User gerald edgar - MathOverflow

Answer by Gerald Edgar for Basic results with three or more hypotheses

2011-03-25T16:02:31Z

Answer by Gerald Edgar for Principal value of integral

2013-05-14T16:42:55Z

Yes the value $\pi/2$ can be obtained like this.

Let $$ f(x):=\frac {1}{x^2}-\frac{\cot(x)}{x} $$

We may compute $$ \int_0^{\pi/2} f(x)\;dx + \sum_{k=1}^\infty \int_0^{\pi/2}\big(f(k\pi+x)+f(k\pi-x)\big)\;dx=\frac{\pi}{2} \tag{1}$$ and this converges.

We can think of (1) as a "rearrangement" of the required integral. But the integrands in (1) are positive: Use $\cot x > 0$ for $0 < x < \pi/2$ and $\cot(k\pi+x) = \cot x$ and $\cot(k\pi-x) = -\cot x$. Also $(1/x) - \cot x$ increases from $0$ on $(0,\pi/2)$, so $f(x) >0$ on $(0,\pi/2)$. Next $$ f(k\pi+x)+f(k\pi-x) = \left(\frac{1}{(k\pi+x)^2}+\frac{1}{(k\pi-x)^2}\right) + \left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)\cot x $$ and each of the two halves is positive on $(0,\pi/2)$. Recall $$ \sum_{k=1}^\infty \left(\frac{1}{(k\pi+x)^2}+\frac{1}{(k\pi-x)^2}\right) = \csc^2 x - \frac{1}{x^2} $$

$$ \sum_{k=1}^\infty\left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)=\frac{1}{x}-\cot x $$

$$ \sum_{k=1}^\infty\left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)\cot x=\frac{\cot x}{x}-\cot^2 x $$

Our answer is the sum of three integrals:

$$ \int_0^{\pi/2} \left[\left(\frac{1}{x^2}-\frac{\cot x}{x}\right)+\left(\csc^2 x-\frac{1}{x^2}\right)+\left(\frac{\cot x}{x}-\cot^2 x\right)\right]dx = \int_0^{\pi/2} 1\;dx = \frac{\pi}{2} $$

Answer by Gerald Edgar for The pth power of a distance function is twice continuously differentiable, for $p>2$?

2013-05-14T15:55:34Z

How about in $\mathbb R$ the open set is $(-2,-1) \cup (1,2)$. Study $\beta(x)$ at $x=0$.

Answer by Gerald Edgar for What fields can be used for an inner product space?

2013-05-02T15:21:15Z

Of course if you insist on condition $\langle \mathbf{x},\mathbf{x}\rangle > 0$, and not merely $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$, then you must have an order.

Let $F$ be a formally real field. Then $$ \langle \mathbf{x}, \mathbf{y}\rangle = \sum_{j=1}^n x_j y_j $$ can be a reasonable inner product on $F^n$. According to an ordering for $F$ (indeed, any ordering, since there may be more than one) we have $\langle \mathbf{x}, \mathbf{x}\rangle > 0$ if $\mathbf x \ne \mathbf 0$.

Another part that you quote is what would be required for metric completeness. Do you want that? If $F$ is a proper subfield of $\mathbb R$, then even the one-dimensional space is not complete.

Something weaker than completeness will be enough to carry out the Gram-Schmidt process. It requires only that square-roots of $\langle \mathbf{x}, \mathbf{x}\rangle$ exist.

Answer by Gerald Edgar for Identity of binomial series with factorial.

2013-04-29T16:13:36Z

Darij's comment...

Tthe truncated exponential series: $$ e_p(z) = \sum_{n=0}^p\frac{z^n}{n!} $$ Your sum is $$ p!x^pe_p(1/x) $$

Answer by Gerald Edgar for Continuous automorphisms of Q*

2013-04-24T14:18:37Z

I assume by "continuous" you mean in the topology inherited from the usual topology on $\mathbb R$. A continuous homomorphism will be uniformly continuous (in the natural uniformm structure for a group). And therefore extend continuously to the complettion, which will be $\mathbb R^\ast$, the nonzero reals under multiplication. Same for the inverse map, since it is assumed continuous as well. So we get an extension that is a continuous automorphism of $\mathbb R^\ast$. Such an automorphism is of the form $x \mapsto x^c$ on $x>0$, where $c$ is a nonzero constant. (And appropriately for $x<0$.) But the only maps of this form that map rationals bijectively to rationals are $c=1$ and $c=-1$, I guess.

Answer by Gerald Edgar for Is function from topological group to metric space Borel?

2013-04-21T19:05:58Z

Let's try this. I'm not using the hypothesis in your second paragraph, so maybe I am missing something.

Suppose $G$ is pseudometrizable but not metrizable. The the closure of ${e}$, (the identity), is a closed subgroup $N$ of $G$. And every open set in $G$ either contains $N$ or is disjoint from $N$. Then this same thing is true for every Borel set. On the other hand, for any two points of $X$, there is an open set that contains one but not the other. So, whatever bijection $f$ we choose, it is not Borel.

Answer by Gerald Edgar for textbooks on asymptotic expansions

2013-04-16T17:20:25Z

http://www.amazon.com/Asymptotics-Summability-Monographs-Surveys-Mathematics/dp/1420070312/

Asymptotics and Borel Summability, O. Costin

Answer by Gerald Edgar for Formal writing: numbers under 10

2013-03-06T13:10:20Z

Also in engineering watch for units... "The two rods had diameter 5 mm." "A 6-volt battery was used."

Answer by Gerald Edgar for Defining definite integral using indefinite integral.

2013-03-05T18:59:48Z

In case the exceptional set $C$ is empty, I have seen this called the Newton integral.

It can exist in cases where the Riemann integral does not exist, also in cases where the Lebesgue integral does not exist. There is an integral (the generalized Riemann integral or gauge integral or Kurzweil integral or Henstock integral) that does have that theorem: if $f = F'$ everywhere, then the integral exists and satisfies $\int_a^b f(x)dx = F(b)-F(a)$. An elementary reference: Generalized Riemann Integral by R. M. McLeod.

If $f$ is Riemann integrable, then it is true that there is a null set $C$ (but not, in general countable) such that $F'(x)=f(x)$ for all $x \in [a,b]\setminus C$ where $$ F(x) = \int_a^x f $$
and this $F$ is of course continuous. The same for Lebesgue integral.

In measure spaces $(X, \mathcal F,\mu)$, the usual thing is not to connect function $f$ to another function $F$ where $F'=f$, but to connect a function $f$ and a signed measure $\nu$ where $$ \int_A f d\mu = \nu(A)\qquad\text{for all $A \in \mathcal F$} $$

Answer by Gerald Edgar for Is there a constructive proof of Cantor–Bernstein–Schroeder theorem ?

2013-03-03T17:48:00Z

Failure of the law of the excluded middle is hard for me to imagine. But here is a try...

That König proof has this:

For any particular $a$, this sequence may terminate to the left or not

But perhaps we also have to consider the case of both: it terminates to the left and also does not terminate to the left... ???

Answer by Gerald Edgar for Extension of measures from the ball sigma-algebra to the borel sigma-algebra

2013-02-28T19:52:30Z

Take a set $X$ of power $\aleph_1$, with the discrete metric where two distinct points have distance $1$. The balls are singletons and the whole space. The ball sigma-algebra is the countable and co-countable sets. Let countable sets have measure zero, co-countable sets have measure 1.

Now all subsets are open, so the Borel sigma-algebra is the power set. There is no extension of this measure!

Answer by Gerald Edgar for Maximal ideals of the algebra of measurable functions

2013-02-19T13:41:30Z

If, instead of all measurable functions, you use the bounded measurable functions, then you probably want to consult the literature known as "lifting" ... for example:

Topics in the Theory of Lifting (Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge)

Answer by Gerald Edgar for problems from the scottish book

2013-02-18T01:20:05Z

The book version edited by Daniel Mauldin (from 1982) has commentaries on the problems as of that date.

Answer by Gerald Edgar for Second order difference implies differentiability

2013-02-16T20:12:11Z

I am assuming you mean $|f(x+2h)-2f(x+h)+f(x)|\le |h|^{3/2}$.

Well, what if $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$? Certainly your inequality holds then. But (according to the Axiom of Choice) there are badly discontinuous functions like this.

Answer by Gerald Edgar for Good examples of random variables whose image is not a measurable set?

2013-01-29T19:25:13Z

An analytic set that is not a Borel set...see this post from long ago.

Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.

Answer by Gerald Edgar for Mean value theorems for the Haar integral?

2013-01-28T18:06:30Z

Say $\mu$ is a Borel probability measure on a connected set $A$ in a topological space. Let $f : A \to \mathbb R$ be continuous. Then the mean value $\int_A f\;d\mu$ is equal to $f(a)$ for some $a \in A$. Proof: the mean value is between the sup of all values and the inf of all values, so (by connectedness) it is a value of the function.

Of course the desired result fails for non-connected sets. Even in the two-point group we get a counterexample.

Answer by Gerald Edgar for Translation of a non-standard analysis formulation

2013-01-23T22:05:12Z

How about this ...

Let's say a finite set $V \subseteq \mathbb N$ is a Feldmann set iff there are infinitely many $n$ such that for all $u \in V$, $n+u$ is prime.

For example, $\{0,2\}$ is a Feldmann set iff there are infinitely many twin primes.

Your equivalent statment: there are arbitrarily large Feldmann sets.

Answer by Gerald Edgar for Apollonian gasket and the degree of convergence

2013-01-17T02:09:03Z

This critical value is $\alpha_0 \approx 1.3056867$ ... For $\alpha > \alpha_0$ the series converges, for $\alpha < \alpha_0$ it diverges.

Before the inexpensive computer, it was difficult to tell whether the critical value is ${}> 1$ or not.

Boyd, David W. The sequence of radii of the Apollonian packing. Math. Comp. 39 (1982), no. 159, 249–254.

http://www.ams.org/mathscinet-getitem?mr=658230

added

mentioned in the comments... arbitrarily packed disks, not necessarily touching as in Apollonian packings. The critical value (= dimension of the residual set) is shown to be ${}> 1.02$.

Larman, D. G. On the Besicovitch dimension of the residual set of arbitrarily packed disks in the plane. J. London Math. Soc. 42 1967 292–302.

http://www.ams.org/mathscinet-getitem?mr=209982

Answer by Gerald Edgar for Measurable sets and Valuation Theory

2012-12-30T13:25:15Z

edit, January 13

Write $|\cdot|$ for the extended $2$-adic absolute value. I am assuming you mean $A = \{ (x,y) : |x|<1, |y|<1\}$, but since you say ``valuation'' maybe that is not right. Anyway, your set $A$ is simply related to my set $A$, perhaps by taking complements or multiplying by a constant.

Note that $A$ is a group under addition, since $|-x| = |x|$ and $|x+y| \le \max\{|x|,|y|\}$.

Assume (for purposes of contradiction) that $A$ is measurable.

If $A$ has positive measure we get a contradiction: indeed, the set $A - A$ contains a neighborhood of zero (for the usual topology). But $A-A = A$ and $A+A=A$, so $A$ is the whole plane, which is false.

Now consider sets $A_n = 2^{-n}A = \{(2^{-n}x,2^{-n}y): (x,y) \in A\}$. These sets are also groups under addition. The map $(x,y) \mapsto 2^{-n}(x,y)$ is an affine bijection, so all sets $A_n$ are Lebesgue measurable. But also note that multiplication is continuous with respect to $|\cdot|$, and $|2^n| \to 0$, and $A$ is a neighborhood of zero for the $|\cdot|$ topology. So for any $(x,y) \in \mathbb R^2$ there is $n$ so that $2^n(x,y) \in A$, and that means $(x,y) \in A_n$. Thus $$ \bigcup_{n=1}^\infty A_n = \mathbb R^2 . $$ A union of measurable sets. So some $A_n$ has positive measure. Get a contradiction as before.

Answer by Gerald Edgar for RFC for definite integral connection to second derivative

2013-01-12T12:43:57Z

Hint ... write $u(x) = f''(x)$, so that the condition is $$ \int_0^T u(t) g(x,y)dt = \int_0^x\left[\int_0^y u(s) ds\right]dy $$

Answer by Gerald Edgar for Are All Irrational Elementary Numbers Conjectured to Be Normal?

2013-01-07T22:21:27Z

Perhaps connected to general conjectures published here ... "On the Random Character of Fundamental Constant Expansions", David H. Bailey and Richard E. Crandall
Experiment. Math. Volume 10, Issue 2 (2001), 175-190.

Answer by Gerald Edgar for How has "what every mathematician should know" changed?

2013-01-07T18:11:38Z

In Littlewood's Miscellany there is an essay "A Mathematical Education" where he describes the situation before 1907.

Answer by Gerald Edgar for Which metric spaces have this superposition property?

2013-01-04T02:04:49Z

Like Euclidean geometry, also hyperbolic geometry has this extension property: an isometry defined on any subset extends to an isometry of the whole space. As I recall from long ago, in the book
Busemann & Kelly Projective Geometry and Projective Metrics
it is shown (among that class of geometries) there are very few of these spaces.

Answer by Gerald Edgar for Streamlined probability measure for tossing infinitely many coins

2012-12-31T22:25:04Z

This version is due to Emile Borel ...

Sequence of heads & tails encoded as 0s and 1s, then sequence is taken to represent a number in $[0,1]$ in its binary expansion. The measure is Lebesgue measure.

So you still need to know that Lebesgue measure exists.

Answer by Gerald Edgar for Approximating erf by tanh

2012-12-31T17:37:42Z

First, $$\begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align}$$ Subtract: $$ \mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt $$ So it suffices to show that this integrand is positive. It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.

Answer by Gerald Edgar for Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$

2012-12-30T13:42:03Z

beginning

Start with the unit cube $E$ in $\mathbb R^3$ and the unit interval $[0,1]$ in $\mathbb R$. Choose an injective map $\phi : E \to [0,1]$. This is the remaining question: do this so that $\phi$ is Riemann integrable. That is, the set of discontinuities has measure zero.

Then cover $\mathbb R^3$ by disjoint unit cubes, make them disjoint by including boundary points in only one of the possible cubes. Call these $F_n, n \in \mathbb N$. Choose disjoint intervals $I_n$ going to zero fast enough. Define $f$ using $\phi$ with translation and dilation to map $F_n$ injectively into $I_n$. If the original $\phi$ has set of discontinuities of measure zero, then this pieced-together function $f$ does too, since its set of discontinuities can be at most the discontinuities of a copies of $\phi$ in the interiors of the $F_n$, together with the boundary planes of the $F_n$. So $f$ is (improperly) Riemann integrable.

Answer by Gerald Edgar for Math for a cake

2012-12-29T20:22:09Z

Maybe just make the cake in the shape of a golden rectangle, and use two colors of icing to show the decomposition into a square and a smaller golden rectangle.

Answer by Gerald Edgar for a monotone relation for s-numbers

2012-12-27T19:55:52Z

I just tried a few random matrices... $$ \begin{align} &A=\begin{bmatrix} -0.1 & -0.4\cr -0.4&0\end{bmatrix}, \qquad B=\begin{bmatrix} 1.5 & -0.5 + i\cr -0.5 - i& 3.5\end{bmatrix}, \cr &\|A+iB\| = \frac{7}{2}+\frac{\sqrt{61}}{10}\approx 4.28 \cr &\|2A+iB\| = \frac{7}{2}+\frac{\sqrt{29}}{10}\approx 4.04 \end{align} $$

Answer by Gerald Edgar for Lebesgue differentiation theorem beyond Euclidean spaces

2012-12-26T14:10:49Z

A standard reference on derivation theory:
Hayes & Pauc, Derivation and Martingales (Springer 1970)

(plug) there is a chapter on derivation in:
Edgar & Sucheston, Stopping Times and Directed Processes (Cambridge Univ Pr 1992)

Comment by Gerald Edgar

2013-05-15T15:51:55Z

Needs some motivation or reason anyone should care about it. Perhaps some interpretation involving a random walk or ??.

Comment by Gerald Edgar

2013-05-14T21:48:19Z

So the set $\{\delta_x : x \in [0,1]\}$ is not compact; it is discrete.

Comment by Gerald Edgar

2013-05-14T14:22:06Z

@Carlo: that is perhaps a sensible definition. Can you cite a textbook that uses that definition?

Comment by Gerald Edgar

2013-05-14T14:18:58Z

$0$ is always in the spectrum (for a compact operator on an infinite-dimensional Hilbert space). For some such operators $0$ is an eigenvalue, but for others it is not.

Comment by Gerald Edgar

2013-05-14T12:18:46Z

Since there are multiple poles, I do not know what "principal value" means. Perhaps you can provide a definition?

Comment by Gerald Edgar

2013-05-13T16:59:48Z

For your first paragraph, specify that $S$ is a finite set. Then vertex and edge make sense.

Comment by Gerald Edgar

2013-05-13T16:56:45Z

@Lutz... you are right. Because andy has the wrong definition for the weak* topology. The usual way to define it is to use only continuous $Z$. If we use all measurable $Z$, as andy does, then you get a much stronger topology. Moreover, andy didn't even say that $Z$ should be bounded. So, let's give him a chance to say whether he wants to correct the definition.

Comment by Gerald Edgar

2013-05-13T13:28:35Z

Indeed, in Davide's example, the map $x \mapsto \delta_x$ is a homeomorphism from $[0,1]$ onto $\mathcal P$.

Comment by Gerald Edgar

2013-05-12T22:37:29Z

$p(z)=z^8+k$, for $k$ large, has no zeros inside a large square. But still $|p(0)|$ smaller than the specified points on the unit square.

Comment by Gerald Edgar

2013-05-12T16:48:17Z

Try polynomial $z^8+2$.

Comment by Gerald Edgar

2013-05-12T13:10:48Z

Denominator 16 ... maybe it is somehow related to these: en.wikipedia.org/wiki/…

Comment by Gerald Edgar

2013-05-12T12:39:09Z

If convexity is not required, then there are simpler counterexamples. The unit cube is the limit of a sequence of finite sets.

Comment by Gerald Edgar

2013-05-12T12:37:32Z

If you mean for $L$ to be convex, add that to the text of the question. Then we can use the fact that the boundary of a convex set has measure zero.

Comment by Gerald Edgar

2013-05-10T13:50:21Z

Let's hope we get an answer here. Things on Mathworld (and Wikipedia, and so on) stated without citation are not entirely reliable...

Comment by Gerald Edgar

2013-05-10T13:27:03Z

@Aerik: It is my experience that in elementary courses (like calculus) it is a Bad Idea to deviate from the textbook in any way.