B \cos A \cos B + 2\sin^2 A \sin^2 B$$ Cancelling terms, your goal is to prove $$\sin^2A + \sin^2B - \sin^2 A\cos^2 B-\cos^2 A \sin^2B = 2\sin^2 A \sin^2 B$$ Equivalently, $$\sin^2A(1 - \cos^2B) + \sin^2B(1 - \cos^2A) = 2\sin^2 A \sin^2 B$$ This last equation is true since $1 - \cos^2 = \sin^2$. So going backwards in the above steps, one obtains the original equation $\sin^2A + \sin^2B - \sin^2C = 2\sin A \sin B \sin C$ as desired.
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$\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin(B+C)\sin(B-C)$ either using the identity $\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$
or $\sin^2B-\sin^2C=\frac{1}{2}(2\sin^2B-2\sin^2C)=\frac{1}{2}(1-\cos2B-(1-\cos2C))=\frac{1}{2}(\cos2C-\cos2B)=-\frac{1}{2}2\sin(B+C)\sin(C-B)=\sin(B+C)\sin(B-C)$
as $\sin(-x)=-\sin(x)$
Now, $\sin(B+C)=\sin(180^\circ-A)=\sin{A}$
So, $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin{A} \sin(B-C)$
$=\sin{A}(\sin{A}+\sin(B-C))$
$=\sin{A}(\sin(B+C)+\sin(B-C))$ replacing $\sin{A}$ with $\sin(B+C)$
$=\sin{A}(2\sin{B}\cos{C})$
$=2\sin{A}\sin{B}\cos{C}$
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Fernando especially likes this response. – Fernando Martinez Aug 1 '12 at 16:39
@FernandoMartinez, thanks. Trigonometry is all about formulae & identities. – lab bhattacharjee Aug 1 '12 at 17:55
– lab bhattacharjee Aug 2 '12 at 16:06
Use:
• $\sin(180 - \alpha) = \sin(\alpha)$, $\cos(180 - \alpha) = -\cos(\alpha)$
• $\sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \sin(\beta) \cos(\alpha)$
• $\sin^2(\alpha) + \cos^2(\alpha)=1$
Start with solving for $C$, then use identities from the first bullet. Then expand $\sin(A+B)$ and $\cos(A+B)$ using the identity from the second bullet, and its companion for $\cos(\alpha+\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha)\sin(\beta)$.
Then simplify the algebraic expressions and use the identity from the third bullet.
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Sasha would I start on the right or left side sorry if its a dumb question.... – Fernando Martinez Jul 31 '12 at 15:36
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http://mathhelpforum.com/calculus/176766-integration-substituion-partial-factions.html
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# Thread:
1. ## Integration by substituion and partial factions
Hi, I'm trying to solve this problem by substitution and partial fractions (both methods stipulated in the question).
$\int\frac{1}{\sqrt{x}(1-\sqrt{x})(2-\sqrt{x})}\ dx$
so i substitute $u=\sqrt{x}$ so dx=2u du and form the new integral:
$\int\frac{2u}{2u^2(1-u)(2-u)}\ du$ but am a) not sure if this is correct and b) when I try and break this up into partial fractions i get some numerators equal to 0 :S
2. Originally Posted by flashylightsmeow
Hi, I'm trying to solve this problem by substitution and partial fractions (both methods stipulated in the question).
$\int\frac{1}{\sqrt{x}(1-\sqrt{x})(2-\sqrt{x})}\ dx$
so i substitute $u=\sqrt{x}$ so dx=2u du and form the new integral:
$\int\frac{2u}{2u^2(1-u)(2-u)}\ du$ but am a) not sure if this is correct and b) when I try and break this up into partial fractions i get some numerators equal to 0 :S
After simplification, you are left with this $2\int \frac{1}{(1-u)(2-u)}du$, try breaking 1/(1-u)(2-u) again.
3. Here is what I purpose:
Reduce your integral to $\int \frac {1}{u(u-1)(u-2)}$, then perform partial fractions.
4. Ah, ok. But one thing, doesnt the simplication leave the denominator as u(1-u)(2-u)?
5. Yeah, that's what I got, too.
6. Yes, mathaddict miswrote. The u in the numerator cancels one u in the denominator. Now you need A, B, C, so that
$\frac{2}{u(u-1)(u-2)}= \frac{A}{u}+ \frac{B}{u-1}+ \frac{C}u-2}$. That should be easy.
7. Originally Posted by tttcomrader
Here is what I purpose:
Reduce your integral to $\int \frac {1}{u(u-1)(u-2)}$, then perform partial fractions.
This does not work.
Notice that if you let $\displaystyle u = \sqrt{x}$ then $\displaystyle du = \frac{1}{2\sqrt{x}}\,dx$.
So $\displaystyle \int{\frac{1}{\sqrt{x}(1 - \sqrt{x})(2 - \sqrt{x})}\,dx} = 2\int{\left[\frac{1}{(1-\sqrt{x})(2 - \sqrt{x})}\right]\frac{1}{2\sqrt{x}}\,dx}$
$\displaystyle = 2\int{\frac{1}{(1 - u)(2 - u)}\,du}$
which can then be solved using partial fractions. Mathaddict is correct.
8. So the $\sqrt{x}$ that is part of dx doesn't form part of the new integral!?
9. $\displaystyle \frac{1}{2\sqrt{x}}\,dx$ is replaced with $\displaystyle du$.
Alternatively, $\displaystyle \frac{1}{2\sqrt{x}}$ is replaced with $\displaystyle \frac{du}{dx}$, so that $\displaystyle \frac{1}{2\sqrt{x}}\,dx = \frac{du}{dx}\,dx = du$.
But yes, the derivative that you find does not become part of the resulting $\displaystyle u$ integral.
10. Ah that does make sense. Many thanks!
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http://mathoverflow.net/questions/8741?sort=oldest
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Justifying a theory by a seemingly unrelated example
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Here is a topic in the vein of Describe a topic in one sentence and Fundamental examples : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model.
A classical example is Galois theory for solving polynomial equations.
Any examples for homological algebra ? For Fourier analysis ? For category theory ?
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I think that this should be community wiki, for reasons similar to those given by others in the comments at Fundamental Examples. – Jonas Meyer Dec 13 2009 at 11:06
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@Harry: I think your position concerning honesty is rather extreme. – S. Carnahan♦ Dec 13 2009 at 16:21
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I've converted this question to wiki, but please make such questions community wiki in the future. See the FAQ and the discussion on meta.MO (meta.mathoverflow.net/discussion/6) for more on when a question should be community wiki. – Anton Geraschenko♦ Dec 13 2009 at 17:13
27 Answers
[In front of a blackboard, in an office at Real College]
Skeptic: And why should I care about holomorphic functions?
Holomorphic enthusiast:$\;$ Can you compute $\quad$ $\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}$ ? Here $a$ is one of your cherished real numbers, but not an integer.
Skeptic: Well, hm...
Holomorphic enthusiast, nonchalantly: Oh, you just get
$$\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}=\pi^2 cosec^2 \pi a$$
It's easy using residues.
Skeptic: Well, maybe I should have a look at these "residues".
Holomorphic enthusiast (generously): Let me lend you this introduction to Complex Analysis by Remmert, this one by Lang and this oldie by Titchmarsh. As Hadamard said: "Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe".You can look for a translation at Mathoverflow. They have a nice list of mathematical quotations, following a question there.
Skeptic: Mathoverflow ??
Holomorphic enthusiast (looking a bit depressed) : I think we should have a nice long walk together now. [Exeunt]
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There's an error in either the top display or the bottom, as one of these has a minus and one of them has a plus. It doesn't change the answer, since we're integrating over the integers, but I figured you should know. – Harry Gindi Dec 13 2009 at 16:01
@ Harry: I don't know what you mean.The denominator is the square of a+n in both cases.There are no "minus" signs. – Georges Elencwajg Dec 13 2009 at 17:35
oh, sorry, I have my jsmath font rendering at 100% instead of 130%, and the plus sign is invisible. – Harry Gindi Dec 13 2009 at 17:44
I can't reproduce the invisible plus sign. Even when I set jsMath to render math at 50%, both plus signs are perfectly visible. – Anton Geraschenko♦ Dec 14 2009 at 0:13
The plus sign didn't disappear, the vertical cross disappeared so it looked like a minus sign. – Harry Gindi Dec 14 2009 at 18:49
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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.
Also, differential Galois theory for differential equations.
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The Jones polynomial: a knot invariant that originally came from subfactor theory.
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Can you elaborate, please? Thanks! – Jose Brox Dec 13 2009 at 11:57
To be honest, I'm not able to do a good job of explaining this example without more time than I currently have to study up. I was hoping that community wiki justified my light post; I'd be happy (but by no means expect) to have someone more knowledgeable elaborate. Otherwise I may do so late in the week. – Jonas Meyer Dec 14 2009 at 3:15
mathoverflow.net/questions/544/…, last paragraph, contains some elaboration. – Jonas Meyer Feb 10 2010 at 15:37
The spectral theory of commutative Banach algebras led to an elegant proof, due to Gelfand, of the following (previously difficult) theorem of Wiener: If $f$ is a nowhere vanishing complex valued function on the unit circle whose Fourier coefficients are absolutely summable, then the Fourier coefficients of $1/f$ are also absolutely summable.
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Complex analysis and the theory of spaces of analytic functions on open subsets of $\mathbb{C}$ are used to prove the Riemann Mapping Theorem, which has the nice topological consequence that there is a unique homeomorphism class of (nonempty) simply connected open subsets of $\mathbb{C}$.
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Root systems and Dynkin diagrams for classification matters (see both Root system and ADE classification at Wikipedia).
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This example doesn't strike me as "seemingly unrelated." I think something more in the spirit of the question is a question that doesn't require any knowledge of, say, Lie groups to state, but which is easiest to solve using the classification of (insert adjective here) Lie groups. – Qiaochu Yuan Dec 13 2009 at 19:03
The Laplace transform to systematically solve homogeneous ODE's with constant coefficients by transforming them in polynomial equations (and then transforming the solution back).
Solving a differential equation by the Laplace transform
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There are a number of algebraic theorems which are easier to prove using topology. The best known is probably the fundamental theorem of algebra, but there are others. For example, an $n\times n$ matrix with positive entries always has an eigenvector all of whose enetries are nonnegative. (The matrix defines a continuous function from the standard simplex to itself which always has a fixed point.) I learned about this from Elmer Rees.
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Can the fundamental theorem of algebra be proven not using topology/trascendent methods? Mabe it is not made easier by topology but *possible :) – Mariano Suárez-Alvarez Dec 13 2009 at 16:35
There are analytic and algebraic proofs of the FTA, but not as simple as the topological one. Perhaps someone here knows the history better and can put the three proofs in chronological order? – José Figueroa-O'Farrill Dec 13 2009 at 16:38
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All the "algebraic" proofs I know of depend on things like the fact that odd-degree real polynomials have a real root (that is, on the mean value theorem). – Mariano Suárez-Alvarez Dec 13 2009 at 17:16
You'd need to use some special fact about the real numbers though, since it is essentially a special property of \$\mathbb{C}=\mathbb{R}+i\mathbb{R}. I think the impressive thing is that all that is needed is just the (really quite weak) MVT plus some algebra (elementary Sylow theorem and Galois theory stuff). – Thomas Bloom Dec 13 2009 at 17:20
There were "algebraic" proofs that predated Gauss, but they're really technical (they aim to be constructive!) and have wide gaps. I think they can be patched up, but not without some pretty heavy machinery (at least Galois theory). Are the topological and analytic proofs that different? The complex-analytic proof I know is basically an application of Cauchy's formula, which is a corollary of Stokes' theorem... Unfortunately I'm not that familiar with the topological proof, but Gauss' proof is more or less topological in nature. Fully analytical proofs I think require Cauchy. – Harrison Brown Dec 13 2009 at 21:47
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The Poincaré conjecture is an example of a purely topological statement which apparently cannot be proved only by topological means.
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Many topologists were arguing the reverse -- the Poincare conjecture is naturally an analytic statement. It's a misleading coincidence (due to Moise) that you can think of it as a purely topological or combinatorial statement. Classification of 2-manifolds (uniformization) was first proven using analysis. The high-dimensional Poincare conjecture (Smale) used basics of ODEs, if anything it was remarkable for how little analysis his proof used. – Ryan Budney Dec 13 2009 at 21:39
Yet certainly there were many topologists trying to prove the Poincaré conjecture by purely topological means. – José Figueroa-O'Farrill Jan 20 2010 at 11:09
Many geometric problems cannot be solved without hard analysis. Perhaps the best known example is the Calabi Conjecture proved by Yau.
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Let us call "division algebra over $\mathbb R$" a finite-dimensional vector space $A$ equipped with a bilinear map $A \times A \to A: (a,b) \mapsto a \bullet b$ , such that $a\bullet b=0$ implies $a=0$ or $b=0$. ( Associativity is not required).
Examples : the reals, the complexes, the real quaternions and the octonions of Graves-Cayley.
Any such division algebra must necessarily have dimension 1,2,4 or 8 (as in the examples). This was proved indepently in 1958 by Kervaire and Milnor using Bott's periodicity theorem, a fantastic result in algebraic topolgy which had just been proved.
To the best of my knowledge there is still no purely algebraic proof of this theorem on possible dimensions of real division algebras, although the statement is completely algebraic and elementary.
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If you're a combinatorialist and you want to know the asymptotics of a sequence $a_n$ with a nice generating function $A(z) = \sum_{n \ge 0} a_n z^n$, the very first thing you should do is find out if $A$ is meromorphic, since then one can analyze the asymptotics of $a_n$ using its poles. Even if $A$ isn't meromorphic, if one has sufficiently good information about its singularities then there are transfer theorems that translate information about the behavior of $A$ near its poles to the behavior of $a_n$ for large $n$. In other words, combinatorialists (and by extension computer scientists) should learn complex analysis.
For example, let $E_n$ be the number of alternating permutations on $n$ letters. Then $E(z) = \sum_{n \ge 0} E_n z^n = \sec z + \tan z$ is meromorphic with poles $z = \frac{\pi}{2} + 2k \pi, k \in \mathbb{Z}$. The dominant singularity is at $z = \frac{\pi}{2}$ and one now knows without doing any other computations that $E_n \sim n! \left( \frac{2}{\pi} \right)^n$. Even better one can write down an exact series converging to $E_n$ with one term for each pole. The corresponding expansion of the Bernoulli numbers $B_n$ gives the classical evaluation of the zeta function at even integers.
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Another example for this (using Tauberian Theorems) is www2.math.uni-paderborn.de/fileadmin/Mathematik/…. – norondion Jan 16 2010 at 13:20
One problem that one can solve with Fourier analysis very easily is the isoperimetrical inequality and the corresponding characterization of the circle. Of course, this can also be done in many. many other ways, but using Fourier series it becomes particularly simple.
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Suppose you are interested in random walks on an extremely structured graph such as a hypercube graph or a cycle graph. If your graph happens to be the Cayley graph of an abelian group $G$, as in both of the above examples, then it is easy to describe the behavior of random walks on it because the eigenvectors of the adjacency matrix are precisely the characters of $G$ and the eigenvalues depend in a simple way on the characters; in other words, you should learn about the discrete Fourier transform.
Edit: Some elaboration. Let $G$ be a finite abelian group with $|G| = n$. A character of $G$ is a homomorphism $G \to \mathbb{C}$, and it is a basic fact of character theory that the characters form a basis of the space of functions $G \to \mathbb{C}$; this is the discrete Fourier transform. Now let $\mathbf{A}(G)$ be the adjacency matrix of a Cayley graph of $G$ using generators ${ s_1, ... s_k }$. The group $G$ acts on the space of functions $G \to \mathbb{C}$ by sending a function $f : G \to \mathbb{C}$ to $f(gx)$. Call this representation $\rho$; then (and this is the important connnecting observation) one may regard $\mathbf{A}(G)$ as the linear operator $\displaystyle \sum_{i=1}^{k} \rho(s_i)$.
Proposition: Let $\chi_j : G \to \mathbb{C}$ be a character of $G$. Then $\chi_j$ is an eigenvector of $\mathbf{A}(G)$ with eigenvalue $\displaystyle \sum_{i=1}^{k} \chi_j(s_i)$, and these are all the eigenvectors.
Proof. Just observe that $\rho(s_i) \chi_j(g) = \chi_j(s_i g) = \chi_j(s_i) \chi_j(g)$. The fact that these exhaust the set of eigenvectors follows from the basic fact cited above.
For example, the cycle graph $C_n$ is the Cayley graph of the cyclic group $\mathbb{Z}/n\mathbb{Z}$ with generators ${ 1, -1 }$, so its eigenvectors are just the rows of the discrete Fourier transform matrix on $\mathbb{Z}/n\mathbb{Z}$ and its eigenvalues are $e^{ \frac{2\pi i k}{n} } + e^{- \frac{2\pi ik}{n} } = 2 \cos \frac{2\pi k}{n}$. (Note that I have implicitly identified the space of functions $G \to \mathbb{C}$ with the free vector space on the elements of $G$ in the usual way.)
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The connection with the "Fourier discrete transform" is not quite clear to me. Could you elaborate ? – Ewan Delanoy Dec 13 2009 at 20:35
I've added some elaboration. I hope it helps. – Qiaochu Yuan Dec 13 2009 at 21:19
One set of problems to which homological algebra applies surprisingly well is those posed by topology, as evinced by algebraic topology :P
This is of course quite anhistorical...
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This is not an answer as much as a request for answers. Here are some topics for which I don't know an example of this sort, but would really like to see one (and nobody seems to have done them yet). Feel free to edit this post if good examples appear for some of these topics, or if you have a topic to add to the list. They are in no particular order,
• Symplectic geometry (I'd really like to know...)
• Algebraic geometry (there should be a ton of stuff here, it's so diverse)
• Category theory
• Homological algebra (is there an example simpler than some monstrous calculation?)
• Group theory (so far, all that comes to my mind is that it applies to Galois theory, which in turn solves equations. But I think there should be a ton of simpler and nicer applications, shouldn't there?)
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Homological algebra is hallowed by its applications in algebraic topology (singular homology) and algebraic geometry (sheaf cohomology). Singular homology is, e.g., a fundamental tool in the classification of manifolds. Sheaf cohomology can be used to prove the Riemann-Roch theorem for algebraic curves which allows one, e.g., find coordinates for elliptic curves. – Lennart Meier Dec 13 2009 at 21:14
"Algebraic geometry" means a lot of different things to different people. What part and what level of the theory would you want a motivating problem for? – Qiaochu Yuan Dec 13 2009 at 22:06
Ilya, your comment on "group theory" suggests to me that you want all mathematics to be motivated by applications to number theory! Don't you think that, say, the symmetries of platonic solids are a subject of basic mathematical interest? – HW Jan 18 2010 at 6:18
@Qiaochu: Well, the simpler the example, and the more of algebraic geometry needed to understand it, the better. – Ilya Grigoriev Jan 19 2010 at 3:07
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@Henry. No, I definitely do not want all mathematics to be motivated by number theory. On the other hand, I don't know anything about symmetries of platonic solids that excites me too much. But if you have an example that is exciting, I'd be very interested to see it. More to the point, if it's simple to understand, interesting, and requires group theory to answer, by all means write it down in this thread! – Ilya Grigoriev Jan 19 2010 at 3:10
Set theory provides by far the easiest proof of the existence of transcendental numbers -- just show that the algebraic numbers and the integers can be put into 1-1 correspondence, but the real numbers and the integers can not. Liouville's proof isn't too hard, but it's nowhere near as elegant.
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Right -- and not only that, if you follow the proof through, you get an explicit example of a transcendental number. That's because (1) you can enumerate the algebraic numbers explicitly, and (2) the usual proof of the uncountability of the reals gives you, for any list of reals, an explicit real not on the list. But I don't know how many times I've read that the set theory proof doesn't give you an explicit transcendental number. – Tom Leinster Dec 24 2009 at 0:59
One can use the machinery of the fundamental group and of covering spaces to easily prove that any subgroup of a free group must be free.
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Well for a long time there was no proof of the Burnside theorem avoiding Representation theory. Now there are methods to proof it without Representation theory, but they are still a lot harder then the original representation theoretic one.
http://en.wikipedia.org/wiki/Burnside_theorem
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Mathematical logic can be motivated by other areas of math in at least two different ways:
[1] It allows you to formulate (and prove) results about the unsolvability of certain problems. These are obviously essential, since they tell you that you shouldn't spend too much time trying to solve those problems, which are often very natural problems.
For example, as a group theorist, you might often want to know whether two particular groups given by generators and relations are isomorphic. It would be nice to have some set of tools that allowed you to solve the problem mechanically, but no such tools exist (Novikov's Theorem).
Or, as a number theorist, you might wish for a set of tools allowing you to decide effectively whether a given polynomial equation has integer solutions. This is ruled out by the Davis-Putnam-Robinson-Matiyasevich Theorem.
[2] It can give you easier proofs of theorems in seemingly unrelated subfields. I hope somebody can provide/confirm examples here...for example, I think Gödel's Compactness Theorem gives some mileage in algebraic geometry (Nullstellensatz?), and nonstandard analysis can simplify a number of proofs (Tychonoff's Theorem?). (Although nonstandard analysis isn't exactly mathematical logic, the fact that proofs of standard results using nonstandard analysis can be trusted is a theorem of logic.)
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Lowenheim--Skolem gives a very quick proof of Ax's Theorem that every injective polynomial map from $\mathbb{C}^n$ to itself is surjective. – HW Dec 13 2009 at 23:24
This example is more closely related to a question of mine, but I'll give it here anyway.
A graded poset $P$ is Sperner if no antichain is larger than the largest level $P_i$ of $P$. This property is named after Sperner, who proved that the Boolean posets $B_n$ of subsets of ${ 1, 2, ... n }$ are Sperner. the Boolean posets $B_n$ are also rank-symmetric because they satisfy $(B_n)_i = (B_n)_{n-i}$, i.e. ${n \choose i} = {n \choose n-i}$ and unimodal because the sequence $(B_n)_i$ is at first increasing and then decreasing.
Let $G$ be a group acting on ${ 1, 2, ... n }$. Then $G$ acts on $B_n$ in the obvious way by order- and grade-preserving automorphisms. Define the quotient poset $B_n/G$ in the obvious way, which inherits the grading on $B_n$. It's not hard to see that $B_n/G$ is also rank-symmetric.
Theorem: $B_n/G$ is unimodal and Sperner.
The only known proofs of this theorem are algebraic (the one I know uses linear algebra), and even for special cases a purely combinatorial proof is not known. For example, in the case that $n = ab$ is a product of two positive integers and $G = S_b \wr S_a$ we recover the poset $L(a, b)$ of Young diagrams that fit into an $a \times b$ box; then the above theorem implies that the coefficients of the q-binomial coefficient ${a + b \choose b}_q$ are unimodal. This was first proven by Sylvester in 1878, and a combinatorial proof was not found until 1989. A combinatorial proof of the Sperner property is still not known. This example is all the more remarkable because the statement that $L(m, n)$ is Sperner requires almost no mathematics to understand.
Edit: I should probably provide a reference. This material is from some notes on algebraic combinatorics by Stanley.
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"The only proof I know uses linear algebra": I haven't thought this through all the way, but I would think that $B_n/G$ would inherit a weighting of its elements from the LYM inequality, from which the fact that $B_n/G$ is Sperner would follow from some easy counting argument. Does this not work? Or is it what you were referring to? – Harrison Brown Dec 13 2009 at 23:14
Ah, wait, I think I see why that doesn't work. Never mind, then. – Harrison Brown Dec 13 2009 at 23:17
Sometimes you want to understand a group $G$, but the only thing you know is that there is an extension $1 \to A \to G \to E \to 1$. If everything is abelian, $G$ corresponds to an element in $Ext^1(E,A)$. If at least $A$ is abelian, then $E$ acts on $N$ by conjugation and $G$ corresponds to an element in $H^2(E,A)$. Thus the classification of groups naturally leads to cohomology groups, which have a rich theory.
There is also a topological motivation: Which spheres act freely on finite groups?
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As for category theory, I don't think that there is a motivating example which has not already a category theoretic flavor. The leading theme is to unify and then generalize constructions resp. arguments, which come up in all areas of mathematics. Historically, natural transformations were introduced for the foundations of homology theory of topological spaces. But to start with an easy example, you may observe that for abelian groups $A,B,C$ there is a canonical isomorphism $(A \oplus B) \oplus C \cong A \oplus (B \oplus C)$, which reminds you of other associativity results such as $(X \cup Y) \cup Z \cong X \cup (Y \cup Z)$ for sets (here $\cup$ means disjoint union). Within category theory, you can see what's the real content of this: direct sum and disjoint unions are examples of coproducts, and coproducts are always associative. Even more striking, Yoneda's Lemma, which lies at the heart of foundations of category theory, tells you that the case of sets already settles the general case!
But category theory is more than just a language, it also provides general constructions: Assume you want to approximate a theory with another theory. This may be formalized by finding an adjunction between two categories. Freyds/Special Adjoint Functor Theorem tell you when this is possible. Although in some situations you can't write down the adjunction, the only thing you need is to know that it exists. For example, what is the categorical coproduct of an infinite family of compact hausdorff spaces? Can you write it down without using Stone-Cech?
There are also somewhat global motivations: Some Theories behave like other theories, and thus you may develope a theory for a large class of categories at once: monodial categories, topological categories, algebraic categories, locally presentable categories, etc. Of course, the same is true for other notions of category theory (functors, natural transformations, types of morphisms, etc.).
But if one has not heard of category theory before, the first motiviation should be to think in categories (in the colloquial sense). For example the set, which underlies a group, really differs from the group. In almost every book and lecture, this is absorbed by abuse of notation. The existence of bases in vector spaces is no reason at all to restrict linear algebra to vector spaces of the form $K^{(B)}$. Similarily, vector bundles should not be defined as bundles which are locally isomorphic to some $\mathbb{R}^n \times X$, which most topologists still ignore! Rather, it is first of all a vector space object in the category of bundles over $X$.
Let's conclude with an example which both introduces functors and algebraic geometry: Assume you have a system of polynomial equations $f_1(x)=...=f_n(x)=0$ in $m$ variables defined over $\mathbb{Z}$ and you want to study the solutions in arbitrary rings at once, using a single mathematical object, e.g. having in mind some local-global results of algebraic number theory. So for every ring $R$, we put $F(R) = \{x \in R^m : f_1(x)=...=f_n(x)=0\}$. Observe that for every ring homomorphism $R \to S$ there is a set map $F(R) \to F(S)$ and that this is compatible with composition of homomorphisms. This exactly means that $F$ is a functor from the category of rings to the category of sets. Algebraic geometry studies functors which locally look like the functor above.
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For me personally, the first time I "got" Fourier analysis was when I understood how it could be used to prove Roth's theorem on arithmetic progressions (that any dense set of integers contains an arithmetic progression of length 3). It can be proved in several ways, but when you see this way you immediately realize that the technique is likely to be useful for many other problems. And in fact, Roth's theorem can also be used to justify Szemerédi's regularity lemma (which is not quite the same as justifying a whole theory, but it is a very useful technique) in a similar way.
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I thought that it was the other way round, that Szemerédi's regularity lemma was one of the tools used in the original proof of Szemerédi's theorem for progressions of arbitrary length ? – Ewan Delanoy Jan 16 2010 at 14:54
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Well, yes, but nowadays we realise that the regularity lemma and the Fourier-analytic method are closely related, so that the fact that they are both useful for tasks such as finding arithmetic progressions. In particular, one can prove the regularity lemma by a spectral decomposition of the adjacency matrix (a generalisation of the Fourier decomposition of a function). There is also a similarity of technique between the basic idea of Roth's argument (density increment argument) and the basic idea behind the regularity lemma (energy increment argument). – Terry Tao Jan 18 2010 at 5:07
Problem: You need to multiply large numbers, with $10^9$ digits (or take products of power series). Your computer doesn't have the ability to do $10^{18}$ calculations.
Solution: Recognize multiplication of a power series as a convolution. Take a discrete Fourier transform of the digit sequences, multiply, and apply the inverse Fourier transform. Then perform the carries. This should take under $10^{12}$ calculations. The Fast Fourier Transform takes about $n \log n$ calculations.
The point is not that this is a fast algorithm or a clever trick. It's that you start out with a basic question about integers you can explain to someone comfortable with grade school math, and you end up dealing with complex or at least real numbers, characters of $\mathbb Z/n$, and properties of convolutions.
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Problem: Suppose you care about the real world and objects you can hold in your hands. Show that any flexible polyhedron maintains a constant volume while it is flexed. This was known as the Bellows Conjecture.
Solution: With a little commutative algebra, you can prove that 12*volume is an algebraic integer in $\mathbb Q$ adjoin the lengths of the sides. Any continuous function from $\mathbb R$ to a countable set is constant. In fact, the volume is a root of a single polynomial.
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Proving the termination of Goodstein sequences (a problem in natural numbers) via arithmetic on infinite ordinals.
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http://mathhelpforum.com/advanced-algebra/157067-geometry-hard-conic-problem-print.html
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# Geometry...hard conic problem
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• September 22nd 2010, 07:23 AM
zhupolongjoe
Geometry...hard conic problem
Identify the conic with equation x^2+3xy+4y^2-7=0, and find its center/vertex and axis.
My A matrix is (1 3/2; 3/2 4), which gives eigenvalues 1/2(5+3sqrt(2)) and 1/2(5-3sqrt(2))....these don't seem right as in every other problem, it comes out much nicer. Am I doing something wrong?
• September 22nd 2010, 12:39 PM
Opalg
Quote:
Originally Posted by zhupolongjoe
Identify the conic with equation x^2+3xy+4y^2-7=0, and find its center/vertex and axis.
My A matrix is (1 3/2; 3/2 4), which gives eigenvalues 1/2(5+3sqrt(2)) and 1/2(5-3sqrt(2))....these don't seem right as in every other problem, it comes out much nicer. Am I doing something wrong?
You have the correct eigenvalues, and you could continue to find the eigenvectors that way. But there is another method which seems to work better in this case.
When you rotate the axes through an angle $\theta$, x becomes $x\cos\theta - y\sin\theta$ and y becomes $\x\sin\theta + y\cos\theta$. Make those substitutions in the equation $x^2+3xy+4y^2-7=0$, to get $(x\cos\theta - y\sin\theta)^2 + 3(x\cos\theta - y\sin\theta)(\x\sin\theta + y\cos\theta) + 4(\x\sin\theta + y\cos\theta)^2-7=0$. If you choose $\theta$ so that the xy-term is zero then you will have found the directions of the principal axes. The coefficient of xy is $6\sin\theta\cos\theta +3(\cos^2\theta-\sin^2\theta) = 3\sin2\theta + 3\cos2\theta.$ This will be zero when $\tan2\theta=-1.$ So $2\theta = -\pi/4$, or $\theta=-\pi/8$. That gives you the orientation of the conic, and you should then be able to complete the question.
The numbers that come in the eigenvalue/eigenvector equations are related to the trig functions of $\pi/8$, which is what makes them messy.
• September 22nd 2010, 03:23 PM
zhupolongjoe
But the professor wants us to use the eigenvalue transformation technique to practice. So I've got it down to ((5+3sqrt(2))*x'^2+(5-3sqrt(2))*y'^2)/14=1.
So I believe this to be an ellipse with center (0,0). Is that correct? Also would the major axis be x=0 and the minor be y=0?
• September 22nd 2010, 11:53 PM
Opalg
Quote:
Originally Posted by zhupolongjoe
But the professor wants us to use the eigenvalue transformation technique to practice. So I've got it down to ((5+3sqrt(2))*x'^2+(5-3sqrt(2))*y'^2)/14=1.
So I believe this to be an ellipse with center (0,0). Is that correct? Also would the major axis be x=0 and the minor be y=0?
Ellipse – yes. Centre (0,0) – yes. But the major axis is x'=0, not x=0, and similarly for the minor axis. You need to give the direction of the line x'=0 in terms of the original coordinates x and y (which according to my method of solution should give a line making an angle $\pi/8$ with the original x-axis).
All times are GMT -8. The time now is 05:10 AM.
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http://crypto.stackexchange.com/questions/1058/how-can-rainbow-tables-be-used-for-a-dictionary-attack/3062
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# How can rainbow tables be used for a dictionary attack?
I'm putting together a password policy for my company. I very much want to avoid requiring complex passwords, and would much rather require length.
The maximum length I can enforce is 14 characters. I can calculate that 14 random lower case characters is stronger than 8 characters using anything on the keyboard. However I'm suggesting that people use phrases, song titles, things like that.
I don't think I need to protect against someone guessing passwords since our system will lock you out after 5 failures. I imagine I'm protecting against someone stealing the hashes of our passwords. Thus I imagine the mode of attack, if there was one, would be via rainbow tables.
I think 14 random lower case characters is reasonably safe against rainbow tables (we have Windows Server 2008 which I understand eliminates the LM compatibility weakness). However if phrases are used, there are many less possibilities than random characters would have.
Does anyone know if rainbow tables can be designed in conjunction with dictionary words, say by tokenizing them?
Along with this, I believe that NTLM passwords are hashed but not salted - does anyone know if that is true? If it turns out that they're salted, then I think I have little to worry about.
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## 4 Answers
Finding a decent explanation of rainbow tables was something I struggled with, so firstly I'll cover what they are. I will get to your question in the end. My sources for this are this guide and the wikipedia article.
Why can't I just use a big bucket of hashes?
Firstly the naive way to build a reverse lookup table is this. Let's say we want to generate all the hashes for every 8-digit password in existence, using the set `[A-Za-z0-9]`. In that case there are 62 unique characters and with that, using the counting function $n^r$ where $n$ is the possible number of outputs and $r$ is the number of choices we can make, then there are $218340105584896$ possible strings in this output space. Navely storing this data we can take a 8 character string as taking 8 bits per character (so each string costs 64 bits) plus a separator character plus the 256-bit output of say sha256, the total cost is then $218340105584896(64+1+256) = 70087173892751616$ bits. Converting that to bytes, i.e. $\frac{70087173892751616}{8*(1024)^3} \approx 8159220$ gigabytes.
Two notes on this:
1. It only considers exactly 8-character passwords. If you want to consider passwords containing 4-8 entries, you need $62^4+62^5+62^6+62^7+62^8$.
2. It assumes you're actually going to store that data in raw form. There are probably better ways to handle this.
So the first problem is storage. Above, we simply computed the total output space of a hash function.
What is a rainbow table?
The idea behind rainbow tables is to offset the space issue. To this end, let's define a few things: firstly we have a search domain we'll call $\mathbb{P}$ and a hash output domain we'll call $\mathbb{H}$. Then we have a hash we want to invert we'll define as $\mathcal{H}: \mathbb{P} \rightarrow \mathbb{H}$. i.e. the hash function takes an element of the search domain and produces a value in the hash output domain.
We then introduce a concept called chaining. To do this, consider we could define a function that maps the other way fairly trivially; let's call it $\mathcal{R}: \mathbb{H} \rightarrow \mathbb{P}$. In a rainbow table, a chain starts with a starting value and applies $\mathcal{H}$ then a $\mathcal{R}$ alternately, but always in pairs such that when done you end up with the first and last elements $p_0, p_k \in \mathbb{P}$. This is what you store.
A rainbow table is slightly more complicated than using the same $\mathcal{R}$ for each pair; this has problems relating to collisions. If two chains produce the same value they converge, meaning we waste time spend computing said chains - this is a chain collision. I had trouble visualizing this, so drawing a diagram:
````a_1 ----> a_2 ----> a_3 / b_2 --> a_4 / b_3 --> a_5 / b_4 ---> a_6 / b_5
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b_1 --------------------- -----> b_6.
````
Instead, a set of functions $\{\mathcal{R}_0, \ldots, \mathcal{R}_{k-1}\}$ are applied, one for each pair of the chain. SO when chains merge using this setup, they always produce the same final value and can then be de-duplicated, saving space - detection of wasted space is also much easier on generation.
Then:
• Generating rainbow tables: pick a length $k$ and define the functions $\mathcal{R}_{0}, \ldots, \mathcal{R}_{k-1}$. Then, for a given input $p \in \mathbb{P}$ we compute $c_0 = p, c_{n+1} = \mathcal{R}_{n-1}(r_{n}), r_{n} = \mathcal{H}(c_{n})\;\;(n=0,1,2,\ldots, k)$. These $c$ form a chain $C$. We compute our chains and for each chain we store just the pair $(c_0, c_k)$.
• Searching a rainbow table. Now, assume we want to inverse a hash value $h$. To do this we run through this process, starting with $i=k-1$:
1. Generate chain for $h$ starting at $R_{i}$
2. Using the end-value of the above chain, search our list of end values for computed chains. If we find a matching end-value, compute its chain term at a time (we can do this because we know the start value). If we find $h$ as say $r_n$ in that chain then the corresponding $c_n$ value is the inverse of $h$. Stop. If we don't find the end value in the chain list, carry on. If we don't find the hash in a generated chain where we did find the chain value, carry on.
3. If $i \neq 0$ do $i=i-1$ and go back to 1.
4. If we get here, we haven't found the inverse.
Right so exactly what is the benefit?
It's a space/time trade off. Specifically, the reverse lookup table takes a lot of space. This is a scheme that takes less space, but requires more time per lookup to work. Since the size of a complete reverse table is prohibitive for most people, the increased computational cost is generally preferable. Storage space is reduced massively, but is actually harder to calculate as it depends on $k$, and the $\mathcal{R}$ you use.
Clearly, also, we have different options with rainbow tables in terms of the length of $k$. The longer $k$, the fewer the number of total chains before all elements in $\mathbb{P}$ are covered. However, that also increases the run time of a lookup.
Oh no, now I have no idea how salt comes into this?!
Salt increases the size of $\mathbb{P}$ by increasing the $r$ in $n^r$. This makes both an inverse hash list astronomically big and also increases the size and computation time required for a rainbow table, too.
An attacker then has two options:
1. Produce a rainbow table specific to a given salt, making it invalid for a different salt.
2. Produce a huge rainbow table.
And what about slow hashing functions?
So far we've mostly talked about space as a consideration, without regards to the time taken to look up a function. Most cryptographic hashes are designed to be fairly quick, so this is ultimately feasible to do for say MD5.
Now what happens if we choose $\mathcal{H}$ which we know takes approximately a second to calculate each hash? Assuming there are no shortcuts that remove this additional time cost, the giant inverse table will take $218340105584896$ seconds, or approximately $6923519$ years. Your rainbow table is going to take a long time to generate too - assuming you never have collisions and cover the whole domain, just as long as a hash reverse lookup, plus added cost to search, depending on the length of $k$.
Combining both is a fairly effective defense against a rainbow table, making it both specific to a given salt and expensive to generate and use.
Why do password policies mention things like character classes, e.g. must have an upper case, must have a piece of punctuation?
We defined $\mathbb{P}$ as the set `[A-Za-z0-9]`. If you add punctuation into the mix you increase the size of $\mathbb{P}$ again, and so increase the size of the rainbow table (number of chains needed) again. Password length requirements also do this.
So dictionaries?
The whole premise behind a rather famous XKCD comic is the idea of information entropy. To tread roughshod over a fairly interesting area of science (sorry!) basically what you say is that whilst the total permutations of $\mathbb{P}$ are large, actually, quite a lot of those are totally meaningless to humans and we would not, given a choice, use them. Said XKCD comic was saying that actually, if you make certain judgements about the likely format of passwords, using information entropy as a measure of uncertainty within these formats then longer pass-phrases actually score better than shorter complicated passwords.
There is no reason you cannot produce a rainbow table using a set of reduction functions that takes this sort of assumption into account.
A dictionary is just a simplified version of this guessing - namely you're making the assumption that the thing you are inverting is actually a known dictionary word. You could also generate a rainbow table with known dictionary pass-phrases.
In both cases you are reducing $\mathbb{P}$ which by extension decreases the size of the rainbow table and the time to perform a lookup; however, such a technique is susceptible to the fact your approximate representation of a password just might not be right.
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The count of passwords is strange: it assumes that the same character may not appear twice in a given password. But what prevents that ? A more common count would be 62^6, close to 56 billions instead of your 44 billions. Storage cost is also overestimated: in a table of hashes for lookups, one would sort the pairs (password,hash) by hash value, keeping only as many bits of the hash value as necessary. The total cost would be closer to the cost of simply storing the passwords themselves, maybe with one or two extra bytes. About 400 GB, I would say. – Thomas Pornin Oct 26 '11 at 12:49
@ThomasPornin ah, thanks, I've used the wrong formula, looked that bit up quite hastily. Will fix. Also, yes, you could compress it down, but even so, a 400GB table for just 6-digit passwords is huge. And we haven't added in 5 and 4 digit passwords either; it only counts passwords of exactly 6 digits. – Antony Vennard Oct 26 '11 at 12:53
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About salts: the effect of a salt can be seen as: we do not have one hash function, we have one per salt value. There is not a single hash function but a family of hash function, and the salt value tells you which family member is actually used. When you build a table, you do it for a given hash function; it cannot be used for any other. – Thomas Pornin Oct 26 '11 at 12:54
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You could include the complexity: a rainbow table covers a set of N potential passwords. Let's call t the average chain length. Then, the building cost of the table is about 1.7*N (yes, it is 70% more expensive than a simple exhaustive search on the set). The storage cost is N/t elements of size at least log N (but not necessarily much bigger). Attacking a password has cost about t^2/2 hash computations, and t lookups (a "lookup" is when you are actually looking for data in the harddisk). – Thomas Pornin Oct 26 '11 at 13:00
400 GB is cheap, I can have 5 times that much storage space for 100\$. You may want to crank up the numbers a bit by considering 8-character passwords; they would look more impressive. – Thomas Pornin Oct 26 '11 at 13:08
show 2 more comments
Rainbow tables are just a hyped-up name for tables of precomputed hash values with some trickery to allow for handling huuuuge tables in less huge storage space (e.g. mere terabytes).
Precomputed tables, including rainbow tables, are utterly defeated by salting. Assuming you used a proper password hashing process, one which includes a salt and can be configured to be as slow as appropriate (hint: it is called bcrypt), the one remaining weakness is about an attacker "trying" passwords to attack a single hashed password (assuming he got a copy of it). The salt prevents the villain from attacking several passwords in parallel; he has to pay the full price of trying out the potential password for each hashed password he wants to attack. Moreover, the configurable slowness can make each try arbitrarily expensive for the attacker (this has a cost: it also makes it expensive for you; hence the "configurable" part: you make it as expensive as is still tolerable in your context).
Now, if you are stuck with existing software, some of the advice above may fail to be applicable. Remember that password hashing is a second line of defence: the attacker should not be able to access the hashed passwords; password hashing is there to prevent an attacker who did obtain unauthorized read access from upgrading his attack into user impersonation and associated write access.
The main issue with "password policies" is that they are like herding cats (the cats are a metaphor for the users). Users are human beings, so they are just bad at choosing passwords because they cannot do good random in their head. That's consubstantial to being human. By enforcing policies, you are trying to weed out some cases of extremely bad randomness that humans can invent with alarming creativity. In my view, a better "policy" would be to provide a tool which does create uniformly random passwords, so that users may run it and get "strong passwords". One scheme looks like this: passwords consisting in two lowercase letters, then two digits, then two lowercase letters, then two digits. Just 8 characters, reasonably easy to memorize, yet a bit more than 32 bits of entropy. 32 bits of entropy are enough to deter attackers if you also have an appropriate hashing process, as alluded to above (bcrypt !).
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Of course, users have more trouble remembering generated passwords than remembering ones they came up with. Passphrases might be better if you weren't so length limited. There are tools to check for really really bad passwords too. – erjiang Oct 26 '11 at 16:06
Rainbow tables can be used with words out of a dictionary rather than letters out of a charset.
The ophcrack vista liveCD is an example. In contains two dictionaries and tries combinations of words as well as modifications. For example the main dictionary contains "house" and "boat" and the second dictionary has "2010" "2011" "january". It will then create passwords like boat2010 or BOAT2010 or h0us3january.
You just need to create a reduction function that picks words from dictionaries and modifications from a set of modifications. Say you have a dictionary of 2^15 words (including an empty word) and you want to crate combinations of up to three words. To reduce a hash to a password you would - take the first 15 bits of the hash to select a first word from the dictionary - take the second 15 bits of the hash to select a second word from the dictionary - take the thrid 15 bits for the last word Then you could use some further bits of the hash to select some modifications (e.g. capitalization, writing the word backward, leet speak) up to your imagination.
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As far as I know, NTLM v2 uses a MD5 HMAC with the user password as the key and the MD4 hash of the concenation of user name and domain.
MD5 itself isn't very secure as a hash (collisions are trivially to find), but it retains some value for applications like this one, since the best know preimage attack has about $2^{123}$ complexity (so almost the full $2^{128}$).
However, Microsoft itself advises against its use in applications (source: Security Considerations for Implementers).
That being said, even without a salt, there are $26^{14}$ different lower-case 14-character passwords. MD5 hashes have a 16-byte output, so a complete rainbow table would hold $$26^{14} \cdot 16\ bytes \approx 2^{69.8}\ bytes = 2^{29.8}\ Tebibytes \approx 900\ million\ Tebibytes.$$
However, if you require your users to use phrases, keep in mind that there are only that much words in the English language. Let's say $10,000$ of them are likely to be used (it's probably less). About two words fit in 14 characters. That's only $100,000,000$ likely passwords.
If HMAC and user name (or hash and salt) are known, a simple brute-force attack will reveal the password.
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http://mathoverflow.net/questions/91183/could-svd-be-used-to-optimize-the-partial-inner-products
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## Could SVD be used to optimize the partial inner-products?
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Suppose a set $N$ of $n$ distinct points in $m-$dimensional space is given in $X\in\mathbb{R}^{n\times m}$. Also, suppose a subset $L\subset N$, `$|L|=l<m<n$`, with $m-$dimensional coordinates in $X_l\in\mathbb{R}^{l\times m}$.
Now, given a inner-product matrix $M=XX_l^T\in\mathbb{R}^{n\times l}$, could the solution $Z$ minimizing `$$||M-ZZ_l^T||~~~\equiv~~~||M-ZZ_l^T||^2,$$` where $||\cdot||$ denotes the Frobenius matrix norm, over all $Z\in\mathbb{R}^{n\times k}$, `$k<m$`, be obtained in a closed form. Note that $Z_l\in\mathbb{R}^{l\times k}$ is a configuration corresponding to the subset $L\subset N$ defined above.
With the Eckart-Young theorem in mind, I'm inclined to think that the solution $Z$ might be $$Z=U\Sigma_k^{1/2},$$ where $\Sigma_k^{1/2}$ is a diagonal matrix containing square roots of only $k$ dominant singular values of $M$, with $U$ containing the corresponding left-singular vectors as its columns. The plain Eckart-Young theorem might be used by showing that $Z_l$ is related to right-singular vectors $V$, as in $$Z_l=V\Sigma_k^{1/2},$$ but I believe the answer might not be easily expressible in a closed-form. Feel free to post an existing theorem that might be helpful for proving the above, either exactly or with certain guarantees on the approximation, ie. "by taking $$Z=U\Sigma_k^{1/2},$$ as above, we are guaranteed that the error wrt $$\min_Z||M-ZZ_l^T||~~~\equiv~~~\min_Z||M-ZZ_l^T||^2$$ is not larger than ..." Also, notes on some related problems already present in the literature will be appreciated.
Note that in case $L=N$, the solution indeed translates to the Eckart-Young theorem, thence with optimal solution being based (now) on spectral decomposition of $M$ relying only on its $k$ dominant eigenvalues, hence with $$Z=E\Lambda_k^{1/2},$$ where $\Lambda_k^{1/2}$ is a diagonal matrix with square roots of only $k$ dominant eigenvalues of $M$, and $E$ is a matrix containing the corresponding eigenvalues.
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Could someone edit my question? The preview seems not to capture everything I wrote. – usero Mar 14 2012 at 16:25
Done. As a rule of thumb, put backticks (...) around formulas when they are displayed incorrectly (as is reminded in the "How to write math" box on the right of every page). – Federico Poloni Mar 14 2012 at 16:30
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http://math.stackexchange.com/questions/200618/what-is-the-remainder-when-4100-is-divided-by-6
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# What is the remainder when $4^{100}$ is divided by 6?
I am trying to find the remainder when $4^{96}$ is divided by 6.
SO using the cyclicity method,
Dividing $4^1$ by 6 gives remainder 4.
Dividing $4^2$ by 6 gives remainder 4.
Dividing $4^3$ by 6 gives remainder 4.
Dividing $4^4$ by 6 gives remainder 4.
Dividing $4^5$ by 6 gives remainder 4.
..
..
So by this method we get the answer as `4`.
But when we reduces this to
$$4^{100} /6$$ to $$2^{200} /6$$
which is equal to
$$2^{199} /3$$
Then by using the cyclicity method:
Dividing $2^1$ by 3 gives remainder 2.
Dividing $2^2$ by 3 gives remainder 1.
Dividing $2^3$ by 3 gives remainder 2.
Dividing $2^4$ by 3 gives remainder 1.
Dividing $2^5$ by 3 gives remainder 2.
..
..
So accrding to this ,We get the answer as 2.
So which one is correct?
And how we are getting two different answers for the same numbers?
Thanks in advance.
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The first answer is right. The last step of your second argument is incorrect. You could (correctly) argue that $4^{100} \equiv 1$ (mod 3) and $4^{100} \equiv 0$ (mod 2), so that $4^{100} \equiv 4$ (mod 6) – Geoff Robinson Sep 22 '12 at 8:34
Geoff : I am not able to get you. I have added a proof of 2nd argument .Can you tell me where i am wrong? – vikiiii Sep 22 '12 at 8:37
3
The error comes in assuming that $2^{200}$ (mod 6) is the same as $2^{199}$ (mod 3). It is true that $2^{200}$ divided by $6$ is the same as $2^{199}$ divide by $3,$ but you are talking about remainders in the question. – Geoff Robinson Sep 22 '12 at 8:41
1
and 4/6 = 2/3, but 4 (mod 6) is not equal to 2 (mod 3). – Deedlit Sep 22 '12 at 8:49
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@GeoffRobinson: You surely meant $\frac{2^{199}-2}{3}$. – celtschk Sep 22 '12 at 9:05
show 4 more comments
## 1 Answer
The first method is correct. Since $4\times 4 \equiv 4 \mod 6$, you can conclude $4^n \equiv 4 \mod 6$ for $n \ge 1$.
The second method is wrong. A simple example is that it suggests the remainder when $8$ is divided by $6$ is the same as the remainder when $4$ is divided by $3$.
If you know that $$2^{2n-1} \equiv 2 \mod 3$$ and $$2^{2n} \equiv 1 \mod 3,$$ then modulo $6$ you can conclude $$2^{2n-1} \equiv 2 \text{ or } 5 \mod 6$$ and $$2^{2n} \equiv 1 \text{ or } 4 \mod 6.$$ You want the latter expression. You can similarly say that since $2^m \equiv 0 \mod 2$ you can conclude $$2^m \equiv 0 \text{ or } 2 \text{ or } 4 \mod 6.$$ Put those two together and you get $$2^{2n} \equiv 4 \mod 6$$ which is the first result.
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http://math.stackexchange.com/questions/207707/finding-all-real-zeros-of-the-polynomial/207710
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Finding all real zeros of the polynomial?
Okay, so I need to find all real zeros in this polynomial...
$$f(x) = 2x^3 + x^2 - 13x + 6$$
I know that the first step is to find the factors of 6 and 2, then see which when multiplied by the other coefficients have them add up to equal zero, but none of the factors I tried came out to zero. Is there an easier way to go about doing this???
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You might apply Cardano's formula. – AD. Oct 5 '12 at 12:43
(Yes, that was a joke..) – AD. Oct 5 '12 at 12:43
4 Answers
Your method will in general not find all real, but only all rational zeroes. If the leading coefficient were 1 instead of 2, all rational zeroes would have to be divisors of 6 (i.e. $\{\pm1, \pm2, \pm3, \pm6\}$). However with a leading coefficient of 2, one should also check halves of these values (i.e. also {$\pm\frac12, \pm\frac32\}$). Plugging in $x=2$, you will find that it is in fact a root. By polynomial division you thus obtain a quadratic for the other roots, which you can solve (or you will happen to find the remaining roots also by trying the above candidates).
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Ugh, okay, for whatever reason I didn't consider halves as factors too. This makes far more sense now. Thanks! – Brandt Oct 5 '12 at 13:46
If we assume we have only rational zeros we may the equation as $$2(x-a)(x-b)(x-c)=0= f(x)$$ with $a,b,c\in\mathbb{Q}$, expanding this leads to relations, which is a guide for guessing.
If there is only one rational solution we have $$(x-a)(2x^2-bx+c)=0= f(x)$$ with $a,b,c\in\mathbb{Q}$.
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find a number x that when you substitute in the polynomial will make it zero: the factor theorem. E.g when i substitute x = 2 , it gives zero. therefore (x-2) is a factor of the polynomial, then you perform polynomial division. when you divide 2x^2+x^2-13x+6 by (x-2) we get 2x^2+5x-3. then factorise 2x^2+5x-3 we get (x+3)(2x-1). therefore zeros are X=2,-3 and 1/2
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$$2x^{3}+x^{2}-13x+6=(x-2)(2x^{2}+5x-3)=(x-2)(x+3)(2x-1)$$ so the zeros are $-3$, $\frac{1}{2}$ and $2$.
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3
-1: While these are the zeroes, this doesn't help the OP learn how to solve problems. – Hurkyl Oct 5 '12 at 12:22
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http://www.physicsforums.com/showpost.php?p=4265478&postcount=1
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View Single Post
## Difference between These Electric Field Formulas
just a quick question. I seem to have come across two very similar formulas for electric fields produced by uniformly charged parallel plates. My book gives the formula as
$$E = \frac{\omega}{\epsilon_0}$$
on the other hand the notes from my prof tell me that
$$E = \frac{\omega}{2\epsilon_0}$$
My book or the notes don't mention the other formulas and this whole concept is still new to me to I'm a bit confused. what is the difference between the two formulas?
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
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http://mathoverflow.net/questions/44446/eisenstein-mod-p-hilbert-modular-forms
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## Eisenstein mod p Hilbert modular forms
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a question regarding non-cuspidal Hilbert modular forms. If one starts with a non-parallel weight for example, it is easy to prove that there are no Eisenstein series of any level, or as is generally stated, all forms are cuspidal. My question is what happens with mod p Hilbert modular forms? Are there (non-zero) non-cuspidal mod p Hilbert modular forms of non-parallel weight? (say at least when one or all the weights are greater than 1).
For classical modular forms, if the weight is greater than 1, the mod p modular forms are exactly the reduction of global modular forms, so the naive answer would be that there are none, but I am not too familiar with mod p Hilbert modular forms...
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## 1 Answer
The partial Hasse invariants $h_1,\ldots,h_d$ are mod $p$ Hilbert modular forms of non-parallel weight whose $q$-expansion at each cusp is equal to 1. The forms $h_1-1,\ldots,h_d-1$ generate the kernel of the $q$-expansion map over $\mathbb{F}_p$. Technically, these forms are of weight $(0,\ldots,0,p,-1,0,\ldots,0)$ or $(0,\ldots,0,p-1,0,\ldots,0)$, at least when $p$ is unramified, but you can always multiply them by some large parallel weight form to get something of everywhere positive weight. As you remarked, they have no characteristic 0 lift on account being non-cuspidal and having non-parallel weight.
If you want a more detailed account of these guys, and mod $p$ Hilbert modular forms in general, I recommend Goren's Lectures on Hilbert Modular Varieties and Modular Forms, especially chapter 5, and Andreatta-Goren's Hilbert Modular Forms: mod p and p-adic aspects, available on Goren's website.
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Just a technical question (while I read Goren's book) does this work for $p=2$ as well? Because if so you would get a form of weight (0,...,0,1,0,...,0) which does not satisfy the usual parity condition (but you might have a form of weight (0,...,0,2,0,..,0) modulo 2) – A. Pacetti Oct 25 2011 at 0:18
$p=2$ works fine. The "usual parity condition" plays no role, as far as I know, in the elementary "definitions and first properties" part of the theory; for example I can take a certain non-algebraic grossencharacter on $GL(1)$ of a CM extension of $F$ and induce it up to get a classical char 0 Hilbert modular form of e.g. weight $(1,2)$ on a real quadratic field. The parity condition only comes in when trying to do arithmetic -- e.g. it's the parity condition you need to get Satake parameters all defined over a number field or to attach a Galois representation. – Kevin Buzzard Oct 25 2011 at 8:07
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[NB even for $p>2$ the partial Hasse invariants don't satisfy the parity condition in general!] – Kevin Buzzard Oct 25 2011 at 8:08
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http://mathoverflow.net/questions/13107?sort=votes
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## Good books on Arithmetic Functions ?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
As, I was studying the Mobius Mu Function, and Gram Series... I got myself some pretty nice books:
Ribenboim - The New Book of Prime Number Records
Apostol - Introduction to Analytic Number Theory
Niven, Zuckerman, Montgomery - An Introduction to the Theory of Numbers
Inwaniec - Analytic Number theory
All of them dealt with the mobius mu function. But none of them dealt with the subject in details other than giving a few theorems and problems...
So I would like to know, If you guys know of some good books that deal exclusively with Arithmetic functions ?
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1
Inwaniec ==> Iwaniec and Kowalski. – Anweshi Jan 27 2010 at 11:28
1
While it doesn't deal exclusively with arithmetic functions, Hardy and Wright's 'An introduction to the theory of numbers' has quite a bit on arithmetic functions and is a bit of a classic within the field. You can check out the table of contents here: books.google.com/… – Ben Linowitz Jan 27 2010 at 12:59
"But none of them dealt with the subject in details other than giving a few theorems and problems." I have to say, I'm completely mystified by this remark. Iwaniec and Kowalski is one of the most complete books on analytic number theory you'll find anywhere. – David Hansen Jul 22 2010 at 3:10
I'm sorry, David, but the OP is right. And to give you just one example of what is not covered in I&K, well, it's Probabilistic Number Theory. They do mention the Erdos-Kac but that's it. Otherwise the subject is very wast and deserves at least 2 books the size of I&K (for the moment we have Elliott's two-volume treatise). Also the flavor of analytic number theory, as done by Luca, Erdos, etc. is not covered in I&K. I think this is what the OP is looking for. – anon Sep 14 2010 at 3:29
## 3 Answers
A MathSciNet search set to Books and with "arithmetic functions" entered into the "Anywhere" field yields 148 matches. Some of the more promising ones:
The theory of arithmetic functions. Proceedings of the Conference at Western Michigan University, Kalamazoo, Mich., April 29--May 1, 1971. Edited by Anthony A. Gioia and Donald L. Goldsmith. Lecture Notes in Mathematics, Vol. 251. Springer-Verlag, Berlin-New York, 1972. v+287 pp.
Narkiewicz, Wƚadysƚaw Elementary and analytic theory of algebraic numbers. Monografie Matematyczne, Tom 57. PWN---Polish Scientific Publishers, Warsaw, 1974. 630 pp. (errata insert).
Babu, Gutti Jogesh Probabilistic methods in the theory of arithmetic functions. Macmillan Lectures in Mathematics, 2. Macmillan Co. of India, Ltd., New Delhi, 1978.
Elliott, P. D. T. A. Arithmetic functions and integer products. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 272. Springer-Verlag, New York, 1985.
Sivaramakrishnan, R. Classical theory of arithmetic functions. Monographs and Textbooks in Pure and Applied Mathematics, 126. Marcel Dekker, Inc., New York, 1989.
Schwarz, Wolfgang; Spilker, Jürgen Arithmetical functions. An introduction to elementary and analytic properties of arithmetic functions and to some of their almost-periodic properties. London Mathematical Society Lecture Note Series, 184. Cambridge University Press, Cambridge, 1994.
Tenenbaum, Gérald Introduction to analytic and probabilistic number theory. Translated from the second French edition (1995) by C. B. Thomas. Cambridge Studies in Advanced Mathematics, 46. Cambridge University Press, Cambridge, 1995.
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Thanks Pete... I'll look into them... – Roupam Ghosh Jan 27 2010 at 9:15
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See also Chandrasekharan, Intro. to Analytic Number Theory, ETH lectures. – Anweshi Jan 27 2010 at 12:45
Actually, Narkiewicz wrote a book called "Number theory". It is quite comprehensive, but there is a treatment of arithmetic functions. – anon Sep 14 2010 at 3:31
### 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.
Isn't it weird that nobody's mentioned so far K. Chandrasekharan's treatise on Arithmetic Functions? OK, they've already mentioned his book on the principles of Analytic Number Theory, yet the book I'm now referring to is
K. Chandrasekharan. Arithmetical Functions. Die Grundlehren der Mathematischen Wissenschaften in Einzeldarstellungen, Band 167, Springer-Verlag.
You should definitely take a look at it, esteemed rpg16! I consider that none of the texts listed above is more to the point than this one.
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There's also this one: McCarthy, Paul J. Introduction to arithmetical functions. Springer, 1986.
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http://mathhelpforum.com/calculus/153267-i-have-problem-understanding-problem-about-limit-function.html
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# Thread:
1. ## I have a problem in understanding a problem about the limit of a function.
The question is as follows:
Use l'Hopital's rule to evaluate
lim [(a^x - b^x) / (e^ax - e^bx)] when x approaches 0. (Notice: e here stands for exponential symbol).
The answer given by the textbook is (lna - lnb) / (a - b). However, I have problem of getting to this answer. In particular, I don't know how to obtain (a - b) in the denominator.
Please show me the steps of getting to answer. thank you in advance
2. Originally Posted by Real9999
The question is as follows:
Use l'Hopital's rule to evaluate
lim [(a^x - b^x) / (e^ax - e^bx)] when x approaches 0. (Notice: e here stands for exponential symbol).
The answer given by the textbook is (lna - lnb) / (a - b). However, I have problem of getting to this answer. In particular, I don't know how to obtain (a - b) in the denominator.
Please show me the steps of getting to answer. thank you in advance
to where did u come with your work, show and we'll see where is the problem that u have
just to note that :
$\displaystyle \frac {d}{dx} ( e^{ax} )=ae^{ax}$
$\displaystyle \frac {d}{dx} ( a^{x} )=a^{x} \ln {a}$
$e^0 =1$
does this helps ? or ? need to go in that more detailed ?
anyway after using L'Hospital's rule (and knowing derivate's from above ) u can easy come to conclusion that result will be
$\displaystyle \frac {\ln{a}-\ln{b}}{a-b}$
3. this can be done by not using that rule, if you want the solution, let me know.
4. Originally Posted by Krizalid
this can be done by not using that rule, if you want the solution, let me know.
but it's clearly stated that he have to use L'Hospital's rule by his concept of question : "Use L'Hospital's rule to evaluate.. " and evaluating this using any another procedure could (although lead to the correct solution) lead to incorrect solution (as for teachers point of view)
5. Originally Posted by yeKciM
but it's clearly stated that he have to use L'Hospital's rule by his concept of question : "Use L'Hospital's rule to evaluate.. " and evaluating this using any another procedure could (although lead to the correct solution) lead to incorrect solution (as for teachers point of view)
Which is why he did not provide the solution but rather asked him if he wants the other solution :P
6. Thank you very much for your replies! thanks to all! And sorry for the late reply as I was sick over the past two days...
I still have a problem in understanding how to come to the (a-b) in the denominator. My procedure of doing this question is to differentiate the original function, [(a^x - b^x) / (e^ax - e^bx)], by using the quotient rules. I came to (e^ax - e^bx)^2, and then to (e^ax - e^bx) in the denominator after the entire calculation by the quotient rule. That is, I still do not understand how to go from (e^ax - e^bx) to (a - b).
Sorry for the mess, I will now learn how to type mathematical symbols.
7. ## Still have a problem.
Oh, I know how to come to the answer now! The problem I had was that I misunderstood the l'Hopital's rule!!! thanks to all!
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http://math.stackexchange.com/questions/263051/dimension-of-the-space-of-all-symmetric-matrices-with-trace-0-and-a-11-0
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# dimension of the space of all symmetric matrices with trace $0$ and $a_{11}=0$,
I want to know the dimension of the space of all symmetric matrices with trace $0$ and $a_{11}=0$,
I can show that the dimension of space of all symmetric matrices $S$ is $n(n+1)/2$, now I give a linear map $T:S\rightarrow\mathbb{R}^2$ by $T(A)=(a_{11},trace(A))$ so the kernel is exactly the space I want, so now it is enough to show the map is surjective so that I can apply rank nulity theorem. am I in right path? in that case $dimKer(T)=n(n+1)/2 -2$
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Yes, I think you are in the right path. Except you may need to consider the trivial case when $A$ is $1\times 1$ matrix. In that case, your $T$ is given by $T(A)=(a_{11},a_{11})$, which is not surjective. – Paul Dec 21 '12 at 5:43
:-o :-o :-o :-o :(, I was given $n\ge 2$ – Taxi Driver Dec 21 '12 at 5:44
1
There is always the tried and tested method of counting choices. You have a choice of every element above the diagonal, which is $\frac{1}{2}(n-1)n$ choices, and a choice for every element except two on the diagonal, which is $n-2$ more choices. Giving a total of $\frac{1}{2}(n^2+n-4)=\frac{1}{2}n(n+1)-2$ choices. – Daniel Rust Dec 21 '12 at 15:52
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http://math.stackexchange.com/questions/85075/how-to-solve-for-the-equation-ax-expbx-c?answertab=votes
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# How to solve for the equation $ax \exp(bx)=c$?
How to solve for the equation $ax \exp(bx)=c$? It is known that $x\geq 0$.
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## 1 Answer
Solving this equation requires the Lambert W-function, which when applied to c gives the solution for a=b=1. Barring trivial cases, I don't think there's any easier way to derive it.
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In general, we can rewrite the equation as $(bx) \exp(bx) = \frac{bc}{a} = d$ (say). The solution for this is $bx = W(d)$, or $x = \frac{1}{b}W(d)$. – Srivatsan Nov 23 '11 at 22:55
Srivatsan's comment ought to be the answer, methinks. – J. M. Nov 24 '11 at 8:33
Thanks, it is helpful. And what about more general equation ax*exp(bx+p)=c? And how to find a solution for W(d), if d>=0 and is supposed to be a real number? – Sergei Nov 25 '11 at 0:14
ax*exp(bx)*exp(p)=c, then bx*exp(bx)=b*(c/a)*exp(-p)=d. Am I right? Let d = 5. How to find the value of W(d)? Taylor series expansion around 0? – Sergei Nov 25 '11 at 0:33
@Sergei: Why not ask a new question? – J. M. Dec 4 '11 at 4:54
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http://physics.stackexchange.com/questions/tagged/vector-fields+special-relativity
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# Tagged Questions
2answers
125 views
### Metric coefficients in rotating coordinates
Let $(t,x,y,z)$ be the standard coordinates on $\mathbb{R}^4$ and consider the Minkowski metric $$ds^2 = -dt^2+dx^2+dy^2+dz^2.$$ I am trying to compute the metric coefficients under the change of ...
2answers
144 views
### Applying $\nabla\times\mathbf{B} = \mu_0\mathbf{J}$ in the presence of magnetic shielding
2012-06-13 - Revised question in experimental format (This is a thought experiment for which RF experts may have an immediate answer.) I'll assume (I could be wrong) the possibility of creating a ...
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http://physics.stackexchange.com/questions/46883/how-deep-is-the-great-red-spot/46886
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# How deep is the Great Red Spot?
The Great Red Spot (GRS) is a very persistent storm system that's easily visible through a telescope on the surface of Jupiter. But what is the three-dimensional structure of the GRS, and how deep into the planet does it extend?
When I first wrote this question I assumed that the GRS must be the end of a linear vortex, which leads to the question of why we can't see the other end elsewhere on the surface, as I discuss next. However, as hwlau pointed out in an answer, this isn't the only possibility, so I've added a section discussing another, more probable, case below.
If the Great Red Spot is the end of a vortex
According to my possibly-flawed understanding of fluid mechanics, you can't have a vortex with only one end, at least in invsicid flow. On the surface of Jupiter, the flow has an enormous Reynolds number on the scale of the GRS, so it's very nearly inviscid. If this is also true in the planet's interior then the other end of GRS vortex should terminate against a surface - but there is no solid surface on Jupiter (at least not anywhere near the upper part of the atmosphere that we can observe), so what plays that role in this situation? Is it just that the viscosity of Jupiter's atmosphere rapidly increases with depth, allowing the Red Spot to be only a shallow surface feature, or does it actually extend deep into the planet's interior?
Wikipedia has some information about deep models of Jupiter's interior. It doesn't mention vortices in this context but it does say these models predict that Jupiter's flow is organised into concentric cylinders parallel to the axis of rotation. A friend pointed out to me that this means the Red Spot vortex should extend into the interior parallel to the rotation axis as well, like this:
As shown in the figure, if the flow were completely inviscid (and this idea is correct), we should expect to see the other end of the vortex as a second storm, rotating in the opposite direction in the north equatorial belt. This doesn't match the observations, so presumably the vortex terminates before this point - but I'd like to know how far into the planet it extends before doing so.
If the Great Red Spot is a closed vortex
hwlau pointed out that the GRS needn't be the end of linear vortex at all, but could instead be a closed vortex oriented parallel to the surface, like a flattened smoke ring, as I've attempted to draw below:
This probably makes a lot more sense than my linear vortex assumption, because the GRS is presumably driven by thermal convection, and this geometry makes sense for a rotating convection cell. (Large storms on Earth have a similar structure.) Intuitively, if the structure of the GRS is like this, it seems like it should be less deep than it is wide - but since the GRS is wider than the Earth, that could still be quite a substantial distance, and I'm interested in whether a number can be put to it.
If the structure of the GRS looks like this, I would also be interested in anything that might be known about the structure of the flow beneath it, and how the two flows might interact.
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Isn't the core of Jupiter 'approximately' a solid? I thought there was metallic hydrogen there. Maybe hydrogen can be metallic without being solid? – QuantumDot Dec 15 '12 at 5:17
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@QuantumDot I believe metallic hydrogen is a liquid. (Many things can be metallic without being solid.) However, it might be an extremely viscous liquid for all I know. – Nathaniel Dec 15 '12 at 5:19
Metallic (i.e., degenerate) hydrogen can be either, but I believe it is fluid at the core temperature of Jupiter. – Chris White Dec 15 '12 at 5:24
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A vortex doesn't have to end on a solid surface. That's only in inviscid flow. When there is viscosity present, the viscous forces can and will kill the vortex. As the vortex get smaller at the bottom (think tornado), the vorticity gets larger and the viscous forces also get larger. You can see this in those soda-bottle tornadoes you can make at home. The vortex doesn't always terminate at a solid boundary, it will dissipate once it gets really thin. – tpg2114 Dec 15 '12 at 7:21
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@tpg2114 that's a good point. After I wrote the question I had a feeling something like that would be the case. But then the question becomes, how viscous is Jupiter's atmosphere in comparison to the scale of the Great Red Spot? I.e. does viscosity kill the vortex close to the surface, or does it end somewhere deep within the planet? I'll update the question to reflect this tomorrow. – Nathaniel Dec 15 '12 at 13:44
show 3 more comments
## 2 Answers
The atmosphere is not a simple two dimension spherical surface. You should also consider the convection up or down the atmosphere.
Lets consider the following situation: A circular cylinder with heat flow into the system at the exactly middle bottom of the cylinder. So there are convection up near the central axis of the cylinder and there are convection down near the outside of the cylinder. The streamline flow smoothly up in the middle and down from the outside. So, when you see from the top, it is like radial line moving out.
As the red spot is more like a spiral than a vortex, now lets imagine, instead of moving up directly, the streamline spiral around the central axis for a little bit less than one turn, so it need to spiral on the top surface in order to complete the remaining turn to make the streamline closed. Now when you view it from top, you should be able to see the spiral like the red spit from the top.
So why there is no vortex created? I think the key point is that the streamline are perfectly moving down along its outside circumference boundary. This feature is also true for other cylindrical shape (or a part of conic). Hence, it is possible to combine them somehow together to form a spherical shell like the atmosphere. Along this imaginary boundary, all streamlines are pointing down and no streamline are pointing parallel to the surface, so there is no new vortex created.
I dont have a good understand about vortex/antivortex, but I think the vortex are created due to the opposite streamline near a point. It also sound strange to me as I never see a pair of tornado in the north and south hemisphere on Earth.
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Regarding your last comment, the difference with the Earth is that there is a solid surface against tornadoes etc. can terminate. The rest of your answer sounds interesting if I understood it correctly - I need to think about it. (And still I'd really like to put a number to the spot's depth, even if it's not strictly speaking a vortex.) – Nathaniel Dec 15 '12 at 6:10
Thanks. But if there is no solid ground, then you can think the convection in a cone in which the heating point is the tip of the cone, so the streamline is still flowing upward in the middle and down along the conical surface. It is again parallel to other cone along this conical surface. So there is way the vortex can be created. – hwlau Dec 15 '12 at 6:18
The main idea is that I show that one "spiral" is a possible in my restricted situation. That is, it is possible in general – hwlau Dec 15 '12 at 6:20
it took me a while to see it, but I guess what you're saying here is that maybe when we look at the red spot we're not seeing the terminating end of a linear vortex, but instead we're seeing a side view of a closed vortex, like a flattened smoke ring. If this is the case then maybe the red spot can be very shallow, even if the fluid below isn't viscous. I appreciate your answer very much, but I hope eventually to hear from someone who knows specifically about the situation on Jupiter, so I'll leave the question open for now. – Nathaniel Dec 17 '12 at 2:54
Yes, it is what I want to say. It is just one possibility, there are more possibility out there. So I think that it is far from the truth, it needs someone have more knowledge of this particular situation to answer you. – hwlau Dec 17 '12 at 3:13
show 1 more comment
I may have found at least a partial answer to my own question in this review paper:
Vasavada, A.R.; Showman, A. (2005). "Jovian atmospheric dynamics: An update after Galileo and Cassini". Reports on Progress in Physics 68(8)
The thickness of Jupiter’s vortices is unknown, but dynamical arguments suggest that they are thin compared with a planetary radius. The rapid evolution and short dynamical lifetimes of most Jovian vortices argue against them being Taylor columns that penetrate through the molecular envelope. Furthermore, the non-zonal airflow that occurs in these vortices would violate the Taylor–Proudman theorem, especially for the largest vortices such as the Great Red Spot, if they were Taylor columns. Upper tropospheric and lower stratospheric temperature measurements imply that most large vortices extend only 2–4 scale heights (40–80 km) above the clouds (Conrath et al 1981). A recent series of multi-layer, quasi-geostrophic numerical simulations by Dritschel and colleagues (Dritschel and de la Torre Juarez 1996, Dritschel et al 1999, Reinaud et al 2003) show that three-dimensional, geostrophically balanced vortices tend to be baroclinically unstable if their thickness exceeds their width by a factor greater than $∼f/N$, where $f$ is the Coriolis parameter and $N$ is the Brunt–Väisälä frequency. In the sub-cloud troposphere of Jupiter, the latent heating associated with condensation of water might produce a characteristic Brunt–Väisälä frequency ∼0.002 $\text{s} ^{−1}$. If so, vortices such as the Great Red Spot and White Ovals probably extend less than ∼500 km below the clouds.
(Original emphasis.) I don't understand all the terminology above, but in context I interpret it as saying that the GRS is probably not the end of a linear vortex extending deep into the planet but instead is a fairly shallow closed vortex that doesn't penetrate into the metallic hydrogen layer. (500 km is a big number but is quite small compared to the observable dimensions of the red spot, which are of the order 10,000 km.) But at the same time the language is quite hedged and I get the impression that this is not a closed case.
In general it's interesting how little is known about the flows in the interior of jupiter, and from skimming this paper it seems that opinion is pretty divided about whether the features we can observe are confined to a shallow surface layer, or whether they're only the visible part of much deeper structures. It seems like a very interesting field with plenty of room left for surprises.
Addendum: I randomly came across this nice demonstration video, which shows the formation of flattened "pancake vortices" in stratified, rotating flows. It makes the claim in passing that the GRS is such a structure. This demonstration makes the idea that the GRS is not particularly deep seem very plausible to me.
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http://quant.stackexchange.com/tags/auto-correlation/hot
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# Tag Info
## Hot answers tagged auto-correlation
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### Who cares about autocorrelation?
Just about every introductory Econometrics class teaches that the violations of BLUE ("Best Linear Unbiased Estimator" -- the properties of linear least squares) are invalid standard errors in the case in the heteroscedasticity, so while your parameter estimates are still valid ("unbiased") your inference may be off invalid estimates (!!) in the presence ...
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### Auto-correlation of GBM
W.l.o.g we use the initial condition $S(0)=1$ and define $\gamma:=\mu-\frac{\sigma^2}{2}$. Hence we have the dynamics $$S_t=e^{\gamma t +\sigma W_t}$$ By definition $Cov(X,Y)=E((X-E(X))(Y-E(Y))=E(XY)-E(X)E(Y)$, where the last equality follows from the linearity of the expectation. Note $\gamma t+\sigma W_t$ is normal distributed with mean $\gamma t$ and ...
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### Is there a closed-form solution for the partial autocorrelation function of a Markov regime-switching process?
Apparently yes, (I haven't verified the math but have no reason to doubt it). For this simple case you can find a closed form in the following paper: Jeff A. BILMES: What HMM can do The closed form is given on part 4.4 of the paper but the whole thing is worth reading as it clearly shows the main properties of these models. You can also note that ...
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### Why do long-term equity return forecast models use dependent observations?
Generally we use models that go so far out in a comparative sense, not as an absolute decision. You are definitely do some good reading but I believe you are thinking about these models in the wrong way. I think (and correct me if I'm wrong) you are looking at creating or finding the perfect "crystal ball" model that will predict returns/risk premiums etc. ...
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### Why do long-term equity return forecast models use dependent observations?
Couple points I like to make: There exists no reliable model that can even predict future price returns (risk premiums, excess returns, whatever you want to call it) beyond a year, run as fast as you can if you hear from someone who claims he can predict risk premiums 10 years out, whether reliably or not. It makes zero sense and clearly comes from either ...
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### Squared and Absolute Returns
Simple...because you are interested in deviations from a metric, and not whether it deviates above or below. The very definition of volatility is a "measure of deviation". Squaring returns or using the absolute values just eases the calculation to arrive at a deviation measure. Otherwise volatility would have to be calculated in other ways as positive and ...
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### Squared and Absolute Returns
To simplify, consider the errors rather than the returns. The variance is effectively the average of the squared errors, while absolute deviation is the average of the absolute errors. So plotting the squared errors or absolute errors over time could give an indication of whether the variance or absolute deviation is constant over time. Since variance is ...
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### How do you correct Max Draw-Down for auto-correlation?
Thanks gappy for your precise response. However the answer to this auto-correlation is much more important than an academic discussion of which portfolio performance ratio is best. Auto-correlation distorts max draw-down calculations raising the question of whether the (positive) auto-correlation will continue in the future producing large draw-downs, or ...
3
### Are shorter holding period strategies better?
This is a partial explanation in that trading strategies with longer horizons have higher information ratios, t-statistics, slope coefficients, and R^2 in general. In other words, if information ratios for both strategies are identical then the longer-term trading strategy is already worse. John Cochrane illustrates how longer horizons have higher t-stats ...
3
### Who cares about autocorrelation?
Autocorrelation is usually a problem when you are doing the analysis of your error terms. When you build a model, you expect that the error term will have non significant autocorrelation. It is simple to understand: If your error term still have autocorrelations it certainly means that you are missing some information that could be introduced in your model. ...
3
### Time Series Regression with Overlapping Data
This question was ultimately answered on Cross Validated Here are a couple of articles that deal with this subject: Britten-Jones and Neuberger, Improved inference and estimation in regression with overlapping observations Harri & Brorsen, The Overlapping Data Problem
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### Good reference on sample autocorrelation?
Three good references are the Asymptotic theory for econometricians, H. White Stochastic Limit Theory, Davidson Asymptotic Theory of Statistical Inference for Time Series, Taniguchi and Kakizawa They are roughly in order of complexity. The crux of the matter is to balance the requirements of finiteness of higher moments of $X$ with its dependence ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://stats.stackexchange.com/questions/tagged/validation+bayesian
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# Tagged Questions
1answer
67 views
### Bayesian $\chi^2$ discrepancy
We define the $\chi^2$ discrepancy as $$T(y,\theta)=\sum_i\frac{(y_i-E(y_i|\theta))^2}{\mathrm{var}(y_i|\theta)}$$ (definition from Gelman). How do we apply this if $\theta$ is a random variable, and ...
0answers
99 views
### Validation of Bayesian hierarchical model [duplicate]
Possible Duplicate: Validation techniques for hierarchical model I already posted this to Math.SE, but I realized this would probably be a better place to post to. I have a hierarchical ...
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http://math.stackexchange.com/questions/tagged/function-fields+finite-fields
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Tagged Questions
1answer
172 views
Is a field perfect iff the primitive element theorem holds for all extensions, and what about function fields
Let $L/K$ be a finite separable extension of fields. Then we have the primitive element theorem, i.e., there exists an $x$ in $L$ such that $L=K(x)$. In particular, the primitive element theorem ...
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http://mathoverflow.net/questions/27670/independence-of-pa-implies-independence-of-pa-union-all-true-pi-1-statements/27709
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## Independence of PA implies independence of PA union all true $\Pi_1$ statements
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Prove that if a statement is independent of Peano Arithmetic (PA), then it's also independent of PA$_1$, where PA$_1$ is the union of the set of axioms in PA and the set of all true $\Pi_1$ statements.
This claim appears in this paper as Corollary 3. Ben-David attributes this theorem to "the folklore of proof theory". I want to see a proof.
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Welcome to MathOverflow. The convention here is to state one's questions in the form of a question (preferably also with polite language), rather than textbook-exercise-style in the form of a command as you have. You can edit your question by clicking on 'edit'. – Joel David Hamkins Jun 10 2010 at 12:11
Context would help too since the statement to be proven is clearly false. (E.g. Con(PA) is a true Pi_1 statement.) Perhaps there is some restriction on the statement in question. – François G. Dorais♦ Jun 10 2010 at 12:16
The paper that Wang Zirui is referring to is presumably this one: cs.technion.ac.il/~shai/ph.ps.gz, which he links to in his other question. Look at page 3 and footnote 2. I am not sure what they mean, but they do rule out Con(PA) and fixed-point-lemma self-referential statements. It isn't clear (at least upon quick perusal) whether they are making a strict mathematical claim or an empirical observation. – Joel David Hamkins Jun 10 2010 at 12:52
@Joel David Hamkins: Right, I mean Corollary 3 and possibly Lemma 6. They do use Corollary 3 to prove the main result, Corollary 6. What I want is to figure out the proof for Corollary 3 so that I can understand the main result. Many thanks in advance. – Zirui Wang Jun 10 2010 at 14:59
## 2 Answers
The claim you have asked us to prove is not true. If PA is consistent, then by the Incompleteness Theorem there are $\Pi_1$ statements that are independent of PA, such as Con(PA), which can be seen to be $\Pi_1$ when expressed in the form "no number is the code of a proof of a contradiction in PA". Thus, if PA is consistent, then Con(PA) is a statement that is independent of PA but provable in $PA_1$, so it is a counterexample to your claim.
Perhaps a more striking counterexample would be $\neg\text{Con(PA)}$, which is independent of PA, but refutable in $\text{PA}_1$. More generally, any statement having complexity $\Sigma_1$ or $\Pi_1$ that is independent of PA will be a counterexample to your claim, since such statements are settled by $\text{PA}_1$.
Perhaps the folklore result you meant to ask about is the following?
Theorem. If a $\Pi_1$ statement is independent of PA, then it is true.
Proof. If a $\Pi_1$ statement $\sigma$ is independent of PA, then it is true in some model $M\models PA$. The standard model $\mathbb{N}$ is an initial segment of $M$, and since the statement $\sigma$ is $\Pi_1$, it has the form $\sigma = \forall n \varphi(n)$, where $\varphi$ has only bounded quantifiers. Since $\sigma$ holds in $M$, it holds for all standard $n$ in $M$ and hence $\sigma$ is true in the standard model. In other words, it is true. QED
Note that the proof that $\sigma$ is true is not a proof in PA, but rather in a theory, such as ZFC, that is able to theorize about models of PA. So another way to view the theorem is as the claim that if ZFC can prove that a given $\Pi_1$ statement is independent of PA, then ZFC can also prove that it is true.
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+1. Of course the theorem only uses the fact that the statement is consistent with PA. That brings out an interesting modality: if S is Pi^0_1 then Con(S) implies S. So for an effective theory T, Con(Con(T)) implies Con(T). – Carl Mummert Jun 10 2010 at 13:07
You are certainly right, but unfortunately it's not what I'm asking. The theorem you guessed is also wrong. Could you please refer to Corollary 3 in Ben-David's paper? I refer to Corollary 3, or rather how it's used to prove Corollary 6. Thanks. – Zirui Wang Jun 10 2010 at 15:07
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The paper explains what is meant by "current approaches", i.e., specifically what they indicate in Lemma 6, the model theoretic approach common to the usual independence proofs (Paris-Harrington-Kanamori-McAloon, Kirby-Paris,...). The key is indeed Lemma 5, for which a reference is given in the paper. Of course, as stated, the corollary is false. With the interpretation of "current approaches" given by Lemma 6, the proof of Corollary 6 is obvious. If Mr. Zirui still has doubts, I would suggest first studying the 3 classical proofs I just mentioned, and then rereading Joel's answer. – Andres Caicedo Jun 10 2010 at 15:48
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As others have pointed out, the assertion is false. What the authors mean by calling it "folklore" is that virtually all the known techniques for proving a "natural" statement (like the Paris-Harrington theorem) independent of PA also prove the stronger result that the statement is independent of PA$_1$. Thus they are heuristically arguing that proving that $P\ne NP$ is independent of PA would require a new technique. Well, stated that way, that's not really news, but I think their ideas are still interesting.
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http://math.stackexchange.com/questions/172873/subgroups-between-s-n-and-s-n1
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# Subgroups between $S_n$ and $S_{n+1}$
Lets look at $S_n$ as subgroup of $S_{n+1}$. How many subgroups $H$, $S_{n} \subseteq H \subseteq S_{n+1}$ there are ?
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1
What can you say about an element of $H$ which is not in $S_n$? – Qiaochu Yuan Jul 19 '12 at 15:33
Stabilizers in primitive (hence 2-transitive) groups are maximal. – Steve D Jul 19 '12 at 18:25
## 1 Answer
None.
Let $S_n<H<S_{n+1}$ and suppose that $H$ contains a cycle $c$ involving $n+1$, say $$c=(a,\cdots,b,n+1).$$ Then by composing to the left with a suitable permutation $\sigma\in S_n<H$ such that $\sigma(a)=b$ we have $$\sigma c=\sigma^\prime (b,n+1)$$ where $\sigma^\prime(n+1)=(n+1)$, i.e. $\sigma^\prime\in H$. Thus the transposition $(b,n+1)$ is in $H$. Of course we may assume that $b=1$ and so all transpositions $(1,2)$, $(1,3)$, ..., $(1,n+1)$ are in $H$. These transpositions are known to generate $S_{n+1}$.
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Sorry I used a non-standard notation for cycles (the commas aren't usually there) but I couldn't find a way to manage spaces in this TeX environment. – Andrea Mori Jul 19 '12 at 15:39
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If by none you mean two... (the inclusions are not necessarily proper on either end!) – Qiaochu Yuan Jul 19 '12 at 15:40
1
I often put in ~'s for cycle notation. e.g. `$(1~2~3)$`=$(1~2~3)$ – anon Jul 19 '12 at 15:40
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@QiaochuYuan: of course you are right. The "none" incipit was chosen for dramatic effects :) I guess that I should say that I was impliciting assuming that $H$ is not one of the trivial possibilities. – Andrea Mori Jul 19 '12 at 15:45
@anon: thanks for the tip. – Andrea Mori Jul 19 '12 at 15:46
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http://math.stackexchange.com/questions/188529/what-is-wrong-with-this-approach-to-find-expected-value-of-distance-between-x
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# What is wrong with this approach to find Expected value of distance between $X, Y \in$ Uniform $(0, 1)$?
Let $X$ and $Y$ be i.i.d. random variables with Uniform $(0, 1)$ continuous distribution.
The problem is to find the expected value of the distance between X and Y.
My reasoning was, for all $(x, y) \in X \times Y$, any distance must be $0 \le |x - y| \le 1$, so we can consider two cases, $0 \le y - x \le 1$ and $0 \le x - y \le 1$.
So to find the expectated value of the distance, I would take the sum of the two integrals,
$$\int_0^1 \int_{y-1}^1 (y - x) \; dx \; dy$$
$$\int_0^1 \int_{x-1}^1 (x- y) \; dy \; dx$$
The first and second integrals each give $1/3$, so the answer I reached is $2/3$, whereas the correct answer for expected distance is $1/3$ and uses integrals over different limits of integration. I know what the correct answer says, but where am I going wrong here?
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## 3 Answers
The setting up of the integrals was not correct. If $X$ and $Y$ are our random variables, we want to find $$\iint_{S} |x-y|\,dxdy,$$ where $S$ is the $1\times 1$ square. Draw that square. Now it is useful to break up this integral into two parts, by drawing the line $y=x$, which slices our square into two parts: (i) where $y\ge x$ and (ii) where $y\le x$.
To do (i), express as an iterated integral. If you wish to integrate first with respect to $x$, note that $x$ travels from $0$ to $y$. The travels of $x$ can be seen clearly in the picture. So for (i) we need $$\int_{y=0}^1\left(\int_{x=0}^y(y-x)\,dx \right)\,dy.$$
Because of symmetry, (ii) will give the same result.
There are a couple of problems with your integrals. First note that there has not even been an integration with respect to $y$. Secondly, the setting up of the limits has no clear connection to the geometry.
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+1 yes those limits (i, ii) are the same given in the answer (not mine). What I'd like to find out, however, is why the setup of $0 \le (y - x) \le 1$ and $0 \le (x - y) \le 1$ cannot work or what is wrong with that reasoning. If it's possible to use that approach then how would you setup the proper integrals to express that? – T. Webster Aug 29 '12 at 21:25
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@T.Webster: For the second integral, it is true that $0\le x-y\le 1$. But if you want to express as an iterated integral, and want to integrate first with respect to $x$, $x$ will travel from $y$ to $1$, and not from $y$ to $y+1$. For if we travel to $y+1$, we are beyond our square. Personal note: After a zillion years of teaching this sort of stuff, I still have to draw a picture. Not for the sake of the students, but because I can't get the answer right, and certainly can't be sure I am right, without a picture. – André Nicolas Aug 29 '12 at 21:40
I did try to draw a picture the first time, but until I looked more closely didn't see that the math will just not work out with my choice of integration. While the expression $0 \le |x - y| \le 1$ is how I intuitively saw the answer, I don't see any upper bound on the integral that would make sense using the inequality $|x - y| \le 1$ as a lower bound, since e.g. $(1 - y, y)$ can't work, since it would always span a length of $y - (1 - y) = 1$. – T. Webster Aug 29 '12 at 23:28
Edit: my previous answer was actually not correct, so I removed it completely.
I had said that you need to divide your integral results by 2, since they each represented half of the domain space. This is not correct; if the integrals' bounds are set up correctly and we are actually integrating over the correct areas, the fact that they individually account for only half of the domain will naturally be accounted-for.
The only error in your original approach is the setup of the integrals. So, draw the image of $[0,1]\times[0,1]$ as a visual aid, and work from there. The integrals corresponding to your two cases should be $$\int_{0}^{1}\int_{0}^{y}(y-x)dxdy$$ for the upper triangle, where $y>x$, and $$\int_{0}^{1}\int_{0}^{x}(x-y)dydx$$, where $x>y$. By symmetry these integrals will have the same value, so double the calculation from each. (each has value 1/6).
Sorry for the confusion.
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thanks and +1. I like your answer because to me it helps explain where I went wrong from closer to my perspective. Could the conditions of domain space you mentioned on domain space (e.g. $P(x < y)=1/2$) be satisfied with the upper limit on the inner integrals? If so, what is the upper limit there? (Please see I updated my post because I left off the proper dx, dy from carelessness typing, but really had them in my work). – T. Webster Aug 29 '12 at 21:18
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Apparently I was careless in typing my integrals as well (I had no upper limits!). The upper limits should both be 1 here as well. – Kirk Boyer Aug 29 '12 at 21:26
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Actually the limits on my inner integrals did not reflect the image I was describing at all. I've corrected them as well. – Kirk Boyer Aug 29 '12 at 22:20
I really appreciate your answers, but I'm only allowed to choose one. – T. Webster Aug 29 '12 at 23:23
Don't worry about it :D I was wrong at first anyway, lol. I just wanted to make sure I didn't have a misleading answer that sounded correct. I upvoted Andre's answer, myself. – Kirk Boyer Aug 29 '12 at 23:45
This is not really an answer to the question "Where am I going wrong here?" but rather an alternative way of approaching the problem to be solved that gives the answer quite easily.
$Z = |X-Y|$ is a non-negative random variable taking on values in $[0,1]$, and hence $$E[Z] = \int_0^{\infty} P\{Z > z\}\,\mathrm dz = \int_0^1 P\{Z > z\}\,\mathrm dz.$$ But, for $0 \leq z \leq 1$, $P\{Z > z\}$ is the probability that the random point $(X,Y)$ lies in one of two right-triangular regions of area $\frac{1}{2}(1-z)^2$ each. This is easiest to figure out if you follow André Nicolas's suggestion to draw a sketch: the triangles have vertices $(z,0), (1,0), (1,1-z)$ and $(0,z), (0,1),(1-z,1)$ respectively. Hence, $$E[Z] = \int_0^1 P\{Z > z\}\,\mathrm dz = \int_0^1(1-z)^2\,\mathrm dz = \left. \frac{-1}{3}(1-z)^3\right|_0^1 = \frac{1}{3}.$$
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http://mathoverflow.net/questions/16266/bounding-roots-of-a-polynomial-with-rouches-theorem/16385
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## bounding roots of a polynomial with Rouche’s Theorem
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose `f(z) = z^n - k [ z^(n-1) + ... + z + 1 ]` where n is a positive integer and k is a real constant such that nk<1. I have shown that a root of this polynomial must satisy |z|<1, but I want a slightly better bound such as 1-k. This seems plausible from computational results but is difficult to prove. I am trying to use Rouche's theorem to do this but finding an appropriate bounding function is difficult. Is there any other result about holomorphic functions that may help?
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Note that f(z)*(z-1) = z^{n+1} - (1+k) z^n + k. This polynomial has the same roots as f(z), plus a root at z = 1. This might be a better function to look at. – Michael Lugo Feb 24 2010 at 16:10
@FC: Yes, I realized that after asking. See my answer. (I am going to delete my above comments now, since they have more or less become parts of the answer. Late comers to the discussion may notice that FC did indeed answer a question of mine, now gone.) – Harald Hanche-Olsen Feb 24 2010 at 22:00
## 4 Answers
Summary of the discussion: Using the triangle inequality, one sees that that $|f(z)|\ge f(|z|)$, and so the root of largest absolute value is the positive real root $z_k$. Differentiating $f(z)(z-1)$, one gets a bound:
$$z_k < \frac{1 + k}{1 + n^{-1}}.$$
When $k \rightarrow 1/n$, the largest real root approaches $1$ (by continuity, since $f(1) = 1 - nk$). Thus any bound must involve $n$.
The OP complains that he wants something better. It is pointed out that as $k \rightarrow 1/n$, the quantity $1 - z_k$ is asymptotic to
$$\frac{2(1 - kn)}{(1 + n)}.$$
The OP then complains that he wants a bound in $n$ and $k$ (which was already given). The OP askes whether the asymptotic above was found in the following way: "Are you simply using the fact that the root would occur roughly twice as far as the turning point?" No --- mathematics was used at this point.
The OP says that he simply wants an upper bound on the real part of each root. Since the real part of the real root $z_k$ is itself, this question has already been answered. The asymptotic result shows it is impossible to impove this bound significantly.
It's hard to tell if the problem with the OP's repeated questions involve English, Mathematics, or both. In either case, this has already wasted 15 minutes of my time. To paraphrase Zagier, that's the equivalent of 15 days of the OP's time. Feel free to edit this post to make it more "civil".
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Out of curiosity: what is the original phrase of Zagier you are paraphrasing? – Mariano Suárez-Alvarez Feb 25 2010 at 22:11
There is a saying about Zagier (although I don't know if it originates with him) that one day of his time is a week of anybody else's time. – Emerton Feb 25 2010 at 23:39
### 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 about any further boundings, but n = 3 and k = 1/4, or polynomial $4z^3 - z^2-z-1 = 0$ has a solution (1/12 + 1/12 (235 - 6 Sqrt[1473])^(1/3) + 1/12 (235 + 6 Sqrt[1473])^(1/3)), whose absolute value is ~ 0.868877, which is greater than 1-k. Other {n,k} pairs are {2,3}, {4,6}, and {5,6}.
EDIT I noticed that if the roots are multiplied by nk, then as k goes from 0 to 1/n, the largest root in absolute value (which happens to be the largest root) goes from 0 to about 1. So I suppose that the roots are bound in the range (0, 1/n).
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Thanks for pointing this out. 1-k was only a suggestion. 1-k^2 would suffice! – Josh Feb 24 2010 at 15:43
I got the same bound via a different method. I looked at fixing the complex part of z and taking the partial derivative of g(z) with respect to the real part. The first turning point to the left of the root at z=1 is at 1-(1-kn)/((1+n) = (1+k)/(1+n^(-1)).
The fact that this is where the turning point lies motivates the fact that there is still some more room to push to the left of this value where a root cannot lie.
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I would like a bound in terms of k and/or n. One which would work whether or not k is close to 1/n. My method was different in the sense that I took partial derivatives which would tell you the effect of fixing different values for the complex part.
Are you simply using the fact that the root would occur roughly twice as far as the turning point? I would like a better lower bound of |1-z_k| where |z_k| is the largest root with positive real part (better than (1-kn)/(1+n) I mean).
I actually simply want to find an upper bound on the real part of any root. i.e find an upper bound for p where z=p+iq is a root and p,q are real.
However it seems more natural to consider |z|. I hope this makes more sense.
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http://math.stackexchange.com/questions/287405/working-with-exponent-on-series
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# Working with exponent on series
Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum. $$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$ $$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4})}^n$$
So $a= (-1)^n\cdot 3^{-2}$ and $r=\frac{3}{4}$ and the sum is given by $$(-1)^n\cdot 3^{-2}\cdot \frac{1}{1-\frac{3}{4}}$$
Solving this I'm getting the result as $\frac{4}{9}$ witch I know Is incorrect because WolframAlpha is giving me another result.
So were am I making the mistake?
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The ratio of your geometric series is $r=-3/4$, not $3/4$. The term $a$ must be a constant, it can not depend on $n$. – julien Jan 26 at 16:30
Please use markdown rather than LaTeX for text formatting ("Geometric series"). There's fairly extensive help on markdown syntax if you click the little ? above the right side of the text-entry box. – Jonathan Christensen Jan 26 at 16:32
you cannot exract $(-1)^n$ because the variable $n$ is involved. Instead you must leave it at the quotient $\frac 34$... – Gottfried Helms Jan 26 at 16:33
## 3 Answers
The objective here is to transform your sum into a sum of the form:
$$\sum_{n=1}^\infty ar^{n-1}$$
$$\text{Transformation: }\quad\quad\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-3)^{n-1}}{4^{n-1}} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\left(\frac{-3}{4}\right)^{n-1}$$
Hence $a = -\dfrac{1}{12}$ and $r = -\dfrac{3}{4}.\quad$ Now use the fact that
$$\sum_{n=1}^\infty ar^{n-1} = \dfrac{a}{1 - r} = -\left(\frac{1}{12}\right)\cdot \left(\frac{1}{1 - (-\frac{3}{4})}\right)$$
Simpilfy, and then you are done!
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There is something wrong about the second term of your equalities. – julien Jan 26 at 16:39
Not correct, because $3^{n-2}=3^{n-1}3^{-1}=\frac{1}{3}3^{n-1}$ and not $3^{n-2}=3*3^{n-1}=$ In addition, wrong handling of the term $(-1)^n$ – Tomas Jan 26 at 16:42
@amWhy Thanks. Since $n=1$ you transform the $r^{n-1}$. But if $n=4$? It's still valid what you have done? – Favolas Jan 26 at 16:44
Thanks, Thomas! You beat me to it... – amWhy Jan 26 at 16:47
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@Babak Hahaha...you're certainly not under zero! ;-) – amWhy Jan 26 at 19:56
show 5 more comments
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}=\sum\limits_{n=1}^\infty (-1)^n\frac{1}{9}\left(\frac{3}{4}\right)^n=\frac{1}{9}\sum\limits_{n=1}^\infty \left(-\frac{3}{4}\right)^n= \frac{1}{9}\left(\frac{1}{1+3/4}-1\right)$$
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No, it sould be 1/9 instead 9 – Tomas Jan 26 at 16:43
Yes i miss that. Thanks – Adi Dani Jan 26 at 16:49
You have
$$\begin{align} \sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} &= \sum_{n=1}^{\infty} \frac{(-1)(-1)^{n-1}\frac{1}{3}3^{n-1}}{4\cdot 4^{n-1}} \\ &= \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-1)^{n-1}3^{n-1}}{4^{n-1}}\\ &= \sum_{n=1}^{\infty} \frac{-1}{12}\left(\frac{-3}{4}\right)^{n-1} \end{align}$$ So you have $a = \frac{-1}{12}$ and $r = \frac{-3}{4}$.
Note the key thing here that both $a$ and $r$ are constants/numbers. They do not depend on $n$. The idea is that you rewrite your series so that it is of exactly the form $$\sum_{n=1}^{\infty} a r^{n-1}$$ where again $a$ and $r$ are constants/numbers.
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http://mathhelpforum.com/calculus/10889-find-limit-given-sequence.html
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# Thread:
1. ## find the limit of the given sequence.
If the limit exists, find the limit of the sequence:
heres the equation: http://img402.imageshack.us/img402/5907/untitledav2.jpg
any help with this one please.
2. Observe the inequalities,
$\ln (1+e^n)\geq \ln (e^n)=n$
$\ln (1+e^n)\leq \ln (e^n+e^n)=\ln (2e^n)=\ln 2+\ln e^n=n+\ln 2$
Thus,
$\frac{1}{\ln (1+e^n)} \leq \frac{1}{n}$
$\frac{1}{\ln(1+e^n)} \geq \frac{1}{n+\ln 2}$
Thus, multiply by $n>0$,
$\frac{n}{n+\ln 2}\leq \frac{n}{\ln (1+e^n)}\leq \frac{n}{n}=1$
Now,
$\lim_{n\to \infty} \frac{n}{n+\ln 2}=1$
$\lim_{n\to \infty} 1=1$
Thus, by the squeeze theorem,
$\lim_{n\to \infty} \frac{n}{\ln (1+e^n)}=1$
3. okay, i see that we used the squeeze theorem, however,
in the second part of 'Observe the inequalities' where it = n + ln2
why did we raise 1 to e^n?
also, we used 1/n for a bound in the squeeze theorem, is this because you recognized it as a series? if so, which one.
thanks for all the help!
4. Originally Posted by rcmango
okay, i see that we used the squeeze theorem, however,
in the second part of 'Observe the inequalities' where it = n + ln2
why did we raise 1 to e^n?
also, we used 1/n for a bound in the squeeze theorem, is this because you recognized it as a series? if so, which one.
thanks for all the help!
Because,
$1\leq e^n$
Thus,
$1+e^n\leq e^n+e^n=2e^n$
5. okay, i can see as both sides of the squeeze theorem converge to 1.
then, also the middle equation converges to 1.
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http://math.stackexchange.com/questions/259459/prove-that-the-intersection-of-two-equivalence-relations-is-an-equivalence-relat?answertab=active
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# Prove that the intersection of two equivalence relations is an equivalence relation.
I am reading this chapter of the Book of Proof, and I'm stuck at the Exercise 10 of section 11.2. It is as follows.
Suppose $R$ and $S$ are two equivalence relations on a set $A$. Prove that $R \cap S$ is also an equivalence relation.
Thanks for helps!
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Which of the three conditions are you finding hard to verify? – Isomorphism Dec 15 '12 at 19:02
## 3 Answers
Hint: Use the fact that $R$ and $S$ are EQUIVALENCE relations on THE SAME set, and hence both must be reflexive, symmetric, and transitive on that set.
Then use the definition of set intersection: $R\cap S$ is the set of all pairs of elements in the set such that $(x, y) \in R$ AND $(x, y) \in S$ or, put differently, $(x, y) \in R\cap S \iff (x, y)\in R$ and $(x, y) \in S$.
Try to figure out what elements must necessarily be in $R\cap S$ and check to see that they must then be in both $R$ and $S$.
Another approach would be to use an indirect proof with the hints above:
"Given $R$ and $S$ are equivalence relations on a set $A$, suppose for the sake of contradiction, that $R\cap S$ is NOT an equivalence relation...". If not an equivalence relation, then $R\cap S$ fails to be reflexive and/or fails to be symmetric, and/or fails to be transitive. If you can work towards a contradiction (that this assumption must contradict the fact that both $R$ and $S$ are equivalence relations), then you are done.
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Hint (did this in school):
Let's have 2 relations $$R- antisymmetric$$ $$S-antisymmetric$$ I had to prove that $R \cap S$ is also antisymmetric.
$$P=R \cap S$$ $$(x,y) \in P$$ $$\implies (x,y) \in R \cap S$$ $$\implies (x,y) \in R \wedge (x,y) \in S$$ $$\implies ((y,x) \notin R \wedge (y,x) \notin S) \Rightarrow(x \ne y)$$ $$\implies (y,x) \notin R \cap S$$ $$\implies (y,x) \notin P$$
PS: Someone who know how to align text to left please edit :) Thanks
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For the solution of this exercise, you have to show that $R \cap S$ keeps the three properties of equivalence relations (reflexive, symmetric and transitive).
This means that for each x you have to show that $<x,x> \in R \cap S$ and for each pair$<x,y> \in R \cap S$, you have to show that $<y,x> \in R \cap S$ and for each pairs $<x,y>, <y,z> \in R \cap S$ you have to show that $<x,z> \in R \cap S$
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http://mathhelpforum.com/trigonometry/80074-inverse-function-multiple-x-terms.html
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# Thread:
1. ## Inverse Function with multiple x terms.
Hi I'm trying to find the inverse function of $f(x) = 8+x^2+tan{\frac{\pi*x}{2}}$ where $-1< x <1$ and $f^{-1}(8)$
The main place i am having trouble is I don't know how to rearrange for x when there is more than 1 x term. The best I've done is; by letting $f(x)= y$, get the equation $y-8=x^2+tan{\frac{\pi*x}{2}}$which doesn't seem to be a great deal better.
Any ideas would be greatly appreciated.
2. Originally Posted by mattty
Hi I'm trying to find the inverse function of $f(x) = 8+x^2+tan{\frac{\pi*x}{2}}$ where $-1< x <1$ and $f^{-1}(8)$
The main place i am having trouble is I don't know how to rearrange for x when there is more than 1 x term. The best I've done is; by letting $f(x)= y$, get the equation $y-8=x^2+tan{\frac{\pi*x}{2}}$which doesn't seem to be a great deal better.
Any ideas would be greatly appreciated.
please state the problem in its entirety. i suspect that what you were asked to find is something along the lines of $\frac d{dx}f^{-1}(8)$, in which case, finding the inverse function is not necessary. if all you are after is $f^{-1}(8)$, finding the inverse function is still not necessary
3. If , where , find .
That is the question verbatim. If finding the inverse function is not necessary then would you mind giving any hints as to the path i should be taking then? It wouldn't be as simple as letting $f(x)=8$ would it? and then since we know that x is between -1 and 1 sub 0 in and $f^{-1}=0$
If it's that simple I will quite happily kick myself
4. Originally Posted by mattty
If , where , find .
That is the question verbatim. If finding the inverse function is not necessary then would you mind giving any hints as to the path i should be taking then? It wouldn't be as simple as letting $f(x)=8$ would it? and then since we know that x is between -1 and 1 sub 0 in and $f^{-1}{\color{red}(8)}=0$
If it's that simple I will quite happily kick myself
start kicking
5. le sigh, thanks a bunch
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http://mathhelpforum.com/advanced-algebra/58204-modern-alg.html
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# Thread:
1. ## modern alg
Let R be acommutative ring and let b be a fixed element of R. Prove that the set b = {r is in R and r = cb for some element c in R} is an ideal of R.
2. Originally Posted by wvlilgurl
Let R be acommutative ring and let b be a fixed element of R. Prove that the set b = {r is in R and r = cb for some element c in R} is an ideal of R.
Define $(b) = \{ rb | r\in R\}$. To show $(b)$ is an ideal of $R$ you need to show if $x,y\in (b)$ then $x\pm y \in (b)$, and also that if $r\in R$ then $r(b),(b)r \subseteq (b)$. These should be clear by definition.
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http://mathoverflow.net/questions/61449/recovering-a-set-of-discrete-values-closed
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## Recovering a set of discrete values [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a set of positive integers $a$, where $\displaystyle\sum_i{a_i}=700$ and $\displaystyle\sum_i{a_i^2}=1212$.
Is there a unique decomposition of $a$ and, if so, how do I recover it?
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I'm afraid your question is not appropriate for MathOverflow (please see the FAQ). You are likely to get help if you ask the nice people at math.stackexchange.com Before you do so, you might want to ask yourself just how strong a constraint two equations will put on a potentially large number of variables. – S. Carnahan♦ Apr 12 2011 at 19:40
Also, there is no set of positive integers that will work. A multiset might. Gerhard "Ask Me About System Design" Paseman, 2010.04.12 – Gerhard Paseman Apr 12 2011 at 20:44
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http://www.illustrativemathematics.org/standards/hs
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Illustrative Mathematics
Content Standards: High School
hs_standards_nav_table fragment rendered at 2013-05-18 17:34:39 +0000
NUMBER AND QUANTITY ALGEBRA FUNCTIONS GEOMETRY STATISTICS AND PROBABILITY
hs_standards_body fragment rendered at 2013-05-18 17:34:39 +0000
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### Illustrated Standards
• #### Extend the properties of exponents to rational exponents.
N-RN: Extend the properties of exponents to rational exponents.
1. Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define $5^{1/3}$ to be the cube root of $5$ because we want $(5^{1/3})^3 = 5^{(1/3)3}$ to hold, so $(5^{1/3})^3$ must equal $5$.
2. Rewrite expressions involving radicals and rational exponents using the properties of exponents.
• #### Use properties of rational and irrational numbers.
N-RN: Use properties of rational and irrational numbers.
1. Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational.
• #### Reason quantitatively and use units to solve problems.
N-Q: Reason quantitatively and use units to solve problems.
1. Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. ${}^{\huge\star}$
2. Define appropriate quantities for the purpose of descriptive modeling. ${}^{\huge\star}$
3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. ${}^{\huge\star}$
• #### Perform arithmetic operations with complex numbers.
N-CN: Perform arithmetic operations with complex numbers.
1. Know there is a complex number $i$ such that $i^2 = -1$, and every complex number has the form $a + bi$ with $a$ and $b$ real.
2. Use the relation $i^2 = -1$ and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
3. $(+)$ Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers.
• #### Represent complex numbers and their operations on the complex plane.
N-CN: Represent complex numbers and their operations on the complex plane.
1. $(+)$ Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number.
2. $(+)$ Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example, $(-1 + \sqrt{3} i)^3 = 8$ because $(-1 + \sqrt3 i)$ has modulus $2$ and argument $120^\circ$.
3. $(+)$ Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints.
• #### Use complex numbers in polynomial identities and equations.
N-CN: Use complex numbers in polynomial identities and equations.
1. Solve quadratic equations with real coefficients that have complex solutions.
2. $(+)$ Extend polynomial identities to the complex numbers. For example, rewrite $x^2 + 4$ as $(x + 2i)(x - 2i)$.
3. $(+)$ Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
• #### Represent and model with vector quantities.
N-VM: Represent and model with vector quantities.
1. $(+)$ Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., $\textbf{v}$, $|\textbf{v}|$, $||\textbf{v}||$, $v$).
2. $(+)$ Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point.
3. $(+)$ Solve problems involving velocity and other quantities that can be represented by vectors.
• #### Perform operations on vectors.
N-VM: Perform operations on vectors.
1. $(+)$ Add and subtract vectors.
1. Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes.
2. Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum.
3. Understand vector subtraction $\textbf{v} - \textbf{w}$ as $\textbf{v} + (-\textbf{w})$, where $-\textbf{w}$ is the additive inverse of $\textbf{w}$, with the same magnitude as $\textbf{w}$ and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise.
2. $(+)$ Multiply a vector by a scalar.
1. Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as $c(v_x, v_y) = (cv_x, cv_y)$.
2. Compute the magnitude of a scalar multiple $c\textbf{v}$ using $||c\textbf{v}|| = |c|v$. Compute the direction of $c\textbf{v}$ knowing that when $|c|{v} \neq 0$, the direction of $c\textbf{v}$ is either along $\textbf{v}$ (for $c > 0$) or against $\textbf{v}$ (for $c < 0$).
• #### Perform operations on matrices and use matrices in applications.
N-VM: Perform operations on matrices and use matrices in applications.
1. $(+)$ Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network.
2. $(+)$ Multiply matrices by scalars to produce new matrices, e.g., as when all of the payoffs in a game are doubled.
3. $(+)$ Add, subtract, and multiply matrices of appropriate dimensions.
4. $(+)$ Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation, but still satisfies the associative and distributive properties.
5. $(+)$ Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.
6. $(+)$ Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors.
7. $(+)$ Work with $2 \times2$ matrices as a transformations of the plane, and interpret the absolute value of the determinant in terms of area.
• #### Interpret the structure of expressions.
A-SSE: Interpret the structure of expressions.
1. Interpret expressions that represent a quantity in terms of its context. ${}^{\huge\star}$
1. Interpret parts of an expression, such as terms, factors, and coefficients.
2. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret $P(1+r)^n$ as the product of $P$ and a factor not depending on $P$.
2. Use the structure of an expression to identify ways to rewrite it. For example, see $x^4 - y^4$ as $(x^2)^2 - (y^2)^2$, thus recognizing it as a difference of squares that can be factored as $(x^2 - y^2)(x^2 + y^2)$.
• #### Write expressions in equivalent forms to solve problems.
A-SSE: Write expressions in equivalent forms to solve problems.
1. Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. ${}^{\huge\star}$
1. Factor a quadratic expression to reveal the zeros of the function it defines.
2. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.
3. Use the properties of exponents to transform expressions for exponential functions. For example the expression $1.15^t$ can be rewritten as $(1.15^{1/12})^{12t} \approx 1.012^{12t}$ to reveal the approximate equivalent monthly interest rate if the annual rate is $15\%$.
2. Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments. ${}^{\huge\star}$
• #### Perform arithmetic operations on polynomials.
A-APR: Perform arithmetic operations on polynomials.
1. Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
• #### Understand the relationship between zeros and factors of polynomials.
A-APR: Understand the relationship between zeros and factors of polynomials.
1. Know and apply the Remainder Theorem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x - a$ is $p(a)$, so $p(a) = 0$ if and only if $(x - a)$ is a factor of $p(x)$.
2. Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
• #### Use polynomial identities to solve problems.
A-APR: Use polynomial identities to solve problems.
1. Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$ can be used to generate Pythagorean triples.
2. $(+)$ Know and apply the Binomial Theorem for the expansion of $(x + y)^n$ in powers of $x$ and $y$ for a positive integer $n$, where $x$ and $y$ are any numbers, with coefficients determined for example by Pascal's Triangle.The Binomial Theorem can be proved by mathematical induction or by a com- binatorial argument.
• #### Rewrite rational expressions.
A-APR: Rewrite rational expressions.
1. Rewrite simple rational expressions in different forms; write $\frac{a(x)}{b(x)}$ in the form $q(x) + \frac{r(x)}{b(x)}$, where $a(x)$, $b(x)$, $q(x)$, and $r(x)$ are polynomials with the degree of $r(x)$ less than the degree of $b(x)$, using inspection, long division, or, for the more complicated examples, a computer algebra system.
2. $(+)$ Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
• #### Create equations that describe numbers or relationships.
A-CED: Create equations that describe numbers or relationships.
1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. ${}^{\huge\star}$
2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. ${}^{\huge\star}$
3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. ${}^{\huge\star}$
4. Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm's law $V = IR$ to highlight resistance $R$. ${}^{\huge\star}$
• #### Understand solving equations as a process of reasoning and explain the reasoning.
A-REI: Understand solving equations as a process of reasoning and explain the reasoning.
1. Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
2. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.
• #### Solve equations and inequalities in one variable.
A-REI: Solve equations and inequalities in one variable.
1. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
2. Solve quadratic equations in one variable.
1. Use the method of completing the square to transform any quadratic equation in $x$ into an equation of the form $(x - p)^2 = q$ that has the same solutions. Derive the quadratic formula from this form.
2. Solve quadratic equations by inspection (e.g., for $x^2 = 49$), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as $a \pm bi$ for real numbers $a$ and $b$.
• #### Solve systems of equations.
A-REI: Solve systems of equations.
1. Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
2. Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
3. Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line $y = -3x$ and the circle $x^2 + y^2 = 3$.
4. $(+)$ Represent a system of linear equations as a single matrix equation in a vector variable.
5. $(+)$ Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension $3 \times 3$ or greater).
• #### Represent and solve equations and inequalities graphically.
A-REI: Represent and solve equations and inequalities graphically.
1. Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).
2. Explain why the $x$-coordinates of the points where the graphs of the equations $y = f(x)$ and $y = g(x)$ intersect are the solutions of the equation $f(x) = g(x)$; find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where $f(x)$ and/or $g(x)$ are linear, polynomial, rational, absolute value, exponential, and logarithmic functions. ${}^{\huge\star}$
3. Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes.
• #### Understand the concept of a function and use function notation.
F-IF: Understand the concept of a function and use function notation.
1. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If $f$ is a function and $x$ is an element of its domain, then $f(x)$ denotes the output of $f$ corresponding to the input $x$. The graph of $f$ is the graph of the equation $y = f(x)$.
2. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context.
3. Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. For example, the Fibonacci sequence is defined recursively by $f(0) = f(1) = 1$, $f(n+1) = f(n) + f(n-1)$ for $n \ge 1$.
• #### Interpret functions that arise in applications in terms of the context.
F-IF: Interpret functions that arise in applications in terms of the context.
1. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. ${}^{\huge\star}$
2. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function $h(n)$ gives the number of person-hours it takes to assemble $n$ engines in a factory, then the positive integers would be an appropriate domain for the function. ${}^{\huge\star}$
3. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. ${}^{\huge\star}$
• #### Analyze functions using different representations.
F-IF: Analyze functions using different representations.
1. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. ${}^{\huge\star}$
1. Graph linear and quadratic functions and show intercepts, maxima, and minima.
2. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.
3. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
4. Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.
5. Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude.
2. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.
1. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context.
2. Use the properties of exponents to interpret expressions for exponential functions. For example, identify percent rate of change in functions such as $y = (1.02)^t$, $y = (0.97)^t$, $y = (1.01)^{12t}$, $y = (1.2)^{t/10}$, and classify them as representing exponential growth or decay.
3. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum.
• #### Build a function that models a relationship between two quantities.
F-BF: Build a function that models a relationship between two quantities.
1. Write a function that describes a relationship between two quantities. ${}^{\huge\star}$
1. Determine an explicit expression, a recursive process, or steps for calculation from a context.
2. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model.
3. Compose functions. For example, if $T(y)$ is the temperature in the atmosphere as a function of height, and $h(t)$ is the height of a weather balloon as a function of time, then $T(h(t))$ is the temperature at the location of the weather balloon as a function of time.
2. Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms. ${}^{\huge\star}$
• #### Build new functions from existing functions.
F-BF: Build new functions from existing functions.
1. Identify the effect on the graph of replacing $f(x)$ by $f(x) + k$, $k f(x)$, $f(kx)$, and $f(x + k)$ for specific values of $k$ (both positive and negative); find the value of $k$ given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.
2. Find inverse functions.
1. Solve an equation of the form $f(x) = c$ for a simple function $f$ that has an inverse and write an expression for the inverse. For example, $f(x) =2 x^3$ or $f(x) = (x+1)/(x-1)$ for $x \neq 1$.
2. Verify by composition that one function is the inverse of another.
3. Read values of an inverse function from a graph or a table, given that the function has an inverse.
4. Produce an invertible function from a non-invertible function by restricting the domain.
3. $(+)$ Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents.
• #### Construct and compare linear, quadratic, and exponential models and solve problems.
F-LE: Construct and compare linear, quadratic, and exponential models and solve problems.
1. Distinguish between situations that can be modeled with linear functions and with exponential functions. ${}^{\huge\star}$
1. Prove that linear functions grow by equal differences over equal intervals, and that exponential functions grow by equal factors over equal intervals.
2. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another.
3. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another.
2. Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). ${}^{\huge\star}$
3. Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. ${}^{\huge\star}$
4. For exponential models, express as a logarithm the solution to $ab^{ct} = d$ where $a$, $c$, and $d$ are numbers and the base $b$ is 2, 10, or $e$; evaluate the logarithm using technology. ${}^{\huge\star}$
• #### Interpret expressions for functions in terms of the situation they model.
F-LE: Interpret expressions for functions in terms of the situation they model.
1. Interpret the parameters in a linear or exponential function in terms of a context. ${}^{\huge\star}$
• #### Extend the domain of trigonometric functions using the unit circle.
F-TF: Extend the domain of trigonometric functions using the unit circle.
1. Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle.
2. Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle.
3. $(+)$ Use special triangles to determine geometrically the values of sine, cosine, tangent for $\pi/3$, $\pi/4$ and $\pi/6$, and use the unit circle to express the values of sine, cosines, and tangent for $\pi - x$, $\pi + x$, and $2\pi - x$ in terms of their values for $x$, where $x$ is any real number.
4. $(+)$ Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions.
• #### Model periodic phenomena with trigonometric functions.
F-TF: Model periodic phenomena with trigonometric functions.
1. Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. ${}^{\huge\star}$
2. $(+)$ Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed.
3. $(+)$ Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context. ${}^{\huge\star}$
• #### Prove and apply trigonometric identities.
F-TF: Prove and apply trigonometric identities.
1. Prove the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ and use it to find $\sin(\theta)$, $\cos(\theta)$, or $\tan(\theta)$ given $\sin(\theta)$, $\cos(\theta)$, or $\tan(\theta)$ and the quadrant of the angle.
2. $(+)$ Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.
• #### Experiment with transformations in the plane
G-CO: Experiment with transformations in the plane
1. Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc.
2. Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs. Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch).
3. Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.
4. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments.
5. Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another.
• #### Understand congruence in terms of rigid motions
G-CO: Understand congruence in terms of rigid motions
1. Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.
2. Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent.
3. Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions.
• #### Prove geometric theorems
G-CO: Prove geometric theorems
1. Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.
2. Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to $180^\circ$; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.
3. Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
• #### Make geometric constructions
G-CO: Make geometric constructions
1. Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.
2. Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.
• #### Understand similarity in terms of similarity transformations
G-SRT: Understand similarity in terms of similarity transformations
1. Verify experimentally the properties of dilations given by a center and a scale factor:
1. A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged.
2. The dilation of a line segment is longer or shorter in the ratio given by the scale factor.
2. Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides.
3. Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar.
• #### Prove theorems involving similarity
G-SRT: Prove theorems involving similarity
1. Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.
2. Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
• #### Define trigonometric ratios and solve problems involving right triangles
G-SRT: Define trigonometric ratios and solve problems involving right triangles
1. Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.
2. Explain and use the relationship between the sine and cosine of complementary angles.
3. Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. ${}^{\huge\star}$
• #### Apply trigonometry to general triangles
G-SRT: Apply trigonometry to general triangles
1. $(+)$ Derive the formula $A = 1/2 ab \sin(C)$ for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side.
2. $(+)$ Prove the Laws of Sines and Cosines and use them to solve problems.
3. $(+)$ Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).
• #### Understand and apply theorems about circles
G-C: Understand and apply theorems about circles
1. Prove that all circles are similar.
2. Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
3. Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle.
4. $(+)$ Construct a tangent line from a point outside a given circle to the circle.
• #### Find arc lengths and areas of sectors of circles
G-C: Find arc lengths and areas of sectors of circles
1. Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector.
• #### Translate between the geometric description and the equation for a conic section
G-GPE: Translate between the geometric description and the equation for a conic section
1. Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.
2. Derive the equation of a parabola given a focus and directrix.
3. $(+)$ Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant.
• #### Use coordinates to prove simple geometric theorems algebraically
G-GPE: Use coordinates to prove simple geometric theorems algebraically
1. Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point $(1, \sqrt{3})$ lies on the circle centered at the origin and containing the point $(0, 2)$.
2. Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).
3. Find the point on a directed line segment between two given points that partitions the segment in a given ratio.
4. Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula. ${}^{\huge\star}$
• #### Explain volume formulas and use them to solve problems
G-GMD: Explain volume formulas and use them to solve problems
1. Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.
2. $(+)$ Give an informal argument using Cavalieri's principle for the formulas for the volume of a sphere and other solid figures.
3. Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. ${}^{\huge\star}$
• #### Visualize relationships between two-dimensional and three-dimensional objects
G-GMD: Visualize relationships between two-dimensional and three-dimensional objects
1. Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects.
• #### Apply geometric concepts in modeling situations
G-MG: Apply geometric concepts in modeling situations
1. Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder). ${}^{\huge\star}$
2. Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot). ${}^{\huge\star}$
3. Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios). ${}^{\huge\star}$
• #### Summarize, represent, and interpret data on a single count or measurement variable
S-ID: Summarize, represent, and interpret data on a single count or measurement variable
1. Represent data with plots on the real number line (dot plots, histograms, and box plots). ${}^{\huge\star}$
2. Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. ${}^{\huge\star}$
3. Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). ${}^{\huge\star}$
4. Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. ${}^{\huge\star}$
• #### Summarize, represent, and interpret data on two categorical and quantitative variables
S-ID: Summarize, represent, and interpret data on two categorical and quantitative variables
1. Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. ${}^{\huge\star}$
2. Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. ${}^{\huge\star}$
1. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models.
2. Informally assess the fit of a function by plotting and analyzing residuals.
3. Fit a linear function for a scatter plot that suggests a linear association.
• #### Interpret linear models
S-ID: Interpret linear models
1. Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. ${}^{\huge\star}$
2. Compute (using technology) and interpret the correlation coefficient of a linear fit. ${}^{\huge\star}$
3. Distinguish between correlation and causation. ${}^{\huge\star}$
• #### Understand and evaluate random processes underlying statistical experiments
S-IC: Understand and evaluate random processes underlying statistical experiments
1. Understand statistics as a process for making inferences about population parameters based on a random sample from that population. ${}^{\huge\star}$
2. Decide if a specified model is consistent with results from a given data-generating process, e.g., using simulation. For example, a model says a spinning coin falls heads up with probability $0.5$. Would a result of $5$ tails in a row cause you to question the model? ${}^{\huge\star}$
• #### Make inferences and justify conclusions from sample surveys, experiments, and observational studies
S-IC: Make inferences and justify conclusions from sample surveys, experiments, and observational studies
1. Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. ${}^{\huge\star}$
2. Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling. ${}^{\huge\star}$
3. Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are significant. ${}^{\huge\star}$
4. Evaluate reports based on data. ${}^{\huge\star}$
• #### Understand independence and conditional probability and use them to interpret data
S-CP: Understand independence and conditional probability and use them to interpret data
1. Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). ${}^{\huge\star}$
2. Understand that two events $A$ and $B$ are independent if the probability of $A$ and $B$ occurring together is the product of their probabilities, and use this characterization to determine if they are independent. ${}^{\huge\star}$
3. Understand the conditional probability of $A$ given $B$ as $P(\mbox{$A$ and $B$})/P(B)$, and interpret independence of $A$ and $B$ as saying that the conditional probability of $A$ given $B$ is the same as the probability of $A$, and the conditional probability of $B$ given $A$ is the same as the probability of $B$. ${}^{\huge\star}$
4. Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. ${}^{\huge\star}$
5. Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. ${}^{\huge\star}$
• #### Use the rules of probability to compute probabilities of compound events in a uniform probability model
S-CP: Use the rules of probability to compute probabilities of compound events in a uniform probability model
1. Find the conditional probability of $A$ given $B$ as the fraction of $B$'s outcomes that also belong to $A$, and interpret the answer in terms of the model. ${}^{\huge\star}$
2. Apply the Addition Rule, $P(\mbox{$A$ or $B$}) = P(A) + P(B) - P(\mbox{$A$ and $B$})$, and interpret the answer in terms of the model. ${}^{\huge\star}$
3. $(+)$ Apply the general Multiplication Rule in a uniform probability model, $P(\mbox{$A$ and $B$}) = P(A)P(B|A) = P(B)P(A|B)$, and interpret the answer in terms of the model. ${}^{\huge\star}$
4. $(+)$ Use permutations and combinations to compute probabilities of compound events and solve problems. ${}^{\huge\star}$
• #### Calculate expected values and use them to solve problems
S-MD: Calculate expected values and use them to solve problems
1. $(+)$ Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions. ${}^{\huge\star}$
2. $(+)$ Calculate the expected value of a random variable; interpret it as the mean of the probability distribution. ${}^{\huge\star}$
3. $(+)$ Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes. ${}^{\huge\star}$
4. $(+)$ Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value. For example, find a current data distribution on the number of TV sets per household in the United States, and calculate the expected number of sets per household. How many TV sets would you expect to find in 100 randomly selected households? ${}^{\huge\star}$
• #### Use probability to evaluate outcomes of decisions
S-MD: Use probability to evaluate outcomes of decisions
1. $(+)$ Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values. ${}^{\huge\star}$
1. Find the expected payoff for a game of chance. For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant.
2. Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident.
2. $(+)$ Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator). ${}^{\huge\star}$
3. $(+)$ Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game). ${}^{\huge\star}$
hs_illuminated_illustrables fragment rendered at 2013-05-18 17:34:41 +0000
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http://en.wikibooks.org/wiki/Data_Mining_Algorithms_In_R/Classification/SVM
|
# Data Mining Algorithms In R/Classification/SVM
## Introduction
Support Vector Machines (SVMs) are supervised learning methods used for classification and regression tasks that originated from statistical learning theory [1]. As a classification method, SVM is a global classification model that generates non-overlapping partitions and usually employs all attributes. The entity space is partitioned in a single pass, so that flat and linear partitions are generated. SVMs are based on maximum margin linear discriminants, and are similar to probabilistic approaches, but do not consider the dependencies among attributes [2].
Traditional Neural Network approaches have suffered difficulties with generalization, producing models which overfit the data as a consequence of the optimization algorithms used for parameter selection and the statistical measures used to select the best model. SVMs have been gaining popularity due to many attractive features and promising empirical performance. They are based on the Structural Risk Minimization (SRM) principle [3] have shown to be superior to the traditional principle of Empirical Risk Minimization (ERM) employed by conventional Neural Networks. ERM minimizes the error on the training data, while SRM minimizes an upper bound on the expected risk. This gives SRM greater generalization ability, which is the goal in statistical learning [4]. According to [5], SVMs rely on preprocessing the data to represent patterns in a high dimension, typically much higher than the original feature space. Data from two categories can always be separated by a hyperplane when an appropriate nonlinear mapping to a sufficiently high dimension is used.
A classification task usually involves training and test sets which consist of data instances. Each instance in the training set contains one target value (class label) and several attributes (features). The goal of a classifier is to produce a model able to predict target values of data instances in the testing set, for which only the attributes are known. Without loss of generality, the classification problem can be viewed as a two-class problem in which one's objective is to separate the two classes by a function induced from available examples. The goal is to produce a classifier that generalizes well, i.e. that works well on unseen examples. The below picture is an example of a situation in which various linear classifiers can separate the data. However, only one maximizes the distance between itself and the nearest example of each class (i.e. the margin) and for that is called the optimal separating hyperplane. It is intuitively expected that this classifier generalizes better than the other options [4]. The basic idea of SVM classifier uses this approach, i.e. to choose the hyperplane that has the maximum margin.
Figure 1: Example of separating hyperplanes
## Algorithm
Let D be a classification dataset with n points in a d-dimensional space D = {(xi, yi)}, with i = 1, 2, ..., n and let there be only two class labels such that yi is either +1 or -1. A hyperplane h(x) gives a linear discriminant function in d dimensions and splits the original space into two half-spaces:
$h(x) = w^Tx + b = w_1x_1 + w_2x_2 + ... + w_dx_d + b\,$,
where w is a d-dimensional weight vector and b is a scalar bias. Points on the hyperplane have h(x) = 0, i.e. the hyperplane is defined by all points for which wTx = -b.
According to [2], if the dataset is linearly separable, a separating hyperplane can be found such that for all points with label -1, h(x) < 0 and for all points labeled +1, h(x) > 0. In this case, h(x) serves as a linear classifier or linear discriminant that predicts the class for any point. Moreover, the weight vector w is orthogonal to the hyperplane, therefore giving the direction that is normal to it, whereas the bias b fixes the offset of the hyperplane in the d-dimensional space.
Given a separating hyperplane h(x) = 0, it is possible to calculate the distance between each point xi and the hyperplane by:
$\delta_i = \tfrac{y_ih(x_i)}{||w||}$
The margin of the linear classifier is defined as the minimum distance of all n points to the separating hyperplane.
$\delta^*=\min_{x_i}\{\tfrac{y_ih(x_i)}{||w||}\}$
All points (vectors x*i) that achieve this minimum distance are called the support vectors for the linear classifier. In other words, a support vector is a point that lies precisely on the margin of the classifying hyperplane.
In a canonical representation of the hyperplane, for each support vector x*i with label y*i we have that $y^*_ih(x^*_i) = 1$. Similarly, for any point that is not a support vector, we have that $y_ih(x_i) > 1\,$, since, by definition, it must be farther from the hyperplane than a support vector. Therefore we have that $y_ih(x_i) \geq 1,\ \forall x_i \in D$.
The fundamental idea behind SVMs is to choose the hyperplane with the maximum margin, i.e. the optimal canonical hyperplane. To do this, one needs to find the weight vector w and the bias b that yield the maximum margin among all possible separating hyperplanes, that is, the hyperplane that maximizes $\tfrac{1}{||w||}$. The problem then becomes that of solving a convex minimization problem (notice that instead of maximizing the margin $\tfrac{1}{||w||}.$, one can obtain an equivalent formulation of minimizing $||w||$) with linear constraints, as follows:
Objective Function
$min \tfrac{||w||^2}{2}$
Linear Constraints
$y_ih(x_i) \geq 1,\ \forall x_i \in D$
This minimization problem can be solved using the Lagrange multiplier method, which introduces a Lagrange multiplier α for each constraint:
$\alpha_i(y_ih(x)-1) = 0\ \text{with}\ \alpha_i \geq 0$
This method states that αi = 0 for all points that are at a distance larger than $\tfrac{1}{||w||}$ from the hyperplane, and only for those points that are exactly at the margin, i.e. the support vectors, αi > 0. The weight vector of the classifier is obtained as a linear combination of the support vectors, while the bias is the average of the biases obtained from each support vector [2].
SVMs can handle linearly non-separable points, where the classes overlap to some extent so that a perfect separation is not possible, by introducing slack variables εi for each point xi in D. If 0 ≤ εi < 1, the point is still correctly classified. Otherwise, if εi > 1, the point is misclassified. So the goal of the classification becomes that of finding the hyperplane (w and b) with the maximum margin that also minimizes the sum of slack variables. A methodology similar to that described above is necessary to find the weight vector w and the bias b.
SVMs can also solve problems with non-linear decision boundaries. The main idea is to map the original d-dimensional space into a d’-dimensional space (d’ > d), where the points can possibly be linearly separated. Given the original dataset D = {xi, yi} with i = 1,...,n and the transformation function Φ, a new dataset is obtained in the transformation space DΦ = {Φ(xi), yi} with i = 1,...,n. After the linear decision surface is found in the d’-dimensional space, it is mapped back to the non-linear surface in the original d-dimensional space [2]. To obtain w and b, Φ(x) needn't be computed in isolation. The only operation required in the transformed space is the inner product Φ(xi)TΦ(xj), which is defined with the kernel function (K) between xi and xj. Kernels commonly used with SVMs include:
• the polynomial kernel:
$K(x_i,x_j) = (x_i^Tx_j+1)^q$, where $q$ is the degree of the polynomial
• the gaussian kernel:
$K(x_i,x_j) = e^{-\frac{||x_i-x_j||^2}{2\sigma^2}}$, where $\sigma$ is the spread or standard deviation.
• the gaussian radial basis function (RBF):
$K(x_i,x_j)=e^{-\gamma||x_i-x_j||^2},\, \gamma\geq 0$
• the Laplace Radial Basis Function (RBF) kernel:
$K(x_i,x_j)=e^{-\gamma||x_i-x_j||},\, \gamma\geq 0$
• the hyperbolic tangent kernel:
$K(x_i,x_j)=tanh(x_i^Tx_j+offset)$
• the sigmoid kernel:
$K(x_i,x_j)=tanh(ax_i^Tx_j+offset)$
• the Bessel function of the first kind kernel:
$K(x_i,x_j)=\left( \tfrac{Bessel_{v+1}^n(\sigma||x_i-x_j||)} {(||x_i-x_j||)^{-n(v+1)}} \right)$
• the ANOVA radial basis kernel:
$K(x_i,x_j)= \left( \sum_{k=1}^n e^{-\sigma(x_i^k - x_j^k)^2} \right)^d$
• the linear splines kernel in one dimension:
$K(x_i,x_j)=1+x_ix_jmin(x_i,x_j)-\tfrac{x_i+x_j}{2}min(x_i,x_j)^2 + \tfrac{min(x_i,x_j)^3}{3}$
According to [6], the Gaussian and Laplace RBF and Bessel kernels are general-purpose kernels used when there is no prior knowledge about the data. The linear kernel is useful when dealing with large sparse data vectors as is usually the case in text categorization. The polynomial kernel is popular in image processing, and the sigmoid kernel is mainly used as a proxy for neural networks. The splines and ANOVA RBF kernels typically perform well in regression problems.
## Available Implementations in R
R [7] is a language and environment for statistical computing and graphics. There are five packages that implement SVM in R [6]:
• e1071 [8]
• kernlab [9]
• klaR [10]
• svmpath [11]
• shogun [12]
This documentation will focus on the e1071 package because it is the most intuitive. For information on the others, see the references cited above and the report of [6].
### e1071 package
The e1071 package was the first implementation of SVM in R. The svm() function provides an interface to libsvm [13], complemented by visualization and tuning functions. libsvm is a fast and easy-to-use implementation of the most popular SVM formulation of classification (C and $\nu$), and includes the most common kernels (linear, polynomial, RBF, and sigmoid). Multi-class classification is provided using the one-against-one voting scheme. It also includes the computation of decision and probability values for predictions, shrinking heuristics during the fitting process, class weighting in the classification mode, handling of sparse data, and cross-validation.
The R implementation is based on the S3 class mechanisms. It basically provides a training function with standard and formula interfaces, and a predict() method. In addition, a plot() method for visualizing data, support vectors, and decision boundaries is provided. Hyperparameter tuning is done using the tune() framework, which performs a grid search over specified parameter ranges.
#### Installing and Starting the e1071 Package
To install e1071 package in R, type
``` install.packages('e1071',dependencies=TRUE)
```
and to start to use the package, type
``` library(e1071)
```
#### Main Functions in the e1071 Package for Training, Testing, and Visualizing
Some e1071 package functions are very important in any classification process using SVM in R, and thus will be described here.
The first function is svm(), which is used to train a support vector machine. Some import parameters include:
• data: an optional data frame containing the variables in the model. If this option is used, the parameters x and y described below, aren't necessary;
• x: a data matrix, a vector, or a sparse matrix that represents the instances of the dataset and their respective properties. Rows represent the instances and columns represent the properties;
• y: a response vector with one label for each row (instance) of x;
• type: sets how svm() will work. The possible values for classification are: C, nu and one (for novelty detection);
• kernel: defines the kernel used in training and prediction. The options are: linear, polynomial, radial basis and sigmoid;
• degree: parameter needed if the kernel is polynomial (default: 3);
• gamma: parameter needed for all types of kernels except linear (default: 1/(data dimension));
• coef0: parameter needed for polynomial and sigmoid kernels (default: 0);
• cost: cost of constraint violation (default: 1). This is the ‘C’-constant of the regularization term in the Lagrange formulation;
• cross: specifies the cross-validation. A k > 0 is necessary. In this case, the training data is performed to assess the quality of the model: the accuracy rate for classification;
• probability: logical indicating whether the model should allow for probability predictions.
An example of svm() usage is given below:
``` library(MASS)
data(cats)
model <- svm(Sex~., data = cats)
```
The first two commands specify the usage of the cats dataset, which contains 144 instances, 2 numerical attributes for each instance ("Bwt" and "Hwt"), and the class for each instance (attribute "Sex"). The instance class can be "F", for female, or "M", for male. In the third command, the parameter "Sex~." indicates the attribute (column) of the dataset to be used as instance classes.
For information on the parameters of the model and on the number of support vectors, type:
``` print(model)
summary(model)
```
The result of the summary command is shown below:
``` Call:
svm(formula = Sex ~ ., data = cats)
Parameters:
SVM-Type: C-classification
SVM-Kernel: radial
cost: 1
gamma: 0.5
Number of Support Vectors: 84
( 39 45 )
Number of Classes: 2
Levels:
F M
```
To see the built model with a scatter plot of the input, the plot() function can be used. This function optionally draws a filled contour plot of the class regions. The main parameters of this function are listed below:
• model: an object of class svm data, which results from the svm() function;
• data: the data to visualize. It should be the same data used for building the model in the svm() function;
• symbolPalette, svSymbol, dataSymbol, and colorPalette: these parameters control the colors and symbols used to represent support vectors and the other data points.
The following command will produce the below graph, in which support vectors are shown as ‘X’, true classes are highlighted through symbol color, and predicted class regions are visualized using colored background.
``` plot(model,cats)
```
The predict() function predicts values based on a model trained by svm. For a classification problem, it returns a vector of predicted labels. Detailed information about its usage can be obtained with the following command.
``` help(predict.svm)
```
Let us first divide the cats dataset into a train and a test set:
``` index <- 1:nrow(cats)
testindex <- sample(index, trunc(length(index)/3))
testset <- cats[testindex,]
trainset <- cats[-testindex,]
```
Now we run the model again using the train set and predict classes using the test set in order to verify if the model has good generalization.
``` model <- svm(Sex~., data = trainset)
prediction <- predict(model, testset[,-1])
```
The -1 is because the dependent variable, Sex, is in column number 1.
A cross-tabulation of the true versus the predicted values yields (the confusion matrix):
``` tab <- table(pred = prediction, true = testset[,1])
```
If you type tab, you will see the confusion matrix like is shown below:
``` true
pred F M
F 10 8
M 6 24
```
With this information, it is possible to compute the sensitivity, the specificity and the precision of the model to the test set.
Model accuracy rates can be computed using the classAgreement() function:
``` classAgreement(tab)
```
The tune() function can be used to tune hyperparameters of statistical methods using a grid search over the supplied parameter ranges.
``` tuned <- tune.svm(Sex~., data = trainset, gamma = 10^(-6:-1), cost = 10^(1:2))
summary(tuned)
```
These commands will list the best parameters, the best performance, and details of the tested parameter values, as shown below.
``` Parameter tuning of `svm':
- sampling method: 10-fold cross validation
- best parameters:
gamma cost
0.1 100
- best performance: 0.1566667
- Detailed performance results:
gamma cost error dispersion
1 1e-06 10 0.2600000 0.1095195
2 1e-05 10 0.2600000 0.1095195
3 1e-04 10 0.2600000 0.1095195
4 1e-03 10 0.2600000 0.1095195
5 1e-02 10 0.2833333 0.1230890
6 1e-01 10 0.1788889 0.1359264
7 1e-06 100 0.2600000 0.1095195
8 1e-05 100 0.2600000 0.1095195
9 1e-04 100 0.2600000 0.1095195
10 1e-03 100 0.2833333 0.1230890
11 1e-02 100 0.1788889 0.1359264
12 1e-01 100 0.1566667 0.1014909
```
## Case Study
In this section we use a dataset to breast cancer diagnostic and apply svm in it. The svm model will be able to discriminate benign and malignant tumors.
### The DataSet
The dataset can be downloaded at [1]. In this dataset there are 569 instances and 32 attributes for each instance. The first attribute is the identification of instance, the second is the label for the instance class, which can be M (malignant tumor) or B (benign tumor). The following 30 attributes are real-valued input features that are computed from a digitized image of a fine needle aspirate (FNA) of a breast mass. Finally, there are 357 benign instances and 212 malignant instances in dataset.
In order to read the dataset, after downloading it and saving it, type in R:
```dataset <- read.csv('/home/myprofile/wdbc.data',head=FALSE)
```
'/home/myprofile/' is the path where the dataset was saved.
### Preparing the DataSet
Let us now divide at random the dataset in two subsets, one with about 70% of the instances to training, and another with around the remaining 30% of instances to testing:
```index <- 1:nrow(dataset)
testindex <- sample(index, trunc(length(index)*30/100))
testset <- dataset[testindex,]
trainset <- dataset[-testindex,]
```
### Choosing Parameters
Now, we will use the tune() function to do a grid search over the supplied parameter ranges (C - cost, $\gamma$ - gamma), using the train set. The range to gamma parameter is between 0.000001 and 0.1. For cost parameter the range is from 0.1 until 10.
It's important to understanding the influence of this two parameters, because the accuracy of an SVM model is largely dependent on the selection them. For example, if C is too large, we have a high penalty for nonseparable points and we may store many support vectors and overfit. If it is too small, we may have underfitting [14].
Notice that there aren't names for the columns (attributes) in the database . Then, R considers default names for them, as such V1, to the first column, V2 to the second and so on. It's possible to check this typing:
```names(dataset)
```
Then, as the class label is the second column of the dataset, the first parameter to tune() function will be V2:
```tuned <- tune.svm(V2~., data = trainset, gamma = 10^(-6:-1), cost = 10^(-1:1))
```
The results are showed with the following command:
```summary(tuned)
```
```
Parameter tuning of `svm':
- sampling method: 10-fold cross validation
- best parameters:
gamma cost
0.001 10
- best performance: 0.02006410
- Detailed performance results:
gamma cost error dispersion
1 1e-06 0.1 0.36333333 0.05749396
2 1e-05 0.1 0.36333333 0.05749396
3 1e-04 0.1 0.36333333 0.05749396
4 1e-03 0.1 0.30064103 0.06402773
5 1e-02 0.1 0.06256410 0.04283663
6 1e-01 0.1 0.08512821 0.05543939
7 1e-06 1.0 0.36333333 0.05749396
8 1e-05 1.0 0.36333333 0.05749396
9 1e-04 1.0 0.28314103 0.05862576
10 1e-03 1.0 0.05506410 0.04373139
11 1e-02 1.0 0.02756410 0.02188268
12 1e-01 1.0 0.03256410 0.02896982
13 1e-06 10.0 0.36333333 0.05749396
14 1e-05 10.0 0.28314103 0.05862576
15 1e-04 10.0 0.05500000 0.04684490
16 1e-03 10.0 0.02006410 0.01583519
17 1e-02 10.0 0.02256410 0.01845738
18 1e-01 10.0 0.05532051 0.04110686
```
### Training The Model
In order to build a svm model to predict breast cancer using C=10 and gamma=0.001, which were the best values according the tune() function run before, type:
```model <- svm(V2~., data = trainset, kernel="radial", gamma=0.001, cost=10)
```
To see the results of the model, as the number of support vectors is necessary type:
```summary(model)
```
The result follows:
```Call:
svm(formula = V2 ~ ., data = trainset, kernel = "radial", gamma = 0.001, cost = 10)
Parameters:
SVM-Type: C-classification
SVM-Kernel: radial
cost: 10
gamma: 0.001
Number of Support Vectors: 79
( 39 40 )
Number of Classes: 2
Levels:
B M
```
### Testing the Model
Now we run the model again the test set to predict classes.
```prediction <- predict(model, testset[,-2])
```
The -2 is because the label column to intance classes, V2, is in the second column.
To produce the confusion matrix type:
```tab <- table(pred = prediction, true = testset[,2])
```
The confusion matrix is:
``` true
pred B M
B 103 6
M 0 61
```
This means that there are 103 benign instances in test set and all of them were predicted as benign instances. On the other hand, there are 67 malign instances in test set, 61 were predicted rightly and 6 as benign instances.
Let:
• TP: true positive, i.e. malign instances predicted rightly
• FP: false positive, i.e. benign instances predicted as malign
• TN: true negative, i.e. benign instances predicted rightly
• |N|: total of benign instances
• |P|: total of malign instances
$sensitivity=\frac{TP}{|P|}$
$specificity=\frac{TN}{|N|}$
$precision=\frac{TP}{TP+FP}$
For this problem we have:
$sensitivity=\frac{61}{61+6}=0.91$
$specificity=\frac{103}{103}=1$
$precision=\frac{61}{61+0}=1$
The classification results are suitable.
## References
1. V. Vapnik, Statistical learning theory. Wiley, New Tork (1998).
2. ↑ M. J. Zaki and W. Meira Jr. Fundamentals of Data Mining Algorithms. Cambridge University Press, 2010.
3. S. R. Gunn, M. Brown and K. M. Bossley, Network performance assessment for neurofuzzy data modeling. Intelligent Data Analysis, volume 1208 of Lecture Notes in Computer Science (1997), 313.
4. ↑ S. R. Gunn. Support vector machines for classification and regression.Tech. rep., University of Sothampton, UK, 1998.
5. R. O. Duda, P. E. Hart and D. G. Stork, Pattern Classification. Ed.Wiley-Interscience, 2000.
6. ↑ A. Karatzoglou, D. Meyer and K. Hornik, Support vector machine in R. Journal of Statistical Software (2006).
7. R Development Core Team (2005). R: A Language and Environment for Statistical Computing. R Foundation for Statistical Computing, Vienna, Austria. ISBN 3-900051-07-0, URL http://www.R-project.org/.
8. E. Alpaydin. Introduction to machine learning. MIT Press, 2004.
|
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|
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|
http://math.stackexchange.com/questions/94559/multiplication-on-torus
|
# Multiplication on Torus
I am looking at two examples of the $n$-torus. Specifically, the cases where $n = 1$ and $n = 2$, that is, $S^1$ and $S^1 \times S^1$. I am trying to see if there is a continuous multiplication with identity element on these two spaces. The unit circle has identity element 1 in the complex plane and consists of all complex numbers $z$ such that $|z| = 1$. I am wondering if a continuous multiplication can be defined on the torus in the same way as the circle but using component-wise multiplication. Does this work?
-
4
– Dylan Moreland Dec 27 '11 at 22:18
1
More specifically, if $G$ and $H$ are topological groups, then the set $G\times H$ has a natural topology and a natural group structure, and it is fairly easy to prove that the group operation is continuous under the topology. – Thomas Andrews Dec 27 '11 at 22:26
1
Another way to think of $S^1\times S^1$ as a topological group is to think of it as $\mathbb R^2/\mathbb Z^2 \cong (\mathbb R/\mathbb Z)^2$ – Thomas Andrews Dec 27 '11 at 22:27
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http://mathematica.stackexchange.com/questions/9126/how-to-get-the-boolean-value-of-an-inequality-involving-an-interpolatingfunction?answertab=votes
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# How to get the boolean value of an inequality involving an InterpolatingFunction?
Here's the code:
yan = FunctionInterpolation[x^2, {x, -1, 1}];
FullSimplify[yan[x] > -1, -1 < x < 1]
Needless to say, what I expect to see in the output is "True", but FullSimplify doesn't seem to work. What function should I turn to?
@J.M. @belisarius @acl
yan = FunctionInterpolation[x^2, {x, -1, 1}];
MinValue[{yan[x], -1 < x < 1},x]>-1
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Huh? "seems not to work" is correct? I just delete it because of my language sense…OK, let me add it back. – xzczd Aug 6 '12 at 5:55
"doesn't seem to work" as you added is more standard, but in my opinion "seems not to work" is also acceptable and understandable. – Mr.Wizard♦ Aug 6 '12 at 6:10
In fact I've become confused after I searched the Internet, so I turned to the standard form to be on the safe side 囧. – xzczd Aug 6 '12 at 6:47
## 2 Answers
The following is basically the same @acl did, but using the package InterpolatingFunctionAnatomy which (in principle) will behave better than peeking at the internal structures when Mma version changes.
Needs["DifferentialEquationsInterpolatingFunctionAnatomy"];
yan = FunctionInterpolation[x^2, {x, -1, 1}];
yin = InterpolatingPolynomial[Transpose[Flatten /@
{InterpolatingFunctionCoordinates@yan,
InterpolatingFunctionValuesOnGrid@yan}], x];
FullSimplify[yin > -1, -1 < x < 1]
(*
True
*)
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very nice. how come I have less votes than you though, despite being first and explaining details?! Not fair! :) – acl Aug 5 '12 at 19:38
@acl That was because I forgot to upvote your answer. Easy to correct! :D – belisarius Aug 5 '12 at 19:40
really, I was joking! – acl Aug 5 '12 at 19:48
1
In fact, you don't need to load the package if you remember the actual syntax being used internally; in this case, it's yan["Coordinates"] and yan["ValuesOnGrid"] that can be used directly. Still, this works only because the original function was well approximated by a polynomial. In general, a polynomial interpolant can be more oscillatory than the piecewise polynomial interpolant used by InterpolatingFunction[]; be careful! – J. M.♦ Aug 5 '12 at 23:48
@J.M. yes, it was intended just for this case – belisarius Aug 5 '12 at 23:52
show 5 more comments
You can do this by explicitly constructing the InterpolatingPolynomial corresponding to yan, and then using FullSimplify:
yin = InterpolatingPolynomial[Transpose[Flatten /@ {yan[[3]],yan[[4]]}],x];
FullSimplify[yin > -1, -1 < x < 1]
(*True*)
Why does this work? Because yan actually has a list of points:
FullForm[yan]
so I can extract them with Transpose[Flatten /@ {yan[[3]],yan[[4]]}] and use them to construct a polynomial, which does the same thing as the interpolation function but which FullSimplify can now handle.
Maybe there's a better way to construct the InterpolatingPolynomial but this works.
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lang-mma
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http://mathoverflow.net/questions/63651/a-linear-algebra-question/63659
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## A Linear Algebra Question
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$ be a symmetric square matrix with integer coefficients and $M_k$ the matrix obtained by deleting the k-th line and k-th column. If det(M)=0 does it follow that $\det(M_kM_j)$ is a square?
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6
I would like to know what led you to consider this question. Who knows, this may even suggest an angle of attack. – Thierry Zell May 1 2011 at 21:31
I guess the answer is yes for $3\times 3$ matrices! – Andres Caicedo May 1 2011 at 21:35
2
Well, 2 by 2 anyway. – Will Jagy May 1 2011 at 22:10
2
The Kirchhoff matrix tree theorem uses the fact that det($M_i$)=det($M_j$), so it's true for Laplacian matrices, at least. (en.wikipedia.org/wiki/…) – Peter Shor May 1 2011 at 22:18
## 2 Answers
Yes one gets the square of the determinant of the matrix where one deletes row j and column k. That follows from Dodgson's condensation formula for determinants.
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10
(Also known as Lewis Carroll, inventor of cats) – Mariano Suárez-Alvarez May 1 2011 at 23:09
Inventor of cats? – Igor Rivin May 2 2011 at 0:36
My guess is the Cheshire Cat in Alice in Wonderland. en.wikipedia.org/wiki/Cheshire_Cat – Will Jagy May 2 2011 at 1:53
### 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.
Abdelmalek's answer was posted while I was finishing up this, but I decided to post it anyway since it gives insight into why the statement is indeed true.
Let $M = (m_{i,j})\in Mat^{n\times n}(\mathbb{Z})$ and define $D_i := Det(M_i)$
Since det(M) = 0, one has that some row of M is linearly dependent on the others, WLOG say the last row so that $m_{n,j} = \sum_{k=1}^{n-1}a_im_{k,j}$ for some $a_k's$; furthermore since $M$ is integer-valued, upon scaling one can assume the $a_k$'s are integers.
Now $M_n = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots\\m_{1,n-1}&m_{2,n-1}&\ldots&m_{n-1,n-1}\end{pmatrix}$
Lets compare $M_{n-1}$ to $M_n$:
$M_{n-1} = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&m_{1,n}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&m_{2,n}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&m_{n,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&m_{n,n}\end{pmatrix}$
$= \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_1m_{1,1}+a_2m_{1,2}+\ldots a_{n-2}m_{1,n-2}+a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_1m_{1,2}+a_2m_{2,2}+\ldots a_{n-2}m_{2,n-2}+a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_1m_{1,n-2}+a_2m_{2,n-2}+\ldots a_{n-2}m_{n-2,n-2}+a_{n-1}m_{n-1,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&a_1m_{1,n}+a_2m_{2,n}+\ldots a_{n-2}m_{n-2,n}+a_{n-1}m_{n-1,n}\end{pmatrix}$
Now since a determinant is unchanged by subtracting a multiple of one column from another, for each $i$ from 1 to $n-2$ subtract $a_i$ times the $i^{th}$ column from the last column, this leaves the following matrix with the same determinant as $M_{n-1}$:
$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_{n-1}m_{n-1,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&a_{n-1}m_{n-1,n}\end{pmatrix}$
Now do the same thing along the rows of this matrix, giving the following matrix with the same determinant as $M_{n-1}$
$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_{n-1}m_{n-1,n-2}\\a_{n-1}m_{1,n-1}&a_{n-1}m_{2,n-1}&\ldots&a_{n-1}m_{n-2,n-1}&a_{n-1}^2m_{n-1,n-1}\end{pmatrix}$.
Note that this matrix is obtained from $M_n$ by multiplying the last row by $a_{n-1}$ and then multiplying the last column by $a_{n-1}$, hence one has $D_{n-1} = a_{n-1}^2D_n$. A similar argument holds for the other $D_k$, thus $det(M_jM_k) = D_jD_k = a_{j}^2a_{k}^2D_n^2$ is indeed always a square.
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Do we even need that scaling? – Thierry Zell May 1 2011 at 23:25
@Thierry: I think you are right; the scaling is probably superfluous; I just mentioned it to make absolutely sure the $a_k$'s are integers for the conclusion in the final line. – ARupinski May 1 2011 at 23:27
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http://mathoverflow.net/questions/74480?sort=newest
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## More questions involving characteristic 2 theta series identities
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In my answer to my earlier question, "Existence of certain identities involving characteristic 2 thetas", I established some curious identities when the thetas have prime "level" congruent to 1 mod 4 or to 3 mod 8. This question concerns the case when the level is 7 mod 8.
I reprise notation from earlier questions. l is an odd prime and [j] is the sum of the x^(n^2), where n runs over the integers congruent to j mod l; we view the "theta series" [j] as elements of Z/2[[x]]. F is the power series x+x^9+x^25+x^49+x^81..., G=F(x^l) and H=G(x^l). My identities involve G,H and the various [j].
There is evidently a unique C in Z/2[[x]], having constant term 0, with C^2+C=G+H. I showed that when l is 1 mod 4 or 3 mod 8 (or when l=7), then C can be written explicitly as a polynomial in the [j]. Here is what the computer suggests when l=7 mod 8 and is < 50. First some notation. If (r,s,t) is a triple of integers, we define C(r,s,t) to be the sum of the power series [rj][sj][tj] where j runs from 1 to (l-1)/2. Define C(r,s,t,u) similarly. (When l is 3 mod 8, I showed that C is C(1,1,t) where t^2 is congruent to -2 mod l).
(1) When l=7, I can show that C=C(1,1,1,2)+C(1,2,3)
(2) When l=23 I think that C=C(3,3,1,2)+C(1,3,6)
(3) When l=31 I think that C=C(3,3,2,3)+C(2,3,7) (In my original post I wrote C(2,5,8), but C(2,3,7)=C(2,5,8))
(4) When l=47 I think that C=C(3,3,2,5)+C(2,3,9)
(Note that the sum of the squares of 3,3,2 and 5 is 47, etc.)
QUESTION 1: Can one establish the truth of (2),(3) and (4)? Kevin Buzzard explained to me that it's enough to show that the power series expansions agree up to a certain exponent, but I'm not sure what that exponent is, and I doubt that I have the computer power.
QUESTION 2: Are there identities like those above for l>50? And if so, what are these identities explicitly?
EDIT: Let V be the space spanned by the C(r1,r2,r3,r4) with r1=r2 and l dividing the sum of the squares of r1,r2,r3 and r4, together with the C(s1,s2,s3) with l dividing the sum of the squares of s1,s2 and s3. When l=7 mod 16 I can use Jacobi's 4-square theorem to show that C is in V. It's then possible to prove identities like those of (2) above by exploiting the geometry of of Spec R where R is the subring of Z/2[[x]] generated by the theta series [j].
-----One can show that an element of V has at most l(l-1)(l+1)/6 poles, counted with multiplicity, on the obvious projective completion of this curve. So if it has a zero of large enough order at the origin, it vanishes. I applied this technique for various l congruent to 7 mod 16; the results boggled my mind. It's only necessary to use 2 terms in the power series expansion of each theta series. When l=23, I got (2) above.
When l=71, I found that C=C(3,3,2,7)+C(5,6,9)
When l=103, I got 5 different expressions for C! Explicitly:
a----C(3,3,6,7)+C(2,9,11)
b----C(7,7,1,2)+C(5,9,10)
c----C(5,5,2,7)+C(1,3,14)
d----C(3,3,2,9)+C(6,7,11)
e----C(1,1,1,10)+C(1,6,13)
It seems possible to me that in general, for l=7 mod 8, one gets h/4 formulae of this sort where h is the class-number of Q(Root(-2l)). I've discussed the case l=31 in the comment to ARupinski. When l=47, I can show that C(3,3,2,5)+C(2,3,9)=C(1,1,3,6)+C(3,6,7). So if (4) above holds, there's a second formula for C in this case, just as in the case l=31. But I can't prove that C is in V when l=15 mod 16.
UPDATE__ Suppose l is 7 mod 8; consider the vectors W in Z^3 with(W,W)=2l. There is a group of order 48 operating on the set of such W by permutation and sign change of co-ordinates; the group operates without fixed points. So if there are 12h such W there are h/4 orbits under the group action.
----Ira Gessel's calculations, carried out for l<1500, indicate that there is an involution, O-->O' on the set of orbits, which has the following property. Let O be any of the (h/4) orbits and (r1,r2,r3) be a representative of O with r1 even (so that r2 and r3 are odd). Then if (s1,s2,s3) is a representative of O', we have the explicit identity C((r1)/2,(r1)/2,(r2+r3)/2,(r2-r3)/2)+C(s1,s2,s3)=C.
----But to know what these conjectured(but true beyond possibility of doubt) equations are for l>1500, we need to describe the involution. Franz Lemmermeyer suggested that the involution comes from an involution on a set of equivalence classes of quadratic forms of discriminant -8l. This is surely the case; I'll explain what the involution on the forms is, and how to transfer it to the orbits.
----Consider positive quadratic forms rx^2+2sxy+ty^2 with s^2-rt=-2l. Gauss showed that these fall in exactly h equivalence classes under the action of SL_2(Z), where 12h is the number of W with (W,W)=2l; we'll be interested in GL_2 equivalence however. Since rt=2l+s^2, we find that mod 16, rt is 2,7,14 or 15. This can be used to show that one of the following possibilities must occur:
a.--- Every non-zero n represented by the form is the product of an integer that is 1 or 7 mod 8 by a power of 2.
b.---Every non-zero n represented by the form is the product of an integer that is 3 or 5 mod 8 by a power of 2.
----In the first case we say that the form is in the principal genus, while in the second that it is in the non-principal genus. There are (h/4) GL_2 classes in the non-principal genus. Furthermore there is an involution on this set of classes taking the class of rx^2+2sxy+2ty^2 to the class of 2rx^2+2sxy+ty^2. I'll call this involution "composition with 2x^2+ly^2".
----I now describe a map from the set of (h/4) orbits to the set of (h/4) classes. The map can be shown to be onto, and so is bijective. When we transfer composition with 2x^2+ly^2 to the set of orbits, we get our desired involution; one which is in complete accord with Gessel's calculations. Suppose (W,W)=2l. Let W# consist of all elements of Z^3 orthogonal to W. We attach to W the class of the form (xU+yV,xU+yV), where U and V are a basis of W#. This class is evidently independent of the choice of basis; one can show that it consists of forms of discriminant -8l and lies in the non-principal genus. This gives the desired map from orbits to classes of forms; as I've indicated it is bijective.
EXAMPLE____Take l=1567, and W=(3,25,50) so that (W,W)=2l. Let O be the orbit of W. I'll calculate O', and write down the conjectured equations coming from O and O'. A basis for W# consists of U=(0,2,-1) and V=(25,1,-2). Then (U,U)=5, (U,V)=4, (V,V)=630, and a form attached to O is 5x^2+8xy+630y^2. Composition with 2x^2+1567y^2 takes this to 10x^2+8xy+315y^2. So we seek U' and V' with (U',U')=10, (U',V')=4, and (V',V')=315. Take U'=(3,1,0). A little experimenting, writing 315 as a sum of 3 squares, shows that we should take V'=(5,-11,13). Then W' which is orthogonal to U' and V' can be taken to be their vector product (13,-39,-38). So O' is the orbit of (13,38,39). And one of our predicted expressions for C is C(25,25,11,14)+C(13,38,39), while another is C(19,19,13,26)+C(3,25,50).
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Extending your comment that $3^2+3^2+2^2+5^2 = 47$ etc., is there any reason you can see as to why the sum of the squares of the arguments of the second function is always twice the I-value except in case (3) where it is triple the I-value? – ARupinski Sep 5 2011 at 5:06
In that case, it also could be twice the l-value; multiplication by 12 takes(2,5,8) to(-7,-2,3). So C(2,5,8)=C(2,3,7) when l=31. But I have no idea what happens in general. When l is 7 mod 16, I think I see how to write C as a sum of various C(r1,r2,r3) and various C(s1,s2,s3,s4). But my argument fails when l is 15 mod 16. – paul Monsky Sep 5 2011 at 12:18
@ARupinski---Quadratic forms arguments(Jacobi's 4 square theorem is the key!) will give (2) above. When l=31, similar arguments show that C(3,3,2,3)+C(2,3,7) is equal to C(1,1,2,5)+C(1,5,6). But I can't use these arguments to prove (3); the trouble is that 31=15 mod 16. – paul Monsky Sep 6 2011 at 23:54
## 1 Answer
I can now, with less computer calculation than I'd feared, answer Question 1. (I'll say more about Question 2 later).
Lemma:__ Let V be the vector space over Z/2 spanned by the C(r1,r2,r3) and the C(s1,s2,s3,s4). If an element of V has its power series expansion divisible by x^(l^2), it is 0.
To see this, let K be an algebraic closure of Z/2, S' be the subring of K[[x]] generated over K by the [j] and L be the field of fractions of S'. I can show that L/K is the function field of a curve, that there are exactly l(l-1)(l+1)/24 valuation rings in L/K that don't contain S', and that each of [1],...,[l-1] has a simple pole at each of them. So an element of V has at most l(l-1)(l+1)/6 poles in L/K, counted with multiplicity.
Also, the localization of S' at the maximal ideal generated by [1],...,[l-1] is dominated by exactly (l-1)/2 valuation rings in L/K. I can show that for each r prime to l there is an automorphism of L/K taking [j] to [rj] for each j, and that these automorphisms act transitively on this set of valuation rings. Now the elements of V are fixed by these automorphisms. So an element of V whose power series expansion is divisible by x^(l^2) has zeros of order at least l^2 at each of these valuation rings. Since (l^2)(l-1)/2 > l(l-1)(l+1)/6, such an element must vanish.
Suppose now that l=23. Since l=7 mod 16, my answer to my question "Existence of certain identities..." shows that C is in V. The lemma then shows that to prove 2. it's enough to show that C(3,3,1,2)+C(1,3,6)+C is divisible by x^529 in Z/2[[x]]. Now C(3,3,1,2)+C(1,3,6) is a sum of monomials in the [j]. Replacing each [j] by the sum of the first 2 terms in its power series expansion only modifies the sum by something divisible by x^529, and we're reduced to an easy computer calculation.
Establishing 3. and 4. is harder since we don't know in advance that C is in V. I'll use:
Theorem____Suppose there is an element R of V such that R+C is divisible by x^(d+1) where d=l(l+1)(l+1). Then R=C.
To see this, recall that F=x+x^9+x^25+..., that G=F(x^l), that H=G(x^l) and that C^2+C=G+H. Now there is a symmetric degree l+1 2-variable polynomial P over Z/2 with P(F,G)=0; furthermore P(z,H) is monic in z of degree l+1. This is discussed in my question "What's known about the mod 2 reduction...". Suppose a^l=1. Replacing x by ax^l in the identity P(F,G)=0, we get a root of P(z,H)=0 of the form ax^l+... This gives us l distinct roots, with G among them. By symmetry P(H,G)=0. So H(x^l) is still another root, and P(z,H) factors into linear factors over K[[x]].
Now R^2+R+G+H=(R+C)^2+ (R+C), and so is divisible by x^(d+1). Since P(G,H)=0, P(R^2+R+H,H) is divisible by x^(d+1). Also R^2+R has poles of order at most 8 at each valuation ring in L/K that doesn't contain S', while H has poles of order at most 12. It follows that P(R^2+R+H,H) has at most (l+1)l(l-1)(l+1)/2 =d(l-1)/2 poles, counted with multiplicity, in L/K. Arguing as in the proof of the lemma we see that it has more than d(l-1)/2 zeros, counted with multiplicity. So it vanishes, and R^2+R+H is a root of P(z,H)=0. Examining the roots of this equation we see that R^2+R+H can only be G. So R^2+R=G+H, and R=C.
Suppose now that l=31. To prove 3. we now see that it suffices to show that C(3,3,2,3)+C(2,3,7)+C is divisible by x^(186^2), since 186^2 >(31)(32)(32). This is carried out as in the case l=23, but now we have to use the first 12 terms in the power series expansion of each [j] rather than just the first 2. The treatment of 4. is similar, but now we show divisibility by x^(329^2) using the first 14 terms in the power series expansion of each [j].
EDIT___(UPDATE ON QUESTION 2)
Ira Gessel has now carried out computer calculations for all l<1500; here's a summary of his remarkable results. Consider the triples (r1,r2,r3) where the squares of r1,r1,r2 and r3 sum to l. To avoid duplicates, normalize each such triple so that r1,r2 and r3 are positive with r2>r3. Ira finds that for each such triple there is a unique second (normalized) triple (s1,s2,s3) such that the power series C(r1,r1,r2,r3)+C(2s1,s2+s3,s2-s3)+C is divisible by x^(l^2). Furthermore (r1,r2,r3)-->(s1,s2,s3) is an involution on the set of normalized triples with at most 1 fixed point.
When l=7 mod 16, the argument I gave above when l=23 then shows that for each normalized triple (r1,r2,r3) and the corresponding (s1,s2,s3) we have the identity C=C(r1,r1,r2,r3)+C(2s1,s2+s3,s2-s3); note that the squares of 2s1, s2+s3, and s2-s3 sum to 2l. So, for example, when l=1447 we get 17 distinct formulae for C.
When l=15 mod 16, it seems certain that once again C(r1,r1,r2,r3) and C(s1,s2,s3) sum to C. This could be proved by extending Ira's calculations to prove divisibility by x^(d+1) where d=l(l+1)(l+1), as in my treatment of l=31 and l=47. But I think this extension is unnecessary, and that one may deduce the identities simply from the divisibility by x^(l^2) established by Ira. If my idea for demonstrating this works out I'll post it as a comment.
But great mysteries remain. Why should this all be true? And can one describe the mysterious involution (r1,r2,r3)-->(s1,s2,s3) explicitly? By the way, it's known that the number of normalized triples is odd or even according as n is 7 or 15 mod 16. The proof of this goes back to Hasse; one shows that the number of triples is h/4 where h is the class number of Q(Root(-2l)) and uses results of Gauss on representations by sums of 3 squares, together with some genus theory for binary quadratic forms.
-
In the theorem I state above I can now show that d can be replaced by l(l+1). As a result, to verify the analogues of (3) and (4) for any l that is 15 mod 16, it's enough to use 2 terms in the power series expansion of each [j]. And so Ira's calculations which led to h/4 identities for each l<1500 and congruent to 7 mod 16, extended a minute bit further, will do the same for l that are 15 mod 16. Here's a proof of my assertion. If x^(d+1) divides R+C, it divides (R+C)(R+C+1)=R^2+R+G+H. Now my (self-accepted) answer to "Existence of certain identities.." shows that (to be continued) – paul Monsky Sep 28 2011 at 14:58
R^2+R+G+H lies in S', is stabilized by the automorphisms [j]-->[rj] of S', and has poles of order no more than 12 at each of the l(l-1)(l+1)/24 valuation rings in L/K that do not contain S'. If it's divisible by x^(d+1) it has at least (d+1)(l-1)/2 zeros counted with multiplicity. So it has more zeros than poles, and must vanish. It follows that R=C. – paul Monsky Sep 28 2011 at 15:06
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http://www.ats.ucla.edu/stat/sas/dae/ztnb.htm
|
# Welcome to the Institute for Digital Reseach and Education
Institute for Digital Reseach and Education Home
Help the Stat Consulting Group by giving a gift
### SAS Data Analysis Examples
Zero-Truncated Negative Binomial
Zero-truncated negative binomial regression is used to model count data for which the value zero cannot occur and when there is evidence of over dispersion .
Please Note: The purpose of this page is to show how to use various data analysis commands. It does not cover all aspects of the research process which researchers are expected to do. In particular, it does not cover data cleaning and verification, verification of assumptions, model diagnostics and potential follow-up analyses.
#### Examples of zero-truncated negative binomial
Example 1. A study of the length of hospital stay, in days, as a function of age, kind of health insurance and whether or not the patient died while in the hospital. Length of hospital stay is recorded as a minimum of at least one day.
Example 2. A study of the number of journal articles published by tenured faculty as a function of discipline (fine arts, science, social science, humanities, medical, etc). To get tenure faculty must publish, i.e., there are no tenured faculty with zero publications.
Example 3. A study by the county traffic court on the number of tickets received by teenagers as predicted by school performance, amount of driver training and gender. Only individuals who have received at least one citation are in the traffic court files.
#### Description of the data
Let's pursue Example 1 from above.
We have a hypothetical data file, ztp.sas7bdat with 1,493 observations available here . The variable describing length of hospital visit is stay. The variable age gives the age group from 1 to 9 which will be treated as interval in this example. The variables hmo and died are binary indicator variables for HMO insured patients and patients who died while in hospital, respectively. These are the same data as were used in the ztp example.
Let's look at the data.
```proc means data=mylib.ztp;
var stay;
run;
The MEANS Procedure
Analysis Variable : stay Length of Stay
N Mean Std Dev Minimum Maximum
--------------------------------------------------------------------
1493 9.7287341 8.1329081 1.0000000 74.0000000
--------------------------------------------------------------------
proc univariate data=mylib.ztp noprint;
histogram stay / midpoints = 0 to 80 by 2 vscale = count;
run;
proc freq data=mylib.ztp;
tables age hmo died;
run;
The FREQ Procedure
Age Group
Cumulative Cumulative
age Frequency Percent Frequency Percent
--------------------------------------------------------
1 6 0.40 6 0.40
2 60 4.02 66 4.42
3 163 10.92 229 15.34
4 291 19.49 520 34.83
5 317 21.23 837 56.06
6 327 21.90 1164 77.96
7 190 12.73 1354 90.69
8 93 6.23 1447 96.92
9 46 3.08 1493 100.00
hmo
Cumulative Cumulative
hmo Frequency Percent Frequency Percent
--------------------------------------------------------
0 1254 83.99 1254 83.99
1 239 16.01 1493 100.00
died
Cumulative Cumulative
died Frequency Percent Frequency Percent
---------------------------------------------------------
0 981 65.71 981 65.71
1 512 34.29 1493 100.00
```
#### Analysis methods you might consider
Before we show how you can analyze these data with a zero-truncated negative binomial analysis, let's consider some other methods that you might use.
• Zero-truncated Negative Binomial Regression - The focus of this web page.
• Zero-truncated Poisson Regression - Useful if there is no overdispersion in the zero truncated variable. See the Data Analysis Example for ztp.
• Negative Binomial Regression - Ordinary negative binomial regression will have difficulty with zero-truncated data. It will try to predict zero counts even though there are no zero values.
• Poisson Regression - The same concerns as for negative binomial regression, namely, ordinary poisson regression will have difficulty with zero-truncated data. It will try to predict zero counts even though there are no zero values.
• OLS Regression - You could try to analyze these data using OLS regression. However, count data are highly non-normal and are not well estimated by OLS regression.
#### Zero-truncated negative binomial regression using proc nlmixed
In order to use proc nlmixed to perform truncated negative binomial regression, we must supply it with a likelihood function. The probability that an observation has count $$y$$ under the negative binomial distribution (without zero truncation) is given by the equation: $P(Y=y) = {y+\frac{1}{\alpha}-1 \choose \frac{1}{\alpha}-1}\left(\frac{1}{1+\alpha\mu}\right)^{\frac{1}{\alpha}}\left(\frac{\alpha\mu}{1+\alpha\mu}\right)^y,$ where $$\alpha$$ is the overdispersion parameter and $$\mu$$ is the mean of the negative binomial distribution. With zero truncation, we calculate the probability that $$Y=y$$ conditional on $$Y>0$$, that is, that $$Y$$ is observed as 0 values are not observed. The probability of a zero count under the negative binomial distribution is $$P(Y=0) = \left(\frac{1}{1+\alpha\mu}\right)^{\frac{1}{\alpha}}$$. The conditional probability is then: $P(Y=y|Y>0) = \frac{P(Y=y)}{P(Y>0)} = \frac{P(Y=y)}{1-P(Y=0)} = {y+\frac{1}{\alpha}-1 \choose \frac{1}{\alpha}-1}\left(\frac{1}{1+\alpha\mu}\right)^{\frac{1}{\alpha}}\left(\frac{\alpha\mu}{1+\alpha\mu}\right)^y\frac{1}{1- \left(\frac{1}{1+\alpha\mu}\right)^{\frac{1}{\alpha}}}.$ The log-likelihood function for the zero-truncated negative binomial distribution is thus: $\mathcal{L}=\sum\limits_{i=1}^nlog\Gamma\left(y + \frac{1}{\alpha}\right) - log\Gamma\left(y+1\right) - log\Gamma\left(\frac{1}{\alpha}\right) - \frac{1}{\alpha}log(1+\alpha\mu) + ylog(\alpha\mu) - ylog(1 + \alpha\mu) - log\left(1- \left(1+\alpha\mu\right)^{-\frac{1}{\alpha}}\right).$ In negative binomial regression, we model $$log(\mu)$$, the log of the mean (expected counts), as a linear combination of a set of predictors: $log(\mu) = \beta_0 + \beta_1x_1 + \beta_2x_2 + \beta_3x_3$ We supply the last two equations to proc nlmixed to model our data using a zero-truncated negative distribution. Additionally, proc nlmixed does not support a class statement, so categorical variables should be dummy-coded before running the analysis. Unlike other SAS procs, by default the first group is the reference group by default in proc nlmixed.
```proc nlmixed data = mylib.ztp;
log_mu = intercept + b_age*age + b_died*died + b_hmo*hmo;
mu = exp(log_mu);
het = 1/alpha;
ll = lgamma(stay+het) - lgamma(stay + 1) - lgamma(het) - het*log(1+alpha*mu)
+ stay*log(alpha*mu) - stay*log(1+alpha*mu) - log(1 - (1 + alpha * mu)**-het);
model stay ~ general(ll);
run;
The NLMIXED Procedure
Specifications
Data Set MYLIB.ZTP
Dependent Variable stay
Distribution for Dependent Variable General
Optimization Technique Dual Quasi-Newton
Integration Method None
Dimensions
Observations Used 1493
Observations Not Used 0
Total Observations 1493
Parameters 5
Parameters
intercept b_age b_died b_hmo alpha NegLogLike
1 1 1 1 1 10136.7274
Iteration History
Iter Calls NegLogLike Diff MaxGrad Slope
1 3 5203.75757 4932.97 1718.332 -825332
2 6 5130.65185 73.10572 212.6078 -12208.4
3 8 4922.88698 207.7649 1701.184 -735.733
4 9 4862.95248 59.9345 176.3689 -177.538
5 11 4851.81702 11.13546 393.0774 -13.9647
6 12 4838.27102 13.546 179.7832 -7.96192
7 16 4788.46175 49.80926 168.3697 -26.6674
8 17 4774.94754 13.51421 105.3687 -117.309
9 18 4759.72531 15.22222 77.4436 -25.9074
10 20 4755.95435 3.77096 85.88275 -22.2361
11 22 4755.3438 0.610557 39.18804 -2.65095
12 24 4755.29354 0.050252 30.83521 -0.14278
13 26 4755.28066 0.012889 3.944229 -0.03589
14 28 4755.28014 0.000512 0.44716 -0.00416
15 29 4755.27964 0.0005 0.195745 -0.00109
16 31 4755.27962 0.000028 0.007496 -0.00006
17 33 4755.27962 1.109E-7 0.030916 -4.12E-7
NOTE: GCONV convergence criterion satisfied.
The SAS System 09:40 Monday, June 4, 2012 5
The NLMIXED Procedure
Fit Statistics
-2 Log Likelihood 9510.6
AIC (smaller is better) 9520.6
AICC (smaller is better) 9520.6
BIC (smaller is better) 9547.1
Parameter Estimates
Standard
Parameter Estimate Error DF t Value Pr > |t| Alpha Lower Upper Gradient
intercept 2.4083 0.07198 1493 33.46 <.0001 0.05 2.2671 2.5495 -0.00749
b_age -0.01569 0.01311 1493 -1.20 0.2314 0.05 -0.04140 0.01002 -0.03092
b_died -0.2178 0.04616 1493 -4.72 <.0001 0.05 -0.3083 -0.1272 0.000972
b_hmo -0.1471 0.05922 1493 -2.48 0.0131 0.05 -0.2632 -0.03090 -0.00018
alpha 0.5663 0.03123 1493 18.13 <.0001 0.05 0.5050 0.6276 -0.00461```
The output looks very much like the output from an OLS regression:
• Close to the top is the iteration log giving the values of the log likelihoods starting with a model that has no predictors. The last value in the log (4755.27962) is the final value of the negative log likelihood for the full model and is repeated below.
• Next comes a number of fit statistics, which can be used to compare the fit of nested models.
• Below the header you will find the zero-truncated negative binomial coefficients for each of the variables along with standard errors, t-values, p-values and 95% confidence intervals for each coefficient.
• Below that, is the the overdispersion parameter alpha along with its standard error, t-value, p-value, and 95% confidence interval.
Looking through the results we see the following:
• The value of the coefficient for age, -.01569, suggests that the log count of stay decreases by .01569 for each unit increase in age group. This coefficient is not statistically significant.
• The coefficient for hmo, -.1471, is significant and indicates that the log count of stay for HMO patient is .1471 less than for non-HMO patients.
• The log count of stay for patients that died while in the hospital was .2178 less than those patients that did not die.
• The value of the constant (intercept), 2.4083 is log count of the stay when all of the predictors equal zero.
• The estimate for alpha is .5663. For comparison, a model with an alpha of zero is equivalent to a zero-truncated poisson model. In this model, alpha is statistically different from zero, suggesting that the negative binomial model is a better choice than a poisson model.
We can also use estimate statments to help understand our model. For example we can predict the expected number of days spent at the hospital across age groups for the two hmo statuses for patients who died. The estimate statement for proc nlmixed works slightly differently from how it works within other procs. Here, each parameter must be explicitly multiplied by the value at which is to be held for that estimate statment. Additionally, because we would like to predict actual number of days rather than log number of days, we need to exponentiate the estimate.
```proc nlmixed data = mylib.ztp;
log_mu = intercept + b_age*age + b_died*died + b_hmo*hmo;
mu = exp(log_mu);
het = 1/alpha;
ll = lgamma(stay+het) - lgamma(stay + 1) - lgamma(het) - het*log(1+alpha*mu)
+ stay*log(alpha*mu) - stay*log(1+alpha*mu) - log(1 - (1 + alpha * mu)**-het);
model stay ~ general(ll);
estimate 'age 1 died 1 hmo 0' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 0);
estimate 'age 1 died 1 hmo 1' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 1);
estimate 'age 3 died 1 hmo 0' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 0);
estimate 'age 3 died 1 hmo 1' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 1);
estimate 'age 5 died 1 hmo 0' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 0);
estimate 'age 5 died 1 hmo 1' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 1);
estimate 'age 7 died 1 hmo 0' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 0);
estimate 'age 7 died 1 hmo 1' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 1);
estimate 'age 9 died 1 hmo 0' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 0);
estimate 'age 9 died 1 hmo 1' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 1);
run;
<**SOME OUTPUT OMITTED**>
Additional Estimates
Standard
Label Estimate Error DF t Value Pr > |t| Alpha Lower Upper
age 1 died 1 hmo 0 8.8010 0.6291 1493 13.99 <.0001 0.05 7.5668 10.0351
age 1 died 1 hmo 1 7.5974 0.6545 1493 11.61 <.0001 0.05 6.3135 8.8813
age 3 died 1 hmo 0 8.5290 0.4385 1493 19.45 <.0001 0.05 7.6688 9.3893
age 3 died 1 hmo 1 7.3627 0.5212 1493 14.13 <.0001 0.05 6.3404 8.3850
age 5 died 1 hmo 0 8.2655 0.3256 1493 25.39 <.0001 0.05 7.6268 8.9042
age 5 died 1 hmo 1 7.1352 0.4498 1493 15.86 <.0001 0.05 6.2530 8.0174
age 7 died 1 hmo 0 8.0101 0.3430 1493 23.35 <.0001 0.05 7.3373 8.6830
age 7 died 1 hmo 1 6.9147 0.4540 1493 15.23 <.0001 0.05 6.0242 7.8052
age 9 died 1 hmo 0 7.7627 0.4586 1493 16.93 <.0001 0.05 6.8630 8.6623
age 9 died 1 hmo 1 6.7011 0.5200 1493 12.89 <.0001 0.05 5.6811 7.7211```
The expected stay for non-HMO patients in age group 1 who died was 8.8010 days, while it was 7.5974 days for HMO patients in age group 1 who died.
We can also test whether the effect of HMO is significant at each level of age for patients who died. We can simply subtract the two estimates that vary by hmo at each level of age.
```proc nlmixed data = mylib.ztp;
log_mu = intercept + b_age*age + b_died*died + b_hmo*hmo;
mu = exp(log_mu);
het = 1/alpha;
ll = lgamma(stay+het) - lgamma(stay + 1) - lgamma(het) - het*log(1+alpha*mu)
+ stay*log(alpha*mu) - stay*log(1+alpha*mu) - log(1 - (1 + alpha * mu)**-het);
model stay ~ general(ll);
estimate 'age 1 died 1 hmo 0 vs 1' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 0) -
exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 1);
estimate 'age 3 died 1 hmo 0 vs 1' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 0) -
exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 1);
estimate 'age 5 died 1 hmo 0 vs 1' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 0) -
exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 1);
estimate 'age 7 died 1 hmo 0 vs 1' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 0) -
exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 1);
estimate 'age 9 died 1 hmo 0 vs 1' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 0) -
exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 1);
run;
<**SOME OUTPUT OMITTED**>
Additional Estimates
Standard
Label Estimate Error DF t Value Pr > |t| Alpha Lower Upper
age 1 died 1 hmo 0 vs 1 1.2036 0.4698 1493 2.56 0.0105 0.05 0.2820 2.1251
age 3 died 1 hmo 0 vs 1 1.1664 0.4511 1493 2.59 0.0098 0.05 0.2816 2.0512
age 5 died 1 hmo 0 vs 1 1.1303 0.4350 1493 2.60 0.0095 0.05 0.2770 1.9837
age 7 died 1 hmo 0 vs 1 1.0954 0.4215 1493 2.60 0.0094 0.05 0.2686 1.9222
age 9 died 1 hmo 0 vs 1 1.0616 0.4103 1493 2.59 0.0098 0.05 0.2568 1.8664
```
The effect of hmo is significant for each age group tested.
It may be illustrative for us to plot the predicted number of days stayed as a function of age and hmo status. To do this, we must tell SAS to save this table of predicted values as a dataset. Tables and graphics produced by procedures are given names upon creation. We will need the name of this prediction table to tell SAS to save it. Place ods trace on and ods trace off statements around the procedure which produced this table to obtain its name. Output from the ods trace statements is located in the log, not the output.
```ods trace on;
proc nlmixed data = mylib.ztp;
log_mu = intercept + b_age*age + b_died*died + b_hmo*hmo;
mu = exp(log_mu);
het = 1/alpha;
ll = lgamma(stay+het) - lgamma(stay + 1) - lgamma(het) - het*log(1+alpha*mu)
+ stay*log(alpha*mu) - stay*log(1+alpha*mu) - log(1 - (1 + alpha * mu)**-het);
model stay ~ general(ll);
estimate 'age 1 died 1 hmo 0' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 0);
estimate 'age 1 died 1 hmo 1' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 1);
estimate 'age 3 died 1 hmo 0' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 0);
estimate 'age 3 died 1 hmo 1' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 1);
estimate 'age 5 died 1 hmo 0' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 0);
estimate 'age 5 died 1 hmo 1' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 1);
estimate 'age 7 died 1 hmo 0' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 0);
estimate 'age 7 died 1 hmo 1' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 1);
estimate 'age 9 died 1 hmo 0' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 0);
estimate 'age 9 died 1 hmo 1' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 1);
run;
ods trace off;
<***SOME OF THE LOG OMITTED***>
Output Added:
-------------
Name: AdditionalEstimates
Label: Additional Estimates
Template: Stat.NLM.AdditionalEstimates
Path: Nlmixed.AdditionalEstimates
-------------
NOTE: PROCEDURE NLMIXED used (Total process time):
real time 0.23 seconds
cpu time 0.17 seconds
105 ods trace off;
```
Towards the end of the log we find the name of this table, which as expected by its heading in the output above, is "AdditionalEstimates". We can now tell SAS to save this output table as the dataset "mylib.addest" using an ods output statement.
```ods output AdditionalEstimates = mylib.addest;
proc nlmixed data = mylib.ztp;
log_mu = intercept + b_age*age + b_died*died + b_hmo*hmo;
mu = exp(log_mu);
het = 1/alpha;
ll = lgamma(stay+het) - lgamma(stay + 1) - lgamma(het) - het*log(1+alpha*mu)
+ stay*log(alpha*mu) - stay*log(1+alpha*mu) - log(1 - (1 + alpha * mu)**-het);
model stay ~ general(ll);
estimate 'age 1 died 1 hmo 0' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 0);
estimate 'age 1 died 1 hmo 1' exp(intercept * 1 + b_age * 1 + b_died * 1 + b_hmo * 1);
estimate 'age 3 died 1 hmo 0' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 0);
estimate 'age 3 died 1 hmo 1' exp(intercept * 1 + b_age * 3 + b_died * 1 + b_hmo * 1);
estimate 'age 5 died 1 hmo 0' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 0);
estimate 'age 5 died 1 hmo 1' exp(intercept * 1 + b_age * 5 + b_died * 1 + b_hmo * 1);
estimate 'age 7 died 1 hmo 0' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 0);
estimate 'age 7 died 1 hmo 1' exp(intercept * 1 + b_age * 7 + b_died * 1 + b_hmo * 1);
estimate 'age 9 died 1 hmo 0' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 0);
estimate 'age 9 died 1 hmo 1' exp(intercept * 1 + b_age * 9 + b_died * 1 + b_hmo * 1);
run;```
Now we can use this predicted values for plotting. We need to add actual values of age and hmo to the dataset for plotting as well.
```data mylib.addest;
set mylib.addest;
input age hmo;
datalines;
1 0
1 1
3 0
3 1
5 0
5 1
7 0
7 1
9 0
9 1
;
run;```
Finally, we use proc sgplot to plot our predicted number of days stayed as well as 95% confidence interval bands. The predicted values, lines connecting them, and confidence interval bands are all specified separately within the same proc sgplot. The group option will produce separate points, lines, and bands by the grouping variable.
```proc sgplot data = mylib.addest;
title 'Predicted number of days stayed (with 95% CL) by age and hmo status for patients who died';
band x = age lower = lower upper = upper / group=hmo;
scatter x= age y = estimate / group = hmo;
series x = age y = estimate / group = hmo;
run; ```
#### Things to consider
• Count data often use exposure variable to indicate the number of times the event could have happened. You can incorporate exposure into your model by including a log-linear term for exposure in the log-likehood function specification.
• It is not recommended that zero-truncated negative binomial models be applied to small samples. What constitutes a small sample does not seem to be clearly defined in the literature.
#### References
• Cameron, A. Colin and Trivedi, P.K. (2009). Microeconometrics using stata. College Station, TX: Stata Press.
• Cameron, A. Colin and Trivedi, P.K. (1998). Regression analysis of count data. Cambridge, UK: Cambridge University Press.
• Hilbe, J. M. (2007). Negative binomial regression. Cambridge, UK: Cambridge University Press.
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http://mathoverflow.net/questions/16672?sort=newest
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## Sheaves as full reflective subcategories
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello everyone.
My question is concerned with the following statement.
"Having a grothendieck topology on a category C is equivalent to having a full reflective subcategory Sh(C) in the category PSh(C) of presheaves, whose reflection is left exact."
What i need is a reference for this containing a proof. I tried google but could not find anything besides citations of this result.
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## 3 Answers
I have seen a reference for this fact, and I think it was in Artin's book on Grothendieck Topologies. I have no copy available to check this right now.
Before I found that reference, I wrote up a little treatment for my own benefit; I took the "full reflective subcategory" idea as the definition of a Grothendieck topos, then proved that all such come from Grothendieck topologies. It's in section 3.7 of http://www.math.uiuc.edu/~rezk/homotopy-topos-sketch.pdf
The proof goes like this. That a Grothendieck topology gives rise to a full reflective subcategory with left-exact reflection is standard. If you're given such a reflective subcategory $D\subseteq Psh(C)$, consider all the sieves, i.e., monomorphisms $f:S\to h_X$ where $h_X$ is the representable functor determined by $X\in C$. Call $f$ a covering sieve if $Lf$ is an isomorphism, where $L: Psh(C)\to D$ is the left adjoint. You then show (i) the collection of covering sieves is a Grothendieck topology $\tau$, and (ii) sheaves for $\tau$ are exactly those presheaves isomorphic to objects of $D$. Both (i) and (ii) require using the fact that $L$ is left exact. (ii) is equivalent to the statement: (ii') for all $f:X\to Y$ in $Psh(C)$, $Lf$ is iso if and only if $L_\tau f$ is iso (where "$L_\tau$" is sheafification with respect to $\tau$.) It's convenient to prove (ii') first for monomorphisms $f$, and then for epimorphisms $f$.
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thank you very much. this is exactly what i needed. plus the rest of your notes seem interesting as well. – Garlef Wegart Feb 28 2010 at 16:20
just wanted to say, nice notes! – B. Bischof Mar 1 2010 at 2:33
### 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 mentioned references and some more are at nLab: category of sheaves.
For instance the book by Kashiwara-Shapira has a useful account. When I myself learned this stuff I found it useful to read Mac-Lane/Moerdijk in parallel to Kashiwara/Shapira. The former has more of the topos-theoretic picture, the latter more of the homotopy-theoretic picture.
As we know from Rezk and Lurie, it*s both these aspects taken together that give the full picture.
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To add to what Charles wrote, another reference is Mac Lane and Moerdijk's Sheaves in Geometry and Logic. They prove something a bit more general, involving Lawvere-Tierney topologies on a topos. For the purposes of understanding what I'm about to write, it's not necessary to know what a Lawvere-Tierney topology is.
Mac Lane and Moerdijk's book contains the following two results:
1. Let $\mathcal{E}$ be a topos. Then the subtoposes of $\mathcal{E}$ (i.e. the reflective full subcategories with left exact reflectors) correspond canonically to the Lawvere-Tierney topologies on $\mathcal{E}$.
2. Let $\mathbf{C}$ be a small category. Then the Lawvere-Tierney topologies on $\mathbf{Set}^{\mathbf{C}^{\mathrm{op}}}$ correspond canonically to the Grothendieck topologies on $\mathbf{C}$.
Result 1 is almost part of Corollary VII.4.7. The "almost" is because they don't go the whole way in proving the one-to-one correspondence, but I guess it's not too hard to finish it off. (Edit: it also appears as Theorem A.4.4.8 of Johnstone's Sketches of an Elephant, where Lawvere-Tierney topologies are called local operators.) Result 2 is Theorem V.4.1.
I agree with the point of view that Charles advocates. When I started learning topos theory I got bogged down in detailed stuff about Grothendieck topologies, and it all seemed pretty technical and unappealing. It wasn't until years later that I learned the wonderful fact that Charles mentions: an elementary topos is Grothendieck iff it's a subtopos of some presheaf topos. I wish someone had told me that in the first place!
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http://physics.stackexchange.com/questions/30288/waves-on-water-generated-by-a-falling-object/30301
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waves on water generated by a falling object
Let an object of mass $m$ and volume $v$ be dropped in water from height $h$, and $a$ be the amplitude of the wave generated. What is the relation between $a$ and $h$. How many waves are generated? What is the relation between the amplitudes of successive waves? Does it depend on the shape of the particle?
Assume the particle is spherical. What would be the shape of the water that rises creating first wave?
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– David Zaslavsky♦ Jun 18 '12 at 21:53
I'm planning a simulation project .. I'm not in hurry!! We will discuss this later after my exam. – experimentX Jun 18 '12 at 22:08
3 Answers
This is crude.
Maybe there can be an energy approach. Initially the mass has potential energy $T=m g h$. At the point of peak splash-back lets assume all the energy has been transferred to the water with peak potential energy related to the radial wave height function $y(r)=?$. A small volume of water a distance $r$ from impact has differential volume ${\rm d}V = y(r) 2\pi r {\rm d}r$ .
The potential energy of the small volume of water is ${\rm d}T = \rho g \frac{y}{2} {\rm d}V$ where $\rho$ is density of water. The total energy is thus:
$$T = \int_0^\infty \rho g \frac{y(r)^2}{2} 2\pi r {\rm d} r$$
Putting a nice smooth wave height function of $$y(r) = Y \exp(-\beta\, r) \left(\cos(\kappa\, r) +\frac{\beta}{\kappa} \sin(\kappa\, r)\right)$$ with $Y$ a height coefficient. This has the properties of ${\rm d}y/{\rm d}r=0$ at $r=0$ with $y(0)=Y$.
$$T = \frac{\pi Y^2 g \rho \left(9 \beta^4+2 \beta^2 \kappa^2+\kappa^4\right)}{8 \beta^2 \left( \beta^2+\kappa^2 \right)^2 } = m g h$$
So wave height should be $$Y = \propto \sqrt{ \frac{h m}{\rho g \pi }}$$
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– AlanSE Jun 18 '12 at 20:40
Interesting, although this equation looks empirical. – ja72 Jun 18 '12 at 20:41
The question is empirical only due to the fact that we can't answer it exactly. The link I gave is for an empirical equation, but the point I was trying to establish is that it has basically h, m, rho, and g equivalents present and on the same side of the fraction. – AlanSE Jun 18 '12 at 20:52
It looks empirical to me because of the fractional powers. – ja72 Jun 19 '12 at 0:18
There isn't any simple relationship between the size of the splash and the impact velocity. Modelling splashes turns out to be a surprisingly hard thing to do. The problem is that the response of the water to the falling object is described by the Navier-Stokes equations, and apart from a few simple cases these are fiendishly difficult to solve.
I had a quick Google and found various school experiments where pupils measured splash height as a function of impact velocity. The nearest I found to a comprehensive description is this PhD thesis.
If you have a look on Youtube there are loads of slow motion videos of splashes.
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thank you ... :) – experimentX Jun 18 '12 at 10:06
I want just to add a thing. I think that the only reasoning we can do without solving the Navier-Stokes equations, nor doing experiments, is a dimensional analysis calculation.
(I suggest these notes sec 3.6 pag.83 for a complete treatment of this approach.)
The physical parameters we have are the density of water, the density of the ball (we have its mass and volume), the speed of the ball hitting the surface of the flow (via the conservation of kinetic energy, assuming no air resistance), and the viscosity of water. So:
$$\rho_b=m/V,\ \ mgh=\frac{1}{2}mv_b^2 \Longrightarrow v_b=\sqrt{2gh}$$
The only group of these parameters having the dimensions of a lenght is:
$$\left(\frac{\rho_b}{\rho_w}\right)^\alpha\frac{\mu}{v_b}=\left( \frac{m}{\rho_wV}\right)^\alpha \frac{\mu}{\sqrt{2gh}}$$
with arbitrary $\alpha$. So we can't estimate neither the value of the amplitude, nor the frequency of the wave with only this argument, and we should solve the Navier-Stokes equations in this particular case (as said in the previous answer, this is possible only in a few special cases). Therefore, we can just state the dependence on some physical parameters, saving the arbitrariness about the power of the non-dimensional number $\rho_b/\rho_w$.
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@ja72 so $Y$ doesn't depend on the viscosity of the fluid? Is this reasonable? – usumdelphini Jun 18 '12 at 20:07
I think the last comment was directed to the other answer, but I think the need or the lack of need for viscosity is an interesting question. I would suggest that surface tension might actually matter more than viscosity. The mental image the OP has in mind is probably in a regime where surface tension matters a great deal. Plus, when we are dealing with large scale drops into water where surface tension doesn't matter much, does it make the kind of waves we have in mind in the first place? – AlanSE Jun 18 '12 at 20:25
Actually it was, but I can't comment the other answer, I don't know why :( – usumdelphini Jun 18 '12 at 20:32
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http://math.stackexchange.com/questions/293356/difference-between-inclusion-exclusion-problems
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# Difference between inclusion-exclusion problems
There are two situations.The first is a bakery which has three type of doughnuts, {6*chocolate , 6*cinnamon, 3*plain}. How many options do they have for a box of 12 doughnuts?
The second question is what is the number of ways 10 people can be placed in 6 rooms? My actual question is not the answer for those questions but rather the first step. They are both problems in the chapter of Inclusion-exclusion. The first step is to count the total possible ways that each can occur without restrictions.
In my mind in both situations I can think of having a multiset with 3 or 6 elements respectively with infinite amounts of each. The second problem was worked out in class and the total without restrictions was 6^10. Clearly, this makes sense because each person could choose any room.
The professor made a point that $$\binom{10+(6-1)}{10}$$ was not the answer. Even though it seems to me that this would be valid. Couldn't one think of the rooms as a set of "infinite" rooms which you are making 10 choices from, one for each person? I think I'm having a hard time understanding where the order comes from that turns the total into a permutation rather than a combination.
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## 1 Answer
You can think of the first problem as distributing $12$ choices amongst $3$ kinds of doughnuts. This is like distributing $12$ identical marbles amongst $3$ labelled boxes: it doesn’t matter which two choices go in the plain doughnut box, so to speak.
You cannot think of the second problem as distributing $10$ identical marbles amongst $6$ boxes, however, because the ‘marbles’ (i.e., the people) are not identical: it matters which two people you put in Room $3$, for instance.
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http://stats.stackexchange.com/questions/45982/identifying-the-communicating-classes-and-stating-whether-they-are-period-aperi
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# Identifying the communicating classes and stating whether they are period, aperiodic, recurrent or transient
I don't understand what the answers say
The transition matrix is
$$\begin{pmatrix} 0.5 & 0.5 & 0 \\ 0 & 0.5 & 0.5\\ 0 & 0 & 1 \end{pmatrix}$$
In the answers it says:
$C_1 = \{3\}, T = \{1,2\}$, state 3 is aperiodic because $p_{33} > 0$ and recurrent. States 1 and 2 are aperiodic and transient.
I understand the bits about the individual states, but what does $C_1 = \{3\}, T = \{1,2\}$ mean? The lecturer hasn't explained this or written this in the notes. Is this saying the closed state is 3 and the other communicating classes are 1 and 2?
EDIT: C is the set of irreducible closed classes and T is the set of transient classes
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– whuber♦ Dec 15 '12 at 21:32
## 2 Answers
$T = \{1,2\}$ almost certainly means that the set of transient states has as elements the states $1$ and $2$.
$C_1 = \{3\}$ probably means that one of irreducible closed sets has as its only element elements the state $3$. If there were more than one irreducible closed set (not this example) then you might see $C_2$ or others.
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The states that communicate are said to be in the same class. Periodicity is a class property. So the states in the same class will share this property, or the states will have the same period. Here neither of the states is communicating. So, you will need to check all the 3 states for their periodicity separately. That is also another explanation why the Markov Chain is not reducible.
As Henry pointed out, $T$ is the set of transient states only. $C$ contains only one state and it is a closed set. The subscript 1 comes here, because there is only one such closed set.
$p_{33}>0$ means $p_{33}^{1}>0$ (which is the 1st step transition probability). In fact as $p_{33}=1$, so the last row always remains as it is. Thus, $p_{33}^{n}>0$ always for all $n$. So the period, $d(3)=gcd\{n:p_{33}^{n}>0\}=1$
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http://en.wikibooks.org/wiki/Control_Systems/Stability
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Control Systems/Stability
| | | | |
|----------------------------------------------------------|-------------------------|--------------|-------------|
| ← Nichols Charts | State-Space Stability → | Glossary | |
Stability
When a system becomes unstable, the output of the system approaches infinity (or negative infinity), which often poses a security problem for people in the immediate vicinity. Also, systems which become unstable often incur a certain amount of physical damage, which can become costly. This chapter will talk about system stability, what it is, and why it matters.
The chapters in this section are heavily mathematical, and many require a background in linear differential equations. Readers without a strong mathematical background might want to review the necessary chapters in the Calculus and Ordinary Differential Equations books (or equivalent) before reading this material. Negativeness of any coefficient of a characteristic polynomial indicates that the system is either unstable or at most marginally stable. If any coefficient is zero/negative then we can say that the system is unstable.
BIBO Stability
A system is defined to be BIBO Stable if every bounded input to the system results in a bounded output over the time interval $[t_0, \infty)$. This must hold for all initial times to. So long as we don't input infinity to our system, we won't get infinity output.
A system is defined to be uniformly BIBO Stable if there exists a positive constant k that is independent of t0 such that for all t0 the following conditions:
$\|u(t)\| \le 1$
$t \ge t_0$
implies that
$\|y(t)\| \le k$
There are a number of different types of stability, and keywords that are used with the topic of stability. Some of the important words that we are going to be discussing in this chapter, and the next few chapters are: BIBO Stable, Marginally Stable, Conditionally Stable, Uniformly Stable, Asymptoticly Stable, and Unstable. All of these words mean slightly different things.
Determining BIBO Stability
We can prove mathematically that a system f is BIBO stable if an arbitrary input x is bounded by two finite but large arbitrary constants M and -M:
$-M < x \le M$
We apply the input x, and the arbitrary boundaries M and -M to the system to produce three outputs:
$y_x = f(x)$
$y_M = f(M)$
$y_{-M} = f(-M)$
Now, all three outputs should be finite for all possible values of M and x, and they should satisfy the following relationship:
$y_{-M} \le y_x \le y_M$
If this condition is satisfied, then the system is BIBO stable.
For a SISO linear time-invariant (LTI) system is BIBO stable if and only if $g(t)$ is absolutely integrable from [0,∞] or from:
$\int_{0}^{\infty} |g(t)| \,dt \leq M < {\infty}$
Example
Consider the system:
$h(t) = \frac{2}{t}$
We can apply our test, selecting an arbitrarily large finite constant M, and an arbitrary input x such that -M < x < M.
As M approaches infinity (but does not reach infinity), we can show that:
$y_{-M} = \lim_{M \to \infty} \frac{2}{-M} = 0^-$
And:
$y_M = \lim_{M \to \infty} \frac{2}{M} = 0^+$
So now, we can write out our inequality:
$y_{-M} \le y_x \le y_M$
$0^- \le x < 0^+$
And this inequality should be satisfied for all possible values of x. However, we can see that when x is zero, we have the following:
$y_x = \lim_{x \to 0} \frac{2}{x} = \infty$
Which means that x is between -M and M, but the value yx is not between y-M and yM. Therefore, this system is not stable.
Poles and Stability
When the poles of the closed-loop transfer function of a given system are located in the right-half of the S-plane (RHP), the system becomes unstable. When the poles of the system are located in the left-half plane (LHP), the system is shown to be stable. A number of tests deal with this particular facet of stability: The Routh-Hurwitz Criteria, the Root-Locus, and the Nyquist Stability Criteria all test whether there are poles of the transfer function in the RHP. We will learn about all these tests in the upcoming chapters.
If the system is a multivariable, or a MIMO system, then the system is stable if and only if every pole of every transfer function in the transfer function matrix has a negative real part. For these systems, it is possible to use the Routh-Hurwitz, Root Locus, and Nyquist methods described later, but these methods must be performed once for each individual transfer function in the transfer function matrix.
Poles and Eigenvalues
Note:
Every pole of G(s) is an eigenvalue of the system matrix A. However, not every eigenvalue of A is a pole of G(s).
The poles of the transfer function, and the eigenvalues of the system matrix A are related. In fact, we can say that the eigenvalues of the system matrix A are the poles of the transfer function of the system. In this way, if we have the eigenvalues of a system in the state-space domain, we can use the Routh-Hurwitz, and Root Locus methods as if we had our system represented by a transfer function instead.
On a related note, eigenvalues and all methods and mathematical techniques that use eigenvalues to determine system stability only work with time-invariant systems. In systems which are time-variant, the methods using eigenvalues to determine system stability fail.
Transfer Functions Revisited
We are going to have a brief refesher here about transfer functions, because several of the later chapters will use transfer functions for analyzing system stability.
Let us remember our generalized feedback-loop transfer function, with a gain element of K, a forward path Gp(s), and a feedback of Gb(s). We write the transfer function for this system as:
$H_{cl}(s) = \frac{KGp(s)}{1 + H_{ol}(s)}$
Where $H_{cl}$ is the closed-loop transfer function, and $H_{ol}$ is the open-loop transfer function. Again, we define the open-loop transfer function as the product of the forward path and the feedback elements, as such:
$H_{ol}(s) = KGp(s)Gb(s)$
Now, we can define F(s) to be the characteristic equation. F(s) is simply the denominator of the closed-loop transfer function, and can be defined as such:
[Characteristic Equation]
$F(s) = 1 + H_{ol} = D(s)$
We can say conclusively that the roots of the characteristic equation are the poles of the transfer function. Now, we know a few simple facts:
1. The locations of the poles of the closed-loop transfer function determine if the system is stable or not
2. The zeros of the characteristic equation are the poles of the closed-loop transfer function.
3. The characteristic equation is always a simpler equation than the closed-loop transfer function.
These functions combined show us that we can focus our attention on the characteristic equation, and find the roots of that equation.
State-Space and Stability
As we have discussed earlier, the system is stable if the eigenvalues of the system matrix A have negative real parts. However, there are other stability issues that we can analyze, such as whether a system is uniformly stable, asymptotically stable, or otherwise. We will discuss all these topics in a later chapter.
Marginal Stablity
When the poles of the system in the complex S-Domain exist on the complex frequency axis (the vertical axis), or when the eigenvalues of the system matrix are imaginary (no real part), the system exhibits oscillatory characteristics, and is said to be marginally stable. A marginally stable system may become unstable under certain circumstances, and may be perfectly stable under other circumstances. It is impossible to tell by inspection whether a marginally stable system will become unstable or not.
We will discuss marginal stability more in the following chapters.
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http://math.stackexchange.com/questions/202339/eigenvector-with-integer-coordinates
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# Eigenvector with integer coordinates
Let be $A$ a matrix, $n\times n$ with integer coordinates, and such that $A$ have an eigenvector $\underline{n}$ with real coordenates and eigenvalue 1. My problem is: how to show that $A$ has an eigenvector with integer coordinates (with eigenvalue 1)
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## 1 Answer
If we bring $A-I$ to reduced row echelon form (RREF) then the entries will be rational. From the RREF we can read off an eigenvector with rational coordinates. Since any multiple of an eigenvector is still an eigenvector corresponding to the same eigenvalue, it follows that we can clear denominators to produce an eigenvector with integer coordinates.
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Forgive me if I'm ignorant, but I do not quite understand your answer, would you be more clear? – user27456 Sep 25 '12 at 22:40
Which part is confusing to you? – EuYu Sep 25 '12 at 22:43
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http://mathhelpforum.com/algebra/35968-another-variation.html
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# Thread:
1. ## another variation
find the required value by setting up the general equation and then evaluating.
find y when x=10 if y varies directly as x and y=20 when x=8
thanks!
2. Originally Posted by mamajen
find the required value by setting up the general equation and then evaluating.
find y when x=10 if y varies directly as x and y=20 when x=8
thanks!
When y varies directly as x, we have the equation $y = kx$. The first thing to do is solve for k. Substituting in y = 20 and x = 8 we have $20 = 8k$, or $k = \frac{20}{8} = \frac{5}{2}$.
Given that $y = \frac{5}{2}x$, can you determine y when x = 10?
3. yes i can get it from there. thanks. I was not thinking to find k first!
4. Icemanfan, hope you don't mind me just adding something.
mamajen, In the other question you asked, I said what "proportional" mean and told you about putting the k there and then working out what k was. Well, "varies directly as" just means "is proportional to", so the two questions are quite similar.
5. thanks iceman. i remembered that part but didn't know where to put all the equal to #'s. then when I saw he used the 2nd y= to find k I knew where I was going
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http://math.stackexchange.com/questions/271156/how-to-show-that-a-certain-function-is-in-lp
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# How to show that a certain function is in $L^p$?
How can I show that the function $F(x)= \dfrac{|x| ^{-n+1}} { \log \frac{1}{|x|} }$, for $0 < |x| \leqslant \large\frac{1}{2}$ and $F(x)=0$, if $|x|>\large\frac{1}{2}$, is in $L^p(\mathbb{R}^n)$ for $p \leqslant \large\frac{n}{n-1}$?
I managed to bound $(F(x))^p$ by $|x|^{-a}$ for some $a<n$ in order to use a corollary from G.Folland's book that states:
If $|f(x)| \leqslant C|x|^{-a}$ on B for some $a<n$, then $f \in L^1(B)$, where $B= \left\{{x \in \mathbb{R}^n: |x|<c }\right\}$ and $\,c,\, C\,$ are positive constants.
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## 1 Answer
For radial function on $\Bbb R^n$, that is, which can be written by $f(x)=g(|x|)$ where $|\cdot|$ is the euclidian norm, we have the following: for $0<a<b$,
$$\int_{\{x,a<|x|<b\}}f(x)dx=C\int_a^bg(r)r^{n-1}dr,$$ where $C$ is an universal constant (which doesn't depend on $f$). So we have to see whether $$\int_0^{1/2}\frac{(r^{1-n})^p}{(-\log r)^p}r^{n-1}dr$$ is convergent. Work with the integral $\int_\varepsilon^{1/2}\frac{1}{(-\log r)^p}r^{(n-1)(1-p)}dr$, and do the substitution $s=-\log r$ to get an easier integral.
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http://en.wikipedia.org/wiki/Out_of_phase
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# Phase (waves)
From Wikipedia, the free encyclopedia
(Redirected from Out of phase)
Jump to: navigation, search
Phase in sinusoidal functions or in waves has two different, but closely related, meanings. One is the initial angle of a sinusoidal function at its origin and is sometimes called phase offset or phase difference. Another usage is the fraction of the wave cycle which has elapsed relative to the origin.[1]
## Formula
The phase of an oscillation or wave refers to a sinusoidal function such as the following:
$\begin{align} x(t) &= A\cdot \cos( 2 \pi f t + \varphi ) \\ y(t) &= A\cdot \sin( 2 \pi f t + \varphi ) = A\cdot \cos\left( 2 \pi f t + \varphi - \frac{\pi}{2}\right) \end{align}$
where $\scriptstyle A\,$, $\scriptstyle f\,$, and $\scriptstyle \varphi\,$ are constant parameters called the amplitude, frequency, and phase of the sinusoid. These functions are periodic with period $\scriptstyle T = \frac{1}{f}\,$, and they are identical except for a displacement of $\scriptstyle \frac{T}{4}\,$ along the $\scriptstyle t\,$ axis. The term phase can refer to several different things:
• It can refer to a specified reference, such as $\scriptstyle \cos( 2 \pi f t)\,$, in which case we would say the phase of $\scriptstyle x(t)\,$ is $\scriptstyle \varphi\,$, and the phase of $\scriptstyle y(t)\,$ is $\scriptstyle \varphi\,-\, \frac{\pi}{2}\,$.
• It can refer to $\scriptstyle \varphi\,$, in which case we would say $\scriptstyle x(t)\,$ and $\scriptstyle y(t)\,$ have the same phase but are relative to their own specific references.
• In the context of communication waveforms, the time-variant angle $\scriptstyle 2 \pi f t \,+\, \varphi$, or its principal value, is referred to as instantaneous phase, often just phase.
## Phase shift
Illustration of phase shift. The horizontal axis represents an angle (phase) that is increasing with time.
Phase shift is any change that occurs in the phase of one quantity, or in the phase difference between two or more quantities.[1]
$\scriptstyle \varphi\,$ is sometimes referred to as a phase shift or phase offset, because it represents a "shift" from zero phase. But a change in $\scriptstyle \varphi\,$ is also referred to as a phase shift.
For infinitely long sinusoids, a change in $\scriptstyle \varphi\,$ is the same as a shift in time, such as a time delay. If $\scriptstyle x(t)\,$ is delayed (time-shifted) by $\scriptstyle \frac{1}{4}\,$ of its cycle, it becomes:
$\begin{align} x\left(t - \frac{1}{4} T\right) &= A\cdot \cos\left(2 \pi f \left(t - \frac{1}{4}T \right) + \varphi \right) \\ &= A\cdot \cos\left(2 \pi f t - \frac{\pi}{2} + \varphi \right) \end{align}$
whose "phase" is now $\scriptstyle \varphi \,-\, \frac{\pi}{2}$. It has been shifted by $\scriptstyle \frac{\pi}{2}$ radians.
## Phase difference
In-phase waves
Out-of-phase waves
Left: the real part of a plane wave moving from top to bottom. Right: the same wave after a central section underwent a phase shift, for example, by passing through a glass of different thickness than the other parts. (The illustration on the right ignores the effect of diffraction, which would make the waveform continuous away from material interfaces and would add increasing distortions with distance.).
Phase difference is the difference, expressed in electrical degrees or time, between two waves having the same frequency and referenced to the same point in time.[1] Two oscillators that have the same frequency and different phases have a phase difference, and the oscillators are said to be out of phase with each other. The amount by which such oscillators are out of step with each other can be expressed in degrees from 0° to 360°, or in radians from 0 to 2π. If the phase difference is 180 degrees (π radians), then the two oscillators are said to be in antiphase. If two interacting waves meet at a point where they are in antiphase, then destructive interference will occur. It is common for waves of electromagnetic (light, RF), acoustic (sound) or other energy to become superposed in their transmission medium. When that happens, the phase difference determines whether they reinforce or weaken each other. Complete cancellation is possible for waves with equal amplitudes.
Time is sometimes used (instead of angle) to express position within the cycle of an oscillation. A phase difference is analogous to two athletes running around a race track at the same speed and direction but starting at different positions on the track. They pass a point at different instants in time. But the time difference (phase difference) between them is a constant - same for every pass since they are at the same speed and in the same direction. If they were at different speeds (different frequencies), the phase difference is undefined and would only reflect different starting positions. Technically, phase difference between two entities at various frequencies is undefined and does not exist.
• Time zones are also analogous to phase differences.
A real-world example of a sonic phase difference occurs in the warble of a Native American flute. The amplitude of different harmonic components of same long-held note on the flute come into dominance at different points in the phase cycle. The phase difference between the different harmonics can be observed on a spectrograph of the sound of a warbling flute.[2]
## In-phase and quadrature (I&Q) components
See also: Quadrature phase and Constellation diagram
The term in-phase is also found in the context of communication signals:
$A(t)\cdot \sin[2\pi ft + \varphi(t)] = I(t)\cdot \sin(2\pi ft) + Q(t)\cdot \underbrace{\sin\left(2\pi ft + \begin{matrix} \frac{\pi}{2} \end{matrix} \right)}_{\cos(2\pi ft)}$
and:
$A(t)\cdot \cos[2\pi ft + \varphi(t)] = I(t)\cdot \cos(2\pi ft) + Q(t)\cdot \underbrace{\cos\left(2\pi ft + \begin{matrix} \frac{\pi}{2} \end{matrix}\right)}_{-\sin(2\pi ft)},$
where $\scriptstyle f\,$ represents a carrier frequency, and
$\begin{align} I(t) &\equiv A(t)\cdot \cos\left(\varphi(t)\right) \\ Q(t) &\equiv A(t)\cdot \sin\left(\varphi(t)\right) \end{align}$
$\scriptstyle A(t)\,$ and $\scriptstyle \varphi(t)\,$ represent possible modulation of a pure carrier wave; e.g., $\scriptstyle \sin(2\pi ft)\,$ (or $\scriptstyle \cos(2\pi ft)$). The modulation alters the original $\scriptstyle \sin\,$ (or $\scriptstyle \cos\,$) component of the carrier, and creates a (new) $\scriptstyle \cos\,$ (or $\scriptstyle \sin\,$) component, as shown above. The component that is in phase with the original carrier is referred to as the in-phase component. The other component, which is always 90° ($\scriptstyle \frac{\pi}{2}$ radians) "out of phase", is referred to as the quadrature component.
## Phase coherence
Coherence is the quality of a wave to display well defined phase relationship in different regions of its domain of definition.
In physics, quantum mechanics ascribes waves to physical objects. The wave function is complex and since its square modulus is associated with the probability of observing the object, the complex character of the wave function is associated to the phase. Since the complex algebra is responsible for the striking interference effect of quantum mechanics, phase of particles is therefore ultimately related to their quantum behavior.
## References
1. ^ a b c Ballou, Glen (2005). Handbook for sound engineers (3 ed.). Focal Press, Gulf Professional Publishing. p. 1499. ISBN 0-240-80758-8.
2. Clint Goss; Barry Higgins (2013). "The Warble". Flutopedia. Retrieved 2013-03-06.
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http://nrich.maths.org/260/solution?nomenu=1
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"Hello. I'm Kim Jinhyuna from Kingston-Grammar School. I would like to inform you that I have worked out the question 'Upsetting Pitagoras'. The mathematical problem has an infinite number of solutions. Let $x=1$ and $y=2$. Then $${1\over x^2}+{1\over y^2}= {1\over 1^2}+{1\over 2^2} = 1{\cdot}25 = {1\over 0{\cdot}8}.$$ So $z^2 = 0{\cdot}8$, and $z = \sqrt{0{\cdot}8}$ which gives $z=\pm 0{\cdot}894$ to 3 significant figures. For another solution, let Let $x=3$ and $y=4$. Then $${1\over x^2}+{1\over y^2}= {1\over 3^2}+{1\over 4^2} = 1{\cdot}736\ldots = {1\over 5{\cdot}76}.$$ So $z^2 = 5{\cdot}76$, and $z = \sqrt{5{\cdot}76}$ which gives $z=\pm 2{\cdot}4$ (exactly). However, I think that the problem comes with the assumption that $x$, $y$ and $z$ are all integers, in which case one answer is $x=30$, $y=40$ and $z=24$; that is $${1\over 30^2}+{1\over 40^2} = {1\over 24^2}.$$ If we multiply both sides by $4$ we get $${1\over 15^2}+{1\over 20^2} = {1\over 12^2}.$$ I'm looking forward to more tough and hard questions."
If $a^2 + b^2 = c^2$ then $${1\over b^2c^2} + {1\over a^2c^2} = {1\over a^2b^2}.$$ So every Pythgorean triple gives rise to a solution to our problem. The smallest solution of this type arises from $a=3, b=4, c=5$ and gives $${1\over 20^2} + {1\over 15^2} = {1\over 12^2}.$$ Notice we have not proved that there is no other way of producing solutions but a simple computer program to make an exhaustive check of values of $x, y$ and $z$ up to these values will prove that there are no smaller solutions.
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http://divisbyzero.com/2009/10/06/tennenbaums-proof-of-the-irrationality-of-the-square-root-of-2/
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# Division by Zero
A blog about math, puzzles, teaching, and academic technology
Posted by: Dave Richeson | October 6, 2009
## Tennenbaum’s proof of the irrationality of the square root of 2
Yesterday I came a across a new (new to me, that is) proof of the irrationality of $\sqrt{2}$. I found it in the paper “Irrationality From The Book,” by Steven J. Miller, David Montague, which was recently posted to arXiv.org.
Apparently the proof was discovered by Stanley Tennenbaum in the 1950′s but was made widely known by John Conway around 1990. The proof appeared in Conway’s chapter “The Power of Mathematics” of the book Power, which was edited by Alan F. Blackwell, David MacKay (2005).
It is a proof by contradiction. Suppose $\sqrt{2}=a/b$ for some positive integers $a$ and $b$. Then $a^2=2b^2=b^2+b^2$. Geometrically this means that there is an integer-by-integer square (the pink $a\times a$ square below) whose area is twice the area of another integer-by-integer square (the blue $b\times b$ squares).
Assume that our $a\times a$ square is the smallest such integer-by-integer square.
Now put the two blue squares inside the pink square as shown below. They overlap in a dark blue square.
By assumption, the sum of the areas of the two blue squares is the area of the large pink square. That means that in the picture above, the dark blue square in the center must have the same area as the two uncovered pink squares. But the dark blue square and the small pink squares have integer sides. This contradicts our assumption that our original pink square was the smallest such square. It must be the case that $\sqrt{2}$ is irrational.
[Note: the squares in the pictures almost work. They are $17\times 17$ and $12\times 12$. As Conway points out, $17^2=289\approx 288=2\cdot 12^2$. Indeed $\sqrt{2}=1.4142\ldots\approx 1.41666\ldots=17/12$.]
If you want to see more examples, look at Miller and Montague’s paper “Irrationality From The Book.” They extend this idea to give geometric proofs that $\sqrt{3}$, $\sqrt{5}$, $\sqrt{6}$, and $\sqrt{10}$ are irrational.
Also, Cut-the-Knot has 19 proofs of the irrationality of $\sqrt{2}$ (including this one).
### Like this:
Posted in Math, Teaching | Tags: David Montague, Euclid, irrational, John Conway, proof by contradiction, square root of 2, Steven J. Miller
## Responses
1. I just read (literally a few short hours ago) an almost identical proof (except done with a right triangle) in The Calculus of Friendship by Steven Strogatz. Thanks for posting this version!
By: Dan M on October 6, 2009
at 4:31 pm
2. I am Stanley Tennenbaum’s youngest son and partially responsible for getting the word out about this proof after he died in May 2006, in Princeton.
Sadly, John Conway got a number of facts wrong in the book version of this proof and, previously, in some of his online versions of it.
There is absolutely no question that my father discovered this in the early 50s. Note the website:
http://www.cancerdisaster.org
Soon, thousands of pages of his unpublished math and educational views will appear online, some polished, most of it notes, albeit a lot taken from diary entries and other handwritten sources.
There will also be audio and video.
Note the two quotes by Kurt Goedel that appear on the very bottom right hand side of the above site, taken from two letters of recommendation. Only one page of this site is up now, but that should change in a few weeks.
The true story of my father is wild beyond imagination. He was an extremely controversial figure, but probably the closest person to Goedel, both personally and professionally, certainly from the late 60s until Goedel died. This is well known among the top logicians and mathematicians many of whom met Goedel through my dad (indeed, Conway himself, by his own words, on videotape at a conference in my father’s honor/memory, met Goedel in a meeting arranged by my dad).
I am now 53 and witnessed him showing this beautiful proof to hundreds of people during my lifetime, including Hans Bethe, the famous physicist, when we met up with him in Cornell, in 1978.
Many people knew about it, and many did not. Again, sadly, it appears that Conway, since he had not heard about it, thought that my dad JUST found it, because my father told Conway the proof in the early 90s. But many others saw it many many decades earlier. My brother remembers it from the 50s and so does Prof. Nerode, from Cornell who recently told me in no uncertain terms that it came from the early 50s when my dad was at the University of Chicago.
More interesting, however, than any of the above talk about dates and famous people, are Goedel’s amazing conversation notes that he took during conversations with my dad. When they are finally fully translated they will be fascinating beyond measure because the range of topics discussed was extraordinarily broad.
Thanks for posting it — not simply for the attribution to my father (although he certainly deserves it) — but for getting the word out about a proof that is simply beautiful. It can inspire others, especially young people, to learn about mathematics.
My dad thought that MOST of math will be reduced, at some point, to proofs of this nature, where the essential ideas are crystal clear and the proofs are radically simplified.
I can still remember him repeating to me:
“Always a picture, always a picture…..”
Peter Tennenbaum October 26, 2009
Rochester, NY
By: Peter Tennenbaum on October 26, 2009
at 12:35 pm
• Peter,
Thank you for sharing all of this. Please let me know by email when your website is live. It sounds like it will be a wonderful resource.
Dave
By: Dave Richeson on October 26, 2009
at 1:11 pm
3. Prof. Richeson,
I would like to point out that the paper “Irrationality From The Book,” by Steven J. Miller, David Montague, which was recently posted to arXiv.org has just been updated (see version 2); it is now far more accurate, historically.
Conway did indeed publicize this proof, but he did this 40 years after it was discovered and AFTER my father had ALREADY publicized it himself, albeit orally, by telling hundreds of people, mostly students, but many, many mathematicians. It is a shame that he did not publish it himself before he died. But he died suddenly, without any warning and he had spoken often about publishing it, especially as he got older.
It is also a shame that more people did not tell others about it after hearing it from my dad. Forty years is a long time and, if they had, then the whole mathematical community would surely have seen this proof long ago and it probably would have already appeared in many books and many papers.
Another point:
Your reference to Conway, above, with respect to the size of the squares, in my father’s proof, makes little sense, at least to me. Moreover, from my point of view, Conway’s OWN use of these numbers “17 and 12″ makes little sense in his chapter titled “The Power of Mathematics” in the book “Power”.
Why?
Well, first of all, it is indeed true that 17/12 is a close approximation to the square root of two. But so what?
Since the rationals are everywhere dense on the real line, it is hardly surprising that some of them make good approximations to the square root of two or, for that matter, any real number whether irrational or transcendental.
What IS surprising is that fabulous approximations exist when the denominators are small compared to the degree of accuracy that is obtained.
Neither you nor Conway note the truly interesting part: these fabulous rational approximations are convergents to the continued fraction approximation to the square root of two.
If you could show some connection between the above proof and the well-known (periodic) continued fraction approximation to the square root of two, then it would make sense to include something about 17/12, above. Otherwise, I do not see the point — it has little to do with the proof, unless it had some historical interest.
17/12 is simply one of an infinite number of close (for the size of the denominator) rational approximations to the square root of two.
On the other hand, 22/7 is a famous historical example of a truly great rational approximation to pi. It’s value is:
3.1428…
while pi itself is (approximately):
3.14159265358…
Of course, 103993/33102 is far better than 22/7:
3.1415926530…
Interested readers might like to know more about continued fractions. The best intuitive, geometrical device that I know of for “seeing” them is the use of “Kline’s String”.
Look at the function y = x*SQRT(2) for positive x and y. This is simply a line in the plane. Note, however, that this line CANNOT intersect ANY integral lattice point, otherwise y/x would be a rational number equal to the square root of two and we know that such a number cannot exist.
But this line does come CLOSE to integral lattice points.
Imagine now, that the integral lattice points have tiny, tiny pegs in them and that the line, y = x*SQRT(2), is a thin, thin, string anchored at its end — infinitely far away. Now, holding it taut, move this string at the origin, upwards, along the y axis. The string will come to rest against those pegs — those integral lattice points — that are closet to it, and you will obtain a polygonal figure. Note that coordinates of these pegs — the lattice points that the string touches, will be excellent rational approximations to the square root of two, albeit their value will be higher than the actual square root of two. The farther up the polygonal figure you go the numbers get larger but the approximations to the square root of two become fantastically accurate.
You can move the string down the y axis and, similarly, the string will butt up against those lattice points that are excellent rational approximations to the square root of two, albeit they will be lower in value.
Indeed, these pegs — the integral lattice points that the string rests upon in both directions — will be EXACTLY the convergents computed from the continued fraction expansion to the square root of two.
Kline’s String is simply a beautiful GEOMETRIC way of seeing why the string will find these integral lattice points (the rational approximations).
It is rather mind boggling to think that the function y = x*SQRT(2) is an infinitely long line, yet it NEVER passes through an integral lattice point.
Prof. Richeson: Perhaps you will find the time to write a short intro to continued fractions and also include a nice graphical example of Kline’s string.
Meantime, interested readers can consult a truly wonderful book written decades ago by Davenport, titled “The Higher Arithmetic” (Number Theory):
When published, a review in the Bulletin of the American Mathematical Society stated that:
“Although this book is not written as a textbook but rather as a work for the general reader, it could certainly be used as a textbook for an undergraduate course in number theory and, in the reviewer’s opinion, is far superior for this purpose to any other book in English.”
The book explains continued fractions and Kline’s String far better than my awkward prose, above. In short, I hope I have done more good than harm by “publicizing” its existence.
Note, finally, that even though Kline’s string is in famous books, such as Davenport’s, many mathematicians — at least in my experience — are completely unaware of this truly wonderful geometric way of “seeing” the convergents to an irrational or transcendental number.
Peter Tennenbaum November 4, 2009
By: Peter Tennenbaum on November 4, 2009
at 4:24 am
4. ERRATUM to the above post of Peter Tennenbaum’s. To wit:
I believe that “one should practice what one preaches” and, in this spirit, I readily admit to a stupid mistake in my attribution, above, concerning “Klein’s String”. I repeatedly misspelled it as “Kline’s String”.
Reader’s should note that this is a discovery of Felix Klein, the renowned mathematician (see, for example: http://en.wikipedia.org/wiki/Felix_Klein).
My version of H. Davenport’s “The Higher Arithmetic” was published in 1960 by Harper and Row in the United States of America. There, it says that “This book was first published in 1952 in the Mathematical and Physical Sciences division, edited by Sir George P. Thomson, of the Hutchinson & Company Limited, London.”
On page 114, reference 12, at the end of Chapter IV on Continued Fractions, Davenport writes:
“See Klein’s “Ausgewählte Kapitel der Zahlentheorie” (Teubner, 1907), pp. 17-25.”
When I wrote my post I was exhausted and under tremendous stress — enough to crush a large diamond into dust. I have been replicating, under a life and death situation (my wife’s rare ovarian cancer), the work of Novocure, an Israeli company, that is conducting Phase III trials at major institutions around the world of a revolutionary new cancer therapy — therapy now being given to patients suffering from glioblastoma, an aggressive brain cancer with an extremely poor prognosis. Note that Phase I breast cancer trials are pending.
Novocure’s work is based on EXTERNALLY placed capacitive conducting electrodes that, under the right conditions (applied frequency, geometric orientation, field strength, etc.), can make cancer cells explode into benign particles that are harmlessly reabsorbed into the body. Very roughly speaking, the technique works via a type of resonance effect although the actual bio-electrical-physics are far more complicated.
That said, I regard the above excuse as A POOR EXCUSE FOR MY EXTREMELY SHODDY SCHOLARSHIP. It is, however, an opportunity to show that I too make mistakes in attribution and, more importantly, that there is truly revolutionary work being done in cancer that’s been effective on EVERY cell type yet tested, in Petri dishes, rodents, and now human beings. Further, the therapy appears to be non-toxic, although it can be combined with various semi-toxic chemotherapeutic regimens to make it more efficacious.
A cursory Google search turned up little on “Klein’s String” but I obtained many more hits using the terms:
Klein continued fraction geometrical interpretation
Sadly, Wolfram’s Mathworld (http://mathworld.wolfram.com/ContinuedFraction.htmlhas) has a write-up that completely obfuscates the original idea (at least to me and my extremely limited mind). I doubt if Klein’s fabulous “string” will ever disappear from the modern literature but it would be an intellectual crime if less and less people were given an opportunity to experience the thrill of seeing it. To quote Davenport:
“A striking geometrical interpretation of the continued fraction for an irrational number was given by Klein in 1895″
In a sense, therefore, Klein’s String and Tennenbaum’s Proof are related.
I now sign off (for good, finally!) with a tremendous “Thank You” you to everyone who presented my father’s original work in a reasonable way, and to Steven J. Miller and David Montague for their truly beautiful extension of this line of development.
Peter Tennenbaum November 6, 2009
By: Peter Tennenbaum on November 6, 2009
at 2:10 am
5. Terrific!
By: Richard Elwes on December 20, 2010
at 10:06 am
6. Given that “being irrational” is defined as “not being rational” (in other words “being rational implies false”), this is not a proof by contradiction. I’ve never encountered a proof of a negative statement that was not purely constructive.
By: igor_d on April 30, 2012
at 8:58 am
7. Dear Peter,
I knew your dad, in 1980, when I was at IAS. He was a true mensch, one of the kindest, most caring people that I have met.
Penny Smith
By: penny on May 1, 2012
at 8:24 am
8. Beautiful! It would speed up comprehension of the proof if it spelled out why the sides of the remaining pink and dark blue square must be integers. I had to think about that for a minute.
By: Jon on January 24, 2013
at 11:00 pm
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http://mathoverflow.net/questions/104144/identifying-subgroups-of-the-modular-group-via-permutation-representations-on-cos
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Identifying Subgroups of the Modular Group via Permutation Representations on Cosets
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose we have a known group G and an unknown subgroup H. The permutation representation of G on the cosets of H gives a permutation group C, which is known. Is it possible to identify the generators of H using this information? If so, how? If not, what further information is needed?
(For background, my G is the modular group PSL(2,Z); H is a not-necessarily-congruence subgroup of G. C is sometimes called a "cartographic group").
Many thanks!
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In what sense is $C$ known if you don't know what $H$ is (and hence don't know what the cosets of $H$ are)? Do you only know $C$ as an abstract group? Do you know its action on the cosets as an abstract group action? – Qiaochu Yuan Aug 6 at 20:43
2 Answers
Choose a set of generators for $G$. Draw the Cayley graph of the action of $G$ on $G/H$. Choose a set of generators for the fundamental group of the graph. For each generator of the fundamental group, multiply the generators of $G$ corresponding to the edges together in order. This will be a set of generators for $H$.
Proof: Each element of $H$ is a product of the generators of $G$ which, in the action of $G$ on $G/H$, preserves the identity coset. Thus it's a product of generators that forms a closed loop of the Cayley graph. Homotoping that loop to a composition of the generators of the fundamental group does not change the element it refers to. You now have an expression for each element of $H$ in terms of the supposed generators of $H$, so they are in fact the generators of $H$.
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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.
(I wanted to post this a comment on @Will Sawin's answer, but I don't have the rep yet. Also it turned out to be convenient to flesh it out a little longer than comment.)
Will's answer assumes that you know a fixed element of $G/H$ that you can use as your basepoint for the fundamental group, namely the identity coset $H$. If you don't know that, you can't recover $H$ exactly, you can only recover $H$ up to conjugacy. (The permutation actions of $G$ on $G/H$ and $G/H'$ are permutationally isomorphic iff $H$ and $H'$ are conjugate subgroups.)
For example, in the case of $PSL(2,\mathbb{Z})$, this means that subgroups-up-to-conjugacy of index $n$ are in bijective correspondence with pairs of permutations $(a,b)$ on $[n]$ such that $a^{2} = b^{3} = 1$ (picking some natural generators of $PSL(2,\mathbb{Z}) \cong \mathbb{Z}_2 * \mathbb{Z}_3$)
and $\langle a, b\rangle$ acts transitively on $[n]$, up to the relation $(a,b) \sim (a',b')$ if there is a permutation $\pi \in S_{n}$ such that $\pi a \pi^{-1} = a'$ and $\pi b \pi^{-1} = b'$ (permutational isomorphism).
Subgroups (not up to conjugacy, but on the nose) are classified by the same pairs, but where the $\pi$ that we are allowed to conjugate by must fix, say, $1 \in [n]$ (corresponding to the identity coset of $H$ in $G/H$).
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http://math.stackexchange.com/questions/142728/understanding-fatous-lemma
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# Understanding Fatou's lemma
I want to prove that (without using Fatou's lemma)
for every $k \in N$ let $f_k$ be a nonnegative sequence $f_k(1),f_k(2),\ldots$
$$\sum^\infty_{n=1}\liminf_{k \to \infty} f_k(n) \le \liminf_{k \to \infty} \sum^\infty_{n=1}f_k(n)$$
Can you give some hint for me about that? hat
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Thank you for help. I got this question. Then I wonder why this inequality do for only liminf not for limsup either? – japee May 8 '12 at 17:32
## 1 Answer
Hints:
• Fix an integer $N$, and show that $$\sum_{n=1}^N\liminf_{k\to +\infty}f_k(n)\leq \liminf_{k\to +\infty}\sum_{n=1}^Nf_k(n).$$
• Show that $\sum_{n=1}^Nf_k(n)\leq \sum_{n=1}^{+\infty}f_k(n)$.
• Conclude, still using that all the terms are non-negative.
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http://physics.stackexchange.com/questions/34990/aharonov-bohm-effect-and-flux-quantization-in-superconductors/35017
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# Aharonov-Bohm Effect and Flux Quantization in superconductors
Why is the magnetic flux not quantized in a standard Aharonov-Bohm (infinite) solenoid setup, whereas in a superconductor setting, flux is quantized?
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## 2 Answers
According to Wigner, the wave function of a quantum particle can be multivalued, i.e., can acquire a nontrivial phase around a closed loop. A phase is nontrivial when it cannot be removed using a gauge transformation by $e^{i \alpha(\theta)}$, with a true function $\alpha$, i.e., $\alpha(2\pi) = \alpha(0)$. The wave functions having this property are sections of nontrivial line bundles over the configuration manifold.
The reason that a wave function is not required to be a true function is because its overall phase and magnitude are nonphysical (if one defines quantum expectations as:)
$<X> = \frac{\int \Psi \hat{X} \Psi}{\int \Psi \Psi}$
Such wave functions arise when the configuration manifold is not simply connected with a nontrivial cohomology group $\mathcal{H}^{1}(M,\mathbb{R})$ (This is the case of the circle). In this case, there will exist vector potentials on the manifold which are not the gradients of a true function on the manifold. $A \ne d\alpha(\theta)$. with, $\alpha(2\pi) =\alpha(0)$. However, there is no need for the flux to be quantized as the wave function needs not be a true function on the configuration manifold. On the contrary, if the flux had been quantized, then no Aharonov-Bohm effect would not have observed. A quantization condition occurs when $\mathcal{H}^{2}(M,\mathbb{R})$ (The Dirac quantization condition), but this is the case of a particle moving on a sphere rather than on a circle.
However, this is not the case in superconductivity. The difference between the two situations lies in the fact that the "macroscopic wave function" of a superconductor is not a "wave function". i.e., it is not the coordinate representation of a state vector in a Hilbert space. It is a quantum field describing Goldstone bosons (Cooper pair) of the superconducting phase (usually called an order parameter). The modulus of the macroscopic wave function $|\Psi(\theta)|^2$ describes the number density operator of the Goldstone bosons. Its two point functions describe the (long range) correlations. This quantum field couples minimally to electromagnetism, and this is the reason why its equation of motion is similar to the Schrodinger equation of a particle coupled to electromagnetism. But the main difference this field is a true scalar field and not a section of a line bundle. This gives us the reason why the phase it acquires in a full loop should vanish because otherwise for example, its correlation functions would depend on how many times the circle was wrapped.
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Clarifying question: is it safe to say that mod-squared of the wavefunction must be single-valued? since, as you say, the phase is unphysical? – QuantumDot Sep 6 '12 at 2:56
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Yes, this is exactly the definition of multivaluedness. Take for example the "funtion" $e^{\frac{i\theta}{2}}$ on the circle, it is multiple valued since it takes two different values at $\theta = 2\pi$ and $\theta = 4\pi$ which are the same physical point. Its modulus is a true function on the circle. Of course, the modulous operation cancels only a single global phase and if the wave function is a superposition, the relative phases will still exist. This is the reason why the wave function "feels" the topology in the Aharonov-Bohm effect. – David Bar Moshe Sep 6 '12 at 7:20
Ok, thanks for the help! – QuantumDot Sep 7 '12 at 0:25
The whole Aharonov-Bohm effect – a nontrivial phase – is actually due to nothing else than a deviation from the flux quantization rule. The angle we may measure as the shift of the interference pattern in the Aharonov-Bohm effect is
$$\Delta\phi = \frac{q\Phi_B}{\hbar}, \quad \Phi_B \equiv\int d\vec S\cdot \vec B$$
So the conventional flux quantization rule is $\Delta \phi = 2\pi k$ for $k\in{\mathbb Z}$ which means nothing else than "the solenoid behaves as exactly as if there were no solenoid".
That's how we derive the flux quantization in the first place. The magnetic monopoles, for example (an important third situation I am adding where the flux quantization deserves to be discussed), have to obey the Dirac quantization rule which is equivalent to the flux quantization rule for the flux through the surface surrounding them. And they have to obey it exactly for the Dirac string – a semi-infinite line starting from the magnetic monopole where $\vec A$ is inevitably singular and which must exists because ${\rm div}\,\vec B\neq 0$ anymore – to be unobservable. The Dirac string is nothing else than an Aharonov-Bohm solenoid, however one that we know to be unobservable because there's no matter over there and we required the location of the Dirac string to be a pure convention.
Note that there is a difference between the surfaces above. The flux quantization (which doesn't hold) in the Aharonov-Bohm effect counts the flux through an open, disk-shaped region; in the Dirac quantization rule, it's a closed surface, a sphere around a monopole. It's only the latter flux through a closed surface that has to be quantized.
Now, in superconductor, electron pairs act as bosons that effectively produce a complex classical scalar field $\Psi$. Its charge is $2e$ because it is composed of electron pairs. Now, the vacuum expectation value of $|\Psi|^2$ is a nonzero constant but the phase of $\Psi$ is arbitrary. In particular, when you study how the phase of $\Psi$ changes if you encircle the boundary of a ring, you will find out it comes back to the original phase but it may "wind around zero" $w$ times, a winding number, and this integer $w$ exactly measures the magnetic flux in the units of the superconductor magnetic flux (which is $1/2$ times the minimum magnetic monopole dual to the electron).
This condition that "the phase of $\Psi$ has to return to itself" is mathematically equivalent to what we used in the magnetic-monopole, Dirac quantization discussion: its phase is changing as much as when an electron was encircling the Dirac string (or solenoid) in the previous two examples. A difference is that now, an electron pair must be allowed to peacefully encircle the ring without changing the wave function – because it's still the same state. So the phase is changing twice as quickly and the allowed unit of flux is $1/2$ of what it was before.
In the superconductor case, the phase must return to its original value (after the electron pair makes a round trip around the ring) because this situation is close to the "Dirac string". In particular, we require that there is no observable effect of the material inside the disk simply because there's nothing – there's no solenoid etc. – inside the ring. So much like for the Dirac string, the matter inside the ring has to be invisible – there isn't any – which means that the wave function has to return to its original value after a 360-degree rotation by the electron pair.
Summary
One could just dismiss these three situations as entirely different situations but the key maths is still analogous in the three situations, with some differences:
• the Dirac string or the interior of the superconducting ring must act as if there were nothing, so the wave functions must return to themselves, and therefore impose the quantization rule; it's important to distinguish the open/closed topology of the surfaces over which the fluxes are measured
• the Aharonov-Bohm solenoid contains stuff so there's no valid proof that the solenoid has to be invisible to the electrons running around it, and indeed, their interference pattern is allowed to shift as a result
• one must be careful about the diffferent elementary charges, $e$ (or $e/3$ if quarks are included) in the Dirac string and/or the Aharonov-Bohm solenoid, and $2e$ in the superconducting case
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http://mathhelpforum.com/calculus/164684-how-you-evaluate-indefinite-integral-lnx.html
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# Thread:
1. ## How to you evaluate an indefinite integral with lnx?
antiderivative(ln5x/x)dx
i know ln5x(dx)=1/5x but not sure how to go about solving this
2. This is hard to read, is it
$\displaystyle \int{\frac{\ln{(5x)}}{x}\,dx}?$
If so, $\displaystyle \int{\frac{\ln{(5x)}}{x}\,dx} = \int{\ln{(5x)}\,\frac{1}{x}\,dx}$.
Now make the substitution $\displaystyle u = \ln{(5x)}$ so that $\displaystyle \frac{du}{dx} = \frac{1}{x}$.
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http://stats.stackexchange.com/questions/3730/pearsons-or-spearmans-correlation-with-non-normal-data?answertab=votes
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# Pearson's or Spearman's correlation with non-normal data
I get this question frequently enough in my statistics consulting work, that I thought I'd post it here. I have an answer, which is posted below, but I was keen to hear what others have to say.
Question: If you have two variables that are not normally distributed, should you use Spearman's rho for the correlation?
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## 4 Answers
Pearson's correlation is a measure of the linear relationship between two continuous random variables. It does not assume normality although it does assume finite variances and finite covariance. When the variables are bivariate normal, Pearson's correlation provides a complete description of the association.
Spearman's correlation applies to ranks and so provides a measure of a monotonic relationship between two continuous random variables. It is also useful with ordinal data and is robust to outliers (unlike Pearson's correlation).
The distribution of either correlation coefficient will depend on the underlying distribution, although both are asymptotically normal because of the central limit theorem.
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Pearson's $\rho$ does not assume normality, but is only an exhaustive measure of association if the joint distribution is multivariate normal. Given the confusion this distinction elicits, you might want to add it to your answer. – user603 Oct 19 '10 at 7:42
@kwak. Good point. I'll update the answer. – Rob Hyndman Oct 19 '10 at 7:45
Is there a source that can be quoted to support the above statement (Person's r does not assume normality)? We're having the same argument in our department at the moment. – user8891 Feb 1 '12 at 14:52
@RobHyndman In the field of financial time series (for example when trying to learn about correlations between stock returns), would you recommend Pearson correlation or rank based correlations? Wikipedia is pretty strongly against Pearson but their source is dubious. – Jase Dec 27 '12 at 5:57
Don't forget Kendall's tau! Roger Newson has argued for the superiority of Kendall's τa over Spearman's correlation rS as a rank-based measure of correlation in a paper whose full text is now freely available online:
Newson R. Parameters behind "nonparametric" statistics: Kendall's tau,Somers' D and median differences. Stata Journal 2002; 2(1):45-64.
He references (on p47) Kendall & Gibbons (1990) as arguing that "...confidence intervals for Spearman’s rS are less reliable and less interpretable than confidence intervals for Kendall’s τ-parameters, but the sample Spearman’s rS is much more easily calculated without a computer" (which is no longer of much importance of course). Unfortunately I don't have easy access to a copy of their book:
Kendall, M. G. and J. D. Gibbons. 1990. Rank Correlation Methods. 5th ed. London: Griffin.
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I'm also a big fan of Kendall's tau. Pearson is far too sensitive to influential points/outliers for my taste, and while Spearman doesn't suffer from this problem, I personally find Kendall easier to understand, interpret and explain than Spearman. Of course, your mileage may vary. – Stephan Kolassa Oct 19 '10 at 7:44
From an applied perspective, I am more concerned with choosing an approach that summarises the relationship between two variables in a way that aligns with my research question. I think that determining a method for getting accurate standard errors and p-values is a question that should come second. Even if you chose not to rely on asymptotics, there's always the option to bootstrap or change distributional assumptions.
As a general rule, I prefer Pearson's correlation because (a) it generally aligns more with my theoretical interests; (b) it enables more direct comparability of findings across studies, because most studies in my area report Pearson's correlation; and (c) in many settings there is minimal difference between Pearson and Spearman correlation coefficients.
However, there are situations where I think Pearson's correlation on raw variables is misleading.
• Outliers: Outliers can have great influence on Pearson's correlations. Many outliers in applied settings reflect measurement failures or other factors that the model is not intended to generalise to. One option is to remove such outliers. Univariate outliers do not exist with Spearman's rho because everything is converted to ranks. Thus, Spearman is more robust.
• Highly skewed variables: When correlating skewed variables, particularly highly skewed variables, a log or some other transformation often makes the underlying relationship between the two variables clearer (e.g., brain size by body weight of animals). In such settings it may be that the raw metric is not the most meaningful metric anyway. Spearman's rho has a similar effect to transformation by converting both variables to ranks. From this perspective, Spearman's rho can be seen as a quick-and-dirty approach (or more positively, it is less subjective) whereby you don't have to think about optimal transformations.
In both cases above, I would advise researchers to either consider adjustment strategies (e.g., transformations, outlier removal/adjustment) before applying Pearson's correlation or use Spearman's rho.
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Updated
The question asks us to choose between Pearson's and Spearman's method when normality is questioned. Restricted to this concern, I think the following paper should inform anyone's decision:
It's quite nice and provides a survey of the considerable literature, spanning decades, on this topic -- starting from Pearson's "mutilated and distorted surfaces" and robustness of distribution of $r$. At least part of the contradictory nature of the "facts" is that much of this work was done before the advent of computing power -- which complicated things because the type of non-normality had to be considered and was hard to examine without simulations.
Kowalski's analysis concludes that the distribution of $r$ is not robust in the presence of non-normality and recommends alternative procedures. The entire paper is quite informative and recommended reading, but skip to the very short conclusion at the end of the paper for a summary.
If asked to choose between one of Spearman and Pearson when normality is violated, the distribution free alternative is worth advocating, i.e. Spearman's method.
Previously ..
Spearman's correlation is a rank based correlation measure; it's non-parametric and does not rest upon an assumption of normality.
The sampling distribution for Pearson's correlation does assume normality; in particular this means that although you can compute it, conclusions based on significance testing may not be sound.
As Rob points out in the comments, with large sample this is not an issue. With small samples though, where normality is violated, Spearman's correlation should be preferred.
Update Mulling over the comments and the answers, it seems to me that this boils down to the usual non-parametric vs. parametric tests debate. Much of the literature, e.g. in biostatistics, doesn't deal with large samples. I'm generally not cavalier with relying on asymptotics. Perhaps it's justified in this case, but that's not readily apparent to me.
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No. Pearson's correlation does NOT assume normality. It is an estimate of the correlation between any two continuous random variables and is a consistent estimator under relatively general conditions. Even tests based on Pearson's correlation do not require normality if the samples are large enough because of the CLT. – Rob Hyndman Oct 19 '10 at 1:46
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I am under the impression that Pearson is defined as long as the underlying distributions have finite variances and covariances. So, normality is not required. If the underlying distributions are not normal then the test-statistic may have a different distribution but that is a secondary issue and not relevant to the question at hand. Is that not so? – user28 Oct 19 '10 at 1:47
@Rob, @Srikant: True, I was thinking of significance testing. – ars Oct 19 '10 at 1:59
@Srikant: I'm not sure it's a "secondary issue". You can compute anything after all -- it's the inference that matters. @Rob: your "if" qualifier is key here -- it seems to me that's central to this question. We can justify a whole lot with asymptotic hand waving; exceptions matter. – ars Oct 19 '10 at 5:54
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@Rob: Yes, we can always come up with workarounds to make things work out roughly the same. Simply to avoid Spearman's method -- which most non-statisticians can handle with a standard command. I guess my advice remains to use Spearman's method for small samples where normality is questionable. Not sure if that's in dispute here or not. – ars Oct 19 '10 at 23:43
show 5 more comments
## protected by Community♦Apr 12 '12 at 14:05
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://mathoverflow.net/revisions/70143/list
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## Return to Question
3 Changed the title to make the question more explicit.
# ToricWhatisthefanofthetoric blow-up of $\mathbb{P}^3$ along the union of two intersecting lines?
2 deleted 9 characters in body
Is there a good way to find the fan and polytope of the blow-up of $\mathbb{P}^3$ along the union of two invariant (and so intersecting ) lines?
Everything I find in the literature is for blow-ups along smooth invariant centers.
Thanks!
1
# Toric blow-up of $\mathbb{P}^3$ along the union of two intersecting lines
Is there a good way to find the fan and polytope of the blow-up of $\mathbb{P}^3$ along the union of two invariant (and so intersecting) lines?
Everything I find in the literature is for blow-ups along smooth invariant centers.
Thanks!
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http://mathoverflow.net/questions/103837?sort=newest
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## Is there a categorical treatment of dynamical systems?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $X$ be a set and $(T,\cdot)$ an abelian group. Is there a category of $T$-dynamical systems on $X$ which yields useful information about $X$ and $T$?
More precisely, is there a category whose objects are dynamical systems, i.e., $\phi:X \times T \to X$ such that $\phi(x,t + s) = (\phi(\phi(x,t),s)$? And if so,
what are morphisms $\phi \to \phi'$ of $T$-dynamical systems on $X$?
I assume that in general this would involve imposing structure on $X$: for example, a topology so that one could consider homotopically perturbing $\phi$. Mostly, I am interested in asking when two such dynamical systems may be considered equivalent, and what it would take to have functors from $T$-dynamical systems on $X$ to $T'$-dynamical systems on $X'$, and to have natural transformations of those functors, etc.
I promise I've done (some) homework by looking at this. But note that this document only provides candidates for equivalent dynamical systems which presumably only accounts for isomorphisms in the desired category rather than all morphisms.
This may be too basic a question for the folks here; in this case I will delete it.
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The usual thing in dynamics if to fix a topological semigroup $T$ and looks at the category whose objects are actions of $T$ on a compact spaces and with maps the $T$-equivariant continuous maps. Isomorphisms in this category are commonly called conjugacies and epimorphisms are called factor maps. – Benjamin Steinberg Aug 3 at 3:37
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Perhaps you might change the question slightly. Is there a category of T-dynamical systems on X which gives useful information on the systems? Whether or not it is non-trivial as a category is not really that important. For some $(X,T)$ it might be trivial (in some sense)... and that might correspond to a useful class of dynamics. As a dynamical system is a group action, there is always a category of T-sets, and you can restrict to the full subcategory of those whose underlying set is X. – Tim Porter Aug 3 at 5:05
Dear Tim, I've made the edits. Thank you for your suggestion. Benjamin, I was aware of factor maps, but those are too specific since they require topological notions. In general, that framework wouldn't even accomodate shift maps on infinite strings in finitely many symbols since the space isn't compact, right? – Vidit Nanda Aug 3 at 17:04
## 3 Answers
For every category $\mathfrak C$ and every monoid $\mathcal S$, you can define the category of $\mathcal S$-flows on $\mathfrak C$ as follows:
-- Objects are pairs $(X, \alpha:\mathcal S\to \mathrm{End}(X))$, where $\alpha$ is a monoid homomorphism (that is, multiplication in $\mathcal S$ becomes composition of morphisms in $\mathfrak C$ and the neutral element goes to identity);
-- Morphisms in the category between two objects $(X,\alpha)$ and $(X',\alpha')$ are just morphisms $\phi:X\to X'$ that commute with the actions, that is $\alpha'(s)\phi=\phi\alpha(s)$ for all $s\in \mathcal S$.
Clearly an isomorphism in this category is just a morphism that is an isomorphism in $\mathfrak C$.
This is standard and probably does not answer completely your question as you probably want a weaker form of equivalence than really isomorphism in the category of $\mathcal S$-flows. Anyway I think this is a good starting point.
Just to understand what this category is, I can give you some examples. If you consider $\mathbb N$-flows on the category of modules $\mathrm {Mod}(R)$ over a ring $R$, you just obtain the category of modules over the polynomial ring $R[X]$. In fact, it is a classical way of looking at modules over $R[X]$ as $R$-modules with a distinguished $R$-linear endomorphism acting on them, that is, discrete-time dynamical systems.
If you take $\mathbb Z$-flows you obtain the ring of Laurent polynomials $R[X^{\pm 1}]$. This is easily generalized to $\mathbb N^k$ and $\mathbb Z^k$, giving rise to polynomials in $k$ commuting variables. This point of view is generally adopted by K. Schmidt in his book "Dynamical Systems of Algebraic Origin". In fact, the general approach there is to study dynamical systems of the form $(G,\phi_1,\dots,\phi_k)$, where $G$ is a compact abelian topological group and $\phi_1,\dots,\phi_k$ are commuting topological automorphisms of $G$. This is the category of $\mathbb Z^k$-flows on the category of compact abelian groups. Via Pontryagin duality, this category can be seen to be dual to the category of $\mathbb Z^k$-flows on discrete Abelian groups, that, by what we said above, is exactly $\mathrm{Mod}(\mathbb Z[X_1^{\pm 1},\dots,X_k^{\pm 1}])$.
Generalizing more, you can easily prove that $\mathcal S$-flows on $\mathrm {Mod}(R)$ are the category of modules over the monoid ring $R[\mathcal S]$.
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An action of a group $T$ on a set $X$ defines the action groupoid $T\times X \rightrightarrows X$ (If $T$ is a semigroup then $T\times X \rightrightarrows X$ is a category). Thinking of dynamical systems this way suggests that morphisms are functors between action groupoids. If there is a topology on $T$ and $X$ and the action is continuous, then the action groupoid is a topological groupoid. You may then take your morphisms to be continuous functors. However, this is not the best one can do. A better notion of morphism between topological groupoid is that of a bibundle also known as a ‘Hilsum–Skandalis map.’ If you go down this road you end up with bicategories since the composition of bibundles is associative only up to isomorphism (see this paper by Christopher Schommer-Pries for a nice discussion of the issues).
If your dynamical systems are vector fields on manifolds then the morphisms are smooth maps intertwining the vector fields. That is, if $X$ is a vector field on $M$, $Y$ a vector field on $N$ then a morphism from $(M,X)$ to $(N, Y)$ is a smooth map $f:M\to N$ with $Y \circ f = Df \circ X$.
There is also a large body of literature on the category of labelled transition systems and on the category of Petrie nets...
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Note a map from a terminal object is a fixed point and a monomorphism is an invariant subsystem. In the case where the category in question is manifolds and vector fields, a map from a circle with the constant vector field is a periodic orbit. – Eugene Lerman Aug 4 at 0:27
See "J. de Vries, Topological transformation groups 1, A cat.approach" and similar works about topological group actions from a categorical perspective.
(http://oai.cwi.nl/oai/asset/12605/12605A.pdf or search for "Topological transformation groups" (any author) in http://repository.cwi.nl/).
THese is a account of it in the last article in Springer Lecture notes in MAth n. 540 (catagorical topology).
ABout homotopical question of Topological transformation groups see the Tammo Tom Dieck book (http://www.amazon.com/Transformation-Groups-Gruyter-Studies-Mathematics/dp/3110097451)
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http://math.stackexchange.com/questions/26542/what-are-the-differences-between-fiber-bundle-and-sheaf/26572
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# What are the differences between fiber bundle and sheaf?
They are similar.All contain a map and can define the section,the fiber of the fiber bundle is just like the stalk of the sheaf.But what are the differences between them,maybe sheaf is more abstract and can break down, the fibre bundle is more geometric and must keep itself continue. Any other differences?
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Fibre bundles look the same locally at any point of your space. This does not have to be true for sheaves. – anonymous Mar 12 '11 at 10:54
I have been told (but do not understand) that there is an adjunction between the category of presheaves on a topological space and the category of bundles over the space, which restricts to an equivalence of categories between the category of sheaves and the category of étale bundles. – Zhen Lin Mar 12 '11 at 15:40
a fiber bundle is a local product, and a sheaf of (say rings) is a way of associating rings to the open sets of a space so that inclusions of open sets induce homomorphisms of the respective rings. theyre just different things – yoyo Mar 12 '11 at 16:13
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@yoyo: There are many kinds of sheaves other than sheaves of rings. A fibre bundle gives rise to a sheaf of sections, from which (in reasonable circumstances, and when endowed with the appropriate extra structure) the bundle can be recovered. So, while they are different things, it is not a matter of them being just different things. – Matt E Mar 12 '11 at 21:11
## 1 Answer
If $(X,\mathcal{O}_X)$ is a ringed topological space, you can look at locally free sheaves of $\mathcal{O}_X$-modules on $X$.
If $\mathcal{O}_X$ is the sheaf of continuous functions on a topological manifold (=Hausdorff and locally homeomorphic to $\mathbb{R}^n$), or the sheaf of smooth functions on a smooth manifold, you get fiber bundles (the sheaf associated to a fiber bundle is the sheaf of "regular" (=continuous or smooth here) sections).
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1
... and conversely any fiber bundle has a natural sheaf associated to it, namely the sheaf of sections of the bundle. – Matt Mar 12 '11 at 19:53
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http://mathoverflow.net/questions/33088/entropy-of-a-general-prob-measure/33090
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## Entropy of a general prob. measure [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
How is entropy of a general probability measure defined?
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I can suggest en.wikipedia.org/wiki/Differential_entropy – falagar Jul 23 2010 at 13:10
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Did you try Google and/or Wikipedia? – Qiaochu Yuan Jul 23 2010 at 15:14
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I am casting the final vote to close, because even though the original question has generated some interesting details, it wasn't asked with enough background context., nor with indication of what the questioner already knows or has read. – Yemon Choi Jul 23 2010 at 20:31
## 3 Answers
The Entropy of a function $f$ with respect to a measure $\mu$ is
$$Ent_{\mu}(f)=\int f \log f d\mu - \int f d\mu \log(\int f d\mu )$$
The entropy of a probability distribution $P$ with respect to $\mu$ is given by $Ent(\frac{dP }{d\mu })$. I a not aware of a general definition that would not implie a reference (here it is $\mu$) measure ...
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It is not. If a probability measure on $\mathbb{R}$ is absolutely continuous and has density $f$, then "entropy" usually refers to the differential entropy, defined in the Wikipedia page falagar linked to. If the probability measure has discrete support, entropy is defined by an analogous formula, given in this Wikipedia page. In the most classical treatments, these are the only situations covered at all.
However, both of these are special cases of the more general notion of relative entropy that Helge and robin girard pointed out: in the continuous case the reference measure ($\nu$ in Helge's notation, $\mu$ in robin's notation) is Lebesgue measure, and in the discrete case the reference measure is counting measure on the support.
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Entropy is not defined for a single probability measure!
Entropy is a relative thing, you define between two measure $\mu$ and $\nu$. Then the entropy is defined by $$\mathcal{E}(\mu,\nu) = \begin{cases} - \int w(x) \log(w(x)) d\nu(x),& d \mu = w d\nu; \\ -\infty, & otherwise.\end{cases},$$ I might have messed up the signs. $w$ is the Radon-Nikodym derivate of $\mu$ with respect to $\nu$.
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http://math.stackexchange.com/questions/218904/sheaf-on-a-stein-variety-such-that-h1x-mathcalf-neq-0
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# Sheaf on a Stein variety such that $H^{1}(X, \mathcal{F}) \neq 0$
I would like to find a non-coherent sheaf on a Stein variety $X$ such that $H^{1}(X, \mathcal{F}) \neq 0$. Does anyone know any example?
Thank you!
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## 1 Answer
Just take $\mathcal F=\mathcal O^*_X$, the sheaf of invertible holomorphic functions.
Then $H^{1}(X, \mathcal{F}) =H^{1}(X, \mathcal O^*_X) =Pic(X)$, the Picard group consisting of isomorphism classes of holomorphic line bundles on $X$.
Now from the exponential exact sequence $0\to \mathbb Z\to \mathcal O_X\to \mathcal O_X^*\to 0$ you immediately get by taking the long exact sequence of cohomology that the first Chern class yields a group isomorphism $$H^{1}(X, \mathcal O_X) =0 \to H^{1}(X, \mathcal O^*_X) \xrightarrow {c_1} H^2(X,\mathbb Z)\to H^{2}(X, \mathcal O_X)=0$$
So a very simple example of Stein manifold with $H^{1}(X, \mathcal O^*) \neq 0$ is $\mathbb C^*\times \mathbb C^*$ since $$H^{1}(\mathbb C^*\times \mathbb C^*, \mathcal O^*)\stackrel {c_1}{\cong} H^{2}(\mathbb C^*\times \mathbb C^*, \mathbb Z)=\mathbb Z$$
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So nice @Georges, but how do you know that $H^{2}(\mathbb{C}^{\ast} \times \mathbb{C}^{\ast}, \mathbb{Z}) = \mathbb{Z}$? It seems come from some topological information of the torus, and I am not familiar with that. Could you explain or give me any reference? Thank you! – rla Oct 23 '12 at 11:40
– Georges Elencwajg Oct 23 '12 at 12:09
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http://physics.stackexchange.com/questions/21902/how-was-this-pressure-pulse-propagation-speed-be-derived/21947
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# How was this pressure pulse propagation speed be derived?
Some lecture notes I was reading through claimed that a pressure pulse propagates through a liquid-filled tube (blood in a vein) with the speed
$$c=\sqrt{\frac{A}{\rho}\frac{dP}{dA}},$$
where $A$ is the cross-sectional area, $\rho$ the density and $P$ the pressure (which pressure?).
How can this result be derived? Why would the pressure vary with area? I suppose I'm missing some assumptions or misunderstand the expression.
FYI, the expression is also written, without explanation of the variables, as
$$c=\sqrt{\frac{Eh}{2\rho R}}$$
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## 1 Answer
The sound speed $c_s$ of any fluid in which the fluid pressure $P$ is a function of the fluid density $\rho$ is going to be given by $c_s^2 = \frac{dP}{d\rho}$. If that expression is not already familiar to you, it's a good exercise to write down the fluid equations for conservation of mass and momentum in one dimension and then write down what happens when the density and velocity are perturbed from their equilibrium, static values. You'll get the wave equation for the perturbed values with the wave speed given by the above equation.
If $\rho$ can be treated as a function of $A$, then applying the chain rule gives $c_s = \sqrt{\frac{dP}{dA} \,\frac{dA}{d\rho}}$. If $A$ is proportional to $\rho$ with the same proportionality constant everywhere, you get the expression you've asked about.
Even if the relationship between $A$ and $\rho$ is not exactly that simple, you'd expect to get the same answer to order of magnitude since $A$ and $\rho$ should change by values comparable to themselves.
EDIT: It turns out that you can get the same expression for the sound speed if $A$ is inversely proportional to $\rho$. In that case both $\frac{dP}{dA}$ and $\frac{dA}{d\rho}$ are negative, which might be useful to keep in mind. And an inverse relationship is exactly what you'd expect if the flow is steady or nearly so ($\rho$ and $u$ at each position do not vary with time). In that case conservation of mass implies that $\rho u A$ is a constant at each position. And if $u$ is not varying much with position, then you indeed get $A \propto 1/\rho$.
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http://mathhelpforum.com/differential-equations/186532-dst-differential-equation.html
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# Thread:
1. ## dst differential equation
I've got an answer for this question but i'd like to check it against someone else and see the working used
Attached Thumbnails
2. ## Re: dst differential equation
Originally Posted by jonnyl
I've got an answer for this question but i'd like to check it against someone else and see the working used
$\displaystyle \begin{align*} \frac{d^2x}{dt^2} &= -\frac{dx}{dt} + e^{-t} + 3 \\ \frac{d^2x}{dt^2} + \frac{dx}{dt} &= e^{-t} + 3 \end{align*}$
You need to solve the homogeneous equation $\displaystyle \frac{d^2x}{dt^2} + \frac{dx}{dt} = 0$.
The characteristic equation is
$\displaystyle \begin{align*} m^2 + m &= 0 \\ m(m + 1) &= 0 \\ m = 0 \textrm{ or }m &= -1 \end{align*}$
which means the solution to the homogeneous equation is
$\displaystyle x_c = Ae^{0t} + Be^{-t} = A + Be^{-t}$.
Now, using the method of undetermined coefficients to solve the nonhomogeneous equation, guess $\displaystyle x_p = C_1t^2 + C_2t + C_3 + C_4t\,e^{-t}$, then $\displaystyle \frac{dx}{dt} = 2C_1t + C_2 + C_4e^{-t} - C_4t\,e^{-t}$ and $\displaystyle \frac{d^2x}{dt^2} = 2C_1 - C_4e^{-t} - C_4e^{-t} + C_4t\,e^{-t}$.
Substitute each of these into the DE $\displaystyle \frac{d^2x}{dt^2} + \frac{dx}{dt} = e^{-t} + 3$ and equate like coefficients to solve for $\displaystyle C_1, C_2, C_3, C_4$.
3. ## Re: dst differential equation
Originally Posted by jonnyl
I've got an answer for this question but i'd like to check it against someone else and see the working used
If you have an answer, you can check it simply by substituting it into the DE and seeing if it works.
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I understand this is a homework problem. I don't expect anyone to do my homework for me, but help me understand what I am doing wrong. The problem goes: Two shuffleboard disks of equal mass, one ...
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### Where can I find the equations for “quasi” elastic collisions?
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### Would it help if you jump inside a free falling elevator?
Imagine you're trapped inside a free falling elevator. Would you decrease your impact impulse by jumping during the fall? When?
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http://physics.stackexchange.com/questions/28842/calculating-kinetic-energy?answertab=active
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# Calculating kinetic energy?
Would this be a valid equation to calculate kinetic energy created from a drop from a height:
$$E_{kinetic} ~=~ v_{vertical}tmg$$
Velocity multiplied by time gives distance. Distance multiplied by gravitational force acting on it provides kinetic energy. Would this equation be valid?
-
The answer to the question(v1) is Yes if $v_{vertical}=\frac{gt+0}{2}$ denotes the average velocity during the drop. – Qmechanic♦ May 23 '12 at 15:39
Although, what you might be getting at would be that $E_{kinetic}=\int v_{vertical}mg \cdot dt$, which is identically $E_{kinetic}=mgh$. If you don't know calculus, the equation $dE_{kinetic}=v_{vertical}mg \cdot dt$ might make more sense. Because v changes over time, you can only say that "h=v*dt" over very small values of "dt", so this gives you a very small portion of energy, and when you add all the tiny portions of energy together you get the correct result. – NeuroFuzzy Sep 8 '12 at 6:55
## 2 Answers
No.
You are correct that the kinetic energy is equal to the change in the potential energy, $mgh$, where $h$ is the distance fallen, but because the object is accelerating $h$ is not simply velocity times time. If the object starts at rest then (ignoring air resistance):
$$h = \frac{1}{2} g t^2$$
so substituting this into $mgh$ gives:
$$E_{kinetic} = \frac{1}{2} m g^2 t^2$$
Note that $gt$ is just the velocity at time $t$, so this expression is the same as:
$$E_{kinetic} = \frac{1}{2} m v^2$$
which may look familiar :-) Note however that the velocity $v$ is a function of time.
-
Thank You John! – user8791 May 23 '12 at 15:40
For a body of mass m in a uniform specific-force field $g$:
$$E_{kinetic} = \frac{1}{2} m g^2 t^2$$
For two mutually interacting bodies of masses M and m: $$E_{kinetic} = \frac{G(M+m)}{r_1}-\frac{G(M+m)}{r_0}$$
where $r_0$ and $r_1$ are the initial and final separations of the two bodies.
Since uniform gravitational fields cannot exist in nature, the first equation is not applicable to this problem. It is a commonly used approximation that fails miserably unless the initial seperation is large, and the change in separation is small.
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http://mathematica.stackexchange.com/questions/14361/how-to-change-coordinates-of-a-differential-operator?answertab=oldest
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# How to change coordinates of a differential operator?
I'm doing a basic quantum mechanics problem and am trying to learn how to do it in Mathematica. Any help would be much appreciated.
$\vec{L} = \vec{x} \times \vec{p}$
where $\vec{x}$ has components:
$\begin{align*} x &= r \sin \theta \cos \phi\\ y &= r \sin \theta \sin \phi\\ z &= r \cos \theta \end{align*}$
and $\vec{p}$ has components:
$p_x = - i \hbar \dfrac{\partial}{\partial x}$
etc.
Show that $L_z = -i \hbar \dfrac{\partial}{\partial \phi}$
-
1
where have you gotten stuck? – acl Nov 9 '12 at 22:41
2
– b.gatessucks Nov 9 '12 at 22:47
## 1 Answer
Here is the Mathematica proof. I'll leave out the prefactor $\hbar/i$ for simplicity. Also, in case this is a homework problem, I decided not to add too many comments to the code. Instead I'll let you figure it out. The basic idea is to do cross products and gradients in spherical coordinates. The calculation shown here actually gives you a way to calculate all angular momentum components (not just the $z$ component).
In terms of notation, I use the spherical coordinates `r, θ, φ` and call their unit vectors `er, eθ, eφ`.
````Needs["VectorAnalysis`"]
Clear[r, θ, φ]
SetCoordinates[Spherical[r, θ, φ]]
(* ==> Spherical[r, θ, φ] *)
{er, eθ, eφ} =
Transpose[JacobianMatrix[{r, θ, φ}]]/
ScaleFactors[{r, θ, φ}]
````
$\left( \begin{array}{ccc} \cos (\phi ) \sin (\theta ) & \sin (\theta ) \sin (\phi ) & \cos (\theta ) \\ \cos (\theta ) \cos (\phi ) & \cos (\theta ) \sin (\phi ) & -\sin (\theta ) \\ -\sin (\phi ) & \cos (\phi ) & 0 \\ \end{array} \right)$
````l =
FullSimplify@
Cross[r er,
Transpose[{er, eθ, eφ}].Grad[f[θ, φ]]]
````
$\left( \begin{array}{c} \cot (\theta ) (-\cos (\phi )) f^{(0,1)}(\theta ,\phi )-\sin (\phi ) f^{(1,0)}(\theta ,\phi )\\ \cos (\phi ) f^{(1,0)}(\theta ,\phi )-\cot (\theta ) \sin (\phi ) f^{(0,1)}(\theta ,\phi )\\ f^{(0,1)}(\theta ,\phi) \end{array} \right)$
This is a list of Cartesian components (because the `Cross` product needed to be done in Cartesian coordinates) - the last entry in the list `l` is the `z` component of the angular momentum.
Here `f` is a test function to which I apply the operator. $f^{(0,1)}(\theta ,\phi)$ is the derivative with respect to the azimuthal coordinate.
The spherical components of the angular momentum (which you didn't ask for, but I mentioned in an earlier version of this answer) are obtained as follows:
````FullSimplify[{er, eθ , eφ}.l]
````
$\left(0,-\csc (\theta ) f^{(0,1)}(\theta ,\varphi ),f^{(1,0)}(\theta ,\varphi )\right)$
Edit for version 9
Although the above solution still works if you load the `VectorAnalysis` package, there are compatibility warnings now because the package has become obsolete with Mathematica version 9. So for this new version it's better to extract the coordinate transformation data in a different way:
````{er, eθ, eφ} =
CoordinateTransformData[{"Spherical" -> "Cartesian"},
"OrthonormalBasisRotation", {r, θ, φ}];
l = FullSimplify@
Cross[r er,
Transpose[{er, eθ, eφ}].Grad[
f[θ, φ], {r, θ, φ},
"Spherical"]]
````
The result is the same as above. Typing the lengthy `"OrthonormalBasisRotation"` expression to define the orthonormal basis vectors of the spherical coordinate system takes some getting used to. It's almost equally easy to re-derive them from first principles - but I wanted to show where the obsolete `JacobianMatrix` and `ScaleFactors` information is hidden now.
-
Thank you for your thorough explanation! This was in fact a homework problem, though I had solved it on paper already. – TaylorR137 Nov 12 '12 at 22:11
lang-mma
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|
http://mathhelpforum.com/advanced-algebra/204145-determine-if-two-vectors-orthogonal.html
|
1Thanks
• 1 Post By Plato
# Thread:
1. ## Determine if two vectors are orthogonal
Hey everyone! I have a question about the orthogonality of two vectors
The question is: It is given: llmll = 4; llnll = square root of 3; (m^,n)= 150 degrees
(i) find the norm of the vector m + 2n
(ii) Determine if the vectors (m+2n) and (-m+n) are orthogonal
I got an answer of 2 for the norm but I don't know what to do for the second part. Do I just substitute the values of m and n into the vectors and if they both equal 0 when multiplied, they are orthogonal? Any help would be greatly appreciated
2. ## Re: Determine if two vectors are orthogonal
Originally Posted by hemster83
Hey everyone! I have a question about the orthogonality of two vectors
The question is: It is given: llmll = 4; llnll = square root of 3; (m^,n)= 150 degrees
(i) find the norm of the vector m + 2n
(ii) Determine if the vectors (m+2n) and (-m+n) are orthogonal
What does the notation (m^,n)= 150 degrees mean?
I have never seem that used.
3. ## Re: Determine if two vectors are orthogonal
Originally Posted by Plato
What does the notation (m^,n)= 150 degrees mean?
I have never seem that used.
It means the angle between m and n equals 150 degrees. I think my prof made it up himself
4. ## Re: Determine if two vectors are orthogonal
Originally Posted by hemster83
It means the angle between m and n equals 150 degrees. I think my prof made it up himself
How very odd is that?
Originally Posted by hemster83
The question is: It is given: llmll = 4; llnll = square root of 3; (m^,n)= 150 degrees
(i) find the norm of the vector m + 2n
(ii) Determine if the vectors (m+2n) and (-m+n) are orthogonal
Well you know that $\frac{{m \cdot n}}{{\left\| m \right\|\left\| n \right\|}} = \cos \left( {\frac{{5\pi }}{6}} \right) = \frac{{ - \sqrt 3 }}{2}$.
From that you can find $m\cdot n~.$
Now $\|m+2n\|^2=(m+2n)\cdot(m+2n)=m\cdot m+4(n\cdot m)+4(n\cdot n)~.$
5. ## Re: Determine if two vectors are orthogonal
I have this same exact question.
Well you know that $\frac{{m \cdot n}}{{\left\| m \right\|\left\| n \right\|}} = \cos \left( {\frac{{5\pi }}{6}} \right) = \frac{{ - \sqrt 3 }}{2}$.
From that you can find $m\cdot n~.$
Now $\|m+2n\|^2=(m+2n)\cdot(m+2n)=m\cdot m+4(n\cdot m)+4(n\cdot n)~.$
Now this is the method I used to find the norm, or length of the vector (m+2n). However, I am still at a loss as to how to test the orthogonality of the two vectors (m+2n) and (-m+n).
Thanks in advance.
6. ## Re: Determine if two vectors are orthogonal
Originally Posted by Kristoffermk3
I have this same exact question.
Now this is the method I used to find the norm, or length of the vector (m+2n). However, I am still at a loss as to how to test the orthogonality of the two vectors (m+2n) and (-m+n).
What is $(m+2n)\cdot(-m+n)=~?$
7. ## Re: Determine if two vectors are orthogonal
Yes, that's what i'm looking for.
I suppose the part that is confusing me is simply that its not asking something as simple as (m dot n). Which is very simple to figure out, but I am confused on how to approach this when they are combinations of vectors such as (m+2n) and (-m+n)...
8. ## Re: Determine if two vectors are orthogonal
Originally Posted by Kristoffermk3
Yes, that's what i'm looking for.
I suppose the part that is confusing me is simply that its not asking something as simple as (m dot n). Which is very simple to figure out, but I am confused on how to approach this when they are combinations of vectors such as (m+2n) and (-m+n)...
$(m+2n)\cdot(-m+n)=-m\cdot m-2 m\cdot n+m\cdot n+2n\cdot n$
9. ## Re: Determine if two vectors are orthogonal
Oh my, that seems so self-evident now.
Thank you very much for your clarification.
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|
http://mathoverflow.net/questions/69980?sort=oldest
|
## Give a restriction to ensure a surgery of a balanced manifold is still balanced.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
M is a balanced compact complex manifold, if I do a surgery on M and get N. My question is under what appropriate conditions can ensure that N is still balanced?
-
1
You may get more answers if you give a bit more background to your question. See mathoverflow.net/howtoask#motivation – jc Jul 11 2011 at 14:24
3
There are at least two totally different definitions of "balanced" for a complex manifold. Also, what do you mean by "surgery"? There are many different things you could mean, most of which do not keep your manifold complex (at least not in a natural way). – Joel Fine Jul 12 2011 at 8:32
## 1 Answer
Maybe you mean a complex manifold $M$ of dimension $n$ is balanced iff it admits a hermitian metric $\omega$ satisfying $d\omega^{n-1}=0$.
In the paper:Metric properties of manifolds bimeromorphic to compact Kahler spaces. (JDG.v37.1993.95-121), L. Alessandrini and G. Bassanelli had proved the following result:
Let $M$ and $N$ be compact complex manifolds and $f:N\longrightarrow M$ be a modification,then 1) $M$ is balanced $\Longrightarrow N$ is balanced. 2) $N$ is balanced and satisfies a cohomological condition (it 's called B in the above paper)$\Longrightarrow M$ is balanced.
For the details,you should read their paper.
In addition,if you mean "balance" in the Kahler-Einstein problem，the following two papers maybe helpful.
a)S.K. DONALDSON:SCALAR CURVATURE AND PROJECTIVE EMBEDDINGS I
b）CLAUDIO AREZZO AND FRANK PACARD：BLOWING UP AND DESINGULARIZING CONSTANT SCALAR CURVATURE KAHLER MANIFOLDS
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http://mathhelpforum.com/advanced-algebra/124601-need-help-matrix-decomposition.html
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# Thread:
1. ## Need help in Matrix Decomposition
Hi all,
I have the product of a 4x3 rectangular matrix and its transpose (the transpose is a 3x4 matrix). The product is a 4x4 symmetric matrix.
How can I find the original 4x3 rectangular matrix, if I know the values of all the elements of the 4x4 product matrix?
If anyone could help me solve this problem, i would realy appreciate it.
Thanks
2. Originally Posted by dimper129
Hi all,
I have the product of a 4x3 rectangular matrix and its transpose (the transpose is a 3x4 matrix). The product is a 4x4 symmetric matrix.
How can I find the original 4x3 rectangular matrix, if I know the values of all the elements of the 4x4 product matrix?
If anyone could help me solve this problem, i would realy appreciate it.
Thanks
The only matrix you have is the 4x4? Can you post this matrix and any others?
3. Originally Posted by pickslides
The only matrix you have is the 4x4? Can you post this matrix and any others?
Yes the only matrix I have is the 4x4 product.
e.g. if
$\left[\begin{matrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \\ a_{10} & a_{11} & a_{12} \end{matrix}\right]\left[\begin{matrix} a_1 & a_4 & a_7 & a_{10} \\ a_2 & a_5 & a_8 & a_{11} \\ a_3 & a_6 & a_9 & a_{12} \end{matrix}\right] = \left[\begin{matrix} 0.0018 & -0.0062 & -0.0009 & 0.0021 \\ -0.0062 & 0.0537 & -0.0027 & 0.0014 \\ -0.0009 & -0.0027 & 0.1142 & 0.0063 \\ 0.0021 & 0.0014 & 0.0063 & 0.1005 \end{matrix}\right]$
then how can I find the values of $a_1, a_2 ...$ upto $a_{12}$
Thanks
4. It makes 10 equations in 12 unkowns like
${a_1}^2 + {a_2}^2 + {a_1}^2 = 0.0018$
$a_1a_4 + a_2a_5 + a_3a_6 = -0.0062$
$a_1a_7 + a_2a_8 + a_3a_9 = -0.0009$
$a_1a_{10} + a_2a_{11} + a_3a_{12} = 0.0012$
${a_4}^2 + {a_5}^2 + {a_6}^2 = 0.0537$
$a_4a_{7} + a_5a_{7} + a_6a_{9} = -0.0027$
$a_4a_{10} + a_5a_{11} + a_8a_{12} = 0.0014$
${a_7}^2 + {a_8}^2 + {a_9}^2 = 0.1142$
$a_7a_{10} + a_8a_{11} + a_8a_{12} = 0.0063$
${a_{10}}^2 + {a_{11}}^2 + {a_{12}}^2 = 0.1005$
we can solve (approximate) this non-linear system of equations in MATLAB using "fsolve" function. But this does not solve my problem because I need exact solution.
Does anyone know how to do it?
Thanks
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http://unapologetic.wordpress.com/2007/12/10/metric-spaces-and-continuity-of-real-valued-functions/?like=1&source=post_flair&_wpnonce=eb8d9d939c
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# The Unapologetic Mathematician
## Metric Spaces and continuity of real-valued functions
Now that we’ve got the real numbers which correspond to our usual notion of magnitudes like distances, let’s refine our concept of a uniform space to take into account this idea of distance.
A metric space is a set equipped with a notion of “distance” in the set. That is, we have a set $M$ and a function $d:M\times M\rightarrow\mathbb{R}$, which assigns a real number to every pair of points in $M$. This function will satisfy the following axioms:
• For all $x,y\in M$, $d(x,y)\geq0$.
• For all $x,y\in M$, $d(x,y)=0$ if and only if $x=y$.
• For all $x,y\in M$, $d(x,y)=d(y,x)$.
• For all $x,y,z\in M$, $d(x,z)\leq d(x,y)+d(y,z)$.
The first says that distances are all nonnegative real numbers. The second says that any point is distance ${0}$ from itself, and only from itself. The third says that the distance between two points doesn’t depend on the order in which we take the points. The last is called the triangle inequality, because if we think of the points as the vertices of a triangle then it’s shorter to go from $x$ to $z$ along the leg connecting those two than to take the detour to $y$.
Notice that these properties line up with those of absolute values. That is, the function $d:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ defined by $d(x,y)=\left|x-y\right|$ will be a distance function on $\mathbb{R}$.
Now any metric space is actually a uniform space. We define an entourage $E_\delta$ for each positive real number $\delta>0$. This $E_\delta$ consists of all the pairs $(x,y)\in M\times M$ with $d(x,y)<\delta$. Since $d(x,x)=0$ each of these will contain the diagonal. The intersection $E_\delta\cap E_{\delta'}$ is the entourage corresponding to the smaller of $\delta$ and $\delta'$. Each $E_\delta$ is its own reflection by the symmetry of the distance function. And the triangle inequality gives our half-size entourages — if $(x,y)\in E_{\frac{\delta}{2}}$ and $(y,z)\in E_{\frac{\delta}{2}}$ then $(x,z)\in E_\delta$.
For the real numbers themselves we should verify that we get back the same uniform structure as we did before. Remember that the uniform structure we got from completing the uniform structure on the rational numbers had an entourage $E_r$ for each positive $r\in\mathbb{Q}$, with $E_r=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\left|x-y\right|^lt;r$. Each one of these shows up in the entourages for the metric structure, by considering $r$ as a positive real number, but does every basic entourage from the metric structure show up as an entourage in the complete uniform structure? It does! The Archimedean property tells us that for any positive $\delta$ we can find a positive rational number $r<\delta$. Then $E_r\subseteq E_\delta$, and so $E_\delta$ is an entourage in the completion of the uniform structure on the rationals.
Let’s look at the neighborhood structure we get from the entourages of the metric structure. A subset $U$ is a neighborhood of $x$ if and only if it contains $E_\delta\left[x\right]$ for some $\delta>0$. That is, it must contain the “open ball” of all $y\in X$ such that $\left|x-y\right|<\delta$.
In $\mathbb{R}$ this means that we have a neighborhood base for each point $x$ consisting of the intervals $(x-\delta,x+\delta)=\{y\in\mathbb{R}|\left|x-y\right|<\delta\}$. Thus a subset $U$ of $\mathbb{R}$ will be open if and only if it contains such a symmetric neighborhood of each of its points, and this will happen if and only if $U$ is the union of a collection of open intervals. Then we can take the intervals $(a,b)=\{x\in\mathbb{R}|a<x<b\}$ as a base for our topology.
As a final coup de grâce, let’s write down explicitly the condition that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous map. We have a neighborhood base of our topology, and we know we only need to check the neighborhood definition of continuity on a neighborhood base.
So, a function $f$ will be continuous at $x$ if and only if for each neighborhood $V\in\mathcal{N}(f(x))$ there is a neighborhood $U\in\mathcal{N}(x)$ with $f(U)\subseteq V$. Translating this all into our explicit language for the real numbers and restricting to neighborhood bases says that a function $f$ is continuous at $x$ if and only if for each $\epsilon>0$ there is a $\delta>0$ so that $\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon$. And we’re back to the old definition of continuity from calculus 1! Then, as usual, we say that $f$ is continuous if the above condition holds for all $x\in\mathbb{R}$.
What about uniform continuity. We can again translate the statements to our special case and check them on the basic entourages. A function $f:\mathbb{R}\rightarrow\mathbb{R}$ will be uniformly continuous if for every $\epsilon>0$ there is a $\delta>0$ so that for all $x,y\in\mathbb{R}$, $\left|x-y\right|<\delta$ implies that $\left|f(x)-f(y)\right|<\epsilon$.
Notice particularly the difference between uniform continuity and continuity. Continuity says that (for all $x\in\mathbb{R}$) (for all $\epsilon>0$ there exists a $\delta>0$) such that ($\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon$). Uniform continuity says that (for all $\epsilon>0$ there exists a $\delta>0$) such that (for all $x\in\mathbb{R}$) ($\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon$). The quantifier for $x$ shows up after the quantifier for $\delta$ in the latter definition. That is, for a uniformly continuous function we can pick the $\delta$ uniformly to apply to all points $x$, while for a merely continuous function we may have to use a different $\delta$ for each point $x$. At first it doesn’t seem to be that big a deal, which always causes a certain amount of confusion in an advanced calculus (undergraduate real analysis) class, but it turns out that being able to choose the same $\delta$ at every point makes a lot of nice things work out that don’t otherwise hold.
[UPDATE]: I’m feeling a little silly that I didn’t mention this before, but the last two definitions immediately port over to any function between metric spaces $X$ and $Y$ by just using the local definitions of “distance” in place of that for $\mathbb{R}$. A function $f:X\rightarrow Y$ is continuous if for all $x\in X$ and $\epsilon>0$ there is a $\delta>0$ so that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$. Similarly, $f$ is uniformly continuous if for all $\epsilon>0$ there is a $\delta>0$ so that for all $x\in X$ $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$.
### Like this:
Posted by John Armstrong | Topology
## 16 Comments »
1. John, could you correct some typos in UPDATE (and also in the text). Because the variable y looks like as a free variable in definitions of the (uniform) continuity.
–
Z. Karno
Comment by Zbigniew Karno | December 11, 2007 | Reply
2. There’s no typo there. The quantifier on $y$ is implicit, just as it is for the definition of continuity, and both definitions in the case of $\mathbb{R}$. I’m not writing well-formed formulæ in a predicate calculus here, and it’s common to informally let a variable range over whatever values make sense. I’m sure I’ve done it a dozen other places before this.
Comment by | December 11, 2007 | Reply
3. [...] to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same [...]
Pingback by | January 3, 2008 | Reply
4. [...] over at The n-Category Café reminded me of an interesting fact I haven’t mentioned yet: a metric space is actually an example of an enriched [...]
Pingback by | February 11, 2008 | Reply
5. [...] if for every there is a so that implies . But this talk of and is all designed to stand in for neighborhoods in a metric space. Picking a defines a neighborhood of the point . All we need is to come up with a notion of a [...]
Pingback by | April 19, 2008 | Reply
6. [...] know about products of topological spaces. We can take products of metric spaces, too, and one method comes down to us all the way from [...]
Pingback by | August 19, 2008 | Reply
7. [...] give the answer to the problem of pointwise convergence. It’s analogous to the notion of uniform continuity in a metric space. In that case we noted that things became nicer if we could choose our the same for every point, [...]
Pingback by | September 5, 2008 | Reply
8. [...] vector which points from one to the other. But a notion of distance is captured in the idea of a metric! So whatever a norm is, it should give rise to a metric by defining the distance as the norm of [...]
Pingback by | April 21, 2009 | Reply
9. [...] There’s an interesting little identity that holds for norms — translation-invariant metrics on vector spaces over or — that come from inner products. Even more interestingly, it [...]
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10. [...] it turns out that is a metric space, so all of the special things we know about metric spaces can come into play. Indeed, inner [...]
Pingback by | September 15, 2009 | Reply
11. [...] real spaces in hand, we can discuss continuous functions between them. Since these are metric spaces we have our usual definition with and and all that: A function is continuous at if and only if [...]
Pingback by | September 16, 2009 | Reply
12. [...] provides a nice way to restate our condition for continuity, and it works either using the metric space definition or the neighborhood definition of continuity. I’ll work it out in the latter case for [...]
Pingback by | December 7, 2009 | Reply
13. [...] as Metric It turns out that a measure turns its domain into a sort of metric space, measuring the “distance” between two sets. So, let’s say is a measure on an [...]
Pingback by | March 24, 2010 | Reply
14. [...] if we define a norm we get a “normed vector space”. This is a metric space, with a metric function defined by . This is nice because metric spaces are first-countable, and [...]
Pingback by | May 12, 2010 | Reply
15. [...] Metric Space of a Measure Ring Let be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that be positive [...]
Pingback by | August 6, 2010 | Reply
16. [...] The metric only tells us about the lengths of tangent vectors; it is not a metric in the sense of metric spaces. However, if two curves cross at a point we can use their tangent vectors to define the angle [...]
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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://mathhelpforum.com/geometry/125786-couple-problems-radius-segment.html
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# Thread:
1. ## A couple of problems, radius and segment
Aijaa.com - 5653302.png
The first problem is a barrel, whose volume is the only thing I know about it and I gotta solve it's measurements. There is a tip that it is a square when looked far from the side of the barrel (both height and width are the same). I have no clue about this one..
The second is a long swimming pool and I need to calculate the volume of the water in the pool, although my drawing skills aren't the greatest, I try my best, so: the pool is 4,8 metres long, it's the size of a half circle and it's radius (r) is 0,18m and diameter is 0,36m. So I need to calculate a segment of the circle first but I don't know how to do that, so please help me?
Thank you in advance, and this time I read the description twice so I didn't miss anything relevant And also a 7 hour deadline on this :X
EDIT: The asnwers are as following: Problem one's measurements are 6,8m x 6,8m and the pool's volume is 96 liters, I just need to know how to get those results..
2. Originally Posted by mazzuli
Aijaa.com - 5653302.png
The first problem is a barrel, whose volume is the only thing I know about it and I gotta solve it's measurements. There is a tip that it is a square when looked far from the side of the barrel (both height and width are the same). I have no clue about this one..
The barrel, then, is a cylinder having h equal to diameter or twice the radius. Since the volume of a cylinder is $\pi r^2 h$, the volume of this cylinder is $\pi r^2 (2r)= 2\pi r^3$.
The second is a long swimming pool and I need to calculate the volume of the water in the pool, although my drawing skills aren't the greatest, I try my best, so: the pool is 4,8 metres long, it's the size of a half circle and it's radius (r) is 0,18m and diameter is 0,36m. So I need to calculate a segment of the circle first but I don't know how to do that, so please help me?
What do you mean by "its size is a half circle"? What is its shape? If a half circle, what measurement of the circle is its length od 4,8 m.?
f
Thank you in advance, and this time I read the description twice so I didn't miss anything relevant And also a 7 hour deadline on this :X
EDIT: The asnwers are as following: Problem one's measurements are 6,8m x 6,8m and the pool's volume is 96 liters, I just need to know how to get those results..
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http://math.stackexchange.com/questions/211129/how-to-compute-the-upper-incomplete-gamma-function-in-matlab
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# How to compute the upper incomplete gamma function in MATLAB?
I need to compute the upper incomplete gamma function $$\Gamma(0,i\pi K)$$ where $i=\sqrt{-1}$ and K is real.
I have some basic idea about what this "special function" is, but am not used to using it. I am confused because MATLAB refuses to compute this for me - I get an error message says that inputs must be real.
The CAS wxMaxima computes it no problem with or without complex or inputs. How can I implement this in MATLAB?
In MATLAB the command I am using is `1 - gammainc(a,z)`
In wxMaxima the command is `gamma_incomplete(a,z)` or `gamma_incomplete_regularized(a,z)`.
EDIT: What I mean to say here is that, in MATLAB, I want to get the same result that gamma_incomplete(a,z) gives in wxMaxima. For example, in wxMaxima the command `gamma_incomplete(0,%i*%pi);` gives the output `0.28114072518757*%i-0.073667912046426` Now what do I have to do in MATLAB to get the same result?
-
## 1 Answer
Your inputs are the wrong way round. Additionally, you should specify that you want the upper tail of the gamma function, rather than subtracting from 1. This gives more accurate values in the case where the real part of $\Gamma(x,a)$ is near to 1.
On Wolfram Alpha you have
$$\Gamma(1,i) = 0.5403 - 0.8415i$$
and in Matlab
````>> gammainc(1i,1,'upper')
ans =
0.5403 - 0.8415i
````
which matches up.
-
This is what I get when I enter pretty much the exact same thing. >> gammainc(i*pi,1,'upper') ??? Error using ==> gammainc Inputs must be real, full, and double or single. – ben Oct 11 '12 at 15:15
On my machine that returns `-1.0000 + 0.0000i`. What version of Matlab do you have? Have you checked that you don't have a variable called `i` in your environment? It's always good practice to declare complex numbers with the `1+2i` format, rather than using `i` on its own. It prevents errors where you accidentally overwrite `i` with some other value (e.g. if you use it in a loop counter). In this example you could write `1i*pi` instead of `i*pi`. – Chris Taylor Oct 11 '12 at 15:17
Version 7.10 R2010a. When I enter "i" in the command line I get the following, so I think the program recognizes it as such: >> i ans = 0 + 1.000000000000000i – ben Oct 11 '12 at 15:20
If you type `edit gammainc` and look at the last line of the header comment to get the revision number, what do you see? For example, I see `% $Revision: 5.17.4.3 $ $Date: 2004/07/05 17:02:00 $`. – Chris Taylor Oct 11 '12 at 15:22
I get: % $Revision: 5.17.4.5$ $Date: 2008/10/31 06:20:55$ – ben Oct 11 '12 at 15:23
show 5 more comments
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http://mathoverflow.net/questions/92013?sort=oldest
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## Willmore minimizers for genus $\geq 2$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
For an immersed closed surface $f: \Sigma \rightarrow \mathbb R^3$ the Willmore functional is defined as $$\cal W(f) = \int _{\Sigma} \frac{1}{4} |\vec H|^2 d \mu_g,$$ where $\vec H$ is the mean curvature vector in $\mathbb R^3$and $g$ is the induced metric.
If $\Sigma$ is closed we have the estimate $$\cal W(f) \geq 4 \pi$$ with equality only for $f$ parametrizing a round sphere.
Recently, the Willmore conjecture was proved (the paper can be found on arxiv), which states that for closed surfaces $\Sigma$ of genus $g \geq 1$ this estimate can be improved: $$\cal W(f) \geq 2 \pi^2$$ with equality only for the Cilfford torus.
Are there any conjectures about the minimizers in the case of genus $g \geq 2$? And what happens if we consider surfaces immersed in some $\mathbb R^n$ instead of $\mathbb R ^3$?
-
Can you give an explici citing of the archiv paper? – drbobmeister Mar 23 2012 at 18:39
1
It is the following paper by Marques and Neves: arXiv:1202.6036 – Sebastian Mar 23 2012 at 18:52
## 2 Answers
First of all, by a result of Bauer and Kuwert, there exists a smooth minimizer of the Willmore functional in the class of compact surfaces with fixed genus g, for any g. They have Willmore functional below $8\pi$ and by a result of Kuwert, Li and Schaetzle, the Willmore functional of the minimzers for genus $g$ tends to $8\pi$ when $g$ goes to infinity. Not much more is known about higher genus surfaces, but there is a vague conjecture, that the minimzers are the so called Lawson surface $\xi_{g,1}.$
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The paper of Lawson where $\xi_{g,1}$ was introduced is here: jstor.org/stable/1970625 – YangMills Mar 23 2012 at 21:53
### 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 remember there is a paper by Kusner named: comparison surfaces for the Willmore problem in which the author conjectured that the Lawson surface(see Sebastian's answer) minimizes the Willmore energy of genus g surface. For surfaces immersed in R^n, it is also conjectured the Clifford torus should be the minimizer, but it seems to me that this is still an open question.
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http://mathhelpforum.com/differential-geometry/79032-real-analysis-test-question.html
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# Thread:
1. ## real analysis test question
Suppose that lim s(sub n) =s, with s >0. Prove that there exists an N in R (real numbers) such that s(sub n) >0 for all n in N.
2. Hello,
Originally Posted by trojanlaxx223
Suppose that lim s(sub n) =s, with s >0. Prove that there exists an N in R (real numbers) such that s(sub n) >0 for all n in N.
$\lim s_n=s$ means that :
$\forall \epsilon>0, \exists N \in \mathbb{N}, \forall n\geq N, |s_n-s|< \epsilon$
Now take $\epsilon$ such that $s-\epsilon$ is positive. You can always find one, such as $\epsilon=\frac s2$
From the inequality $|s_n-s|<\epsilon$, we can say that $s-\epsilon<s_n<s+\epsilon$
Since we defined $\epsilon$ such that $s-\epsilon>0$, it's easy to conclude.
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http://quomodocumque.wordpress.com/2013/01/14/idle-question-cluster-algebras-over-finite-fields-and-spectral-gaps/?like=1&source=post_flair&_wpnonce=8b2dcf09a6
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# Quomodocumque
## Idle question: cluster algebras over finite fields and spectral gaps
Yet another great talk at the JMM: Lauren Williams gave an introduction to cluster algebras in the Current Events section which was perfect for people, like me, who didn’t know the definition. (The talks by Wei Ho, Sam Payne, and Mladen Bestvina were equally good, but I don’t have any idle questions about them!)
This post will be too long if I try to include the definitions myself, and I wouldn’t do as good a job of exposition as Williams did, so it’s good news that she’s arXived a survey paper which covers roughly the same ground as her talk.
Setup for idle question: you can get a cluster algebra comes from a process called “seed mutation” — given a rational function field K = k(x_1, … x_m), a labelled seed is a pair (Q,f) where Q is a quiver on m vertices and f = (f_1, … f_m) is a labelling of the vertices of Q with rational functions in K. For each i, there’s a seed mutation mu_i which is an involution on the labelled seeds; see Williams’s paper for the definition.
Now start with a labelled seed (Q,(x_1, … x_m)) and let T be the set of labelled seeds obtainable from the starting seed by repeated application of seed mutations mu_1, …. m_n for some n < m. (I didn’t think carefully about the meaning of this special subset of n vertices, which are called the mutable vertices.)
It’s called T because it’s a tree, regular of degree n; each vertex is indexed by a word in the n seed mutations with no letter appearing twice in succession.
Anyway, for each vertex of T and each mutable vertex i you have a rational function f_i. The cluster algebra is the algebra generated by all these rational functions.
The great miracle — rather, one of the great miracles — is that, by a theorem of Fomin and Zelevinsky, the f_i are all Laurent; that is, their denominators are just monomials in the original functions x_i.
We are now ready for the idle question!
Let’s take k to be a finite field F_q, and let U be (F_q^*)^m, the rational points of the m-torus over F_q. Choose a point u = (u_1, … u_n) in (F_q^*)^m.
Then for any vertex of T, we can (thanks to the Laurent condition!) evaluate the functions (f_1, …. f_m) at u, getting an element of F_q^m.
So a random walk on the tree T, combined with evaluation at u, gives you a random walk on F_q^m.
Idle question: Is there a spectral gap for this family of walks, independent of q?
Update: As David Speyer explains in the comments, this finite random walk is not in general well-defined. Let me try another phrasing which I think makes sense.
Let t be the endpoint of a length-R random walk on T; then evaluation at (1,1,..1) gives a probability distribution P_{R,N} on (Z/NZ)^m. Let U_N be the uniform distribution on (Z/NZ)^m. Now for each N we can ask about the limit
$\Lambda_N = \lim_{R \rightarrow \infty} ||P_{R,N} - U_{N}||^{1/R}$
(I don’t think it matters what norm we use.)
The idea is that the random walk on the tree should be equidistributing mod N, and the speed of convergence is governed by Λ_N. Then we can ask
Idle question mark 2: Is Λ_N bounded away from 1 by a constant independent of N?
This is a question in “spectral gap” style, which, if I did it right, doesn’t a priori have to do with a sequence of finite graphs.
Motivation: this setup reminds me of a very well-known story in arithmetic groups; you have a discrete group Gamma which comes equipped with an action on an set of objects “over Z” — then reducing mod p for all p gives you a family of actions of Gamma on larger and larger finite sets, and a critical organizing question is: do the corresponding graphs have a spectral gap?
For that matter, what happens if you, say, keep k = C and then evaluate your f_i at (1,1,… 1)? Looking at a bigger and bigger ball in the tree you get bigger and bigger sets of elements of C^m; what do these look like? Do they shoot off to infinity, accumulate, equidistribute…..?
## 9 thoughts on “Idle question: cluster algebras over finite fields and spectral gaps”
1. David Speyer says:
I think studying dynamical systems in cluster algebras is interesting, but I think the precise question about $F_q$ doesn’t work. Your random walk isn’t well defined: If one of the coordinates of your point in $F_q^m$ becomes zero, then we can’t divide by it at the next step.
For example, I took the cluster mutation $(x,y) \mapsto (y, (y^2+1)/x)$. (This is $m=2$, so the tree is just a path; this is cluster type $\tilde{A}_1$.) We can rewrite this as studying the recursion $x_{n+1} = (x_n^2+1)/x_{n-1}$.
Take $q = 5$ and look at the orbit of $(1,1)$. Formally plugging into the Laurent polynomials gives $1, 1, 2, 0, 3, 4, 4, 1, 1, \ldots$ repeating periodically. But directly using the recursion is undefined at $(3^2+1)/0$; you need to look at the general Laurent formulas to resolve this $0/0$ quantity.
2. JSE says:
Exactly! Sorry, I have been chewing on this a bit today but didn’t want to modify the post until I understood things better.
You need to do it over Q_p, and then the Laurent condition tells you that if you start at (1,1) [or anything in Z_p^* x Z_p^*] your orbit stays in Z_p^2, at which point you can reduce mod p. But as you say, it’s not literally a dynamical system on F_p^2. Or rather, it’s not a priori a dynamical system on F_p^2. But note that in the case you write down over F_5, you are indeed tracing a deterministic process on F_p^2; it’s just that the transition rule (0,3) -> (3,4) isn’t specified directly by the cluster mutation. Are there other orbits which contain (for instance) one consecutive sequence reducing to (0,3,4), and another reducing to (0,3,1)? I don’t immediately see why not. I think this is related to the question of whether the entries of the sequence are bounded away from 0.
The (highly preliminary) computations I’ve been doing on some rank-2 cluster algebras suggest that the bounded orbits in this case are periodic mod p^k for every k, have coordinates bounded away from 0, and have closure of Minkowski dimension 1.
Look forward to hearing further thoughts!
3. David Speyer says:
“The (highly preliminary) computations I’ve been doing on some rank-2 cluster algebras suggest that the bounded orbits in this case are periodic mod p^k for every k, have coordinates bounded away from 0, and have closure of Minkowski dimension 1. ”
Kedlaya and Propp did a bunch of computations like this in http://arxiv.org/abs/math/0409535 . You might want to see how your results relate to theirs.
The example I ran is very special because there is a conserved quantity: $(x_{n-1}^2+x_{n}^2+1)/(x_{n-1} x_n)$. So the orbits do not only have Minkowski dimension $\leq 1$, they have Zariski dimension $\leq 1$!
My understanding (but don’t quote me!) is that there is no analogous rational conserved quantity in the “wild” rank two cases. (A rank $2$ cluster algebra is given by a $B$-matrix of the form $\left( \begin{smallmatrix} 0 & b \\ -c & 0 \end{smallmatrix} \right)$. Finite type is $bc \leq 3$, affine type is $bc=4$ and wild type is $bc \geq 5$.) It would be very interesting if the $p$-adic orbit closures were level sets of some sort of non-algebraic conserved quantity.
It would also be interesting to take a look at the “Markov” cluster algebra. The recurrence in question is to start with $(x,y,z)$ and mutate one of the three variables to produce $(\frac{x^2+y^2}{z}, y, z)$. In this example, the tree never folds back on itself, so we could ask more interesting questions about probabilistic distribution along the orbit closures. Here again, there is a conserved quantity: $(x^2+y^2+z^2)/(xyz)$.
In all of the examples we’ve brought up, mutation gives us an isomorphic seed. I think I do not know how to ask the right questions if mutation keeps changing the $B$-matrix. For example, suppose that we start with the $B$-matrix $\left( \begin{smallmatrix} 0 & 3 & -3 \\ -3 & 0 & 3 \\ 3 & -3 & 0 \end{smallmatrix} \right)$. So the first round of mutation changes the signs of two of the $3$‘s and turns the third one into a $6$, the next mutation changes the signs of a $3$ and a $6$ and creates a $15$, and so forth. In this setting, I’m not sure that there is a natural question to ask about orbits or random walks. Could you provide one?
There has been a lot of work on integrable systems and cluster algebras, which is probably relevant. Unfortunately, as far as I can tell, there is no good survey, and I’ve had to pick up a lot of the ideas as folklore. I can tell you about some of them, but it might be more fun to just think about examples like the above.
4. JSE says:
Is the modified version of the Idle Question ok for the last example you give?
Also: that Kedlaya-Propp paper you cite does indeed seem very relevant (and it seems that they explicitly say they don’t know whether the stability phenomena they study occur for Fomin-Zelevinsky type sequences, but that this is a reasonable question.)
5. JSE says:
More relevant info: the Somos 4 sequence of integers can be obtained via mutations from a certain 4-vertex quiver:
http://www1.maths.leeds.ac.uk/~marsh/research_articles/pp31.pdf
and, as Lauren Williams pointed out to me, it was proved by Raphael Robinson in 1992 that this sequence is periodic mod N for every N:
http://www.ams.org/journals/proc/1992-116-03/S0002-9939-1992-1140672-5/S0002-9939-1992-1140672-5.pdf
6. David Speyer says:
Somos-4 should be analyzable very explicitly. There are two obvious symmetries: If $s_n$ solves the Somos-4 recursion, so does $a b^n s_n$ for any nonzero $a$ and $b$. There is also a conserved quantity:
$T=\frac{s_{n-1} s_{n+2}}{s_n s_{n+1}} + \frac{s_n^2}{s_{n-1} s_{n+1}} + \frac{s_{n+1}^2}{s_n s_{n+2}} + \frac{s_n s_{n+1}}{s_{n-1} s_{n+2}}$.
If you fix $T$ and scale out by the $\mathbb{G}_m^2$ symmetries, what remains is a genus one curve, and mutation is translation by a point on that curve.
This was all worked out in the language of $\Theta$ functions by Noam Elkies. I prefer the write up of Hone and Swart, which is in more modern algebraic geometry language.
It is my understanding, but I can’t give you a direct citation, that there are a lot of examples like this one. Namely, take any of the cluster algebras from Goncharov and Kenyon arXiv:1107.5588 . They have dimension $2n+m$, an action of $\mathbb{G}_m^m$ and $n$ conserved quantities. The $n$ conserved quantities can be thought of as the coefficients of curve in a toric surface. Fixing all of the conserved quantities gives (at least rationally) the Jacobian of that curve. (More or less — it might be a twist of the Jacobian or a homogeneous space for it.) There is a sequence of mutations which returns the initial quiver to itself, and gives a translation in the Jacobian. All the Gale-Robinson sequences ($s_n s_{n-k} = s_{n-a} s_{n-k+a} + s_{n-b} s_{n-k+b}$) should fall into this setting.
While this is a beautiful subject, it is my understanding that it falls dramatically short of covering all the quivers where a sequence of mutations returns them to itself.
7. David Speyer says:
Regarding the Idle Question when the quiver doesn’t return to itself: I see two problems.
The first problem is the one that occurs in every case: Until you prove some sort of Robbins stability, you have to ask the question $p$-adically and it isn’t clear that there is any finite $p^k$ such that we get a random walk on the $\mathbb{Z}/p^k$ points. I don’t know how to define spectral gaps in this setting, but maybe you do.
But let me brush that problem under the rug to bring up the second problem: Since the quiver is changing at every step, you don’t have a Marov process. (1) The rule to propagate from stage $n$ to stage $n+1$ is not the same as the rule to propagate from stage $n-1$ to $n$ but, even worse (2) Different points in the stage $n$ cloud will be propagating according to different rules, because they will come from different mutation histories that give different quivers.
Now, you can still ask what the Minkowski dimension of the limit points is, and some questions about probability distributions in the limit. But it seems much less natural to me. Specific questions about the spectral gap don’t seem to make sense to me, but maybe you have a broader understanding of spectral gaps than I do.
8. JSE says:
My modification of the Idle Question was meant to get around those problems — does it? My understanding was that, whether there’s Robbins stability or not, and whether the quiver keeps changing or not, you still have a regular tree with (after setting all variables to 1) an m-tuple of integers at each vertex. If that’s right, then I think IQ2 makes sense even though there are no finite Markov processes in the picture. I only say it makes sense, by the way — not that it is natural!
9. David Speyer says:
Ah, I see. This make sense now. In the examples where the quiver does come back to itself and there is a conserved quantity, then the walk doesn’t equidistribute at all, but in the situations where there is no useful structure, I don’t know where to start.
Cancel
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http://mathoverflow.net/revisions/110140/list
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## Return to Question
2 added 44 characters in body
Hallo,
I am reading the paper "Hyperkaehler structures on the total space of holomorphic cotangent bundles" by D.Kaledin and I am asking if it is possible to embedd a real-analytic Kähler manifold, isometrically, as a special Lagrangian in a Calabi-Yau manifold. Acctualy what I am looking for is the following: Start with a compact real-analytic Kähler manifold $(M, I, \omega)$ and in a neigbourhood of the zero-section in the cotangent bundle $T^{*}M$ there should exists a holomorphic $(n,0)-$form $\Omega$ (with respect to some complex structure on this neigbourhood) and a Kaehler form $\tilde{\omega}$ such that the forms $Im(\Omega)$ and $\tilde{\omega}$ vanishes when restricted to $M$ (the zero section) and $\tilde{\omega}^{2n} = C_{n} \Omega \wedge \bar{\Omega}$ for some constant $C_{n}$ that depends only on $n$. I know that one can do this. But I don't know some references where I can find a explanation of this. Is it sufficient just to read the paper of Kaledin or do I have also to switch to other references? By using Kaledin's paper what ingredients are necessary for a proof of this embedding problem? I am a beginner in Calabi-Yau manifolds and Hyperkaeler manifolds and I would be very thankfull if someone has the answers. I hope for a lot of replys and also hope that this question is not too trivial.
Best Regards, Pavel
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# Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold
Hallo,
I am reading the paper "Hyperkaehler structures on the total space of holomorphic cotangent bundles" by D.Kaledin and I am asking if it is possible to embedd a Kähler manifold as a special Lagrangian in a Calabi-Yau manifold. Acctualy what I am looking for is the following: Start with a compact Kähler manifold $(M, I, \omega)$ and in a neigbourhood of the zero-section in the cotangent bundle $T^{*}M$ there should exists a holomorphic $(n,0)-$form $\Omega$ (with respect to some complex structure on this neigbourhood) and a Kaehler form $\tilde{\omega}$ such that the forms $Im(\Omega)$ and $\tilde{\omega}$ vanishes when restricted to $M$ (the zero section) and $\tilde{\omega}^{2n} = C_{n} \Omega \wedge \bar{\Omega}$ for some constant $C_{n}$ that depends only on $n$. I know that one can do this. But I don't know some references where I can find a explanation of this. Is it sufficient just to read the paper of Kaledin or do I have also to switch to other references? By using Kaledin's paper what ingredients are necessary for a proof of this embedding problem? I am a beginner in Calabi-Yau manifolds and Hyperkaeler manifolds and I would be very thankfull if someone has the answers. I hope for a lot of replys and also hope that this question is not too trivial.
Best Regards, Pavel
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http://mathoverflow.net/questions/66046?sort=votes
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## Which nonlinear PDEs are of interest to algebraic geometers and why?
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Motivation
I have recently started thinking about the interrelations among algebraic geometry and nonlinear PDEs. It is well known that the methods and ideas of algebraic geometry have lead to a number of important achievements in the study of PDEs, suffice it to mention the construction of finite-gap solutions to integrable PDEs (see e.g. this book) and the geometric approach to PDEs developed by A.M. Vinogradov et al. which revolves around the concept of diffiety (the word itself was merged from "differential" and "variety") and which the authors themselves consider, at least to some extent, as a "translation" of ideas from algebraic geometry and commutative algebra to the realm of PDEs, see e.g. these two books.
On the other hand, it appears (as far as my googling skills allow me to tell) that the other way around, i.e., in the applications of nonlinear PDEs in algebraic geometry, the interaction is at least somewhat less intense. I was able to come up with basically just two things: the Novikov conjecture (proved by Shiota) on the relation of the KP equation to the Schottky problem and the applications of the Monge-Ampere equations in Kahler geometry.
Question
Which are the other applications of the nonlinear PDEs in (broadly understood) algebraic geometry? In other words, which nonlinear PDEs are of interest to algebraic geometers and why?
EDIT: It should be obvious, but to play it safe I would like to spell this out loud and clear: please feel free to share not only the already known cases where PDEs have helped the algebraic geometers but also the more open-problem-type cases where, say, there is a PDE that could be of use in algebraic geometry but some crucial bits of information about this PDE (for instance, about the existence of solution(s) with the desired properties) are still missing.
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this must be community wiki – Denis Serre May 26 2011 at 14:57
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@Dennis: I believe that this question is not that trivial and easy to answer, even by experts, so I'd like to keep it unwikified to give people an extra incentive (in the form of gaining rep) to answer. Also, I am not convinced that having a sorted list of answers to this particular question is a right thing to do. – mathphysicist May 26 2011 at 17:37
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Denis with one N, please... – Denis Serre May 27 2011 at 8:08
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@Denis: Sorry, just a typo. Mille pardons! – mathphysicist May 27 2011 at 10:36
## 5 Answers
Strictly speaking, this is not meant as an answer to the question---it's more like a suggestion that you might find it interesting to also ask a related, but different, question.
I would say the most interesting feature of many interactions between nonlinear (especially integrable) PDE and algebraic geometry, from the point of view of an algebraic geometer, is that they show us new structures that nobody knew existed classically.
One instance of this is the story that you're referring to relating the conformal field theory of free fermions, KP, and the geometry of moduli of bundles, mediated (geometrically) by the Sato Grassmannian. This is what leads to Shiota's characterization of the Schottky locus. That's well before my time, but I wouldn't have guessed that most classical algebraic geometers thought of this as a very big advance on the Schottky problem at the time: if you'd like equations for the Schottky locus, or even some nice algebro-geometric description of it, you probably aren't satisfied with what Shiota tells you (I'm a huge fan of this story, so my remark isn't meant to convey my own opinion of this work, rather to guess at what some others might have thought). Algebraic geometers were interested in moduli of curves, bundles, etc. long before this story appeared, but a whole new world emerged from it.
In a different direction, the discovery of integrability in topological string theory (as formulated, for example, in Witten's conjecture, later Kontsevich's theorem) shows us striking new structure, again governed by integrable PDE (KdV, n-KdV) in the intersection theory on moduli spaces of curves (and later 2D Toda in the Gromov-Witten theory of curves). Again, I would argue (as someone who, admittedly, is incredibly far from expert in the subject) that the most interesting part of the story is the amazingly rich structure (involving integrable PDE, matrix models, and intersection theory on moduli spaces) that was revealed by these discoveries.
EDIT: What I wrote initially takes a slightly misleading tone about applications back to classical AG: for example, Krichever's stunning work on the trisecant conjecture, which grows directly out of integrable PDE, is surely something of classical interest!
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One important use of PDEs in algebraic geometry is in so-called "Hitchin-Kobayashi correspondences". The original example of this is the following theorem.
Theorem (Donaldson, Uhlenbeck-Yau) Let $L \to X$ be an ample line bundle over a compact complex manifold and $\omega$ a Kähler metric representing $c_1(L)$. A holomorphic vector bundle $E \to X$ is slope polystable with respect to $L$ if and only if it admits a Hermitian-Einstein metric with respect to $\omega$.
To spell this out in detail: the slope of a bundle $\mu(E)$ is its degree divided by its rank, where degree is $\langle c_1(E) \cup c_1(L)^{n-1}, [X] \rangle$. A bundle is slope stable if all proper coherent subsheaves have strictly smaller slope. A bundle is called slope polystable if it is the sum of stable bundles of equal slope. Meanwhile, a Hermitian metric in $E$ is called Hermitian-Einstein if its curvature $F$ satisfies the equation $(F,\omega) = c \cdot \mathrm{Id}$, for a constant $c$. (Here we are taking the innerproduct on 2-forms, the result being an endomorphism of $E$.) This is a non-linear PDE on the Hermitian metric.
Notice that the slope polystability is purely algebraic - it makes no mention whatsoever of the metric $\omega$. What is remarkable is that this is equivalent to the existence of a solution of a non-linear PDE. The story behind this theorem is quite a long one. It can be seen as an infinite dimensional example of the equivalence between quotients via GIT and symplectic quotients. The PDE plays the role of the moment map.
Since this result was proved there have been many other versions involving bundles with additional data, say, e.g., a Higgs field, which appears both in the definition of stability and the corresponding PDE. The corresponding "Hitchin-Kobayashi correspondences" play important roles in the study of the moduli of the algebraic objects. For example the fact that one can always solve the relevant PDE leads directly to an interesting Kähler metric on the moduli space of stable objects. In the case of Higgs fields, this is a hyperkähler metric. This metric is one of the starting points of the approach to geometric Langlands proposed by Kapustin and Witten. There are also other applications of the PDE point of view here leading to, amongst other things, strong restrictions on the fundamental groups of Kähler manifolds. This subject sometimes goes by the name "Non-abelian Hodge theory". (I should stress that this is a long way from my expertise!)
In a different vein, one version of the Hitchin-Kobayashi correspondence (which is still conjectural) concerns not metrics on bundles but rather metrics on the manifold itself. Many people in Kähler geometry are currently working on understanding both the conjecture and its ramifications. The idea (originally due to Yau, later refined by Donaldson and Tian) is that given an ample line bundle $L \to X$, one should be able to find a so-called "extremal" Kähler metric in $c_1(L)$ if and only if the polarised variety is "stable". Here a metric is "extremal" if it's a critical point of the $L^2$-norm of the curvature tensor, restricted to metrics in $c_1(L)$. This turns out to be equivalent to the gradient of the scalar curvature being holomorphic, a sixth-order fully non-linear PDE. "Simple" examples are Kähler-Einstein metrics (when $L$ is a multiple of the canonical bundle). The correct definition of stability, known as K-polystability, is a little too involved to give neatly here, but it is important to mention that, just as for slope polystabilty of a vector bundle, it is a purely algebraic concept. This whole subject is vast, and I could write about it for pages and pages, but I've probably already said too much for one answer!
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Thanks a lot! Could you recommend some references where the PDEs in question are written down explicitly so that the PDE people could have a first look at them without having to dig through all of the algebro-geometric machinery? – mathphysicist Jun 6 2011 at 17:15
@mathphysicist: Sorry to take so long to get back to you. Good places to read about this are the two articles of Donaldson: "A new proof of a theorem of Narasimhan and Seshadri" in the Journal of Differential Geometry and "Anti-self-dual connections over complex algebraic surfaces and stable vector bundles" in Proceedings of the London Mahtematical Society. Another great article is "The Yang-Mills equations over Riemann surfaces" by Atiyah and Bott, in, I think, Transactions of the Royal Society. – Joel Fine Jun 10 2011 at 13:46
It's a bit embarrassing to be giving this answer when there are many others far better qualified, but: in connection with the minimal model program, there have been efforts to understand canonical models by producing canonical singular Kahler-Einstein metrics via Kahler-Ricci flow. See for example http://arxiv.org/pdf/math.AG/0603064.pdf.
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The complex Monge-Ampère equation gives rise to remarkable metrics on complex line bundles. In that respect, the theorem of Calabi-Yau has very important applications in algebraic variety, particularly for understanding the geometry of complex algebraic varieties with trivial canonical class.
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This is a bit borderline. But the notion of hyperbolicity, which is associated to the well-posedness of the Cauchy problem, led Garding to introduce the concept of hyperbolic polynomials. The theory of HPs is an important part of so-called real algebraic geometry. In particular, the study of lacunas of the fundamental solution of a hyperbolic operator is intimately related to the algebraic topology of real algebraic surfaces.
Edit. A remarkable result about HPs (Helton, Vinnikov, CPAM 2009, conjectured by P. Lax in 1972): let $P(T,X,Y)$ be a homogeneous polynomial of degree $n$, hyperbolic in the $T$-direction. Then there exist Hermitian matrices $H,K$ such that $P(T,X,Y)=c\det(TI_n+XH+YK)$. This is obviously false for polynomials in more than $3$ variables, for instance for $T^2-X^2-Y^2-Z^2$.
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Thanks, Denis! Could you please suggest a good introductory book or survey article on the subject? – mathphysicist May 27 2011 at 15:16
Maybe this Séminaire Bourbaki article of Atiyah: numdam.org/item?id=SB_1966-1968__10__87_0 ? – Michael May 28 2011 at 11:39
Thanks for the references once more! – mathphysicist May 28 2011 at 14:00
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http://www.aimath.org/textbooks/beezer/JCFsection.html
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Jordan Canonical Form
We have seen in Section IS:Invariant Subspaces that generalized eigenspaces are invariant subspaces that in every instance have led to a direct sum decomposition of the domain of the associated linear transformation. This allows us to create a block diagonal matrix representation (Example ISMR4, Example ISMR6). We also know from Theorem RGEN that the restriction of a linear transformation to a generalized eigenspace is almost a nilpotent linear transformation. Of course, we understand nilpotent linear transformations very well from Section NLT:Nilpotent Linear Transformations and we have carefully determined a nice matrix representation for them.
So here is the game plan for the final push. Prove that the domain of a linear transformation always decomposes into a direct sum of generalized eigenspaces. We have unravelled Theorem RGEN at Theorem MRRGE so that we can formulate the matrix representations of the restrictions on the generalized eigenspaces using our storehouse of results about nilpotent linear transformations. Arrive at a matrix representation of any linear transformation that is block diagonal with each block being a Jordan block.
## Generalized Eigenspace Decomposition
In Theorem UTMR we were able to show that any linear transformation from $V$ to $V$ has an upper triangular matrix representation (Definition UTM). We will now show that we can improve on the basis yielding this representation by massaging the basis so that the matrix representation is also block diagonal. The subspaces associated with each block will be generalized eigenspaces, so the most general result will be a decomposition of the domain of a linear transformation into a direct sum of generalized eigenspaces.
Theorem GESD (Generalized Eigenspace Decomposition) Suppose that $\ltdefn{T}{V}{V}$ is a linear transformation with distinct eigenvalues $\scalarlist{\lambda}{m}$. Then
\begin{align*} V&= \geneigenspace{T}{\lambda_1}\ds \geneigenspace{T}{\lambda_2}\ds \geneigenspace{T}{\lambda_3}\ds \cdots\ds \geneigenspace{T}{\lambda_m} \end{align*}
Proof.
Besides a nice decomposition into invariant subspaces, this proof has a bonus for us.
Theorem DGES (Dimension of Generalized Eigenspaces) Suppose $\ltdefn{T}{V}{V}$ is a linear transformation with eigenvalue $\lambda$. Then the dimension of the generalized eigenspace for $\lambda$ is the algebraic multiplicity of $\lambda$, $\dimension{\geneigenspace{T}{\lambda_i}}=\algmult{T}{\lambda_i}$.
Proof.
## Jordan Canonical Form
Now we are in a position to define what we (and others) regard as an especially nice matrix representation. The word "canonical" has at its root, the word "canon," which has various meanings. One is the set of laws established by a church council. Another is a set of writings that are authentic, important or representative. Here we take it to mean the accepted, or best, representative among a variety of choices. Every linear transformation admits a variety of representations, and we will declare one as the best. Hopefully you will agree.
Definition JCF (Jordan Canonical Form) A square matrix is in Jordan canonical form if it meets the following requirements:
1. The matrix is block diagonal.
2. Each block is a Jordan block.
3. If $\rho < \lambda$ then the block $\jordan{k}{\rho}$ occupies rows with indices greater than the indices of the rows occupied by $\jordan{\ell}{\lambda}$.
4. If $\rho=\lambda$ and $\ell < k$, then the block $\jordan{\ell}{\lambda}$ occupies rows with indices greater than the indices of the rows occupied by $\jordan{k}{\lambda}$.
Theorem JCFLT (Jordan Canonical Form for a Linear Transformation) Suppose $\ltdefn{T}{V}{V}$ is a linear transformation. Then there is a basis $B$ for $V$ such that the matrix representation of $T$ with the following properties:
1. The matrix representation is in Jordan canonical form.
2. If $\jordan{k}{\lambda}$ is one of the Jordan blocks, then $\lambda$ is an eigenvalue of $T$.
3. For a fixed value of $\lambda$, the largest block of the form $\jordan{k}{\lambda}$ has size equal to the index of $\lambda$, $\indx{T}{\lambda}$.
4. For a fixed value of $\lambda$, the number of blocks of the form $\jordan{k}{\lambda}$ is the geometric multiplicity of $\lambda$, $\geomult{T}{\lambda}$.
5. For a fixed value of $\lambda$, the number of rows occupied by blocks of the form $\jordan{k}{\lambda}$ is the algebraic multiplicity of $\lambda$, $\algmult{T}{\lambda}$.
Proof.
Before we do some examples of this result, notice how close Jordan canonical form is to a diagonal matrix. Or, equivalently, notice how close we have come to diagonalizing a matrix (Definition DZM). We have a matrix representation which has diagonal entries that are the eigenvalues of a matrix. Each occurs on the diagonal as many times as the algebraic multiplicity. However, when the geometric multiplicity is strictly less than the algebraic multiplicity, we have some entries in the representation just above the diagonal (the "superdiagonal"). Furthermore, we have some idea how often this happens if we know the geometric multiplicity and the index of the eigenvalue.
We now recognize just how simple a diagonalizable linear transformation really is. For each eigenvalue, the generalized eigenspace is just the regular eigenspace, and it decomposes into a direct sum of one-dimensional subspaces, each spanned by a different eigenvector chosen from a basis of eigenvectors for the eigenspace.
Some authors create matrix representations of nilpotent linear transformations where the Jordan block has the ones just below the diagonal (the "subdiagonal"). No matter, it is really the same, just different. We have also defined Jordan canonical form to place blocks for the larger eigenvalues earlier, and for blocks with the same eigenvalue, we place the bigger ones earlier. This is fairly standard, but there is no reason we couldn't order the blocks differently. It'd be the same, just different. The reason for choosing some ordering is to be assured that there is just one canonical matrix representation for each linear transformation.
Example JCF10: Jordan canonical form, size 10.
## Cayley-Hamilton Theorem
Jordan was a French mathematician who was active in the late 1800's. Cayley and Hamilton were 19th-century contemporaries of Jordan from Britain. The theorem that bears their names is perhaps one of the most celebrated in basic linear algebra. While our result applies only to vector spaces and linear transformations with scalars from the set of complex numbers, $\complexes$, the result is equally true if we restrict our scalars to the real numbers, $\real{\null}$. It says that every matrix satisfies its own characteristic polynomial.
Theorem CHT (Cayley-Hamilton Theorem) Suppose $A$ is a square matrix with characteristic polynomial $\charpoly{A}{x}$. Then $\charpoly{A}{A}=\zeromatrix$.
Proof.
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http://mathoverflow.net/questions/92618?sort=oldest
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## What is the original statement of Jung-Abhyankar theorem?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I can find many modification of the Jung-Abhyankar theorem. I can even find a new proof of the theorem (by K. Kiyek and J. L. Vicente). But I cannot find the original statement. Does any one know which paper/book is it in? Where can I access it?
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## 2 Answers
The original papers are accessible online:
• H. W. E. Jung, Darstellung der Funktionen eines algebraischen Körpers zweier unabhängigen Veränderlichen x, y in der Umgebung einer Stelle x = a, y = b. Journal für die reine und angewandte Mathematik 133, 289-314 (1908)
• S. S. Abhyankar, On the ramification of algebraic functions. Amer. J. Math. 77 (1955), 575–592.
Jung's paper is devoted exactly to this result, whereas Abhyankar's gives a more contextualized explanation (his was an unsuccessful attempt to pass to characteristic $p$); I think the version of Abhyankar-Jung in Abhyankar is Theorem 3 (but it might be worth studying the paper carefully).
In Jung's statement, which I reproduce here, the field $K$ is defined as $K=\mathbb{C}(x,y)[z]/(f)$ for some irreducible polynomial $f$ (which is implicitly assumed to involve all three variables).
Man kann Funktionenpaare $u, v$ des Körpers $K$ bestimmen derart, daß $x$ und $y$ gewönliche Potenzreihen von $u$, $v$ werden, die für $u=v=0$ verschwinden, während alle anderen Funktionen von $K$ entweder gewöhnliche Potenzreihen von $u, v$ werden, oder Quotienten solcher. Eine endliche Anzahl solcher Funtionenpaare und Entwicklungen genügt, die Funktionen von $K$ für die ganze Umgebung von $x=0, y=0$ darzustellen.
My translation:
It is possible to determine pairs of functions $u, v \in K,$ such that $x$ and $y$ become usual power series in $u, v$, vanishing for $u=v=0$, while every other function in $K$ is either a usual power series in $u,v$ or a quotient of such. A finite number of such pairs and series is enough to represent all functions of $K$ in a neighborhood of $x=0, y=0$.
This seems to be equivalent, in the formulation usual in more recent papers, to the following (I use $\mathbb{C}\{x\}$ to denote convergent power series):
Let $f\in\mathbb{C}\{x,y\}[z]$ be a monic irreducible Weierstrass polynomial having a discriminant of the form $x^\alpha y^\beta u$, with $\alpha, \beta$ nonnegative integers, and $u\in \mathbb{C}\{x,y\}$ a unit. Then there exist positive integers $n, m$ such that $f$ has all its roots in $\mathbb{C}\{x^{\frac{1}{n}},y^{\frac{1}{m}}\}$.
Abhyankar considers the case of $n$ variables over an algebraically closed field of characteristic zero.
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K may be not a field.I think that you mean qutient field of K – gauss Mar 30 2012 at 13:54
Thanks! The notation did not make much sense, really. – quim Mar 30 2012 at 16:56
Abhyankar's approach seems purely formal. Does the result tell us anything about the convergence of the fractional power series? – ssquidd Apr 1 2012 at 15:09
Abhyankar's motivation (Zariski's!) was trying to generalize to positive characteristic, so he probably didn't care much for convergence. Jung does care about convergence, and I suppose the series are convergent for more variables too, one should look at modern proofs I guess. – quim Apr 10 2012 at 9:21
### 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 fractional power series solutions remain convergent in several variables. In fact, if you replace the ring of convergent power series by any Henselian ring $H$ of power series series satisfying some stability properties, then the fractional power series solution remain in the ring. See for instance http://arxiv.org/abs/1103.2559
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http://meta.cstheory.stackexchange.com/questions/300/how-to-ask-a-good-question/318
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# How to ask a good question
The moderators spend considerable time on editing badly written questions. A possible way for reducing the number of badly written questions is writing down some guidelines (similar to the MO guidelines) on how to ask/write a good question, and possibly add a link to it on the top of the "ask question" page.
This thread is for discussing related issues, specially what should be included in our guidelines?
Related: http://meta.cstheory.stackexchange.com/q/295/186
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no need: on meta it doesn't matter so much anyway. – Suresh Venkat♦ Sep 13 '10 at 21:31
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@Suresh: I think CW is better for threads about formulating guidelines, because we want to encourage people to edit each others texts. – Charles Stewart Sep 14 '10 at 19:23
I agree with Charles – Ryan Williams Sep 14 '10 at 20:30
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I have just realized that there is a boilerplate page called How to Ask and the system FAQ links there. That page is pretty good, although (naturally) it does not cover the issues specific to cstheory.stackexchange.com. – Tsuyoshi Ito Apr 26 '11 at 14:45
@Tsuyoshi, that's nice. Should we add a link to it inside the question? I can also edit the editable part of the site's FAQ to add a link to it increasing the visibility. – Kaveh♦ Apr 27 '11 at 5:58
@Kaveh: From here, probably. From the system FAQ, I am not sure. The system is already long, and adding a link from there may not be very effective (and there is a link from there already). – Tsuyoshi Ito Apr 27 '11 at 10:12
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@Suresh: Not sure what to do, but your first comment now looks as if you are saying “No need to ask a good question.” :) – Tsuyoshi Ito Feb 7 at 18:08
ah. I think it was about whether to make it CW. – Suresh Venkat♦ Feb 7 at 21:22
## 12 Answers
### Come back soon, check if there are comments, and fix your question
No drive-by asking, please; nurture your question. Use other people’s comments as a hint to improve your question (instead of merely posting a reply in another comment). Do not post a question and then leave for a two-week vacation.
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nice one. I've been frustrated by this before, after crafting a beautiful answer and getting a dead silence in return. – Suresh Venkat♦ Sep 17 '10 at 22:56
### Only ask what you care about
When stated explicitly, this may sound obvious. Do not ask what someone cares about, but ask what you care about. If you do not care about a question, either no one really cares or those who care can ask a better question. Posting a question which you do not care about is a waste of time for everyone who reads the question (and possibly tries to answer it). Put differently: Do not ask a question for the sake of asking a question.
A corollary of this is: Try to answer by yourself before posting a question. If you really care about a question, the first thing to do should not be posting the question on a website, but trying to solve it by yourself. This includes looking up literature, formulating it from different angles, considering small cases, special cases and/or variations, and so on; in short, what you normally do to solve a problem. If you do not want to think about the question by yourself, ask yourself whether you really care about the question.
This does not mean that you must not ask a question unless you have proved that you cannot answer it by yourself (which is obviously impossible). However, make some reasonable effort to try to answer your own question before posting it.
Curiosity is a great motivator for science, but you should still make an effort to solve a question by yourself before posting it.
Related guidelines:
• Even if you care about the question, the reader may not know it. Therefore, it is a good idea to include some evidences that you care about it in the question. Doing so also helps the reader to understand the question better and hopefully results in better answers. For more on this point, see this guideline and this guideline.
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This is closely related to Suresh’s answer, Kaveh’s answer and the other answer of mine, but this is more fundamental than them in my opinion. – Tsuyoshi Ito Sep 15 '10 at 14:37
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I like it. Ask what you care about, and ONLY ask what you care about. The ONLY makes sure that you've done due diligence. nice ! – Suresh Venkat♦ Sep 15 '10 at 16:25
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@Tsuyoshi Ito: I share your feeling about not asking questions that I am personally working on. But assume that you are working on a problem and in the middle of your work you need some bound on some quantity to make your proof go through, but you are not an expert on those kind of quantities. In this situation I would post a question asking for bounds on the quantity on cstheory as I would ask the same question from a friend who knows more about those kind of quantities. – Kaveh♦ Sep 15 '10 at 18:07
@Kaveh: I assume that your story is not about throwing a tedious calculation one needs to do at other people, but about asking for calculation which requires some expertise. In that case, I guess nothing is wrong about asking it on the website, as long as it is really what the questioner cares about and he/she has considered about it enough to conclude that it requires something which he/she does not know well. – Tsuyoshi Ito Sep 15 '10 at 21:35
– Tsuyoshi Ito Sep 15 '10 at 21:51
@Tsuyoshi: No, I did not meant asking someone to do the tedious calculations. I think one good way of checking appropriateness of a question is by asking oneself "would I ask it from a colleague offline?". – Kaveh♦ Sep 16 '10 at 2:39
@Kaveh: In that case, I do not see any disagreement between us, unless I am missing something. – Tsuyoshi Ito Sep 16 '10 at 2:44
For records, I removed two comments of mine which had mentioned a shortcoming of this answer because I hope I addressed it in revision 5. – Tsuyoshi Ito Sep 16 '10 at 4:35
### Tell us why you care
There are many open problems, and it's not hard to post them as questions one by one. But if you want someone to care enough to write an answer, it's useful to know where you're coming from, and if you've put in some due diligence. Evidence of this would include
• things you've gleaned from google searchers/wikipedia
• cases where the question can be answered
• identifying what appears to be the key stumbling block
The ideal question is one where some expert probably knows the answer, but you'd normally have no chance to ask them directly, and take advantage of their expertise. If a problem seems to be open, there's probably a good reason for this, and asking it here is unlikely to generate an answer.
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Would this better be "Tell us what you've tried"? Or even, "Show us how much you care"? – Charles Stewart Sep 15 '10 at 8:39
Or maybe even, to make it more obviously complementary to Kaveh's point: "Tell us why we should care", but it would need a bit of reworking to fit that headline. Would you like to make it CW, so we can trample over your nice wording? – Charles Stewart Sep 17 '10 at 8:53
@Charles, done ! trample away... – Suresh Venkat♦ Sep 17 '10 at 14:40
And I like your reformulation – Suresh Venkat♦ Sep 17 '10 at 14:40
### Understand what you really want to ask
This may sound odd but, even if you know that you want to ask something, it is sometimes difficult to pin down exactly what you really want to ask. It is a good idea to make some effort to identify the exact question you want to ask before actually posting a question.
For example, many people incorrectly ask “Is there an algorithm which computes X?” when what they really want to know is “Is there a polynomial-time algorithm which computes X?” or even “X can be computed in time O(n3) in a straightforward way. Is there a faster algorithm?”
Knowing the precise question makes it easier to write a clear and easy-to-understand question, which hopefully makes more people interested in your question. Also, by the time you know the question well enough so that you can state the question clearly, you can sometimes find out the answer by yourself!
Related guidelines:
• Why should you make any effort to ask a question in the first place? See this guideline.
• Asking a precise question also allows one to tell what is an answer and what is not. See this guideline for more on this.
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– Tsuyoshi Ito Sep 17 '10 at 16:22
Yes, I agree. They are very different takes on the issue. – Charles Stewart Sep 17 '10 at 17:25
### Use math rendering
LaTeX is enabled on our site, and there's really no reason not to use it, especially if you're typing in LaTeX style anyway. Alternatively, you can use unicode rendering to get subscripts and superscripts, or even simple HTML constructions. If your question involves a reasonable amount of math notation (even lots of sub/super scripts) and doesn't use one of these approaches, it's liable to get closed or heavily downvoted.
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Use Latex not only in the body of your question but also in its title. – Jukka Suomela Sep 13 '10 at 17:07
– Tsuyoshi Ito Sep 14 '10 at 2:28
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actually no that's fine, as long as it's readable. the problem I guess is that people write latex without doing the $..$ enclosure that actually generates the latex. – Suresh Venkat♦ Sep 14 '10 at 6:00
@Tsuyoshi: Anything is fine as long as the end result looks good. The question that you linked is fairly readable, but I think it would be better if you used italics for symbols. This should be doable in HTML, too. – Jukka Suomela Sep 14 '10 at 8:19
@Suresh, @Jukka: Thanks for the reply and suggestion. I agree that making variables italic is probably better. I will try it next time I post. – Tsuyoshi Ito Sep 14 '10 at 11:54
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Due to ongoing MathJax problems (with mobile devices as well as Firefox, with caches cleared) I'm becoming less excited about using LaTeX notation. On my phone the various MathJax components often time out, leaving error messages instead of formulas, so I'm starting to avoid LaTeX markup for variable names and smaller bits of notation. When LaTeX markup is used in the title or in key parts of the question, what remains is often incomprehensible. – András Salamon Sep 14 '10 at 15:00
I'm another HTML character entities preferer, particularly in qn titles: Mathjax is much flakier, and doesn't have the good property that jsmath has of making the Tex-like source available, either when it fails or in a hover box. Can we change this item to (i) say that html math is ok, and (ii) mention the Mathjax issues? – Charles Stewart Sep 14 '10 at 19:22
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This is not correct. I am able to right click to get the source for mathjax rendering – Suresh Venkat♦ Sep 15 '10 at 1:21
If you agree that using LaTeX is not the only way, I appreciate if you can remove a warning like “If your question [...] doesn't use LaTeX, it's liable to get closed or heavily downvoted.” (By the way, I agree that using LaTeX is a reasonable advice to give in the FAQ because using Unicode characters can be trickier.) – Tsuyoshi Ito Sep 15 '10 at 2:04
edited in the light of the comments, and made it CW. – Suresh Venkat♦ Sep 15 '10 at 4:29
@Suresh I am able to right click to get the source - Hmm. I've looked again: I get a right menu for mathjax completely hidden under Firefox's default right menu. I can get that menu back, and view source, which appears in a new window. This is a horrible UI: when Mathjax fails, as it fairly frequently does, to render a Latex-formula dense page, looking at the source this way will be very time consuming. But switching Mathjax to output Mathml seems like a workaround. @András: have you tried using the Mathml output? – Charles Stewart Sep 15 '10 at 8:26
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When MathJax finds a formula in the document it (1) hides it, (2) renders it, and (3) replaces it by the result of rendering. I believe it's possible to configure MathJax to skip (1), so that if it gets stuck at (2) you still see the TeX source. – Radu GRIGore Sep 15 '10 at 13:00
@Radu: Not that I've seen. – Charles Stewart Sep 15 '10 at 13:18
@Charles: I didn't test, but that's what the comment on preview inside tex2jax inside MathJax.Hub.Config inside MathJax/config/MathJax.js says. – Radu GRIGore Sep 15 '10 at 17:24
### Read your question after you post it
Reading your question before you post it is an obvious way to improve it, but here is a more technical guideline: read your question after you post it, too.
Sometimes a question or part of it is formatted badly because of some technical limitation of the server (or JavaScript or browser or whatever, I do not know). For example, part of a formula can be missing (in particular LaTeX math on this website seems to often have a problem with $\lt$, $\gt$, $\{$ and $\}$). Read your question after you post it to make sure that everything is as you intended.
If you find a problem, please edit the question by yourself or ask for help in a comment to your question.
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I know almost nothing about the internals of the website, so I am sure that this answer is inaccurate. Please feel free to edit it. Also I do not know if asking for help in a comment is the right thing to do (but I cannot think of anything better). – Tsuyoshi Ito Sep 16 '10 at 11:48
Ask a focused question that has a specific goal
MO's FAQ has this beautiful section, which I'll just link to instead of copying: http://mathoverflow.net/howtoask#specific
Some of the questions I vote to close fail this test. As the MO FAQ puts it:
Ask yourself, "If I saw an answer to this question, could I confidently determine whether it tells the asker what she actually wants to know?"
Come to the think of it, the MO FAQ is very well written. Can we copy some of it verbatim?
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I think that that specific section distinguishes MathOverflow from general bulletin boards on mathematics. It would be very good if we can adopt it. (I agree that other sections are good, too.) – Tsuyoshi Ito Sep 15 '10 at 17:42
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I'm all for adopting it in totality, or partially. We should probably check with the MO admins if this is ok with them. for now, it might be sufficient merely to link to their page – Suresh Venkat♦ Sep 16 '10 at 3:54
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I'm quite sure they would have no problem, since I think their FAQ is licensed under "Creative Commons Attribution-Share Alike 3.0 Unported License with attribution required." So we'd just have to copy it and attribute it to them. We should check with them of course, as a matter of courtesy. I guess what we have to do is decide collectively if we want all of that to part of our FAQ too. – Robin Kothari Sep 16 '10 at 4:51
Tell us why you care
Motivate your question. Explain not just why the question is "mathematically interesting" -- why are you personally interested? Is it related to some other problem that you are working on? Were you reading a paper and didn't understand some part of it? ...
Providing partial answers is another way of showing your personal interest in the problem and that you have thought about it. This helps others tailor their answers to what you know, for example, by providing a reference relevant to the application of the answer, or skipping basics you already know. The answer to the question may also not be useful in the way you hoped it would be; if you make your motivation clear, then someone may be able to help you find a more useful way of looking at your problem.
This site is not an encyclopedia of possible good questions. This site is mainly for researchers to help each other about problems they face during their research.
Providing clues about personal motivation helps make questions interesting and easy to understand, and ensures answers are more useful to everyone.
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Yes, yes! I've added two paragraphs, paragraph two about helping answerers help you, and paragraph four providing a summary, which is in some kind of conflict with Kaveh's paragraph three. I've left them like that, so we can keep open how we want to end this point. – Charles Stewart Sep 15 '10 at 8:38
– Tsuyoshi Ito Sep 15 '10 at 14:44
Thanks Charles for improving this answer. I modified your paragraphs a little bit so the main point of it which is personal interest is not lost in the details. Please let me know if you are not happy with my modifications. – Kaveh♦ Sep 15 '10 at 17:53
This might be only me, but the second paragraph (in revision 4) is confusing and difficult to read for me. I guess that this paragraph is listing up some of the things a questioner can include in a question to convince readers that the questioner has personal interest in knowing the answer. If this is correct, using an itemized list might make the paragraph more readable. – Tsuyoshi Ito Sep 15 '10 at 21:42
By the way, how about changing the title from “Tell us why you care 2” to something more descriptive, e.g. “Tell us why you personally care”? – Tsuyoshi Ito Sep 15 '10 at 23:58
I think I see what you mean. There are two points here, and although they are related, having both of them in one answer makes the answer confusing. The first one is about showing personal interest, the second one is about the effect of providing personal motivation in improving the answer one would get. I think posting the second one as another answer can remove the confusion. – Kaveh♦ Sep 16 '10 at 2:56
Simplified sentence construction a little, and added a few phrases. I think a trimmed version of this answer may be more effective. – András Salamon Sep 16 '10 at 15:00
@András: agreed. I like this answer: it makes a lot of good points effectively, but the text as a whole would be more effective if it were more condensed. Aaron's last edit was a good step in this direction; I'll look for more improvements later. – Charles Stewart Sep 17 '10 at 8:47
### Provide background information
Try to make your question self-contained. The users of this site are not machines and are not paid to answer your questions. Be nice to others. Spend some time in explaining the background of your question and the terminology you are using to make it understandable to people not familiar with what you are asking (at least provide a link to a source that would explain the terminology). Make reading your question pleasant to other users of the site.
You might say, “But anyone working on this field should know what XYZ is. If someone does not know what it is, he/she cannot answer my question anyway, so explaining it will not help my question get answered.” That may be correct or incorrect. It may be incorrect because sometimes people from surprisingly different background can give you useful insight if a question is written in an accessible way. However, more importantly, do not be selfish. Let other users learn from your question even if they have absolutely zero probability of being able to answer your question.
Read the related MO guideline.
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3
I edited the answer to include a counterargument to an argument against providing background which I have seen several times. – Tsuyoshi Ito Feb 4 '11 at 10:36
### Make your title your question
Try to be as specific as possible about the title of your question. Avoid using general titles that don't specify your question. Don't use the title as a tag.
Read the corresponding MO guideline.
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Is this really the question you want to ask?
All too often, we think that we need to solve some problem by getting the answer to a particular question, but then it turns out that the answer doesn't help us the way we expected, or there can be no answer at all. This may be because we had some misconceptions about our problem, or because the question we think we need the answer to in fact was overly boad or narrow for the answers to help us.
It may still be the best course to ask the question you haven't yet realised is wrong, but trying to look for alternative questions before asking may help you get more useful answers.
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– Charles Stewart Sep 17 '10 at 9:13
– Tsuyoshi Ito Sep 17 '10 at 11:16
@Tsuyoshi Ito: We do indeed have a problem with overlapping points here. I has not toticed that this overlapped with yours. The solution is to talk about how the points fit together. I'm planning on writing such a post as an answer. – Charles Stewart Sep 17 '10 at 11:27
@Tsuyoshi: You mention formulating it from different angles in the context of trying to solve a problem - this is not quite the same point I am making. This assumes the questioner has been diligent in trying to solve the problem, but for reasons that have eluded them, the question does not match their interest in the problem that led to it. – Charles Stewart Sep 17 '10 at 11:54
I think that you misunderstood my comment. The answer I referred to is not this but this. – Tsuyoshi Ito Sep 17 '10 at 12:56
### Read the MO guidelines "how to ask"
MathOverflow has a very well written guideline about asking questions, all of the items in their guideline applies to this site also. Please read their guideline.
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http://math.stackexchange.com/questions/56433/about-positive-definiteness-of-a-matrix-q-depending-on-matrices-h-and-p?answertab=votes
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# About positive-definiteness of a matrix Q depending on matrices H and P
$H + H^T$ is a positive definite matrix and $P$ is also a positive definite matrix.
Will $Q = PH + H^TP$ be a positive definite matrix?
In my calculations, it is not positive definite. But I read a paper saying that $Q$ should be positive definite. Is it so?
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Nobody's ever done that before! @Fatima, the place you just posted this question is the meta site, which is for discussion of the main site, not for discussion of mathematics. – Qiaochu Yuan Aug 9 '11 at 3:18
1
– Jonas Meyer Aug 9 '11 at 3:46
@Jonas: my apologies. I should have been more precise: I've never seen anyone do that before! – Qiaochu Yuan Aug 9 '11 at 3:48
1
Fatima: What calculations, and what paper? – Jonas Meyer Aug 9 '11 at 4:01
@Jonas: paper is "A new approach to the LQ design from the viewpoint of the inverse regulator problem" by Takao Fujii – Fatima Tahir Aug 9 '11 at 4:43
show 1 more comment
## 2 Answers
You refer to your calculations; does that mean you already have a counterexample? My calculations seem to agree with yours, as seen in the example $H=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ and $P=\begin{bmatrix}1&0\\0&5\end{bmatrix}$.
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yes, I wanted to use the results of the mentioned paper for my work but I came across matrices which do not satisfy the results mentioned in the paper. – Fatima Tahir Aug 9 '11 at 5:54
Fatima, thanks for giving the name of the article, but I do not know precisely what is claimed in the article, and I do not have access to it at the moment. Perhaps there are additional hypotheses there. If you would like to ask further questions on the particular claim made in the article, it would help to provide a little more context, and perhaps an excerpt for those like me who cannot view it. – Jonas Meyer Aug 9 '11 at 6:02
If
$$\mathbf H=\begin{pmatrix}15&9&7\cr-1&9&-8\cr-3&-9&11\cr\end{pmatrix}$$
and
$$\mathbf P=\begin{pmatrix}81&-5&30\cr-5&75&-54\cr30&-54&54\cr\end{pmatrix}$$
then $\mathbf Q$ isn't positive definite, having two positive and one negative eigenvalues. It should be easy to generate other counterexamples...
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How have you generated this one? The numbers seem rather arbitrary. – Patrick Da Silva Aug 10 '11 at 12:53
Yes @Patrick, I just randomly generated some unsymmetric matrix $H$ and perturbed it so that $H+H^T$ is symmetric positive definite... – J. M. Aug 11 '11 at 1:40
Oh =) Okay cool – Patrick Da Silva Aug 11 '11 at 2:21
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http://mathhelpforum.com/calculus/125321-sine-product-formulae-print.html
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# sine product formulae
Printable View
• January 24th 2010, 08:49 PM
bkarpuz
sine product formulae
Dear Friends,
I have a question about the sine product formulae.
As far as I know, Euler thinks the function $\sin$ as a polynomial of $\infty$ order, and writes it by using its roots as
$\sin(x)=a_{0}\prod_{k=-\infty}^{\infty}\big(x-k\pi\big)$
or equivalently
$\sin(x)=b_{0}x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(1)
for some constants $a_{0},b_{0}$.
From the fact that
$\lim_{x\to0}\frac{\sin(x)}{x}=1,$
the constant $b_{0}$ in (1) is computed to be $1$.
Hence
$\sin(x)=x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(2)
My question comes at this point.
How I can be sure that the right-hand side of (2) gives $\sin(x)$, not $\big(\sin(x)\big)^{2}/x$?
• January 24th 2010, 10:33 PM
Drexel28
Quote:
Originally Posted by bkarpuz
Dear Friends,
I have a question about the sine product formulae.
As far as I know, Euler thinks the function $\sin$ as a polynomial of $\infty$ order, and writes it by using its roots as
$\sin(x)=a_{0}\prod_{k=-\infty}^{\infty}\big(x-k\pi\big)$
or equivalently
$\sin(x)=b_{0}x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(1)
for some constants $a_{0},b_{0}$.
From the fact that
$\lim_{x\to0}\frac{\sin(x)}{x}=1,$
the constant $b_{0}$ in (1) is computed to be $1$.
Hence
$\sin(x)=x\prod_{k=1}^{\infty}\bigg(1-\Big(\frac{x}{k\pi}\Big)^{2}\bigg)$........(2)
My question comes at this point.
How I can be sure that the right-hand side of (2) gives $\sin(x)$, not $\big(\sin(x)\big)^{2}/x$?
I'm not too sure how easy this would be to prove, but I'm guessing that what you are saying is that the roots of $\sin(x)$ are precisely those of $\sin^\ell(x)$ for any $\ell>0$. So how do you know that the RHS of (2) isn't the expansion for one of those? Well, assuming that you agree the RHS is the product for something of the form $\sin^\ell(x)$ we must merely note that we arrived at it by evaluating $\lim_{x\to0}\frac{\sin^\ell(x)}{x}$. But, if $0<\ell<1$ this limit diverges and if $\ell>1$ the limit is zero. Can you use this fact to show that the RHS neither diverges nor is zero to conclude that $\ell=1$ and thus the RHS is the product for $\sin(x)$?
• January 24th 2010, 10:44 PM
bkarpuz
Quote:
Originally Posted by Drexel28
I'm not too sure how easy this would be to prove, but I'm guessing that what you are saying is that the roots of $\sin(x)$ are precisely those of $\sin^\ell(x)$ for any $\ell>0$. So how do you know that the RHS of (2) isn't the expansion for one of those?
Up to here everything goes right.
Then I say that the roots of $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ for $\ell\geq1$ are also the roots of $\sin(x)$, and they both sayisfy $\lim_{x\to0}\big(f(x)/x\big)=1$ condition. (Thinking)
How can we decide the RHS of (2) is th product for $\sin(x)$.
There must be something more to answer this.
• January 24th 2010, 10:47 PM
Drexel28
Quote:
Originally Posted by bkarpuz
Up to here everything goes right.
Then I say that the roots of $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ for $\ell\geq1$ are also the roots of $\sin(x)$, and they both sayisfy $\lim_{x\to0}\big(f(x)/x\big)=1$ condition. (Thinking)
How can we decide the RHS of (2) is th product for $\sin(x)$.
There must be something more to answer this.
I disagree $\frac{\sin^\ell(x)}{x^{\ell-1}}\underset{x\to0}{\sim}\frac{x^{\ell}}{x^{\ell-1}}\to0$
• January 24th 2010, 11:05 PM
bkarpuz
Quote:
Originally Posted by Drexel28
I disagree $\frac{\sin^\ell(x)}{x^{\ell-1}}\underset{x\to0}{\sim}\frac{x^{\ell}}{x^{\ell-1}}\to0$
You have a simple mistake I guess, you forgot to divide the function by $x$.
$\frac{\frac{\big(\sin(x)\big)^{\ell}}{x^{\ell-1}}}{x}=\bigg(\frac{\sin(x)}{x}\bigg)^{\ell}\to1$ as $x\to0$.
• February 19th 2010, 05:29 AM
Laurent
Quote:
Originally Posted by bkarpuz
Up to here everything goes right.
Then I say that the roots of $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ for $\ell\geq1$ are also the roots of $\sin(x)$, and they both sayisfy $\lim_{x\to0}\big(f(x)/x\big)=1$ condition. (Thinking)
How can we decide the RHS of (2) is th product for $\sin(x)$.
There must be something more to answer this.
Something like the order of the roots?...
(and advanced complex analysis in order to justify rigorously the argument)
• February 19th 2010, 06:16 AM
bkarpuz
Quote:
Originally Posted by Laurent
Something like the order of the roots?...
(and advanced complex analysis in order to justify rigorously the argument)
Actually, I dont know but I wonder to know the rigorous proof since I could not find it anywhere. Thanks.
• February 19th 2010, 07:07 AM
Laurent
Quote:
Originally Posted by bkarpuz
Actually, I dont know but I wonder to know the rigorous proof since I could not find it anywhere. Thanks.
My remark about the order is a serious one; it doesn't prove that the above infinite product equals the sine function, but allows to discard functions like powers of sine.
Anyway, it is not elementary to make Euler's argument rigorous (note that if you multiply $\sin z$ by $e^{g(z)}$ where $g$ is an entire function, the function is still analytic and has the same zeroes and same orders). As far as I know, the proof that is closest (in intuition, not technique...) to Euler's proof would go along Weierstrass Factorization Theorem (which gives the existence of an analytic function with given zeroes, and its expression as an infinite product). The difficulty (besides the proof of the theorem...) is then to prove that the above entire function $g(z)$ is a constant; you can find a full proof of the sine product formula along these lines in "Functions of one complex variable I" by John B. Conway (Springer, Graduate texts in mathematics), p. 175 in 2nd edition. (His proof uses an expansion of the cotangent function that can be obtained from Residue Theorem. )
• February 19th 2010, 07:49 AM
vince
there's a rigorous derivation of sine's infinite product expansion that relies on the Gamma function, Weierstrass' product formula and the Legendre relation.
• February 19th 2010, 08:30 AM
bkarpuz
Quote:
Originally Posted by vince
there's a rigorous derivation of sine's infinite product expansion that relies on the Gamma function, Weierstrass' product formula and the Legendre relation.
I think you are talking about the Gamma reflection formula
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you check the proof
$\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)},$
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
$\Gamma(z):=\frac{\mathrm{e}^{-\gamma z}}{z}\prod_{k\in\mathbb{N}}\frac{\mathrm{e}^{z/k}}{1+(z/k)},$
where $\gamma$ is the Euler–Mascheroni constant defined by
$\gamma:=\lim_{n\to\infty}\Big(\sum_{k=1}^{n}\frac{ 1}{k}-\int_{1}^{n}\frac{\mathrm{d}x}{x}\Big).$
This is where my interest to the SPF originates. (Itwasntme)
• February 19th 2010, 08:39 AM
vince
Quote:
Originally Posted by bkarpuz
I think you are talking about the Gamma reflection formula
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you trace the proof
$\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)},$
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
$\Gamma(z):=\frac{\mathrm{e}^{-\gamma z}}{z}\prod_{k\in\mathbb{N}}\frac{\mathrm{e}^{z/k}}{1+(z/k)},$
where $\gamma$ is the Euler–Mascheroni constant defined by
$\gamma:=\lim_{n\to\infty}\Big(\sum_{k=1}^{n}\frac{ 1}{k}-\int_{1}^{n}\frac{\mathrm{d}x}{x}\Big).$
This is where my interest to the SPF originates. (Itwasntme)
what do u mean by tracing a proof? given what i see there, you can make it rigorous...no "tracing" needed (Cool)
p.s. use the Legendre relation:
$\Gamma(\frac{x}{2})\Gamma(\frac{x+1}{2})=\frac{\sq rt{\pi}}{2^{x-1}}\Gamma(x)$ and define the function $\Phi(x)=\Gamma(x)\Gamma(1-x)\sin({\pi}x)$and note that $\Phi(x+1)=\Phi(x)$. Also, i came across this proof over the net. Not my own :D
• February 20th 2010, 07:04 AM
vince
Quote:
Originally Posted by bkarpuz
I think you are talking about the Gamma reflection formula
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$
Actually, this is not how SPF (sine product formula) is obtained, it is rather in this way if you check the proof
$\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)},$
and as you have mentioned the proof makes use of Weierstrass' Gamma function expansion
$\Gamma(z):=\frac{\mathrm{e}^{-\gamma z}}{z}\prod_{k\in\mathbb{N}}\frac{\mathrm{e}^{z/k}}{1+(z/k)},$
where $\gamma$ is the Euler–Mascheroni constant defined by
$\gamma:=\lim_{n\to\infty}\Big(\sum_{k=1}^{n}\frac{ 1}{k}-\int_{1}^{n}\frac{\mathrm{d}x}{x}\Big).$
This is where my interest to the SPF originates. (Itwasntme)
If you truly understand this proof, then you'll notice the key to its being resolved is proving that a function defined to be the second derivative of $\log\Phi(x)$, \;where \;\Phi(.)[/tex] as defined above, is a CONSTANT. Key in concluding that is also determining that $\log\Phi(x)$ is periodic! $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ is not periodic. Once you notice this, a contradiction can be constructed.(Wondering) I finally got around to looking at the proof today.:cool:
• February 20th 2010, 07:17 AM
bkarpuz
Quote:
Originally Posted by vince
If you truly understand this proof, then you'll notice the key to its being resolved is proving that a function defined to be the second derivative of $\log\Phi(x)$, \;where \;\Phi(.)[/tex] as defined above, is a CONSTANT. Key in concluding that is also determining that $\log\Phi(x)$ is periodic! $\big(\sin(x)\big)^{\ell}/x^{\ell-1}$ is not periodic. Once you notice this, a contradiction can be constructed.(Wondering) I finally got around to looking at the proof today.:cool:
vince thanks.
I have not seen the proof for SPF, I just know the proof of the Gamma reflection formula.
I will also check the proof of the SPF in the book Laurent referred me to but just postponed it a few days.
Bests.
bkarpuz
• February 20th 2010, 02:19 PM
Laurent
Quote:
Originally Posted by bkarpuz
I have not seen the proof for SPF, I just know the proof of the Gamma reflection formula.
Besides the proof I mentioned, which has a taste of Euler's initial proof, there are plenty other (more elementary) proofs of the formula.
There is a proof starting from the expansion $\pi\cot \pi z=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{1}{z^2-n^2}$ (proved by one mean or another), then noting that $\pi\cot \pi z= \frac{d}{dz}\log \sin \pi z$ so that if we can integrate term by term, we get a series expansion of $\log \sin$ which is equivalent to the sine product formula. A ref (also gives another proof)
There is also a fairly elementary proof I learned from "Analysis by its history" by Hairer and Wanner, the French edition. I looked it up, it is p.64 of the English edition, which you can probably find in any good library ;) (The justification of a convergence step is partly left as an exercise, with a hint) By the way, this book is nice reading; rather simple maths (undergraduate analysis), but put in its historical context. Of course, they mention how Euler came up with the formula!
All times are GMT -8. The time now is 10:26 AM.
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http://mathoverflow.net/questions/52370?sort=oldest
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## Eigenvalues of an “oblique diagonal” matrix
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for guidance about the behavior of powers of a particular matrix (call it $A_n$ for $n\ge2$), which has come up in a counting problem about quantum knot mosaics (a good reference for quantum knot mosaics is here ). Here's a description of the matrix. It has $4^n$ rows and columns. Instead of a traditional diagonal matrix with its non-zero entries on the main diagonal, the non-zero entries of $A_n$ are on an oblique diagonal" of slope $4$, modulo the size of the matrix. More precisely, the non-zero entries occur where $\left\lfloor\frac{\text{column}}4\right\rfloor=\text{row}\text{, mod}\;4^{n-1}$. The matrix has sixteen possibly non-zero values $a_1,\ldots,a_{16}$, arranged as follows (with boxes for visual clarity):
```$\begin{array}{l|lll|lll|}
\text{row}\backslash\text{column}&0&4&\ldots&4^n/2&4^n/2+4&\ldots\\ \hline
0&\boxed{a_1\,a_2\,a_1\,a_2}\\
1&&\boxed{a_1\,a_2\,a_1\,a_2}\\
\vdots&&&\ddots\\ \hline
4^n/8&&&&\boxed{a_3\,a_4\,a_3\,a_4}\\
4^n/8+1&&&&&\boxed{a_3\,a_4\,a_3\,a_4}\\
\vdots&&&&&&\ddots\\ \hline
2\cdot4^n/8&\boxed{a_5\,a_6\,a_5\,a_6}\\
2\cdot4^n/8+1&&\boxed{a_5\,a_6\,a_5\,a_6}\\
\vdots&&&\ddots\\ \hline
3\cdot4^n/8&&&&\boxed{a_7\,a_8\,a_7\,a_8}\\
3\cdot4^n/8+1&&&&&\boxed{a_7\,a_8\,a_7\,a_8}\\
\vdots&&&&&&\ddots\\ \hline
4\cdot4^n/8&\boxed{a_{9}\,a_{10}\,a_{9}\,a_{10}}\\
4\cdot4^n/8+1&&\boxed{a_{9}\,a_{10}\,a_{9}\,a_{10}}\\
\vdots&&&\ddots\\ \hline
5\cdot4^n/8&&&&\boxed{a_{11}\,a_{12}\,a_{11}\,a_{12}}\\
5\cdot4^n/8+1&&&&&\boxed{a_{11}\,a_{12}\,a_{11}\,a_{12}}\\
\vdots&&&&&&\ddots\\ \hline
6\cdot4^n/8&\boxed{a_{13}\,a_{14}\,a_{13}\,a_{14}}\\
6\cdot4^n/8+1&&\boxed{a_{13}\,a_{14}\,a_{13}\,a_{14}}\\
\vdots&&&\ddots\\ \hline
7\cdot4^n/8&&&&\boxed{a_{15}\,a_{16}\,a_{15}\,a_{16}}\\
7\cdot4^n/8+1&&&&&\boxed{a_{15}\,a_{16}\,a_{15}\,a_{16}}\\
\vdots&&&&&&\ddots\\ \hline
\end{array}$```
Since $A_n$ has a straightforward geometrical description with non-zero entries only on a diagonal (albeit an oblique one), the following question seems reasonable:
Is there an elementary way to compute the eigenvalues of this matrix in terms of $a_1,\ldots,a_{16}$ and $n$?
I'm no expert in tensor algebra, but it seems that $A_n$ might be expressed as the tensor product of $A_2$ with $n-1$ copies of another transformation. Even an approximation of the largest eigenvalue would be useful, but it would be best to avoid the power method with the Rayleigh quotient since I'm trying to analyze the powers of the matrix in terms of the eigenvalues, not vice versa. Any insight into the computation of the eigenvalues of $A_n$ would be greatly appreciated and would go a long way in answering a question in the paper referenced above.
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2
Impressive LaTeX {array}! :-) – Joseph O'Rourke Apr 5 2012 at 0:23
Well, as Donald Knuth said, TeX is intended for the creation of beautiful mathematics. – Russell May Apr 5 2012 at 12:32
## 1 Answer
I think that if you move some rows around you get a block diagonal form (this the tensor product from the previous post). Move all the rows that start in col 1 together, rows in col 5 etc. The eigenvalues etc. will be related to the eigenvalues etc. of the blocks.
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2
I agree that there's a permutation matrix P and a block diagonal matrix A' so that the oblique diagonal matrix A is PA'. However, it's not clear how to get the eigenvalues of a product, given the eigenvalues of the factors. – Russell May Apr 6 2012 at 18:44
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http://math.stackexchange.com/questions/154231/calculation-of-the-moments-using-hypergeometric-distribution
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Calculation of the moments using Hypergeometric distribution
Let vector $a\in 2n$ is such that first $l$ of its coordinates are $1$ and the rest are $0$ ($a=(1,\ldots, 1,0, \ldots, 0)$). Let $\pi$ be $k$-th permutation of set $\{1, \ldots, 2n\}$. Define $$g=\left|\sum_{i=1}^n a_{\pi(i)}-\sum_{i=n+1}^{2n}a_{\pi(i)}\right|.$$
Using Hypergeometric distribution calculate /approximate the $q$-th moment $E|g|^q,$ for any $q\ge 2$.
I've got that the $q$-th moment is $$E|g|^q=\sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}.$$ But now I am stuck...
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– David Jun 10 '12 at 13:38
By comparing the last expression to the probability function of the hypergeometric distribution, you see that $E|g|^q=E(2X-l)^q$, where $X$ is $Hypergeometric(2n, l, n)$. Does that help? – MånsT Jun 12 '12 at 19:08
@MansT: Thank you. But I still don't understand how to calculate the sum. Could you elaborate please. – Aleks.K Jun 12 '12 at 22:38
1 Answer
By comparing the last expression to the probability function of the hypergeometric distribution, you see that $E|g|^q=E(2X−l)^q$, where $X$ is $\rm{Hypergeometric}(2n,l,n).$
Therefore $E(X)=\frac{nl}{2n}=l/2=:\mu$. Thus $$E|g|^q=E(2X−l)^q={2}^qE(X-l/2)^q=2^qE(X-\mu)^q.$$
Expressed in words, $E|g|^q$ is $2^q$ times the $q$:th central moment of $X$.
The central moments of the hypergeometric distribution are known and can be computed (preferably not by hand...).
-
Thank you. Is it possible to find a general formula for the q-th central moment of the hypergeometric distribution? Or the upper bound of it? – Aleks.K Jun 14 '12 at 16:01
in your main formula in the RHS you have '\?' is it typo or it suppose to be some math symbol? – Aleks.K Jun 14 '12 at 16:02
@Aleks.K: It was a typo. :) There are some results (1, 2, 3) regarding how to compute hypergeometric moments. They probably go beyond what can be considered homework problems for most courses though... As $n\rightarrow\infty$ you can approximate the hypergeometric distribution with the binomial, but I'm not sure if that helps here. – MånsT Jun 14 '12 at 17:28
@MansT:You said that the hypergeometric distribution can be approximated with the binomial. Does it mean that $E(X-\mu)^q\sim E(Y-E(Y))^q,$ where $Y$ is Binomial? – David Jun 15 '12 at 17:12
@MansT: Is it some possibilities to get an asymptotic upper bound for your $2^qE(X-\mu)^q$? I think one can use consentration for that... – Michael Jul 4 '12 at 14:15
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http://mathhelpforum.com/calculus/130045-center-mass.html
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# Thread:
1. ## Center of Mass
Here is the question:
A solid is formed by rotating the region bounded by the curve and the -axis between and , around the -axis. The volume of this solid is . Assuming the solid has constant density , find and .
To find the moment, I did:
$\int_0^1 x dx$ = $\frac {1} {2}$
As density was constant, I divided the moment by the volume and found to be 1/2.
And I am not sure that is right and so I haven't even attempted to find . All help is much appreciated. Thanks.
2. The volume, when y= y(x) is rotated around the x- axis is (by the disk method) $\pi \int_{x_0}^{x_1} y^2(x)dx$.
In this case, that would be $\pi \int_0^1 e^{-7x} dx$ which gives the volume you quote.
But the moment is NOT just that volume times $\int x dx$. It is, rather, that same integral with another x in the integral: $\pi \int_0^1 xe^{-7x}dx$. Use "integration by parts".
Since this is rotated around the x-axis, it is symmetric with respect to both y and z and it immediately follows that $\overline{y}= 0$ and $\overline{z}= 0$.
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http://www.physicsforums.com/showthread.php?t=106213
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Physics Forums
## Taylor rule of thumb??
When calculating limits by using taylor series is there any easy way to know how many elements that should be included in the taylor series?
if I have
$$\lim_{x\rightarrow\zero} \frac{exp(x-x^2)-Cos2x-Ln(1+x+2x^2)}{x^3}$$
How do I know many terms to include in the taylors series for the exponent, cos and logaritm functions?? Should I include all terms that contains a $$x^3$$??
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
x->0 but for some reason the latex wont include the 0.
When I solve things like this I always notice at the end that I have used to many terms in the taylor series or to few. To many isnt much of a problem obviously except that it clutters up the calculations. But very anoying when I end up with to few.
Recognitions:
Homework Help
Science Advisor
## Taylor rule of thumb??
How many you 'should' include depends on what the question asks for, surely. There is no universal 'correct amount'. If you want to know what the error is for any truncation then you should look up Talyor Remainder and other such things which should be in your book, assuming you have one.
yes I have one, not the best to be honest, but it usualy works pretty good. The answere it gave to this particular question is that is trial and error. This question doesnt involve error estimations in the series. But rather how to get a "feel" for how many terms to include to make limit problems like the one I postes as easily solvable as possible. If you solve that one how many terms would you include?? I first tried by using the first 2 terms of each taylor series but it couldnt be solved that way. So I tried with the first 3 terms of each taylor series and that way I got the correct answere. When I look at the solution they used the first 3 terms from the taylor series for the exponent, the first 2 for the cos function and the first 3 for the logaritm. How can I "tell" right away that I need no more then 2 terms from the cos to solve it?
Recognitions: Gold Member Science Advisor Staff Emeritus Approximating a function by a Taylor's polynomial is just that- an approximation. No approximation, no matter how close it is well tell you exactly what a limit is. In this case, however, since the denominator is x3, writing out the entire Taylor's series, then dividing by x3, you know that all terms in the series past x3 will have an x in them and will go to 0. That's not a "rule", that's basic algebra.
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Homework Help
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Quote by Azael When I look at the solution they used the first 3 terms from the taylor series for the exponent, the first 2 for the cos function and the first 3 for the logaritm. How can I "tell" right away that I need no more then 2 terms from the cos to solve it?
When you're counting the number of terms they used, remember that the x^3 term in the taylor series for cos has a 0 coefficient. The actual number of terms isn't really what's important, it's the size of the error between your taylor polynomial and your function.
In this question, you are dividing by an x^3. Your book should give some explanation about finding how many terms needed so the error/x^3 goes to zero (maybe in terms of big-Oh notation, but possibly not), then the limit will be determined by your taylor polynomials to whatever degree was needed for this size error term.
More precisely, if $$P_n(x)$$ is the nth degree taylor polynomial of f(x) and $$R_n(x)$$ is the error, so $$f(x)=P_n(x)+R_n(x)$$, you want to be able to determine the value of n so that $$\lim_{x\rightarrow 0 }R_n(x)/x^3=0$$. Then $$\lim_{x\rightarrow 0}f(x)/x^3=\lim_{x\rightarrow 0}P_n(x)/x^3$$. You can replace x^3 with any power x^m and it's the same idea (the required n may change of course).
thanks shmoe that clears it up very good :) Yeah we are taught to use the big-Oh notation(or well we call it big Ordo but I guess its the same thing)in problems like these.
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http://mathhelpforum.com/calculus/134938-find-equation-both-lines.html
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# Thread:
1. ## Find the equation of both lines?
Determine the equation of both lines that are tangent to the graph of y=x^2 +4 and pass through the point (1,-2)
So would I simply use the y=mx+b form, m=2, y= -2, and x = 1?
2. Originally Posted by RyGuy
Determine the equation of both lines that are tangent to the graph of y=x^2 +4 and pass through the point (1,-2)
So would I simply use the y=mx+b form, m=2, y= -2, and x = 1?
no ... the point (1,-2) does not lie on the parabola.
3. Originally Posted by skeeter
no ... the point (1,-2) does not lie on the parabola.
Oh right.
4. Originally Posted by RyGuy
Determine the equation of both lines that are tangent to the graph of y=x^2 +4 and pass through the point (1,-2)
So would I simply use the y=mx+b form, m=2, y= -2, and x = 1?
the derivative of the parabola equation gives the slope of all tangent lines to the curve.
You need the equations of the two that pass through (1,-2).
hence, write the derivative to get the slopes of these lines.
then use
$y-y_1=m(x-x_1)$
5. point of tangency on the curve is $(x,x^2+4)<br />$
each tangent line passes thru the point $(1, -2)$
remember how to find the slope between two points?
... also, note that the slope of both tangent lines is y' = 2x
I'll let you take it from here.
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http://unapologetic.wordpress.com/2011/03/30/tangent-vectors-and-coordinates/?like=1&source=post_flair&_wpnonce=bcc517a348
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# The Unapologetic Mathematician
## Tangent Vectors and Coordinates
Let’s say we have a coordinate patch $(U,x)$ around a point $p$ in an $n$-dimensional manifold $M$. We can use the function $f$ to give us some tangent vectors at $p$ called the “coordinate vectors”.
We define the coordinate vector $\frac{\partial}{\partial x^i}(p)\in\mathcal{T}_pM$ as follows: given a smooth function $f\in\mathcal{O}_p$, we define
$\displaystyle\left[\frac{\partial}{\partial x^i}(p)\right](f)=\left[D_i(f\circ x^{-1})\right](x(p))$
Okay, I know that that’s confusing. But all we mean is this: start with a function $f:U\to\mathbb{R}$. We compose it with the inverse of the coordinate map to get $f\circ x^{-1}:x(U)\to\mathbb{R}$, where $x(U)$ is some open neighborhood of the point $x(p)\in\mathbb{R}^n$. Now we can take that $i$th partial derivative of this function and evaluate it at the point $x(p)$.
The first thing we really should check is that it doesn’t matter which representative $f$ we pick. That is, if $f=g$ in some neighborhood of $p$, do we get the same answer? Indeed, in that case $f\circ x^{-1}=g\circ x^{-1}$ in some neighborhood of $x(p)$, and so their partial derivatives are identical. Thus this operation only depends on the germ $f\in\mathcal{O}_p$.
But is it a tangent vector? It’s easy to see that it’s a linear functional, so we just have to check that it’s a derivation at $p$:
$\displaystyle\begin{aligned}\left[\frac{\partial}{\partial x^i}(p)\right](fg)=&\left[D_i((fg)\circ x^{-1})\right](x(p))\\=&\left[D_i((f\circ x^{-1})(g\circ x^{-1}))\right](x(p))\\=&\left[D_i(f\circ x^{-1})\right](x(p))\left[g\circ x^{-1}\right](x(p))\\&+\left[f\circ x^{-1}\right](x(p))\left[D_i(g\circ x^{-1})\right](x(p))\\=&\left[\frac{\partial}{\partial x^i}(p)\right](f)g(p)+f(p)\left[\frac{\partial}{\partial x^i}(p)\right](g)\end{aligned}$
And so we have at least these $n$ vectors at each point $p\in M$. We can even tell that they much be distinct — and even linearly independent — since we can calculate
$\displaystyle\begin{aligned}\left[\frac{\partial}{\partial x^i}(p)\right](x^j)&=\left[\frac{\partial}{\partial x^i}(p)\right](\pi^j\circ x)\\&=\left[D_i(\pi^j\circ x\circ x^{-1})\right](x(p))\\&=\left[D_i(\pi_j)\right](x(p))\end{aligned}$
where $\pi^j$ is the $j$th coordinate projection $\pi^j:\mathbb{R}^n\to\mathbb{R}$. But we know that $D_i(\pi^j)$ is always and everywhere $\delta_i^j$ — it takes the value $1$ if $i=j$ and $0$ otherwise.
Thus $\frac{\partial}{\partial x^i}(p)$ takes a different value on $x^i$ than on all the other $x^j$. Further, any linear combination of the $\frac{\partial}{\partial x^j}(p)$ for $j\neq i$ must take the value $0$ on $x^i$, while $\frac{\partial}{\partial x^i}(p)$ takes the value $1$; we see that none of the coordinate vectors can be written as a linear combination of the rest, and conclude that the dimension of $\mathcal{T}_pM$ is at least $n$.
### Like this:
Posted by John Armstrong | Differential Topology, Topology
## 8 Comments »
1. I’ve always been bothered by the really poor notation here. It’s very tedious,cumbersome, and confusing to wade through all the different senses of parentheses (function/operator application, grouping), operators, functions, and the sets and function spaces in which they live. I know that people like to draw a picture of all the parts living in M versus R^n or R, and that helps. Maybe typographically reminding the reader what space each symbol lives in would help (by typesetting the spaces for each letter on the line above, or using color). Just thinking out loud. Hell, I might right it up with type annotations a la Haskell just to make it clearer to myself lol.
Sorry, I’m a visual thinker, and it helps to see visually just how this definition carries over our knowledge of R^n to M by all the coordinate patches.
Excellent as always, though, thank you.
Comment by Robert | March 31, 2011 | Reply
2. [...] vector space of tangent vectors at . Given a coordinate patch around , we’ve constructed coordinate vectors at , and shown that they’re linearly independent in . I say that they also span the space, [...]
Pingback by | March 31, 2011 | Reply
3. [...] a coordinate patch in a neighborhood of a point in an -dimensional manifold , we get coordinate vectors which form a basis for the tangent space . But this is true of any coordinate patch! If we have [...]
Pingback by | April 1, 2011 | Reply
4. [...] are related for , but within a single coordinate patch we can use the coordinate map to define coordinate vectors at every single point , and this lets us compare vectors at different points by comparing their [...]
Pingback by | April 4, 2011 | Reply
5. [...] whole domain of we can restrict down to a coordinate patch containing — we get a basis of coordinate vectors at . Similarly, if is a coordinate patch around we get a basis of coordinate vectors at . We want [...]
Pingback by | April 6, 2011 | Reply
6. [...] out a segment of the curve that does — then we have a curve . We can also use to define a coordinate basis on , and thus get components of in those coordinates. As usual, we calculate the th component [...]
Pingback by | April 8, 2011 | Reply
7. [...] terms of components, pick a basis of and use it to get a coordinate map on all of . We also get a basis of coordinate vectors for at each point , and in particular at . What does this [...]
Pingback by | April 11, 2011 | Reply
8. [...] Indeed, at each point we can define the coordinate vectors: [...]
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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://www.fields.utoronto.ca/programs/scientific/12-13/algebras_seminar/index.html
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# SCIENTIFIC PROGRAMS AND ACTIVITIES
May 18, 2013
THE FIELDS INSTITUTE FOR RESEARCH IN MATHEMATICAL SCIENCES
20th ANNIVERSARY YEAR
Set Theory and C*-algebras Seminar Logic and Operator Algebras Fields Institute, Room 210 (Map) 10:00 - 11:30 a.m. Organizers: Ilijas Farah (York University), Juris Steprans (York University), Bradd Hart (McMaster University)
### OVERVIEW
This weekly seminar is devoted to applications of logic to operator algebras. In its first semester it will feature three types of talks: (i) introductory talks on operator algebras aimed at logicians, (ii) introductory talks on logic aimed at operator algebraists and (iii) discussions of recent applications of set theory and model theory to C*-algebras and von Neumann algebras.
In addition to these, some of the meetings will be 'working seminars' devoted to addressing open problems or working through the literature.
Upcoming Seminars
TBA
Past Seminars 2012-13
Mar. 26
Dave Penneys
GJS C*-algebras
Guionnet-Jones-Shlyakhtenko (GJS) gave a diagrammatic proof of a result of Popa which reconstructs a subfactor from a subfactor planar algebra. In the process, certain canonical graded *-algebras with traces appear. In the GJS papers, they show that the von Neumann algebras generated by the graded algebras are interpolated free group factors. In ongoing joint work with Hartglass, we look at the
C * -algebras generated by the graded algebras. We are interested in a connection between subfactors and non-commutative geometry, and the first step in this process is to compute the K-theory of these C * -algebras. I will talk about the current state of our work.
Mar. 21
Makoto Yamashita
Deformation of algebras from group 2-cocycles
Algebras with graded by a discrete can be deformed using 2-cocycles on the base group. We give a K-theoretic isomorphism of such deformations, generalizing the previously known cases of the theta-deformations and the reduced twisted group algebras. When we perturb the deformation parameter, the monodromy of the Gauss-Manin connection can be identified with the action of the group cohomology.
Mar. 5
Danny Hay
Feb. 28
Martino Lupini
C*-algebras and omytting types
I will give an introduction to model theory for operator algebras. I will then explain how many important classes of C*-algebras can be characterized by the model-theoretic notion of omitting types. I will conclude presenting some applications to the theory of UHF and AF algebras.
Jan. 14
Martino Lupini
Logic, ultraproducts and central sequence algebras
I will introduce the fundamental notions of the logic for metric structures, among which the notiond of ultraproduct. I will then describe and study in this framework the notion of central sequence algebra of a unital C*-algebra, in particular in relation to its applications in the paper "CENTRAL SEQUENCE C*-ALGEBRAS AND TENSORIAL ABSORPTION OF THE JIANG-SU ALGEBRA"
Dec 19 Bradd Hart, McMaster University
Model theory and operator algebra
The title is a catch all for the thesis that these two subjects have something to do with one another. I will try to make this point by looking at the state of our knowledge of the continuous theory of the hyperfinite II_1 factor and its relationship with the Connes Embedding Problem.
Dec 12 Dave Penneys
II_1 factors and subfactors 2
We will give a broad introduction to II_1 factors, starting with some basic facts, including standard form and the coupling constant. We will then focus on subfactors, and we will aim to discuss some classification results.
Dec 5 Dave Penneys
We will give a broad introduction to II_1 factors, starting with some basic facts, including standard form and the coupling constant. We will then focus on subfactors, and we will aim to discuss some classification results.
Nov 28
Paul McKenney (CMU)
Automorphisms of a corona algebra
I will discuss the automorphisms of $\ell^\infty(CAR) / c_0(CAR)$, and give an overview of the proof that, assuming Todorcevic's Axiom and Martin's Axiom, they are all trivial.
Nov 21
Saeed Ghasemi (York)
Tensor products of SAW* algebras
First I will introduce SAW*-algebras as non-commutative analogous of sub-Stonean spaces in topology and state some of the results about sub-Stonean spaces which have been generalized by G.K. Pedersen to SAW*-algebras. Secondly I will show that there is no surjective *-homomorphism from SAW*-algebras into tensor product of two infinite dimensional C*-algebras using the corresponding result for sub-Stonean spaces, i.e. SAW*-algebras are essentially non-factorizable.
Nov 14
No Seminar
Nov 7
Isaac Goldbring (UIC)
Pseudofinite and pseudocompact metric structures
In classical logic, an L-structure M is said to be pseudofinite if every L-sentence which is true in all finite L-structures is also true in M; equivalently, if an L-sentence is true in M, then it is true in some finite L-structure. The random graph is a pseudofinite structure and pseudofinite fields have proven to be very interesting to model theorists. In joint work with Vinicius Cifu Lopes, we initiate the study of pseudofinite metric structures (in the sense of continuous logic). Due to the lack of negations in continuous logic, the aforementioned equivalence doesn't hold, leading to two separate notions, which we call pseudofinite and strongly pseudofinite. By replacing finite structures by compact structures, we obtain the related notions of pseudocompact and strongly pseudocompact. In this talk, I will discuss some basic properties of these notions as well as many examples, including a connection with von Neumann algebras. I will also discuss some interesting open questions.
Oct. 16 &
Oct. 24
No seminar Oct. 16 due to Fields Medal Symposium
No seminar October 24 due to Workshop on Forcing Axioms and their Applications, October 22-26
Oct. 10
Martino Lupini (York)
Connes spectrum and inner automorphisms of C*-algebras
I will introduce the Connes spectrum for automorphisms of C*-algebras and explain how inner automorphisms can be characterized in terms of it.
Oct. 3
Martino Lupini (York)
Spectrum of one-parameter groups of automorphisms and derivations of C*-algebras
I will introduce the spectrum of a one-paramter group of automorphisms of a C*-algebra, and present its applications to the study of derivations. In particular, I will prove the result of Sakai that every derivation of a simple C*-algebra is inner, and the result of Akemann, Elliott, Pedersen and Tomiyama that every derivation of a separable C*-algebra with continuous trace is inner.
Sept. 26
Tristan Bice (Fields Institute and York)
General applications of set theory to C*-algebras
Sept. 19
Aaron Tikuisis (University of Muenster)
Nuclear dimension and decomposition rank
Sept. 5
Luis Santiago ( University of Oregon)
Jiang-Su algebra
Aug. 29
Zhiqiang Li (University of Toronto)
A basic introduction to KK-theory of C*-algebras
I will present some basic facts about KK-groups, and show calculation of some concrete examples.
Aug. 22
1:30 p.m.
Henning Petzka (University of Toronto)
The Bott projection in the classification of C*-algebras
Vector bundles and Chern classes have been used to construct C*-algebras with rather exotic behavior. In particular, the Bott projection has played a prominent role. We will look on some constructions of `exotic' C*-algebras, - for instance Rordam's simple C*-algebra containing both a finite and a (non-zero) infinite projection,- and the unifying idea behind their constructions.
Aug. 15
Jorge Plazas ( Fields Institute)
An Introduction to Noncommutative Geometry
Noncommutative geometry, largely based on the theory operator algebras, extends the tools of geometry beyond their classical scope leading to deep insights and applications in various areas of mathematics. In this talk we will give a short review of some of the key ideas of the field, introduce some of its techniques and discuss a few examples.
Aug. 8
Luis Santiago (University of Oregon)
An Introduction to Cuntz Semigroup
Aug 1, 2012
Zhiqiang Li
Introduction to KK-theory
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http://nrich.maths.org/1320
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### Amida
To draw lots each player chooses a different upright, the paper is then unrolled, the paths charted and the results declared. Prove that no two paths ever end up at the foot of the same upright?
### Painting Cubes
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Permute It
Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers.
# Ding Dong Bell
##### Stage: 3, 4 and 5
Article by Toni Beardon
The sound of bells
If you wear a different coloured T-shirt every day someone may say that you are "ringing the changes". This expression simply means that you plan for variety rather than keeping things the same. It comes from the the custom of ringing church bells.
In East Anglia, as you look across the fens, villages appear almost like little islands (indeed some of them were islands before the fens were drained) and these villages are dominated by big churches with tall towers. In the past people regulated their lives and passed messages by ringing church bells, which could be heard for miles around, telling the time of day, and giving news of births, marriages and deaths in a parish. The following quotation comes from the ringer's rules from Southhill in Bedfordshire "When mirth and pleasure is on the wing we ring; at the departure of a soul we toll".
Bell ringing is good exercise for the body and mind, the bells are heavy and the bellringers have to remember the changes. Bell ringing often appeals to mathematicians because there is a lot of mathematics to be found in these complicated sequences of changes. Before reading on you might like to try this problem on bellringing.
The mathematics of the changes
The reader is invited to investigate permutations in the ringing of church bells, that is the different orders in which the bells can be rung on a ring of bells. Each bell is mounted on a wheel and controlled by a bellringer pulling on a rope. A bell rings once per revolution. Roughly speaking it rotates 360 degrees from top-dead-centre back to top-dead-centre and strikes (once) when it is about 70% of the way round. The bell rings alternately clockwise and anti-clockwise (roughly 360 degrees each way) and two revolutions constitute a so-called whole pull. Rather confusingly a whole pull (singular) consists of two pulls (plural) on the rope. The bellringer can speed up or slow down his or her bell to alter its turn in the order by one place, but not by more than one. Each bell is rung just once in a sequence (a row), then all the bells are rung again, but in a different order (a permutation of the previous order), then all again in a different order and so on. We say the bells are rung in rounds when the bells are rung in order with the notes descending in pitch from the lightest bell, the treble, to the heaviest, the tenor.
So for example, with just three bells there are six possible orders in which they can ring. Mathematically speaking there is a group of six bell ringing rows. Starting with the row 123 the next could be 213 (interchanging the order of the first two) or 132 (interchanging the order of the second two), but it could not be anything else. When the order in which the first two bells are rung is interchanged we call the change p, and when the order of the second pair are interchanged we call the change q. One way of ringing every posible row once beginning and ending with rounds is to ring the changes p then q then p then q then p and finally q again as illustrated by this braid diagram. Each line shows the order in which the bells are rung and the bellringers remember their 'parts' by remembering the shape and pattern of their strand in the braid.
A circuit diagram provides another way of illustrating these changes. Here we see the change p illustrated by a black line, and q by a blue line. Each node represents one of the six permutations. By traversing the circuit either clockwise or anticlockwise, we see two ways to ring a block denoted by ${(pq)}^3$ and ${(qp)}^3$.
With four bells there are of course many more possibilities; there are twenty four different permutations or orders in which the bells can be rung and there are four bell ringing changes.
Braid diagram 2 Plain Hunt Minumus. The first pair can be transposed and the second pair stay the same, we call this p, the second pair can be transposed and the first pair stay the same, we call this q, the middle pair can be transposed and we call this r, and finally both pairs can be transposed and we call this s. The bellringers' notation records the bells that don't move so what we record as p or X|| is written 34, what we record as q or ||X is written as 12, what we record as r or |X| is written as 14 and what we record as s or XX has the special name 'cross'. Repeating $s$ and $r$ alternately gives the sequence of changes shown in braid diagram 2 which is denoted by ${(sr)}^4$ representing Plain Hunt Minimus. Braid diagram 2 does not give all 24 permutations although it does start and finish with the same row and bellringers describe this as a round block. Braid diagram 3 gives an example where all 24 rows occur once and only once and the sequence starts and ends with the same row.
Now look at the network diagram below which shows all 24 possible permutations in the order of the four bells. This network represents all the sequences of the changes $p$, $q$, $r$ and $s$ for four bells in much the same way as the simple hexagonal network for three bells. Any two adjacent vertices in the network represent permutations for which there is a bell ringing change between one and the other. The four changes $p$, $q$, $r$ and $s$ are shown in four different colours.
Find the vertex labelled 1234. Now follow the green edge which represents the change s from 1234 to 2143. This is the first change shown in braid diagram 2 which shows the sequence $s r s r s r s r$ or ${(sr)}^4$ which starts and ends with 1234. This is shown in the network below by the large circular loop on the bottom right hand side. As we noted before, although this loop starts and ends at the same vertex it does not visit every vertex in the network. However, by traversing the entire network, visiting each vertex once and only once and returning to the stating point, we get an extent on four bells in which each row is rung once and only once.
Circuit diagram showing the 24 permutations of 4 bells
Bellringers are interested in different ways of ringing all possible changes without repeats which is known as an extent and braid diagram 3 shows an example of an extent on four bells. If you start at 1234 in the network diagram above can you trace a path around the network corresponding to braid diagram 3 visiting each vertex in the network once and only once? Start at the vertex representing 1234 then follow the green path for the permutation s to the vertex for 2143, then follow the red path for the permutation r to the vertex for 2413 and so on... and finally return to the starting point along the yellow path for the permutation p from the vertex 2134 to the vertex 1234. This is denoted by $\{{(sr)^3sp\}}^3$. Can you find other Hamiltonian circuits around the network visiting every vertex once and only once, and can you draw their corresponding braid diagrams? For example, to ring what bellringers call Plain Bob, you follow the path ${(sr)}^3s$ and then instead of switching by $p$ from 1324 to 3124 you switch by $q$ from 1324 to 1342 and use $q$ again rather than $p$ later in the sequence giving${\{(sr)^3sq \}}^3$. Each bellringer remembers his or her pattern in the braid and, in particular, the 'dodges', for example where bell number 1 permutes with 3 and then with 3 again at the first occurrence of the permutation $p$ in braid diagram 3. Braid diagram 3 represents an extent on four bells.
In addition to studying the mathematics of bellringing, there is an interest in investigations into the mechanics of church bell ringing as described in this news item from the Cambridge University website of 17th January 2005: "As part of a fourth-year student project on the mechanics of church bell ringing, a "bell tower" has been built in the Mechanics Laboratory. The bell has been instrumented to provide data on the movements of bell and clapper, to compare with a theoretical model. In this picture, the bell is being rung by Dr Frank King of the Computer Laboratory, who is the official University Bellringer. He is being watched by Professor David Newland, past head of the Engineering Department and a keen bellringer. "
A computer program can be written to play this music. You might be able to borrow some hand bells, or you could try it out on any musical instrument, or you could even tune glass bottles, with different amounts of water in them, and 'ring' the bottle-bells by striking them with a teaspoon.
The next article will explain the group theory which underlies these ideas and which is represented by the network diagram.
Further information can be found in Frank King's book Ringing Elementary Minor Methods on Handbells, on the Central Council of Church Bellringers website, in John Harrison's website, the Wikepedia website and many more.
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://nrich.maths.org/2086
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### F'arc'tion
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and paper.
### Do Unto Caesar
At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won \$1 200. What were the assets of the players at the beginning of the evening?
### Plutarch's Boxes
According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have their surface areas equal to their volumes?
# Farey Sequences
##### Stage: 3 Challenge Level:
If I gave you a list of decimals, you might find it quite straightforward to put them in order of size. But what about ordering fractions?
A man called John Farey investigated sequences of fractions in order of size - they are called Farey Sequences.
The third Farey Sequence, $F_3$, looks like this:
It lists in order all the fractions between $0$ and $1$, in their simplest forms, with denominators up to and including $3$.
Here is $F_4$:
Write down $F_5$.
Which extra fractions are in $F_5$ which weren't in $F_4$?
Use $F_5$ to help you complete $F_6$ and $F_7$.
Here are some questions to consider:
There are lots of extra fractions in $F_{11}$ which are not in $F_{10}$.
There are only a few extra fractions in $F_{12}$ which are not in $F_{11}$.
Can you explain why this is the case?
When will you need lots of extra fractions to get the next Farey Sequence?
Will every Farey Sequence be longer than the one before? How do you know?
So far, all the Farey Sequences have contained an odd number of fractions. Can you find a Farey Sequence with an even number of fractions?
In $F_4$, $\frac{3}{4}$ slotted in between $\frac{2}{3}$ and $\frac{1}{1}$. What do you notice about the fractions on either side when you slot in a new fraction?
Choose any three consecutive fractions from a Farey Sequence. Can you find a way to combine the two outer fractions to make the middle one?
To see how these sequences relate to some beautiful mathematical patterns see the pictures in the problem Ford Circles. There is no need to attempt this problem at this stage - it is aimed at older students.
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://physics.stackexchange.com/questions/33377/what-is-the-mechanism-for-fast-scrambling-of-information-by-black-holes?answertab=oldest
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# What is the mechanism for fast scrambling of information by black holes?
Sekino and Susskind have argued that black holes scramble information faster than any quantum field theory in this paper. What is the mechanism for such scrambling?
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## 2 Answers
Fast scrambling presupposes black hole complementarity, which is why it's so mysterious.
An alternative explanation is retrocausality. Black hole singularities exert retrocausal influences. Suppose you have a black hole with a radius of one light-second. A spacecraft carrying some quantum information passes through the event horizon and hits the singularity. Work in Eddington-Finkelstein coordinates. Ingoing light rays take one second to go from the horizon to the singularity. The spacecraft moves slower than that, so it might take two or so seconds to hit the singularity. What hits the singularity at that moment? It is the spacecraft and the quantum information it carries. But also, pair entangled Hawking particle pairs of massless particles are constantly produced across the event horizon. One of them lies outside and escapes the black holes gravitational pull. The other falls in and hits the singularity. Infalling Hawking radiation also hits the singularity at the same instant. Retrocausal influences from the singularity shows up as a fixing of the form of entanglement between everything which will hit the singularity at the same instant, i.e. the internal microstates of the spacecraft and the infalling Hawking radiation. This can be described in the two-state formalism as postselection. However, the infalling Hawking radiation is also entangled with the outgoing Hawking radiation. So, information on the microstates of the spacecraft are transmitted retrocausally to the outgoing Hawking radiation via this mechanism. In flat spacetime, "virtual" particle pair entangled states don't increase in spatial separation. In the warped geometry of a black hole, a pair of "not-so-virtual" massless particles both travelling locally outward at the speed of light across the event horizon will see the spatial distance between them increasing with time. If their initial wavelength is $\lambda$, there will be around $A/\lambda^2$ many of such pairs at any given instant. The smallest wavelengths are cutoff around the Planck scale, and at that scale, there are about $A/\ell_P^2$ of them. Such pairs dominate the counting, but longer wavelengths also contribute. The infalling Hawking particle is travelling outward locally, but is still dragged inward by coordinate dragging. Initially, it is about $\lambda$ inward from the boundary. So, it would take about $R\ln(R/\lambda)$ to hit the singularity. There's a cutoff about the Planck wavelength, so the longest time it would take is around $R\ln(R/\ell_P)$ which is about 100 seconds or so. So, all you need is to collect all the outgoing Hawking radiation starting 100 seconds or so before the craft passes the horizon until one second after (two seconds minus one second) to extract the infalling quantum information. Any earlier Hawking radiation is not needed. Because infalling Hawking radiation hitting the singularity can come in from all directions, outgoing Hawking radition has to be collected from all directions.
The Hayden-Presskill mechanism is all wrong because it assumes collecting all info starting from the moment the craft hits the horizon for about 100 seconds or so after that, with the entire history of all the outgoing hawking radiation since the formation of the black hole needed as context.
If there’s a possibility that once craft is just inside the horizon, it hits something which causes it to recoil locally outward near the speed of light, then it can take longer than two seconds to hit the singularity. In that case, we might still need to collect all outgoing Hawking radiation up till 100 secs later “just in case”.
The black hole information paradox is only so mysterious because most people have an automatic mental block against retrocausality.
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The mechanism by Charging Bull is interesting, but deeply flawed. It is true at any given moment, there are about $A/\ell_P^2$ entangled Hawking pairs across the horizon, and that the curvature of the black hole serves to pull them apart. The particle on the inside will of course hit the singularity. But also note that unless the direction of of the outgoing Hawking particle points nearly exactly outward in direction, it will only move out from the horizon a bit before following a curved trajectory in Eddington-Finkelstein coordinates causing it to fall back through the horizon and also hit the singularity, but at a different moment. Only an average of an order unity of the $A/\ell_P^2$ many entangled pairs lead to an outgoing Hawking particle escaping for good. Those which escape are mostly s-waves, with a few p-waves and d-waves delocalized over all directions. What we get instead are $A/\ell_P^2$ entangled pairs of ingoing Hawking particles hitting the singularity at different moments. Also, it's also true there are $A/\lambda^2$ entangled pairs of wavelength $\lambda$ for the vacuum, but they are really redshifted versions of some of the $A/\ell_P^2$ entangled pairs from an earlier moment due to the redshifting mechanism of Eddington-Finkelstein warping. They are not independent pairs. The other entangled pairs also show up as $\lambda$ wavelength entanglements later after redshifting, but of the nonvacuum sort, a squeezed state sort.
In fact, if we're only interested in pairs of infalling Hawking particles hitting the singularity at different moments, it's not even necessary for them to be produced across the horizon. That's only necessary for entangled pairs with one escaping for good and the other hitting the singularity. The location of pair production of both eventually infalling Hawking particles can be outside the horizon, across it, or inside it.
On the average, we only get an order of unity of outgoing Hawking qubits per unit Schwarzschild time. So, over a time period of $R\ln(R/\ell_P)$, only somelike like $c\ln(R/\ell_P)$ qubits could have escaped where $c$ is of order unity. It's possible for the spacecraft to dump a lot more information than that, and we have Holevo bounds. The problem with the Hayden-Preskill mechanism is, maybe Alice's top secret diary contains $k$ qubits of information. But that doesn't necessarily mean collecting only $ck\ln(R/\ell_P)$ qubits of outgoing radiation will suffice to snoop into her top secret confidential diary. What they overlooked is the microstates of the craft contain a lot more than k qubits worth of info. To suggest that only collecting that few qubits is enough to crack the k chosen diary qubits out of the that many more craft microstate qubits is to violate information causality.
But this suggests another very interesting mechanism. If the craft hits the singularity at time $t$, it hits simultaneously with $A/\ell_P^2$ ingoing Hawking particles. Actually, probably more than that, maybe $R^3/\ell_P^3$ because the location of pair production can be far inside the hole. Retrocausal influences still exist, but most of them relate to pairs of entangled infalling Hawking radiation, both infalling. Via these entanglements, this retrocausal effect zigzags back forward causally again to the singularity at a different moment. This zigzag mechanism can continue for many many iterations in both directions in Eddington-Finkelstein time, which really becomes space inside the black hole. A tiny fraction of the entangled Hawking pairs really have an escaping outgoing Hawking particle. These carry scrambled information about the craft. This information is spread out over a long time both far into the future and far into the past of the moment when the craft plunges into the hole. A very long zigzag of alternating causality and retrocausality carried over many entangled pairs of infalling Hawking particles hitting the singularity at different moments.
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