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## Return to Answer
3 edited body
A simple-minded argument:
Pick any element $g\in G$ of order $p$, the characteristic of $k$, and let $C(g)$ be the centralizer of $g$.
Now $G\setminus C(g)$ is the disjoint union of orbits under the action of the inner automorphism $\iota_g:h\in G\mapsto ghg^{-1}\in G$, and those orbits are of size $p$. The sum of the terms in your sum corresponding to the elements of $g$ in one of those orbits is then zero, for those terms are all equal.
If $h\in C(g)$, then the terms in your sum corresponding to the elements $h$, $gh$, $g^2h$, $\dots$, $g^{p-1}h$ are all equal---because $g$ and $h$ commute, you can take them simutaneously to Jordan canonical form and $g$ has only $1$ as an eigenvalue---so that their sum is also zero.
We have thus partitioned $G$ into, on one hand, the orbits of $\iota_g$ in $G\setminus C(g)$, and, on the other, the cosets of $\langle g\rangle$ in $C(g)$, and checked that the sum of the terms in each part of this partition is zero. Therefore your sum is zero.
NB: Notice that the specific form of the terms in your sum do does not really matter, as long as they it only depend on the eigenvalues of the $g\in G$. Thus, for example, exactly the same reasoning shows that the "other" Molien formula $$\sum_{g\in G}\det(I-t\rho(g))$$ also vanishes.
2 added 285 characters in body
A simple-minded argument:
Pick any element $g\in G$ of order $p$, the characteristic of $k$, and let $C(g)$ be the centralizer of $g$.
Now $G\setminus C(g)$ is the disjoint union of orbits under the action of the inner automorphism $\iota_g:h\in G\mapsto ghg^{-1}\in G$, and those orbits are of size $p$. The sum of the terms in your sum corresponding to the elements of $g$ in one of those orbits is then zero, for those terms are all equal.
If $h\in C(g)$, then the terms in your sum corresponding to the elements $h$, $gh$, $g^2h$, $\dots$, $g^{p-1}h$ are all equal---because $g$ and $h$ commute, you can take them simutaneously to Jordan canonical form and $g$ has only $1$ as an eigenvalue---so that their sum is also zero.
We have thus partitioned $G$ into, on one hand, the orbits of $\iota_g$ in $G\setminus C(g)$, and, on the other, the cosets of $\langle g\rangle$ in $C(g)$, and checked that the sum of the terms in each part of this partition is zero. Therefore your sum is zero.
NB: Notice that the specific form of the terms in your sum do not really matter, as long as they only depend on the eigenvalues of the $g\in G$. Thus, for example, exactly the same reasoning shows that the "other" Molien formula $$\sum_{g\in G}\det(I-t\rho(g))$$ also vanishes.
1
A simple-minded argument:
Pick any element $g\in G$ of order $p$, the characteristic of $k$, and let $C(g)$ be the centralizer of $g$.
Now $G\setminus C(g)$ is the disjoint union of orbits under the action of the inner automorphism $\iota_g:h\in G\mapsto ghg^{-1}\in G$, and those orbits are of size $p$. The sum of the terms in your sum corresponding to the elements of $g$ in one of those orbits is then zero, for those terms are all equal.
If $h\in C(g)$, then the terms in your sum corresponding to the elements $h$, $gh$, $g^2h$, $\dots$, $g^{p-1}h$ are all equal---because $g$ and $h$ commute, you can take them simutaneously to Jordan canonical form and $g$ has only $1$ as an eigenvalue---so that their sum is also zero.
We have thus partitioned $G$ into, on one hand, the orbits of $\iota_g$ in $G\setminus C(g)$, and, on the other, the cosets of $\langle g\rangle$ in $C(g)$, and checked that the sum of the terms in each part of this partition is zero. Therefore your sum is zero.
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http://physics.stackexchange.com/questions/52765/how-does-the-voltage-between-two-charged-sheets-change-if-change-their-distance/52828
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# How does the voltage between two charged sheets change if change their distance
Suppose I have two charged capacitor plates that both are isolated and carry a charge density $D = \frac QA$. According to textbook physics the electric field between them is given by $E=\frac D {\epsilon\epsilon_0}$ and the voltage by $U = Ed = \frac {Dd}{\epsilon\epsilon_0}$ with $d$ the distance between the plates. According to the formula for the voltage from above I could set any voltage between the plates if I just separate them far enough from each other and also the electric field would be constant no matter how far the plates are apart which is also quite counter-intuitive. As far as I remember this is true as long as $d$ is small compared to the size of the charged plates.
But what if this condition no longer holds? What is happening then? Is there another formula for this case that is comparably simple? I would suppose that for very large $d$ the whole thing can be seen as two point charges which would give a $\frac1r$ dependency of the voltage. But what is happening in between?
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## 2 Answers
Here is a simplified approach to this question-I hope it is not too simplistic. Sorry I could not upload the mathematics and the illustrating diagram from my computer file. I need to learn how to do this, or I would appreciate if someone could leave some ideas.
Basically the approach by "MyUserIsThis" is intuitively sound. The analysis is not detailed enough to show how V depends on d (distance between the plates) at small and large d.
Imagine the two parallel plates $P_1$ and $P_2$ with finite Area $A_1$ and $A_2$, carrying eletric charges with uniform densities $D_1$ (charge $+Q_1$) and $D_2$ (charge $-Q_2$) respectively. The plates are placed on top of each other ($P_2$ above $P_1$). We assume uniform densities for simplicity. Now, choose two differential elements: $dA_1 = dx_1dy_1$ on $P_1$ at point ($x_1, y_1, 0$) and $dA_2=dx_2dy_2$ on $P_2$ at point ($x_2, y_2, d$). The potential difference between the plates is given from standard electrostatic theory, leading to the following general but ‘complicated’ double integral on the surfaces $A_1$ and $A_2$
$V(d)=-\frac{D_1D_2}{4\pi \epsilon_o} \int_{A1,A2} dA_1dA_2 \frac{1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+d^2}}$
However, for large $d$-values, due to the small size of the plates, the terms $(x_2-x_1)^2$ and $(y_2-y_1)^2$ are very small compared to $d^2$, so that the above equation reduces to this
$V(d)=-\frac{D_1D_2A_1A_2}{4\pi \epsilon_o} \frac1d= \frac{-Q_1Q_2}{4\pi \epsilon_o} \frac1d$
Therefore, the potential difference drops as $1/d$, which is equivalent to saying that for large $d$, the two plates see each other as point particles of charge $+Q_1$ and $-Q_2$, as mentioned in the previous answer. I hope this adds some clarity to the answer.
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Thanks Benedikt. I appreciate your editing the mathematics of my answer. I hope soon I shall be able to do this myself. – JKL Feb 2 at 0:08
You're very welcome. You will see, writing formula isn't that complicated as in most cases you will need only a few basic commands and the syntax is nearly completely derived from `LaTeX` syntax. Therefore also a quick look in to some short introduction on maths in `LaTeX` could be helpful. Also thanks for the 'complicated' double integral formula, maybe I can reduce it a it in complexity for my special use case. Oh, nearly forgot: if you write an `@name` into your comment, user "name" will be notified that there is a new comment – Benedikt Bauer Feb 2 at 18:24
Comparably simple, no. That condition is there so you can avoid boundary effects in the limits of the sheets. If they're close enough you can approximate their size/distance ratio to infinity and this allows you to suppose the field /by Gauss law) has tu be uniform inside. This is what actually happens:
As you can see the field stops being uniform an perpendicular to the sheet in the borders.
That condition may not be true always, by the way. For example, if you have a couple of plates and you connect them to a battery, then the battery will maintain the voltage difference between the plates, even if you change the distande $d$, by moving charges in and out, until the potential bewtween the plates is the same as in the battery and you get to a stationary point. If instead of connecting a battery you charge them with an amount of charge $\pm q$ on each plate, then what happens is what you have said. By using Gauss law, you can create gaussian surfaces in which to integrate the field and deduce that if the charge is constant, then the field $E$ will be constant, and the potential is what's going to change. This are the gaussian sufaces you could pick (the rectangular pillbox):
If the field is constant and uniform, the potential will be, by definition: $$V=\int_0^d Edx=E\int_0^ddx=Ed$$ And that's the way you get your formula. All of the above can be applied as long as we keep the approximation of very large and close plates, so we can ignore boundary effects.
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Thank you for the elaborate explanation. Unfortunately this doesn't really answer my question. I already had a clue how to come to the simplified formula but I don't know how to calculate this if the simplified formula is not applicable. – Benedikt Bauer Feb 1 at 11:32
1
@BenediktBauer You can't get a general formula in the case in which you can't ignore boundary effects. When you need values of field or voltage in complex cases like that one, you generally use numerical methods (computers) to get numerical values, but you can't get formulas. That's how pictures like the first one in my answer are done, by numerical approximations, that with computers can have huge preccisions. – MyUserIsThis Feb 1 at 12:06
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http://math.stackexchange.com/questions/234339/if-displaystyle-sum-n-0-infty-c-n4n-is-convergent-is-displaystyle-su/234352
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# If $\displaystyle \sum_{n=0}^\infty c_n4^n$ is convergent, is $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ convergent as well?
Please identify the flaw in my reasoning:
$\displaystyle \sum_{n=0}^\infty c_n4^n$ is convergent, so by the ratio test: $\displaystyle \lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}4^{n+1}}{c_n4^n}\right\vert < 1$.
$\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}(4)\right\vert < 1 \Rightarrow \displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert < \frac{1}{4}$
Now, applying the ratio test to $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$:
$\displaystyle \lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}(-4)^{n+1}}{c_n(-4)^n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}(-4)\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert(4)$
Since $\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert < \frac{1}{4}$, $\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert(4) < 1$.
Therefore by the ratio test, $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ is convergent.
Turns out the answer is that we cannot conclude $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ is convergent, so I'm trying to figure out where I took a wrong turn.
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4
If we apply the ration test to $\sum_n\frac 1{n^2}$n we don't have the conclusion you have on the first line. – Davide Giraudo Nov 10 '12 at 18:25
Typo, I of course meant the ratio test. – Davide Giraudo Nov 10 '12 at 18:40
2
Could $c_n$ itself have an alternating sign? – Greg Ros Nov 10 '12 at 18:40
3
The ratio test says that if a certain limit is less than $1$, then a series converges. It does not say that if the series converges, then that limit is less than $1$. – Michael Hardy Nov 10 '12 at 18:46
## 4 Answers
You’re misusing the ratio test. It tells you that if the limit is less that $1$, the series converges; it does not tell you that if the series converges, the limit is less than $1$. Thus, your very first line is wrong.
Here’s an example illustrating how $\displaystyle \sum_{n=1}^\infty c_n(-4)^n$ can fail to converge even when $\displaystyle \sum_{n=1}^\infty c_n4^n$ converges.
Let $c_n=\dfrac{(-1)^n}{n4^n}$; then $$\sum_{n=1}^\infty c_n(-4)^n=\sum_{n=1}^\infty\frac{(-1)^n(-4)^n}{n4^n}=\sum_{n=1}^\infty\frac1n\;,$$
which of course diverges: it’s the harmonic series. But
$$\sum_{n=1}^\infty c_n4^n=\sum_{n=1}^\infty\frac{(-1)^n4^n}{n4^n}=\sum_{n=1}^\infty\frac{(-1)^n}n$$
is the alternating harmonic series, which is convergent.
(If you insist on having the index start at $0$, just let $c_0=0$.)
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+1 : I was about to post this same example. – Michael Hardy Nov 10 '12 at 18:53
Brian M. Scott has posted what may be among the simplest counterexamples. You can create a counterexample by starting with any conditionally convergent series and doing what he did with it.
But the question asks for identification of flaws in the reasoning. Here's one: The ratio test says that if a certain limit is less than $1$, then a series converges. It does not say that if the series converges, then that limit is less than $1$. In fact, there are some cases in which the limiting ratio is $1$ and the series converges: $\sum_n 1/n^2$ is one such case. (There are also some cases where the limiting ratio is $1$ and the series diverges. One such case is $\sum_n 1/n$. The existence of examples of both kinds is precisely the reason why it is said that when the limit is $1$ then the test is inconclusive.)
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Let
$$c_n:=\frac{(-1)^n}{n4^n}\Longrightarrow \sum_{n=1}^\infty c_n4^n=\sum_{n=1}\frac{(-1)^n}{n}\,\,\,\;\;\;\text{converges}$$
yet
$$\sum_{n=1}^\infty c_n(-4)^n=\sum_{n=1}^\infty\frac{1}{n}\,\,\,\;\;\;\;\;\text{diverges}$$
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You can express you series as a power series $$\sum c_nx^n.$$ The radius of convergence of the series can be given by the ratio test. Suppose that radius is $r$. We have $3$ different cases
• If $r$ is $0$, then the series only converges for $x=0$.
• If $r=\infty$, then the series converges for every value of $x$
• If $0<r<\infty$, then the series converges for every values of $x$ in $(-r,r)$
• Now in this case, you need to check if it converges at $x=r,x=-r$. It is possible to converge at $-r$ but not at $r$, however if you converge at $r$, then you also converge at $-r$.
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http://mathhelpforum.com/advanced-statistics/4147-sampling.html
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# Thread:
1. ## Sampling
An unbiased dice is thrown once. The var(x)=35/12. The same dice is thrown 70 times.
a) Find the probability that the mean score is less than 3.3.
b)Find the probability that the total score exceeds 260.
The final answer is not expected but please give working.
2. Originally Posted by kingkaisai2
An unbiased dice is thrown once. The var(x)=35/12. The same dice is thrown 70 times.
a) Find the probability that the mean score is less than 3.3.
b)Find the probability that the total score exceeds 260.
The final answer is not expected but please give working.
I will assume that this is a question where we use the central limit theorem to
justify treating the distribution of the mean score as normal with mean equal
to the mean for a single throw of the die $\mu=3.5$ and variance equal to the variance
of the result of a single throw divided by the sample size $\sigma^2=(35/12)/70=1/24$.
a) Now a mean of less than 3.3 means 230 or less for the sum, so to correct
for the continuity of the normal approximation we use a z-score for a mean
of 230.5/70, which is the z-score for for a mean of 3.2928 so:
$<br /> z=\frac{3.2928-3.5}{1/\sqrt{25}}\approx -1.015<br />$
Looking this up in a normal table or using a normal calculator we have for
a normal distributed RV we have:
$<br /> p(z<-1.015) \approx 0.155<br />$
b) There are at least two equivalent ways to do this, one another is to
observe that the event: "total score exceeds 260.5" is the same as "the mean
exceeds 260.5/70". I will use the latter, then the z-score for a mean of
260.5/70 (again the extra 0.5s are continuity corrections) is:
$<br /> z=\frac{260.5/70-3.5}{1/\sqrt{24}}\approx 1.0848<br />$
Looking this up as before we find:
$<br /> p(z<1.0848) \approx 0.861<br />$
so:
$<br /> p(z>1.0848) = 1-p(z<1.0848) \approx 0.139<br />$.
RonL
Notes:
i. A mean of less than 3.3 means the sum of the 70 throws is less than
3.3x70=231. So as the normal distribution is continuous and the actual
distribution of the sum and mean is discrete we use in the normal model a
sum less than 230.5 to represent an actual sum of 230 or less.
ii. Similarly in part b), a total of more than 260 is modelled as more than 260.5
in the normal approximation.
iii. Given the nature of the way problems are set, there is a significant chance
that the expected answers to the two parts of this problem are the same.
So please check the working to see if there is some slight modification which
would make this so.
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http://physics.stackexchange.com/questions/tagged/graphene+solid-state-physics
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http://en.m.wikipedia.org/wiki/Crunode
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# Crunode
A crunode at the origin of the curve defined by y2−x2(x+1)=0
In mathematics, a crunode (archaic) or node is a point where a curve intersects itself so that both branches of the curve have distinct tangent lines at the point of intersection.
For a plane curve, defined as the locus of points f(x, y) = 0, where f(x, y) is a smooth function of variables x and y ranging over the real numbers, a crunode of the curve is a singularity of the function f, where both partial derivatives $\partial f\over \partial x$ and $\partial f\over \partial y$ vanish. Further the Hessian matrix of second derivatives will have both positive and negative eigenvalues.
## See also
This geometry-related article is a stub. You can help Wikipedia by expanding it. v t e
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http://borisreuderink.nl/
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In the next weeks I hope to add the old content again, and put some of my notes online was well, for which I added math support: $y = X \vec{w}$.
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http://psychology.wikia.com/wiki/Pearson's_chi-square_test?oldid=114657
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# Pearson's chi-square test
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Pearson's chi-square test (χ2) is one of a variety of chi-square tests – statistical procedures whose results are evaluated by reference to the chi-square distribution. It tests a null hypothesis that the relative frequencies of occurrence of observed events follow a specified frequency distribution. The events must be mutually exclusive. One of the simplest examples is the hypothesis that an ordinary six-sided die is "fair", i.e., all six outcomes occur equally often. Chi-square is calculated by finding the difference between each observed and theoretical frequency, squaring them, dividing each by the theoretical frequency, and taking the sum of the results:
$\chi^2 = \sum {(O - E)^2 \over E}$
where:
O = an observed frequency
E = an expected (theoretical) frequency, asserted by the null hypothesis
For example, to test the hypothesis that a random sample of 100 people has been drawn from a population in which men and women are equal in frequency, the observed number of men and women would be compared to the theoretical frequencies of 50 men and 50 women. If there were 45 men in the sample and 55 women:
$\chi^2 = {(45 - 50)^2 \over 50} + {(55 - 50)^2 \over 50} = 1$
If the null hypothesis is true (ie men and women are chosen with equal probability in the sample), the test statistic will be distributed with one degree of freedom. One might expect there to be two degrees of freedom, one for the male count and one for the female. However, in this case there is only one degree of freedom because the male and female count are constrained have a sum of 50 (the sample size), and this constraint reduces the number of degrees of freedom by one. Alternatively, if the male count is known the female count is determined, and vice-versa.
Consultation of the chi-square distribution for 1 degree of freedom shows that the probability of observing this difference (or a more extreme difference than this) if men and women are equally numerous in the population is approximately 0.3. This probability is higher than conventional criteria for statistical significance, so normally we would not reject the null hypothesis that the number of men in the population is the same as the number of women.
Pearson's chi-square is used to assess two types of comparison: tests of goodness of fit and tests of independence. A test of goodness of fit establishes whether or not an observed frequency distribution differs from a theoretical distribution. A test of independence assesses whether paired observations on two variables, expressed in a contingency table, are independent of each other – for example, whether people from different regions differ in the frequency with which they report that they support a political candidate.
Pearson's chi-square is the original and most widely-used chi-square test.
The null distribution of the Pearson statistic is only approximated as a chi-square distribution. This approximation arises as the true distribution, under the null hypothesis, if the expected value is given by a multinomial distribution. For large sample sizes, the central limit theorem says this distribution tends toward a certain multivariate normal distribution. In the special case where there are only two cells in the table, the expected values follow a binomial distribution.
$E =^d \mbox{Bin}(n,p)$
where:
p = probability, under the null hypothesis
n = number of observations in the sample
In the above example the hypothesised probability of a male observation is 0.5, with 100 samples. Thus we expect to observe 50 males.
When comparing the Pearson test statistic against a chi-squared distribution, the above binomial distribution is approximated as a Gaussian (normal) distribution:
$\mbox{Bin}(n,p) \approx^d \mbox{N}(np, np(1-p))$
Let O be the number of observations from the sample that are in the first cell. The Pearson test statistic can be expressed as
$\frac{(O-np)^2}{np} + \frac{(n-O-n(1-p))^2}{n(1-p)}$
Which can in turn be expressed as:
$\left(\frac{(O-np)}{\sqrt{np(1-p)}}\right)^2$
By the normal approximation to a binomial this is the square of one standard normal variate, and hence is distributed as chi-square with 1 degree of freedom.
In the general case where there are $k$ cells in the contingency table, the Normal approximation results in a sum of $k-1$ standard normal variates, and is thus distributed as chi-square with $k-1$ degrees of freedom:
In cases whereby the expected value, E, is found to be small (indicating either a small underlying population probability, or a small number of observations), the normal approximation of the multinomial distribution can fail, and in such cases it is found to be more appropriate to use the G-test, a likelihood ratio-based test statistic. Where the total sample size is small, it is necessary to use an appropriate exact test, typically either the binomial test or (for contingency tables) Fisher's exact test.
A more complicated, but more widely used form of Pearson's chi-square test arises in the case where the null hypothesis of interest includes unknown parameters . For instance we may wish to test whether some data follows a normal distribution but without specifying a mean or variance. In this situation the unknown parameters need to be estimated by the data, typically by maximum likelihood estimation, and these estimates are then used to calculate the expected values in the Pearson statistic. It is commonly stated that the degrees of freedom for the chi-square distribution of the statistic are then $k-1-r$, where $r$ is the number of unknown parameters. This result is valid when the original data was Multinomial and hence the estimated parameters are efficient for minimizing the chi-square statistic. More generally however, when maximum likelihood estimation does not coincide with minimum chi-square estimation, the distribution will lie somewhere between a chi-square distribution with $k-r-1$ and $k-1$ degrees of freedom (See for instance Chernoff and Lehmann 1954).
## References
• Chernoff H, Lehmann E.L. The use of maximum likelihood estimates in $\chi^2$ tests for goodness-of-fit. The Annals of Mathematical Statistics 1954; 25:576-586.
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http://math.stackexchange.com/questions/294436/theta-series-and-riemann-hypothesis
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# Theta series and Riemann Hypothesis
in the paper http://www.fuchs-braun.com/media/dd209bf5c2203a87ffff80a3ffffffef.pdf section 2 ' Hilbert-Polya space' page: 180 the author introduce the Theta series
$$F(\phi(x))= x^{1/2}\sum_{n=0}^{\infty}\phi (nx)$$ x >0
of course if we take the Mellin transform of this we will get the formula
$$G(1/2+is)\zeta(1/2+is)$$
$$G(s)= \int_{0}^{\infty}dtF(\phi(t))t^{s-1}$$
then the author affirms that the function vanishes at the Riemann zeros only
i know all the properties of this function but why does the authoer introduce it ?? he wants to give an spectral interpretation but how does the spectral interpretation appears from this series ??
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http://mathoverflow.net/revisions/36478/list
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## Return to Answer
In characteristic 0, Kadeishvili has a notion of $C_{\infty}$ algebra which models rational homotopy theory. See the last paragraph of the introduction of his paper arXiv:0811.1655. His point of view is to simply consider $A_{\infty}$ algebras whose operations satisfy a certain property with respect to shuffle maps. So your computer doesn't have to remember any new operations, just check that the old ones are right.
In characteristic $p$, things are probably hopeless.
Added Remark: I just want to make clear that this does not give a "trivial proof" that a commutative dga is formal as a commutative dga if the underlying dga is formal in the "non-commutative" sense. The reason is that when you transfer from cochains from cohomology, you are restricted in the kind of morphisms allowed if you are interested in the commutative theory. So, just as in the answers to this question, there is some work to be done if you want results like that (to be completely honest, there is not yet a proof that I completely understand, so declare myself agnostic).
2 added 688 characters in body
In characteristic 0, Kadeishvili has a notion of $C_{\infty}$ algebra which model models rational homotopy theory. See the last paragraph of the introduction of his paper arXiv:0811.1655. His point of view is to simply consider $A_{\infty}$ algebras whose operations satisfy a certain property with respect to shuffle maps. So your computer doesn't have to remember any new operations, just check that the old ones are right.
In characteristic $p$, things are probably hopeless.
Added Remark: I just want to make clear that this does not give a "trivial proof" that a commutative dga is formal as a commutative dga if the underlying dga is formal in the "non-commutative" sense. The reason is that when you transfer from cochains from cohomology, you are restricted in the kind of morphisms allowed if you are interested in the commutative theory. So, just as in the answers to this question, there is some work to be done if you want results like that (to be completely honest, there is not yet a proof that I completely understand, so declare myself agnostic).
1
In characteristic 0, Kadeishvili has a notion of $C_{\infty}$ algebra which model rational homotopy theory. See the last paragraph of the introduction of his paper arXiv:0811.1655. His point of view is to simply consider $A_{\infty}$ algebras whose operations satisfy a certain property with respect to shuffle maps. So your computer doesn't have to remember any new operations, just check that the old ones are right.
In characteristic $p$, things are probably hopeless.
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http://gilkalai.wordpress.com/2010/02/03/a-discrepency-problem-for-planar-configurations/
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Gil Kalai’s blog
## A Discrepancy Problem for Planar Configurations
Posted on February 3, 2010 by
Yaacov Kupitz and Micha A. Perles asked:
What is the smallest number C such that for every configuration of n points in the plane there is a line containing two or more points from the configuration for which the difference between the number of points on the two sides of the line is at most C?
We will refer to the conjecture that C is bounded as Kupitz-Perles conjecture. It was first conjectured that C=1, but Noga Alon gave an example with C=2. It is not known if C is bounded and, in fact, no example with C>2 is known.
Alon’s example
Kupitz himself proved that $C \le n/3$, Alon proved that $C \le K\sqrt n$, Perles showed that $C \le K\log n$, and Rom Pinchasi showed that $C \le K\log \log n$. This is the best known upper bound. ($K$ is a constant.) Pinchasi’s result asserts something a little stronger: whenever you have n points in the plane, not all on a line, there is a line containing two or more of the points such that in each open half plane there are at least $n/2-K\log \log n$ points.
The proof uses the method of allowable sequences developed by Eli Goodman and Ricky Pollack. Another famous application of this method is a theorem of Ugnar asserting that 2n points in the plane which are not on the same line determine at least 2n directions. The method of allowable sequences translates various problems in combinatorial geometry into problems about permutations. For example, there is a famous problem about the number of “halving lines” for planar configurations. This problem is one of the “holy grails” of combinatorial geometry. The translated problem is also quite natural as a problem about permutations. Consider the permutation on {1,2,3,…,n} which send i to n-i. A reduced representation is a presentation of this permutation as a product of $n \choose 2$ adjacent transpositions. The algebraic question is: What is the maximum number of times that a specific transposition (i,i+1) can occur in such a reduced word?
There are many geometric discrepancy problems which are closer in spirit and in methodology to discrepancy problems in combinatorics and number theory such as the Erdos Discrepancy problem, now under polymath5 attack. The Kupitz-Perles conjecture seems of a different nature, but I find it very exciting.
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http://mathoverflow.net/questions/93203?sort=newest
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## Cardinals of transitive permutation groups acting on $\{1,\dots,n\}$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is there a nice description of all divisors of $n!$ which can be realized as cardinals of permutation groups acting transitively on ${1,\dots,n}$?
A necessary condition is of course that such a divisor is a multiple of $n$.
$n$ (cyclic) and $2n$ (dihedral) are always possible but, without mistake on my behalf, $3n$ is not feasible unless $n\equiv 1\pmod 3$ (it can then be realized as a semi-direct product analogous to the dihedral group). can only be realized if $3$ divides the number of invertible integers modulo $n$.
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9
This is almost certainly too difficult to answer in general. What you write about $3n$ is not true. There are transitive groups of order $3n$ for $n=9$ and $n=14$ for example. On the other hand, there are none for $n=10$. The transitive groups have been enumerated for all $n \le 32$ by the way. Could you try asking a more specific question? – Derek Holt Apr 5 2012 at 12:23
Thank you for the correction: A semi-direct product needs of course divisibility by 3 of the number of invertible elements modulo $n$. I think you suggest that the answer is messy! – Roland Bacher Apr 5 2012 at 13:25
## 1 Answer
As Derek suggests in his comment, this question is too difficult to answer in general. However one could limit the question as follows: clearly if $K$ is a transitive permutation group then $|K|$ divides $|M|$ where $M$ is a maximal transitive subgroup of ${\mathrm Sym}(n)$; thus we can ask about the cardinality of a maximal transitive subgroup $M$ of ${\mathrm Sym}(n)$.
The O'Nan-Scott theorem is the main tool here. Roughly speaking it asserts that such a subgroup $M$ is either imprimitive (and hence a wreath product, with order formula easy), or else it is in a bunch of primitive families. Most of these families have a geometric description and, as such, it is easy to calculate their order.
The `difficult' family in this regard is the family of primitive almost simple groups. In this case one basically needs an enumeration of the maximal subgroups of all almost simple groups, which is a difficult problem but one which has received a great deal of attention.
Depending on what level of information you need, there are complete enumerations of maximal subgroups for many of the almost simples (although not all). However there are also some very nice general statements about the possible sizes of maximal subgroups. One example is this paper by Martin Liebeck; there are many others like it (many by Liebeck and his collaborators).
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http://mathhelpforum.com/calculus/208663-evaluate-derivative-using-properties-logarithms-where-needed.html
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4Thanks
• 1 Post By MarkFL
• 2 Post By Prove It
• 1 Post By MarkFL
# Thread:
1. ## Evaluate the derivative using properties of logarithms where needed.
As some added context, I am taking Calculus: Early Transcendental Functions and we are studying "The Natural Logarithm as an Integral"
The problem presented is as follows:
d/dx [ln (x5 sin x cos x)]
I've read through the material and even looked at some of the odd numbered problems that have answers but I just can't seem to get started here. Any advice on how to approach this?
2. ## Re: Evaluate the derivative using properties of logarithms where needed.
I think I would simplify the application of the chain rule a bit by first writing the expression as:
$\frac{d}{dx}\left(\ln(x^5\sin(2x))-\ln(2) \right)$
Now, the derivative of the constant $\ln(2)$ is zero, so we are left with:
$\frac{d}{dx}\left(\ln(x^5\sin(2x)) \right)$
See if you can now apply the rule:
$\frac{d}{dx}\left(\ln(u(x)) \right)=\frac{1}{u(x)}\cdot\frac{du}{dx}$
3. ## Re: Evaluate the derivative using properties of logarithms where needed.
Originally Posted by JDS
As some added context, I am taking Calculus: Early Transcendental Functions and we are studying "The Natural Logarithm as an Integral"
The problem presented is as follows:
d/dx [ln (x5 sin x cos x)]
I've read through the material and even looked at some of the odd numbered problems that have answers but I just can't seem to get started here. Any advice on how to approach this?
Following Mark's example of simplifying the logarithm, you should simplify even further before trying to take the derivative. Write it as $\displaystyle \begin{align*} y = 5\ln{(x)} + \ln{\left[ \sin{(x)} \right]} + \ln{\left[ \cos{(x)} \right]} \end{align*}$ and then apply the much simpler chain rules to each term.
4. ## Re: Evaluate the derivative using properties of logarithms where needed.
Thanks for the quick reply! Here is what I came up with and hopefully I have not confused myself!
= (1/x5 sin2x) * x5
= 1/sin2x
5. ## Re: Evaluate the derivative using properties of logarithms where needed.
Originally Posted by Prove It
Following Mark's example of simplifying the logarithm, you should simplify even further before trying to take the derivative. Write it as $\displaystyle \begin{align*} y = 5\ln{(x)} + \ln{\left[ \sin{(x)} \right]} + \ln{\left[ \cos{(x)} \right]} \end{align*}$ and then apply the much simpler chain rules to each term.
By following that, I get....
Y = 5 ln (x) + ln[sin(x)] + ln[cos(x)]
5 ln (x) + ln[cos(x)] + ln[-sin(x)]
grrr, and I am defintely sure I am confused now! LOL!
6. ## Re: Evaluate the derivative using properties of logarithms where needed.
Let's take a look at both approaches:
My approach:
$\frac{d}{dx}(\ln(x^5\sin(2x)))=\frac{x^5(2\cos(2x) )+5x^4\sin(2x))}{x^5\sin(2x)}=2\cot(2x)+5x^{-1}$
Prove It's approach:
$\frac{d}{dx}(5\ln(x)+\ln(\sin(x))+\ln(\cos(x)))=5x ^{-1}+\frac{\cos(x)}{\sin(x)}-\frac{\sin(x)}{\cos(x)}=$
$2\cot(2x)+5x^{-1}$
My approach has a more difficult application of the chain rule, but no need to apply double-angle identities (at least if you wish to combine the two trig. terms).
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http://mathoverflow.net/questions/7446/intuition-about-schemes-over-a-fixed-scheme/7474
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Intuition about schemes over a fixed scheme
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am taking a first course on Algebraic Geometry, and I am a little confused at the intuition behind looking at schemes over a fixed scheme. Categorically, I have all the motivation in the world for looking at comma categories, but I would like to make sense of this geometrically.
Here is one piece of geometric motivation I do have: A family of deformations of schemes could be thought of as a morphism $X \rightarrow Y$, where the fibers of the morphism are the schemes which are being deformed, and these are indexed by the scheme Y.
This is all well and good, but I am really interested in Schemes over $Spec(k)$ are thought of as doing geometry "over" $k$. I know that their is a nice "schemeification functor" taking varieties over $k$ to schemes over $k$, but this is somewhat unsatisfying. All that I see algebraically is that $k$ injects into the ring of global sections of the structure sheaf of $X$, but this does not seem like much of a geometric condition...
Any help would be appreciated.
EDIT: The answers I have received so far have been good, but I am looking for something more. I will try to flesh out an example that give the same style of answer that I am asking for:
The notion of a k-valued point: Every point $(a,b)$ of the real variety $x^2 + y^2 - 1 = 0$ has a corresponding evaluation homomorphism $\mathbb{R}[x,y] \rightarrow \mathbb{R}$ given by $x \mapsto a$ and $y \mapsto b$. Since $a^2 + b^2 - 1 = 0$, this homomorphis factors uniquely through $\mathbb{R}[x,y]/(x^2 + y^2 -1)$. So the real valued points of the unit circle are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{R}$. Similarly, the complex valued points are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{C}$. Actually, points of the unit circle valued in any field $k$ are going to be in one to one correspondence with homomorphisms from $\mathbb{Z}[x,y]/(x^2 +y^2 -1)$ to $k$.
Dualizing everything in sight, we are saying that the $k$- valued points of the scheme $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$ are just given by homomorphisms from the one point scheme $Spec(k)$ into $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$.
EDIT 2: Csar's comment comes very close to answering the question for me, and I will try and spell out the ideas in that comment a little better. I wish Csar had left his comment as an answer so I could select it.
So it seems to me that the most basic reason to think about schemes over a field $k$ is this:
I already spelled out above why $k$-valued points of a scheme are important. But a lot of the time, a morphism from $Spec(k)$ to $X$ will point to a generic point of $X$, not a closed point. Different morphisms can all point to the same generic point. For instance the dual of the any injection $\mathbb{Z}[x,y] \rightarrow \mathbb{C}$ (of which there are many), will all "point" to the generic point of $\mathbb{Z}[x,y]$.
On the other hand if we are looking at $\mathbb{Q}[x,y]$, this is a $\mathbb{Q}$ scheme. $Spec(\mathbb{Q})$ is also a $\mathbb{Q}$ scheme via the identity map. A morphism of $\mathbb{Q}$ schemes from $Spec(\mathbb{Q})$ to $\mathbb{Q}[x,y]$ will correspond to a closed $\mathbb{Q}$ -valued point! So this is a nice geometric reason for looking at schemes over a fixed field $k$: $k$ morphisms of $Spec(k)$ to $X$ correspond to $k$-valued closed points of $X$.
It will take some work for me to internalize the vast generalization that S-schemes entail, but I think this is a good start. Does everything I said above make sense?
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Let me see if I get your question right. The correspondence you are looking for is {close points of X} <--> {all morphisms k[X]-->k} the latter have to be $k$-homomorphisms, ie, $k-->K[X]-->k$ the composition from $k$ to $k$ has to be the identity. Now let K be a field extension of k. {Points in X with coordinates in K}<-->{all morphism k[V]-->K} again the latter have to be k-homomorphism. Now we generalize the idea of field extension by {T-points in X} whete $T$ is a S-Scheme. Then we copy-paste of the argument above, and we get a functor $T--Hom(T,X)$. This functor represents X. – Csar Lozano Huerta Dec 1 2009 at 18:18
Hmm, this seems closer in spirit to what I am after. I will have to mull things over for a bit. You might have just answered my question. – Steven Gubkin Dec 1 2009 at 18:28
3 Answers
This is going to be perhaps vague, but I'l try to write down the idea. I've been told that in doing scheme theory, most of the time what is studied is not a scheme but a morphism of schemes $f:X\rightarrow S$. The studied of such a maps can be enriched allowing a chance of "base" $S$. Such a scheme $S$ can be crazy, but I'll impose finite properties to the map $f$ to keep everything under control.
Now, as you say, if we have two $S$-schemes $f:X\rightarrow S$ and $Y\rightarrow S$ then we can consider the fiber product of them (which is a categorical product of $X$ and $Y$ over $S$). Such a process can be though of as replacing the base $S$ by the scheme $Y$. In doing so, the new map $f_Y:X_Y\rightarrow Y$ (standing for $p_2:X\times_{S}Y\rightarrow Y$) may be easier to work with than the map itself $f$. This very construction is generalizing the idea of "extending scalars".
Here is a sort of example of a product described above. Given the point $s\in S$ and its residue field $k(s)$ of the local ring $\mathcal{O}_s$ at that point. Then it is well known that $X_s=X\times_S Spec(k(s))$ has an underlying space the "fiber" $f^{-1}(s)$ (as long as $f$ is "finite"). This space can be considered as an algebraic variety over the field $k(s)$. This way $S$ is the scheme parameterizing the varieties $X_s$ some of which a priori may have different ground fields $k(s)$.
On the other hand, back to the fiber product, in considering a variety $X$ over the field $k$ (scheme of finite type over $k$), we can extend the scalars from $k$ to a field extension $K$. In doing so, we may think of such a variety now over a field extension $K$. Here is where the product comes into play. Vaguely enough, this is saying that if you have an equation and you have solved it over the field $k$ now you might ask yourself if you'd be able to solved it again but over a field extension $K$, here you want to perform an extension of scalars and the ideas written above may help out.
Needless to say that we can consider as well $X\times_k Spec(\overline{k})$ and due to the fact that field $k$ may not have been algebraically closed, such a fiber product can be handful.
Now, to get intuition, let's take a look at an example: $$\pi:Spec \mathbb{Z}[i]\rightarrow Spec(\mathbb{Z})$$
Let $X=Spec\mathbb{Z}[i]$ and notice that the fiber $$X_p=X\times_{\mathbb{Z}}\mathbb{F}_p=Spec(\mathbb{Z}[i]\otimes\mathbb{F}_p)$$
Such a fiber is going to have cardinality 2 if $p\equiv 1 mod(4)$ and cardinality 1 if $p\equiv 3 mod(4)$ (this is the fact that $p=a^2+b^2$ where $a,b\in \mathbb{Z}$ if $p\equiv 1mod 4$). These fibers are $X$ reduce mod $p$. Notice that at the generic point we have $X_0=\mathbb{Q}[i]$.
Now the $T-points$ in $X$ where $T=\mathbb{F}_p[i]$, and all that story give rise to the functor which represents the scheme $X$. If I am not wrong, the info that carries such a functor is nothing but that of the fibers of the map $\pi$. So naively it is like having $X$ fibers wise.
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A motivation for the term "over" is that if $k$ is a relatively initial ring, for instance a field, then the algebra and geometry of a scheme "over" $k$ have coefficients in $k$. Long before schemes were defined, you could talk about a curve defined "over" the complex numbers or a ring defined "over" the integers. The realization was that you might as well generalize this concept as much as possible and let $k$ be any scheme.
As you say, it also captures the notion of fibrations or families, and part of the point is that in the end that's not really different. For instance if you have a scheme over the integers $\mathbb{Z}$, is that choosing coefficients or is that a fibration? It's both, because you can specialize to $\mathbb{Z}/p$ for any prime $p$ and that specialization is a fiber.
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I tried to post this as a comment to Csar's post, but the margin was to small to contain it.
I understand what you are trying to say here, but this style of answer isn't what I am exactly looking for. Extension of scalars should not be vague. Here is an explicit example (I am sticking with circles as per my edit.): Say we are looking at rational points of the circle. Then naturally we are interested in $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$. Say we want to instead look at complex valued points. Well $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$ is a scheme over $\mathbb{Q}$, and $\mathbb{C}$ is a scheme over $\mathbb{Q}$ (the dual of the inclusion maps give the appropriate maps here). When you take the fiber product to get a scheme over $Spec(\mathbb{C})$, this corresponds to a pushout in the category of rings, i.e. a tensor product. So the fiber product is $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1) \otimes_\mathbb{Q} \mathbb{C})$ which is isomorphic to $Spec(\mathbb{C}[x,y]/(x^2+y^2-1))$, i.e. exactly what you would expect.
I would like something this explicit explaining the geometric meaning of a scheme over a fixed scheme.
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"I tried to post this as a comment to Csar's post, but the margin was to small to contain it." Haha:) – Jan Weidner Mar 11 2010 at 19:07
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http://mathoverflow.net/questions/65806/big-line-bundle-closed
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## Big line bundle [closed]
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On a Moishezon manifold,is a big line bundle nef?Why?
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Terminology? Motivation? Anything? – David Hansen May 24 2011 at 1:25
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Why would it be? Being big is very much a birational notion. Being nef is very much not. – Sándor Kovács May 24 2011 at 1:35
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p.s.: By the way, let $B$ be an arbitrary big line bundle and $E$ an arbitrary line bundle that has a global section, but which is not nef (i.e., there exists a (compact) curve $C$ such that $E|_C$ has negative degree). Then $B\otimes E^{\otimes m}$ is big for any $m\geq 0$, but not nef for $m\gg 0$. This works on any manifold/variety/scheme/space when the notions big and nef make sense. – Sándor Kovács May 24 2011 at 4:21
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http://mathoverflow.net/questions/1750/can-anyone-give-me-a-good-example-of-two-interestingly-different-ordinary-cohomol/26386
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## Can anyone give me a good example of two interestingly different ordinary cohomology theories?
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An answer to the following question would clarify my understanding of what a cohomology theory is. I know it's something that satisfies the Eilenberg-Steenrod axioms, and I know that those axioms allow you to work out quite a lot. But what sort of thing is not determined by the axioms? In particular, can someone give me a simple example of a space that has different cohomology groups with respect to two different theories? Obviously a trivial answer would be to take coefficients in different rings, so let me add the requirement that the coefficient rings should be the same. And if there's some other condition needed to make the question non-trivial, then add that in too.
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## 6 Answers
For any space that has the homotopy type of a CW complex, its cohomology is determined purely formally by the Eilenberg-Steenrod axioms, so a counterexample is necessarily some reasonably nasty space. Here's an example you can see with your bare hands: consider the space X={1,1/2,1/3,1/4,...,0}. Now 0th singular cohomology is exactly the group of Z-values functions on your space which are constant on path-components, so H^0(X)=X^Z (an uncountable group) naturally for singular cohomology. On the other hand, 0th Cech cohomology computes global sections of the constant Z sheaf, i.e. locally constant Z-valued functions on your space. These must be constant in a neighborhood of 0, so the Cech cohomology H^0(X) is actually free of countable rank, generated (for example) by the functions f_n that are 1 on 1/n, -1 on 1/(n+1), and 0 elsewhere, plus the constant function 1.
I should add that topologists don't actually care about such examples. The point of the Eilenberg-Steenrod axioms is to show that cohomology of reasonable spaces is determined by purely formal properties, and these formal properties are actually much more useful than any specific definition you could give (the only point of a definition is to show that the formal properties are consistent!). What is of interest is when you remove the dimension axiom to get "extraordinary" cohomology theories, which Oscar talks about in his answer.
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Here's a question I don't know the answer to (perhaps I should post it, hmm): What is the class of spaces for which the ES axioms determine ordinary cohomology? We know it includes all spaces with the homotopy type of a CW complex, but is that all, or are there others? – Charles Rezk Oct 21 2009 at 23:14
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@Charles: I think this indeed deserves its own question-page; please post it! – Benoit Jubin Oct 22 2009 at 0:33
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I unfortunately don't have anything of value to add, but I couldn't resist pointing out that "an example you can see with your bare hands" is a wonderful (and apparently original!) mixed metaphor. :) – Harrison Brown Oct 22 2009 at 1:07
Oh, and @Charles: I'd like to second Benoit here, please ask it! – Harrison Brown Oct 22 2009 at 1:08
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An example I recently came across while learning about shape theory is the closed topologist's sine curve. The zero-th singular cohomology (integer coefficients) is Z^2, while the zero-th Cech cohomology (again integer coefficients) is Z.
I can't give a great reason why this is so as I'm just learning about the technicalities myself, but it seems to be a standard example.
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Does Cech cohomology in this setting satisfy all the Eilenberg-Steenrod axioms? – Andreas Holmstrom Oct 21 2009 at 23:01
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Yes. In fact, it actually satisfies a stronger form of excision than singular cohomology. However Cech cohomology fails to satisfy that a weak equivalence of spaces induces an isomorphism on (ohomology. – Eric Wofsey Oct 21 2009 at 23:09
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And, to close the circle: if your cohomology theory does take weak equivalences to isomorphisms, then it is uniquely determined (up to isomorphism). – Charles Rezk Oct 21 2009 at 23:35
It might be interesting to note that the examples given by Eric and Justin also show why it is necessary to use Čech cohomology when treating Alexander duality for arbitrary closed subsets of a closed (and orientable) manifold. If one considers A={1/n | n>=1} as a subset of R/Z=S^1 then its complement is a countable disjoint union of open intervals and its 0-th homology (for any theory satisfying the axioms) is free with a countable basis. So Alexander duality would demand that the 0-th cohomology of A also have a countable basis, which is true for Čech cohomology but not for singular cohomology.
Similarly for for the closed topologist's sine curve as a subset of the two-dimensional sphere. Its complement has the homology of a point, so it itself should have the cohomology of a point, which again is true for Čech cohomology but not for singular cohomology.
This has indeed lead to a remark in the classical textbook by Seifert and Threllfall that while it is possible to define singular homology for arbitrary topological spaces and not just complexes, it is not very useful to do so. But that was in 1934.
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The examples mentioned so far, made the point that Cech cohomology differs from singular cohomology in many examples which are, to the eyes of most topologists, quite pathological. I want to concentrate on an example of an ordinary homology theory, which is probably much less well known than Cech cohomology, but quite interesting in geometric contexts: stratifold homology (see Kreck's Differential Algebraic Topology). It is defined as bordism classes of certain stratified spaces (including manifolds, but much more).
An example, where stratifold homology and singular homology differ is the one-point compactification of the surface of infinite genus (i.e. the infinite connected sum of tori), denoted by $F_\infty^+$. This can be given the structure of a stratifold and therefore possesses a fundamental class, which is non-zero, even after applying the induced map of $F_\infty^+\to F_g$ to the surface of genus g. Such a class does not exist in singular homology (see Differential Algebraic Topology, chapter 20.2). The space $F_\infty^+$ is surely not the first thing one considers, but it has still some geometric appeal (and is a compact set of $\mathbb{R}^3$). I think, there are probably similar phenomena for one-point-compactifications of certain other open manifolds.
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The sheaf cohomology with coefficients in the constant sheaf $Z$ coincides with the singular cohomology or the Cech cohomology for all homologically locally connected hausdorff paracompact spaces (this is the case for finite CW-complexes or locally contractible spaces).
Sheaf cohomology is better behaved than singular cohomology when it comes to dimension. Recall that the cohomological dimension of a compact topological space X is the greatest integer $n$ (or $\infty$) such that $H^n(X)\neq 0$.
Theorem: If $F$ is a closed subset of a compact topological space $X$, then its sheaf cohomological dimension is less than the sheaf cohomological dimension of $X$ (see e.g. Bredon "sheaf theory", II.16).
Surprisingly, this result is false for singular cohomology. The following example is due to Barratt and Milnor (1962). Let X be the union in $R^3$ of countably many spheres of radius 1/n, all tangent to the xy-plane at the origin. The sheaf cohomological dimension of this compact space is 2. But the singular cohomology of this space is non-zero in arbitrarily large degrees. In other words, its singular cohomological dimension is infinite.
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This is really cool! – Lennart Meier May 30 2010 at 8:20
An interesting cohomology theory is the continuous singular cohomology theory with real coefficients introduced, as far as I know, in the seventies by R.Bott, G.D.Mostow and others.
Let a singular cochain be called continuous if it restricts to a real continuous map on the space of singular simplices, endowed with the compact-open topology. It is readily seen that continuous cochains provide a subcomplex of the complex of singular cochains. The (co)homology of the complex of continuous cochains is called continuous cohomology.
As one can expect, continuous cohomology is canonically isometric to singular cohomology for reasonable spaces (for example, for locally contractible metrizable spaces). However, one can construct spaces with non-isomorphic continuous and singular cohomology just by looking at subspaces of the Euclidean plane.
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http://mathhelpforum.com/math-topics/36507-physics-question-print.html
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# Physics question
Printable View
• April 29th 2008, 03:57 AM
Coach
Physics question
How can I calculate the intensity of radiation of the Sun at Earth(at the same distance as the Earth is at). The intenisity of radiation at the Sun is about 64MW/m2. I mean the radiation spreads around so how can I know which portion comes to the Earth?
All help is appreciated.
• April 29th 2008, 04:28 AM
Danshader
well saying that the radiation from the sun all travels at the same speed
, this radiation will cover a spherical shape of area.(just like blowing a balloon)
now the distance of the earth to the sun is about $1.5x10^{11}m$
this means you need to find the surface area of a sphere with a radius of $1.5x10^{11}m$, then you need to find the area of intersection between the sphere and the earth(assume to be in the middle of the earth)
surface area of the sphere:
$<br /> A_1 = 4\pi r_{ES}^2<br />$
where $r_{ES} ^2$ is the radius of the earth and has the value of $1.5x10^{11}$
hence the area of intersection is:
$<br /> A_2 = \pi r _E ^2<br />$
where $r_E ^2$ is the radius of the earth and has the value of $6.4x10^6$. The area of intersection is can be assumed as a circle since the distance of the earth from the sun is very much larger tahn the radius of the earth.
hence the fraction of radiation received by earth is
$\frac{A_2}{A_1}$
All times are GMT -8. The time now is 10:26 PM.
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http://mathhelpforum.com/differential-geometry/188706-topology-x.html
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# Thread:
1. ## Topology on X
Consider the set $X=\{a,b,c\}$
The collection of subsets are $\O,X,\{a,b\},\{b,c\},\{b\}$
The book says the one point set $\{b\}$ is not closed because its complement is not open. What is the complement of $\{b\}\mbox{?}$
2. ## Re: Topology on X
Originally Posted by dwsmith
Consider the set $X=\{a,b,c\}$
The collection of subsets are $\O,X,\{a,b\},\{b,c\},\{b\}$
The book says the one point set $\{b\}$ is not closed because its complement is not open. What is the complement of $\{b\}\mbox{?}$
$\color{blue}X\setminus\{b\}=\{a,c\}$
3. ## Re: Topology on X
Originally Posted by Plato
$\color{blue}X\setminus\{b\}=\{a,c\}$
But that isn't in topology.
4. ## Re: Topology on X
Originally Posted by dwsmith
But that isn't in topology.
That is exactly the point.
The complement of $\{b\}$ is not an open set, not in the topology.
There because the complement of $\{b\}$ is not open means $\{b\}$ is not closed.
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http://cstheory.stackexchange.com/questions/tagged/primal-dual
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# Tagged Questions
The primal-dual tag has no wiki summary.
1answer
250 views
### LP Approximation: Primal relaxation + rounding vs. Dual relaxation. Why is the latter better?
Given any Integer Linear Program (ILP) there are 2 ways to approximate it: Write down ILP, convert to LP by relaxing the integer constraints and round the solution Write down the ILP, convert to LP ...
2answers
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### An intuitive/informal proof for LP Duality?
What would be a good informal/intuitive proof for 'hitting the point home' about LP duality? How best to show that the minimized objective function is indeed the minimum with an intuitive way of ...
1answer
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### Is it enough for linear program constraints to be satisfied in expectation?
In the paper Randomized Primal-Dual analysis of RANKING for Online Bipartite Matching, while proving that the RANKING algorithm is $\left(1 - \frac{1}{e}\right)$-competitive, the authors show that the ...
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### Difference between weak duality and strong duality?
For an optimization problem $(P)$ and its dual $(D)$, I have read about two concepts: Weak Duality, and strong Duality. What I don't understand is how they are different: Weak duality: If $\bar{x}$ ...
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### Primal vs dual decomposition methods
I noticed that dual decomposition methods tend to be preferred over primal ones in the large scale optimization literature (here are some examples: (1), (2)). The reason seems to be, from what I ...
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### Difference between Primal Dual Algorithm for Proper and Uncrossable Functions
Williamson with many of his co-authors had worked on generalized primal dual algorithms on edge weighted graphs considering three types of functions: (1) super-modular functions (2) proper functions ...
1answer
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### Primal Dual model in the continuous domain
The continuous max flow problem is posed as follows : sup $\int_\Omega p_s(x)dx$ subject to : $|p(x)| \le C(x); \forall x \in \Omega$ $p_s(x) \le C_s(x); \forall x \in \Omega$ \$p_t(x) \le ...
1answer
735 views
### Toy Examples for Plotkin-Shmoys-Tardos and Arora-Kale solvers
I would like to understand how the Arora-Kale SDP solver approximates the Goemans-Williamson relaxation in nearly linear time, how the Plotkin-Shmoys-Tardos solver approximates fractional "packing" ...
2answers
245 views
### Necessary and sufficient conditions for the existence of a combinatorial algorithm for a given problem.
Last semester, I took a combinatorial optimization course where the reference book was Combinatorial Optimization by William J. Cook et al. It was very interesting for me to see the relationship ...
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851 views
### Generalization of the Hungarian algorithm to general undirected graphs?
The Hungarian algorithm is a combinatorial optimization algorithm which solves the maximum weight bipartite matching problem in polynomial time and anticipated the later development of the important ...
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### Use of Lagrangian dual information to prove optimalitiy of a solution : Any example?
Can anyone please tell me what is Lagrangian Dual Information and how can it be used to prove the optimality of a solution? I'm talking about the solution to NP-Complete problems. Is it something that ...
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### Online learning: Perceptron updates
It seems that the perceptron updates come from some notion of primal-dual updates for convex programs. Can anyone explain how this is true or point to relevant literature?
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http://math.stackexchange.com/questions/21003/the-elementary-coordinate-geometry-of-polynomials-of-rational-expressions-of-r
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# The elementary coordinate geometry of polynomials? Of rational expressions? Of radicals?
With a few colleagues, we're trying to design an (intermediate) algebra course (US terminology) where we stress the interplay between algebra and geometry. The algebraic topics we would like to cover are (1) linear equation in two variables, (2) quadratic equations in two variables, (3) polynomials in one variable, (4) rational functions in one variable (though we're not sure we want to introduce functions), (5) radicals.
For (1) and (2) there are obvious geometric counterparts: lines and conic sections.
Question: Are there natural geometric counterparts for (3), (4) and (5)? Are there elementary geometric constructions that naturally lead to these algebraic objects?
Side question: Are there (affordable) textbooks or lecture notes out there which have this kind of approach?
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Yes, if you're willing to work with complex numbers there is a beautiful geometry of rational functions. – Qiaochu Yuan Feb 8 '11 at 17:54
@Qiaochu Yuan: Could you elaborate a little? Thanks. – Bart Van Steirteghem Feb 9 '11 at 2:51
– Qiaochu Yuan Feb 9 '11 at 3:10
You want polynomials without defining functions??? – quanta Apr 10 '11 at 7:33
## 3 Answers
This seems rather ambitious for most U.S. college intermediate algebra courses, which typically are not-for-credit remedial courses that lie below the level of college algebra and precalculus courses. Nonetheless, here are some things I've used in precalculus courses that might of use.
To see what the graph of something like $yx^2 + 2y^3 = 3x + 3y$ looks like using a (standard implicit-incapable) graphing calculator, solve for $x$ in terms of $y$ using the quadratic formula (or $y$ in terms of $x$ when possible, but I'm giving an example where we don't have the option of solving for $y$) and then graph both solutions simultaneously as if $x$ and $y$ were switched. That is, if you're using one of the TI graphing calculators, then enter for `y1=` and `y2=` the following:
`y1 = (3+(9-4x(2x^3-3x))^(1/2))/(2x)`
`y2 = (3-(9-4x(2x^3-3x))^(1/2))/(2x)`
(Note for would-be editors: Please don't LaTeX the expressions above, as what I've given is what the calculator input should be.)
To account for the fact that you're graphing the inverse relation, rotate what you see 90 degrees counterclockwise and then reflect the rotated result across the vertical axis. Equivalently, you can reflect what you see about the line $y=x$, but I suspect what I first suggested is easier for students to carry out.
Here's a more elaborate example. Suppose we want to know what the graph of $y^4 - 4xy^3 + 2y^2 + 4xy + 1 = 0$ looks like. (Yes, I know about the Newton polygon method, but let's not go there.) Although this can be solved for $y$ in terms of $x$, it is rather difficult to do so and the result is somewhat difficult to interpret graphically by hand. You'll get the 4 different expressions $y = x \pm \sqrt{x^2 - 1} \pm \sqrt{x^2 - x} \pm \sqrt{x^2 + x}$, where the 4 sign permutations are $(+,+,+),$ $(+,-,-)$, $(-,+,-)$, and $(-,-,+)$. On the other hand, it is easy to solve for $x$ in terms of $y$ and the result is $x = \frac{\left(y^2+1\right)^2}{4y\left(y^2-1\right)}$, which can be graphed by hand using standard methods for graphing rational functions.
For graphs of polynomial functions, especially when given as (or easily put into) factored as linear and real-irreducible quadratics with real coefficients (and probably best to mostly avoid using real-irreducible quadratics, at least at the beginning), you can discuss how their graphs locally look at each $x$-intercept and how their graphs roughly look globally by using "order of contact with the $x$-axis" notions (which you don't have to define precisely, of course) and end behavior. For example, since $y = (x+2)^3 (2x+1)^2 x (3-x)$ has the form $y = (x)^3(2x)^2(x)(-x) + \;$ lower order terms, or $y = -4x^7 + \;$ lower order terms, a zoomed out view of the graph will look like the graph of $y = -4x^7$, so the graph "enters at the upper left of quadrant 2" and "exits at the lower right of quadrant 4". Also, the graph passes through the $x$-axis at $x=-2$ "in a cubic fashion" so that locally at this zero the graph looks like a version (translated and reflected, the latter because in going from left to right the graph passes from positive $y$-values to negative $y$-values) of the graph of $y = x^3$. The same kind of analysis leads to what the graph locally looks like at the other $x$-intercepts, which I'll assume you know what I'm talking about by now since this is (or at least it used to be) a fairly standard topic in precalculus courses.
In the case of rational functions, one topic that could be investigated is linear fractional transformations (as they're called in complex analysis), or quotients of linear (= affine) functions, by looking at them through the lens of precalculus transformations of the graph of $y = \frac{1}{x}$. Although you definitely don't want to consider the general case, here's the general case version, where I'm assuming $c \neq 0$. (The first equality comes from a 1-step long division calculation.)
$\frac{ax+b}{cx+d} = \frac{a}{c} + \frac{b - \frac{ad}{c}}{cx+d}$
$= \frac{a}{c} + \frac{\frac{b}{c} - \frac{ad}{c^2}}{x+\frac{d}{c}}$
$= \frac{a}{c} + \frac{1}{c^2}(bc-ad)\left[ \frac{1}{1 - \left(-\frac{d}{c}\right)}\right]$
Note that this shows the graph of $y = \frac{ax+b}{cx+d}$ can be obtained from the graph of $y = \frac{1}{x}$ by a horizontal translation to the right by $-\frac{d}{c}$ units, followed by a vertical stretch by a factor of $\frac{1}{c^2}(bc-ad)$, followed by a vertical translation up by $\frac{a}{c}$ units. For a possibly useful reference, see Edward C. Wallace, "Investigations involving involutions", Mathematics Teacher 81 (1988), pp. 578-579. [See also these related letters in the Reader Reflections column: Thomas Edwards (MT 83, p. 496), Larry Hoehm (MT 83, p. 496), Andrew Berry (MT 95, p. 406), and Sidney H. Kung (MT 97, pp. 227 & 242).] While I'm on the topic, here's an "intermediate algebra appropriate" situation where a linear fractional transformation arises. For which number or numbers $x$, if any, can we find a number whose sum with $x$ equals its product with $x$? If we denote the desired number by $y$, then the condition becomes $x + y = xy$, or $y = \frac{x}{x-1}$.
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Bart, as linear equations related to lines, some rational functions are related to orbifolds. You can also explain the solutions of r(z)=c where c is given. It might be very instructive to see the solution set of r(z)=c as c changes continuously on the sphere, or more generally, r(z)=L or C where L is an arbitrary line and C is a circle.
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For 5, for square roots, it seems almost too obvious to use the hypotenuse of right triangles. For higher roots, diagonals on cubes of higher dimensions?
For 3, a quadratic in one variable is also a conic section. For higher degrees...so this is for high school right? ...yeah this one isn't obvious.
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http://math.stackexchange.com/questions/234127/natural-isomorphism-in-linear-algebra/306198
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natural isomorphism in linear algebra
Let $\mathsf{C}$ and $\mathsf{D}$ two categories and $\mathcal F,\mathcal G$ two functors $\mathsf{C}\rightarrow\mathsf{D}$. A natural isomorphism from $\mathcal F$ to $\mathcal G$ is the datum of a isomorphism $\nu_X:\mathcal F(X)\rightarrow \mathcal G (X)$ for every $X\in Obj(\mathsf{C})$ such that for every $\alpha\in \operatorname{Hom}(X,Y)$ in $\mathsf{C}$ we have that
$$\mathcal G(\alpha)\circ\nu_X=\nu_Y\circ\mathcal F(\alpha)$$
Now, many books say that a linear isomorphism $f$ between vector spaces is a natural isomorphism if "$f$ doesn't depend from the choice of the basis". I have two questions:
1) What does formally mean the phrase "$f$ doesn't depend from the choice of the basis"?
2) How can I match the two definitions of natural isomorphism?
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2 Answers
Consider the two real $\mathbb{R}^2 \to \mathbb{R}$ defined by
• $f(x,y) = x + y$
• $g(x,y) = x$
Clearly, different values for $y$ will give different values when plugged into $f(x,y)$, but will not give you different values when plugged into $g(x,y)$. We say that the value of $g(x,y)$ "does not depend on $y$", and we say the value of $f(x,y)$ does.
Now consider calculations we might do with a rational number $q$. Here are two examples:
• Write $q=m/n$, define $r = (m+1)/n$
• Write $q=m/n$, define $s = (m+n)/n$
While both calculations are phrased as the $r$ and $s$ being computed are functionally dependent on the value of $q$, that's not quite true; there is another, implicit dependency: a choice of the way we wrote $q=m/n$. In functional terms, we've defined two functions
• $r(q,m,n) = (m+1)/n$ whenever $q = m/n$
• $s(q,m,n) = (m+n)/n$ whenever $q = m/n$
However, the value of $s$ does not depend on the choice of $m$ and $n$ satisfying $q=m/n$! This is important, because it means the value we constructed really and truly can be expressed as a function of $q$ alone; i.e. we can express it as $s(q)$.
These same concerns apply to "higher" constructions; e.g. in the usual construction of an isomorphism bewteen finite dimensional vector spaces $T : V \to V^*$, the isomorphism we construct depends not only on $V$, but on a choice of bases $B$ for $V$ (and a basis for $V^*$, but I'll choose the dual basis to $B$) so rather than getting an isomorphism $T_V : V \to V^*$, we get an isomorphism $T_{V,B} : V \to V^*$, where I've added subscripts to $T$ to indicate the dependence.
It turns out, unfortunately that the notion of a natural transformation doesn't really have anything to do with the above concerns; however, it often comes up in the same context.
Consider the construction above. $V$ is a variable that denotes a finite-dimensional vector space and $B$ a basis on $V$; it is easy to interpret the above construction not as something you do for a particular choice of vector space and basis, but as something you do for all choices. If we define
• S = { (V,B) | V is a finite dimensional vector space and B is a basis for V }
• FinVect = the category of finite-dimensional vector spaces
then we can view $S$ as a discrete category and there are functors
• $F : S \to \mathbf{FinVect}$ defined by $F(V,B) = V$
• $G : S \to \mathbf{FinVect}$ defined by $G(V,B) = V^*$
and a natural transformation
• $\eta : F \to G$ defined by choosing $\eta_{V,B}$ to be arrow constructed by the argument mentioned above.
While $\eta$ is indeed a natural transformation, we usually wouldn't say that $\eta_{V,B} : V \to V^*$ is natural, because the choice of $S$ was somewhat artificial and trivial.
e.g., since $V$ is a variable denoting a vector space, $S$ really should have had a component that looks like $\mathbf{FinVect}$; but to set everything up in terms of categories, functors, and natural transformations, we had to "forget" the vector space structure to get things to work.
The usual isomorphism $I_V : V \to V^{**}$, however, depends only on $V$. And it depends on $V$ in a way that is consistent with the vector space structure: we have
• $F : \mathbf{FinVect} \to \mathbf{FinVect}$ given by $F(V) = V$
• $G : \mathbf{FinVect} \to \mathbf{FinVect}$ given by $G(V) = V^{**}$
• $\eta : F \to G$ given by $\eta_V = I_V$.
Because of this, we say that $V$ and $V^{**}$ depend functorally on $V$, and we say that $\eta$ is natural in $V$.
Constructions that depend on choices can still be natural in this sense, though. For example, the construction of homotopy groups of a nonempty topological space depend on choosing a point in that space, but it depends functorally on the space and the choice of point.
This fact is usually expressed by coming up with a new term for a space and choice of point (an "pointed space"), and expressing homotopy in terms of pointed spaces.
The "colloquial" usage of natural usually wants to imply both of the concerns mentioned above, and usually connotes some degree of aesthetics as well.
However, I'm not sure if the colloquial usage really means to imply the technical meaning of natural I described, or if they just turn out to be highly correlated because things that violate the technical meaning tend to also have some sort of dependence on choices, or are rather aesthetically displeasing.
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Formally, the meaning of 'does not depend on particular choices (e.g., of basis)' means precisely that the functions (or morphisms more generally) form a natural transformation. In the original article where Eilenberg and Mac Lane introduce categories they explicitly say that categories are introduced to define functors and that functors are introduced to define natural transformations.
It is very common in mathematics to use the term 'natural' about a construction (long before natural transformations existed) and in pretty much all of those cases it means that one actually constructs a natural transformation in the formal sense of category theory.
Now to answer your questions. A construction in linear algebra is said to be independent of the choice of basis if the construction uses a basis in the definition but choosing any other basis will yield the same final result. For instance, one can define the determinant of a linear transformation $T:V\to V$, on some finite dimensional vector space $V$, to be the determinant of a representing matrix relative to a basis $B$. One can then show that no matter which basis is chosen the determinant of the representing matrix is always the same. Thus, this concept is independent of the choice of basis.
Incidentally, the determinant gives an example of a natural transformation, see Why determinant is a natural transformation?
Now for question 2, quite often when you are able to construct a linear transformation that does not depend on any particular choice of basis it will be the case that this will actually be a general construction that will give you a whole family of linear transformation that together will form a natural transformation.
You might want to have a look at What is a natural isomorphism? for more examples.
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The fact that confuses me is that the definition of natural transformation involves functors, but then we use the adjective "natural" for morphisms. Clearly in the case of the bidual functor we show that this functor is naturally isomorphic to the identity functor, but when we use a natural isomorphism between vector spaces, where are functors? – Galoisfan Nov 20 '12 at 8:30
2
the functors are then hidden somewhere. Typically, if you have a construction between vector spaces that is 'natural' in some non-formal sense (i.e., no dependence on particular choices) then chances are there are some functors hiding there and that the construction is actually a natural transformation between the functors. Imagine that one would consider the vector space $\mathbb {R}^3$, construct its double dual and show it is naturally isomorphic to it. It's a particular result, but the identity functor and the double dual functor are lurking there. – Ittay Weiss Nov 20 '12 at 8:48
1
If $1$ is the category with one object and just the identity morphism, then any category $\mathbf{C}$ is isomorphic to the functor category $\hom(1, \mathbf{C})$; this makes every arrow a natural transformation in disguise. But we can say something more interesting. When we say "If $V$ is a vector space, then there is a map $V \to V^{**}$", $V$ is a variable that ranges over all vector spaces. .... – Hurkyl Feb 17 at 12:33
1
... Rather than identify $V$ with a specific functor $1 \to \mathbf{Vect}$, we can instead make a statement about all vector spaces at once by identifying $V$ with the identity functor $\text{Ob}\mathbf{Vect} \to \mathbf{Vect}$. Even better, this fact is "functoral in $V$", so we can use all of $\mathbf{Vect}$ as the source category, rather than just its class of objects. So we identify $V$ with the identity functor $1_\mathbf{Vect}$, double dual with a functor $\mathbf{Vect} \to \mathbf{Vect}$, and the claimed arrow is a natural transformation between them. – Hurkyl Feb 17 at 12:34
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http://www.physicsforums.com/showthread.php?p=4186059
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Physics Forums
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## Angle between vector and its transpose
Hi
What is the angle between a vector (e.g. a row vector A) and it's transpose (a column vector) ? I know what transpose means mathematically but what is the intuition?
Thanks guys
Quote by newphysist Hi What is the angle between a vector (e.g. a row vector A) and it's transpose (a column vector) ? I know what transpose means mathematically but what is the intuition? Thanks guys
If you're talking about simple real vectors (e.g. arrows in euclidian space) than the transposed vector is the same as the original. It's merely a matter of notation,whether you write the components in row or column.
In general case there is a distinction between row vectors and column vectors however in such cases it's hard to give concise meaning to the word "angle".
I don't understand the question. Usually in vector spaces like this (well Euclidean Vector Spaces), we define "angle" to be the cosine of the inner product of two vectors. However, a vector and its transpose are not in the same space. (In the sense that one is a 1xn matrix and one is a nx1 matrix.)
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## Angle between vector and its transpose
I agree with Robert.
The question is meaningless.
Mentor Consider ##u = (1, 2, 3)## and $$u^T = \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}$$ Clearly the angle between the two vectors is 90°
I don't think it's meaningless. In introductory classes (where vectors aren't even defined properly) rows and columns are often just different notations for components of vectors.
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Quote by Mark44 Consider ##u = (1, 2, 3)## and $$u^T = \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}$$ Clearly the angle between the two vectors is 90°
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Quote by Dead Boss I don't think it's meaningless. In introductory classes (where vectors aren't even defined properly) rows and columns are often just different notations for components of vectors.
In such classes, the word "transpose" ought to be banned. THAT word should be introduced AFTER proper definition of a vector.
Assuming that question is meaningless, what is the intuition behind taking transpose?
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Quote by newphysist Assuming that question is meaningless, what is the intuition behind taking transpose?
You transpose, in order to make the transpose.
It is a well-defined matrix operation.
Quote by newphysist Assuming that question is meaningless, what is the intuition behind taking transpose?
Are you familiar with the dot product? Well, this can be written like $v^\top w$. If you have some vector $v$ then the matrix that projects onto that vector is given by $vv^\top$ (if you don't understand what projection means, that's OK, but this example probably doesn't make sense.)
Additionally, there are important classes of matricies that are defined using transpose (or complex conjugate transpose which is where you do the transpose and take the complex conjugates of the entries of your matrix, if your matrix has real entries then obviously the complex conjugate transpose is just the transpose.)
For example, if $Q^\top=Q^{-1}$ the matrix is called orthogonal. These matrices preserve angles and lengths of vectors. These are good for numerical applications for that reason, but also it is MUCH easier to compute the transpose than it is to compute the inverse (in a sense, you don't need to "compute" the transpose, your code just needs to "iterate backward" - if you don't understand that part, its OK.)
Another special class is Symmetric Matrices where $M$ is symmetric if $M=M^\top$. These are really nice for several reasons. First, they are diagonalisable by an orthogonal matrix. Since they are diagonalisable, there is an eigenbasis and so you can do a "spectral resolution." Also, the eigenvalues are all real, even if the entries are complex.
This is just *very light* scratching the surface, but there are MANY important topics that involve transposes of matrices and vectors.
Quote by Robert1986 Are you familiar with the dot product? Well, this can be written like $v^\top w$. If you have some vector $v$ then the matrix that projects onto that vector is given by $vv^\top$.
Can you please explain what you mean by above projection example or did you switch $v$ and $w$? I don't see any use of $w$ in your logic.
Thanks
Blog Entries: 2 It's two different examples. The first is an alternative definition for the dot product (that I believe only applies to column vectors, someone correct me if I'm wrong), the second is an orthogonal projection onto a line. To verify this, take two unit vectors and perform the operation described and draw them!
The projection of $w$ onto $v$ is given by $(v\circ w)v$ which is equal to $(v^\top w)v$ which is equal to $(vv^\top)w$. That is, $vv^\top$ is the projection matrix. Note that I have assumed that $v$ has unit length.
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Quote by Dead Boss I don't think it's meaningless. In introductory classes (where vectors aren't even defined properly) rows and columns are often just different notations for components of vectors.
What kind of "introductory class" are you talking about? Every introductory linear algebra class I heard of defines vectors quite clearly. And if you mean physics classes where they just give vectors as "rows and columns", it still is meaningless to talk about "angles" between them. More rigorously the "transpose" of a vector, v, is its "dual" which is not even in the same vector space as v.
Meaningless is a strong word that probably does not help the OP much. To most non-mathematicians who use vectors, the the dual space is implicitly identified with the original space via the dot product. "Vectors" are written either as rows or columns depending on convenience. Why do so many people want to "ban" people from using reasonable notation just because they aren't versed with abstract mathematics? It's like being against differentials, or implicit functions, or Leibniz notation.
Recognitions: Gold Member Science Advisor Staff Emeritus Using that interpretation, you are saying that a "vector" and its "transpose" are just different ways of writing the same vector so the "angle between them" is necessarily 0. (Of course "identifying the dual space with the original space via the dot product" depends on the choice of basis so still doesn't have any meaning as a property of the vector space itself. Oh, and "differentials", "implicit functions", and "Leibniz notation" all have very specific, rigorous, mathematical definitions.)
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http://nrich.maths.org/60
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### Magazines
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### Stairs
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Taking a Die for a Walk
Investigate the numbers that come up on a die as you roll it in the direction of north, south, east and west, without going over the path it's already made.
# Month Mania
##### Stage: 1 and 2 Challenge Level:
When it's the end of January and we need to turn over a page of our calendar, the next page may look a bit like this:
Of course, it probably has a good picture on the page as well, perhaps of a scene from a T.V. show or an animal from your favourite story book.
If we ignore the edge squares which tell you what day of the week it is, then we have a rectangle with twenty eight squares in it like this:-
Well, how about going a bit wild and designing a new shape for these twenty eight squares in which the numbers follow on in some way that is not too difficult to follow? For example, you could put the twenty eight squares into a triangle and decide to go in an up-and-down order like this:-
What other wonderful designs can you come up with?
You can then go further by looking at the way the numbers are arranged and see if you can spot any patterns. For example in the first usual way:-
you may notice things like:
1. If you take any two by two square (like the one in at the top left-hand corner with $1$, $2$, $8$ and $9$ in it), then the opposite corners of that square add up to $10$. Then, if you take the next two by two square along [with $3$, $4$, $10$ and $11$ in] then the opposite corners add up to $14$ and so on.
2. Going down any column the numbers increase by $7$.
Do let us know what you discover in the designs you have created. Can you explain why the patterns occur?
Then ask "I wonder what would happen if ...?''
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://unapologetic.wordpress.com/2007/08/10/
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# The Unapologetic Mathematician
## Internalizations commute
Okay, back in the saddle. We know about monoid objects and group objects, and also the slight variant of abelian group objects. These are functors (preserving some sort of product structure) from the toy categories $\mathrm{Th}(\mathbf{Mon})$, $\mathrm{Th}(\mathbf{Grp})$, and $\mathrm{Th}(\mathbf{Ab})$ to our target category $\mathcal{C}$, respectively. We could also write out all the properties of (unital) rings in terms of a category $\mathrm{Th}(\mathbf{Ring})$ — the “theory of rings” — and then use this to define “ring objects
We know, though, that a ring is a monoid object in the category $\mathbf{Ab}$ of abelian groups. And the category of abelian groups is the category of abelian group objects in $\mathbf{Set}$. That is, $\mathbf{Ab}\cong\mathbf{Set}^{\mathrm{Th}(\mathbf{Ab})}$. And then we get monoid objects (rings) by considering $(\mathbf{Set}^{\mathrm{Th}(\mathbf{Ab})})^{\mathrm{Th}(\mathbf{Mon})}$.
But what is a functor from one category into the category of functors from a second category into a third? What is $(\mathcal{A}^\mathcal{B})^\mathcal{C}$? This is just a functor from the product category $\mathcal{C}\times\mathcal{B}$ to $\mathcal{A}$! Thus $(\mathcal{A}^\mathcal{B})^\mathcal{C}\cong\mathcal{A}^{\mathcal{C}^\mathcal{B}}$. This should look familiar — it says that (to whatever extent it’s well-defined) the category of categories is a closed category!
Now we also know that the product of two categories is “weakly symmetric” — $\mathcal{B}\times\mathcal{C}$ is equivalent to $\mathcal{C}\times\mathcal{B}$. And thus the categories of functors from either one of these to a third category are equivalent — $\mathcal{A}^{\mathcal{C}\times\mathcal{B}}\cong\mathcal{A}^{\mathcal{B}\times\mathcal{C}}$. Putting this together with the “currying” above we find that $(\mathcal{A}^\mathcal{B})^\mathcal{C}\cong(\mathcal{A}^\mathcal{C})^\mathcal{B}$.
So let’s bring this back to internalizations. We find that $(\mathbf{Set}^{\mathrm{Th}(\mathbf{Ab})})^{\mathrm{Th}(\mathbf{Mon})}$ and $(\mathbf{Set}^{\mathrm{Th}(\mathbf{Mon})})^{\mathrm{Th}(\mathbf{Ab})}$ are equivalent. A monoid object in $\mathbf{Ab}$ is equivalent to an abelian group object in $\mathbf{Mon}$.
And we saw evidence of this before. We’ve taken abelian groups and built free monoid objects on them, and we’ve taken monoids and built free abelian group objects on them. And the rings we get from each case are isomorphic to each other.
This whole setup generalizes. For any two algebraic structures we can describe by “theory of” categories, we can add both structures to an object of a category $\mathcal{C}$ in either order. the processes of “internalizing” the two structures into the category $\mathcal{C}$ commute with each other.
Now, for an exercise you can work out the naïve approach mentioned above. Write down the definition of $\mathrm{Th}(\mathbf{Ring})$ from first principles, just like we did for monoids and groups and such. Then show that the resulting category is equivalent to $\mathrm{Th}(\mathbf{Mon})\times\mathrm{Th}(\mathbf{Ab})$.
Posted by John Armstrong | Category theory | 2 Comments
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://math.stackexchange.com/questions/245141/do-the-real-numbers-and-the-complex-numbers-have-the-same-cardinality?answertab=oldest
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# Do the real numbers and the complex numbers have the same cardinality?
So it's easy to show that the rationals and the integers have the same size, using everyone's favorite spiral-around-the-grid.
Can the approach be extended to say that the set of complex numbers has the same cardinality as the reals?
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One can show that $|\mathbb R| = |\mathbb R^2| = |\mathbb C|$ – Stefan Nov 26 '12 at 19:05
3
It's quite sad, but it's easier to write an answer than finding the duplicate. And I am sure this question has been asked before. – Asaf Karagila Nov 26 '12 at 19:08
2
– Brian M. Scott Nov 26 '12 at 19:31
## 3 Answers
Yes.
$$|\mathbb R|=2^{\aleph_0}; |\mathbb C|=|\mathbb{R\times R}|=|\mathbb R|^2.$$
We have if so:
$$|\mathbb C|=|\mathbb R|^2 =(2^{\aleph_0})^2 = 2^{\aleph_0\cdot 2}=2^{\aleph_0}=|\mathbb R|$$
If one wishes to write down an explicit function, one can use a function of $\mathbb{N\times 2\to N}$, and combine it with a bijection between $2^\mathbb N$ and $\mathbb R$.
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Of course. I will show it on numbers in $[0,1)$ and $[0,1)\times[0,1)$. Consider $z=x+iy$ with $x=0.x_1x_2x_3\ldots$ and $y=0.y_1y_2y_3\ldots$ their decimal expansions (the standard, greedy ones with no $9^\omega$ as a suffix). Then the number $f(z)=0.x_1y_1x_2y_2x_3y_3\ldots$ is real and this map is clearly injective on the above mentioned sets. Generalization to the whole $\mathbb C$ is straighforward. This gives $\#\mathbb C\leq\#\mathbb R$. the other way around is obvious.
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4
This requires a bit more work. The map isn’t well-defined until you deal with the $0.4999\dots=0.5000\dots$ issue; if you deal with that straightforwardly, it’a nor surjective. – Brian M. Scott Nov 26 '12 at 19:14
Yes, you are right. However, they all all (complex) rational hence of no interest for the sets of continuum cardinality. I'll add a comment. – tohecz Nov 26 '12 at 19:16
And btw, usually a string with suffix $9^\omega$ is not considered to be an expansion (it is only a representation), in the usual greedy expansions as defined by Rényi in 1957. – tohecz Nov 26 '12 at 19:22
I’ve never seen anyone make a distinction between representation and expansion, and I very much doubt that the distinction can be considered standard; certainly it does not qualify as well-known, so if you use it, you need to explain it. – Brian M. Scott Nov 26 '12 at 19:29
1
And yes, I know that only countably many numbers are affected and that this does not affect the result, but I don’t know that the OP knows this. – Brian M. Scott Nov 26 '12 at 19:33
show 2 more comments
One particularly nice class of bijections from $\Bbb R$ to $\Bbb C = \Bbb R^2$, which is in my opinion a little bit similar to the spiral around the grid, is given by the space-filling curves.
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http://terrytao.wordpress.com/2009/06/21/freimans-theorem-for-solvable-groups/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
Freiman’s theorem for solvable groups
21 June, 2009 in math.CO, math.GR, paper | Tags: Freiman's theorem, Milnor-Wolf theorem, nilpotent groups, solvable groups
I’m continuing the stream of uploaded papers this week with my paper “Freiman’s theorem for solvable groups“, submitted to Contrib. Disc. Math.. This paper concerns the problem (discussed in this earlier blog post) of determining the correct analogue of Freiman’s theorem in a general non-abelian group $G = (G,\cdot)$. Specifically, if $A \subset G$ is a finite set that obeys the doubling condition $|A \cdot A| \leq K|A|$ for some bounded K, what does this tell us about A? Heuristically, we expect A to behave like a finite subgroup of G (or perhaps a coset of such a subgroup).
When G is the integers (with the additive group operation), Freiman’s theorem then tells us that A is controlled by a generalised arithmetic progression P, where I say that one set A is controlled by another P if they have comparable size, and the former can be covered by a finite number of translates of the latter. (One can view generalised arithmetic progressions as an approximate version of a subgroup, in which one only uses the generators of the progression for a finite amount of time before stopping, as opposed to groups which allow words of unbounded length in the generators.) For more general abelian groups, the Freiman theorem of Green and Ruzsa tells us that a set of bounded doubling is controlled by a generalised coset progression $P+H$, i.e. the sum of a generalised arithmetic progression P and a finite subgroup H of G. (Of course, if G is torsion-free, the finite subgroup H must be trivial.)
In this paper we address the case when G is a solvable group of bounded derived length. The main result is that if a subset of G has small doubing, then it is controlled by an object which I call a “coset nilprogression”, which is a certain technical generalisation of a coset progression, in which the generators do not quite commute, but have commutator expressible in terms of “higher order” generators. This is essentially a sharp characterisation of such sets, except for the fact that one would like a more explicit description of these coset nilprogressions. In the torsion-free case, a more explicit description (analogous to the Mal’cev basis description of nilpotent groups) has appeared in a very recent paper of Breulliard and Green; in the case of monomial groups (a class of groups that overlaps to a large extent with solvable groups), and assuming a polynomial growth condition rather than a doubling condition, a related result controlling A by balls in a suitable type of metric has appeared in very recent work of Sanders. In the nilpotent case there is also a nice recent argument of Fisher, Peng, and Katz which shows that sets of small doubling remain of small doubling with respect to the Lie algebra operations of addition and Lie bracket, and thus are amenable to the abelian Freiman theorems.
The conclusion of my paper is easiest to state (and easiest to prove) in the model case of the lamplighter group $G = {\Bbb Z} \rtimes {\Bbb F}_2^\omega$, where ${\Bbb F}_2^\omega = \lim_{n \to \infty} {\Bbb F}_2^n$ is the additive group of doubly infinite sequences in the finite field ${\Bbb F}_2$ with only finitely many non-zero entries, and ${\Bbb Z}$ acts on this space by translations. This is a solvable group of derived length two. The main result here is
Theorem 1. (Freiman’s theorem for the lamplighter group) If $A \subset {\Bbb Z} \ltimes {\Bbb F}_2^\omega$ has bounded doubling, then A is controlled either by a finite subspace of the “vertical” group $\{0\} \times {\Bbb F}_2^\omega$, or else by a set of the form $\{ (n,\phi(n)): n \in P \}$, where $P \subset {\Bbb Z}$ is a generalised arithmetic progression, and $\phi: P \to {\Bbb F}_2^{\omega}$ obeys the Freiman isomorphism property $(n_1,\phi(n_1)) \cdot (n_2, \phi(n_2)) = (n_3,\phi(n_3)) \cdot (n_4,\phi(n_4))$ whenever $n_1,n_2,n_3,n_4 \in P$ and $n_1+n_2=n_3+n_4$.
This result, incidentally, recovers an earlier result of Lindenstrauss that the lamplighter group does not contain a Følner sequence of sets of uniformly bounded doubling. It is a good exercise to establish the “exact” version of this theorem, in which one classifies subgroups of the lamplighter group rather than sets of small doubling; indeed, the proof of this the above theorem follows fairly closely the natural proof of the exact version.
One application of the solvable Freiman theorem is the following quantitative version of a classical result of Milnor and of Wolf, which asserts that any solvable group of polynomial growth is virtually nilpotent:
Theorem 2. (Quantitative Milnor-Wolf theorem) Let G be a solvable group of derived length O(1), let S be a set of generators for G, and suppose one has the polynomial growth condition $|B_S(R)| \leq R^d$ for some d = O(1), where $B_S(R)$ is the set of all words generated by S of length at most R. If R is sufficiently large, then this implies that G is virtually nilpotent; more precisely, G contains a nilpotent subgroup of step O(1) and index $O(R^{O(1)})$.
The key points here are that one only needs polynomial growth at a single scale R, rather than on many scales, and that the index of the nilpotent subgroup has polynomial size.
The proofs are based on an induction on the derived length. After some standard manipulations (basically, splitting A by an approximate version of a short exact sequence), the problem boils down to that of understanding the action $\rho$ of some finite set A on a set E in an additive group. If one assumes that E has small doubling and that the action of A leaves E approximately invariant, then one can show that E is a coset progression, and the action of A can be described efficiently using the generators of that progression (after refining the set A a bit).
In the course of the proof we need two new additive combinatorial results which may be of independent interest. The first is a variant of a well-known theorem of Sárközy, which asserts that if A is a large subset of an arithmetic progression P, then an iterated sumset kA of A for some $k=O(1)$ itself contains a long arithmetic progression. Here, we need the related fact that if A is a large subset of a coset progression, then an iterated subset kA for $k=O(1)$ contains a large coset progression Q, and furthermore this inclusion is “robust” in the sense that all elements the elements of Q have a large number of representations as sums of elements of A. We also need a new (non-commutative) variant of the Balog-Szemerédi(-Gowers) lemma, which asserts that if A has small doubling, then A (or more precisely $A \cdot A^{-1}$) contains a large “core” subset D such that almost all of a large iterated subset kD of D still lies inside $A \cdot A^{-1}$). (This may not look like the usual Balog-Szemerédi lemma, but the proof of the lemma is almost identical to the original proof of Balog and Szemerédi, in particular relying on the Szemerédi regularity lemma.
15 comments
21 June, 2009 at 1:41 pm
Anonymous
Missing “paper” tag. [Fixed, thanks - T.]
21 June, 2009 at 2:00 pm
pierre
Out of curiosity, how do you usually decide which journal to submit to?
Dear Pierre,
http://terrytao.wordpress.com/advice-on-writing-papers/submit-to-an-appropriate-journal/
for my thoughts on these sorts of issues.
22 June, 2009 at 3:26 am
David Speyer
Typo: In the statement of Theorem 1, and in the paragraph before it, I think your semidirect product is backwards. The integers act on $F_2^{\omega}$, right? [Corrected, thanks]
I am very happy to see both this paper and the new paper by Green and Breuillard. I’d like to focus on your Corollary 1.21, and why you truly need the hypothesis that G be totally torsion-free in order to take H=N. I thought your readers might be interested in a few examples over finite fields. In these examples, G is not totally torsion-free (because it is finite!), one cannot take H=N (the conclusion would then be false), and the statement of Corollary 1.21 becomes unfortunately trivial.
These examples show that a subset A of a solvable group G can have small doubling constant and yet be far from being contained in a nilpotent group (or a few cosets thereof).
At the same time, as I will say thereafter, a different kind of reduction to the nilpotent case should still be possible.
Example 1.-
Let $G = SL_3(K)$, where $K = Z/pZ$.
Let
$A = \left\{\left(\begin{matrix} r^k & x & y\\ 0 & s^k & z \\ 0 & 0 & (r s)^{-k}\end{matrix}\right) : x,y,z \in K, -N\leq k\leq N\right\}$,
where r and s are arbitrary elements of $K^*$.
The doubling constant is $<=4$, but A is far from being contained in anything nilpotent.
Example 2.-
Let $G = SL_3(K)$, where $K=Z/pZ$.
Let
$A = \left\{\left(\begin{matrix} r^k & k' & y\\ 0 & r^k & z \\ 0 & 0 & (r)^{-2k}\end{matrix}\right) : y,z \in K, -N\leq k\leq N, -M\leq k'\leq M\right\},$
where r is an arbitrary element in $K^*$.
The doubling constant is $\leq 4$, but, again, A is far from being nilpotent.
———-
It seems to me that one can make, nevertheless, make a general
reduction of the solvable case to the nilpotent case. Moreover, one
should be able to prove such a result with techniques available now.
—————
Very realistic conjecture. Let A be a subset of a solvable group G
of derived length at most l. Then, for any K, either
$|A A A| \geq K |A|$
or
there are subgroups $H_1\triangleleft H_2 \triangleleft \langle A\rangle$ such that
(a) $H_1$ is contained in $(A \cup \{I\} \cup A^{-1})^k$, where k depends only on l,
(b) A is contained in $<= K^C$ cosets of $H_2$, where C depends only on l, and
(c) $H_2/H_1$ is nilpotent.
—————–
I like to require these subgroups to be normal (which you haven't done); that makes the reduction more complete, and makes some of that business of K-approximate groups
automatically true. On the other hand, requiring $H_2 \triangleleft \langle A\rangle$
as opposed to just $H_2 < \langle A \rangle$ will probably make the proof more complicated. (Or do you think that $H_2\triangleleft \langle A\rangle$ is perhaps not even always true?)
The "very realistic conjecture" becomes a much harder conjecture (but is still probably true) if, instead of letting G be a solvable group of derived length <=l, we let G
be any subgroup of $SL_l(K)$ for any field K. I recently proved such a statement for
$SL_3(Z/pZ)$ (http://arxiv.org/abs/0807.2027, Thm 1.1). I am now working on growth in $SL_n(K)$, but I'll probably restrict myself to sets generating $SL_n(K)$ in the near future: I am having difficulties with sets that generate $SO_n$, for example.
Returning to the solvable case: the very realistic conjecture should yield to techniques related to the sum-product theorem. As you know, I really think the sum-product theorem is a special case of a more general statement on groups acting on groups. I prove a statement of that nature in section 3 of my $SL_3$ paper – it may very well be strong enough to use as the main tool to prove the (very realistic) conjecture, but it's certainly not something that solves the problem immediately.
At any rate, you see how the conjecture would reduce the solvable problem more or less completely to what Ben [Green] and Emmanuel [Breuillard] are doing. (In fact, since it has polynomial dependence, it will still be enough if some day a polynomial Freiman-Ruzsa theorem is proven.) I say "more or less" simply because the few cosets of $H_2$ occupied by elements of A may act on $H_2$ in an interesting way by conjugation, and that may force some sets to grow that would not grow otherwise; that still needs to be looked into.
Tell me what you think.
23 June, 2009 at 1:18 pm
Harald
My longer comments are not appearing; do you know what is the matter? Are you receiving them? [Your comment was caught in the automatic spam filter, but I restored it manually. -T.]
Dear Harald,
Thanks for the examples!
I think the conjecture is quite plausible (except perhaps for the polynomial dependence on K in the gap between A and H_2, which is rather ambitious with current technology – note we don’t have polynomial Freiman-Ruzsa even in the abelian case right now). In my paper I manage to control A by a slightly weaker object that I call a “coset nilprogression”. Roughly speaking, you are controlling A by a finite extension $H_2$ of a nilpotent group $N := H_2/H_1$, thus $0 \to H_1 \to H_2 \to N \to 0$, (and nilpotent groups can be viewed as towers of central extensions of the trivial group); coset nilprogressions are a bit more complicated, being a tower in which each stage is either a central extension or a finite extension, but they can come jumbled up in a funny way, in particular the finite stuff doesn’t necessarily come at the very top of the tower, but is sort of scattered through the tower. But it may actually be possible to disentangle this tower and pull all the finite stuff out as a big normal subgroup, as you suggest. I’ll have to think about it.
[Actually, with my nilprogressions, I'm not so much extending things by central groups, but rather by central progressions. This is sort of a fuzzy concept - the language of "approximate group theory" is not yet well developed enough to cleanly deal with extending an approximate group by another approximate group (e.g. extending a nilprogression by an arithmetic progression, somewhat similar to how your examples are operating) - and so the precise formalisation of this concept in my paper (Definition 1.11) is quite yucky, analogous to what the concept of a group extension $0 \to K \to G \to H \to 0$ would look like if one insisted on writing everything in coordinates, thus creating a partial inverse $\phi: H \to G$ and expressing elements of G in the form $k \phi(h)$ for $k \in K, h \in H$.]
25 June, 2009 at 8:15 am
Harald
Dear Terry,
>I think the conjecture is quite plausible (except perhaps for the polynomial dependence >on K in the gap between A and H_2, which is rather ambitious with current technology – >note we don’t have polynomial Freiman-Ruzsa even in the abelian case right now).
One doesn’t need polynomial Freiman-Ruzsa in any shape or form to get polynomial dependence here. In fact, I have already got polynomial dependence for A in a solvable subgroups of SL_3(Z/pZ) – well, actually, for any A in SL_3(Z/pZ). Otherwise I would not have been able to prove a result of the form |A A A| >> |A|^{1+epsilon} for subsets A of SL_3(Z/pZ) generating SL_3(Z/pZ).
In your paper, you are using ideas connected to Freiman-Ruzsa. I posit that these ideas belong to the study of nilpotent groups. The reduction of the solvable case to the nilpotent case can be managed entirely by what may call generalised sum-product technology (though the name is arguably outdated now), without any recourse to Freiman-Ruzsa. One great advantage of this is that we can get polynomial dependence right now.
>But it may actually be possible to disentangle this tower and pull all the finite stuff out >as a big normal subgroup, as you suggest. I’ll have to think about it.
I would be very curious to know what you get.
Best,
Harald
Dear Harald,
I suppose that our current range of polynomially quantitative techniques may indeed be able to get a polynomial solvable-to-nilpotent reduction, but my guess is that it would have to use some rather explicit structure of the solvable group. It isn’t quite the same, but Fisher-Katz-Peng’s paper make a kind of polynomial nilpotent-to-abelian reduction based primarily on the Baker-Campbell-Hausdorff formula.
I did some calculations. It does indeed seem that the methods in my paper give your claim except for the polynomial bounds. The key new point is that if a matrix is both nilpotent and of finite order (with integer coefficients), then it is the identity. As a consequence, when one is extending some coset nilprogression by a proper abelian progression, with the conjugation action of the former on the latter being nilpotent, the action of any finite abelian subgroup of the base coset nilprogression on the proper progression is trivial and so that finite abelian subgroup can be absorbed into the extension group (and can eventually be quotiented out). Combining this with the stuff in my paper and using induction disentangles the tower and eventually gives the claim (I’ll send you more detailed computations at some point).
Dear Terry,
>I suppose that our current range of polynomially quantitative techniques may indeed be >able to get a polynomial solvable-to-nilpotent reduction, but my guess is that it would >have to use some rather explicit structure of the solvable group.
I cannot fully agree. There are already partial statements one can prove without any explicit work. This makes me think that it might be possible to settle the “very realistic conjecture” I stated above without much explicit work at all.
(I’ve sent you something – it’s very preliminary, though.)
[...] weaker version of this lemma was also established by myself [...]
[...] and proving this statement is the noncommutative Freiman theorem problem, discussed in these earlier blog [...]
[...] a “noncommutative Freiman” type theorem in linear groups . As discussed in earlier blog posts, a general question in additive (or multiplicative) combinatorics is to understand the structure of [...]
[...] groups, nilpotent groups, solvable groups, or simple groups of Lie type) are also known; see these previous blog posts for a summary of several of these [...]
[...] of . The Borel group is solvable, and by using tools from additive combinatorics, such as Freiman’s theorem in solvable groups (or the Helfgott-Lindenstrauss conjecture, discussed in the previous quarter’s notes), one [...]
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http://math.stackexchange.com/questions/337099/a-loan-is-being-repaid-by-12-annual-payments-of-3000-determine-the-principal
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# A loan is being repaid by 12 annual payments of 3000…determine the principal and interest portions of the tenth payment
A loan is being repaid by 12 annual payments of 3000 followed by 8 annual payments of 5000. If i=0.10, determine the principal and interest portions of the tenth payment and the 15th payment.
I am not sure on how to find the equation to this problem. Do i use the Prospective Method to find the outstanding loan balance first?
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You need the outstanding balance at the previous payment. Whatever is owed after the 7th payment, 10% of that is the interest portion of the 8th payment. – User58220 Mar 21 at 20:58
## 1 Answer
You need three formulas: the present value of an annuity, the future value of an annuity, and the compound interest formula for a single amount. You could get away with just one of the first two formulas.
You basically have to gather all the money to one time, so that you can move, add and subtract it.
A)Find the present value of the 12 3000 payments (as of 1 year before the first 3000 payment)
B) Find the present value of the 8 5000 payments (as of 1 year before the first 5000 payment)
C) Bring B) 12 years earlier by dividing by $1.10^{12}$
A) + C) is the original amount of the loan
Now to find the debt immediately after the 9th payment.
D) Move the original loan forward 9 years by multiplying by $1.1^{9}$
E) Find the future value of the first nine payments of 3000, as of immediately after the 9th payment and subtract from D). This is the amount owing after the 9th payment. Multiply by 0.10 to find the interest portion of the 10 payment.
The second part is slightly more complicated. You need to move the original debt to Payment 14, accumulate the 12 small payments and then move them to Payment 14, and then find the future value of the first two big payments...
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http://mathoverflow.net/questions/24376/characterizing-the-local-langlands-correspondence
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## Characterizing the Local Langlands Correspondence
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In the p-adic case, is there any hope for a set of conditions on the local Langlands correspondence which would make it unique? In the case of GL(n) this is provided by L and epsilon factors. For classical groups, can one make a precise statement (even conjecturally), using lifting to GL(n)?
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## 1 Answer
This answer will necessarily be full of conjecture, but I'll try to make things more concrete for some classical groups.
For a reductive group $G$, and smooth irreducible representation $(\pi, V)$ of $G$, and L-group homomorphism $\sigma: {}^L G \rightarrow {}^L GL_n$, one may conjecturally associate a local L-function and $\epsilon$-factor $L(\pi, \sigma, s)$ and $\epsilon(\pi, \sigma, s)$.
As $\sigma$ varies over all L-group representations of ${}^L G$, this family of L-functions and $\epsilon$-factors will uniquely determine the (conjectural) Langlands parameter $$\phi_\pi: W' \rightarrow {}^L G,$$ where $W'$ is the Weil-Deligne group of the $p$-adic field. Namely, there will be a unique $\phi_\pi$ such that $$L(\sigma \circ \phi_\pi, s) = L(\pi, \sigma, s), \epsilon(\sigma \circ \phi_\pi, s) = \epsilon(\pi,\sigma,s)$$ where the L and $\epsilon$ on the left sides are Artin L-functions (adapted to the Weil-Deligne group instead of Galois group in a simple way).
In this way, the L- and $\epsilon$-factors determine the Langlands parameter of a smooth irrep of $G$. This partitions the smooth irreps of $G$ into the "L-packets".
Now, for classical groups, one can get away with less information (or at least more computable information) in many cases. I refer to the recent paper of Wee Teck Gan and Takeda, "The Local Langlands Correspondence for Sp(4)" (avaialble on Wee Teck's home page, for example), where they prove that the Langlands parameterization is uniquely determined by L- and $\epsilon$-factors (and $\gamma$-factors) of pairs (i.e. twists by $GL(n)$ with $n \leq 3$) for generic representations and a fact about Plancherel measure for certain nongeneric representations (this fact is analogous to a fact about Shahidi-type L-functions, which are only understood in the generic case).
For any classical group, I imagine one (probably not me) could come up with an analogous set of twists, etc.. to find a reasonable set of L-functions and $\epsilon$-factors, and Plancherel measures to determine the Langlands parameterization. Perhaps this is done by others who prove functoriality for classical groups (P.S., Cogdell, Rallis, Jiang, Kim, Shahidi, and many others).
Refining the local Langlands conjectures (I appreciate Vogan's survey most here), one should be able to parameterize the contents of an L-packet (with parameter $\phi: W' \rightarrow {}^L G$) by the representations of a finite group $S_\phi$. This finite group has a general description, which in some nice cases (split, adjoint perhaps, I don't recall), is that $S_\phi$ is the component group of the centralizer of the image of $\phi$ in the dual group $\hat G$.)
So to pin down the Langlands correspondence, one must characterize this parameterization of the elements of an L-packet as well. For this, it seems that characters are key -- there is certainly no telling the difference with L-functions or $\epsilon$-factors. I believe this parameterization of the elements of an L-packet should be uniquely determined by the stability of an associated sum of character distributions of elements of the L-packet. I tend to avoid characters, but maybe the work of DeBacker and Reeder -- who prove such a stability result for depth zero representations -- is a good place to start.
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Except in special cases the definition of $L(\pi,\sigma,s)$ and $\varepsilon(\pi,\sigma,s)$ assume the LLC. For GL(n) these can be defined independently using Whittaker models. This gives a set of conditions one can impose on the (conjectural) LLC for GL(n) which (if it exists) make it unique. This is now a theorem. What I'm asking about is something similar for other groups. Do not assume LLC exists, and give a set of conditions on it which, if it exists, make it unique. – Jeffrey Adams May 13 2010 at 12:26
A great example in this spirit is the Sp4 paper of Gan-Takeda that I referenced. They used L-functions in the generic case -- when suitable L-functions can be defined without LLC -- and Plancherel measure in the non-generic case -- when suitable L-functions are harder to define (though there are options using Bessel instead of Whittaker models). My personal hope is that someday, someone will be able to define L-functions for generic irreps of general (quasisplit?) groups, without LLC, by understanding the structure of some Hecke algebras. Such a discovery might answer your wishes as well. – Marty May 13 2010 at 15:37
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http://math.stackexchange.com/questions/160919/types-and-typical-sequences
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# Types and Typical sequences
Joint types can often be given in terms of the type of x and a stochastic matrix \begin{equation} V:X\rightarrow Y \end{equation}such that $P_{x,y}(a,b)=P_{x}(a)V(b|a)$ for every $a\in X$ , $b\in Y$. The question is that how can we define $V(b|a)$ as conditional type given x and what is the V-shell of x denoted by $T_{V}(x)$?
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The type of a sequence is usually defined to be empirical frequency counts of a sequence, but you seem to be referring to probability distributions as types - can you clarify? Also, what is a $V$-shell? – sai Jul 7 '12 at 2:50
## 1 Answer
Lets take an example. I will talk in terms of empirical probabilities (Probability or relative frequency observed in a given sequence). Lets have X={a,b} and Y = {c,d}. Lets have the length of the sequence to be 7. Now for a given 7-length sequence of x^7 = (aababba} lets consider y^7 = (dccddcd). More graphically :
````| x^7 = | a | a | b | a | b | b | a |
| y^7 = | d | c | c | d | d | c | d |
````
Conditional type will be a matrix with X as rows and Y|X as columns:
````| Y/X | 'c' | 'd' |
| 'a' | 1/4 | 3/4 |
| 'b' | 2/3 | 1/3 |
````
This table or matrix is V(b|a) is conditional type of a sequence y^7 given a particular x^7. It is defined for each pair {X, Y} and behaves just like a type 'P' will behave for a sequence x^n in isolation. Now V-shell T_V(x) of Y^n given x^n is counterpart of type class of x^n. SO V-shell is a set of all such y^7 which will have same conditional type V(b|a) given a x^7. One such y^7 for above example would be 'cdddccd'.
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http://mathoverflow.net/questions/107086/relation-between-the-wave-front-set-and-the-semiclassical-frequency-set/107169
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## Relation between the wave front set and the semiclassical frequency set
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I need to prove that the wave front set of a distribution (as defined in Hormander's "The analysis of linear partial differential operators I") is equal to the semiclassical frequency set of an h-dependant $L^2$ function minus $R^n\times 0$, and conversely the semiclassical frequency set is the wave front set plus the points of $supp(u)\times 0$, but I don't know how. Some help (or bibliography) on this matter'll be highly appreciated.
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## 1 Answer
Let $u$ be a distribution on some open subset $\Omega$ of $\mathbb R^n$. A point $(x_0,\xi_0)\in \Omega\times\mathbb S^{n-1}$ does not belong to $WF u$ when there exists a neighborhhod $V$ of $x_0$ and a neighborhood $\Gamma$ of $\xi_0$ (in the sphere $\mathbb S^{n-1}$) such that for all $\chi\in C_c^\infty(V)$ and for all $N\ge 0$, $$\sup_{\lambda \ge 1,\ \xi\in \Gamma}\lambda^N\vert\widehat{(\chi u)}(\lambda \xi)\vert<+\infty.$$ This is equivalent to: there exists a neighborhood $V$ $\dots$ for all $N\ge 0$, $$\sum_{\nu\in \mathbb N}2^{2\nu N}\int_{2^\nu\le \vert \xi\vert\le 2^{\nu+1}, \frac{\xi}{\vert\xi\vert}\in \Gamma}\vert\widehat{(\chi u)}(\xi)\vert^2 d\xi<+\infty.$$ This is equivalent to: there exists a neighborhood $V$ $\dots$ for all $N\ge 0$, $$\Vert(\mathbf 1_\Gamma)(D/\vert D\vert)\phi(h D)(\chi u)\Vert_{L^2}=O(h^N),\quad\text{$h\rightarrow 0_+$},$$ where and $\phi\in C_c^\infty(\mathbb R^n)$ supported in a ring {$\eta, 1/2\le \vert\eta\vert\le 2$}.
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http://mathhelpforum.com/trigonometry/200775-vector-problem.html
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1Thanks
• 1 Post By skeeter
# Thread:
1. ## Vector problem
Hello,
I'm having trouble answering the following question. The answer (s) is a given, but I still can't derive the end result. The chapter this question's at in my book, I think the author may want you to return to it with more knowledge after the next few chapters, but I am too stubborn to move on until I have solved the question.
Anyway:
A load of 4.32 N is lifted by two strings making angles of $18^\tiny o$ and $30^ \tiny o$ with the vertical. If for this system the vectors representing the forces form a closed triangle when in equilibrium, calculate the tensions in the strings.
Answer [1.8 N, 2.91 N ]
Now, I've drawn this to scale and it is a scalene triangle inside two right angled triangles. I've made the assumption in that the vertical line will have a tension of 4.32 N because $\sin (90) = 1$. However, I can't see how to set up the equations in order to use the cosine or conversely the sine rule. My best attempt is the following (using trignometric ratios in order to solve for one of the tensions in the ropes which corresponds to 18 degrees):
Let x, y and z equal a decrease in the adjacent, hypotonuse and opposite side respectively.
Then:
$\cos (\theta) = \frac {A}{H}$
$\cos (18) = \frac {4.32 - x}{\left (\frac {4.32}{\cos (18)}\right) - y}$
$4.32 - 0.951(y) = 4.32 - x$
$x = 0.951(y)\ or\ y= \frac{x}{0.951}$
$\sin (\theta) = \frac{O}{H}$
$\sin (18) = \frac {1.4 - z}{\left (\frac{4.32}{\cos(18)} - y \right)}$
$4.32\tan(18) - \sin (18)y = 1.4 - z$
$z = 0.309(y)$
$\tan (\theta) = \frac{O}{A}$
$\tan(18) = \frac {1.4 - z}{4.32 - x}$
$Substituting\ for\ x\ and\ z\ in\ terms\ of\ one\ variable\ only$
$\tan(18) = \frac{1.4 - 0.309y}{4.32 - 0.951y}$
$1.4 - 0.309y = 1.4 - 0.309y$
$0 = 0.618y\ ?$
Sidenote: I do feel how this end result looks, it should have cancelled.
Any help would brilliant.
2. ## Re: Vector problem
vertical components of the two tensions are in equilibrium with the load's weight ...
$T_1 \cos(18) + T_2 \cos(30) = 4.32$
horizontal components of tension are equal and opposite in direction
$T_1 \sin(18) - T_2 \sin(30) = 0$
solve the system of equations ...
$T_1 = \frac{T_2 \sin(30)}{\sin(18)}$
$\frac{T_2 \sin(30)}{\sin(18)} \cdot \cos(18) + T_2 \cos(30) = 4.32$
$T_2 \left[\sin(30) \cot(18) + \cos(30) \right] = 4.32$
$T_2 = \frac{4.32}{\sin(30) \cot(18) + \cos(30)} \approx 1.8 N$
$T_1 \approx 2.91 N$
3. ## Re: Vector problem
@ Skeeter
Aha, I see. I think my mechanics needs a service.
Thanks!
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http://mathoverflow.net/revisions/38798/list
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## Return to Answer
2 added 62 characters in body
I believe
For aperiodic (sometimes also called, full period) strings, the term you are looking for is Lyndon words. These are the (unique lexicographically-least) representative of a full-period necklace (as stated in the comments, a necklace is the equivalence class under cyclic rotation). The number $k(n)$ you ask for is exactly the number of necklaces, and again, as stated in the comments, it is given by `$k(n)=\frac{1}{n}\sum_{d|n}\phi(d)2^{n/d}$`. You can check out a proof for this in S.W.Golomb's book "Shift Register Sequences" (in the 1967 edition, start looking at around page 171 and look for the cycles of $PCR_n$).
1
I believe the term you are looking for is Lyndon words. These are the (unique lexicographically-least) representative of a necklace (as stated in the comments, a necklace is the equivalence class under cyclic rotation). The number $k(n)$ you ask for is exactly the number of necklaces, and again, as stated in the comments, it is given by `$k(n)=\frac{1}{n}\sum_{d|n}\phi(d)2^{n/d}$`. You can check out a proof for this in S.W.Golomb's book "Shift Register Sequences" (in the 1967 edition, start looking at around page 171 and look for the cycles of $PCR_n$).
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http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/69227
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## Can a vector space over an infinite field be a finite union of proper subspaces?
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Can a (possibly infinite-dimensional) vector space ever be a finite union of proper subspaces?
If the ground field is finite, then any finite-dimensional vector space is finite as a set, so there are a finite number of 1-dimensional subspaces, and it is the union of those. So let's assume the ground field is infinite.
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## 9 Answers
You can prove by induction on n that:
An affine space over an infinite field F is not the union of n proper affine subspaces.
The inductive step goes like this: Pick one of the affine subspaces V. Pick an affine subspace of codimension one which contains it, W. Look at all the translates of W. Since F is infinite, some translate W' of W is not on your list. Now restrict all other subspaces down to W' and apply the inductive hypothesis.
This gives the tight bound that an F affine space is not the union of n proper subspaces if |F|>n. For vector spaces, one can get the tight bound |F|≥n by doing the first step and then applying the affine bound.
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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.
Let $V$ be the union $\cup_{i=1}^n V_i$, where the $V_i$ are proper subspaces and the ground field $k$ is infinite. Pick a non-zero vector $x\in V_1$. Pick $y\in V-V_1$, and note that there are infinitely many vectors of the form $x+\alpha y$, with $\alpha\in k^{*}$. Now $x+\alpha y$ is never in $V_1$, and so there is some $V_j$, $j\neq 1$, with infinitely many of these vectors, so it contains $y$, and thus contains $x$. Since $x$ was arbitrary, we see $V_1$ is contained in $\cup_{i=2}^n V_i$; clearly this process can be repeated to find a contradiction.
Steve
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+1. Short and self-contained. I really liked it! – unknown (google) Feb 5 2010 at 10:50
best proof and shortest one – Koushik Jan 21 at 11:55
Here's a reduction to the finite dimensional case. Let F be a finite set of subspaces of X. For each finite dimensional subspace Y of X, let u(Y) be the set of elements Z of F such that Y is contained in Z. By assumption, u(Y) is non-empty for every Y. Since any two finite dimensional subspaces are contained in a third, the intersection of the sets u(Y), as Y runs among all finite dimensional subspaces of X, is non-empty. Hence there is at least one set in F that contains every finite dimensional subspace of X, hence contains X.
For the finite dimensional case, let F be a finite set of subspaces of X. By induction, every codimension 1 subspace of X is contained in some Y from F. But there are infinitely many codimension 1 subspaces, so some Y in F contains more than one such subspace. Any two distinct codimension 1 subspaces span X (if dim X > 1) so Y = X.
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Very nice. For any future reader who gets confused in the same way I did, the assumption in the sentence "By assumption, u(Y) is non-empty for every Y" is the finite-dimensional case. – Anton Geraschenko♦ Oct 1 2009 at 4:12
I recently completed a short expository note on this subject, Covering Numbers in Linear Algebra. See:
http://math.uga.edu/~pete/coveringnumbersv2.pdf
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Anton Geraschenko's comment prompted me to write a new version of this short answer. I'm leaving the old version to make Anton's comment clearer (and also to increase the probability of having at least one correct answer).
NEW VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each (affine) line. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.
Indeed, as pointed out by Anton, the $K$-valued functions on $A$ which are polynomial on each line form obviously a ring $R$. This ring is a domain, because if $f$ and $g$ are nonzero elements of $R$, then there is a line on which none of them is zero, and their product is nonzero on this line.
OLD VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each finite dimensional affine subspace. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.
Indeed, we can assume that $A$ is finite dimensional, in which case the result is easy and well known.
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1
Ahhh +1. It took me a while to understand what's going on here. The confusing statement was "the product of the $f_i$ is nonzero". Here's my expansion of that statement. (1) The set of functions which are polynomial on each finite-dimensional subspace is clearly a ring. (2) Given two non-zero functions of this form, there is a finite-dimensional subspace on which they are both non-zero (a 2-dimensional subspace spanned by two points witnessing the non-zero-ness of the two functions will do). (3) By (2) and the finite-dimensional case, the ring is an integral domain. – Anton Geraschenko♦ Jul 1 2011 at 16:21
@Anton Geraschenko - Thanks for your vote and your comment, so much clearer than my answer! – Pierre-Yves Gaillard Jul 1 2011 at 16:29
For a slightly worse answer for the fin dim case - prove the following - if k is an infinite field then if f is a polynomial in n variables over k there exists a point of k^n x such that f(x) is non zero (proving this really isn't much easier than the actual problem though - I told you it was a worse answer.) Each subspace is mapped to zero by some poly over k, multiplying the polys gives a contradiction.
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I needed this result for a paper I wrote with David Leep ten years ago. Bruce Reznick came up with a nice proof which we included in the paper (Marriage, Magic, and Solitaire, published in the American Math Monthly). I don't think the proof was any better than the ones already given here, and I seriously doubt this was the first time a proof had ever appeared in print, but I wonder if anyone knows an earlier citation.
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2
Well, this is Thm 1.2 in Roman's book Advanced Linear Algebra. I have the 3rd edition from 2008. The 1st edition is from 1993; I don't know if it was in there. He doesn't cite anything. – Mike Benfield Feb 5 2010 at 12:31
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I believe it is one of those problems that turns up again and again over the years. I have seen it posed at least once as a problem in the American Mathematical Monthly (unfortunately I don't remember when, even up to a decade, and I don't know of a good way to search for it), but my guess is that its provenance dates back from well before 1993. – Pete L. Clark Feb 5 2010 at 18:20
This is a late response to the post, but I noticed that the question was not answered in general.
No vector space is the finite union of proper subspaces.
EDIT: In response to my false solution, Phil Hartwig pointed out that $\mathbb{F}$$_{2}^2$ is a vector space that is the union of three proper spaces. Indeed, the "routine" induction was less routine and more nonsensical. I had fixed my proof, only to realize that my solution was much less elegant than Halmos' solution found in his Linear Algebra Problem Book. You can view the page here.
In the class of Banach spaces there is a stronger result:
If $B$ is a Banach space, then $B$ is not the countable union of proper subspaces.
This relies on the fact that a proper subspace of a topological vector space has empty interior. To appeal to your intuition in $\mathbb{R}^3$, every proper subspace (a plane or line through the origin) cannot completely contain an open ball (an open set in the usual norm topology).
Since $B$ is complete (by definition), by Baire's Theorem it is not the countable union of nowhere dense sets. Since proper subspaces are nowhere dense, $B$ is not the countable union of proper subspaces.
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5
The "routine induction" doesn't work. As Anton already noted in his initial post $(\mathbb{F}_2)^2=\{(0,0),(1,0)\}\cup \{(0,0),(0,1)\}\cup \{(0,0),(1,1)\}$. – Philipp Hartwig Jun 29 2011 at 6:18
@Philipp: Under the assumption the OP gave that the field is infinite, your counterexample does not hold. – Asaf Karagila Jun 29 2011 at 13:35
Where is the assumption that the field is infinite used? – Pierre-Yves Gaillard Jun 29 2011 at 13:43
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A vector space $V$ of dimension infinite is always union of a countable set of proper subspaces. Take a Hamel basis $e_\alpha$ . Each vector can be written $v = \sum_\alpha x_\alpha e_\alpha$. Let $V_\alpha$ the subspace of those vectors for which $x_\alpha = 0$ It is clear that $V$ is the union of any sequence $V_{\alpha_n}$ for $n\in \N$ with $\alpha_n$ differents. – juan May 22 at 21:21
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@Adam: proper subspaces of a Banach space can be dense. – George Lowther May 22 at 22:39
show 8 more comments
The finite dimensional case cannot happen by dimension counting (just view everything as affine spaces).
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This argument certainly doesn't suffice by itself; it doesn't explain where the infinitude of the field comes into play. – Qiaochu Yuan May 22 at 20:01
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http://nrich.maths.org/4838/note
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nrich enriching mathematicsSkip over navigation
### Happy Numbers
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Zooming in on the Squares
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
# Litov's Mean Value Theorem
### Why do this problem?
This problem provides a context for genuine discovery and student directed research/investigation. It is an ideal situation for students to work out how to use ICT to speed up the calculations and support them in their quest to rediscover Litov's Theorem.
### Possible approach
"I found this problem called Litov's Mean Value Theorem. I'm hoping we can work out what the theorem is. We can start with any two numbers, say $8$ and $2$. These are the start of a sequence. The rule is that the next number in the sequence is the average of the last two numbers. So what comes next? Why? And then what?...
Invite students to choose their own pair of starting numbers, to calculate the sequence and find its limit. Students could use calculators for this activity. Giving students free choice can result in a lot of information being collected in a short space of time. While this is going on and the results are appearing on the blackboard, ask some students to think about how these calculations could be done on a spreadsheet.
Bring the class together and ask for observations, comments, suggestions and predictions. Demonstrate the use of a spreadsheet for testing these predictions quickly. The use of the computer makes it possible to operate at a new level and the computer shows the limiting process clearly.
"Given all that information would anyone like to check a result or predict what will happen to any pair of numbers?"
There's a chance to discuss whether these sequences will ever actually reach their limits.
Students could test their hypotheses working on paper, or everyone could be given access to spreadsheets. When students are convinced that they know how to find these limits, challenge them to suggest some reasons why the limits behave as they do.
Students could then move on to working on these:
What would happen if sequences were generated from three initial values by:
averaging the last three numbers (i.e. $(a+b+c)/3$)?
adding the last three numbers and dividing by 2 (i.e. $(a+b+c)/2$)?
### Key questions
Can you tell where these numbers are heading?
Does it matter if I swap the two starting numbers around?
What do these long decimals mean? How big is that number, roughly?
### Possible extension
What happens when you have $n$ start numbers and the rule for working out the next number changes to finding the average of the last $n$ numbers?
### Possible support
This problem is a good context for work on organisation skills and calculator competence with opportunities for making conjectures, and refining conjectures.
Laurinda Brown (1983) wrote about using this problem in the classroom: in Mathematics...with a Micro 1, pp.22-25, Waddingham, Jo (ed), Bristol, County of Avon, Resources for Learning Development Unit. The lesson notes above are adapted from her descriptions of using the problem.
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/22658/why-are-inverse-images-more-important-than-images-in-mathematics
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## Why are inverse images more important than images in mathematics?
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Why are inverse images of functions more central to mathematics than the image?
I have a sequence of related questions:
1. Why the fixation on continuous maps as opposed to open maps? (Is there an epsilon-delta definition of open maps in metric spaces?)
2. Is there an inverse-image characterization of homomorphisms in algebraic categories? (What kind of map do you get if you look at a map from a group to another group, where inverse images of subgroups are subgroups?)
3. Inverse images have better set-theoretic properties than the image (for instance, commuting with unions, intersections, etc..) This clearly is a direct consequence of definition of a function. There is an asymmetry in the definition of a function (the domain and codomain behave differently with respect to the function). I think this also has consequences for differences between existence and uniqueness of left and right inverses for one-to-one or onto functions. Why this asymmetry? What are the historical reasons for the asymmetry? Whats sort of mathematics do we have if the definition of a function was purely symmetric? (For instance, f(a) may give multiple values, just like f^-1(a) may have multiple values).
4. Is it accurate to say the definitions for monomorphisms and epimorphisms in category theory correct for the asymmetry? (And hence, the notion of epimorphisms and onto-morphisms in concrete categories don't coincide)
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This is a good community wiki candidate. – Bill Kronholm Apr 27 2010 at 0:45
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The symmetrically defined functions you mention in 3. are just relations. – Qfwfq Apr 27 2010 at 11:28
Would be great if someone could clean up the tags and retag this [soft-question]! – Sonia Balagopalan Apr 27 2010 at 13:16
## 4 Answers
Open sets can be identified with maps from a space to the Sierpinski space, and maps out of a space pull back under morphisms. (In other words, if you believe that the essence of what it means to be a topological space has to do with functions out of the space, you are privileging inverse images over images. A related question was discussed here.) I think essentially this kind of reasoning underlies the basic appearances of inverse images in mathematics. For example, in the category of sets, subsets can be identified with maps from a set to the two-point set, and again these maps pull back under morphisms. This should be responsible for the nice properties of inverse image with respect to Boolean operations.
Your third question was asked, closed, and deleted once; I started a blog discussion about it here.
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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.
Continuity is important not because of its inverse-image-ness, but because the definition corresponds to the geometric notion that it's intending to capture. A continuous function "takes close things to close things". This geometric intuition precedes the notion of open sets, and even the epsilon-delta definition of continuity.
I think the same sort of thing is true of functions in general. The asymmetry is there as a consequence of the idea that the definition is intending to capture; functions weren't originally thought of as a special class of relations, they were thought of as "rules" for manipulating numbers, and the idea of f(x) being unique and f${}^{-1}$(x) not being unique is the only natural way to capture the idea of a "rule" in a more general context.
I don't know that the notions of monomorphisms and epimorphisms really "correct" for the asymmetry, but I don't think it's something that ought to be corrected anyway. Monomorphisms of sets are the same as injective functions, and epimorphisms of sets are the same as surjective functions, and that's the way it should be. In categories like Ring where, say, epimorphisms aren't always surjective, it's because the non-surjective epimorphisms, in some sense, are "surjective as far as ring maps are concerned"; for example, the inclusion of $\mathbb{Z}$ into $\mathbb{Q}$ is epi because a map out of $\mathbb{Q}$ is determined by what it does to $\mathbb{Z}$. I don't think this has much to do with the asymmetry you're describing.
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I think that the first 2 paragraphs of this answer are truly excellent. – teil Apr 27 2010 at 2:22
I understand that continuity is not important because of its inverse-image-ness, but because it corresponds to the geometric notion. But WHY does the inverse-image-ness correspond to this geometric notion? It seems that there are seemingly-unrelated different geometric notions that ALSO corresponds to inverse-image-ness (such as measurable functions). I suppose the question is, why the prevalence of inverse-image-ness? – atonaltensor May 8 2010 at 17:22
Questions 1, 3, and 4 have been very well explained in the other answers, but I have something to remark about Question 2.
Very frequently, objects that are meant to be like spaces will have some kind of algebraic data attached to them. But this algebraic data is attached contravariantly, that is, there's some functorial relationship between your category of objects and the opposite of the category of algebraic structures.
For example:
• Sets and Boolean Algebras. The power-set functor mentioned in Sammy Black's answer actually gives a contravariant functor from sets to Boolean algebras. This functor actually embeds the category of sets into the opposite category of Boolean algebras, so sets may be regarded as Boolean algebras with certain properties, except the maps go the wrong way.
• Schemes and Rings. A scheme is locally isomorphic to an object in the opposite category of commutative rings. In fact, the category of schemes admits a fully-faithful embedding into $Set^{Rng}$, the free cocompletion of $Rng^{op}$. This is called the "functor of points" approach to schemes.
• Compact Hausdorff Spaces and Unital C*-Algebras. There's a contravariant equivalence between the category of compact Hausdorff spaces and the category of C*-algebras with unit.
• Locales and Frames. A frame is a kind of distributive lattice, and is described in a completely algebraic way. It's space-like counterpart, called a locale, is studied in so-called "Pointless Topology" (don't laugh), and the category of locales is defined to be the opposite category of frames. This was inspired by the last example, which is:
• Topological Spaces and their Lattices of Open Sets. To every topological space, there is associated a certain lattice (the lattice of open sets). The requirement is that this association be contravariantly functorial - that is, every map of topological spaces must give rise to a map of lattices in the opposite direction. And that's what we have: a continuous map is one that induces a well-defined inverse-image map taking open sets to open sets.
So the idea that open maps seem to be more straightforward (so to speak) than continuous maps may be a common one, but in fact it seems that we get better categories of spaces if we ask the algebraic data to be contravariant.
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1 Many important properties of topological spaces are preserved by continuous maps (but not necessarily open maps): connectedness and compactness come to mind immediately. But more importantly, the most familiar, natural maps that we can define are continuous, but not necessarily open: polynomials $\mathbb{R}^n \to \mathbb{R}^m$.
2 Inverse image of a subgroup under a homomorphism is a subgroup.
3 There is a contravariant functor from the category Set to itself, mapping a set $X$ to its power set $\mathcal{P}(X)$ and sending the morphism $f:X \to Y$ to the inverse image $f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)$. A "purely symmetric" function would be a symmetric relation on $X \times Y$, no? Functions are, after all, relations with an extra property that deliberately breaks the symmetry!
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As for the 2, I think he's actually asking whether a map of sets, such that the inverse image of every subgroup is a subgroup, is actually a group homomorphism. – Qfwfq Apr 27 2010 at 11:41
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If that's the question, the answer is no. Let $G$ be a cyclic group of prime order - then the requirement that a function $G\rightarrow G$ pull back subgroups to subgroups is very weak (it implies only that no nonzero element maps to zero), and most such functions won't be group homomorphisms. – Owen Biesel Apr 27 2010 at 17:49
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http://mathoverflow.net/revisions/21031/list
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## Return to Question
3 edited tags
2 added model theory tag
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# Ultrafilters vs Well-orderings
This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it.
Question: Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$?
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http://mathoverflow.net/revisions/22604/list
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## Return to Answer
1 [made Community Wiki]
[[ Sorry I missed that the question was also concerned with the question in an algebraic topological context. This answer is only concerned with algebraic geometry.]]
I think the first question is much easier to answer. mdeland has given the Artin-Mumford non-rationality example as one answer. Another is the Atiyah-Hirzebruch example of an even-degree torsion class of a smooth projective variety which is not algebraic, showing that an integral version of the Hodge conjecture is false. This gives examples (and there are others) where torsion can be used to show something about an algebraic variety which one couldn't show without (actually I would say that it is more a question of integral versus rational cohomology even without torsion one can exploit that certain cohomology classes are not divisible by some particular integer). I would say that gives an answer to the first question.
The second is of a very different nature. In algebraic topology torsion (and more general integral cohomology again versus rational cohomology) are enormously important for understanding the homotopy type of a space. Take as an example the spheres. Rationally their homotopy theory is trivial but integrally you have highly non-trivial homotopy groups (this non-triviality does not reflect itself in the cohomology of the spheres but is closely related to spaces derived from the spheres, the pieces of the Postnikov tower). Of course algebraic varieties (over $\mathbb C$, but that is not essential) give homotopy types too but it not always clear what the homotopy type of an algebraic variety tells you about the algebro-geometric structure of the variety (unless you somehow incorporate algebraic topology under algebraic geometry...). There are some examples though: The torsion in the second cohomology group comes directly from the fundamental group and in particular give you abelian étale covers of the variety. The torsion in the third cohomology group tells you about the Brauer group of the variety and in particular corresponds (for some definition of "corresponds") to projective fibrations over the variety. The correspondence is quite indirect however. I would for instance love to know the least relative dimension of a projective fibration over an Enriques surface which realises the element of order $2$ in the third cohomology group or even better a geometric construction of any such fibration. In higher cohomological degrees the situation is even worse (unless one chooses the above incorporation option, higher algebraic stacks could be said to do that).
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http://mathoverflow.net/questions/59724/the-current-state-of-selbergs-orthonormality-conjecture
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## The current state of Selberg’s orthonormality conjecture
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Hello, I would like to know which results have been obtained concerning Selberg's orthonormality conjecture. For example, has it been proven that for every pair of distinct primitive functions of the Selberg class $(F,G)$, $\displaystyle{\sum_{p\leq x}\frac{a_p(F)\overline{a_p(G)}}{p}=o(\log\log x)}$? Thank you in advance.
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## 1 Answer
Orthogonality has been proven for certain pairs of automorphic L-functions. The proof procedes like a proof of the Prime Number Theorem, replacing the logarithmic derivative of $\zeta(s)$ with that of $L(s,\pi\otimes\tilde\pi')$, where $\pi$ and $\pi'$ are automorphic representations.
The only proofs in print are for pairs of representations on $GL_m$ (not necessarily the same $m$ for both representations) over ${\mathbb Q}$. It looks like it applies to any pair where the Rankin-Selberg L-function satisfies:
1) Absolutely convergent Euler product for Re(s)$>1$.
2) The expected meromorphic continuation and functional equation, with at most one simple pole.
3) Some estimate towards the Ramanujan conjecture.
4) Nonvanishing for Re(s)$\ge 1$.
5) Order 1 (in the sense that $\log|f(z)|\ll |z|^{1+\epsilon}$), away from the possible pole.
It would be very exciting if nonvanishing (or order 1) could be proven from the Selberg Class axioms. Without them (i.e. proving orthogonality for the entire Selberg Class), the above proof fails completely, so something entirely different would be necessary.
Ye has the above proof on his website, assuming one of the representations is self-dual. The above proof without the assumption can be found here.
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http://math.stackexchange.com/questions/289000/un0-a-and-zunu-1-how-to-show-that-uznbz-b-where-b-a-a1
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# $u$~$N(0,A)$ and z$|u$~$N(u,1)$ how to show that $u|z$~$N(Bz,B)$ where $B=A/(A+1)$?
$u$~$N(0,A)$ and $z|u$~$N(u,1)$ how to show that $u|z$~$N(Bz,B)$ where $B=A/(A+1)$ ?
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Knowing $f_U$ and $f_{Z\mid U}$ yields $f_{U,Z}$, which yields $f_{U\mid Z}$. Which part of this program have you trouble with? – Did Jan 28 at 16:11
what is $f(U,Z)$ ? $f(U,Z)=?$ – Qbik Jan 28 at 16:26
Never wrote $f(U,Z)$, but $f_{U,Z}$ the joint density of $(U,Z)$. – Did Jan 28 at 16:53
Which part of this program have you trouble with? – Did Jan 28 at 16:53
Somebody is erasing their footprints... – Did Jan 30 at 22:18
show 1 more comment
## 1 Answer
Note that
$$p(\mu|z) \propto p(z|\mu)p(\mu)$$
After you plug it $p(z|\mu)$ and $p(\mu)$ and you will find $p(\mu|z)$ has a pdf that is normal. And you can easily derive its mean and variance.
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but what is $p(z)$ ? $N(0,1)$ ? – Qbik Jan 28 at 16:22
@Qbik There is no $p(z)$ involved here. You only need the conditional distribution of $p(z|\mu)$. – Patrick Li Jan 28 at 16:55
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http://polymathprojects.org/2010/06/29/draft-version-of-polymath4-paper/?like=1&source=post_flair&_wpnonce=671450efa2
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# The polymath blog
## June 29, 2010
### Draft version of polymath4 paper
Filed under: discussion,finding primes — Terence Tao @ 8:36 pm
I’ve written up a draft version of a short paper giving the results we already have in the finding primes project. The source files for the paper can be found here.
The paper is focused on what I think is our best partial result, namely that the prime counting polynomial $\sum_{a < p < b} t^p \hbox{ mod } 2$ has a circuit complexity of $O(x^{1/2-c+o(1)})$ for some absolute constant $c>0$ whenever $1 < a < b < x$ and $b-a < x^{1/2+c}$. As a corollary, we can compute the parity of the number of primes in the interval $[a,b]$ in time $O(x^{1/2-c+o(1)})$.
I’d be interested in hearing the other participants opinions about where to go next. Ernie has suggested that experimenting with variants of the algorithm could make a good REU project, in which case we might try to wrap up the project with the partial result and pass the torch on.
## 33 Comments »
1. Wow! I glanced at it earlier today — it looks very efficiently written. I will try to read it over tomorrow very carefully for errors.
I think what’s in the paper is good enough… and it’s best to wrap it up at this point, but that’s just by 2 cents.
Today (and last week) I discussed several directions in which to take the project with two REU students… if you think it is a good idea, perhaps you might mention a subset of these in the paper:
1. By doing a linear approximation to n/x we get speedups in lattice-point counting… but you can try to do higher-order polynomial approximations instead. I believe this leads to counting lattice points below a polynomial curve. If this can be done efficiently — as efficiently in the linear case — then it will lead to a vast speedup in the parity-of-pi algorithm (though not for the polynomial analoue). Just today we worked on the quadratic case of this, and came up with several ideas (a promising approach we came up with was to try to tile the polynomial region by affine [actually, rotated and translated] copies of certain polynomial sub-regions).
2. I still think the idea of computing $\sum_{a \leq p \leq b} x^{h(p)} \pmod{2,f(x)}$ quickly, where $h$ is a polynomial, is a nice problem. It turns out not to be so straightforward (though I don’t think it will prove to be THAT difficult in the end)… which makes it all the more intriguing.
3. Besides FFTs, there might still be a way to get FFT-level performance using “matrix arithmetic”… which would be good, because these matrix approaches work over any ring. I suggested that maybe the sort of structure one should try is higher-dimensional matrices. Are there Strassen-type algorithms that work for higher-dimensional matrices, that give better and better speedups the more dimensions one uses? I have not been able to track down the appropriate literature on higher-dimensional matrices, though I know it exists. I’m guessing there’s a way to rephrase this strategy somehow in terms of tensors or something. It seems there are many possible generalizations (from the 2D case) that one try can here. One thought for the 3D case was to think of “matrix multiplication” as involving a “trinary form” $F(A,B,C)$, where $A,B,C$ are 3D matrices; and, in general, the “multiplication” of $n$-dimensional matrices should involve $n$-ary forms.
Anyways, if no “higher dimensional” analogue of Strassen exists, then that alone would make a nice project to explore (and maybe it would have lots of applications! — I could imagine it would, especially if you get better and better results the more dimensions you use).
These are just a few of the things we have been thinking about…
Comment by Ernie Croot — June 30, 2010 @ 4:26 am
• These are great new directions to explore, but given that others will be exploring them I am a little hesitant to detail them too much in this current paper in order to give the next group of people a bit more space. But perhaps we can hint at such developments. Incidentally, you are of course free to send me any specific text you would like in the paper (either here, by email, or via Subversion).
Comment by — June 30, 2010 @ 6:20 pm
2. An orthogonal comment, but I wanted to say that I was very pleased to see that a paper had resulted from this discussion. It’s still far too early to say what a “typical” outcome is for a Polymath project, but I now don’t think it’s too early to say that if the discussion gets going then it will almost certainly lead to something publishable. The experience here, and also with the Erdős discrepancy project, has been that even if the “main” problem is not solved, the process almost inevitably leads to new insights and a deeper understanding of that problem, to the point where they are worth publishing. (If EDP is not eventually crowned with success, then there will certainly be a paper or two’s worth of material to be extracted from the discussion.)
Comment by — June 30, 2010 @ 10:17 am
• Well, one of the useful consequences of a Polymath project is that it reveals to some extent what actual mathematical research is like – and in particular, that partial success is in fact far more common than total success once one gets to really non-trivial problems (i.e. ones that are more than just applying a known method to a problem that is well within range of that method). This fact is often obscured by selection bias in the traditional research paradigm…
Comment by — June 30, 2010 @ 6:22 pm
3. Another comment. I see that the author of this paper is D.H.J. Polymath, which doesn’t make all that much sense. I had originally imagined that all polymath papers would simply be by Polymath, but Ryan O’Donnell came up with the DHJ idea for the DHJ papers. Since nothing has yet been published (though there have by now been some references to the DHJ papers in other papers) it’s perhaps not too late to decide what convention we want to adopt. The three I can think of are to drop the DHJ on the DHJ papers and do without initials from now on, to keep DHJ for all papers (which is a bit strange, but could be thought of as marking the fact that the DHJ papers were the first), or to have different initials for each project (so this one might be e.g. FPD, and the discrepancy one might be EDP, or just ED).
Comment by — June 30, 2010 @ 10:45 am
• Well, on some level all pseudonyms are equally artificial. I have a slight preference for using a unified pseudonym so that Polymath can behave (for the purposes of citation indices etc.) like a single prolific author rather than a whole family of one-hit authors, and DHJ is as good a choice as any other. I also have a slight fear that if we make a cleverly contrived nom de plume for each paper, it may make the series look unnecessarily artificial or “gimmicky” in nature. (Imagine for instance if Bourbaki used the name of a different French general for each branch of mathematics, with some clever association that “explained” the choice of each general for each branch; while entertaining, this would no doubt distract from the primary purpose of their exposition.)
Comment by — June 30, 2010 @ 6:32 pm
• I share your preference for a unified pseudonym, partly because it means that Polymath’s papers will straightforwardly appear in the same place on MathSciNet (without the need to remove the initials), the arXiv etc.
Comment by — July 1, 2010 @ 8:15 am
• I emailed Ryan and it turns out that he had originally imagined different initials for each project, to indicate that the sets of authors were different. But he points out that that has already failed to happen with the two DHJ papers (which, although both about aspects of DHJ, were by different sets of authors), and he’s happy to go with the idea of having DHJ Polymath be the author for all polymath papers and thinking of the initials as a reminder of the first project.
Comment by — July 1, 2010 @ 1:18 pm
• I think one pseudonym would be better than many different ones that differ by a couple of letters.
Comment by kristalcantwell — July 4, 2010 @ 8:33 pm
4. Ok, I’ve read over the draft. It is VERY nicely written! I have a few comments, however:
1. In the time complexities $O( \log^{O(1)} N)$, we don’t really need that extra big-O outside — we could just write $(\log N)^{O(1)}$, for instance. In the $O(N^{1/2 + o(1)})$, on the other hand, keeping the big-O is fine, so as to emphasize that it gives an UPPER BOUND.
2. On page 2, I would put a space between “Heath-Brown” and [6]. Also, at the bottom of page 2, there is an extra `.’.
3. On page 3, you write $\psi(x) = \sum_{n < x} \Lambda(n)$, but the $n < x$ should be $n \leq x$. In the sentence “Since, by as mentioned earlier…'', delete `by'.
4. On page 4, in the parenthetical comment “addition, subtraction…''. You might emphasize that we add `division', which I don't think is standard in the definition of “arithmetic circuit complexity''. In finite rings, perhaps it is ok to include it, since one can use the power map to produce multiplicative inverses.
Also, you write $\pi(x;a,q) := |...$, and then forget to put the matching $|$; and, the $q = O(N^{c/10)})$ has an extra $)$.
5. On page 5, you say “…distinct prime powers of $n$''. It should be “… distinct prime power divisors of $n$.''
6. Also on page 5, you write $2^{\omega(n)}\ :=$, when in fact it should just be `$=$', as it is an identity, not a definition (or assignment).
7. On page 8, I think you mean $A \leq n_0 < 2A$, not $A \leq n_0 < A$. Also on page 8, you say as $r$ runs from $0$ to $q$, when you mean $0$ to $q-1$. Also, you say here “$q$th root of unity'', maybe you should add the word “additive'' or reword it altogether.
8. On page 11, under “Possible extensions'', first sentence: I think you want to take the degree of $g$ to be $O(n^{c/4})$ or something, and then have the time be $O(n^{1/2-c/2})$.
Comment by Ernie Croot — July 1, 2010 @ 8:14 pm
• Thanks very much Ernie! I have incorporated the comments and some further corrections and the new draft is now available at
http://polymathprojects.files.wordpress.com/2010/07/polymath.pdf
I guess we should wait to hear from Harald and any other participants before deciding what next to do. I think the paper is close to being ready to be uploaded to the arXiv, and perhaps also submitted to some journal that can handle computational number theory (e.g Mathematics of Computation, which has Igor Shparlinski as one of the editors).
Comment by — July 1, 2010 @ 9:42 pm
• I was stuck in airports for a couple of days, and now I have some urgent paperwork to fill out – I will be back to this shortly.
Comment by Harald Helfgott — July 3, 2010 @ 11:15 pm
5. I was thinking about the writeup, and in particular that you compute $\sum_{n \leq x} \tau(n)t^n$ instead of $\sum_{a \leq n \leq b} \tau(n) t^n$ as I had suggested… and it made me wonder what I was missing. Then, I looked at the draft again, and I think I have found a error in your writeup: At the bottom of page 10, and top of page 11, where you use the geometric series formula, you divide by $t-1$, not $t^m-1$.
Maybe I’m just misreading this, though…
Comment by Ernie Croot — July 2, 2010 @ 3:38 pm
• Oops, that was a silly mistake on my part – I thought I found a neat simplification and strengthening of your argument but that turns out not to work. (I had the vague analogy with Stokes’ theorem and thought every interior sum could be converted into a boundary sum, but now I see why that doesn’t work…)
Comment by — July 2, 2010 @ 4:45 pm
• Here is a repaired version:
http://polymathprojects.files.wordpress.com/2010/07/polymath1.pdf
Comment by — July 2, 2010 @ 5:12 pm
6. [...] is now a thread about the paper mentioned [...]
Pingback by — July 4, 2010 @ 5:47 pm
7. I made a very short list of tiny typos – and now I cannot find it. At any rate, I rather like the paper right now – in particular the third section.
Comment by Harald Helfgott — July 6, 2010 @ 5:42 pm
8. This is very nice. I wonder if there were further insights/problems from the many polymath4 research threads which we should either mention in the paper or perhaps in a post. (Of course, one can simply look at the threads themselves.)
Comment by — July 9, 2010 @ 4:54 pm
9. I just wanted to write that it turns out that there is a simple algorithm to do multipoint polynomial evaluation in any ring very efficiently (one of my REU students looked it up); so the part of the paper where we use the fast matrix multiplication algorithm can be replaced with a literature reference (I asked my student to email it to me — when he does, I will post it here). Basically, it works as follows: Let’s suppose we can multiply together two polynomials of degrees $M$ and $N$, respectively, in time roughly $(M+N)$, times some logs, in some ring $R$. It is easy to see then that we can compute one polynomial mod another just about as fast (times some small constant). Now suppose we have a polynomial $f(x)$ of degree $n = 2^k$ and some points $u_1, ..., u_n$, and we want to evaluate $f(u_1), ..., f(u_n)$. How do we do it quickly? Well, what we can do is define the polynomials
$g_1(x) = (x-u_1) \cdots (x-u_{n/2})$
and
$g_2(x) = (x-u_{n/2+1})\cdots (x-u_n)$,
and then compute
$f_0(x) = f(x) \pmod{g_1(x)}$
and
$f_1(x) = f(x) \pmod{g_2(x)}$,
where the degrees of $f_0$ and $f_1$ are both smaller than $n/2$. Now, it is easy to see that
$f(u_i) = f_0(u_i)$ for $i=1,...,n/2$; and,
$f(u_j) = f_1(u_j)$ for $j=n/2+1,...,n$.
So, we have a “divide and conquer” strategy for doing multipoint evaluation. One detail remains: How do we compute those $g_i(x)$ polynomials quickly as we iterate? Well, what you can do is build up a tree of polynomial products, starting at the leaves. First, you compute the $n/2$ polynomials
$(x-u_1)(x-u_2), ..., (x-u_{n-1})(x-u_n)$
and then you multiply consecutive pairs of *these* together to give the $n/4$ polynomials
$(x-u_1)\cdots (x-u_4), ..., (x-u_{n-3})\cdots (x-u_n)$
and eventually, you get $g_1(x)$ and $g_2(x)$… and in fact all the other polynomials you will need for your divide-and-conquer.
…
Now, to multiply together two polynomials in time about $M+N$ (times some logs) as discussed above, it turns out that there is an algorithm due to Strassen to handle this, which doesn’t use FFTs (it just uses some Karatsuba-type identities); even so, perhaps we can just use the standard FFT algorithm in an obvious way, thinking of the polynomials first as belonging to $Z[x]$ or perhaps $Z[x,y]$, and then modding out by $2$ later on.
Comment by Ernie Croot — August 4, 2010 @ 7:56 pm
• Ok, here is the reference in the literature to the fast multi-point algorithm:
A. Borodin and R. Moenk, “Fast Modular Transforms”, Jour. of Comp. and System Sciences, 8 (1974), 366-386.
And, in case it matters later on in some way, the two students continuing Polymath4 with me are: David Hollis and David Lowry.
Comment by Ernie Croot — August 13, 2010 @ 12:44 am
10. Dear Ernie, sorry for not getting back to you earlier, I’ve been unusually distracted in the last two weeks.
I uploaded a new version mentioning multipoint evaluation and possible future work (see final section) at
http://terrytao.files.wordpress.com/2010/08/polymath.pdf
I think we should be ready to submit the paper to journal as soon as there is consensus to do so. I was thinking of submitting to Mathematics of Computation, if that is OK with everyone…
Comment by — August 20, 2010 @ 7:52 pm
• [Dear Ernie, sorry for not getting back to you earlier, I’ve been unusually distracted in the last two weeks.]
I totally understand, especially given the ICM conference and all.
The wording looks fine to me; and I’m sure David and David would appreciate seeing that they are mentioned (and perhaps when they go to apply for grad school… etc.). It’s looking like we might be able to prove a fast algorithm for $\sum_{a < p < b} t^{f(p)}$, as hoped; I'll know for sure in another two weeks or so (after I return from India, and get to meet with them again).
…
I see that you list the Borodin and Moenk paper, but didn't see that you replaced the use of matrices in the proof of Lemma 3.1 with a reference to it. Now that I think about it, I think I like including the matrix argument instead of just doing a citation. The fact is that everyone is familiar with Strassen's matrix algorithm, while few people know the argument in B-M; and so, the present argument in Lemma 3.1 is more self-contained (given the Strassen matrix alg.), and would be but one or two lines shorter anyways, if we just used the result in B-M (instead of matrices).
…
I would like to look over the paper again carefully before we submit it; but that will be a week or two from now, as I will be leaving for India tomorrow.
Comment by Ernie Croot — August 26, 2010 @ 4:28 am
• OK. I was not sure exactly how to cite the BM paper in the proof of Lemma 3.1, so if you have some suggested wording, I can splice it in. Enjoy India; I didn’t go to the ICM this time around, but I did very much enjoy my last trip to that country…
Comment by — August 26, 2010 @ 4:34 am
11. [...] Polymath4 By kristalcantwell There is a new draft for the paper for Polymath4. It is here. It may be submitted to Mathematics of Computation. For more information see this post. [...]
Pingback by — August 20, 2010 @ 8:42 pm
12. Hello.
I’m sorry for this off topic (first) post. I was just reading through the wiki page for this project, and noticed that the solution to a toy problem hinged on a conjecture about Sylvester’s sequence. I’m fairly sure that I’ve proved it quite simply, but I may be ignoring some complication. I put my proof up on the wiki page.
Again, apologies for the off topic post (and if there’s somewhere better for me to out this kind of stuff, please let me know).
Thanks
-Zomega
Comment by Zomega — August 23, 2010 @ 8:42 pm
• Dear Zomega,
Thanks for your edit. Unfortunately, there is more to a natural number n being squarefree than simply not being a square; one must also show that n is not divisible by any square larger than 1, see http://en.wikipedia.org/wiki/Square-free_integer .
Comment by — August 23, 2010 @ 8:45 pm
• I realize that now. My apologies for making such a gigantic error. I’ll check my work more carefully in the future.
-Zomega
Comment by Zomega — August 24, 2010 @ 12:57 am
13. Sorry for the long delay in writing about the last draft… I think it is pretty much ready to go; however, I found a few more small typos and things:
1. Page 7, there are two of’s.
2. Page 7, “coefficients O(x)” –> “coefficients of size O(x)”.
3. page 7, “If we restrict to the range… the second term”, I think should be “third term”. And “third term” should be “fourth term”.
4. Page 8, “It suffices to show… in time” –> “… that we can compute in time.”
5. Pag 10, near the bottom: You write that $n_0 \geq x^{1/2-c}$, but all we have is $n \geq x^{1/2-c}$; however, since $n = n_0 + \ell q + r$, where $ell,q,r$ are all “small”, this isn’t much of an issue.
Comment by Ernie Croot — September 20, 2010 @ 3:38 pm
• Thanks Ernie! I’ll submit it now to the arXiv and to Math. Comp. Always a good feeling when a project has reached completion…
Comment by — September 20, 2010 @ 9:18 pm
• The paper is now on the arXiv at
http://arxiv.org/abs/1009.3956
and submitted to Mathematics of Computation.
Comment by — September 22, 2010 @ 3:21 am
14. I wanted to the answer to a slightly different but simple question:
Is there an algorithm which on input $n$, finds a prime larger than $n$, with running time bounded by a polynomial in $\log n$?
Comment by — November 18, 2010 @ 4:39 pm
15. Hello. Is this wonderfully number theoretic thread (and the original problem) still open? Just wondering.
Comment by Anonymous — December 23, 2011 @ 8:37 am
16. Hello, I was reading http://arxiv.org/pdf/1009.3956v3.pdf and have some comments.
One of your key identities EQ2.1
2^w(n)= sum(d^2|n)of mobius(d) * tau(n/d^2)
can be generalized. Let m be any integer with m>=0.
2^(m*w(n)) =
m-fold-sum( d1^2|n, d2^2|n, …, dm^2|n )of
mobius(d1)*mobius(d2)*…*mobius(dm) * tau(n/d1^2) * tau(n/d2^2) *…* tau(n/dm^2).
PROOF SKETCH:
It is easy to see this identity is valid in the following easy cases:
(i) m=0 when it is just 1=1, we agree 0-fold-sum is 1.
(ii) m=1 when it is your old identity.
(iii) n=squarefree when only summand comes from d1=d2=…=dm=1.
(iv) n=p^k=prime power when all dj=1 or p:
If k=0 we get n=1 and 2^(m*w(1))=2^0=1=1.
If k=1 we get 2^m=2^m.
If k>=2 we get 2^m=([k+1]-[k-1])^m=2^m.
Now use coprime-multiplicativity to see the easy cases imply validity for all cases.
QED.
This would enable you to count primes in [a,b] not mod 2, but in fact mod 2^m,
for any desired m>=0. If m>log2(b-a+1) this would count the primes, full stop.
Can this be made efficient? I have not tried to figure that out.
Warren D Smith, warren.wds AT gmail.com
Comment by — March 23, 2012 @ 11:58 pm
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http://mathhelpforum.com/calculus/185059-coming-up-exponential-equation.html
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# Thread:
1. ## Definite Integrals
I need to come up with an exponential equation which fulfils the following criteria;
-is an increasing function
-is concave up
-pass through the points (0,3) and (4,11)
-has a definite integral of 22 between x=0 and x=4.
2. ## Re: Coming up with an exponential equation
Why are you so sure the exponential function has base $e$? We know the exponential function is increasing so the base >1.
3. ## Re: Coming up with an exponential equation
Hello, asaver!
I need an exponential equation which fulfils the following criteria:
. . [1] is an increasing function
. . [2] is concave up
. . [3] pass through the points (0,3) and (4,11)
. . [4] has a definite integral of 22 between x=0 and x=4.
Because of the $y$-intercept $(0,3)$
. . you believe the function has the form: $f(x) \,=\,3e^x$
But it could have the form: . $f(x) \,=\,e^x + 2$
You had best begin with the general exponential function: . $f(x) \:=\:ae^{bx} + c$
[1] and [2] tells us that $a > 0$
[3] says $f(0) = 3$ and $f(4) = 1.$
. . We have: . $\begin{Bmatrix} a + c &=& 3 \\ ae^{4b} + c &=& 11 \end{Bmatrix}$
[4] gives us: . $\int^4_0\left(ae^{bx} + c\right)\,dx \:=\:22$
We have three equations in three variables $\{a,b,c\}.$
. . Solve the system.
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http://mathoverflow.net/questions/92206/what-properties-make-0-1-a-good-candidate-for-defining-fundamental-groups/92223
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## What properties make $[0,1]$ a good candidate for defining fundamental groups?
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The title essentially says it all. Consider the category $\mathfrak{Top}_2$ of triples $(J,e_0,e_1)$ where $J$ is a topological space, and $e_i \in J$. There is an obvious generalization of the definition of homotopic maps. Suppose we have selected $(J,e_0,e_1)\in \mathfrak{Top}_2$. We could say that two continuous maps $f,g:X\to Y$ are "$J$-homotopic" if there is a continuous map $h:X\times J\to Y$ such that $h(x,e_0) = f(x)$ and $h(x,e_1) = g(x)$. We could then define $\pi_1 (X,x)$ to be the set of continuous maps $f:J\to X$ satisfying $f(e_0)=f(e_1)=x$, with $J$-homotopic maps identified. Here in order to define composition of paths in the naive way, we need to have picked some homeomorphism continuous map from $J$ to $({0}\times J\cup {1}\times J)/(({0},e_1) = (1,e_0))$, taking $e_0$ to $e_0\times 0$ and $e_1$ to $e_1\times 1$. I have two questions:
1. Can $([0,1],0,1)$ be characterized as an object in $\mathfrak{Top}_2$ in a purely categorical manner?
2. When is $\pi_1 (X,x)$ a group? For that matter, when is $\pi_1 (X,x)$ associative?
Essentially, the question comes down to: what properties of $[0,1]$ are needed in order to do homotopy theory?
-
1
To define multiplication of paths, you don't need a homeomorphism. You just need a map from $J$ to that quotient (taking $e_0$ and $e_1$ to the right things). – Tom Goodwillie Mar 25 2012 at 22:32
1
I think really the circle is the correct object for defining the fundamental group. What is important is that it is a cogroup object in the category of pointed spaces if you allow coassociativity up to homotopy. – Benjamin Steinberg Mar 26 2012 at 0:35
2
The question supposes that fundamental groups were defined before the unit interval... – Jon Beardsley Mar 26 2012 at 1:56
5
You don't need the unit interval at all to define the fundamental group, at least for "nice spaces, since then you can characterize the fundamental group(oid) of $X$ in terms of its covering spaces. Unfortunately, I don't know how to recognize when a space is "nice" (e.g., locally path connected and semi-locally simply connected) without using the unit interval. – Charles Rezk Mar 26 2012 at 15:48
1
I'd just like to put in a plea for Dedecker, Paul and Valderrama, Jerko, Graphes et cographes sur une cat\'egorie abstraite. Application `a l'homotopie, C. R. Acad. Sci. Paris S\'er. A-B, 262, 1966, A377--A380, for an early discussion of an "interval object". – Ronnie Brown May 3 2012 at 10:27
show 8 more comments
## 6 Answers
The answer to 1 is yes. For the purpose of this answer, a bipointed space is a topological space $J$ equipped with distinct closed points $e_0$ and $e_1$. As you say, for any bipointed space $J = (J, e_0, e_1)$, we can form a new bipointed space $J \vee J$ by taking the disjoint union of two copies of $J$, identifying the first $e_1$ with the second $e_0$, and giving the resulting space the obvious pair of basepoints.
Theorem: In the category of bipointed spaces $J$ equipped with a map $J \to J \vee J$, the terminal object is the bipointed space $([0, 1], 0, 1)$ equipped with the map "multiplication by 2" from $[0, 1]$ to $[0, 1] \vee [0, 1] \cong [0, 2]$.
Or informally: $[0, 1]$ has the structure needed in order to be able to define and compose paths, and is universal as such.
The theorem is proved here, and is a variant of a result of Peter Freyd's (which characterized the interval set-theoretically and order-theoretically, but not topologically). The idea that $[0, 1]$ is universal with the structure needed for homotopy theory is expanded on in these talk slides.
-
1
Thanks! This is exactly the sort of thing I was looking for. – Daniel Miller Mar 26 2012 at 11:22
7
+1, it is the perfect answer to this question. Let me remark something which is even more striking (at least for me), but in fact more easy to prove: $([0,1],0,1)$ is the terminal bipointed set! Here, a bipointed set is just a set equipped with two distinct elements. Why striking? The category of bipointed sets is very, very basic and set-theoretic, but the unit interval $[0,1]$, or equivalently $\mathbb{R} \cup \{\pm \infty\}$ is the basic analytical object. But we can characterize this in terms of a simple universal property which only talks about sets ... – Martin Brandenburg Mar 26 2012 at 19:49
8
@Guillaume: I assume Martin didn't mean what he wrote. As you say, [0,1] (with its endpoints) isn't the terminal bipointed set; it's the terminal bipointed set J equipped with a map $J \to J \vee J$. (As before, "bipointed" must be taken to entail distinct basepoints.) Martin's observation is indeed very nice, and appears to have been first observed by Peter Freyd. – Tom Leinster Mar 26 2012 at 22:02
1
Unfortunaltey one cannot edit comments. Yes, I meant it is the terminal coalgebra with respect to the smash product endofunctor on bipointed sets. – Martin Brandenburg Mar 27 2012 at 8:26
2
More like a join than a smash. – Todd Trimble Mar 27 2012 at 23:45
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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.
Rather than answering the question, I want to claim that it's the wrong question.
There is a way to define fundamental group(oid)s and other homotopical notions that makes no reference to the unit interval or any other bipointed space. This goes by names like "shape theory" or "Cech homotopy" or "toposic homotopy". The basic idea is that a "path" consists of a chain of overlapping open sets---a purely topological notion. There's a nice paper by John Kennison called "What is the fundamental group?" which approaches this idea in an elementary way (no toposes required).
This sort of "homotopy" can "see" things which homotopy defined in terms of the unit interval cannot. For instance, the Warsaw circle has trivial $\pi_1$ in the traditional sense, but Cech homotopy can see that it contains a loop. Same goes for the "long circle" (close up the long line into a circle).
One can then simply observe that if your space admits lots of maps into it from the unit interval (e.g. is locally path-simply-connected or whatever), then the fundamental group(oid) as defined in terms of open sets can equivalently be defined using maps out of the unit interval. So from this point of view, there is nothing fundamental about the unit interval; it's just that a lot of the spaces we care about do indeed contain lots of "paths", so that for them we can give a simpler definition of homotopy in terms of paths.
-
Very inspiring, thank you! – Martin Brandenburg Mar 26 2012 at 19:42
I only have a partial answer for 1. and a hopefully non-confusing answer to 2.
To start with, let us work with the fundamental groupoid, which is more, ahem, fundamental and better suited to generalisation. In particular, we can consider the set $\pi^J(X,a,b)$ of homotopy classes (rel endpoints) of maps $(J,0,1) \to (X,a,b)$, which is more natural in the setting you outline. This is, assuming it isn't empty, a torsor for the groups $\pi^J(X,a,a)$ and $\pi^J(X,b,b)$, so you're not really losing too much. But the more important structure is the whole groupoid.
The unit interval is at least weakly initial in the category of path-connected bipointed spaces and homotopy classes of maps (and we always have a torsor as above). If you don't assume path-connected, then the two-point set (with any of its topologies) can be allowed, but is completely useless in measuring homotopy. This is an important fact using $[0,1]$, and this can't be derived from formal homotopy theory. One could define $J$-connectedness for other bipointed spaces $J$, but the utility of such a definition is debatable unless you put in extra conditions, like making it a cylinder object.
The 'reason' we get a fundamental groupoid is that $[0,1]$ is an $A_\infty$ topological cogroupoid, namely a groupoid object in $Top^{op}$, up to homotopy, and then coherence of that up to homotopy, and so on, all the way up. Woah, I hear you say, that's a bit extreme. But it is true, and we can just focus on the first few layers.
First, we have a cocomposition $[0,1] \to [0,1]\sqcup_{1,0}[0,1]$ and a coidentity $[0,1] \to \ast$. Then instead of coassociativity, which would be the equality of the to obvious maps $$[0,1] \to [0,1]\sqcup_{1,0}[0,1]\sqcup_{1,0}[0,1],$$ we have a homotopy between these two maps. We also have a map $$[0,1] \sqcup_{1,0}[0] \to [0,1]$$ expressing the identity on the right, and a similar one on the left. Again, these aren't equal to the identity maps of $[0,1]$, but are homotopic to them. And again, we have coinverses up to homotopy. The choices of all these homotopies aren't important (although you can look up representatives in any book on algebraic topology), because the spaces of such homotopies are contractible.
When we want to involve another space and actually get $\Pi_1(X)$, what we do is hom this topological $A_\infty$-cogroupoid into the space $X$, and get an $A_\infty$-groupoid, and then we truncate it to a groupoid, by quotienting out by these homotopies that we have chosen (but remember the choices are unimportant). It is important that $[0,1]$ is path-connected, because this makes the $A_\infty$-cogroupoid contractible in certain technical ways which are important for generalisations to higher categories (most of the ideas in this answer come from Todd Trimble's work). For instance, in my thesis I defined a certain sort of fundamental bigroupoid which could be applied to topological stacks, and I relied heavily on the $A_\infty$-cogroupoid structure, because it was the only way I could prove I even had a fundamental bigroupoid (I confess I did have much more complicated interval objects than here).
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Nice answer, but I don't feel that you are getting to the heart of my question. Showing that the unit interval behaves well "up to homotopy" is presupposing the fact that we have defined homotopy using the unit interval. If we, so to speak, we were not already using $[0,1]$ to do homotopy theory, then how could we identify the relevant categorical properties of $[0,1]$? By categorical here, I mean expressible in the category of topological spaces and continuous maps (not up to homotopy). – Daniel Miller Mar 26 2012 at 1:26
You need to be able to define 'contractible' somehow in order to make a lot of this machinery to work. In order to do that, you need some sort of model category (not necessarily a Quillen one) or some sort of category with a cylinder object. The unit interval can be defined as a terminal coalgebra of an endofunctor of the category of intervals, see the examples section at ncatlab.org/nlab/show/terminal+coalgebra, but this is not a homotopy-theoretic construction. Note also that it assumes linearly ordered, not merely bipointed as you have done. – David Roberts Mar 26 2012 at 1:40
I find the notation $[0, 1] \times_{0, 1} [0, 1]$ confusing, since it suggests a pullback; I expect what you mean is a pushout which glues the endpoint 1 of the first interval to the endpoint 0 of the second. On a different note, I think the linear orderedness might be a slight red herring, e.g., the unit interval in the Dedekind reals in any Grothendieck topos is the terminal coalgebra for the sort of double-gluing construction outlined in the first sentence of this comment (if done constructively correctly; see the Elephant), as a functor on bipointed objects. – Todd Trimble Mar 26 2012 at 2:05
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It's not true that the two-point set is completely useless for measuring homotopy. "Homotopies" using the Sierpinski space (the non-discrete, non-codiscrete topology on the two-point space) detect exactly the specialization ordering. And there is a map from [0,1] to the Sierpinski space, so any specialization inequality induces a path in the traditional sense. For the "usual" sorts of spaces, the specialization ordering is boring, but for (e.g.) finite topological spaces, it contains all the homotopy information. – Mike Shulman Mar 26 2012 at 5:18
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I'd like to draw attention to the paper: Dedecker, Paul and Valderrama, Jerko, Graphes et cographes sur une cat\'egorie abstraite. Application `a l'homotopie, C. R. Acad. Sci. Paris S\'er. A-B,262, 1966, A377--A380, as an early discussion of an interval object. – Ronnie Brown May 7 2012 at 10:08
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The above answers explain what is needed for the definition of a fundamental group to make sense. Let me try to answer the question from a different angle and explain what properties of the interval are needed for this notion to be reasonably well-behaved.
I believe that the key property is that, intuitively speaking, "small pieces of $I$ look just like the whole thing". More precisely, the interval can be subdivided arbitrarily finely into smaller intervals, i.e. given any open cover $\mathcal{U}$ of $I$ there is a sequence $0 = t_0 < \ldots < t_m = 1$ such that for every $i$ we have $[t_{i - 1}, t_i] \subseteq U$ for some $U \in \mathcal{U}$. In some sense this is a stronger version of the observation that gluing two intervals yields a space that is again homeomorphic to the interval. (It is interesting that the universal property of the theorem mentioned by Tom Leinster already implies the "strong version" of the subdivision property even though it is stated purely in terms of the "weak version".) This is easily proven using the Lebesgue's Lemma and is the starting point of standard techniques for calculating with fundamental groups like the path lifting property for coverings or the Van Kampen Theorem. A similar property of cubes leads to similar techniques for higher homotopy groups.
I cannot think of any other space with a property of this kind we could use in place of $I$. However, it would be interesting to see if there is some analogy between this standard approach to the fundamental group and approaches to something like Čech fundamental group (which I am unfamiliar with).
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You may have a look at http://mathoverflow.net/questions/80777/what-is-a-continuous-path for a very much related discussion.
My opinion, in two words, is that the main property of $[0,1]$ is that one can glue the intervals and obtain basically the same thing. I think that one can define the fundamental group as soon as one has a way to describe paths using an ordered set for which the concatenation of intervals is well-behaved. Let me give an explicit example: in A-homotopy theory of graphs, one uses the natural numbers as ordered set and defines the continuous paths to be finitely supported functions from $\mathbb N$ to the graph with the property that $f(i-1)$ is a neighbor of $f(i)$. Then defines homotopy equivalence and gets a group that has several good properties. If you are interested, you may find details in Chapter 2 of http://arxiv.org/abs/1111.0268
I never really went through the details but I am confident that one can do something similar also using the hyperreal numbers as ordered set and define for instance the fundamental group of *$\mathbb R^2$ and *$\mathbb R^3$. I think that there is some hope to prove that they are not homeomorphic. This fact does not seem easy to prove using classical notions, as you can see here http://mathoverflow.net/questions/86562/non-standard-algebraic-topology
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You may be interested in a paper entitled "The Big Fundamental Group, Big Hawaiian Earrings, and Big Free Groups" by J. Cannon and G. Conner. In this paper, they talk about "big intervals" and use them to define a "big fundamental group". A "big interval" is a totally-ordered set (with the order topology) which is compact and connected (equivalently, a linear continuum with first and last point). It is true that [0,1] is terminal in the category of (non-degenerate) big intervals. (Which also means that it is initial since for big intervals I and J, a monomorphism $I\to J$ exists precisely when an epimorphism $J\to I$ exists).
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For example the end-compactified long line? Though I don't think this has an interval (=preserving endpoints) map from the unit interval... – David Roberts Mar 7 at 7:12
No, if that's the kind of map you want, no there's no such map. There is an order-preserving map (even that preserve endpoints), and there are plenty of continuous maps, but no continuous map that preserves endpoints, no. – Keith Penrod Mar 8 at 0:47
Also, I suppose I'm abusing the words "initial" and "terminal". For every big interval there is a continuous (and order-preserving) map to [0,1] preserving endpoints, but it is not unique. – Keith Penrod Mar 8 at 2:25
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http://mathoverflow.net/questions/33995/is-the-only-known-universe/34015
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## Is {Ø,{Ø},{Ø,{Ø}}, … } the only known universe?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In the first pages of SGA4 I read
[...] Cependant le seul univers connu est l'ensemble des symboles du type {Ø,{Ø},{Ø,{Ø}}, ... } etc. (tous les éléments de cet univers sont des ensembles finis et cet univers est dénombrable). En particulier, on ne connaît pas d'univers qui contienne un élément de cardinal infini. [...]
(the sole known universe is like {Ø,{Ø},{Ø,{Ø}}, ... }, and we don't know any universe with a infinite cardinal).
Mais, c'est vrai? I wonder if during all these years somebody discovered a universe "bigger" than that exhibited by Grothendieck.
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I bring Grothendieck into this because the quotation I wrote comes from SGA... I'm only asking if, from 1963 to now, someone found out a universe different from {Ø,{Ø},{Ø,{Ø}}, ... }. – tetrapharmakon Jul 31 2010 at 10:02
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The question looks very interesting to me. I don't see Ryan Budney's point. – Pierre-Yves Gaillard Jul 31 2010 at 10:24
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This question is very vague. What do you mean by "discover"? I think it is likely that you're taking Grothendieck's quotation a bit too literally. Sure there are other universes, for example a model of ZFC. Or the initial topos. Or the effective topos. OR a model of ZFC + measurable cardinals. And there are permutation models. And so on. – Andrej Bauer Jul 31 2010 at 10:26
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You should understand Grothendieck as saying "I am not sure infinite sets actually exist". This of course is a matter of opinion, but most mathematicians don't have a problem with the existence of the set of natural numbers. – Andrej Bauer Jul 31 2010 at 10:32
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I don't understand where all these querulous comments are coming from. Like PYG, I think the question is perfectly clear. And indeed, it seems to be understood and answered correctly below (modulo some trivial quibbling about the empty set). But also the sentence "I assume you are referring to Grothendieck universes" in the answer seems to suggest that there is some doubt in the matter -- but the OP refers to a specific passage from SGAIV and this pasage is talking about [what are now called] Grothendieck universes. So what's the problem? – Pete L. Clark Jul 31 2010 at 11:45
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## 5 Answers
The universe that Grothendieck intends to suggest by his notation is known in set theory as HF, the class of hereditarily finite sets, the sets that are finite and have all elements finite and elements-of-elements, and so on (the transitive closure should be finite). The set HF is the same as $V_\omega$ in the Levy hiearchy, and can be built by starting with the emptyset and iteratively computing the power set, collecting everything together that is produced at any finite stage. This is the smallest nonempty transitive set that is closed under power set. It satisfies all the Grothendieck universe axioms, except that it doesn't have any infinite elements, since none appear at any finite stage of this consrtruction.
There is an interesting presentation of this universe by a simple relation on the natural numbers. Namely, define $n\ E\ m$ if the $n^{\rm th}$ bit in the binary expansion of $m$ is $1$. The structure $\langle\mathbb{N},E\rangle$ is isomorphic to $\langle HF,{\in}\rangle$ by the map `$\pi(n)=\{\pi(m)\,|\,m\,E\,n\}$`, which set-theorists will recognize as the Mostowski collapse of $E$.
Since HF doesn't have any infinite elements, it is a rather impoverished universe for many applications of that concept. And so we naturally seek larger universes. But the difficulty is that we cannot prove they exist. The difficulty is not one of "discovery," but rather just that we can prove that the hypothesis of the existence of a univese containing infinite sets is too strong for us to prove from our usual axioms. The reason is, as has been remarked in some of the other answers and comments, all other Grothendieck universes have the form $H_\kappa$, the hereditarily size less than $\kappa$ sets, for an inaccessible cardinal $\kappa$. So this is just like HF, which is $H_\omega$, but on a higher level, and in this sense, these higher universes are not so mysterious. They are intensely studied in set theory, a part of the research effort in large cardinals.
In this MO answer, I mention a number of weaker universe concepts that we can prove exist, and which I believe serve most of the uses of the universe concept in category theory, if one wanted to care more about such set theoretic issues.
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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.
Let me rephrase part of what Joel David Hamkins and Anon already said, but without mentioning inaccessible cardinals:
A Grothendieck universe strictly bigger than the one in the question would be a model of ZFC. (More precisely, it would become a model once we interpret the membership symbol of ZFC as actual membership.) So the existence of such a Grothendieck universe would imply the consistency of ZFC. G"oedel's second incompleteness theorem implies that ZFC cannot prove the consistency of ZFC. Therefore ZFC cannot prove the existence of a Grothendieck universe.
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Clear and concise, I understood what was the "problem". Thanks. – tetrapharmakon Jul 31 2010 at 22:17
I assume you are referring to Grothendieck universes.
The existence of a bigger Grothendieck universe is equivalent to the existence of an inaccessible cardinal, which cannot be proved from ZFC, because it implies the consistency of ZFC.
There is a smaller example of a Grothendieck universe: the empty set. This is the only other Grothendieck universe that can be proven to exist in ZFC.
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A universe is nonempty by definition. – Pierre-Yves Gaillard Jul 31 2010 at 11:00
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Thank you. I think it should be noticed that in SGA4 an universe is defined nonempty, but in the sequent Appendix (redige' par N. Bourbaki, pages 185--...) the empty set is accepted as a universe... It seems like a problem, isn't it?! – tetrapharmakon Jul 31 2010 at 12:32
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Dear tetrapharmakon: I just looked at the article you took the quotation from. I'd have been surprised if Grothendieck and Verdier had overlooked the empty set. I don't know why Grothendieck and Verdier excluded the empty set, nor why Bourbaki included it. – Pierre-Yves Gaillard Jul 31 2010 at 12:50
Thanks for your answer. I don't know why Grothendieck and Verdier excluded the empty set, nor why Bourbaki included it. Excuse me but I can't understand what is your point: what "philosophical" position shall I adopt about the empty set? Must I follow Grothendieck&Verdier or Bourbaki? – tetrapharmakon Jul 31 2010 at 17:56
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Dear tetrapharmakon: Is it better to include the non-emptiness condition into the definition of a universe, or not to include it? I don't have the slightest idea. I don't think it's an important question. – Pierre-Yves Gaillard Jul 31 2010 at 19:27
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Just to make perfectly sure: Grothendieck is absolutely not questioning the existence of infinite sets in this quotation. (He had, and has, some eccentricities, but not in this direction!)
Remember that "universe" is a technical term for a certain type of set, essentially one which has maximally nice closure properties. The universe he is talking about corresponds to the cardinal $\aleph_0$, a countably infinite set whose elements may themselves be identified with the finite cardinals (as is a standard operating procedure since von Neumann, although those who don't think that much about infinite sets can and often do safely forget this point). He is not discussing universes as models of formal set theory or anything like that, so the idea that "internally" in this countably infinite universe, infinite sets do not exist, is not at all what he is getting at. Rather, since he has written down an example of an infinite set, we can conclude (from this passage alone, notwithstanding the rest of his work) that he believes in and is comfortable working with infinite sets.
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There exist plenty of other universe. Recall that most of the times one proves the (relative) consistency of some axiom independent from ZF, one actually builds a model that satisfy that axiom. So for instance, the method of forcing invented by Cohen enables to list a infinite number of (elementary) different universes. Of course the universe you mentioned is much more tangible and intuitive then the one built with forcing, but if you accept AC, then they have the same dignity.
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Once again, I believe you are confusing the notion of universe studied in set theory with the (strictly narrower) notion of universe Grothendieck is discussing here, since called Grothendieck universes. See en.wikipedia.org/wiki/Grothendieck_universe and en.wikipedia.org/wiki/Universe_(mathematics). – Pete L. Clark Jul 31 2010 at 13:13
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http://math.stackexchange.com/questions/186958/multipliers-on-h1
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# multipliers on $H^{1}$
I'm begining to study the hardy space $H^{p}(\mathbb{R}^{n})$. First recall that a $L^{\infty}$ function is called a $H^{1}$ multiplier if the associated operator $T_{m}(f)=\mathcal{F^{-1}}(m\hat{f})(x)$ is a bounded operator on $H^{1}$. I want to know how to prove that $$m_{1}(\xi)=\hat{\mu}(\xi)$$(where $\mu$ is the unit measure on $S^{n-1}$) $$m_{2}(\xi)=|\xi|^{i\beta}$$($\beta$ is a positive real number) are multipliers on $H^{1}$.
I think these are basic examples about $H^{1}$ multipliers, but I can't find its proof right now, so any comments or references are welcome
EDIT：similar to the case on $L^{p}$,there are many types(integral or differential form) of theorems to prove that a particular function belongs to $\mathcal{M}_{p}$(all the multipliers on $H^{p}$).Here is one:
if $m(\xi)\in C^{k}$,where $k=[\frac{n}{2}]+1$,satisfies $$|D^{\alpha}m(\xi)|\leq \frac{C}{(1+|\xi|)^{|\alpha|}},\quad |\alpha|\leq k$$ then m is $H^{1}$ multiplier.
Applying this to $m_{2}(\xi)$,we immediatly get the result.But for $m_{1}(\xi)$,since we have $\hat{\mu}(\xi)=J_{\frac{n}{2}-1}(\xi)|\xi|^{-(\frac{n}{2}-1)}$,and due to the asymptotic behavour of the bessel function,it seems the theorem above doesn't fit for this case.
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– sun Sep 10 '12 at 14:43
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http://mathhelpforum.com/calculus/69470-technical-calculus-problem.html
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Thread:
1. technical calculus problem
I hate that my first post on this forum is a question!! These are two questions I can not figure out:
1. Find the volume of the solid of revolution obtained by rotating R, defined by the bounds given in each exercise, about the y-axis.
Y^2=4X, 4X-3Y-4=0
2. Find the volume of the solid obtained by revolving the region bounded by y=square root of X , x=4
a. the x axis
b. the y axis
c the line x=5
d. the line y=3
I don't just need the answer, I need to understand what I need to do to get the answer. Thanks for any and all help.
2. Originally Posted by patrickg
I hate that my first post on this forum is a question!! These are two questions I can not figure out:
1. Find the volume of the solid of revolution obtained by rotating R, defined by the bounds given in each exercise, about the y-axis.
Y^2=4X, 4X-3Y-4=0
did you graph the region? if not, do so.
then, turn your graph so that the positive x-axis is pointing up. do you see it now? we can use the disk (washer) method to find the volume. of course, we integrate with respect to y.
the volume is given by:
$V = \pi \int_a^b \Bigg[ \left( \frac 34y + 1 \right)^2 - \left( \frac {y^2}4 \right)^2 \Bigg]~dy$
$a$ and $b$ (with $b > a$) are the y-coordinates for the points of intersection of the two graphs--i leave them for you to find
(the shell method could work, but we would need two integrals, and that's too much trouble)
2. Find the volume of the solid obtained by revolving the region bounded by y=square root of X , x=4
a. the x axis
b. the y axis
c the line x=5
d. the line y=3
I don't just need the answer, I need to understand what I need to do to get the answer. Thanks for any and all help.
you are missing one of the bounds with this problem. is it also bounded by the x-axis?
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http://math.stackexchange.com/questions/109523/lower-central-series/109528
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# Lower central series
Let $G$ be a finitely generated group and let $\gamma_i$ be the $i$th group in the lower central series. Could you help me to prove that for every $n$ and for every $i$ with $1\leq i\leq n$, $\gamma_i/\gamma_n$ is finitely generated?
The hint that I have is to use the identities:
$[a,bc]=[a,b][a,c][[c,a],b]$
$[ab,c]=[b,c][[c,b],a][a,c]$
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## 1 Answer
First, prove inductively that $\gamma_n/\gamma_{n+1}$ is finitely generated.
Let $g_1,\ldots,g_k$ be a generating set for $G$.
We can now proceed by induction on $m$. The result is true if $m=1$, since $\gamma_1/\gamma_2 = G^{\rm ab}$ is finitely generated by $\overline{g_1},\ldots,\overline{g_k}$.
Assume now that $\gamma_m/\gamma_{m+1}$ is finitely generated, and let $c_1,\ldots,c_r$ be elements of $\gamma_m$ that generate modulo $\gamma_{m+1}$.
Since $\gamma_{m+1} = [\gamma_m,G]$, it is generated by elements of the form $[x,g]$ with $x\in\gamma_m$, $g\in G$.
Use the second of your identities that any $[x,g]$ is congruent, modulo $\gamma_{m+2}$, to a product of commutators of the form $[c_i,g]$ and their inverses.
Then use the first identity to show that any commutator of the form $[c_i,g]$ can be written, modulo $\gamma_{m+2}$, as a product of commutators of the form $[c_i,g_j]$ and their inverses.
Conclude that $\gamma_{m+1}/\gamma_{m+2}$ is finitely generated.
Then you can use the fact that each of $\gamma_i/\gamma_{i+1}$, $\gamma_{i+1}/\gamma_{i+2},\ldots,\gamma_{n-1}/\gamma_n$ are finitely generated to conclude that $\gamma_i/\gamma_{n}$ is finitely generated.
(If you want to be really ambitious, there is an onto map from $G^{\rm ab}\otimes G^{\rm ab}\otimes\cdots\otimes G^{\rm ab}$ ($n$ factors) to $\gamma_n/\gamma_{n+1}$ via $a_1\otimes\cdots\otimes a_n\mapsto [a_1,a_2,\ldots,a_n]$).
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Oops! Thanks for catching my erroneous (now deleted) other answer. (Which, incidentally, falsely claimed that the $\gamma_i$ themselves were finitely generated.) – Cam McLeman Feb 15 '12 at 5:03
@CamMcLeman: No problem: it's a common (and very tempting) error. – Arturo Magidin Feb 15 '12 at 5:15
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http://mathhelpforum.com/algebra/112698-log-problem.html
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Thread:
1. log Problem
There is this problem that I just can not begin to solve $\log_{y}x=2$, $5y = x + 12\log_{x}y$ this is a simultaneous equation. Can anyone help me solve this equation.
2. Originally Posted by scrible
There is this problem that I just can not begin to solve $\log_{y}x=2$, $5y = x + 12\log_{x}y$ this is a simultaneous equation. Can anyone help me solve this equation.
Notice that $\log_{y}(x)=2 \Rightarrow y^2=x$
Substitute $y=x^{\frac{1}{2}}$ into the second equation.
3. Well to continue this problem I am having trouble working out $5(x^\frac{1}{2})=x + 12\log_{x}(x^\frac{1}{2})$ can some one please help me with this.
4. Originally Posted by scrible
Well to continue this problem I am having trouble working out $5(x^\frac{1}{2})=x + 12\log_{x}(x^\frac{1}{2})$ can some one please help me with this.
Can you see why $log_x(x^{\frac{1}{2}}) = \frac{1}{2}$?
After you do that, you should be able to solve the rest.
5. Originally Posted by Defunkt
Can you see why $log_x(x^{\frac{1}{2}}) = \frac{1}{2}$?
After you do that, you should be able to solve the rest.
In the problem do I have to change the base?
6. Originally Posted by scrible
In the problem do I have to change the base?
It looks like you are not very familiar with logarithms. Try to read and learn the logarithm rules (yes one must know them by heart) and solve some basic but efficient problems, and then come back and try to solve this question. It will be by far easier than it is for you now.
7. Originally Posted by Bacterius
It looks like you are not very familiar with logarithms. Try to read and learn the logarithm rules (yes one must know them by heart) and solve some basic but efficient problems, and then come back and try to solve this question. It will be by far easier than it is for you now.
Do you know of any sights I can go to with examples like these? The book is not explaining it well enough for me. I learn by examples.
8. Originally Posted by scrible
Do you know of any sights I can go to with examples like these? The book is not explaining it well enough for me. I learn by examples.
So do I. This is why I usually refer to Wikipedia for formal lessons, and when I'm fed up with theory I can move on to the heavily-detailed examples (though sometimes I can't find any so I look for another website). And then, I can get on with the exercises in the academic department of Wikipedia when they are sufficiently reliable (let's not forget it is Wikipedia ...) and when they do exist.
For example :
- Lessons : http://en.wikipedia.org/wiki/List_of...mic_identities
- Examples : http://people.hofstra.edu/Stefan_Wan...pic1/logs.html
- Exercises : [pick any suitable website for exercises]
9. Originally Posted by Bacterius
So do I. This is why I usually refer to Wikipedia for formal lessons, and when I'm fed up with theory I can move on to the heavily-detailed examples (though sometimes I can't find any so I look for another website). And then, I can get on with the exercises in the academic department of Wikipedia when they are sufficiently reliable (let's not forget it is Wikipedia ...) and when they do exist.
For example :
- Lessons : List of logarithmic identities - Wikipedia, the free encyclopedia
- Examples : Properties of Logarithms
- Exercises : [pick any suitable website for exercises]
so far I have tried
$5(x^\frac{1}{2}) = x + 12\log_{x}(x^\frac{1}{2})$
${\frac{1}{2}}\log5x= \log x +6\log_{x}x$
and then I am stock right here. Did I do it OK so far? I want to add but I don't know if they both have to be the same base to do that. Could someone please help me with the working of this so that I can use it to solve the other problem like it in the book?
10. It's far simpler than you're making it out to be:
By the definition of a logarithm, $log_x(x)=1$.
Why? -- we know that $x^{log_x(a)} = a$. So $x^{log_x(x)} = x$ and therefore $log_x(x)=1$.
Using this and the fact that $log_b(a^n) = n\cdot log_b(a)$ for any $a,b,n$, we get:
$5\sqrt{x} =x + 12log_x(x^{\frac{1}{2}}) \Rightarrow 5\sqrt{x} = x + 12\cdot \frac{1}{2} \cdot \log_x(x) \Rightarrow$ $5\sqrt{x} = x + 6 \Rightarrow 25x = x^2 + 12x + 36 \Rightarrow x^2-13x+36=0$
I believe you can solve that.
11. Originally Posted by Defunkt
It's far simpler than you're making it out to be:
By the definition of a logarithm, $log_x(x)=1$.
Why? -- we know that $x^{log_x(a)} = a$. So $x^{log_x(x)} = x$ and therefore $log_x(x)=1$.
Using this and the fact that $log_b(a^n) = n\cdot log_b(a)$ for any $a,b,n$, we get:
$5\sqrt{x} =x + 12log_x(x^{\frac{1}{2}}) \Rightarrow 5\sqrt{x} = x + 12\cdot \frac{1}{2} \cdot \log_x(x) \Rightarrow$ $5\sqrt{x} = x + 6 \Rightarrow 25x = x^2 + 12x + 36 \Rightarrow x^2-13x+36=0$
I belive you can solve that.
Thanks a million. You know I keep going over the log laws, it is the application of the laws that is killing me. Do you know of any web sight with tricky examples like these?
12. Originally Posted by scrible
Thanks a million. You know I keep going over the log laws, it is the application of the laws that is killing me. Do you know of any web sight with tricky examples like these?
I don't, sorry, but you should try looking for some on google.
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http://cs.stackexchange.com/questions/6582/integer-lp-formulation-and-the-existence-of-a-solution
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# Integer LP formulation and the existence of a solution
A film producer is seeking actors and investors for his new movie. There are $n$ available actors; actor $i$ charges $s_i$ dollars. For funding, there are $m$ available investors. Investor $j$ will provide $p_j$ dollars, but only on the condition that certain actors $L_j \subseteq \{1,2,...,n\}$, are included in the cast (all actors $i \in L_j$ must be chosen in order to receive funding from investor $j$). The producer's profit is the sum of the payments from investors minus the payments to actors. The goal is to maximize this profit.
1. Express this problem as an integer linear program in which the variables take on values on [0,1]
2. Show that there must in fact be an integral optimal solution (as is the case, for example, with maximum flow and bipartite matching).
I am lost on both parts for this problem.
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4
A hint. You probably want a variable, $x_i$, that is $1$ if actor $i$ is involved, and $0$ otherwise, and a similar variable $y_j$ for investor $j$. Now, start writing down inequalities, objective functions, equations, etc. – Peter Shor Nov 9 '12 at 14:05
david, is there a reason why you completely removed the problem definition? – Nicholas Mancuso Nov 10 '12 at 1:28
3
@David: People invested time to come up with a solution for your problem, but then you make their answer useless by completely changing your question. I strongly advise to put back the necessary part of the original question to make Nicolas answer meaningful. The question in the present form is not a real question. – A.Schulz Nov 10 '12 at 9:57
@A.Schulz Feel free to roll back to a reasonable revision. If there are copyright issues here, david should speak up. – Raphael♦ Nov 12 '12 at 11:23
## 1 Answer
Express this problem as an integer linear program in which the variables take on values on [0,1]
$$\begin{align} \text{max: } &\sum_i^m y_i \cdot p_i - \sum_j^n x_j \cdot s_j &\\ \text{subject to: } &x_i \geq y_j &\forall i \in L_j \\ &0 \leq x_i \leq 1 &\forall i \\ &0 \leq y_j \leq 1 &\forall j \end{align}$$
We see that if we select an investor $y_j$ we must select all of the investor's actors $x_i$ for $i \in L_j$. Our profit is the residual income left over from paying actors.
Show that there must in fact be an integral optimal solution (as is the case, for example, with maximum flow and bipartite matching)
If our constraint matrix $A$ is totally unimodular, then the relaxed LP is sufficient to give a 0-1 solution.
Sketch: Observe that $A$ is an $N \times M$ matrix where $N = n + m$ and $M = \sum_j^m |L_j|$. This defines an incidence matrix of a bipartite graph where each row corresponds to an edge $(y_j, x_i)$ with coefficients $(-1, 1)$. We see that any cycle in the graph defined by $A$ will be balanced. That is, for any cycle $C$, $\prod_{e \in C: e=(u,v)} A_{e, v} = 1$. The proof for this can be done using the fact that any cycle in a bipartite graph is even. The corollary to this is that the incidence matrix for any balanced signed graph is totally unimodular.
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1
The $x_i \leq 1$ and the $y_j \leq 1$ constraints should give q non-zero objective function in the dual. – Peter Shor Nov 10 '12 at 2:33
D'oh, that's right. I was just using $x_i \geq y_j$. – Nicholas Mancuso Nov 10 '12 at 17:22
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http://gauravtiwari.org/tag/set/
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# MY DIGITAL NOTEBOOK
A Personal Blog On Mathematical Sciences and Technology
Home » Posts tagged 'set'
# Tag Archives: set
## A Trip To Mathematics: Part II Set(Basics)
Friday, November 4th, 2011 05:55 / 3 Comments
# Introduction
In English dictionary, the word Set has various meanings. It is often said to be the word with maximum meanings (synonyms). But out of all, we should consider only one meaning: ”collection of objects” — a phrase that provides you enough clarity about what Set is all about. But It is not the exact mathematical definition of Set . The theory of Set as a mathematical discipline rose up with George Cantor, German mathematician. It is said that Cantor was working on some problems in Trigonometric series and series of real numbers, which accidently led him to recognise the importance of some distinct collections and intervals. And he started developing Set Theory. Well, we are not here to discuss the history of sets; but Mathematical importance.
Cantor defined the set as a ‘plurality concieved as a unity’ (many in one; in other words, mentally putting together a number of things and assigning them into one box).
Mathematically, a Set $S$ is ‘any collection’ of definite, distinguishable objects of our universe, concieved as a whole. The objects (or things) are called the elements or members of the set $S$. Some sets which are often termed in real life are, words like ”bunch”, ”herd”, ”flock” etc. The set is a different entity from any of its members.
For example, a flock of birds (set) is not just only a single bird (member of the set). ‘Flock’ is just a concept with no material existence but ‘Bird’ or ‘birds’ are real.
# Representing sets
Sets are represented in two main ways:
1. Standard Method: In this method we use to write all elements of a set in a curly bracket ( { } ).
For example:
Flock of Birds := {Bird-1, Bird-2, …, Bird-100,…}
or, $F:= \{B_1, B_2, \ldots, B_{100}, \ldots \}$ is a set.
Here I have used first capital letter of each term to notate the example mathematically. We read this set as, A set F is defined by a collection of objects $B_1, B_2, \ldots$ etc.
2. Characteristic Method: In this method, we write a representative element and define that by a characteristic property. A characteristic property of a set is a property which is satisfied by each member of that set and by nothing else.
For example, above set of Flock of birds can also be written as:
$\mathrm{ F := \{ B : B \ is \ a \ bird \} }$
which has the same meaning at a wider extent. We read it as: ”A set F is defined by element B such that B is a bird.”
# Standard Sets
Some standard sets in Mathematics are:
Set of Natural Numbers: It includes of the numbers, which we can count, viz. $\mathrm{ \{0,1,2,3,4,5,6,7, \ldots \}}$. The set of natural numbers is denoted by $\mathbb{N}$.
Set of Integers: Integers includes of negatives of natural numbers and natural numbers itself. It is denoted by $\mathbb{Z}$. $-5, -4, 1, 2, 0$ …all are integers. The rigorous definition of integers be discussed in fourth part of the series.
Set of Rational Numbers: Rational numbers are numbers which might be represented as $\frac{p}{q}$, where p and q both are integers and relatively prime to each other and q not being zero. The set of rational numbers is represented by $\mathbb{Q}$ and may include elements like $\frac{2}{3}, \frac{-5}{7}, 8$. The characteristic notation of the set of rational numbers is $\mathbb{Q} := \{ \dfrac{p}{q};/ p,q \in \mathbb{Z}, \ (p,q) \equiv 1 \ q \ne 0 \}$. The rigorous dicussion about rational numbers will be provided in fourth part of the series.
Empty Set: It is possible to conceive a set with no elements at all. Such a set is variously known as an empty set or a void set or a vacuous set or a null set.
An example of emptyset is the set $\{\mathrm{x:\ x \ is \ an \ integer \ and \ x^2=2} \}$, since there exists no integer which square is 2 —the set is empty. The unique empty set is denoted by $\emptyset$.
Unit Set: A set with only one element is called the unit set. {x} is a unit set.
Universal Set: A set which contains every possible element in the universe, is a universal set. It is denoted by $U$.
# Two Sets
Let $A$ and $B$ be two sets. We say that $A$ is a subset of $B$ (or $B$ is superset of $A$ or $A$ is contained in $B$ or $B$ contains $A$) if every element of $A$ is also an element of set $B$. In this case we write, $A \subseteq B$ or $B \supseteq A$ respectively, having the same meaning .
Two sets are equal to each other if and only if each is a subset of the other. Subset word might be understood using ‘sub-collection’ or ‘subfamily’ as its synonyms.
Operations on Sets:
As Addition, Subtraction, Multiplication and Division are the most common mathematical operations between numbers; Union, Intersection, Complement, Symmetric difference, Cartesian Products are the same between sets.
UNION OF SETS
If A and B are two sets, then their union (or join) is the set, defined by another set $S \cup T$ such that it consists of elements from either A or B or both. If we write the sets A and B using Characteristic Method as,
$\mathrm{A:= \{x : x \ is \ an \ element \ of \ set \ A\}}$.
and,$\mathrm{B:=\{x : x \ is \ an \ element \ of \ set \ B\}}$
then the union set of A and B is defined by set J such that
$\mathrm{J := \{x: x \ is \ an \ element \ of \ set \ A \ or \ set \ B \ or \ both \}}$.
For practical example, let we have two sets:
$A:= \{1,2,3,r,t,y\}$ and $B:=\{3,6,9,r,y,g,k\}$ be any two sets; then their union is $A \cup B :=\{ 1,2,3,6,9,r,t,y,g,k \}$.
Note that it behaves like writting all the elements of each set, just caring that you are not allowed to write one element twice.
Here is a short video explaining Unions of Sets:
INTERSECTION OF SETS
Intersection or meet of two sets A and B is similarly defined by ‘and’ connective. The set {x: x is an element of A and x is an element of B} or briefly $\mathrm { \{ x: x \in A \wedge x \in B \}}$. It is denoted by $A \cap B$ or by $A \cdot B$ or by $AB$.
For example, and by definition, if A and B be two sets defined as,
$A:=\{1,2,3,r,t,y\}$
$B:=\{3,6,9,r,g,k\}$
then their intersection set, defined by $A \cap B:= \{3,r\}$.
In simple words, the set formed with all common elements of two or more sets is called the intersection set of those sets.
Here is a video explaining the intersection of sets:
If, again, A and B are two sets, we say that A is disjoint from B or B is disjoint from A or both A and B are mutually disjoint, if they have no common elements. Mathematically, two sets A and B are said to be disjoint iff $A \cap B := \emptyset$ .
If two sets are not disjoint, they are said to intersect each other.
PARTITION OF A SET
A partition set of a set X is a disjoint collection of non-empty and distinct subsets of X such that each member of X is a member of exactly one member (subset) of the collection.
For example, if $\{q,w,e,r,t,y,u\}$ is a set of keyboard letters, then $\{ \{q,w,e\}, \{r\}, \{t,y\},\{u\}\}$ is a partition of the set and each element of the set belongs to exactly one member (subset) of partition set. Note that there are many partition sets possible for a set. For example, $\{\{q,w\}, \{e,r\},\{t,y,u\}\}$ is also a partition set of set $\{q,w,e,r,t,y,u\}$.
A Video on Partition of set:
COMPLEMENT SET OF A SET
The complement set $A^c$ of a set $A$ is a collection of objects which do not belong to $A$. Mathematically, $A^c := \{x: x \notin A \}$.
The relative complement of set $A$ with respect to another set $X$ is $X \cap A^c$ ; i.e., intersection of set $X$ and the complement set of $A$. This is usually shortened by $X-A$, read X minus A. Thus, $X-A := {x : x \in X \wedge x \notin A}$, that is, the set of members of $X$ which are not members of $A$.
The complement set is considered as a relatative complement set with respect to (w.r.t) the universal set, and is called the Absolute Complement Set.
A Video on Complement of A Set:
SYMMETRIC DIFFERENCE
The symmetric difference is another difference of sets $A$ and $B$, symbolized $A \Delta B$, is defined by the union of mutual complements of sets $A$ and $B$, i.e., $A \Delta B := (A -B) \cup (B-A) = B \Delta A$.
# Theorems on Sets
1. $A \cup (B \cup C) = (A \cup B) \cup C$
2. $A \cap (B \cap C) = (A \cap B) \cap C$
3. $A \cup B= B \cup A$
4. $A \cap B= B \cap A$
5. $A \cup (B \cap C)= (A \cup B) \cap (A \cup C)$
6. $A \cap (B \cup C)= (A \cap B) \cup (A \cap C)$
7. $A \cup \emptyset= A$
8. $A \cap \emptyset= \emptyset$
9. $A \cup U=U$
10. $A \cap U=A$
11. $A \cup A^c=U$
12. $A \cap A^c=\emptyset$
13. If $\forall A \ , A \cup B=A$ $\Rightarrow B=\emptyset$
14. If $\forall A \ , A \cap B=A \Rightarrow B=U$
15. Self-dual Property: If $A \cup B =U$ and $A \cap B=\emptyset \ \Rightarrow B=A^c$
16. Self Dual: ${(A^c)}^c=A$
17. ${\emptyset}^c=U$
18. $U^c= \emptyset$
19. Idempotent Law: $A \cup A=A$
20. Idempotent Law: $A \cap A =A$
21. Absorption Law: $A \cup (A \cap B) =A$
22. Absorption Law: $A \cap (A \cup B) =A$
23. de Morgen Law: ${(A \cup B)}^c =A^c \cap B^c$
24. de Morgen Law: ${(A \cap B)}^c =A^c \cup B^c$
# Another Theorem
The following statements about set A and set B are equivalent to one another
1. $A \subseteq B$
2. $A \cap B=A$
3. $A \cup B =B$
I trust that we are familiar with the basic properties of complements, unions and intersections. We should now turn to another very important concept, that of a function. So how to define a function? Have we any hint that can lead us to define one of the most important terms in mathematics? We have notion of Sets. We will use it in an ordered manner, saying that an ordered pair.
First of all we need to explain the the notion of an ordered pair. If $x$ and $y$ are some objects, how should we define the ordered pair $(x,y)$ of those objects? By another set? Yes!! The ordered pair is also termed as an ordered set. We define ordered pair $(x,y)$ to be the set $\{\{x,y\},\{x\}\}$. We can denote the ordered pair $(x,y)$ by too, if there is a desperate need to use the small bracket ‘( )’ elsewhere.
So, note that Ordered Pairs
$(x,y) := \{\{x,y\},\{x\}\}$
and $(y,x) :=\{\{y,x\},\{y\}\} =\{\{x,y\},\{y\}\}$ are not identical. Both are different sets.
You might think that if ordered pair can be defined with two objects, then why not with three or more objects. As we defined ordered pair (ordered double, as a term) $(x,y)$, we can also define $(x,y,z)$, an ordered triple. And similarly an ordered $n$ -tuple $(x_1, x_2, \ldots x_n)$ in general such that:
$(x,y):=\{\{x,y\},\{x\}\}$
$(x,y,z):=\{\{x,y,z\},\{x,y\},\{x\}\}$
$(x_1, x_2, x_3, \ldots x_n) := \{\{x_1, x_2, x_3, \ldots x_n\}, \{x_1, x_2, x_3, \ldots x_{n-1}\}, \ldots, \{x_1, x_2\}, \{x_1\}\}$.
Another important topic, which is very important in process to define function (actually in process to define ordered pair) is Cartesian Product (say it, Product, simply) of two sets. Let $A$ and $B$ be two sets. Then their Product (I said, we’ll not use Cartesian anymore) is defined to be the (another) set of an ordered pair, $(a,b)$, where $a$ and $b$ are the elements of set $A$ and set $B$ respectively. Mathematically; the product of two sets $A$ and $B$
$A \times B := \{(a,b) : a \in A, \ b \in B\}$.
Note that $A \times B \ne B \times A$.
The name as well as the notation is suggestive in that if $A$ has $m$ elements, $B$ has $n$ elements then $A \times B$ indeed has $mn$ elements.
We see that if we product two sets, we get an ordered pair of two objects (now we’ll say them, variables). Similarly if we product more than two sets we get ordered pair of same number of variables. For example:
$X \times Y := (x,y); x \in X, y \in Y$.
$X \times Y \times Z :=(x,y,z); x \in X, y \in Y, z \in Z$. etc.
The sets, which are being product are called the factor sets of the ordered pair obtained. When we form products, it is not necessary that the factor sets be distinct. The product of the same set $A$ taken $n$ times is called the $n$ -th power of $A$ and is denoted by $A^n$. Thus, $A^2$ is $A \times A$. $A^3$ is $A \times A \times A$. And so on.
Now we are ready to define functions. The next part of this series will focus on functions.
Keep Reading and Commenting.
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## Into the Mandelbrot set
Friday, September 30th, 2011 20:10 / 3 Comments
The Mandelbrot set is a particular mathematical set of points, whose boundary generates a distinctive and easily recognizable two-dimensional fractal shape. More detail on Wikipedia..
Mandelbrot Fractal
###### Related articles
• The Real 3D Mandelbrot Set (christopherolah.wordpress.com)
• Benoit Mandelbrot Dies (jamesclementcook.wordpress.com)
26.740278 83.888889
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http://nrich.maths.org/2364/solution
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### Rationals Between
What fractions can you find between the square roots of 56 and 58?
### Root to Poly
Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.
### Consecutive Squares
The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?
# Lost in Space
##### Stage: 4 Challenge Level:
Derek (no school given) offered the following insights into this problem: - and nicely explained too - thank you Derek.
From observation, we compile the table below:
n-row pyramid Ways to count 1-2-3-...
1 1
2 3
3 7
4 15
... ...
The pattern can be deduced by considering route backwards, i.e. from n back to 1. The general n-row pyramid is shown below:
Starting at the orange square, we assert that there are 0 routes leading to the final square:
1 step earlier, at the bright yellow squares, there is 1 route leading to the final square:
1 step earlier, at the light yellow squares, the situation becomes slightly different.The top, left and right squares all have 1 route leading to the final square. However, the middle squares have 2 routes as they each connect to 2 bright yellow squares (each of which are routes leading to the final square):
1 step earlier, the top, left and right squares all have 1 route whilst the middle squares have 3 routes as they each connect to 2 light yellow squares, bearing a total of 3 routes. We repeat this reasoning and get the following pattern:
It is clear that these are the binomial coefficients. 2 of Pascal's triangles have been arranged together and the numbers along the two slanted edges represent how many ways to count 1-2-3-...-n. For example, in the diagram of the 6-row pyramid above, we see that there are 1+5+10+10+5+1+5+10+10+5+1 = 63 ways to count 1-2-3-4-5-6.
We further notice that 1, 5, 10, 10, 5, 1 are the binomial coefficients $5 \choose {k}$ $,k = 1, 2, 3, \cdots, 5$. We generalise by claiming that the number of ways of counting $1-2-3-...-n$ is twice (because there are two Pascal's triangles) the sum of all the binomial coefficients of the $n-1^{th}$ row minus 1 (because of the overlapped 1 down the central column) which can be written as $2\sum_{k=0}^{n-1}$ ${n-1}\choose{k}$ $- 1$.
Editor's comment: at this point it may be worth mentioniing that the sum of the binomial coefficients are powers of 2. Why?
Chen of the Chinese High School took a slightly different approach (similar to the one offered by Andrei of Tudor Vianu National College):
We shall define $T(n)$ to be the number of ways to count 1 - n in a triangular array of 1 - n. We shall solve this problem by establishing a recurrence relation, and hence finding a formula for $T(n)$.
When $n=1, T(1)=1$ and when $n=2, T(2)=3$. We observe that in a triangular array of 1-3, the number of ways to get to the number '2' directly above the number '3' is given by $T(2)$ and the number of ways to get to the '2' on the left of '3' can be easily computed to be 2.
Similarly, there are 2 ways to get to the '2' on the right of the '3'.
Hence, $T(3)=T(2)+2^2=7$ For $T(4)$, we observe again that the number of ways to get to the '3' directly above of the '4' is given to be $T(3)$.
Furthermore, the number of ways to get to the '3' to the left of the '4' is the sum of the number of ways to get to the '2's adjacent to the '3'.
Since by the previous result, there are 2 ways of getting to the '2', there are 2+2=4 ways of getting to the '3' to the left of the '4', and similarly for the '3' to the right of the '4'. Hence, $T(4)=T(3)+2(2^2)=T(3)+2^3$ Extending this argument to T(n), we obtain the recurrence $T(n)=T(n-1)+2^{(n-1)}$.
However, since $T(n-1)=T(n-2)+2^{(n-2)}$, which is also equal to $T(n-3)+2^{(n-3)}+2^{(n-2)}$.
Thus, it is not difficult to see that $T(n)= 1+2+2^2+2^3+...+2^{(n-1)} or 2^n - 1$
Editor's comment: The final result can be verified by summing the geometric series with first term 1 and common ratio 2. You can test your understanding of this proof using the proof sorter at http://www.nrich.maths.org/public/viewer.php?obj_id=1398.
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://en.wikipedia.org/wiki/L%c3%a9vy_arcsine_law
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# Lévy arcsine law
In probability theory, the Lévy arcsine law, found by Paul Lévy (1939), states that for a a Wiener process (which models Brownian motion) the proportion of the time that the process is positive is a random variable whose probability distribution is the arcsine distribution. That distribution has a cumulative distribution function proportional to arcsin(√x).
Suppose W is the standard Wiener process. For every T > 0, let
$m(T) = m\{\, t \in [0,T]\,:\, W(t) > 0 \,\}$
be the measure of the set of times t between 0 and T when W(t) > 0. Then for every x ∈ [0, 1],
$\Pr\left( \frac{m(T)}{T} \le x \right) = \frac{2}{\pi}\arcsin\left(\sqrt{x}\right) .$
This result is also sometimes called the "first arcsine law". The two other arcsine laws are concerned with: the time (between 0 and 1) at which W(t) attains its maximum, and the largest time t* such that W(t) remained positive after t*. There are thus three arcsine laws.
## References
• Lévy, Paul (1939), "Sur certains processus stochastiques homogènes", 7: 283–339, ISSN 0010-437X, MR 0000919
• Rogozin, B. A. (2001), "Arcsine law", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4
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http://mathematica.stackexchange.com/questions/tagged/graphs-and-networks
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# Tagged Questions
Questions about handling graphs in Mathematica, graph theory, graph visualization, GraphPlot, the built-in Graph type and the Combinatorica` package.
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### Why my Mathematica 9.0.1 wont work with some samples of Mathematica in Graph Theory?
I updated my Mathematica recently to 9.0.1 cause the previous version was giving me freaky result especially in FindMinimumCostFlow and ...
1answer
106 views
### How can I draw a polygon from a set of angles?
In recreational mathematics, polytans are polygons formed by edge-connecting isosceles right triangles. Order-n polytans are those constructed from n such triangles. My question is this: Given a ...
0answers
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### I have problem with FindMinimumCostFlow and CostValue [closed]
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0answers
33 views
Here is my ...
1answer
69 views
### How can I change the Graph required data such as EdgeWeight,VertexWeight,EdgeCapacity,VertexCapacity in the Graph?
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2answers
131 views
### How can I give graph required data with GUI?
Consider that I have the simple graph of ...
2answers
93 views
### Convert Graph to Graphics
In Mathematica 9, a graph is returned as an object with head Graph: In[8]:= CompleteGraph[8] // Head Out[8]= Graph ...
1answer
114 views
### How can I conveniently call igraph through RLink?
Mathematica has lots of functions for working with graphs, but this functionality is not yet as mature as the igraph library. It is useful to have access to igraph from Mathematica, both for ...
1answer
69 views
### Adjacency Matrix for an Undirected Graph
Here is an undirected graph ...
0answers
83 views
### shortest paths in probabilistically weighted edges? [duplicate]
I understand Mathematica can find shortest paths. However, is it also able to find the most probable paths? I mean this is a case where edges are given probabilistic wights (from 0 to 1) and the most ...
1answer
54 views
### How can I stop DiscreteMarkovProcess[] from relabeling vertices?
I'm attempting to calculate the mean first passage time between two vertices, $v_1$ and $v_2$, provided some undirected graph $G$. However, I noticed that running the following script: ...
0answers
58 views
### What algorithm is Mathematica using for FirstPassageTimeDistribution[]?
What algorithm is Mathematica 9 using for FirstPassageTimeDistribution[]? Is it attempting to use PseudoInverse[] on the ...
2answers
52 views
### How can I efficiently and uniformly sample the set of vertices a fixed edge-wise distance away from a chosen vertex?
I have a large graph $G$, which may be either directed or undirected. How would I use DepthFirstScan[] or BreadthFirstScan[] to ...
2answers
84 views
### Draw network graph and get the coordinates of each node
I would like to draw an undirected graph by specifying the nodes and the edges among them. This is currently done by, for example: ...
1answer
93 views
### How can I extract coordinates of the nodes of my graph?
I have this output for my graph. ...
3answers
94 views
### Drawing a complete graph of an ego network
I have an ego network (a network of a person (ego) and his friends). It is drawn with the ego at the center surrounded by the friends. It might be possible that all the friends are connected to each ...
0answers
53 views
### Fast calculation of commute distances on large graphs (i.e. fast computation of the pseudo-inverse of a large Laplacian / Kirchhoff matrix)
I have a large, locally connected and undirected graph $G$ with $\approx 10^4$ vertices and $\approx 10^5$ to $\approx 10^6$ edges. Moreover I can bound the maximum vertex degree as $Q_{max}$. I ...
0answers
79 views
### Caching of Graph Graphics
In this answer, Simon Woods concludes that when you display an expression with head graph more than once, something is getting stored somewhere. I looked into it too and found out this only happens if ...
3answers
110 views
### Converting Mesh in Density Plots to a Graph
Consider the following line of code: ...
0answers
81 views
### Counting paths of a certain length between a source and sink vertex
I have a graph $G$, which may be directed or not, and I was wondering if there was an efficient way of using, say, BreadthFirstScan[] and FindShortestPath[] to count the number of paths between some ...
1answer
86 views
### How do I create a graph with randomly directed edges starting from a undirected graph?
I have an undirected graph $G$. How would I randomly assign directions to some fraction $p$ of the edges? How might I do this efficiently / quickly provided I have a large graph structure? For ...
2answers
172 views
### How to specify VertexCoordinates as geographic coordinates of cities?
I have been very busy these days with giving the VertexCoordinates of the cities the geographic coordinates from ...
1answer
78 views
### How to get a list of 3-connected graphs?
I'd like to use GraphData to get a list of graphs with vertex connectivity at least 3. I know GraphData["Nonplanar", 6] gives ...
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35 views
### What is the best way to save a large graph object, allowing it to be reloaded quickly in a new kernel?
I generated a very large graph, too large to simply display and manipulate with copy/paste operations. How can I somehow save the graph to a file to allow it to be rapidly reloaded after I quit out ...
1answer
68 views
### double arrow in GraphPlot
Is it possible to combine two edges into one with a double arrow? For example: ...
0answers
50 views
### How can I recover a graph from a graphical output in a notebook after resetting the kernel?
I have an old Mathematica notebook with an output for a graph structure $G$ I would like to recover. However, the routine for building up the graph is stochastic, so I can't simply rerun the code. ...
1answer
163 views
### Making FindShortestPath a little bit sloppy [duplicate]
I have a dense graph, and I'd like to find multiple "almost shortest" paths from a source vertex, $v_s$, to a sink vertex, $v_s$, on an undirected graph $G$. How can I repeatedly run ...
2answers
198 views
### Is there something akin to “SubgraphIsomorphismQ” in Mathematica 9?
Provided two unlabeled graphs, $G$ and $H$, I would like to test where $H$ is a subgraph of $G$. In other words, I'd like to test whether we can prune some fixed number of vertices or edges from $G$ ...
1answer
164 views
### Dynamic visualization of Graph[] on a webpage
How can we embed a dynamic version of a Graph object on a web page, similar to the one showcased on the networkx home page? This sort of dynamic visualization can ...
2answers
136 views
### Graph does not evaluate some options when not displayed
How can I force Graph to evaluate all its options without actually displaying the graph itself? Evaluating the following code ignores the ...
1answer
83 views
### Weighted Kirchhoff Matrix?
I have a weighted graph and I want its graph Laplacian matrix (what Mathematica calls the Kirchhoff matrix in the unweighted case). Is there an easy way to get this? For example, the command: ...
1answer
62 views
### Is it possible for me to explicitly specify a point list for SpatialGraphDistribution?
The function RandomGraph[SpatialGraphDistribution[n, r]] generates a random geometric graph over $[0,1]^2$ where vertices are connected if they are within a ...
1answer
74 views
### Determining whether two k-chromatic graphs are equivalent (not simply isomorphic) using IsomorphicGraphQ?
In a previous question of mine, I asked whether Mathematica's built-in routines could determine an isomorphism for two $k$-chromatic graphs, Determining whether two $k$-chromatic graphs are isomorphic ...
2answers
154 views
### Determining whether two $k$-chromatic graphs are isomorphic (respecting vertex coloration)
Consider the case where I have two $k$-chromatic graphs $G_1$ and $G_2$, i.e. two graphs where individual vertices can be colored with one of a set of $k$ total colors, and I would like to determine ...
2answers
155 views
### Create a planar graph from a set of random points
A planar graph is a graph embedded in the plane in such a way that the edges intersect at vertices. This is an example of a planar graph: g = GridGraph[{3, 3}] ...
1answer
47 views
### How does SocialMediaData determine vertex size
How does SocialMediaData determine the size of graph vertices, e.g. on what does the vertex size depend in, say, a Facebook friend network graph?
2answers
103 views
### HighlightGraph with style
In this picture you see a graph. The shortest path between vertex 5 and vertex 3 is highlighted in red. Instead I want blue highlighted edges of thickness 10. ...
0answers
59 views
### Tooltip not working in web deployed cdf
I deployed a CDF file in a web page. I used both Mouseover and Tooltip functions. They both work in the raw CDF file. The ...
2answers
85 views
### How do I divide the columns of a matrix by the sum of its elements?
I am trying to create a transition matrix for a network. In order to do this, I need to sum down the column (the out degree), and then divide the column by the out degree in order to normalize it. ...
2answers
121 views
### Is there something wrong with VertexConnectivity
As I noticed in the documentation VertexConnectivity is defined as the following The vertex connectivity of a graph $g$ is the smallest number of vertices whose ...
1answer
229 views
### How can I draw the transition diagram of a Markov chain?
I apologize for my heaviness as I am a Mathematica novice. Is there a kind soul who can tell me how to draw diagrams like those shown on page 158, 160 and 161 of this document.Chaine de Markov I ...
1answer
108 views
### Adjacency List representation of a graph
I am looking to draw a graph knowing a simple adjacency representation of the graph like: A = {{1, 3}, {2, 3}, {3, 4, 5}, {4, 5}, {5}} Where ...
3answers
363 views
### Travelling salesman with start and end points for 30 points
I am doing a route optimisation for delivery vehicles and failing dismally. Please see problem statement below. My current solution uses FindShortestTour, but this does not have defined start and end ...
0answers
63 views
### Extracting Facebook Managed accounts (pages) in Mathematica?
I want to use Mathematica to explore various relationships on my Facebook pages. I've been able to successfully extract information on my personal Facebook page using the build in Social Media ...
1answer
54 views
### combining NeighborhoodGraph and RandomGraph
I have Combinatorica and GraphUtilities loaded in version 9 of Mathematica. If I then set: ...
2answers
49 views
### CommunityGraphPlot doesn't size
When I need to plot a 'large' graph with CommunityGraphPlot the output is always sized at the Mathematica default value. The Graph object itself can be resized with ImageSize option. However that ...
3answers
147 views
### Export Graph to PDF
I know that I exported Graphs to PDFs before but it doesn't seem to work any more. I tried K5 = GraphData[{"Complete", 5}] Export["K5.PDF", K5] And I get a PDF ...
2answers
81 views
### The same value of VertexSize for vertices in different graphs causes different vertices size
I generated some graphs. The smallest vertices in the 1st graph should be the same size as the smallest vertices in the second graph. The smallest vertices have the same size (...
1answer
78 views
### How to get VertexDelete to work?
I'm using Mathematica 9 and want to delete a single vertex and all associated edges on a graph but can't seem to get the function VertexDelete function to work. My code is below. I declare a graph ...
2answers
123 views
### Number of edges to each polygon of a graph constructed from an image
I have an image which looks like from How to convert an image to a graph and get the positions of the edges? which by using MorphologicalGraph becomes a graph ...
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http://mathoverflow.net/questions/118226?sort=votes
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## Seeing topological (geom.) properties of the space via corresponding C^*-algebra
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Compact Hausdorff spaces bijectively correspond to C^*-algebras with identity. One needs to consider the algebra of continuous functions C(X) to go in one direction and spectrum to go in the other. (See e.g. Wikipedia). The situation is similar to algebraic geometry - affine manifolds correspond to commutative algebras... Basic skill in alg.geom. is to recast algebraic properties in geometric and vice versa e.g. projective modules - vector bundles... (the dictionary is lengthy).
I wonder about similar correspondence in C^*-algebra setup. In particular:
Question 1: if space "X" is topological manifold (i.e. locally R^n), is there some "nice" way to recognize it via C^*-algebra of continuous function ? (... is there non-commutative version ? ... )
Question 2: if "X" is smooth manifold, is there nice way to recognize it and define sub-algebra of smooth functions entirely in terms of C^*-algebra ? (... is there non-commutative version ? ... )
Question 3 Is it possible to characterize the set of all measures on "X" in terms of C(X) ? (... is there non-commutative version ? ... )
If you have further comments how interesting algebraic properties can be recasted in topological or vice versa, you are welcome to post.
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Surely we have had some version of this question before on MO? (It usually comes with someone enthusiastically quoting an NCG dictionary while not quite paying enough attention to the issues for non-compact spaces...) – Yemon Choi Jan 6 at 21:45
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And surely to answer your Q2 one should look for extra structure on your $C^\ast$-algebra, not just "entirely in terms of $C^\ast$-algebra" - as in the case of a smooth manifold being a topological space with extra structure – Yemon Choi Jan 6 at 21:47
@Yemon all spaces are compact in my question. I'm primarily interested in the specific questions above, not in general dictionary, but if someone wants to share something interesting I would not be against. – Alexander Chervov Jan 6 at 21:49
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It is by now fairly well understood how to give an operator algebraic characterization of a Riemannian manifold (well, maybe only some of them). However, it is not fair to call it a C*-algebraic characterization because part of the data needed is a suitable dense subalgebra of the given commutative C*-algebra. As far as I know, nobody has figured out the right axioms for a noncommutative topological manifold, or even a noncommutative smooth manifold. – Paul Siegel Jan 7 at 0:17
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There are lots of related (identical?) MO questions: mathoverflow.net/questions/21168/…, mathoverflow.net/questions/82871/…, mathoverflow.net/questions/100461/…, etc. – Martin Brandenburg Jan 7 at 1:31
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## 2 Answers
Question 1: The topological $n$-manifold property is equivalent to every point of $X$ having a neighborhood homeomorphic to $B^n$, the closed unit ball in $\mathbb{R}^n$. The existence of such a neighborhood $x\in B^n_x \subset X$ induces the surjective algebra homomorphisms `$C(X) \to C(B_x^n) \cong C(B^n) \to C(\{x\})\cong \mathbb{R}$` (actually, extremal epimorphisms, I think). The Gelfand duality between compact Hausdorff topological spaces and commutative real $C^*$ algebras ensures that the existence of surjective homomorphisms $C(X) \to C(B^n) \to \mathbb{R}$ (the first map should be an extremal epimorphism, while the second should correspond to the quotient by the maximal ideal of an interior point of $B^n$) implies the existence of continuous maps `$\{x\} \to B^n \to X$`, where $x$ maps to an interior point of $B^n$ and $B^n$ is embedded in $X$, and hence a neighborhood of $x$ in $X$. Having such such algebra homomorphisms for each $x\in X$ characterizes $X$ as a topological manifold.
Question 2: $C^\infty(X)$ for a compact manifold $X$ is not a $C^*$ algebra. It is at the very least a Fréchet algebra, where multiplication satisfies an extra convexity condition (though I'm fuzzy on the details). It is at least clear that one must leave the category of `$C^*$` to characterize it. A point that non-commutative geometry centered discussions of this questions seem to be ignoring is that there already exists an algebraic characterization of $C^\infty(X)$ that has nothing to do with non-commutative geometric spectral triples. In my understanding, such a characterization can be found in an article by Michor and Vanžura (arXiv:math/9404228).
Question 3: As already mentioned in Vahid's answer, this is the content of the Riesz representation theorem. The topological vector space dual to $C(X)$ is the space of signed Radon measures on $X$. $C(X)$ is partially ordered by pointwise comparison, which also induces a partial order on its dual. The positive cone in the space of signed Radon measures consists of the positive Radon measures.
I cannot say anything about non-commutative versions of the above answers. But, since these correspondences are heavily based on lots of non-trivial maximal ideals, and such ideals are likely to be absent in non-commutative algebras, they probably do not translate directly.
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Thank you very much for your informative answer ! So you reduce a question to characterization of C(B^n), is it possible to give some somewhat "intrinsic" characterization of this algebra ? (It seems Michor&K uses similar idea). – Alexander Chervov Jan 7 at 12:18
B is open disk ? $C(B)$ is continuous functions on vanishing at inf, correct ? – Alexander Chervov Jan 7 at 12:35
As written in the first sentence, $B^n$ is the closed ball, which includes the boundary and hence is compact. Otherwise applying the standard Gelfand duality is problematic. So, $C(X)$ is a `$C^*$` algebra without imposing any restrictions on behavior at the boundary. As for an intrinsic characterization, how about a topological completion of $\mathbb{\R}[x_1,\ldots,x_2]$ in the norm of uniform convergence on $B^n$? – Igor Khavkine Jan 7 at 13:56
Then what about intrinsic characterization of R[x_1,...x_n] ? and its completion (so we need norm, norm is sup_{ball}, so again we need ball)... – Alexander Chervov Jan 7 at 16:13
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@Branimir, from what I can see after a quick look, the answer is yes. Another book that covers smooth algebras and the geometric spaces dual to them is Models for smooth infinitesimal analysis by Moerdijk & Reyes. – Igor Khavkine Jan 8 at 0:00
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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.
The answer to your first (and second) question is negative, because commutative C*-algebras only reflect global features of the underlying space.
For the third question: we have the Riesz representation theorem which says: For locally compact and Hausdorff topological space $X$, there is and isometric isomorphism between the dual of $C_0(X)$ and the space of radon measures on $X$. See Theorem 7.17 in Folland's book "Real analysis".
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I think it is Radon "charges", not measures. Charges - need not be positive, e.g. difference of two measures in general is "charge". And measures are not linear spaces in general. In Q3 I am interested how to distinguish this positive part among all "charges" ? I do not quite understand the first sentence/argument. – Alexander Chervov Jan 6 at 22:08
It is complex Radon measures. There is a bijective correspondence between open subsets of $X$ and closed two sided ideals of $C_0(X)$. But we cannot consider closed two sided ideals as a substitute for open sets in noncommutative $C^*$-algebras, because there are simple $C^*$-algebras, for example $M_n(\mathbb{C}$ for every $n\in \mathbb{N}$. Then all simple $C^*$-algebras should be considered as a noncommutative space with one point which is obviously naive. – Vahid Shirbisheh Jan 6 at 22:32
Having very few open sets is not the same as having few points, and even if you do have few "points" that doesn't mean the "noncommutative space" is close to trivial. See for instance the crossed product algebras that arise from non-proper group actions – Yemon Choi Jan 6 at 23:46
@Yemon Choi: Assuming hausdorffness of $X$ having few open sets means having few points. For the noncommutative case I used the word "naive" to say it is not a suitable point of view. – Vahid Shirbisheh Jan 6 at 23:55
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Vahid, as I'm sure you know, one of the whole points of NCG is to find replacements for non-Hausdorff quotient or moduli spaces. Cf the non-Hausdorffness of Zariski topology in classical algebraic geometry. So arguments invoking the Hausdorffness of X seem beside the point when discussing NC generalizations, we know that is too restrictive. – Yemon Choi Jan 7 at 0:26
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http://math.stackexchange.com/questions/148759/quasi-finite-maps-to-quasi-projective-varieties
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# quasi-finite maps to quasi-projective varieties?
Say we are given a smooth complex algebraic variety $Y$ which is quasi-projective, and $X$ a second complex manifold together with a holomorphic map $f:X\rightarrow Y$ which is of finite fibers, is it true that $X$ remains quasi-projective? Is the condition of smoothness essential, namely can we do the same for analytic maps from a complex analytic space $X$ to a quasi-projective variety $Y$? I consider a very ample line bundle $L$ on $Y$ that realizes an embedding of $Y$ into some projective space, but I do not see how $f^*L$ should be expected to do the same thing.
Many thanks!
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## 2 Answers
If you assume that $X \to Y$ is finite, with $X$ a complex analytic space, then $Y$ a quasi-projective variety implies that $X$ is. This is a general version of Riemann's existence theorem, maybe due to Grauert and Remmert when $X$ and $Y$ are normal and to Grothendieck in general (perhaps with additional assumption that $X \to Y$ is unramified). (But I'm not sure if this is the correct attribution, and you should also probably look in the literature for the most general precise statements.)
If you just assume finite fibres, there is not much to say without making more restrictions. E.g. suppose that $X$ is an open subset of $Y$ (in the complex topology). Then the fibres of $X \to Y$ are finite (either singletons or empty), but $X$ won't be quasi-projective unless it is actually Zariski open in $Y$.
Since $X$ is smooth when $Y$ is, this shows that smoothness is not really the issue here.
Here are some speculations:
I don't know how the details would go, but I imagine that you can stratify $Y$ according to the nature of the fibres of the map $X \to Y$ (i.e. consider the strata along which the fibres of $X \to Y$ are of some fixed cardinality, counted with multiplicity). My guess would then be that $X$ is quasi-projective if and only this stratification is an algebraic stratification of $Y$. (The idea is that $X \to Y$ will be finite when restricted to the strata, so Riemann existence would apply. The details could be tricky, though, and my guess might be a little too naive.)
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thanks a lot!. Do you know where I should go to for this general form of Riemann's existence theorem? – turtle May 23 '12 at 18:48
@turtle: Dear turtle, I just typed "Riemann existence theorem Grauert" into google, and got many relevant links. Did you try that? Also, you can look in Grothendieck's Bourbaki seminar on flat descent (I think it is the first of the FGAs), or one of the Grauert--Remmert books on complex analytic geometry. But I would try googling first. Regards, – Matt E May 24 '12 at 11:52
@turtle: Dear turtle, Also, my googling suggests that perhaps my assertion is a little too strong. Maybe if one doesn't assume $X$ and $Y$ are normal, then one has to assume that the map is not just finite, but also unramified. But you should probably do a careful literature search to get the bottom of the most precise general form that is true. Regards, – Matt E May 24 '12 at 11:56
thanks a great deal for your reply! – turtle May 29 '12 at 18:46
The answer to your question is yes. If the map $f: X \rightarrow Y$ is quasi-finite, i.e. it has finite fibers, and $Y$ is quasi-projective, then $X$ must be quasi-projective. By Stein factorization, $f$ factors into an open immersion followed by a finite morphism, so it suffices to treat the case when $f$ is finite, since an open subset of a quasi-projective is again quasi-projective. Furthermore, by composing with an embedding of $Y$ into a projective compactification $\bar{Y}$ you may assume that $Y$ is projective.
In this case, where $f: X \rightarrow Y$ is a finite map with $Y$ projective, take an ample line bundle $L$ on $Y$ and show that $f^*L$ is ample on $X$. This is pretty easy, e.g. Seshadri's criterion.
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Dear Parsa, $X$ is assumed to be merely a complex manifold. Any (complex) open subset of $Y$ is a complex manifold if $Y$ is smooth, and most of these are not quasi-projective varieties (because they are typically not Zariski open). Regards, – Matt E May 29 '12 at 23:00
@MattE Yes, I was working in the algebraic category. – Parsa May 31 '12 at 15:30
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http://mathoverflow.net/questions/105767/non-convex-optimization/105799
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## non convex optimization
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi there, In my studies I come up with this nonconvex optimization problem argmin |Ax|_2+lamda*|x|_1 subject to x'x=1 where cost function is nonsmooth but convex and the constrant in nonconvex. I tries subgradient projection method for convex constraints but the global solution is not my desired solution. My question is that I should solve this problem hurestically or there is a reliable method for this nonconvex optimization problem?
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are you missing a $b$ in the $Ax$ term? – S. Sra Aug 29 at 11:40
## 2 Answers
You can have a look of these papers： 1. Jonathan H. Manton, Optimization algorithms exploiting unitary constraints. 2. Zaiwen Zai and Wotao Yin, A feasible method for optimization with orthogonality constraints.
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Thanks for these papers. – unknown (google) Aug 29 at 20:01
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Conceptually, for algorithm design, the following version of the problem might be amenable to a larger number of techniques:
\begin{equation*} \min_x\quad\|Ax-b\|^2\quad\text{s.t.}\quad \|x\|_1 \le \gamma,\quad\|x\|=1. \end{equation*}
There are two reasons behind this reformulation:
1. The objective function is now differentiable, so without further ado you can invoke the Gradient-Projection method, which under reasonable assumptions can be guaranteed to converge.
2. This formulation makes it easy to use Alternating-Projection approaches.
Of course, several other numerical ideas also apply. For example, to get a good solution, you could start with $\gamma$ very large so that the $\ell_1$ constraint essentially disappears; then solve the problem exactly, and then gradually tighten $\gamma$.
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This trick seems interesting but I have no explicit data term in misfit functional and I am solving system of homogenous equations and data term is in the A matrix. So I have problem in using Alternating-Projection approaches with no b in the misfit functional. – unknown (google) Aug 29 at 20:00
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http://dsp.stackexchange.com/questions/2317/what-is-the-sparse-fourier-transform
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# What is the sparse Fourier transform?
MIT has been making a bit of noise lately about a new algorithm that is touted as a faster Fourier transform that works on particular kinds of signals, for instance: "Faster fourier transform named one of world’s most important emerging technologies". The MIT Technology Review magazine says:
With the new algorithm, called the sparse Fourier transform (SFT), streams of data can be processed 10 to 100 times faster than was possible with the FFT. The speedup can occur because the information we care about most has a great deal of structure: music is not random noise. These meaningful signals typically have only a fraction of the possible values that a signal could take; the technical term for this is that the information is "sparse." Because the SFT algorithm isn't intended to work with all possible streams of data, it can take certain shortcuts not otherwise available. In theory, an algorithm that can handle only sparse signals is much more limited than the FFT. But "sparsity is everywhere," points out coinventor Katabi, a professor of electrical engineering and computer science. "It's in nature; it's in video signals; it's in audio signals."
Could someone here provide a more technical explanation of what the algorithm actually is, and where it might be applicable?
• The paper: "Nearly Optimal Sparse Fourier Transform" (arXiv) by Haitham Hassanieh, Piotr Indyk, Dina Katabi, Eric Price.
• Project website - includes sample implementation.
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## 3 Answers
The idea of the algorithm is this: assume you have a length $N$ signal that is sparse in the frequency domain. This means that if you were to calculate its discrete Fourier transform, there would be a small number of outputs $k \ll N$ that are nonzero; the other $N-k$ are negligible. One way of getting at the $k$ outputs that you want is to use the FFT on the entire sequence, then select the $k$ nonzero values.
The sparse Fourier transform algorithm presented here is a technique for calculating those $k$ outputs with lower complexity than the FFT-based method. Essentially, because $N-k$ outputs are zero, you can save some effort by taking shortcuts inside the algorithm to not even generate those result values. While the FFT has a complexity of $O(n \log n)$, the sparse algorithm has a potentially-lower complexity of $O(k \log n)$ for the sparse-spectrum case.
For the more general case, where the spectrum is "kind of sparse" but there are more than $k$ nonzero values (e.g. for a number of tones embedded in noise), they present a variation of the algorithm that estimates the $k$ largest outputs, with a time complexity of $O(k \log n \log \frac{n}{k})$, which could also be less complex than the FFT.
According to one graph of their results (reproduced in the image below), the crossover point for improved performance with respect to FFTW (an optimized FFT library, made by some other guys at MIT) is around the point where only $\frac{1}{2^{11}}$-th to $\frac{1}{2^{10}}$-th of the output transform coefficients are nonzero. Also, in this presentation they indicate that the sparse algorithm provides better performance when $\frac{N}{k} \in [2000, 10^6]$.
These conditions do limit the applicability of the algorithm to cases where you know there are likely to be few significantly-large peaks in a signal's spectrum. One example that they cite on their Web site is that on average, 8-by-8 blocks of pixels often used in image and video compression are almost 90% sparse in the frequency domain and thus could benefit from an algorithm that exploited that property. That level of sparsity doesn't seem to square with the application space for this particular algorithm, so it may just be an illustrative example.
I need to read through the literature a bit more to get a better feel for how practical such a technique is for use on real-world problems, but for certain classes of applications, it could be a fit.
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1
+1. Excellent Explanation. – Dipan Mehta May 8 '12 at 13:35
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So it's basically a lossy FFT? Like an MP3 encoder? – endolith May 8 '12 at 14:21
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@endolith: I'm not sure that I would put it that way. Maybe more analogous to a pruned FFT algorithm that only calculates a subset of the outputs. The claim is that if the input signal is $k$-sparse, then the $k$ outputs are calculated exactly. – Jason R May 8 '12 at 16:22
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@endolith: MP3 compression is lossy because the coefficients are quantized ; not because only the top k coefficients are kept. Sparse FFT = "what is the k-coefficients representation minimizing the difference with the input signal". Coding of a mp3 frame = "what are the quantized coefficients and quantization levels that minimize the (perceptual) error given a budget of N bits for storing the coefficients & scale factors". – pichenettes May 9 '12 at 8:51
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When they are thrown away, this is a side-effect of quantization (the value is rounded to 0) – pichenettes May 9 '12 at 17:08
show 2 more comments
I have looked through the paper and I think I got the general idea of the method. The "secret souse" of the method is how to get sparse representation of input signal in frequency domain. The previous algorithms used kind of brute force for location of dominant sparse coefficient. This method use instead technique which called "space recovery" or "compressed sensing" wiki article is here The exact method of sparse recovery used here looks similar to 'hard thresholding" - one of the dominant sparse recovery methods.
PS technique of sparse recovery/compressed sensing and connected to it L1 minimization used a lot in modern signal processing and especially in connection with Fourier transform. In fact it's a must to know for modern signal processing. But before Fourier transform was used as one of the methods for solution of sparse recovery problem. Here we see opposite - sparse recovery for Fourier transform.
Good site for overview compressed sensing: nuit-blanche.blogspot.com/
PPS answer to previous comment - if input signal is not exactly sparse it's lossy.
Feel free to correct me if I got method wrong.
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The FFT paper isn't compressed sensing. The idea behind compressed sensing is that you can recover sparse data from sparse random samples drawn from a different domain (eg recover sparse images from random sparse frequency data (ie MRI)). While this can reduce acquisition time, it increases computational cost. The FFT paper is different in that you have all you data in both domains and the goal is to make a computation happen quickly. – rcompton May 18 '12 at 7:43
You are wrong about compressed sensing. – mirror2image Jun 12 '12 at 10:30
Can you elaborate? – rcompton Jun 13 '12 at 8:39
Compressed sensing is a huge area with blurry edges, which include/connected to not only recovery per se but similar areas $L_p$ regularization, minimum complexity pursuits etc. Originally it' a sparsity constrained problem $Ax=y$, x in $R^m$, $y in$R^n$, m>>n with constraint$\|x\|_0 < k\$, but later it become a lot more brad. Start with reading wiki – mirror2image Jul 15 '12 at 5:56
– rcompton Jul 22 '12 at 5:16
show 4 more comments
I haven't read the paper on sFFT, but my feeling is that the idea of fasten the FFT behind is exploiting the prior of k-sparsity. Hence, one do not have to compute all entries of FFT coefficients, instead, only computing k of them. So that's why for k-sparse signal, the complexity is O(klog n) instead of O(nlog n) for conventional FFT.
Any way, regarding to the comments of @rcmpton, by saying "The idea behind compressed sensing is that you can recover sparse data from sparse random samples drawn from a different domain (eg recover sparse images from random sparse frequency data (ie MRI))." The question is what is "sparse random samples"? I think it might be samples collected by randomly projecting the sparse data to some lower (measurement) subspace.
And as I understood, the theoretical framework of compressive sensing is majorly comprised of 3 issues, sparsity, measurement, and recovery. By sparsity, it pertains to seeking sparse representations for certain class of signals, which is the task of dictionary learning. By measurement, it pertains to seeking a efficient way (computational efficiency and recoverable) to measure the data (or projecting data to lower measurement space), which is the task of measurement matrix design, such as random Gaussian matrix, structured random matrix, .... And by recovery, is the sparse regularized linear inversion problems, l0, l1, l1-l2, lp, l-group, blabla..., and the resulted algorithms are various, Matching pursuit, soft thresholding, hard thresholding, basis pursuit, bayesian, ....
It is true that "cs is minimization of L1 norm", and L1 norm is a basic principle for cs, but cs is not only minimization of L1 norm. Besides the above 3 parts, there are also some extensions, like structured (group, or model) compressive sensing, where structured sparsity is also exploited, and is proved to largely improve the recovery ability.
As a conclusion, cs is a big step in sampling theory, providing an efficient way to sample signals, provided that these signals are sparse enough. So, cs is a sampling theory, anyone who is going to use it as some technique for classification or recognition is misleading the principle. And occasionally, I find some paper titled with "compressive sensing based .....", and I think the principle of such paper is exploiting l1-minimization instead of cs and it's better to use "l1-minimization based ....".
If I am wrong, correct me please.
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Welcome to DSP.SE This is a great contribution. – Phonon♦ Mar 6 at 10:35
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http://en.m.wikibooks.org/wiki/Floating_Point/Special_Numbers
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# Floating Point/Special Numbers
## Special Numbers
There are several special numbers specified in the IEEE 754 standard. These include representations for zero, infinity, and Not-A-Number (NaN).
## Zero
A floating point number is said to be zero when the exponent and the significand are both equal to zero. This is a special case, because we remember that the significand is always considered to be normalized. This means that $1 < m < 2$, and there is an implied "1." before the significand. If we look at the following equation:
$(1.0 + m) \times b^e$
And we plug in our values for m and e:
$(1.0 + 0) \times b^0 = 1$
We say that whenever the exponent is zero, we have a special class of numbers called "denormalized numbers". We will discuss denormalized numbers more later.
### Negative Zero?
Notice that our definition of zero doesnt have any mention of the sign bit. This means that we can have both a positive zero, and a negative zero, depending on the sign value. What is the difference between the two? In general, there is no difference between positive zero and negative zero, except that the sign bit can propagate though arithmetic operations.
## Infinity
When the exponent of the number is the maximum value, that is, when the exponent is all 1 bits, we have another special class of numbers. This means that regular numbers may never use the maximum exponent value for representing numbers. If the exponent is the maximum value, and the significand is zero, we say that this special case is called "Infinity". Notice that we can have negative infinity and positive infinity values, depending on the sign bit.
## NaN
If we have a maximum exponent value, and a non-zero significand, we have a very special number known as NaN. NaN stands for "Not a Number", and occurs when we perform certain illegal operations, such a division by zero. NaN can be signed, but the sign rarely matters. NaN numbers cannot be used meaningfully in calculations.
## Cases
Type Exp Fraction Sign
Positive Zero 0 0 0
Negative Zero 0 0 1
Denormalised numbers 0 non zero any
Normalised numbers $1..2^e-2$ any any
Infinities $2^e-1$ 0 any
NaN $2^e-1$ non zero any
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http://mathoverflow.net/questions/104003?sort=newest
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## Is there an explicit example of a coefficient sheaf for which hard Lefschetz fails?
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Recall that if $X$ is a projective algebraic complex manifold, $L$ is a semisimple $\mathbb C$-local system on $X$ of geometric origin (roughly speaking, this means that $L$ is a cohomology sheaf $R^if_*\mathbb C$ for some algebraic morphism $f:Y\to X;$ see BBD for the precise definition), and $\eta\in H^2(X,\mathbb C)$ is an ample class, then $$\eta^i\cup-:H^{\dim X-i}(X,L)\to H^{\dim X+i}(X,L)$$ is an isomorphism. This also holds when the projective variety $X$ is allowed to have singularities and $L$ is a perverse sheaf (again semisimple of geometric origin), appropriately shifted.
I'd like to know an example for which this fails; of course, $L$ is no longer of geometric origin. Over finite field, as long as $L$ is assumed semisimple, hard Lefschetz always holds, and conjecturally $L$ is of geometric origin.
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## 1 Answer
The good news (or is it bad news?) is that there is no counterexample in the semisimple case. Simpson proved that hard Lefschetz holds for any semisimple local system on a smooth complex projective variety. To be more precise, apply theorem 1 and lemma 2.6 of his paper on Higgs bundles and local systems. More recently, Sabbah and Mochizuki have extended this to some (or perhaps all, I can't quite remember) semisimple perverse sheaves in accordance with a conjecture of Kashiwara.
Addendum Here is a counterexample for nonsemisimple local systems (which was something that I had wondered about myself). Let $X$ be a smooth projective curve with genus $g>1$. Let $\pi=\pi_1(X)$. Let $\mathbb{Q}_\rho$ denote a nontrivial rank one $\pi$-module with character $\rho$, and $\mathbb{Q}$ the trivial module. We can see, using Euler characteristics, that `$$Ext^1_\pi(\mathbb{Q},\mathbb{Q}_\rho)\cong H^1(X,\mathbb{Q}_\rho)\not=0$$` Thus we can form a nonsplit extension `$$0\to \mathbb{Q}_\rho \to L\to \mathbb{Q}\to 0$$` We necessarily have $H^0(X,L) = H^0(\pi, L)=0$. On the other hand, by Poincaré duality `$$H^2(X,L) = H^0(X,L^*)^*= H^0(\pi,L^*)^*\not=0$$` So $H^0(X,L)\not= H^2(X,L)$, i.e. hard Lefschetz fails.
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Thanks for the answer, Donu. A somehow related question: what about varieties over $\overline{\mathbb F}_p?$ Namely $X$ is projective (smooth or not) and $L$ is semisimple local system (or perverse sheaf). Without being of geometric origin, they won't have a model over a finite subfield, so that the usual proof breaks. – shenghao Aug 5 at 15:57
Shenghao, I don't know too much about this case, but it may be worth looking at Drinfeld's paper "On a conjecture of Kashiwara", which (unless I've misunderstood) provides some evidence. – Donu Arapura Aug 5 at 16:28
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http://unapologetic.wordpress.com/2008/06/30/spanning-sets/?like=1&source=post_flair&_wpnonce=1ba466d084
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# The Unapologetic Mathematician
## Spanning sets
Today I just want to point out a dual proposition to the one I refined last week. At that time we stated that any linearly independent set can be expanded to a basis. This followed from the fact that a basis is a maximal linearly independent set. But a basis is also a minimal spanning set, and that will lead to our flipped result.
First, a basis must be a spanning set, and if we remove any vector it can no longer be a spanning set. For if the remaining vectors spanned, then the removed vector could be written as a finite linear combination of them, and this would contradict the linear independence of the basis.
On the other hand, if a spanning set $\left\{e_i\right\}$ is minimal it must be linearly independent. For if it were not, we could write out a nontrivial finite linear combination
$0=\sum r^ie_i$
For some index — say $i_0$ — we have $r^{i_0}\neq0$. Then we can rearrange this to write
$e_{i_0}=\frac{1}{r^{i_0}}\sum\limits_{i\neq i_0}r^ie_i$
and so we may omit the vector $e_{i_0}$ from the set and the remaining vectors still span. This contradicts the assumed minimality of the original spanning set.
So a basis is a minimal spanning set. This leads us to the proposition that any spanning set may be narrowed to a basis. And the proof is again the same technique we used to show that every vector space has a basis. It’s just that this time we flip the whole thing over. Now we consider the set of subsets of our spanning set which span the vector space, and we use Zorn’s lemma to show that this must contain a minimal spanning set. This will then be a basis contained in our original spanning set.
Of course, as usual, in the finite-dimensional case we don’t need Zorn’s lemma, so the squeamish can relax in that case.
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 5 Comments »
1. [...] Any vector in the image of can be written as for some vector . If we pick a basis latex V\$, then we can write . Thus the vectors span the image of . And thus they contain a basis for the image. [...]
Pingback by | July 1, 2008 | Reply
2. I am afraid, we cannot “flip the whole thing over” and use Zorn’s lemma for the spanning subsets, since we cannot assert that every chain is bounded from below. Indeed, the chain of the sets {n,n+1,…} (n=1,2,…) consists of spanning subsets of R, but does not posess a lower bound (its intersection is empty).
Comment by Alexander E. Gutman | September 11, 2009 | Reply
3. That seems to be right. Luckily, the finite-dimensional case still holds fine, and that’s all that we need for most of the sequel.
Comment by | September 11, 2009 | Reply
4. [...] let’s say is some finite collection of vectors which span (it doesn’t matter if they’re linearly independent or not). Let be a linear [...]
Pingback by | January 19, 2010 | Reply
5. [...] so we don’t really want zero to be in our collection . We also may as well assume that spans , because it certainly spans some subspace of and anything that happens off of this subspace is [...]
Pingback by | January 20, 2010 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://www.physicsforums.com/showthread.php?t=614724
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Physics Forums
Law of Supply vs Law of Remuneration
If prices increase, supply will increase. This is the Law of Supply. So if your hourly wage were suddenly increased, then according to this law, you'd be willing to work more hours because each of those hours suddenly offers you more income. Because the higher wage has made working more valuable, you decide to work more hours.
But consider this: if your hourly wage were increased, you are now able to make the same amount of money working fewer hours. You exploit this to your advantage, and decide to work fewer hours, spending the rest on leisure, since you still make the same total salary as before. This is the Law of Remuneration.
The two laws are both logical yet they contradict one another, thus they cannot be both right at the same time. Which law is correct and why?
I've been studying this for a while and it turns out that both laws are right, but at different times.
But why? What is the point at which the laws neutralize each other?
BiP
PhysOrg.com social sciences news on PhysOrg.com >> Evolution of lying>> Exhaustive computer research project shows shift in English language>> US white majority to linger if immigration slows
Mentor
Blog Entries: 4
Quote by Bipolarity If prices increase, supply will increase. This is the Law of Supply. So if your hourly wage were suddenly increased, then according to this law, you'd be willing to work more hours because each of those hours suddenly offers you more income. Because the higher wage has made working more valuable, you decide to work more hours. But consider this: if your hourly wage were increased, you are now able to make the same amount of money working fewer hours. You exploit this to your advantage, and decide to work fewer hours, spending the rest on leisure, since you still make the same total salary as before. This is the Law of Remuneration. The two laws are both logical yet they contradict one another, thus they cannot be both right at the same time. Which law is correct and why? I've been studying this for a while and it turns out that both laws are right, but at different times. But why? What is the point at which the laws neutralize each other? BiP
There is no such thing as a "law of supply". It's "supply and demand". Reference to a "law of supply" is BS.
Even this website contradicts itself although it states
(Economists do not really have a “law” of supply, though they talk and write as though they do.)
http://www.econlib.org/library/Enc/Supply.html
Sigh, so much garbage on the internet.
Quote by Evo There is no such thing as a "law of supply". It's "supply and demand". Reference to a "law of supply" is BS. Even this website contradicts itself although it states http://www.econlib.org/library/Enc/Supply.html Sigh, so much garbage on the internet.
OK then mathematically why is it that
$\frac{dS}{dY}$ change sign at some value of Y where Y is the wage rate and S is the number of hours worked?
BiP
Mentor
Blog Entries: 4
Law of Supply vs Law of Remuneration
Quote by Bipolarity OK then mathematically why is it that $\frac{dS}{dY}$ change sign at some value of Y where Y is the wage rate and S is the number of hours worked? BiP
The number of hours worked is dictated by the employer, not the employee. There is no such thing as a "Law of remuneration". Please post a link to a reputable source.
Quote by Evo The number of hours worked is dictated by the employer, not the employee. There is no such thing as a "Law of remuneration". Please post a link to a reputable source.
I do not have links sorry as I use a textbook. My text uses the term to describe the situation in labor economics where \frac{dS}{dY} is negative. It is "Introductory Economics" by the British author G.F. Stanlake. Chapter 3 of the fifth edition describes this law. In any case my question is not so much about the terminology but rather I am looking for an explanation of why the sign of \frac{dS}{dY} changes at a critical value of Y.
It has something to do with utility maximization but I do not understand why. For instance, if you increase the hourly wage, then the marginal utility of working can only increase, thus hours worked can only be expected to increase. I see no reason for the marginal utility of working to decrease as the wage rate goes up.
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Quote by Bipolarity I do not have links sorry as I use a textbook. My text uses the term to describe the situation in labor economics where \frac{dS}{dY} is negative. It is "Introductory Economics" by the British author G.F. Stanlake. Chapter 3 of the fifth edition describes this law. In any case my question is not so much about the terminology but rather I am looking for an explanation of why the sign of \frac{dS}{dY} changes at a critical value of Y. It has something to do with utility maximization but I do not understand why. For instance, if you increase the hourly wage, then the marginal utility of working can only increase, thus hours worked can only be expected to increase. I see no reason for the marginal utility of working to decrease as the wage rate goes up. BiP
This is absolute nonsense in the real world, it just doesn't work that way. You do realize that most professionals are salaried and don't get paid by the hour? If you are an hourly worker, then your employer decides how many hours you work.
I don't find anything anywhere that backs up what you say, so perhaps you have misunderstood what the author is saying.
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Quote by Bipolarity If prices increase, supply will increase. This is the Law of Supply. So if your hourly wage were suddenly increased, then according to this law, you'd be willing to work more hours because each of those hours suddenly offers you more income. Because the higher wage has made working more valuable, you decide to work more hours. But consider this: if your hourly wage were increased, you are now able to make the same amount of money working fewer hours. You exploit this to your advantage, and decide to work fewer hours, spending the rest on leisure, since you still make the same total salary as before. This is the Law of Remuneration. The two laws are both logical yet they contradict one another, thus they cannot be both right at the same time. Which law is correct and why? I've been studying this for a while and it turns out that both laws are right, but at different times. But why? What is the point at which the laws neutralize each other? BiP
If changes are gradual, would you ever get to the top part of the curve? At the far right point of the curve the marginal cost of labour is infinite. Production stops when marginal cost equals marginal revenue. The second point I think is what Evo is alluding to. A person’s hourly rate may depend on the amount of hours they work. A company may not be willing to pay a part time employee as much because the company may want to fully utilize office resources. For instance, say office space was short. The company would have to consider how practical it was for part time employees to share a desk. Your scenario is plausible I think if someone’s skills are very hard for the company to come by. In that case then maybe they would be more willing to hire them on a part time basis.
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Quote by Bipolarity I see no reason for the marginal utility of working to decrease as the wage rate goes up. BiP
In most cases I think it is not the work that brings the utility but what a person gains from the work. For instance: wages, purpose, social relations, etc. When calculating marginal utility don’t use hours worked as your dependent variable.
Quote by Bipolarity If prices increase, supply will increase. This is the Law of Supply. So if your hourly wage were suddenly increased, then according to this law, you'd be willing to work more hours because each of those hours suddenly offers you more income. Because the higher wage has made working more valuable, you decide to work more hours. But consider this: if your hourly wage were increased, you are now able to make the same amount of money working fewer hours. You exploit this to your advantage, and decide to work fewer hours, spending the rest on leisure, since you still make the same total salary as before. This is the Law of Remuneration. The two laws are both logical yet they contradict one another, thus they cannot be both right at the same time. Which law is correct and why? I've been studying this for a while and it turns out that both laws are right, but at different times. But why? What is the point at which the laws neutralize each other? BiP
They aren't laws but income and substitution effect. Check http://www.youtube.com/watch?v=R-VCNJKsvwg
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http://www.physicsforums.com/showthread.php?t=457426
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Physics Forums
## Unit tangent and normal vectors
1. The problem statement, all variables and given/known data
r(t)=ti+t^2j
Find the velocity, speed, acceleration, unit tangent, and unit normal vectors.
2. Relevant equations
Velocity=r'(t)
Speed=magnitude of r'(t)
Acceleration=r''(t)
Unit tangent=r'(t)/magnitude of r'(t)
Unit normal=d/dt[unit tangent]/magnitude of d/dt[unit tangent]
3. The attempt at a solution
Velocity=i+2tj
Speed=$\sqrt{1^2+(2t)^2} = \sqrt{1+4t^2}$
Acceleration=2j
Unit tangent=$$\frac{i+2tj}{\sqrt{1+4t^2}}$$
I'm pretty sure that's all right so far. I get mixed up in the algebra at the unit normal.
For d/dt[unit tangent] I have $$\frac{2j\sqrt{1+4t^2}-\frac{1}{2}(1+4t^2)^{-1/2}(8t)}{1+4t^2}$$.
Is that correct? How do I take the magnitude of that mess? I can't really see a way to simplify it.
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your unit normal still has to have a direction doesn't it? if you find the i and j components, then you can find the magnitude the same way you found the magnitude of r'(t)
That's the problem, I have no idea how to simplify it to something where I can square the components, add them, and take the square root.
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## Unit tangent and normal vectors
hi cdotter!
Quote by cdotter Unit normal=d/dt[unit tangent]/magnitude of d/dt[unit tangent]
nooo!
the unit normal is simply the unit vector perpendicular to the unit tangent!
Quote by tiny-tim hi cdotter! nooo! the unit normal is simply the unit vector perpendicular to the unit tangent!
Maybe I called it the wrong thing...the principal unit normal vector?
http://www.ltcconline.net/greenl/cou...ns/tannorm.htm
I just found this trick to simplify it so maybe that will work:
"Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor"
Quote by cdotter Maybe I called it the wrong thing...the principal unit normal vector? http://www.ltcconline.net/greenl/cou...ns/tannorm.htm I just found this trick to simplify it so maybe that will work: "Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor"
haha funny thing is I was literally just reading that link. I wasn't going to send it to you because it is the exact question that you are given.
Quote by dacruick haha funny thing is I was literally just reading that link. I wasn't going to send it to you because it is the exact question that you are given.
I didn't realize the exact same problem was there until you said it, I just saw the trick/technique.
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Quote by cdotter Maybe I called it the wrong thing...the principal unit normal vector?
ah!
as you can see, my definition also works (and is much quicker) …
the only difference is that my method comes up with two unit normal vectors (oppsotie each other),
the one you need is "is the unique vector that points into the curve", ie the one towards the centre of curvature (the concave side)
for a 3D curve, you do need the book's method, to decide which one is the principal vector
but for a 2D curve like this, just choosing the concave side is enough!
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Introductory Physics Homework 12
Calculus & Beyond Homework 4
Advanced Physics Homework 8
Calculus & Beyond Homework 5
Introductory Physics Homework 2
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http://mathhelpforum.com/differential-equations/94470-linear-dependence.html
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# Thread:
1. ## Linear dependence
Hey
Are sin(x)cos(x) and sin(2x) linearly dependent or independent on the interval (0,1)? I say dependent since sin(2x) = 2sin(x)cos(x) which is simply twice the first function, but according to my book it's independent. Who's wrong?
Thanks
2. Originally Posted by Pulsar06
Hey
Are sin(x)cos(x) and sin(2x) linearly dependent or independent on the interval (0,1)? I say dependent since sin(2x) = 2sin(x)cos(x) which is simply twice the first function, but according to my book it's independent. Who's wrong?
Thanks
It looks like you're right.
There are two ways to do this:
1) Form a linear combination and find $\alpha,\beta$ such that $\alpha\sin x\cos x+\beta \sin\left(2x\right)=0$. If $\alpha,\beta$ are not both zero, then its dependent; otherwise, its independent. In our case, if $\alpha=-2$ and $\beta=1$, then we have shown that the two functions are linearly dependent.
2) Use the Wronksian. If $\exists\,x\in\left(0,1\right):W=0$, then it's dependent.
So our Wronskian is $W=\begin{vmatrix}\tfrac{1}{2}\sin\left(2x\right) & \sin\left(2x\right)\\ \cos\left(2x\right) & 2\cos \left(2x\right)\end{vmatrix}=\sin\left(2x\right)\c os\left(2x\right)-\cos\left(2x\right)\sin\left(2x\right)=0$. Thus its dependent $\forall\,x\in\left(0,1\right)$
3. Thanks. I had trouble believing it was an error, since this is the fifth edition of the book, but I suppose it could be a new problem.
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http://www.physicsforums.com/showthread.php?t=187054
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Physics Forums
## Relativistic Energy
1. The problem statement, all variables and given/known data
An Omega- particle has rest energy 1672 MeV and mean lifetime 8.2X10-11 s. It is created and decays in a particle track detector and leaves a track 24mm long. What is the total energy of the Omega- particle?
2. Relevant equations
E=E0/Sqrt(1-v^2/c^2)
E0=1672 MeV
Lorrentz Equations
3. The attempt at a solution
I took the distance and divided by time and found the V would be .967c. Once that was used to then find E. The answer was off. The correct answer was 2330 MeV. I don't know how they got that. My answer was 7678 MeV.
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Quote by lylos 1. The problem statement, all variables and given/known data An Omega- particle has rest energy 1672 MeV and mean lifetime 8.2X10-11 s. It is created and decays in a particle track detector and leaves a track 24mm long. What is the total energy of the Omega- particle? 2. Relevant equations E=E0/Sqrt(1-v^2/c^2) E0=1672 MeV Lorrentz Equations 3. The attempt at a solution I took the distance and divided by time and found the V would be .967c. Once that was used to then find E. The answer was off. The correct answer was 2330 MeV. I don't know how they got that. My answer was 7678 MeV.
Did you take into account that a moving particle lives longer due to Einstein's time dilation?
Eugene.
## Relativistic Energy
Quote by meopemuk Did you take into account that a moving particle lives longer due to Einstein's time dilation? Eugene.
I thought that to be the case. Considering the life of the particle is given as 8.2X10^-11 s, I figured that would be the time the particle sees. The 24mm would be the distance it travels that the observer sees. Now connecting these two is where I had issues. Should I assume that the 8.2x10^-11 s is the life of the particle that we observe?
Quote by lylos I thought that to be the case. Considering the life of the particle is given as 8.2X10^-11 s, I figured that would be the time the particle sees. The 24mm would be the distance it travels that the observer sees. Now connecting these two is where I had issues. Should I assume that the 8.2x10^-11 s is the life of the particle that we observe?
Normally, "mean lifetime" is supposed to be in the particle's rest frame. You were asked to find particle's energy with respect to the observer. The track length is also measured in the observer's frame. Therefore, you should also use the (dilated) lifetime in the same (observer's) frame of reference.
Eugene.
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Quote by lylos I thought that to be the case. Considering the life of the particle is given as 8.2X10^-11 s, I figured that would be the time the particle sees. The 24mm would be the distance it travels that the observer sees. Now connecting these two is where I had issues. Should I assume that the 8.2x10^-11 s is the life of the particle that we observe?
Something doesn't seem right about one of the numbers in your problem statement. The light-travel time for 24 mm is 8.01x10^-11 sec. If the particle appears time-dilated in the lab frame, shouldn't the light-travel time for its track length be longer than the rest lifetime of the particle? [The lifetime you quote for the omega-minus appears correct; are you sure about the track length?]
Quote by dynamicsolo Something doesn't seem right about one of the numbers in your problem statement. The light-travel time for 24 mm is 8.01x10^-11 sec. If the particle appears time-dilated in the lab frame, shouldn't the light-travel time for its track length be longer than the rest lifetime of the particle? [The lifetime you quote for the omega-minus appears correct; are you sure about the track length?]
I don't see any contradiction here. You, basically, need to solve a system of two equations. Denoting d=24mm the traveled distance, $t_0$= 8.01x10^-11 sec the lifetime at rest, and
$$t = t_0 (1 - v^2/c^2)^{-1/2}$$
the lifetime of the moving particle, we can find the velocity from equation
$$d = vt$$
This velocity should be lower than the speed of light, of course.
Eugene.
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Quote by meopemuk I don't see any contradiction here. You, basically, need to solve a system of two equations. Denoting d=24mm the traveled distance, $t_0$= 8.01x10^-11 sec the lifetime at rest, and $$t = t_0 (1 - v^2/c^2)^{-1/2}$$ the lifetime of the moving particle, we can find the velocity from equation $$d = vt$$ This velocity should be lower than the speed of light, of course. Eugene.
Isn't $$t_0$$ the rest lifetime of the particle? It's the observed time in the lab-frame that should be longer ($$\gamma$$ > 1). The difficulty is that, in the expression $$t = t_0 (1 - v^2/c^2)^{-1/2}$$, it is $$t_0$$ that equals 82 picoseconds, while the lab-frame time is about 80 picoseconds.
The muon illustration is representative. Its rest lifetime is 2.2 microseconds; if it were to travel at the speed of light, it would decay in a distance of 0.66 km. The fact that muons from high-altitude cosmic ray showers are detected on the ground is taken to indicate the effect of time dilation, in which $$\gamma$$ is in the range of at least 50 to 100, giving it an Earth-frame lifetime sufficient to travel some tens of kilometers, in accordance with d ~ ct. In the rest frame of the muon, it is the distance from its point of creation to the ground that appears contracted by a factor of at least 50 to 100, so it only needs to cover a fraction of a kilometer to the apparently approaching ground, again in accordance with d ~ ct.
If the omega were observed in the lab frame to have a lifetime of 82 picoseconds and covers 24 mm in that time, then its lab-frame velocity would be 0.9763 c, giving
$$\gamma$$ = 4.617 . But isn't it the moving omega which has the "dilated" lifetime? We would then find a rest lifetime of 82/4.617 = 17.8 picoseconds. If it's the rest lifetime that is "dilated", we should see a lifetime of (4.617)·82 = 379 picoseconds for the lab-frame lifetime and the track would be
(379 picoseconds)(0.9763)(2.998x10^10 cm/sec) = 11.1 cm.
I think there is something in the problem statement that is not right. (The lifetime issue would be fine if this were the charmed-omega, but the rest energy is wrong.)
(After some further thought...) I finally decided to break down and "reverse engineer" the given answer. When you find gamma from the ratio total energy/rest energy, solve for v, time-dilate the rest lifetime, and now calculate the track length using
d = v · ( $$\gamma$$$$t_0$$ ) , you get 23.87 mm.
SO, the issue appears to be that the precision of the track length is not given at a sufficiently high level to resolve the needed values accurately. The observed track length is too close to d = c$$t_0$$ to get a clear result at low precision.
To sum up, $$t_0$$ in the dilation equation is the rest lifetime and the method meopemuk describes is otherwise correct in principle. However, the measurement given for the track length needs to be given to at least three significant figures (four would be better) in order to find v with enough precision to solve the problem.
Hi dynamicsolo, If I solve my two equations (with two unknowns v and t) with respect to v, I obtain $$v = \frac{d}{\sqrt{t_0^2 + d^2/c^2}}$$ Upon substitution of numerical values, I got $v \approx 2 \cdot 10^{8}$ m/s which looks reasonable. Eugene.
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Quote by meopemuk Hi dynamicsolo, If I solve my two equations (with two unknowns v and t) with respect to v, I obtain $$v = \frac{d}{t_0^2 + d^2/c^2}$$ Upon substitution of numerical values, I got $v \approx 2 \cdot 10^{8}$ m/s which looks reasonable. Eugene.
... presumably with a square root in the denominator. I get 0.6967 c for the velocity from the analysis I made, so I agree with your velocity (which I get, with the formula correction indicated).
I think I realize the other mistake that was being made from the beginning. The track length in the lab frame is d = $$\beta\gamma$$·c·$$t_0$$. [This is the observed velocity in the lab frame, u = (beta)·c , times the observed lifetime in the lab frame, which is the time-dilated lifetime (gamma)·t0.] This gives
$$\beta\gamma$$ = 2.4 cm / (82 picoseconds)(2.998x10^10 cm/sec)
= 2.4 / 2.458 = 0.9763 .
So, lylos, the 0.9763 we saw back at the start of this thread is not (beta) = v/c , but $$\beta\gamma$$ , which must be solved for v. (I think particle physicists have a name for this quantity, but I forget what that is.)
I still hold to my earlier posts. The track length must be given to enough precision to determine $$\beta\gamma$$ accurately.
Quote by dynamicsolo ... presumably with a square root in the denominator.
Yes, of course. I corrected my post. Thanks.
Eugene.
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http://math.stackexchange.com/questions/159493/can-there-be-a-function-thats-even-and-odd-at-the-same-time/159497
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# Can there be a function that's even and odd at the same time?
I woke up this morning and had this question in mind. Just curious if such function can exist.
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– Rory O'Kane Jun 17 '12 at 18:54
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You might find it interesting that I often used to ask this as an extra credit question on precalculus tests when even/odd function properties were covered, typically worth an extra 3 points on a 100 point scale (so a score of 103/100 was possible). I'd usually get about 2 to 5 students getting the extra points (out of a total of maybe 25-35 students) in a U.S. college precalculus class, and about half the class getting the extra points in U.S. honors level high school classes I used to teach. – Dave L. Renfro Jun 18 '12 at 15:56
## 6 Answers
Others have mentioned that $f(x)=0$ is an example. In fact, we can prove that it is the only example of a function from $\mathbb{R}\to \mathbb{R}$ (i.e a function which takes in real values and outputs real values) that is both odd and even. Suppose $f(x)$ is any function which is both odd and even. Then $f(-x) = -f(x)$ by odd-ness, and $f(-x)=f(x)$ by even-ness. Thus $-f(x) = f(x)$, so $f(x)=0.$
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Of course, one could argue that restrictions of the constant $0$ function to different domains symmetric about the origin are different functions, set-theoretically speaking. – Cameron Buie Jun 17 '12 at 15:30
@CameronBuie That is true, I will make my answer more precise to indicate this. Thank you. – Ragib Zaman Jun 17 '12 at 15:31
Funny, I never thought of f(x) = 0 as a possibility. Thanks for the answers everyone! – bodacydo Jun 17 '12 at 21:06
If $K$ is a field of characteristic 2, every function $K\to K$ is both even and odd.
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i'm sorry, wouldn't that be "unequal to 2"? – akkkk Jun 17 '12 at 15:31
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@Auke: No. I won't spoil the joke by spelling it out, sorry. – Harald Hanche-Olsen Jun 17 '12 at 15:40
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Actually, you don't even need a field, any ring of characteristic 2 will do. – Ilmari Karonen Jun 17 '12 at 15:42
@HaraldHanche-Olsen, oh, I am sorry, I misread your answer, you are completely right :) nice one – akkkk Jun 17 '12 at 15:43
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This is a wonderful answer! – Edward Hughes Jun 17 '12 at 23:52
show 4 more comments
Yes. The constant function $f(x) = 0$ satisfies both conditions.
Even: $$f(-x) = 0 = f(x)$$
Odd: $$f(-x) = 0 = -f(x)$$
Furthermore, it's the only real function that satisfies both conditions:
$$f(-x) = f(x) = -f(x) \Rightarrow 2f(x) = 0 \Rightarrow f(x) = 0$$
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Hint $\rm\ f\:$ is even and odd $\rm\iff f(x) = f(-x) = -f(x)\:\Rightarrow\: 2\,f(x) = 0.\:$ This is true if $\rm\:f = 0,\:$ but may also have other solutions, e.g. $\rm\:f = n\:$ in $\rm\:\mathbb Z/2n =\:$ integers mod $\rm 2n,$ where $\rm\: -n \equiv n.$
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+1, but note that your last $\iff$ applies (in the backwards, i.e. 'if' direction) only to $f(x) = -f(x)$, and not to the part where $f(-x)$ equals both of them. – ShreevatsaR Jun 17 '12 at 18:04
Yes, I meant to write $\:\Rightarrow\:$ but it was lost in editing. Now fixed. Thanks. – Gone Jun 17 '12 at 18:19
Suppose $f$ odd an even. Let $x \in D$ ( D is set definition of $f$) then you have : $f(x)=f(x)=-f(x)$. What can you conclude about $f$ ?
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As other people have mentioned already, the real function $f(x)$ which maps every real number to zero (i.e.$f(x) = 0 \space \forall x \in \mathbb{R}$) is both even and odd because $$f(x) - f(-x) = 0 \space \space , f(x)+f(-x) = 0\space \forall x \in \mathbb{R} .$$ Also it is the only function defined over $\mathbb{R}$ to possess this property.
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http://unapologetic.wordpress.com/2007/04/06/tensor-products-of-abelian-groups/?like=1&_wpnonce=34114a6dfa
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# The Unapologetic Mathematician
## Tensor products of abelian groups
Often enough we’re going to see the following situation. There are three abelian groups — $A$, $B$, and $C$ — and a function $f:A\times B\rightarrow C$ that is linear in each variable. What does this mean?
Well, “linear” just means “preserves addition” as in a homomorphism, but I don’t mean that $f$ is a homomorphism from $A\times B$ to $C$. That would mean the following equation held:
$f(a+a',b+b')=f(a,b)+f(a',b')$
Instead, I want to say that if we fix one variable, the remaining function of the other variable is a homomorphism. That is
$f(a,b+b')=f(a,b)+f(a,b')$
$f(a+a',b)=f(a,b)+f(a',b)$
$f(a+a',b+b')=f(a,b)+f(a,b')+f(a',b)+f(a',b')$
we call such a function “bilinear”.
The tensor product is a construction we use to replace bilinear functions by linear functions. It is defined by yet another universal property: a tensor product of $A$ and $B$ is an abelian group $T$ and a bilinear function $t:A\times B\rightarrow T$ so that for every other bilinear function $f:A\times B\rightarrow C$ there is a unique linear function $\bar{f}:T\rightarrow C$ so that $f(a,b)=\bar{f}(t(a,b))$. Like all objects defined by universal properties, the tensor product is automatically unique up to isomorphism if it exists.
So here’s how to construct one. I claim that $T$ has a presentation by generators and relations as an abelian group. For generators take all elements of $A\times B$, and for relations take all the elements $(a+a',b)-(a,b)-(a',b)$ and $(a,b+b')-(a,b)-(a,b')$ for all $a$ and $a'$ in $A$ and $b$ and $b'$ in $B$.
By the properties of such presentations, any function of the generators — of $A\times B$ — defines a linear function on $T$ if and only if it satisfies the relations. That is, if we apply a function $f$ to each relation and get ${}0$ every time, then $f$ defines a unique function $\bar{f}$ on the presented group. So what does that look like here?
$f(a+a',b)-f(a,b)-f(a',b)=0$
$f(a,b+b')-f(a,b)-f(a,b')=0$
So a bilinear function $f$ gives rise to a linear function $\bar{f}$, just as we want.
Usually we’ll write the tensor product of $A$ and $B$ as $A\otimes B$, and the required bilinear function as $(a,b)\mapsto a\otimes b$.
Now, why am I throwing this out now? Because we’ve already seen one example. It’s a bit trivial now, but catching it while it’s small will help see the relation to other things later. The distributive law for a ring says that multiplication is a bilinear function of the underlying abelian group! That is, we can view a ring as an abelian group $R$ equipped with a linear map $m:R\otimes R\rightarrow R$ for multiplication.
## 24 Comments »
1. [...] on tensor products and direct sums We’ve defined the tensor product and the direct sum of two abelian groups. It turns out they interact very [...]
Pingback by | April 17, 2007 | Reply
2. [...] products of modules The notion of a tensor product also extends to modules, but the generalization is not quite as straightforward as it was for [...]
Pingback by | April 30, 2007 | Reply
3. The third condition in the definition of linearity follows from the first two… I think.
Comment by Oingo | December 8, 2007 | Reply
4. I think you mean the third condition in the definition of bilinearity. And I wasn’t asserting that it’s independent, but rather drawing the distinction between the previously defined condition of linearity and this condition of bilinearity.
Comment by | December 10, 2007 | Reply
5. [...] map defined by . Note carefully that this is linear as a function of the pair . It’s not bilinear — linear in each of and separately — which would mean bringing in the tensor [...]
Pingback by | July 24, 2008 | Reply
6. Hi, everyone:
This is my first time here. Great site.
I wonder what the general bilinear map t as above is ,
in the case where A=R^n and B=R^m , as Abelian groups,
i.e., what bilinear map would make the diagram:
R^n x R^m –t–> R^n (x) R^m
|
|f
\|/
etc. commute. Anyone know?
C
Comment by Bacile | March 10, 2009 | Reply
7. It’s pretty straightforward, Bacile. The universal bilinear map from $\mathbb{R}^n\times\mathbb{R}^m$ to $\mathbb{R}^n\otimes\mathbb{R}^m$ sends the pair $(u,v)$ to the tensor product $u\otimes v$.
Comment by | March 10, 2009 | Reply
• O.K., say we are tensoring R^n and R^m as V.Spaces
over R
Am I wrong to assume that once we choose specific
V.Spaces to tensor , that (x) has to be a specific
map?.
Let me give an example for n=m=2, with the only
bilinear map I can think of : inner-product in R^2 :
= xx’+yy’
This is a bilinear map from R^2xR^2 –>R
Now, we must find a linear map L: R^2(x)R^2 –>R
that makes the diagram commute, so that we must have:
L ( (x,y)(x)(x’,y’))= xx’+yy’ =
Now, in order for L to have this property, (x)
must be a specific map on (a,b)(x)(c,d) in R^2(x)R^2.
or do we just appeal to the (valid, I agree) argument
that shows the existence of this (x) making the diagram
commute.
I agree that (x) is an abstract universal bilinear map,
but I think that once we choose the V.Spaces to tensor,
(x) must become a specific map.
Maybe another way of looking at it is that we can
use the fact that R^2(x)R^2 ~ R^4 . Then it seems we
could use this isomorphism to pullback any map
R^2(x)R^2–>R into a map R^4–>R
Hope this is not too far out.
Anyway, thanks for any replies.
Comment by Bacile | March 11, 2009 | Reply
8. I just told you what specific map it was…
Okay, let’s try this. Say your vector space $V$ has a basis $\left\{e_i\right\}$ and $W$ has basis $\left\{f_j\right\}$. Then you have a basis for $V\otimes W$ given by $\left\{e_i\otimes f_j\right\}$. So the pair $(e_i,f_j)$ gets sent to $e_i\otimes f_j$, and we extend by bilinearity.
Say, for instance, we have the vectors $v=v^ie_i$ and $w=w^jf_j$. Then the pair $(v,w)$ gets sent to $(v^ie_i)\otimes(w^jf_j)=v^iw^j(e_j\otimes f_j)$. In your example, you’ve got the vectors $\begin{pmatrix}x\\y\end{pmatrix}$ and $\begin{pmatrix}x'\\y'\end{pmatrix}$. As a pair, they get sent to the linear combination
$\displaystyle xx'\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}+xy'\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}+yx'\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}+yy'\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}$
Now, we can choose an isomorphism $\mathbb{R}^2\otimes\mathbb{R}^2\cong\mathbb{R}^4$ with the correspondence
$\displaystyle\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}\leftrightarrow\begin{pmatrix}{1}\\{0}\\{0}\\{0}\end{pmatrix}$
$\displaystyle\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}\leftrightarrow\begin{pmatrix}{0}\\{1}\\{0}\\{0}\end{pmatrix}$
$\displaystyle\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}\leftrightarrow\begin{pmatrix}{0}\\{0}\\{1}\\{0}\end{pmatrix}$
$\displaystyle\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}\leftrightarrow\begin{pmatrix}{0}\\{0}\\{0}\\{1}\end{pmatrix}$
Then the pair of vectors above is send to the vector
$\displaystyle\begin{pmatrix}xx'\\xy'\\yx'\\yy'\end{pmatrix}$
Comment by | March 11, 2009 | Reply
9. [...] the space itself. Foremost among these is the idea of a bilinear form. This is really nothing but a bilinear function to the base field: . Of course, this means that it’s equivalent to a linear function [...]
Pingback by | April 14, 2009 | Reply
10. [...] scalars and , and for any index . Equivalently, by the defining universal property of tensor products, this is equivalent to a linear function — a linear functional on . That is, the space of [...]
Pingback by | October 22, 2009 | Reply
11. Hi everyone. I have a little problem… Let G be an abelian group and consider G\otimes_{\Z}\Q where \Q is the ring of rational numbers and \Z the one of integers. Take an element g\otimes q with g\in G and q\in \Q (it easily seen that any element of G\otimes_{\Z}\Q has this form).
Is it true that g\otimes q=0 if and ONLY IF g is a torsion element?
the if part is trivial but I have some truble with the other implication (in any case I’m quite sure it is true). Clearly you have to suppose q\neq 0.
Comment by Simone | October 28, 2009 | Reply
12. The if part isn’t even true. Let $q=1$ and even if $g$ is torsion the tensor isn’t zero.
Comment by | October 28, 2009 | Reply
• of course it is! let n be the order of g, then
g\otimes 1=g\otimes n/n=ng\otimes 1/n= 0 \otimes 1/n = 0
If, instead of 1, you put a generic q in the previous line you have the proof of the if part…
Comment by Simone | October 28, 2009 | Reply
• Furthermore it is not difficult to prove a more general statement:
Let G be a torsion abelian group and D a divisible abelian group, then G\otimes D=0.
Even this new statement can be generalized (it is true not only for abelian groups but for modules over a PID).
Comment by Simone | October 28, 2009 | Reply
13. Ah, yes, sorry. It’s been a long time since I’ve bothered with torsion elements.
Can you prove the contrapositive to handle the only-if part?
Comment by | October 28, 2009 | Reply
14. FACT 1
Let G be a torsion free group, then G embedds into G\otimes Q. (G\otimes Q is the divisible (i.e. injective) envelope of G)
FACT 2
Let G be an abelian group, H<G, then H\otimes\Q is a subvector space of G\otimes\Q.
Using this two facts it is possible to prove the result… in fact if x is a torsion free element of G, then x generates a group isomorphic to Z. Thanks to fact 1
—>\otimes Q= Q is injective and thanks to fact 2 we have the inclusion
—>\otimes Q—>G\otimes Q.
This proves that x\otimes 1 is non-zero.
I think this can work but in any case it uses deeper results than needed… I really think that must exist a more elementary proof…
Comment by Simone | October 28, 2009 | Reply
• I don’t know why but there is something missing in the message the maps were:
—> \otimes Q
and
—> \otimes Q—>G\otimes Q
Comment by Simone | October 28, 2009 | Reply
• ok… I cannot write that simbol…. denot with (x) the subgroup generated by x…
(x) —> (x)\otimes Q
and
(x)—>(x) \otimes Q—>G\otimes Q
Comment by Simone | October 28, 2009 | Reply
15. In HTML, text contained within angle brackets is interpreted as markup. You have to write < and > to get < and >
And I don’t really see what your problem is with this proof. It gets right to the heart of the matter, that the behavior of elements under the $\mathbb{Z}$-action is entirely a matter of the subgroup they generate. Tensoring with $\mathbb{Q}$ kills torsion subgroups and preserves non-torsion subgroups.
A really non-elementary proof would use something like a structure theorem for abelian groups, which this doesn’t come close to using.
Comment by | October 28, 2009 | Reply
• Here is the proof I was looking for! (I wrote it)
If \$g\$ is torsion, let \$n\$ be the order of \$g\$, then \$g\otimes q=ng\otimes q/n=0\$. On the other hand suppose, looking for a contradiction, that \$g\$ is not torsion, then \$\left\langle g\right\rangle\cong \Z\$ and so \$G\otimes_{\Z}\Q\supseteq\left\langle g\right\rangle\otimes_{\Z}\Q\cong \Q\neq 0\$. Let now \$ag\otimes p\$, with \$p\in \Q\$ and \$a\in\N\setminus\{0\}\$, be a non-trivial element of \$\left\langle g\right\rangle\otimes_{\Z}\Q\$ (we can suppose that it has this form), let \$p=\frac{p_1}{p_2}\$ and \$q=\frac{q_1}{q_2}\$ with \$p_1,p_2,q_1,q_2\in\Z\setminus \{0\}\$, then
\$\$ag\otimes p=ap_2q_1g\otimes \frac{1}{p_2q_2} \ \ \Rightarrow \ \ 0\neq ap_2q_1(g\otimes q)=a(g\otimes \frac{q_1}{q_2})=a(g\otimes q)\$\$
and so \$g\otimes q\$ is non-trivial contradicting our hypothesis. Then \$g\$ is torsion.
Comment by Simone | October 28, 2009 | Reply
16. Fact is not difficult to prove, it is essentially for definition of tensor product but fact 1 is less trivial. But now I see… Fact 1 is not really needed, in fact
< x > is isomorphic to Z whenever x is torsion free, then
< x > \otimes Q is isomorphic to Q because Z\otimes G=G for every abelian group G.
With this remark I like that proof:) thanks
Comment by Simone | October 28, 2009 | Reply
17. can tensor product of two abelian groups be group?
Comment by z | January 8, 2011 | Reply
• The tensor product of two abelian groups is another abelian group, so yes.
Comment by | January 8, 2011 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathoverflow.net/questions/27126?sort=oldest
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## Interpreting the Famous Five equation
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
$$e^{\pi i} + 1 = 0$$
I have been searching for a convincing interpretation of this. I understand how it comes about but what is it that it is telling us?
Best that I can figure out is that it just emphasizes that the various definitions mathematicians have provided for non-intuitive operations (complex exponentiation, concept of radians etc.) have been particularly inspired. Is that all that is behind the slickness of the Famous Five equation?
Any pointers?
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8
$\exp:\mathfrak{lie}(S^1) \rightarrow S^1 \hookrightarrow \mathbb{C}$ is a pretty natural map. Anyway, I think that this is off-topic. Voting to close. – Steve Huntsman Jun 5 2010 at 3:01
15
I regard this identity as the definition of pi. – Qiaochu Yuan Jun 5 2010 at 3:40
20
Voting to close? That's unbelievable... – J. H. S. Jun 5 2010 at 5:13
5
@Qiaochu Yuan: I laughed out loud when I saw your comment, and then spent the next five minutes trying to figure out whether you were serious. And failed. – Vectornaut Jun 7 2010 at 18:39
5
By the way, where does the term "famous five" come from? Mathematicians don't call it that. Does it come from some popularization? – Gerald Edgar Jun 18 2010 at 19:01
show 4 more comments
## 5 Answers
$$e^{i\pi}=\lim\limits_{N\to\infty}\left(1 + \frac{iπ}{N}\right)^N$$
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19
This may be the best answer I've seen on all of Math Overflow. – Kevin O'Bryant Jun 5 2010 at 3:15
4
Thanks, Kevin, that's very kind of you. The credit really goes to Wikipedia :) – Andrey Rekalo Jun 5 2010 at 4:00
1
Specifically, here: en.wikipedia.org/wiki/Euler%27s_identity – Dan Ramras Jun 5 2010 at 4:18
1
Yes. And the 3D vizualization of Euler's formula is also instructive en.wikipedia.org/wiki/… – Andrey Rekalo Jun 5 2010 at 4:26
11
Really nice. Reminds me of a cute argument that appears in the superb book on Complex Analysis by Tristan Needham. As Professor Needham says, an individual putting forward an answer of the type "well that's just a definition" is giving a low blow to one of Euler's greatest contributions to Math. – J. H. S. Jun 5 2010 at 5:11
show 1 more comment
### 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'd like to add something to the visual answer above (or below, or wherever it ends up). It was not until I was into my 40s that I realized that there was an intuitive way of understanding that $e^{i\pi}=-1$, as opposed to the power-series derivation that seems a bit too formal somehow. (What I'm about to say comes into the category of thing that I happened to think of for myself but it's so natural that obviously many other people will have had the same thought and I absolutely do not wish to make some ludicrous priority claim. I haven't read Needham's book, but I dare say he has the same argument.)
Why should it be that the limit of $(1+i\pi/N)^N$ is equal to -1? To answer this, let's think about what multiplication by $1+i\pi/N$ does. Well, $1+i\pi/N$ has modulus very close indeed to 1 (by which I mean that it has modulus $1+O(N^{-2})$), and argument very close indeed to $\pi/N$. Therefore, multiplication by $1+i\pi/N$ is approximately rotation by $\pi/N$. So if you do it N times, then the result is approximately rotation by $\pi$, which is multiplication by -1. The approximations are good enough that one can make this argument rigorous fairly easily.
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12
I really do like these two answers, but I cannot help thinking that they should be accompanied by an equally intuitive explanation of why `$\lim_{n\to\infty}(1+x/n)^n=e^x$`. Which, in turn, boils down to a question of understanding the exponential function in the first place. I am not saying this is hard to achieve, but it is something I rarely see done well. – Harald Hanche-Olsen Jun 5 2010 at 13:35
8
I agree. My answer to that is that it's fairly easy to prove that the function f(x) = lim (1 + x/n)^n has the property f(x+y) = f(x)f(y). – gowers Jun 5 2010 at 15:17
1
One may also interpret the limit as convergence of solutions to Euler's numerical scheme $$\frac{\phi_n-\phi_{n-1}}{\pi/n}=i\phi_{n-1},\quad \phi_0=1,$$ to a solution of the ODE $$\dot{\phi}=i\phi,\ t\in(0,\pi].$$ – Andrey Rekalo Jun 5 2010 at 16:00
1
The explanation given by Conway and Guy in The Book of Numbers is also in the spirit of these two answers. – Todd Trimble Jun 5 2010 at 16:38
The answers so far give interpretations of the exponential as a limit of discrete approximations. An alternative interpretation is that any continuous map that takes addition to multiplication on the complex line and takes reals to reals has a purely imaginary kernel isomorphic to the integers. The constant $e$ arises from a normalization for which unit speed paths on the imaginary axis are taken to unit speed paths on the unit circle, and $\pi$ then shows up as a path length. One way to emphasize the additive-multiplicative relationship is to expand the formula as: $e^{\pi i-0i} = -1/1$.
Here is a more formal treatment: Both $(\mathbb{C}^\times, \times)$ and $(\mathbb{C}, +)$ are one dimensional analytic groups, and the latter is simply connected, so there is a universal covering homomorphism $\exp: \mathbb{C} \to \mathbb{C}^\times$ from the additive group to the the multiplicative group. The homomorphism is unique once we choose a normalization, e.g., by demanding that it is analytic and satisfies the differential equation $\partial_z \exp = \exp$. The differential equation can be related to the homomorphism, after choosing coordinates, by considering the respective formal group laws, or just reasoning heuristically with infinitesimals.
Claim 1: The function $\exp$ takes purely imaginary numbers to elements of unit norm.
Proof: The function $\exp$ takes additive inverses to reciprocals (because it is a homomorphism), and complex conjugates to complex conjugates (because it is defined over the reals). By composing, we find that reflection in the imaginary axis is taken to unit circle inversion, and fixed points are taken to fixed points.
Remarks: Note that the only part of the normalization we used here was the fact that $\partial_z \exp$ is a real multiple of $\exp$. The "defined over the reals" bit may be unsatisfying to some, but the conjugation behavior can be verified directly by expanding as a power series that converges everywhere, and checking that the coefficients are real. One can also prove the claim by more direct methods, such as applying the above differential equation to grind out the identity $\frac{\partial}{\partial y} | \exp(iy) |^2 = 0$.
Claim 2: $\exp(\pi i) = -1$.
Proof: By combining the previous claim concerning unit norms with the differential equation $\partial_z \exp = \exp$, we conclude that $\exp$ takes any unit speed path on the imaginary axis to a unit speed path on the unit circle. We have $\exp(0) = 1$ by the homomorphism assumption, and the length of a minimal path from $1$ to $-1$ on the unit circle is $\pi$.
Remarks: Depending on how $z \mapsto e^z$ is defined, one may still have to check that it agrees with $\exp$, but this isn't a big deal. I tried to avoid choosing square roots of minus one as much as possible, but the statement of the identity makes it a bit difficult to maintain such discipline.
-
Just wanted to include this excellent illustration of Euler's formula
(it really deserves to be shown here in its own right and not just as a link in one of the comments):
Source: http://en.wikipedia.org/wiki/Euler_formula
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Consider the exponential map for the Lie group $U(1)$.
The formula $e^{\pi i}$ means: starting from the identity element $1$, keep traveling "in the direction of $i$" (Which is "going up" at first, because the imaginary axis is of course going up. But this direction "changes" on the $\mathbb{R}^2$-plane as you move! So eventually you will be moving among the circumstance of the unit circle), and walk a distance of $\pi$.
Where do you end up with? You end up traveling half a circle (distance = $\pi$) and reach $-1$. Therefore $e^{\pi i} = -1$.
BTW, I think it is not a very good idea to simply regard the formula as the definition of $\pi$ and consider it trivial. Because then one have to explain why this $\pi$ is the same $\pi$ in the usual definition using the circumstance / area of a circle.
So, the real question is not about $\pi$, but about why something seemingly "only related to a circle" will have anything to do with an algebraic expression involving $e$ and $i$.
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http://math.stackexchange.com/questions/292984/show-lbrace-phix-n-rbrace-is-a-submartingale-if-lbrace-x-n-rbrace-is
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# Show $\lbrace\phi(X_{n})\rbrace$ is a submartingale if $\lbrace X_{n}\rbrace$ is
If $\lbrace X_{n}\rbrace$ is a submartingale with respect to filtration $\lbrace \mathcal{F}_{n}\rbrace$, and $\phi:\mathbb{R}\to\mathbb{R}$ is convex and increasing, then $\lbrace\phi(X_{n})\rbrace$ is a submartingale, too.
1. By convexity of $\phi$ and the fact that $X_{n}$ is $\mathcal{F}_{n}$-measurable, $\phi(X_{n})$ is $\mathcal{F}_{n}$-measurable, thus the new process is adapted to $\lbrace \mathcal{F}_{n}\rbrace$.
2. To verify that $\mathbb{E}(\phi(X_{n+1})|\mathcal{F}_{n})\geq\phi(X_{n})$, Jensen's inequality can be applied.
3. Now my question: How to show that $\mathbb{E}(\phi(X_{n})^{+})<\infty$?
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You need integrability as another condition. Otherwise consider some random variable with zero mean, but infinite third moment, and $\phi(x) = x^2 1_{x \geq 0}$. – Thomas Feb 4 at 13:45
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http://mathhelpforum.com/pre-calculus/59385-composition-functions.html
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# Thread:
1. ## Composition & Functions
hey all, need help with a problem
Question: Use composition to show that f(x) = (x + 3) / (x - 2) and g(x) = (2x - 3) / (x - 1) are inverse of each other.
Any help would be greatly appreciated, thanks
-Laconian
2. Originally Posted by Laconian
Question: Use composition to show that f(x) = (x + 3) / (x - 2) and g(x) = (2x - 3) / (x - 1) are inverse of each other.
You are being asked to verify that f(g(x)) = x and that g(f(x)) = x. For the first of those, you need to see whether $\frac{\frac{2x-3}{x-1} +3} {\frac{2x-3}{x-1} - 2} = x$. But before you spend any time trying to simplify that compound fraction, I suggest that you look at the question again, because as it stands, there's a mistake. Those two functions are not inverses of each other. (In fact the 3s in the two numerators ought to have the same sign, either both + or both -.)
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http://www.physicsforums.com/showthread.php?p=4277442
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Physics Forums
## Thin wall Pressure Vessel(cylinder)
cylindrical pressure vessel with inner pressure P.
Take a infinitesimal cube on the inner-wall of the vessel. When calculating the normal stress along the radial direction it is assumed that that it is equal to P on the both the negative $\widehat{r}$ and positive $\widehat{r}$ directions.
For an outer wall infinitesimal cube it is assumed that the normal stresses in radial directions are zero.
I was told in class that the normal stress in the radial direction will decrease linearly from P to 0 along the thin wall.
My question is this: If the pressure is slowly decreasing in radial direction shouldn't there be a shear force to compensate for the fact that that normal stress is decreasing. So for the infinitesimal cube on the inner wall I picture the face in negative $\widehat{r}$ direction to have Pressure P and the face on the positive $\widehat{r}$ direction to have pressure P-dP. This change in pressure should cause a shearing force on the 4 faces normal to the $\widehat{r}$ direction in the negative $\widehat{r}$ direction.
Why is this shear force neglected. If it is because it is so small compared to the pressure stresses so it is neglected then can someone explain to me why it is small. It can't reason why this force would be of such small magnitude when the shear forces must compensate for this Pressure over the small distance of the thin wall.
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
Recognitions: Gold Member It is not neglected in the thick wall solution, so look up the thick wall solution and see how it reduces to the thin wall solution when the ratio of the thickness to the radius decreases. From the thick wall solution, you can calculate all the stress components at all locations. Play with the results. chet
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http://nrich.maths.org/777
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### Real(ly) Numbers
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?
### How Many Solutions?
Find all the solutions to the this equation.
### On the Road
Four vehicles travelled on a road with constant velocities. The car overtook the scooter at 12 o'clock, then met the bike at 14.00 and the motorcycle at 16.00. The motorcycle met the scooter at 17.00 then it overtook the bike at 18.00. At what time did the bike and the scooter meet?
# More Parabolic Patterns
##### Stage: 4 and 5 Challenge Level:
The illustration shows the graphs of twelve functions. Three of them have equations
$y = x^2$
$x = y^2$
$x = -y^2 + 2$
Use a graphic calculator or a graph drawing program to sketch these three graphs and then to locate them in this illustration. Use the clues given in this information to help you to find the equations of all the other graphs and to draw the pattern of 12 graphs for yourself. For your solution send in the equations you have found with an explanation of how you did it.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://mathhelpforum.com/discrete-math/23769-proving-bijective-functions.html
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# Thread:
1. ## Proving bijective functions
I have 2 I am struggling with:
1. Let f:A to B. Prove that f is a bijection iff for all b that is an element of B, there exists exactly one a that is an element of A f(a)=f(b).
2. Prove: If f:A to B is a bijection and g:B to C is a bijection, then the composition of g of f is a bijection.
2. Originally Posted by IIuvsnshine
I have 2 I am struggling with:
1. Let f:A to B. Prove that f is a bijection iff for all b that is an element of B, there exists exactly one a that is an element of A f(a)=f(b).
2. Prove: If f:A to B is a bijection and g:B to C is a bijection, then the composition of g of f is a bijection.
You appear to have a typo. Shouldn't it read:
"Let f:A to B. Prove that f is a bijection iff for all b that is an element of B, there exists exactly one a that is an element of A f(a)=b."?
I'm not sure how much of a help this will be as I don't know what problems you are having with these, but in general to show a function is bijective you need to show that it is both injective and surjective. My advice is to simply apply the definition of each.
For example, the function in problem 1 is clearly surjective, since for every element b of B there exists (at least) one element of A such that f(a) = b.
-Dan
3. Yes, I did have a typo.
My concern is the proof (lol).
Do I need to approach this by cases?
For the 1st part I have:
Suppose f is a bijection. Then f is injective and surjective.
Case 1: Suppose f is injective. Then for all a,b that is an element of A, f(a)=f(b).
Case 2: Suppose f is surjective. Then for all of b that is an element of B, there exists an a such that f(b) = a.
I'm not sure I'm headed down the correct path.
4. A bijection is surjective injection.
A function is a surjection if and only if each element of the final set, in this case B, is the image of some element in the initial set A.
A function is an injection if and only if no two of its pairs have the same second term.
From the given does the function f have those two properties?
5. I understand the definitions...I'm struggling with wording for the proof.
6. Its an $\Leftrightarrow$ statement so you have 2 directions to go in.
7. Originally Posted by IIuvsnshine
I understand the definitions
Well that is not at all clear to me that you do understand how to use them.
You are given that each b is the image of some a therefore f is onto.
You are given that each b is the image of exactly one a therefore f is one-to-one.
8. It is sometimes difficult to state exactly what the meaning of words typed into a thread.
I do understand the definitions. I apologize if that is not clear to you. I also understand that i'm given a b that is an image of a such that f is onto, and that i'm given a b that is the image of exactly one a such that f is one-to-one.
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http://physics.stackexchange.com/questions/53473/how-local-is-the-stress-tensor
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# How local is the stress tensor?
I am confused by the definition of the stress tensor in a crystal (let's say a semi-conductor), I don't see how it could be "more local" than over an unit cell. I know that in field theory the stress tensor, or more precisely the whole stress-energy tensor, is local function, a function of the point $\vec{x}$ but I don't understand the interpretation you can give to shearing a point. Plus, in the lattice case I stressed (pun semi-intended), you shouldn't be able to look at displacements or deformations on smallest scales than your elementary cell?
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## 2 Answers
An interesting question.
You are right, the stress in a crystalline solid, or any solid, is treated by engineers as a macroscopic property of matter assuming matter is a continuous medium. It is given in terms of the external forces acting on the solid per unit area at some direction. Hence the distinction of $\sigma_{xy}, \sigma_{yz}$ etc. This goes with the definition of stress
$\sigma_{ij}={\frac {dF_i}{dA_j}}$
where $dF_i, dA_j$ are Cartesian tensors (not like the general contravariant and covariant tensors in general tensor analysis as in GR.) This derivative is meaningful locally and, as said above, treats the solid as a continuous medium.
In order to define stress in terms of the crystal structure you can relate it via the strain tensor, which can be defined in terms of relative infinitesimal displacements of the atoms in the cell. There are a couple of ways of doing this, one of which is the Lagrangian description. In this, the coordinates $(x_1,x_2,x_3)$ of the atoms in the unrestrained state are taken as the independent variables, while $(u_1,u_2,u_3)$ are the relative displacements of the atoms, and they are the dependent variables. This leads to the following definition for the strain tensor (Langrangian strain)
$\eta_{ij}={\frac 1 2}( {\frac {\partial u_i}{\partial xj}}+{\frac {\partial u_j}{\partial x_i}}+\Sigma {\frac {\partial u_r}{\partial x_i}} {\frac {\partial u_r}{\partial x_j}} )$
where the summation is over the index $r$. Note that this is a function of the coordinates of the atoms in the lattice, so $\eta_{ij}$ is a locally defined quantity. Macroscopically strain and stress relate via Young’s modulus $E$ in the relation $\sigma=\epsilon E$ (Hooke’s law,) which would be ok for an isotropic material, in which direction of application of the force is irrelevant. For anisotropic materials, such as crystalline solids in condensed matter physics, this relation is generalised to the following
$\sigma_{ij}=\Sigma C_{ijkl}\eta_{kl}$
and the summation is assumed over the $kl$ indices. The coefficients $C_{ijkl}$ are the second order ‘elastic constants’ or elastic stiffness coefficients of the material, and they are fourth order tensors, defined as derivatives of the elastic energy of the material, i.e. the potential energy of the atoms in the crystal lattice due to their relative displacements. I think this is what TMS meant in his comment. I hope this helps to understand the notion of locality of stress in crystalline solids.
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You right, you can't, when we defining stress-energy tensor like that we suppose that we have everywhere solid, continuous matter inside the object.
And when we want to incorporate the discrete structure of crystal, we usually doing that in a "smooth" way by using the function of Radial distribution function of atoms in crystal cells.
This can be justified by noting that we are usually interested in the potentials between the atoms when we try to study physical properties of the crystal, not the atom's locations itself, which in its turn can't be determined precisely as you know and makes no sense.
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I don't know the term "radial destitution". Maybe autocorrect played a trick on you? – Rafael Reiter Feb 10 at 9:26
Yes, corrected thx. – TMS Feb 10 at 10:07
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http://en.wikibooks.org/wiki/Digital_Circuits/Flip-Flops
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# Digital Circuits/Flip-Flops
A flip-flop is a device very like a latch in that it is a bistable multivibrator, having two states and a feedback path that allows it to store a bit of information. The difference between a latch and a flip-flop is that a latch is asynchronous, and the outputs can change as soon as the inputs do (or at least after a small propagation delay). A flip-flop, on the other hand, is edge-triggered and only changes state when a control signal goes from high to low or low to high. This distinction is relatively recent and is not formal, with many authorities still referring to flip-flops as latches and vice versa, but it is a helpful distinction to make for the sake of clarity.
There are several different types of flip-flop each with its own uses and peculiarities. The four main types of flip-flop are : SR, JK, D, and T.
## SR Flip-flops
File:Flipflop SR type.svg
Logic diagram of RS flip-flop
An SR(Set/Reset) flip-flop is perhaps the simplest flip-flop, and is very similar to the SR latch, other than for the fact that it only transitions on clock edges. While as theoretically valid as any flip-flop, synchronous edge-triggered SR flip-flops are extremely uncommon because they retain the illegal state when both S and R are asserted. Generally when people refer to SR flip-flops, they mean SR latches.
| S | R | Qnext | Comment |
|-----|-----|------------|------------|
| | | | |
| 0 | 0 | 0 | Hold state |
| 0 | 1 | 0 | Reset |
| 1 | 0 | 1 | Set |
| 1 | 1 | Metastable | |
| Q | Qnext | S | R |
|-----|---------|-----|-----|
| | | | |
| 0 | 0 | 0 | X |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | X | 0 |
## D flip-flop
Symbol for a D flip-flop.
The D flip-flop is the edge-triggered variant of the transparent latch. On the rising (usually, although negative edge triggering is just as possible) edge of the clock, the output is given the value of the D input at that moment. The output can be only change at the clock edge, and if the input changes at other times, the output will be unaffected.
| Clock | D | Qnext | Comment |
|-----------|-----|---------|----------------|
| | | | |
| | 0 | 0 | Express D at Q |
| | 1 | 1 | Express D at Q |
| Otherwise | X | Qprev | Hold state |
D flip-flops are by far the most common type of flip-flops and some devices (for example some FPGAs) are made entirely from D flip-flops. They are also commonly used for shift-registers and input synchronisation.
## JK Flip-flop
Symbol for a JK flip-flop
The JK flip-flop is a simple enhancement of the SR flip-flop where the state J=K=1 is not forbidden. It works just like a SR flip-flop where J is serving as set input and K serving as reset. The only difference is that for the formerly “forbidden” combination J=K=1 this flip-flop now performs an action: it inverts its state. As the behavior of the JK flip-flop is completely predictable under all conditions, this is the preferred type of flip-flop for most logic circuit designs. But there is still a problem i.e. both the outputs are same when one tests the circuit practically. This is because of the internal toggling on every propagation elapse completion. The main remedy is going for master-slave jk flip-flop,this flip-flop overrides the self(internal) recurring toggling through the pulsed clocking feature incorporated.
| J | K | Qnext | Comment |
|-----|-----|---------|------------|
| | | | |
| 0 | 0 | Qprev | Hold state |
| 0 | 1 | 0 | Reset |
| 1 | 0 | 1 | Set |
| 1 | 1 | Qprev | Toggle |
| Q | Qnext | J | K | Comment |
|-----|---------|-----|-----|------------|
| | | | | |
| 0 | 0 | 0 | X | Hold state |
| 0 | 1 | 1 | X | Set |
| 1 | 0 | X | 1 | Reset |
| 1 | 1 | X | 0 | Hold state |
## T flip-flops
A circuit symbol for a T-type flip-flop: T is the toggle input and Q is the stored data output.
A T flip-flop is a device which swaps or "toggles" state every time it is triggered if the T input is asserted, otherwise it holds the current output. This behavior is described by the characteristic equation:
$Q_{next} = T \oplus Q = T\overline{Q} + \overline{T}Q$
and can be described either of the following tables:
| T | Q | Qnext | Comment |
|-----|-----|---------|------------|
| | | | |
| 0 | 0 | 0 | Hold state |
| 0 | 1 | 1 | |
| 1 | 0 | 1 | Toggle |
| 1 | 1 | 0 | |
| Q | Qnext | T |
|-----|---------|-----|
| | | |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
When T is held high, the toggle flip-flop divides the clock frequency by two; that is, if clock frequency is 4 MHz, the output frequency obtained from the flip-flop will be 2 MHz. This 'divide-by' feature has application in various types of digital counters. A T flip-flop can also be built using a JK flip-flop (J & K pins are connected together and act as T) or D flip-flop (T input and Qprev are connected to the D input through an XOR gate). d t
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http://www.twelf.org/wiki/Judgment
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# Judgment
### From The Twelf Project
In the context of this wiki, we use the word judgment (or judgement) to refer to a relation that is defined inductively by a collection of inference rules. The judgments as types principle is a name for the methodology by which judgments are represented in LF.
## A judgment in standard notation
For example, we can define a judgment that a natural number is even. The judgement $\mathsf{even}(n)$ holds when $\texttt{}n$ is even. It is inductively defined by the following inference rules:
${\qquad} \over {\mathsf{even}(\mathsf{zero})}$ ${\mathsf{even}(n)} \over {\mathsf{even}(\mathsf{succ}(\mathsf{succ}(n)))}$
## Judgments as types
A judgment is represented in LF using the judgments as types methodology: we represent a judgment with an LF type, where the inhabitants of this type correspond exactly to derivations of the judgement.
For example, we represent the judgment $\mathsf{even}(n)$ using the following signature:
```even : nat -> type.
even-z : even z.
even-s : {N:nat} even N -> even (s (s N)).```
The first declaration says that even is a family of types indexed by a nat. This means that for every term N : nat, there is a type even N. Note that the syntax -> is overloaded: it is used to classify both type-level families and term-level functions. We then use this type family to define the types of two term constants.
The first term constant, even-z, has type even z. This constant represents the derivation that consists of the first inference rule above, which concludes $\mathsf{even}(\mathsf{zero})$.
The second term constant even-s, corresponds to the second inference rule above, which, for any $\texttt{}n$, constructs a derivation of $\mathsf{even}(\mathsf{succ}(\mathsf{succ}(n)))$ from a derivation of $\mathsf{even}(n)$. To encode this inference rule, the constant even-s is given a dependent function type.
For example, the LF term
```even-s z even-z
```
represents the derivation
${\overline{\mathsf{even}(\mathsf{zero})}} \over {\mathsf{even}(\mathsf{succ}(\mathsf{succ}(\mathsf{zero})))}$
The term even-s (s (s z)) (even-s z even-z) represents a derivation that 4 is even, and so on.
## See also
• Hypothetical judgements can be represented in LF in a higher-order manner, using LF binding to represent hypotheses.
• The introductions to Twelf discuss how judgments are represented in LF in more detail.
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|
http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics/83726
|
## Probing a manifold with geodesics
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Supposed you stand at a point $p \in M$ on a smooth 2-manifold $M$ embedded in $\mathbb{R}^3$. You do not know anything about $M$. You shoot off a geodesic $\gamma$ in some direction $u$, and learn back the shape of the full curve $\gamma$ as it sits in $\mathbb{R}^3$. (One could imagine a vehicle traveling along $\gamma$, sending back $xyz$-coordinates at regular time intervals; assume $t \rightarrow \infty$.) For example, if the geodesic happens to be closed, your probe might return the blue curve left below:
(Based on an image created by Mark Irons.)
I would like to know what information one could learn about $M$ from such geodesic probes. I am interested in the best case rather than the worst case. For example, you might learn that $M$ is unbounded, if you are lucky enough to shoot a geodesic to infinity. In particular,
Are there circumstances (a manifold $M$, a point $p$, directions $u$) that permit one to definitively conclude that the genus of $M$ is nonzero, by shooting (perhaps many) geodesics from one fixed (well-chosen) point $p$?
I believe that, if one knew all the geodesics through every point, then there are natural circumstances under which the metric is determined [e.g., "Metric with Ergodic Geodesic Flow is Completely Determined by Unparameterized Geodesics." Vladimir Matveev and Petar Topalov. Electronic Research Announcements of the AMS. Volume 6, Pages 98-101, 2000]. But I am more interested what can be determined from a single point $p$ (and many directions $u$). Thanks for thoughts/pointers!
(Tangentially related MO question: Shortest-path Distances Determining the Metric?.)
-
5
I thought that on a compact $M$, other than a sphere, the typical geodesic is dense in $M$. This would imply that if you really have the image of all of $[0,\infty)$ under a geodesic path then you know $M$ exactly and can thus deduce its genus and any other property you care about. – Noam D. Elkies Nov 22 2011 at 17:35
6
Having two orthogonal circles of different radii (like on the torus) already tells you something. You do not even need geodesics; if two closed curves on the manifold have just one intersection and it is transversal, that certainly should ring some bells though I'd better leave it to the many topologists here to tell what and how exactly can be derived from it. :). – fedja Nov 22 2011 at 18:17
@Noam: Perhaps you mean, "other than homeomorphic to a sphere," rather than precisely a sphere. Because Zoll's surface has the property that every geodesic is closed and simple (mathoverflow.net/questions/28622). I am surprised to learn that typical geodesics are dense. I thought there would generally be "unreachable" sections. Glad to have my faulty intuition corrected! – Joseph O'Rourke Nov 22 2011 at 21:35
@fedja: Excellent point (thanks!), but I am uncertain what conclusions could be drawn... – Joseph O'Rourke Nov 22 2011 at 21:39
1
By recording the geodesics intersections you can collect information on the cut locus of the surface, which in turn provide you with insight into the topology of the surface. – Dror Atariah Nov 23 2011 at 11:52
show 2 more comments
## 2 Answers
There's a different kind of answer to this that you might be interested in: Suppose that, when you fire off a probe along a unit speed geodesic starting at $p\in M$, you record the direction $\theta$ in which you sent it, and the probe reports the inertial forces it is experiencing, i.e., it sends back a running report on the curvature and torsion of the curve it is traveling along. Thus, you get to record these two data as functions $\kappa(t,\theta)$ and $\tau(t,\theta)$ of $t$, the time since the probe was launched, and $\theta$, the direction in which it was sent.
The question, then, is "Can you recover the metric on the surface $M$ from the data $\kappa$ and $\tau$?"
Not surprisingly, the answer is yes in the generic situation. If, for example, you are in the situation in which $\tau\not=0$, you find (by computing with the structure equations) that the induced metric must be of the form $$g = dt^2 + f(t,\theta)^2\ d\theta^2,$$ where $$f(t,\theta) = \frac{\sqrt{|\tau(t,\theta)|}}{2\ \tau(t,\theta)} \int_0^t\frac{\kappa_\theta(\rho,\theta)}{\sqrt{|\tau(\rho,\theta)|}}\ d\rho\ .$$
(Obviously, there will be some singularity issues at places where $\tau$ vanishes, but, generically, this isn't a problem. The case in which $\tau$ vanishes identically, such as when $p$ is a pole of rotational symmetry of the surface $M$, has to be treated separately.)
Since you can recover the curvature from $f$ by the formula $K = -f_{tt}/f$, you could detect when, for example, the surface $M$ is locally convex, and so forth.
As for computing the Euler characteristic, since $K\ dA = -f_{tt}\ dt\wedge d\theta = -d\left(\ f_t\ d\theta\ \right)$, it follows that, if you could figure out the star-shaped domain (in good cases, of the form $0\le t < T(\theta)$ for some piecewise differentiable $2\pi$-periodic function $T$ ) that maps one-to-one and onto the complement of the cut locus of $p$, then you could compute the Euler characteristic as $$\chi(M) = \frac{-1}{2\pi}\int_0^{2\pi} f_t\bigl(T(\theta),\theta\bigr)\ d\theta.$$
How numerically stable all these calculations are, I don't know. (I also don't know how hard it would be to figure out or approximate $T$.) Since you know that the result is an integer, though, you might be able to tolerate a reasonable amount of numerical error.
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@Robert: What a cool idea! Thanks! – Joseph O'Rourke Dec 17 2011 at 19:20
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Fix one $u$, its resulting geodesic probe, and vary $u$ slightly. Comparing the two geodesics at the same arclength values yields a close approximation to the solution to the Jacobi equation along the first curve. Consequently, you can approximate the curvature $K$ along the first curve. Stepping around the unit circle in the tangent space based at $p$ in this way, I have $N$ geodesic probes, and along each an approximation to the curvature. Taking $N$ large, I get an approximation to the curvature of the whole surface. Now you can also approximate the area element on the surface from your skeleton of curves, so you get an approximation of $\int K dA$ and hence the surface's genus.
This seems a bit of a cheat. Your map is the exponential map from the point $p$. I am coming close to saying: pull back the metric on the surface to the tangent space via the exponential map. Now you have a description of the whole surface.
A comment on the dense vs. Zoll business. If the surface is compact then in all circumstances we have Poinare recurrence: any geodesic through $p$ returns to an arbitrarily small neighborhood of $p$.
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3
@Richard: Actually, your last remark is not true as you have stated it. It's easy to construct a compact surface with a geodesic that $\alpha$-limits to one closed geodesic and $\omega$-limits to another, and such a geodesic won't pass through any point $p$ such that the geodesic returns arbitrarily close to $p$. Poincaré recurrence does, of course, apply to the (volume preserving) geodesic flow on the unit tangent bundle of a compact Riemannian surface, but its interpretation in terms of return of geodesics on the surface is different from what you have written. – Robert Bryant Dec 16 2011 at 20:34
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http://math.stackexchange.com/questions/tagged/stochastic-integrals+statistics
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# Tagged Questions
1answer
65 views
### Readings necessary to understand Ito Integrals?
I searched for this question but couldn't find a direct answer. Basically I want to understand (and possibly compute some simple instances of) the Ito integral. I am coming from a physics background ...
2answers
207 views
### Distribution of Maximum of Sum of Sum of Gaussians
Let $X_i$ be a sequence of i.i.d. standard normal random variables. Let $Y_i=\sum_{k=1}^iX_k$ and $Z_i=\sum_{k=1}^iY_k$. I am interested in upper and lower bounds for \$P(\sup_{1\leq i\leq m}|X_i|\leq ...
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http://www.nag.com/numeric/CL/nagdoc_cl23/html/G05/g05kac.html
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NAG Library Function Documentnag_rngs_basic (g05kac)
1 Purpose
nag_rngs_basic (g05kac) returns a pseudorandom number taken from a uniform distribution between $0$ and $1$.
2 Specification
#include <nag.h>
#include <nagg05.h>
double nag_rngs_basic (Integer igen, Integer iseed[])
3 Description
nag_rngs_basic (g05kac) returns the next pseudorandom number from a uniform $\left(0,1\right)$ generator.
The particular generator used to generate random numbers is selected by the value set for the input argument igen. Consult the g05 Chapter Introduction for details of the algorithms that can be used.
The current state of the chosen generator is saved in the integer array iseed which should not be altered between successive calls. Initial states are set or re-initialized by a call to nag_rngs_init_repeatable (g05kbc) (for a repeatable sequence if computed sequentially) or nag_rngs_init_nonrepeatable (g05kcc) (for a non-repeatable sequence).
nag_rngs_uniform (g05lgc) may be used to generate a vector of $n$ pseudorandom numbers which, if computed sequentially using the same generator, are exactly the same as $n$ successive values of this function. On many machines nag_rngs_uniform (g05lgc) is likely to be much faster.
4 References
Knuth D E (1981) The Art of Computer Programming (Volume 2) (2nd Edition) Addison–Wesley
5 Arguments
1: igen – IntegerInput
On entry: must contain the identification number for the generator to be used to return a pseudorandom number and should remain unchanged following initialization by a prior call to nag_rngs_init_repeatable (g05kbc) or nag_rngs_init_nonrepeatable (g05kcc).
2: iseed[$4$] – IntegerCommunication Array
On entry: contains values which define the current state of the selected generator.
On exit: contains updated values defining the new state of the selected generator.
None.
Not applicable.
8 Further Comments
The generator with the smallest period that can be selected is the basic generator. The period of the basic generator is ${2}^{57}$.
Its performance has been analysed by the Spectral Test, see Section 3.3.4 of Knuth (1981), yielding the following results in the notation of Knuth (1981).
| | | |
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|--------------------------------------------------------------------------------------------|
| $n$ | ${\nu }_{n}$ | Upper bound for ${\nu }_{n}$ |
| $2$ | $3.44×{10}^{8}$ | $4.08×{10}^{8}$ |
| $3$ | $4.29×{10}^{5}$ | $5.88×{10}^{5}$ |
| $4$ | $1.72×{10}^{4}$ | $2.32×{10}^{4}$ |
| $5$ | $1.92×{10}^{3}$ | $3.33×{10}^{3}$ |
| $6$ | $593$ | $939$ |
| $7$ | $198$ | $380$ |
| $8$ | $108$ | $197$ |
| $9$ | $67$ | $120$ |
The right-hand column gives an upper bound for the values of ${\nu }_{n}$ attainable by any multiplicative congruential generator working modulo ${2}^{59}$.
An informal interpretation of the quantities ${\nu }_{n}$ is that consecutive $n$-tuples are statistically uncorrelated to an accuracy of $1/{\nu }_{n}$. This is a theoretical result; in practice the degree of randomness is usually much greater than the above figures might support. More details are given in Knuth (1981), and in the references cited therein.
Note that the achievable accuracy drops rapidly as the number of dimensions increases. This is a property of all multiplicative congruential generators and is the reason why very long periods are needed even for samples of only a few random numbers.
9 Example
This example prints the first five pseudorandom numbers from a uniform distribution between $0$ and $1$, generated by nag_rngs_basic (g05kac) after initialization by nag_rngs_init_repeatable (g05kbc).
9.1 Program Text
Program Text (g05kace.c)
None.
9.3 Program Results
Program Results (g05kace.r)
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http://mathoverflow.net/questions/42446/metric-spaces-as-algebraic-systems/50457
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## metric spaces as algebraic systems
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Let $(X, {\mathrm{dist}})$ be a metric space. In the paper by Kramer, Shelah, Tent and Thomas , they define an algebraic system $A(X)$ as the set $X$ with countably many binary relations $R_\alpha$, for all positive rational $\alpha$: $(x,y)\in R_\alpha$ iff ${\mathrm{dist}}(x,y)<\alpha$. Is this the first paper where this algebraic system was defined?
Update 1: I need it because in my paper, I want to call these algebraic systems KSTT-systems. They satisfy axioms 0) $(x,x)\in R_\alpha$ for every $\alpha$, 1) $(x,y)\in R_\alpha$ iff $(y,x)\in R_\alpha$, 2) $(x,y)\in R_\alpha, (y,z)\in R_\beta\to (x,z)\in R_{\alpha+\beta}$, 3) $(x,y)\in R_\alpha\to (x,y)\in R_\beta$ for every $\beta\ge \alpha$. The original metric space $X$ can be elementary defined inside $A(X)$, and for every KSTT-system $A$, and an element $o\in A$, one can canonically (elementary) define a pointed metric space. This can be used to show that (modulo Continuum Hypothesis), for every metric space $X$ and every asymptotic cone $C$ of $X$, $C$ is isometric to any ultralimit of $C$. Thus, the only question remains: whether it is appropriate to call these KSTT-systems or somebody has introduced them earlier.
Update 2: I guess I was not clear enough. I need to know who was the first to consider metric spaces as algebraic systems with countable set of relations (as above). I am not interested in equivalent categories.
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Note that the fourth author is Simon Thomas, who is here at MathOverflow. – Joel David Hamkins Oct 31 2010 at 12:38
Joel: I did ask Simon Thomas, of course. – Mark Sapir Oct 31 2010 at 12:41
I think what you have described is either a nearness relation or some encoding of a metrizable uniformity. – Michael Blackmon Feb 5 2011 at 20:03
@Michael: I do not know what "metrizable uniformity" is. Is it an algebraic system? The whole and only point of the construction is to view metric spaces as algebraic systems with countable signature (so that one can use model theory to treat metric spaces). – Mark Sapir Feb 6 2011 at 2:00
## 3 Answers
Hi, I just want to add that your definition is very similar to Bourbaki's "uniformity" and "entourages."
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There is some similarity. But the main difference (besides the fact that the axioms are different) is that Bourbaki do not consider a uniformity as an algebraic system. And the set of entourages is allowed to be arbitrary large, i.e., say, uncountable. – Mark Sapir Oct 17 2010 at 6:34
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
you KSTT-System is like a axioms for a monoidal symmetric category (order). Do a look in : http://www.tac.mta.ca/tac/reprints/articles/1/tr1abs.html
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I cannot find it there. Are these categories algebraic structures? – Mark Sapir Oct 31 2010 at 12:40
I have to believe that this construction was previously known, though I can't point to a precise reference. The maps that preserve the relations are exactly the non-expansive maps. The category of metric spaces with non-expansive maps is a reasonably standard category (for example, Adamek, Herrlich, and Strecker include it as one of the standard examples, Met). It occurred to me years ago that you could write this category in terms of relations as above, so it must have occurred to many people over the years, and someone must have written it down somewhere. I did a couple of Google searches, but I didn't find anything directly relevant. The category was first introduced by John Isbell, if that helps.
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I also thought that this was standard but the problem is with a precise reference. The idea to view metric spaces as algebraic structures is useful when one wants to apply model theory (that is how it is used in [KSTT]. The idea is quite natural but the question is who invented it first. – Mark Sapir Dec 27 2010 at 1:08
Maybe contact Adamek? The axiomatization shows that the category of extended metric spaces with non-expansive maps is locally presentable. ("Extended", because you have to allow the metric to take on infinite values. This corresponds to pairs of points (x,y) where none of the predicates hold.) Since Ademek included this example in his category theory text, and is also an expert on local presentability, he might know if there's a published form of the result. – arsmath Dec 27 2010 at 23:01
There's a published form of a somewhat analogous presentation for Banach spaces -- totally convex spaces -- in the category theory literature, so it's a natural place to look. (It's not exactly analogous because in that case the signature includes one infinitary operation.) – arsmath Dec 27 2010 at 23:04
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http://mathoverflow.net/questions/80391?sort=oldest
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## Topological weak mixing and $\omega$-linearly-independent sequences generated by composition operators
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A research problem on which I am currently working requires a construction in topological dynamics of the following type:
Let $T \colon X \to X$ be a continuous transformation of a compact metric space which contains at least two points, and let $(a_n)$ be an absolutely summable real sequence which is not the zero sequence. When can we guarantee that there exists a continuous function $f \colon X \to \mathbb{C}$ such that $\sum_{n=1}^\infty a_n f \circ T^n$ is not a constant function?
The above question would seem to amount to a question about the spectral behaviour of the composition operator $U_T \colon C(X) \to C(X)$ defined by $U_Tf(x):=f(T(x))$. It is easy to show that the eigenvalues of $U_T$ form a subgroup of the unit circle. If $U_T$ has an eigenvalue which is not a root of unity then the eigenvalues are dense in the unit circle, so given a fixed sequence $(a_n)$ an easy Fourier analysis argument allows us to choose an eigenfunction $f$ for which $\sum_{n=1}^\infty a_n f \circ T^n$ is not constant. On the other hand, in some cases where $U_T$ has a root of unity as an eigenvalue, there are sequences such that the series converges to a constant for all $f$: for example, if $X$ contains just two points then the sequence given by $a_1=a_2=1$ and $a_n=0$ for $n \geq 3$ has this property. The case of most interest, then, is that in which $U_T$ has no eigenvalues except $1$ and no eigenfunctions other than the constant function, which is referred to as topological weak mixing. Specifically, I ask:
Is topological weak mixing of $T$ sufficient to guarantee the existence of $f$ in the first question?
This suggests to me the following more general functional-analytic question, which (by considering the action of $U_T$ on the quotient of $C(X)$ modulo the subspace of constant functions) would be sufficient for a positive answer to the above:
Let $L$ be a bounded linear operator acting on an infinite-dimensional Banach space $B$, with the spectrum of $L$ being $\{1\}$ and the norm of $L$ being $1$. When does there exist $x \in B$ such that the sequence $\{L^nx \colon n \geq 0\}$ is $\omega$-linearly independent, i.e. for all nonzero absolutely summable sequences $(a_n)$, the sum $\sum_{n=1}^\infty a_n L^nx$ is nonzero?
Thanks in advance!
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1
I don't think your last question states what you want--the identity operator is a counterexample. – Bill Johnson Nov 8 2011 at 17:33
This also suggests that under your definition, the identity map $T=\iota$ is topological weak mixing. I thought that latter property was usually defined as $f\in C(X), f\circ T=\lambda f \implies f\in\mathbb C 1$ (I think?) That is, $T$ is topological weak mixing if the only eigenvectors of $U_T$ are the constants. If $U_T$ acts on $C(X)$ quotiented by the constants, then do you want that $U_T$ has no eigenvectors? – Matthew Daws Nov 8 2011 at 18:20
Thanks Bill, Matthew: I've edited the question to address these mistakes. – Ian Morris Nov 8 2011 at 22:05
## 1 Answer
For each $k > 0$, `$F_k = \{ x \in X \mid T^k(x) = x \}$` is closed, hence if $T$ is not periodic then for each $N > 0$, $G_N = X \setminus (\cup_{k \leq N} F_k)$ is open, and non-empty.
If we are given $(a_n)$ absolutely summable, (for which we will assume $a_0 = 1$ by scaling and shifting, which is fine since $T$ is not periodic), take $N > 0$ such that $\sum_{n > N} |a_n| < 1$. Since $G_N$ is open and non-empty there exists a non-empty, open subset $O \subset G_N$ such that $T^n(O) \cap O = \emptyset$ for all $1 \leq n \leq N$.
If we consider a continuous function $f$ such that $\| f \|_\infty \leq 1$, $f$ has support contained in $O$, and $f$ obtains the values $\pm 1$ within $O$, then `$\sum_{n = 0}^N a_nf \circ T^n$` agrees with $f$ on $O$. Hence `$\sum_{n = 0}^\infty a_n f \circ T^n$` has both positive and negative values and is therefore not constant.
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Thanks, that's a great answer. In some cases it is a little nontrivial to choose the open set $O$ in such a way that it contains two distinct points: for example, if $X$ is the one-point compactification of $\mathbb{Z}$, $T(n):=n+1$ for $n \in \mathbb{Z}$ and $T$ fixes the point at infinity, then $O$ cannot simply by a ball around a point $x$. However I think it is sufficient to let $O$ be the set of all points sufficiently near to a set of the form $\{x\}\cup \{T^{N+1}x\}$. – Ian Morris Nov 9 2011 at 12:34
That's right $O$ should have two points. If $G_N$ is discrete then you can even take equality $O = \{ x \} \cup \{ T^{N + 1}(x) \}$, where $x \in G_{2N + 2}$. – Jesse Peterson Nov 9 2011 at 14:27
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http://mathoverflow.net/questions/25836/manifolds-whose-isometry-group-is-pati-salam
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## Manifolds whose isometry group is Pati-Salam?
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By the Pati-Salam group I refer to SU(2) x SU(2) x SU(4). It can be obtained as the group of isometries of the 8 dimensional manifold $S^3 \times S^5$, but I wonder if this is the only 8 dimensional manifold having this group of isometries.
This particular manifold is interesting because a quotient by any U(1) will produce a 7 dimensional manifold whose isometry group is the unbroken standard model group, as pointed out by Witten time ago. But my particular curiosity comes because Non Commutative Geometry gets the Pati Salam group from a different setup: the finite algebra `$M_2(H) \oplus M_4(C)$`, related perhaps to deformations of even spheres.
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## 1 Answer
$S^3 \times S^5$ has isometry group $SO_4(\mathbb{R}) \times SO_6(\mathbb{R})$, which has $SU(2) \times SU(2) \times SU(4)$ as a four-fold cover. Since it appears that you aren't worrying too much about central terms, we can replace $S^3$ with $\mathbb{R}P^3$, $S^5$ with $\mathbb{R}P^5$, or take a quotient by a diagonal group of order 2.
I'm pretty sure these are the only connected choices, because we can characterize homogeneous orbits by the stabilizers of points. In this case, you need a closed subgroup of Pati-Salam of dimension at least 13 whose intersection with each factor group is not the whole factor. There just aren't that many subgroups of suitably large dimension: we need a diagonally embedded $SU(2)$ (possibly with a central translate) to get dimension at least 3 in the first factors, and we need Spin(5) in the last factor to get dimension at least 10. This forces the orbits to be connected components.
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Indeed, asking for the classification of the non simply connected cases should be a very different question, albeit an interesting one. For instance, 3 dim lens spaces seem to do a interpolation from $S^3$ to $S^2 \times S^1$, with all the spaces, except $S^3$ itself, having a SU(2)xU(1) group. – arivero May 27 2010 at 0:04
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http://conservapedia.com/Stokes'_Theorem
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# Stokes' Theorem
### From Conservapedia
Stokes' Theorem states that the line integral of a closed path is equal to the surface integral of any capping surface for that path, provided that the surface normal vectors point in the same general direction as the right-hand direction for the contour:
$\oint_C \vec F \cdot \vec{\mathrm{d}r}\ = \iint_S (\nabla \times \vec F) \cdot \vec{\mathrm{d}S},$
Intuitively, imagine a "capping surface" that is nearly flat with the contour. The curl is the microscopic circulation of the function on tiny loops within that surface, and their sum or integral results in canceling out all the internal circulation paths, leaving only the integration over the outer-most path.[1] This remains true no matter how the capping surface is expanded, provided that the contour remains as its boundary.
Sometimes the circulation (the left side above) is easier to compute; other times the expresses the surface integral of the curl of vector field is easier to computer (particularly when it is zero).[2]
Stated another way, Stokes' Theorem equates the line integral of a vector fields to a surface integral of the same vector field. For this identity to be true, the direction of the vector normal n must obey the right-hand rule for the direction of the contour, i.e., when walking along the contour the surface must be on your left.
This is an extension of Green's Theorem to surface integrals, and is also the analog in two dimensions of the Divergence Theorem. The above formulation is also called as the "Curl Theorem," to distinguish it from the more general form of the Stokes' Theorem described below.
Stokes' Theorem is useful in calculating circulation in mechanical engineering. A conservative field has a circulation (line integral on a simple, closed curve) of zero, and application of the Stokes' Theorem to such a field proves that the curl of a conservative field over the enclosed surface must also be zero.
## General Form
In its most general form, this theorem is the fundamental theorem of Exterior Calculus, and is a generalization of the Fundamental Theorem of Calculus. It states that if M is an oriented piecewise smooth manifold of dimension k and ω is a smooth (k−1)-form with compact support on M, and ∂M denotes the boundary of M with its induced orientation, then
$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega\!\,$,
where d is the exterior derivative.
There are a number of well-known special cases of Stokes' theorem, including one that is referred to simply as "Stokes' theorem" in less advanced treatments of mathematics, physics, and engineering:
• When k=1, and the terms appearing in the theorem are translated into their simpler form, this is just the Fundamental Theorem of Calculus.
• When k=3, this is often called Gauss' Theorem or the Divergence Theorem and is useful in vector calculus:
$\iiint_R (\nabla \cdot \vec w)\ \mathrm{d}V = \iint_S \vec w \cdot \vec{\mathrm{d}A}\,$
Where R is some region of 3-space, S is the boundary surface of R, the triple integral denotes volume integration over R with dV as the volume element, and the double integral denotes surface integration over S with $\vec{\mathrm{d}A}$ as the oriented normal of the surface element. The $\nabla \cdot$ on the left side is the divergence operator, and the $\cdot$ on the right side is the vector dot product.
• When k=2, this is often just called Stokes' Theorem:
$\iint_S (\nabla \times \vec w) \cdot \vec{\mathrm{d}A} = \oint_E \vec w \cdot \vec{\mathrm{d}l}\,$
Here S is a surface, E is the boundary path of S, and the single integral denotes path integration around E with $\vec{\mathrm{d}l}$ as the length element. The $\nabla \times$ on the left side is the curl operator.
These last two examples (and Stokes' theorem in general) are are the subject of vector calculus. They play important roles in electrodynamics. The divergence and curl operations are cornerstones of Maxwell's Equations.
Stokes' Theorem is a lower-dimension version of the Divergence Theorem, and a higher-dimension version of Green's Theorem. Green’s Theorem relates a line integral to a double integral over a region, while Stokes' Theorem relates a surface integral of the curl of a function to its line integral. Stokes' Theorem originated in 1850.
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http://math.stackexchange.com/questions/221998/monge-ampere-like-equation
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# Monge-Ampere like equation
I'd like to know whether there has been work on Monge-Ampere like equations of the form $$\det u'(x)=f(x,u(x)),$$ where $u:\mathbb{R}^n\to \mathbb{R}^n$ is $C^1$ and $f$ is a positive function.
Under the additional assumption $u=\nabla v(x)$, this becomes a special case of the Monge-Ampere equation.
What is known about uniqueness and/or existence for appropriate boundary conditions?
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http://conan777.wordpress.com/2011/05/29/a-report-from-the-workshop-in-geometric-topology-utah-part-1/
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# Area 777
Conan Wu's mathematical blog
## A report from the Workshop in Geometric Topology @ Utah (part 1)
May 29, 2011
I went to Park City this passed week for the Workshop in Geometric Topology. It was a quite cool place filled with ski-equipment stores, Christmas souvenir shops, galleries and little wooden houses for family winter vacations. Well, as you may have guessed, the place would look very interesting in summer. :-P
As the ‘principal speaker’, Professor Gabai gave three consecutive lectures on his ending lamination space paper (this paper was also mentioned in my last post). I would like to sketch some little pieces of ideas presented in perhaps couple of posts.
Classification of simple closed curves on surfaces
Let $S_{g,p}$ denote the (hyperbolic) surface of genus $g$ and $p$ punchers. There is a unique geodesic loop in each homotopy class. However, given a geodesic loop drew on the surface, how would you describe it to a friend over telephone?
Here we wish to find a canonical way to describe homotopy classes of curves on surfaces. This classical result was originally due to Dehn (unpublished), but discovered independently by Thurston in 1976. For simplicity let’s assume for now that $S$ is a closed surface of genus $g$.
Fix pants decomposition $\mathcal{T}$ of $S$, $\mathcal{T} = \{ \tau_1, \tau_2, \cdots, \tau_{3g-3} \}$ is a disjoint union of $3g-3$ ‘cuffs’.
As we can see, any simple closed curve will have an (homology) intersection number with each of the cuffs. Those numbers are non-negative integers:
Around each cuff we may assign an integer twist number, for a cuff with intersection number $n$ and twist number $z$, we ‘twist’ the curve inside a little neighborhood of the cuff so that all transversal segments to the cuff will have $z$ intersections with the curve.
Negative twists merely corresponds to twisting in the other direction:
Theorem: Every simple closed curve is uniquely defined by its intersection number and twisting number w.r.t each of the cuffs.
Conversely, if we consider multi-curves (disjoint union of finitely many simple closed curves) then any element in $\mathbb{Z}^{3g-3} \times \mathbb{Z}_{\geq 0}^{3g-3}$ describes a unique multi-curve.
To see this we first assume that the pants decomposition comes with a canonical ‘untwisted’ curve connecting each pairs of cuffs in each pants. (i.e. there is no god given ’0′ twist curves, hence we have to fix which ones to start with.)
In the example above our curve was homotopic to the curve $((1,2), (2,1), (1,-4))$.
In other words, pants decompositions (together with the associated 0-twist arcs) give a natural coordinate chart to the set of homotopy class of (multi) curves on a surface. i.e. they are perimetrized by $\mathbb{Z}^{3g-3} \times \mathbb{Z}_{\geq 0}^{3g-3}$.
For the converse, we see that any triple of integers can be realized by filling the pants with a unique set of untwisted arcs:
In fact, this kind of parametrization can be generalized from integers to real numbers, in which case we have measured laminations instead of multi-curves and maximal train trucks on each pants instead of canonical untwisted arcs. i.e.
Theorem: (Thurston) The space of measured laminations $\mathcal{ML}(S)$ on a surface $S$ of genus $g$ is parametrized by $\mathbb{R}^{3g-3} \times \mathbb{R}_{\geq 0}^{3g-3}$. Furthermore, the correspondence is a homeomorphism.
Here the intersection numbers with the cuffs are wrights of the branches of the train track, hence it can be any non-negative real number. The twisting number is now defined on a continuous family of arcs, hence can be any real number, as shown below:
As we can see, just as in the case of multi-curves, any triple of real numbers assigned to the cuffs can be realized as the weights of branches of a train track on the pants.
40.343599 -74.651774
### Like this:
This entry was posted on May 29, 2011 at 11:22 pm and is filed under Uncategorized.
Tags: closed geodesic, David Gabai, Dehn, lamination, measured lamination, surface, Thurston, train track
### 6 Responses to “A report from the Workshop in Geometric Topology @ Utah (part 1)”
1. Tejas Says:
June 14, 2011 at 4:52 pm
Did you draw all those figures ? For a blog post, thats impressive!
• 777 Says:
June 14, 2011 at 7:31 pm
hehe…yes I did~ In fact images are a main attraction of this blog :-P
2. vidyasagarmm Says:
November 29, 2011 at 3:28 am
Hey, can you please let me know how you managed to such beautiful pictures?
• 777 Says:
November 29, 2011 at 5:47 pm
Hi there~ thanks for the comment~
As said before, I pretty much just ‘hand-draw’ them in adobe illustrator…:-P
3. April 10, 2012 at 11:48 am
[...] A report from the Workshop in Geometric Topology @ Utah (part 1) [...]
4. April 22, 2012 at 12:47 am
[...] A report from the Workshop in Geometric Topology @ Utah (part 1) [...]
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http://mathhelpforum.com/trigonometry/82587-find-trigonometric-equation.html
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# Thread:
1. ## Find a Trigonometric equation
Give an equation for a cosine funtion that has a maximum at (pi/4, 8), followed immediately by another maximum at (9pi/4,8), and has a minimum at value of 4.
This is what I did
Amplitude = (8-4)/2
= 2
Vertical shift = 8-2 =6
So far I got this
y = 2cos (x -? )+6
How do I find the phase shift
2. The cosine function normally has consecutive maximums at $x = 0$ and $x = 2\pi$. Your function has consecutive maximums at $x = \frac{\pi}{4}$ and $x = \frac{9\pi}{4}$. This means the period is $\frac{9\pi}{4} - \frac{\pi}{4} = 2\pi$, which is normal for the cosine function. So what is the phase shift when you move $(0, 8)$ to $(\frac{\pi}{4}, 8)$?
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http://mathoverflow.net/questions/67978/additive-form-of-hilbert-90-for-schemes
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## Additive form of Hilbert 90 for schemes?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
First, I am by no means well-versed on cohomology so I apologize if this is too elementary.
I have been going through some basics of etale cohomology, with my ultimate goal being an understanding of some basic applications. I have gone through the Kummer and Artin-Schreier sequences, and wanted to get an idea for how these sequences can help us classify $\mathbb{Z}/n$ and $\mathbb{Z}/p$ torsors.
I found a short exposition by Artin that said $H^1_{et}(X,\mathbb{G}_m)=Pic(X)$, and this was was labelled as Hilbert 90. This presumably has something to do with $\mathbb{G}_m$ being related to $\mathcal{O}_X^*$. Can someone tell me what the additive version of this is, ie what $H^1_{et}(X,\mathbb{G}_a)$ is equal to? Also if anyone has a reference for this being worked out that would be great as well.
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## 1 Answer
`$H^1_{ét}(X, \mathbf{G}_a) = H^1_{Zar}(X, \mathcal{O}_X)$`, see Milne, Étale Cohomology III.§3. (as for the étale cohomology of any coherent sheaf)
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Ah thank you. I did not know about that comparison theorem for coherent sheaves. – Randy Reddick Jun 16 2011 at 18:16
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http://en.wikipedia.org/wiki/Lagrange_multipliers
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# Lagrange multiplier
(Redirected from Lagrange multipliers)
Figure 1: Find x and y to maximize $f(x,y)$ subject to a constraint (shown in red) $g(x,y)=c$.
Figure 2: Contour map of Figure 1. The red line shows the constraint $g(x,y)=c$. The blue lines are contours of $f(x,y)$. The point where the red line tangentially touches a blue contour is our solution.Since $d_1>d_2$, the solution is a maximization of $f(x,y)$
In mathematical optimization, the method of Lagrange multipliers (named after Joseph Louis Lagrange) is a strategy for finding the local maxima and minima of a function subject to equality constraints.
For instance (see Figure 1), consider the optimization problem
maximize $f(x, y) \,$
subject to $g(x, y) = c.\,$
We need both $f$ and $g$ to have continuous first partial derivatives. We introduce a new variable ($\lambda$) called a Lagrange multiplier and study the Lagrange function (or Lagrangian) defined by
$\Lambda(x,y,\lambda) = f(x,y) + \lambda \cdot \Big(g(x,y)-c\Big),$
where the $\lambda$ term may be either added or subtracted. If $f(x_0, y_0)$ is a maximum of $f(x,y)$ for the original constrained problem, then there exists $\lambda_0$ such that $(x_0,y_0,\lambda_0)$ is a stationary point for the Lagrange function (stationary points are those points where the partial derivatives of $\Lambda$ are zero, i.e. $\nabla\Lambda = 0$). However, not all stationary points yield a solution of the original problem. Thus, the method of Lagrange multipliers yields a necessary condition for optimality in constrained problems.[1][2][3][4][5] Sufficient conditions for a minimum or maximum also exist.
## Introduction
One of the most common problems in calculus is that of finding maxima or minima (in general, "extrema") of a function, but it is often difficult to find a closed form for the function being extremized. Such difficulties often arise when one wishes to maximize or minimize a function subject to fixed outside conditions or constraints. The method of Lagrange multipliers is a powerful tool for solving this class of problems without the need to explicitly solve the conditions and use them to eliminate extra variables.
Consider the two-dimensional problem introduced above:
maximize $f(x, y) \,$
subject to $g(x, y) = c. \,$
We can visualize contours of f given by
$f(x, y)=d \,$
for various values of $d$, and the contour of $g$ given by $g ( x, y ) = c$.
Suppose we walk along the contour line with $g = c$. In general the contour lines of $f$ and $g$ may be distinct, so following the contour line for $g = c$ one could intersect with or cross the contour lines of $f$. This is equivalent to saying that while moving along the contour line for $g = c$ the value of $f$ can vary. Only when the contour line for $g = c$ meets contour lines of $f$ tangentially, do we not increase or decrease the value of $f$ — that is, when the contour lines touch but do not cross.
The contour lines of f and g touch when the tangent vectors of the contour lines are parallel. Since the gradient of a function is perpendicular to the contour lines, this is the same as saying that the gradients of f and g are parallel. Thus we want points $(x,y)$ where $g(x,y) = c$ and
$\nabla_{x,y} f = - \lambda \nabla_{x,y} g$,
where
$\nabla_{x,y} f= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$
and
$\nabla_{x,y} g= \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)$
are the respective gradients. The constant $\lambda$ is required because although the two gradient vectors are parallel, the magnitudes of the gradient vectors are generally not equal.
To incorporate these conditions into one equation, we introduce an auxiliary function
$\Lambda(x,y,\lambda) = f(x,y) + \lambda \cdot \Big(g(x,y)-c\Big),$
and solve
$\nabla_{x,y,\lambda} \Lambda(x , y, \lambda)=0.$
This is the method of Lagrange multipliers. Note that $\nabla_{\lambda} \Lambda(x , y, \lambda)=0$ implies $g(x,y)=c$.
The constrained extrema of $f$ are critical points of the Lagrangian $\Lambda$, but they are not local extrema of $\Lambda$ (see Example 2 below).
One may reformulate the Lagrangian as a Hamiltonian, in which case the solutions are local minima for the Hamiltonian. This is done in optimal control theory, in the form of Pontryagin's minimum principle.
The fact that solutions of the Lagrangian are not extrema also poses difficulties for numerical optimization. This can be addressed by computing the magnitude of the gradient, as the zeros of the magnitude are necessarily local minima, as illustrated in the numerical optimization example.
## Handling multiple constraints
A paraboloid, some of its level sets (aka contour lines) and 2 line constraints.
Zooming in on the levels sets and constraints, we see that the two constraint lines intersect to form a "joint" constraint that is a point. Since there is only one point to analyze, the corresponding point on the paraboloid is automatically a minimum and maximum. Yet the simplified reasoning presented in sections above seems to fail because the level set definitely appears to "cross" the point and at the same time its gradient is not parallel to the gradients of either constraint. This shows we must refine our explanation of the method to handle the kinds of constraints that are formed when we have more than one constraint acting at once.
The method of Lagrange multipliers can also accommodate multiple constraints. To see how this is done, we need to reexamine the problem in a slightly different manner because the concept of “crossing” discussed above becomes rapidly unclear when we consider the types of constraints that are created when we have more than one constraint acting together.
As an example, consider a paraboloid with a constraint that is a single point (as might be created if we had 2 line constraints that intersect). The level set (i.e., contour line) clearly appears to “cross” that point and its gradient is clearly not parallel to the gradients of either of the two line constraints. Yet, it is obviously a maximum and a minimum because there is only one point on the paraboloid that meets the constraint.
While this example seems a bit odd, it is easy to understand and is representative of the sort of “effective” constraint that appears quite often when we deal with multiple constraints intersecting. Thus, we take a slightly different approach below to explain and derive the Lagrange Multipliers method with any number of constraints.
Throughout this section, the independent variables will be denoted by $x_1,x_2,\ldots ,x_N$ and, as a group, we will denote them as $p=\left( x_1,x_2,\ldots ,x_N \right)$. Also, the function being analyzed will be denoted by $f\left( p \right)$ and the constraints will be represented by the equations $g_{1}\left( p \right) = 0,g_{2}\left( p \right) = 0,\,\,\ldots ,g_{M}\left( p \right) = 0$.
The basic idea remains essentially the same: if we consider only the points that satisfy the constraints (i.e. are in the constraints), then a point $\left( p,f\left( p \right) \right)$ is a stationary point (i.e. a point in a “flat” region) of f if and only if the constraints at that point do not allow movement in a direction where f changes value.
Once we have located the stationary points, we need to do further tests to see if we have found a minimum, a maximum or just a stationary point that is neither.
We start by considering the level set of f at $\left( p,f\left( p \right) \right)$. The set of vectors $\left\{ v_{L} \right\}$ containing the directions in which we can move and still remain in the same level set are the directions where the value of f does not change (i.e. the change equals zero). Thus, for every vector v in $\left\{ v_{L} \right\}$, the following relation must hold:
$\frac{df}{dx_{1}}v_{x_{1}}+\frac{df}{dx_{2}}v_{x_{2}}+\,\,\,\cdots \,\,\,+\frac{df}{dx_{N}}v_{x_{N}}=0$
where the notation $v_{x_{K}}$ above means the $x_{K}$-component of the vector v. The equation above can be rewritten in a more compact geometric form that helps our intuition:
$\begin{matrix} \underbrace{\begin{matrix} \left[ \begin{matrix} \frac{df}{dx_{1}} \\ \frac{df}{dx_{2}} \\ \vdots \\ \frac{df}{dx_{N}} \\ \end{matrix} \right] \\ {} \\ \end{matrix}}_{\nabla f} & \centerdot & \underbrace{\begin{matrix} \left[ \begin{matrix} v_{x_{1}} \\ v_{x_{2}} \\ \vdots \\ v_{x_{N}} \\ \end{matrix} \right] \\ {} \\ \end{matrix}}_{v} & =\,\,0 \\ \end{matrix}\,\,\,\,\,\,\,\,\,\,\,\,\text{is the same as writing}\,\,\,\,\,\,\,\,\nabla f\,\,\,\centerdot \,\,\,\,v\,\,=\,\,\,0.$
This makes it clear that if we are at p, then all directions from this point that do not change the value of f must be perpendicular to $\nabla f\left( p \right)$ (the gradient of f at p).
Now let us consider the effect of the constraints. Each constraint limits the directions that we can move from a particular point and still satisfy the constraint. We can use the same procedure, to look for the set of vectors $\left\{ v_{C} \right\}$ containing the directions in which we can move and still satisfy the constraint. As above, for every vector v in $\left\{ v_{C} \right\}$, the following relation must hold:
$\Delta g=\frac{dg}{dx_{1}}v_{x_{1}}+\frac{dg}{dx_{2}}v_{x_{2}}+\,\,\,\cdots \,\,\,+\frac{dg}{dx_{N}}v_{x_{N}}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\nabla g\,\,\centerdot \,\,\,v\,\,=\,\,\,0.$
From this, we see that at point p, all directions from this point that will still satisfy this constraint must be perpendicular to $\nabla g\left( p \right)$ .
Now we are ready to refine our idea further and complete the method: a point on f is a constrained stationary point if and only if the direction that changes f violates at least one of the constraints. (We can see that this is true because if a direction that changes f did not violate any constraints, then there would be a “legal” point nearby with a higher or lower value for f and the current point would then not be a stationary point.)
### Single constraint revisited
For a single constraint, we use the statement above to say that at stationary points the direction that changes f is in the same direction that violates the constraint. To determine if two vectors are in the same direction, we note that if two vectors start from the same point and are “in the same direction”, then one vector can always “reach” the other by changing its length and/or flipping to point the opposite way along the same direction line. In this way, we can succinctly state that two vectors point in the same direction if and only if one of them can be multiplied by some real number such that they become equal to the other. So, for our purposes, we require that:
$\nabla f\left( p \right)=\lambda \, \nabla g\left( p \right) \qquad \Rightarrow \qquad \nabla f\left( p \right)-\lambda \, \nabla g\left( p \right)\,\,=\,\,0.$
If we now add another simultaneous equation to guarantee that we only perform this test when we are at a point that satisfies the constraint, we end up with 2 simultaneous equations that when solved, identify all constrained stationary points:
$\begin{cases} g\left( p \right)=0 & \text{means point satisfies constraint} \\ \nabla f\left( p \right)-\lambda \, \nabla g\left( p \right) = 0 & \text{means point is a stationary point}. \end{cases}$
Note that the above is a succinct way of writing the equations. Fully expanded, there are $\text{N}+\text{1}$ simultaneous equations that need to be solved for the $\text{N}+\text{1}$ variables which are $\lambda$ and $x_1,\ x_2, \ldots, x_N$:
$\begin{align} g\left( x_1, x_2, \ldots , x_N \right) & =0 \\ \frac{df}{dx_1}\left( x_1, x_2, \ldots, x_N \right) - \lambda \frac{dg}{dx_1}\left( x_1, x_2,\ldots , x_N \right) & = 0 \\ \frac{df}{dx_2}\left( x_1, x_2, \ldots , x_N \right) - \lambda \frac{dg}{dx_2}\left( x_1, x_2, \ldots, x_N \right) & = 0 \\ & {}\ \ \vdots \\ \frac{df}{dx_N}\left( x_1, x_2, \ldots x_N \right) - \lambda \frac{dg}{dx_N}\left( x_1, x_2, \ldots, x_N \right) & = 0. \end{align}$
### Multiple constraints
For more than one constraint, the same reasoning applies. If two or more constraints are active together, each constraint contributes a direction that will violate it. Together, these “violation directions” form a “violation space”, where infinitesimal movement in any direction within the space will violate one or more constraints. Thus, to satisfy multiple constraints we can state (using this new terminology) that at the stationary points, the direction that changes f is in the “violation space” created by the constraints acting jointly.
The violation space created by the constraints consists of all points that can be reached by adding any linear combination of violation direction vectors—in other words, all the points that are “reachable” when we use the individual violation directions as the basis of the space. Thus, we can succinctly state that v is in the space defined by $b_1, b_2, \ldots ,b_M$ if and only if there exists a set of “multipliers” $\lambda_1, \lambda_2, \ldots , \lambda_M$ such that:
$\sum\limits_{k=1}^M \lambda_k b_k = v$
which for our purposes, translates to stating that the direction that changes f at p is in the “violation space” defined by the constraints $g_1,g_2, \ldots, g_M$ if and only if:
$\sum_{k=1}^M \lambda_k \nabla g_k (p) = \nabla f(p) \quad \Rightarrow \quad \nabla f(p) - \sum_{k=1}^M {\lambda_k \nabla g_k (p)} = 0.$
As before, we now add simultaneous equation to guarantee that we only perform this test when we are at a point that satisfies every constraint, we end up with simultaneous equations that when solved, identify all constrained stationary points:
$\begin{align} g_1(p) & = 0 && \text{these mean the point satisfies all constraints} \\ g_2(p)& =0 \\ & \ \ \vdots \\ g_M(p) &= 0 \\ & \\ \nabla f(p) - \sum_{k=1}^M {\lambda_k \, \nabla g_k (p)} & = 0 && \text{this means the point is a stationary point}. \\ \end{align}$
The method is complete now (from the standpoint of solving the problem of finding stationary points) but as mathematicians delight in doing, these equations can be further condensed into an even more elegant and succinct form. Lagrange must have cleverly noticed that the equations above look like partial derivatives of some larger scalar function L that takes all the $x_1, x_2, \ldots, x_N$ and all the $\lambda _1, \lambda_2, \ldots, \lambda _M$ as inputs. Next, he might then have noticed that setting every equation equal to zero is exactly what one would have to do to solve for the unconstrained stationary points of that larger function. Finally, he showed that a larger function L with partial derivatives that are exactly the ones we require can be constructed very simply as below:
$\begin{align} & {} \quad L\left( x_1, x_2, \ldots , x_N, \lambda_1, \lambda_2, \ldots, \lambda _M \right) \\ & = f\left( x_1, x_2, \ldots, x_N \right) - \sum\limits_{k=1}^M {\lambda_k g_k\left( x_1, x_2, \ldots , x_N \right)}. \end{align}$
Solving the equation above for its unconstrained stationary points generates exactly the same stationary points as solving for the constrained stationary points of f under the constraints $g_1,g_2, \ldots, g_M$.
In Lagrange’s honor, the function above is called a Lagrangian, the scalars $\lambda_1, \lambda_2, \ldots, \lambda_M$ are called Lagrange Multipliers and this optimization method itself is called The Method of Lagrange Multipliers.
The method of Lagrange multipliers is generalized by the Karush–Kuhn–Tucker conditions, which can also take into account inequality constraints of the form h(x) ≤ c.
## Interpretation of the Lagrange multipliers
Often the Lagrange multipliers have an interpretation as some quantity of interest. For example, if the Lagrangian expression is
$L(x_1, x_2, ...;\lambda_1, \lambda_2, \dots) = f(x_1, x_2, ...) + \lambda_1(c_1-g_1(x_1, x_2, ...))+\lambda_2(c_2-g_2(x_1, x_2, ...))+\dots$
then
$\frac{\partial L}{\partial {c_k}} = \lambda_k.$
So, λk is the rate of change of the quantity being optimized as a function of the constraint variable. As examples, in Lagrangian mechanics the equations of motion are derived by finding stationary points of the action, the time integral of the difference between kinetic and potential energy. Thus, the force on a particle due to a scalar potential, $F=-\nabla V$, can be interpreted as a Lagrange multiplier determining the change in action (transfer of potential to kinetic energy) following a variation in the particle's constrained trajectory. In control theory this is formulated instead as costate equations.
Moreover, by the envelope theorem the optimal value of a Lagrange multiplier has an interpretation as the marginal effect of the corresponding constraint constant upon the optimal attainable value of the original objective function: if we denote values at the optimum with an asterisk, then it can be shown that
$\frac{\text{d} f(x_1^*(c_1, c_2, \dots), x_2^*(c_1, c_2, \dots), ...)}{\text{d} c_k} = \lambda_k^*.$
For example, in economics the optimal profit to a player is calculated subject to a constrained space of actions, where a Lagrange multiplier is the change in the optimal value of the objective function (profit) due to the relaxation of a given constraint (e.g. through a change in income); in such a context $\lambda^*$ is the marginal cost of the constraint, and is referred to as the shadow price.
## Sufficient conditions
Main article: Hessian matrix#Bordered Hessian
Sufficient conditions for a constrained local maximum or minimum can be stated in terms of a sequence of principal minors (determinants of upper-left-justified sub-matrices) of the bordered Hessian matrix of second derivatives of the Lagrangian expression.[6]
## Examples
### Example 1
Fig. 3. Illustration of the constrained optimization problem
Suppose we wish to maximize $f(x,y)=x+y$ subject to the constraint $x^2+y^2=1$. The feasible set is the unit circle, and the level sets of f are diagonal lines (with slope -1), so we can see graphically that the maximum occurs at $(\sqrt{2}/2,\sqrt{2}/2)$, and that the minimum occurs at $(-\sqrt{2}/2,-\sqrt{2}/2)$.
Using the method of Lagrange multipliers, we have $g(x,y)-c=x^2+y^2-1$, hence
$\Lambda(x, y, \lambda) = f(x,y) + \lambda(g(x,y)-c) = x+y + \lambda (x^2 + y^2 - 1)$.
Setting the gradient $\nabla_{x,y,\lambda} \Lambda(x , y, \lambda)=0$ yields the system of equations
$\begin{align} \frac{\partial \Lambda}{\partial x} &= 1 + 2 \lambda x &&= 0, \\ \frac{\partial \Lambda}{\partial y} &= 1 + 2 \lambda y &&= 0, \\ \frac{\partial \Lambda}{\partial \lambda} &= x^2 + y^2 - 1 &&= 0, \end{align}$
where the last equation is the original constraint.
The first two equations yield $x=-1/(2\lambda)$ and $y=-1/(2\lambda)$, where $\lambda \neq 0$. Substituting into the last equation yields $1/(4\lambda^2)+1/(4\lambda^2)=1$, so $\lambda = \mp 1/\sqrt{2}$, which implies that the stationary points are $(\sqrt{2}/2,\sqrt{2}/2)$ and $(-\sqrt{2}/2,-\sqrt{2}/2)$. Evaluating the objective function f at these points yields
$f(\sqrt{2}/2,\sqrt{2}/2)=\sqrt{2}\mbox{ and } f(-\sqrt{2}/2, -\sqrt{2}/2)=-\sqrt{2},$
thus the maximum is $\sqrt{2}$, which is attained at $(\sqrt{2}/2,\sqrt{2}/2)$, and the minimum is $-\sqrt{2}$, which is attained at $(-\sqrt{2}/2,-\sqrt{2}/2)$.
### Example 2
Fig. 4. Illustration of the constrained optimization problem
Suppose we want to find the maximum values of
$f(x, y) = x^2 y \,$
with the condition that the x and y coordinates lie on the circle around the origin with radius √3, that is, subject to the constraint
$g(x,y) = x^2 + y^2 = 3. \,$
As there is just a single constraint, we will use only one multiplier, say λ.
The constraint g(x, y)-3 is identically zero on the circle of radius √3. So any multiple of g(x, y)-3 may be added to f(x, y) leaving f(x, y) unchanged in the region of interest (above the circle where our original constraint is satisfied). Let
$\Lambda(x, y, \lambda) = f(x,y) + \lambda (g(x, y)-3) = x^2y + \lambda (x^2 + y^2 - 3). \,$
The critical values of $\Lambda$ occur where its gradient is zero. The partial derivatives are
$\begin{align} \frac{\partial \Lambda}{\partial x} &= 2 x y + 2 \lambda x &&= 0, \qquad \text{(i)} \\ \frac{\partial \Lambda}{\partial y} &= x^2 + 2 \lambda y &&= 0, \qquad \text{(ii)} \\ \frac{\partial \Lambda}{\partial \lambda} &= x^2 + y^2 - 3 &&= 0. \qquad \text{(iii)} \end{align}$
Equation (iii) is just the original constraint. Equation (i) implies $x = 0$ or λ = −y. In the first case, if x = 0 then we must have $y = \pm \sqrt{3}$ by (iii) and then by (ii) λ = 0. In the second case, if λ = −y and substituting into equation (ii) we have that,
$x^2 - 2y^2 = 0. \,$
Then x2 = 2y2. Substituting into equation (iii) and solving for y gives this value of y:
$y = \pm 1. \,$
Thus there are six critical points:
$(\sqrt{2},1); \quad (-\sqrt{2},1); \quad (\sqrt{2},-1); \quad (-\sqrt{2},-1); \quad (0,\sqrt{3}); \quad (0,-\sqrt{3}).$
Evaluating the objective at these points, we find
$f(\pm\sqrt{2},1) = 2; \quad f(\pm\sqrt{2},-1) = -2; \quad f(0,\pm \sqrt{3})=0.$
Therefore, the objective function attains the global maximum (subject to the constraints) at $(\pm\sqrt{2},1)$ and the global minimum at $(\pm\sqrt{2},-1).$ The point $(0,\sqrt{3})$ is a local minimum and $(0,-\sqrt{3})$ is a local maximum, as may be determined by consideration of the Hessian matrix of $\Lambda(x,y,0)$.
Note that while $(\sqrt{2}, 1, -1)$ is a critical point of $\Lambda$, it is not a local extremum. We have $\Lambda(\sqrt{2} + \epsilon, 1, -1 + \delta) = 2 + \delta(\epsilon^2 + (2\sqrt{2})\epsilon)$. Given any neighborhood of $(\sqrt{2}, 1, -1)$, we can choose a small positive $\epsilon$ and a small $\delta$ of either sign to get $\Lambda$ values both greater and less than $2$.
### Example 3: Entropy
Suppose we wish to find the discrete probability distribution on the points $\{x_1, x_2, \ldots, x_n\}$ with maximal information entropy. This is the same as saying that we wish to find the least biased probability distribution on the points $\{x_1, x_2, \ldots, x_n\}$. In other words, we wish to maximize the Shannon entropy equation:
$f(p_1,p_2,\ldots,p_n) = -\sum_{j=1}^n p_j\log_2 p_j.$
For this to be a probability distribution the sum of the probabilities $p_i$ at each point $x_i$ must equal 1, so our constraint is $g(\vec{p})$ = 1:
$g(p_1,p_2,\ldots,p_n)=\sum_{j=1}^n p_j.$
We use Lagrange multipliers to find the point of maximum entropy, $\vec{p}^{\,*}$, across all discrete probability distributions $\vec{p}$ on $\{x_1,x_2, \ldots, x_n\}$. We require that:
$\left.\frac{\partial}{\partial \vec{p}}(f+\lambda (g-1))\right|_{\vec{p}=\vec{p}^{\,*}}=0,$
which gives a system of n equations, $k ={1,\ldots,n}$, such that:
$\left.\frac{\partial}{\partial p_k}\left\{-\sum_{j=1}^n p_j \log_2 p_j + \lambda \left(\sum_{j=1}^n p_j - 1\right) \right\}\right|_{p_k=p^*_k} = 0.$
Carrying out the differentiation of these n equations, we get
$-\left(\frac{1}{\ln 2}+\log_2 p^*_k \right) + \lambda = 0.$
This shows that all $p^*_k$ are equal (because they depend on λ only). By using the constraint ∑j pj = 1, we find
$p^*_k = \frac{1}{n}.$
Hence, the uniform distribution is the distribution with the greatest entropy, among distributions on n points.
### Example 4: numerical optimization
Lagrange multipliers cause the critical points to occur at saddle points.
The magnitude of the gradient can be used to force the critical points to occur at local minima.
With Lagrange multipliers, the critical points occur at saddle points, rather than at local maxima (or minima). Unfortunately, many numerical optimization techniques, such as hill climbing, gradient descent, some of the quasi-Newton methods, among others, are designed to find local maxima (or minima) and not saddle points. For this reason, one must either modify the formulation to ensure that it's a minimization problem (for example, by extremizing the square of the gradient of the Lagrangian as below), or else use an optimization technique that finds stationary points (such as Newton's method without an extremum seeking line search) and not necessarily extrema.
As a simple example, consider the problem of finding the value of $x$ that minimizes $f(x)=x^2$, constrained such that $x^2=1$. (This problem is somewhat pathological because there are only two values that satisfy this constraint, but it is useful for illustration purposes because the corresponding unconstrained function can be visualized in three dimensions.)
Using Lagrange multipliers, this problem can be converted into an unconstrained optimization problem:
$\Lambda(x,\lambda)=x^2+\lambda(x^2-1).$
The two critical points occur at saddle points where $x=1$ and $x=-1$.
In order to solve this problem with a numerical optimization technique, we must first transform this problem such that the critical points occur at local minima. This is done by computing the magnitude of the gradient of the unconstrained optimization problem.
First, we compute the partial derivative of the unconstrained problem with respect to each variable:
$\frac{\partial \Lambda}{\partial x}=2x+2x\lambda$
$\frac{\partial \Lambda}{\partial \lambda}=x^2-1.$
If the target function is not easily differentiable, the differential with respect to each variable can be approximated as
$\frac{\partial \Lambda}{\partial x}\approx\frac{\Lambda(x+\epsilon,\lambda)-\Lambda(x,\lambda)}{\epsilon}$,
$\frac{\partial \Lambda}{\partial \lambda}\approx\frac{\Lambda(x,\lambda+\epsilon)-\Lambda(x,\lambda)}{\epsilon}$,
where $\epsilon$ is a small value.
Next, we compute the magnitude of the gradient, which is the square root of the sum of the squares of the partial derivatives:
$h(x,\lambda)=\sqrt{(2x+2x\lambda)^2+(x^2-1)^2}\approx\sqrt{\left(\frac{\Lambda(x+\epsilon,\lambda)-\Lambda(x,\lambda)}{\epsilon}\right)^2+\left(\frac{\Lambda(x,\lambda+\epsilon)-\Lambda(x,\lambda)}{\epsilon}\right)^2}.$
(Since magnitude is always non-negative, optimizing over the squared-magnitude is equivalent to optimizing over the magnitude. Thus, the ``square root" may be omitted from these equations with no expected difference in the results of optimization.)
The critical points of $h$ occur at $x=1$ and $x=-1$, just as in $\Lambda$. Unlike the critical points in $\Lambda$, however, the critical points in $h$ occur at local minima, so numerical optimization techniques can be used to find them.
## Applications
### Economics
Constrained optimization plays a central role in economics. For example, the choice problem for a consumer is represented as one of maximizing a utility function subject to a budget constraint. The Lagrange multiplier has an economic interpretation as the shadow price associated with the constraint, in this example the marginal utility of income. Other examples include profit maximization for a firm, along with various macroeconomic applications.
### Control theory
In optimal control theory, the Lagrange multipliers are interpreted as costate variables, and Lagrange multipliers are reformulated as the minimization of the Hamiltonian, in Pontryagin's minimum principle.
## See also
• Karush–Kuhn–Tucker conditions: generalization of the method of Lagrange multipliers.
• Lagrange multipliers on Banach spaces: another generalization of the method of Lagrange multipliers.
• Dual problem
• Lagrangian relaxation
## References
1. Bertsekas, Dimitri P. (1999). Nonlinear Programming (Second ed.). Cambridge, MA.: Athena Scientific. ISBN 1-886529-00-0.
2. Vapnyarskii, I.B. (2001), "Lagrange multipliers", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 .
• Lasdon, Leon S. (1970). Optimization theory for large systems. Macmillan series in operations research. New York: The Macmillan Company. pp. xi+523. MR 337317.
• Lasdon, Leon S. (2002). Optimization theory for large systems (reprint of the 1970 Macmillan ed.). Mineola, New York: Dover Publications, Inc. pp. xiii+523. MR 1888251.
3. Hiriart-Urruty, Jean-Baptiste; Lemaréchal, Claude (1993). "XII Abstract duality for practitioners". Convex analysis and minimization algorithms, Volume II: Advanced theory and bundle methods. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences] 306. Berlin: Springer-Verlag. pp. 136–193 (and Bibliographical comments on pp. 334–335). ISBN 3-540-56852-2. MR 1295240.
4. Lemaréchal, Claude (2001). "Lagrangian relaxation". In Michael Jünger and Denis Naddef. Computational combinatorial optimization: Papers from the Spring School held in Schloß Dagstuhl, May 15–19, 2000. Lecture Notes in Computer Science 2241. Berlin: Springer-Verlag. pp. 112–156. doi:10.1007/3-540-45586-8_4. ISBN 3-540-42877-1. MR 1900016.
5. Chiang, Alpha C., Fundamental Methods of Mathematical Economics, McGraw-Hill, third edition, 1984: p. 386.
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I'm trying to animate a torus being cut by a plane, so that as the plane enters the torus the part in front of the plane disappears (the idea is to show the cross section). That much I can do, but I ...
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http://www.proofwiki.org/wiki/General_Binomial_Theorem
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# Binomial Theorem
From ProofWiki
(Redirected from General Binomial Theorem)
## Theorem
Let $X$ be one of the set of numbers $\N, \Z, \Q, \R, \C$.
Let $x, y \in X$.
Then:
$\displaystyle \forall n \in \Z_{\ge 0}: \left({x + y}\right)^n = \sum_{k \mathop = 0}^n {n \choose k} x^{n-k} y^k$
where $\displaystyle {n \choose k}$ is $n$ choose $k$.
Let $\left({R, +, \odot}\right)$ be a ringoid such that $\left({R, \odot}\right)$ is a commutative semigroup.
Let $n \in \Z: n \ge 2$. Then:
$\displaystyle \forall x, y \in R: \odot^n \left({x + y}\right) = \odot^n x + \sum_{k \mathop = 1}^{n-1} \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right) + \odot^n y$
where $\displaystyle \binom n k = \frac {n!} {k!\ \left({n-k}\right)!}$ (see Binomial Coefficient).
If $\left({R, \odot}\right)$ has an identity element $e$, then:
$\displaystyle \forall x, y \in R: \odot^n \left({x + y}\right) = \sum_{k \mathop = 0}^n \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right)$
Let $\alpha \in \R$ be a real number.
Let $x \in \R$ be a real number such that $\left|{x}\right| < 1$.
Then:
$\displaystyle \left({1 + x}\right)^\alpha = \sum_{n \mathop = 0}^\infty \frac {\alpha^{\underline n}} {n!} x^n = \sum_{n \mathop = 0}^\infty \frac 1 {n!}\left(\prod \limits_{k \mathop = 0}^{n-1} \left({\alpha - k}\right)\right) x^n$
where $\alpha^{\underline n}$ denotes the falling factorial.
That is:
$\displaystyle \left({1 + x}\right)^\alpha = 1 + \alpha x + \frac {\alpha \left({\alpha - 1}\right)} {2!} x^2 + \frac {\alpha \left({\alpha - 1}\right) \left({\alpha - 2}\right)} {3!} x^3 + \cdots$
Let $\alpha$ be a multiindex, indexed by $\left\{{1, \ldots, n}\right\}$ such that $\alpha_j \ge 0$ for $j = 1, \ldots, n$.
Let $x = \left({x_1, \ldots, x_n}\right)$ and $y = \left({y_1, \ldots, y_n}\right)$ be ordered tuples of real numbers.
Then:
$\displaystyle \left({x + y}\right)^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}$
where $\displaystyle {n \choose k}$ is a binomial coefficient.
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http://crypto.stackexchange.com/tags/sha1/hot?filter=month
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# Tag Info
## Hot answers tagged sha1
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### SHA-1:Is there any mathematical result that gives us the minimum number of 1's in a 160-bit SHA-1 hash output?
No, theoretically a SHA1 hash can be any 160-bit value, including the string of 160 zeroes. As for your second question, if we fudge a little bit and consider SHA1 a truly random function this becomes the same question as the following: If we flip 160 coins, what is the probability that at least 128 of them will be heads? Solution is left as an exercise ...
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### SHA-1:Is there any mathematical result that gives us the minimum number of 1's in a 160-bit SHA-1 hash output?
Is there any mathematical result that gives us the minimum number of 1's in a 160-bit SHA-1 hash output? A good (secure) hash function has output that is uniformly and evenly distributed and shouldn't be distinguishable from random value. Chi-squared tests of several hash functions So the minimum number of possible ones is $0$ and the maximum ...
1
### Is Base64(SHA1(GUID)) still unique like the original GUID?
Base64 provides a 1:1 transform from input to output (and back again if desired). So if you take a set of unique items and base64 encode all of them they will all be unique. So the question becomes if you run a GUID through SHA1, will the resulting hash have the same uniqueness as the GUID? The answer is practically - yes; theoretically not quite. Multiple ...
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### Is Base64(SHA1(GUID)) still unique like the original GUID?
Assuming no cosmetics, the length of a GUID is 32 bytes so better question would be "What's the collision probability of SHA1 with 32 bytes of input?" I'm sure someone else will answer with the exact statistics but the answer to your question is yes, it's pretty unique (an attacker has a negligible probability of success). Note that I've completely ignored ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://mathoverflow.net/questions/32478?sort=votes
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## A nice explanation of what is a smooth (l-adic) sheaf?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I would like to understand this concept. It seems to be important (for the theory of perverse sheaves), yet I don't know any nice exposition of the properties of smooth sheaves.
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Lisse (and more generally, constructible) $\ell$-adic sheaves are nicely explained (their properties, link with $\pi_1$-representations, etc.) in the standard references on etale cohomology (Milne's book, Freitag-Kiehl book, SGA4,...), so can you clarify what it is that you wish to understand which is not adequately addressed in such places? Lisse sheaves are important for plenty of things more basic than perverse sheaves as well, so the question seems a bit too brief as presently written. – BCnrd Jul 19 2010 at 13:06
I agree with Brian. Let me however add the following reference which might be useful: springer.com/mathematics/algebra/book/…. – Daniel Larsson Jul 19 2010 at 14:05
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Mikhail, though lisse sheaves are nicest, it is constructible ones that admit the widest range of operations (e.g., excision, "noetherian" characterization at "finite level", preservation under higher direct images with proper support -- as well as under higher direct images in many important cases, as explained in Deligne's SGA 4 1/2 "Th. finitude..."), so to prove things about lisse sheaves often one needs a version for constructible ones and then apply "specialization criterion" for constructible to be lisse. This comes up in proof of the smooth and proper base change thm, for example. – BCnrd Jul 20 2010 at 13:00
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If you are not comfortable with the topological theory of locally constant and constructible sheaves (on varieties over $\mathbb C$, say, as explained in e.g. Borel et al.'s book on Intersection Homology) then I would suggest that you learn this theory first before learning the $\ell$-adic theory. Once you know the former, the $\ell$-adic theory will seem much more motivated (see e.g. Donu Arapura's answer below), and it becomes safe to treat it as a black box to a large extent. On the other hand, if you don't know the former, then the latter will not make much sense. – Emerton Jul 21 2010 at 0:52
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Dear Brian, One doesn't have to allow non-Zariski closed stratifications to be in the topological setting (and I'm not advocating that one should). Rather, I am advocating that it makes sense to understand the case of arguments with sheaves (constructible along Zariski closed subsets) in the usual topology on complex varieties before passing to the etale setting. Results such as proper base-change, and local acyclicity of smooth morphisms, are simpler to understand in this setting, and (personally) I find the resulting intuition very important for understanding the etale setting. – Emerton Jul 22 2010 at 18:36
show 8 more comments
## 1 Answer
I'll give an answer, only because I'm interested in chasing down these references myself. But all I'm doing is assembling references. I assume that BCnrd will keep me honest. [July 21: I've added some remarks about constructibility, to makes this more useful (at least to me).]
Since I'm a complex geometer rather an arithmetic one, let me start with the first case for intuition. If $X_{an}$ is a (connected) complex variety endowed with the classical topology then one knows that representations of the usual $\pi_1(X_{an},x)$ correspond to locally constant sheaves on $X_{an}$. This is classical. A good source of examples are as follows: if $f:Y\to X$ is a surjective smooth proper map, then it is topologically a fibre bundle (Ereshmann). Therefore $R^if_*\mathbb{Z}$ is locally constant. The corresponding $\pi_1(X)$-module is the monodromy representation. The most general statement one can make, without making any assumptions on $f$, is that the proper direct image $R^if_!\mathbb{Z}$ is constructible. Note that constructibility can mean different things in the topological world. The best notion (from my point of view) is what is sometimes called algebraic constructibility: there exists a partition of the base into Zariski locally closed strata such that the restrictions of the sheaf are locally constant. The only reference that I know which takes this viewpoint is Verdier, Classe d'homologie associée à un cycle. If people are aware of other sources, please let me know.
Remarkably, the analogous results hold in the $\ell$-adic case, although for different reasons. Let $X$ be variety over some field. A lisse (resp. constructible) $\ell$-adic sheaf is now a prosheaf `$$\ldots \mathcal{F}_n\to \mathcal{F}_{n-1}\ldots $$` on the etale site $X_{et}$ such that each item above is a locally constant (resp. constructible) $\mathbb{Z}/\ell^n$-module etc. (see Freitag-Kiehl, pp 118-131, for the precise conditions). For lisse sheaves, each $\mathcal{F}_n$ gives a representation of the etale fundamental group $$\pi_1^{et}(X,x)\to GL_N(\mathbb{Z}/\ell^n)$$ ($x$ a geom. pt.). So passing to the limit, we get a continuous representation $$\pi_1^{et}(X,x)\to GL_N(\mathbb{Z}_\ell)$$ This constuction is an equivalence [FK,p 286].
The corresponding result that `$R^if_*\mathbb{Z}_\ell$` is lisse, when $f$ is smooth, proper and surjective, should follow from Theorem 20.2 of Milne "Lectures on etale cohomology" from his website. The contrucibility of `$R^if_!\mathbb{Z}_\ell$` would follow from SGA4 exp XIV 1.1 (It ought to be in [FK,M], but I probably didn't look hard enough.)
When $X$ is defined over $\mathbb{C}$, one can compare cohomology for the classical and etale topologies with general coefficients by applying SGA4 exp XVI 4.1 and taking inverse limits. A more general comparison result for the "6 operations" is given in [Beilinson-Bernstein-Deligne p 150], but the proof seems a bit sketchy. Remark added July 22: Unfortunately, this part of the story appears to be inadequately addressed in the literature. See BCnrd's comment below.
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Thank you! Yet (as far as I remember) people also talk about lisse sheaves on big (etale) sites. Does there exist a similar explanation for this? – Mikhail Bondarko Jul 21 2010 at 0:59
Is $f:Y\to X$ also assumed to be surjective? – Victor Protsak Jul 21 2010 at 2:00
Yes, absolutely. I've fixed that. – Donu Arapura Jul 21 2010 at 2:22
Donu, not obvious that analytification of lisse $\mathbf{Z}_{\ell}$-sheaf, as inv. system of finite loc. constant sheaves on $X(\mathbf{C})$, comes from loc. constant sheaf of finite-rank $\mathbf{Z}_{\ell}$-mods. Need open cover "trivializing" all levels of analytified inv. system. Smooth case OK; o/w need work on topology of $X(\mathbf{C})$. Crucial in pf of relative $\ell$-adic Artin comparison isom (with constr. coeffs), for which there's no published pf! (SGA4 has finite coeffs, SGA5 nada, BBD misses it.) Deligne has nifty triangulation-free pf via alterations (private communication). – BCnrd Jul 22 2010 at 2:27
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Brian, thanks for clarifying. It would be nice if someone (hint, hint) wrote something more complete about this. – Donu Arapura Jul 22 2010 at 11:15
show 1 more comment
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http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm
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We use MathJax
## How To Construct a Delta-Epsilon Proof
The proof, using delta and epsilon, that a function has a limit will mirror the definition of the limit. Therefore, we first recall the definition:
$\lim\limits_{x\to c} f(x)=L$ means that for every $\epsilon>0$, there exists a $\delta>0$, such that for every $x$, the expression $0< |x-c|<\delta$ implies $|f(x)-L|<\epsilon$.
Each phrase of the definition contributes to some aspect of the proof. Specifically:
• The phrase "for every $\epsilon >0$ " implies that we have no control over epsilon, and that our proof must work for every epsilon.
• The phrase "there exists a $\delta >0$ " implies that our proof will have to give the value of delta, so that the existence of that number is confirmed. Typically, the value of delta will depend on the value of epsilon.
• The phrase "such that for every $x$" implies that we cannot restrict the values of $x$ any further than the next restriction provides.
• The phrase "the expression $0< |x-c| < \delta$ " is the starting point for a series of implications (algebra steps) which will conclude with the final statement. The expression $|x-c| < \delta$ means that the values of $x$ will be close to $c$, specifically not more than (nor even equal to) delta units away. The expression $0 < |x-c|$ implies that $x$ is not equal to $c$ itself.
• The phrase "implies $|f(x)-L| < \epsilon$ " is the conclusion of the series of implications. Once this statement is reached, the proof will be complete.
Upon examination of these steps, we see that the key to the proof is the identification of the value of delta. To find that delta, we typically begin with the final statement $|f(x)-L| < \epsilon$, and work backwards until we reach the form $|x-c| < \delta$. So let's consider some examples. Linear examples are the easiest. Non-linear examples exhibit a few other quirks, and we will demonstrate them below also.
### Example using a Linear Function
Prove, using delta and epsilon, that $\lim\limits_{x\to 4} (5x-7)=13$.
We will place our work in a table, so we can provide a running commentary of our thoughts as we work.
$|f(x)-L| < \epsilon$ Before we can begin the proof, we must first determine a value for delta. To find that delta, we begin with the final statement and work backwards.
$|(5x-7)-13| < \epsilon$ We substitute our known values of $f(x)$ and $L$.
$|5x-20| < \epsilon$ Our short-term goal is to obtain the form $|x-c| < \delta$. So we begin by simplifying inside the absolute value.
$|5(x-4)|<\epsilon$$|5||x-4|<\epsilon$ $|x-4|<\dfrac{\epsilon}{5}$ In these three steps, we divided both sides of the inequality by 5. Most often, these steps will be combined into a single step. However, when the slope of the linear function is negative, you may want to do the steps separately so as to avoid incorrectly handling the negative sign.
$|x-c|<\delta$$\delta=\dfrac{\epsilon}{5}$ We now recall that we were evaluating a limit as $x$ approaches 4, so we now have the form $|x-c| < \delta$. Therefore, since $c$ must be equal to 4, then delta must be equal to epsilon divided by 5 (or any smaller positive value).
Now we are ready to write the proof. Once again, we will provide our running commentary.
Proof.
Suppose $\epsilon >0$ has been provided. This is always the first line of a delta-epsilon proof, and indicates that our argument will work for every epsilon.
Define $\delta=\dfrac{\epsilon}{5}$. Since the definition of the limit claims that a delta exists, we must exhibit the value of delta. We use the value for delta that we found in our preliminary work above.
Since $\epsilon >0$, then we also have $\delta >0$. The definition does place a restriction on what values are appropriate for delta (delta must be positive), and here we note that we have chosen a value of delta that conforms to the restriction.
Now, for every $x$, the expression $0< |x-c| < \delta$ implies This is the next part of the wording from the definition of the limit.
$|x-4| < \dfrac{\epsilon}{5}$ We replace the values of $c$ and delta by the specific values for this problem. From here on, we will be basically following the steps from our preliminary work, but in reverse order.
$|5x-20| < \epsilon$ We multiplied both sides by 5. If the slope of the original function was negative, we may want to do this using more steps, so as to introduce the negative sign correctly.
$|(5x-7)-13| < \epsilon$ Now we break the expression into the two parts we need to exhibit, the original function and the limit value.
Therefore, $\lim\limits_{x\to 4} (5x-7)=13$. Since we began with $c = 4$, and we obtained the above limit statement, we have met all of the requirements of the definition of the limit, and obtained our final result.
Q.E.D. This is an abbreviation for the Latin expression "quod erat demonstrandum", which means "which was to be demonstrated". Some authors will include it to denote the end of the proof.
### Example using a Non-Linear Function
Prove, using delta and epsilon, that $\lim\limits_{x\to 5} (3x^2-1)=74$.
$|f(x)-L| < \epsilon$ Before we can begin the proof, we must first determine a value for delta. To find that delta, we begin with the final statement and work backwards.
$|(3x^2-1)-74| < \epsilon$ We substitute our known values of $f(x)$ and $L$.
$|3x^2-75| < \epsilon$ Our short-term goal is to obtain the form $|x-c| < \delta$. However, with non-linear functions, it is easier to work toward solving for $x$ by itself, then introduce the value of $c$. So we begin by simplifying inside the absolute value.
$-\epsilon < 3x^2-75 < \epsilon$ With non-linear functions, the absolute values will have to be removed, since the allowable delta-distances will be different on the two sides of the value $x=c$.
$75-\epsilon < 3x^2 < 75+\epsilon$ $25-\dfrac{\epsilon}{3} < x^2 < 25+\dfrac{\epsilon}{3}$ $\sqrt{25-\dfrac{\epsilon}{3}} < x < \sqrt{25+\dfrac{\epsilon}{3}}$ In these three steps, we solve for the variable $x$, by first adding 75 to each expression, then dividing each expression by 3, and finally taking the square root of each expression. The square root function is increasing on all real numbers, so the inequality does not change direction. If you are using a decreasing function, the inequality signs will switch direction. Notice that the two ends of the inequality are no longer opposites of one another, which means that absolute values could not be used to write these as a single inequality. Also, the left hand expression can be undefined for some values of epsilon, so we must be careful in defining epsilon.
$-5+\sqrt{25-\dfrac{\epsilon}{3}} < x-5 < -5+\sqrt{25+\dfrac{\epsilon}{3}}$ Since our short-term goal was to obtain the form $|x-c|<\delta$, and the value of $c$ is 5, we subtract 5 from each expression.
$|x-c|<\delta$$-\delta < x-c < \delta$ Since the two ends of the expression above are not opposites of one another, we cannot put the expression back into the form $|x-c|<\delta$. Therefore, we shall expand this absolute value expression instead.
$\delta=\min\left\{5-\sqrt{25-\dfrac{\epsilon}{3}},-5+\sqrt{25+\dfrac{\epsilon}{3}}\right\}$ There are two candidates for delta, and we need delta to be less than or equal to both of them. Therefore, we will require that delta be equal to the minimum of the two quantities. Notice that since the left-end expression was equivalent to negative delta, we used its opposite in our definition of delta. However, since the first candidate is undefined for $\epsilon > 75$, we will need to handle the "large epsilon" situation by introducing a second, smaller epsilon in the proof.
Now, we are ready to write the proof.
Proof.
Suppose $\epsilon >0$ has been provided. This is always the first line of a delta-epsilon proof, and indicates that our argument will work for every epsilon.
Let $\epsilon_2=\min\{\epsilon, 72\}$. To avoid an undefined delta, we introduce a slightly smaller epsilon when needed. In this example, the value of 72 is somewhat arbitrary, but does need to be smaller than 75.
Define $\delta=\min\left\{5-\sqrt{25-\dfrac{\epsilon}{3}},-5+\sqrt{25+\dfrac{\epsilon}{3}}\right\}$. Since the definition of the limit claims that a delta exists, we must exhibit the value of delta. We use the value for delta that we found in our preliminary work above, but based on the new second epsilon. Therefore, this delta is always defined, as $\epsilon_2$ is never larger than 72.
Since $\epsilon_2 >0$, then we also have $\delta >0$. The definition does place a restriction on what values are appropriate for delta (delta must be positive), and here we note that we have chosen a value of delta that conforms to the restriction. The result is not real obvious, but can be seen as follows. Inside the square root expressions above, when subtracting from 25, the square root will be slightly smaller than 5, so the first delta candidate is positive. When adding to 25, the square root in the second candidate will be slightly larger than 5, so the second delta candidate is also positive. Therefore, their minimum is also positive.
Now, for every $x$, the expression $0 < |x-c| < \delta$ implies This is the next part of the wording from the definition of the limit.
$-\delta < x-c < \delta$ $-5+\sqrt{25-\dfrac{\epsilon_2}{3}} < x-5 < -5+\sqrt{25+\dfrac{\epsilon_2}{3}}$ When we have two candidates for delta, we need to expand the absolute value inequality so we can use both of them. Then we replace the values of c and delta by the specific values for this problem. From here on, we will be basically following the steps from our preliminary work, but in reverse order.
$\sqrt{25-\dfrac{\epsilon_2}{3}} < x < \sqrt{25+\dfrac{\epsilon_2}{3}}$ $25-\dfrac{\epsilon_2}{3} < x^2 < 25+\dfrac{\epsilon_2}{3}$ $75-\epsilon_2 < 3x^2 < 75+\epsilon_2$$-\epsilon_2 < 3x^2-75 < \epsilon_2$ We added 5 to each expression, then squared each expression, then multiplied each by 3, then subtracted 75.
$|3x^2-75| < \epsilon_2 \le \epsilon$ Now we recognize that the two ends of our inequality are opposites of each other, so we can write the result as a single absolute value inequality. Furthermore, $\epsilon_2$ is always less than or equal to the original epsilon, by the definition of $\epsilon_2$.
$|(3x^2-1)-74| < \epsilon$ Then we rewrite our expression so that the original function and its limit are clearly visible.
Therefore, $\lim\limits_{x\to 5} (3x^2-1)=74$. Having reached the final statement that $|f(x)-L| < \epsilon$, we have finished demonstrating the items required by the definition of the limit, and therefore we have our result.
Q.E.D. Our proof is complete.
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http://math.stackexchange.com/questions/6145/what-is-the-name-of-this-lattice?answertab=active
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# What is the name of this lattice?
Suppose we have an atom at every point with integer coordinates in $\mathbb{R}^d$. Take a ($d-1$)-dimensional hyperplane going through $\mathbf{0}$ and orthogonal to $(1,1,1,\ldots)$. What is the name of the lattice formed by atoms in that plane?
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– Rahul Narain Oct 5 '10 at 21:20
Thanks, that's what I was looking for. Search for "permutohedral" seems to give CS papers exclusively, so I'd guess $A_d$ is the mathematical name – Yaroslav Bulatov Oct 5 '10 at 21:53
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– yasmar Oct 5 '10 at 22:51
Oh, as Rahul already mentions the "permutohedral" lattice is $dA^*{d-1}$, presumably the astrisk is significant. – yasmar Oct 5 '10 at 23:05
Yes, it denotes the "dual" of the original lattice. – Robin Chapman Oct 6 '10 at 7:04
## 1 Answer
It's called $A_{d-1}$. See http://en.wikipedia.org/wiki/Root_system#An .
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To be precise, $A_{d-1}$ doesn't consist of all integer vectors in the hyperplane, only those with zeros as coordinates except for one +1 and one -1. To obtain the whole lattice one can take integer-coefficient linear combinations of the vectors in $A_{d-1}$, but I don't know if that has a name of its own. – Hans Lundmark Oct 6 '10 at 5:38
Hans, the name is "the $A_{d-1}$ lattice" :-) – Robin Chapman Oct 6 '10 at 6:53
@Hans Lundmark: A_{d-1} is an object with more structure than a lattice, namely it comes with a distinguished set of generators. But as far as it refers to the entire root system it refers to all of the integer vectors in the hyperplane in question. – Qiaochu Yuan Oct 6 '10 at 9:44
OK, I take it back. But to my defence, I can say that I was just reading the definition of the $A_n$ root system in the Wikipedia article that you yourself linked to. :-) However, as that article also says, the terminology doesn't seem completely fixed ("some authors omit condition this or that in the definition"), and of course if one talks about the $A_n$ lattice it's clear what is meant, so I shouldn't have been so categorical in my statement. – Hans Lundmark Oct 7 '10 at 8:53
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http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s11aac.html
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# NAG Library Function Documentnag_arctanh (s11aac)
## 1 Purpose
nag_arctanh (s11aac) returns the value of the inverse hyperbolic tangent, $\mathrm{arctanh}x$.
## 2 Specification
#include <nag.h>
#include <nags.h>
double nag_arctanh (double x, NagError *fail)
## 3 Description
nag_arctanh (s11aac) calculates an approximate value for the inverse hyperbolic tangent of its argument, $\mathrm{arctanh}x$.
For ${x}^{2}\le \frac{1}{2}$ the function is based on a Chebyshev expansion.
For $\frac{1}{2}<{x}^{2}<1$,
$arctanhx = 1 2 ln 1+x 1-x .$
## 4 References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
## 5 Arguments
1: x – doubleInput
On entry: the argument $x$ of the function.
Constraint: $\left|{\mathbf{x}}\right|<1.0$.
2: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_REAL_ARG_GE
On entry, $\left|{\mathbf{x}}\right|$ must not be greater than or equal to 1.0: ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
The function has been called with an argument greater than or equal to 1.0 in magnitude, for which arctanh is not defined. The result is returned as zero.
## 7 Accuracy
If $\delta $ and $\epsilon $ are the relative errors in the argument and result, respectively, then in principle
$ε ≃ x 1 - x 2 arctanhx δ .$
That is, the relative error in the argument, $x$, is amplified by at least a factor
$x 1 - x 2 arctanhx$
in the result. The equality should hold if $\delta $ is greater than the machine precision ($\delta $ due to data errors etc.), but if $\delta $ is simply due to round-off in the machine representation then it is possible that an extra figure may be lost in internal calculation round-off.
The factor is not significantly greater than one except for arguments close to $\left|x\right|=1$. However, in the region where $\left|x\right|$ is close to one, $1-\left|x\right|\sim \delta $, the above analysis is inapplicable since $x$ is bounded by definition, $\left|x\right|<1$. In this region where arctanh is tending to infinity we have
$ε ∼ 1 / lnδ$
which implies an obvious, unavoidable serious loss of accuracy near $\left|x\right|\sim 1$, e.g., if $x$ and 1 agree to 6 significant figures, the result for $\mathrm{arctanh}x$ would be correct to at most about one figure.
None.
## 9 Example
The following program reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.
### 9.1 Program Text
Program Text (s11aace.c)
### 9.2 Program Data
Program Data (s11aace.d)
### 9.3 Program Results
Program Results (s11aace.r)
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http://mathoverflow.net/questions/27261/groups-related-to-sum-of-squares-function
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## Groups related to sum of squares function?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I will begin by saying that $k=3$ might be a very specific case and this question is useless. Even if that is the case, I would like to know...
The sum of squares function $r_k(n)$ is very famous. It counts the number of ways $n$ can be written as a sum of $k$ squares. In the case of $k=3$, when $n$ is squarefree and not $7\mod{8}$, $r_k(n)$ is related to the class number of $\mathbb{Q}(\sqrt{-n})$. In the next (at least) two odd cases the function is still related to arithmetic constants of quadratic fields. C.f. "On the Representation of a Number as the Sum of any Number of Squares, and in Particular of five or seven", Hardy, 1918.
### Question
Are the numbers $r_k(n)$ known to be related to special groups, like when $k=3$?
### Smaller Question
Is there a book with an in-depth account of these numbers and their arithmetic significance? (more than expressing them as coefficients of a modular form and proving bounds and (lots of) relations...)
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1
There's the book Sums of Squares of Integers by Moreno and Wagstaff: books.google.co.uk/… but I'm not sure that does exactly what you ask for. – Robin Chapman Jun 6 2010 at 18:31
hmm, for k=4 it is the sum of the divisors of n not divisible by 4. This comes from the fact that the generating function is a modular form on a suitable congruence subgroup of SL_2(Z). So at least it has some connection to an algebraic object. – Tobias Kildetoft Jun 6 2010 at 22:34
@Tobias: All the numbers $r_k(n)$ come from modular forms. Koblitz's book Introduction to Elliptic Curves and Modular Forms contains information on the theta function and its powers. – Dror Speiser Jun 7 2010 at 6:42
Dror, I simply cite the Chan-Krattenthaler article (mat.univie.ac.at/~kratt/artikel/SquareSurvey.html): "These formulas (for the number of representations by triangular numbers $t_{4s^2}(n)$ and $t_{4s(s+1)}(n)$ - WZ) follow from a conjectural affine denominator formula for simple Lie superalgebras..." Isn't that a sufficient algebraic context? The situation with squares is very similar (although I am not sure that they are directly related to Kac-Wakimoto). – Wadim Zudilin Jun 7 2010 at 7:30
Is there any reference for this? That is the only statement in the article about Lie superalgebras, and the reference they give is only about the structure, and not the comment. – Dror Speiser Jun 7 2010 at 20:15
show 1 more comment
## 3 Answers
In my view, it depends a little what you mean by "related," but I don't see at first glance any natural group whose order is r_k(n) for any k other than 3. Loosely speaking, representations of a form of rank m by the genus of a form of rank n are related to the set of double cosets
H(Q) \ H(A_f) / H(Zhat)
where H is a form of SO_{n-m} (this can be found e.g. in my paper with Venkatesh "Local-global principles..." but is certainly known to others). When H is abelian (i.e. when n-m = 2) this is naturally a group, otherwise not.
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Little harm in repeating this, Dror was quite pleased with your paper "Local-global principles..." and torsors, see the (currently) sixth comment after my answer. – Will Jagy Jun 11 2010 at 1:57
### 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.
Steve Milne sent me a copy, and a pdf, of his "Infinite Families of Exact Sums of Squares Formulas, Jacobi Elliptic Functions, Continued Fractions, and Schur Functions" which is an entire issue of The Ramanujan Journal: vol. 6, no. 1, March 2002. There is also a two-page preface by George Andrews. I admit, the main focus is dimension $4 n^2$ or $4 ( n^2 + n).$ But at 143 pages and 259 references, there might be something you like. Email me if you would like the pdf, it is not obvious to me that there was implied permission for me to post it on any websites.
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Here is the link to the paper: dx.doi.org/10.1023/A:1014865816981. There are simpler proofs of the result (probably not mentioned in Milne's paper) by D. Zagier [*Math. Res. Lett.* 7 (2000) 597--604], K. Ono [*J. Number Theory* 95 (2002) 253--258], and H.H. Chan and C. Krattenthaler [*Bull. London Math. Soc.* 37 (2005) 818--826]. – Wadim Zudilin Jun 7 2010 at 3:46
A paper on five, seven and nine squares is Shaun Cooper's "Sums of five, seven and nine Squares" also in the Ramanujan Journal vol. 6 (2002) 469-490. – Robin Chapman Jun 7 2010 at 6:43
You might also want to have a look at the very nice paper of Bodo Lass , Démonstration de la conjecture de Dumont, C. R. Math. Acad. Sci. Paris 341 (2005), no. 12, 713--718. – Roland Bacher Jun 7 2010 at 8:42
Thanks, the articles are very impressive! I looked over them and reading more carefully now. Just a note, none mention any arithmetic groups. – Dror Speiser Jun 7 2010 at 20:13
As you like groups, you might look at Ellenberg and Venkatesh arxiv.org/pdf/math/0604232 and the follow-up by Schulze-Pillot, arxiv.org/pdf/0804.2158 although in these the groups may not be the type you want and these focus on existence of representations, rather than counting them when existence is already known. Oh, well. My earlier comment stands as far as my own limited background, I haven't seen anything on five or seven squares that screams number field. – Will Jagy Jun 7 2010 at 20:44
show 3 more comments
For general odd $k=2m+1$ one can still compute the value of the singular series as Hardy does and obtain a formula similar to those for $k=5,7$, involving the value of an $L$-series with quadratic character at $s=m$. A relation to special groups as in the case $k=3$ is not visible from this, just a relation to the arithmetic of quadratic number fields.
For $k \ge 9$, however, the genus of the sum of $k$ squares contains more than one integral equivalence class and by Siegel's Massformel (mass formula) evaluation of the singular series gives the average of the representation numbers for the equivalence classes in the genus and not the representation number of the individual form. (The genus of an integral quadratic form $q$ consists of those forms which have the same signature and are integrally equivalent modulo $m$ for all integral $m$.) Of course this doesn't exclude the possibility of finding a closed formula by other means, as was the case for the special dimensions of the form $4m^2$ or $4(m^2+m)$ in the work of Milne mentioned in Jagy's answer. To my knowledge at present no such formula is known for an odd number $k$ of variables.
Concerning references: The standard reference for sums of squares is still Grosswald's book. Good references for more general questions concerning the arithmetic of quadratic forms are the books of B. Jones, Y. Kitaoka, O. T. O'Meara, G. Shimura and (in german) of M. Eichler and of M. Kneser.
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It's been a long time since I looked at Grosswald's book, so I may misremember this, but doesn't he look at the sums of $\textit{positive}$ squares? – Victor Protsak Jun 10 2010 at 22:08
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http://mathoverflow.net/revisions/13481/list
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## Return to Answer
2 added 11 characters in body
In fact this is a difficult question in general, I believe. For example, if $X$ is a closed subvariety of ${\mathbb P}^n$, then asking for the minimal number of affine opens that cover the complement ${\mathbb P}^n$ P}^n\setminus X$is the same as asking for the minimal number of hypersurfaces whose (set-theoretic) intersection equals$X\$. I believe this question is open in general.
1
In fact this is a difficult question in general, I believe. For example, if $X$ is a closed subvariety of ${\mathbb P}^n$, then asking for the minimal number of affine opens that cover the complement ${\mathbb P}^n$ is the same as asking for the minimal number of hypersurfaces whose (set-theoretic) intersection equals $X$. I believe this question is open in general.
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http://quant.stackexchange.com/questions/tagged/joint-probability
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# Tagged Questions
The joint-probability tag has no wiki summary.
1answer
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I have two stocks, A and B, that are correlated in some way. If I know (hypothetically) that stock A has a 60% chance of rising tomorrow, and I know the joint probability between stocks A and B, how ...
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http://mathoverflow.net/questions/84394?sort=oldest
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## Ideal class groups and extension of number fields
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
[I already posted this question on stackexchange a while ago, but did not get any response: http://math.stackexchange.com/questions/93437/ideal-class-groups-and-extension-of-number-fields]
Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be schemes and `$ f:X \to Y$ be a morphism. Suppose $f^\#:\mathcal{O}_Y \to f_*\mathcal{O}_X $`is injective. Then so is the restriction $\mathcal{O}_Y^* \to f_*\mathcal{O}_X^*$ and we can complete to an exact sequence $1 \to \mathcal{O}_Y^* \to f_*\mathcal{O}_X^* \to \mathcal{Q} \to 1$. From this we get an exact sequence in cohomology starting as `$ 1 \to H^0(\mathcal{O}_Y^*) \to H^0(f_*\mathcal{O}_X^*) \to H^0(\mathcal{Q}) \to H^1(\mathcal{O}_Y^*) \to H^1(f_*\mathcal{O}_X^*) \to H^1(\mathcal{Q}) $`.
Question(s)
Consider the case where $L/K$ is an extension of number fields with rings of integers $\mathcal{O}_L$, $\mathcal{O}_K$, $X = Spec(\mathcal{O}_L)$, $Y = Spec(\mathcal{O}_K)$ and $f$ induced from the inclusion $\mathcal{O}_K \to \mathcal{O}_L$.
1. How do $H^1(f_*\mathcal{O}_X^*)$ and $H^1(\mathcal{O}_X^*) = Pic(X)$ relate? (In particular, are they equal?)
2. Can we describe $\mathcal{Q}$ explicitely? In particular, is there a good expression for $Q = H^0(\mathcal{Q})$? Does $H^1(\mathcal{Q}) = 0$?.
Examples/Motivation
The motivation for considering the above is to end up with an interesting exact sequence relating the Picard and unit groups of $X$ and $Y$ via $Q$. Nice results are obtainable for example if $Y$ is the spectrum of a Dedekind domain and $X$ is an open subset or if $Y$ is the spectrum of a one-dimensional integral domain, $X$ its normalisation, and some finiteness conditions hold. These two cases are worked out "by hand" in Neukirch, algebraic number theory, sections 1.11 and 1.12.
A start on question 1
This is probably not the most elegant method, but the five-term exact sequence from the Leray spectral sequence starts as `$ 1 \to H^1(f_*\mathcal{O}_X^*) \to H^1(\mathcal{O}_X) \to \Gamma(Y, R^1f_*\mathcal{O}_X^*) $`. A sufficient condition for `$ Pic(X) = H^1(f_*\mathcal{O}_X^*) $` is thus that the stalks of the first higher direct image are trivial, which is easily seen to be equivalent to $Pic((f_*\mathcal{O}_X)_p) = 0$ for all $p \in Y$. This does hold for an open immersion of number rings, but it is not clear to me if it applies to a normalisation or to an extension of number fields.
It is also easy to see that $H^1(\mathcal{F}) = H^1(f_* \mathcal{F})$ for all $\mathcal{F}$ if $f_*$ is exact, for example a closed immersion. This does not seem applicable here either, though.
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1
Pushforwards are exact along a finite morphism if you work in the etale topology (or Nisnevich, but not Zariski). So if you're working in the etale topology, the answer to (1) is always "yes," without much thought. In fact, even in the Zariski topology, in the special case at hand, one is in good shape: the stalks of R^1 f_* O_X* are given by Picard groups of semilocal rings of X; such rings are always PIDs (as X is Dedekind), so R^1 f_* O_X* does indeed vanish. – Bhargav Dec 27 2011 at 16:46
Thanks, these are two very helpful things to know. – Tom Bachmann Dec 28 2011 at 15:38
## 1 Answer
if your aim is really to relate $Pic(Y)$ to $Pic (X)$ it is probably a good idea to pybass the use of $\mathcal Q$ and to consider instead the more powerful technique of introducing the relative Picard scheme (functor) $Pic_{X/Y}$. There are plenty of references, among them :
Néron models.
Bosch; Lütkebohmert ; Raynaud,
Ergebnisse der Mathematik und ihrer Grenzgebiete Springer-Verlag, Berlin, 1990.
for instance :
chapter 8 page 204 proposition 4 if $f:X\to Y$ is quasi-compact and quasi-separated and $f_*(\mathcal O_X)=\mathcal O_Y$ then there is an exact sequence :
$$0 \to Pic (Y) \to Pic (X) \to Pic_{X/Y}(Y) \to Br(Y) \to Br(X)$$
If you are interested, the standard is FGA (Grothendieck's Fondations de la Géométrie Algébrique), and the modern version in
The Picard scheme
Steven L. Kleiman
http://arxiv.org/abs/math/0504020
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Well my aim is "really" to learn about algebraic geometry and algebraic number theory; so I just "follow all leads". I'll look at the reference(s) soon. – Tom Bachmann Dec 28 2011 at 15:44
In fact the exact sequence you quote is the five-term exact sequence of the Leray spectral sequence for the Etale topology. Reading Kleiman now; this material seems very interesting and a good way to learn about lots of stuff. I gladly accept your answer. – Tom Bachmann Dec 30 2011 at 16:18
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http://www.physicsforums.com/showthread.php?p=4172122
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Physics Forums
## Derivation of normal strain
http://imgur.com/SnHyP
What are the mathematical steps and assumptions to reach the conclusion that length(ab) ≈ dx + ∂u/∂x*dx ?
If you consider the the squares of the gradients to be negligible, you still have a square root and multiplication by the constant "2". What other assumptions do we make to derive the final equation?
*Edit, I should have posted this in calculus, I apologize.
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
You were right to post this in engineering : it is an engineering issue. The statement $${\rm{length(ab)}} = \sqrt {{{\left( {dx + \frac{{\partial {u_x}}}{{\partial x}}dx} \right)}^2} + {{\left( {\frac{{\partial {u_y}}}{{\partial x}}dx} \right)}^2}}$$ is nothing more than pythagoras theorem for the horizontal and vertical components of ab. I assume you are comfortable with this. Now AB was originally horizontal and it is stated that the strain is very small. Thus the angles alpha and beta in the diagram are very small. If beta is very small then the hypotenuse (ab) is very nearly the same as the horizontal component, which is $$dx + \frac{{\partial {u_x}}}{{\partial x}}dx$$ Thus $${\rm{length(ab)}} \approx dx + \frac{{\partial {u_x}}}{{\partial x}}dx$$
OP, I think your question is simply: how did they go from: $length(ab)= \sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx\right)^2+\left(\frac{\partial u_y}{\partial x}dx\right)^2}$ to $length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx$ ? The math doesn't work, I agree. From a geometric point-of-view, as Studiot suggested, they assume that $length(ab)$ is equal to its horizontal projection, for small deformations. However, I'm not sure that I buy that, to be honest. In terms of the actual physics, and in looking at the provided diagram, I can tell you that if it were only a simple shear, you could get the angle change $\alpha + \beta$ but there has to be some sort of homogeneous (axial) deformation in order for BOTH $dx$ and $dy$ to change lengths. For example, one way to arrive at the apparent deformed shape would be: 1) apply a homogenous deformation (e.x. to the right, of magnitude ($length(ab)-dx$) -- i.e. $\sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx\right)^2+\left(\frac{\partial u_y}{\partial x}dx\right)^2}-dx$) 2) apply a simple shear (e.x. to the right, of amount $\alpha + \beta$) 3) apply a rigid body rotation (e.x. counter-clockwise, of amount $\alpha$) Does that make sense? You can play with this though. Take 1) to be zero. No deformation to the right means $length(ab)=dx$. 2) and 3) still apply - and so we have a simple shear and a rigid body rotation. We should still get $length(ab)=dx$ in this case under either a small shear or a large shear. However, due to the rigid body rotation, $\frac{\partial u_x}{\partial x}dx$ in their diagram would be nonzero and so their expression $length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx$ is not equal to $dx$. This doesn't mean that they are wrong, but I cannot immediately justify approximating $length(ab)$ as its horizontal projection, for the general case that they are showing. In other words, I don't like their expression $length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx$ unless someone can prove to me that it agrees with more advanced solid mechanics.
## Derivation of normal strain
OP, I think your question is simply: how did they go from: ........ to ...........
I thought this at first but came to the conclusion that the source material authors were simply replacing or substituting a simpler calculation, not simplyfying a more complicated one.
This is not unusual, for instance the substitution of the chord for the arc or the other way in circular calculations of small angle.
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http://math.stackexchange.com/questions/227032/how-many-iterations-can-i-have-in-a-year
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# How many iterations can I have in a year
I am no math wiz at all.
I have an 8 day set iteration that I want to run as many times in a year as I can. I can run 2 at a time an I can start 2 more every 3rd day.
So day 1 I start 2 iterations, day 3 I start 2 more, Day 9 I start a 3rd, day 11 I start a 4th. How many can I run per year?
EDIT to clarify and add context-
I'll try to make it more of a word problem. I brew 10 gallons of beer at a time and I have 4 fermenters. Each batch ferments for 6 days. After the beer ferments it has to clear in another tank of which I have 2 and the beer has to sit in this clearing tank for 2 days.
so I have 4 beers in fermentation but I can only move 2 into clearing tanks for 2 days, then after that I can move the other two into the clearing tanks.
At the point where I empty a fermenter I want to refill it.
How many times can I do all this in a year, and whats the equation?
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1
Is it a leap year? If you start on day 1, does it end on day 8 or day 9? – Ben Voigt Nov 1 '12 at 20:34
I don't want to worry about leap year. – TJ Sherrill Nov 1 '12 at 21:49
## 1 Answer
If you can only run $2$ at a time, how can you start more on day $3$? You have four running on day $4$. It sounds like you start $2$ on days $1,9,17,25 \ldots$, which are the days of the form $8k+1$. You have to start them before day $357$. What is the greatest $8k+1$ that is less than $357$?
Added: if you can have four batches in process but have only two clearing tanks, you should start two each on day $1, 3, 9, 11, \ldots , 8k+1, 8k+3.$ They finish on day $8k+8$ and $8k+10$. So find the highest $k$ that each series is done within the year. In this case, for each series it is $44$, so you can make $4 \cdot 44=176$ batches in a year.
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edit question, to clarify, thanks for your help. – TJ Sherrill Nov 1 '12 at 21:50
+1 Ross, in part for helping me understand the problem statement! Just curious: would there be any advantage to starting $1$ iteration each $8K+1^{\text{st}}$ day ($k\geq 0$), and another every $8k+3^{\text{rd}}$ day? I guess I'd need to "do the math"! – amWhy Nov 1 '12 at 21:52
@amWhy: Generally you want to start them as early as possible so they stop as soon as possible. It may not matter. If you were asking about 12 days, for example, you could start two processes any time up to the 5th day and they would finish by day 12. Earlier wouldn't help here. But waiting never gets more done. – Ross Millikan Nov 2 '12 at 1:25
thanks for the help – TJ Sherrill Nov 2 '12 at 2:48
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http://www.cfd-online.com/Wiki/Rayleigh_number
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Home > Wiki > Rayleigh number
# Rayleigh number
### From CFD-Wiki
The Rayleigh number is the ratio of the buoyancy force to the viscous force in a medium.
$Ra = \frac{g \alpha \Gamma d^4}{\nu \kappa}$
where $g =$ acceleration due to gravity, $\alpha =$ coefficient of thermal expansion, $\Gamma =$ background temperature gradient, $d =$ length scale of flow, $\nu =$ kinematic viscosity and $\kappa =$ thermal diffusivity
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