## Math Genius: Bounding secant lines on cumulative distribution functions

Let $$F : mathbb Rto[0,1]$$ be the CDF of a continuous random variable, and let $$(a_n)$$ be a sequence converging to $$0$$ from above. Does there always exist an integrable function $$g : mathbb Rto [0,infty)$$ such that
$$frac{F(x + a_n) – F(x)}{a_n} leq g(x)$$
for all $$xinmathbb R$$ and $$ninmathbb N$$? In other words, I’m looking for an integrable dominating function for the slopes of these secant lines.

I know $$lim_{ntoinfty} frac{F(x + a_n) – F(x)}{a_n} = f(x)$$ where $$f = F’$$ and $$f$$ is integrable so $$f$$ is close to being my bounding function but I don’t think I’m guaranteed that $$f$$ is always greater than the terms I’m trying to bound. Can I “enhance” $$f$$ so that it’s always greater than the secant lines while preserving integrability?

## Math Genius: determining the sign of a power series

Considering $$a_n(v)$$ as a real sequence such that $$| a_n(v) | leq 1$$, and for some constant $$v > 1$$ we have:
$$begin{cases} a_n(v) < 0 & n leq v, \ a_n(v) > 0 & n > v. end{cases}$$

Under which condition(s), the following power series is positive?

$$begin{equation} sum_{n=1}^{infty} a_n(v)x^n, quad (x>0) end{equation}$$

I tried to decompose the series into $$sum_{n=1}^{lfloor v rfloor} a_n(v)x^n + sum_{n=lfloor v rfloor +1 }^{infty} a_n(v)x^n$$.

Now, the first series is negative, and the second one is positive, but have no idea about the next steps.

Update 1:
Numerical analyses suggest that there exist thresholds like $$v^*$$ and $$x^*$$ such that for $$v and $$x>x^*$$, the power series is positive.

Update 2:
The following properties can be shown:

let $$f(v,x) = sum_{n=1}^{lfloor v rfloor} a_n(v)x^n$$, and $$g(v,x) = sum_{n=lfloor v rfloor +1 }^{infty} a_n(v)x^n$$, then:

• For $$x>0$$, the series represented in $$g(v,x)$$ is convergent.
• $$f(v,x)$$ is decreasing in both $$v$$ and $$x$$.
• $$g(v,x)$$ is decreasing in $$v$$, and increasing in $$x$$.

## Math Genius: Is there a function with infinite integral on every interval?

Could give some examples of nonnegative measurable function \$f:mathbb{R}to[0,infty)\$, such that its integral over any bounded interval is infinite?

The easiest example I know is constructed as follows. Let \$q_{n}\$ be an enumeration of the rational numbers in \$[0,1]\$. Consider \$\$g(x) = sum_{n=1}^{infty} 2^{-n} frac{1}{|x-q_{n}|^{1/2}}.\$\$
Since each function \$dfrac{1}{|x-q_{n}|^{1/2}}\$ is integrable on \$[0,1]\$, so is \$g(x)\$ [verify this!]. Therefore \$g(x) < infty\$ almost everywhere, so we can simply set \$g(x) = 0\$ in the points where the sum is infinite.

On the other hand, \$f = g^{2}\$ has infinite integral over each interval in \$[0,1]\$. Indeed, if \$0 leq a lt b leq 1\$ then \$(a,b)\$ contains a number \$q_{n}\$, so \$\$int_{a}^{b} f(x),dx geq int_{a}^{b} frac{1}{|x-q_{n}|},dx = infty.\$\$ Now in order to get the function \$f\$ defined at every point of \$mathbb{R}\$, simply define \$f(n + x) = f(x)\$ for \$0 leq x lt 1\$.

See exercise 26 (c) on p. 327 here.

## Math Genius: Prove that there are infinitely many numbers between two real numbers. (example from Hardy’s book)

I know that this kind of question has been answered before in many places, but I’d like a proof with a little spin on it. In his book “A course of pure mathematics”, Hardy defines a real number \$alpha\$ as a section composed by a lower class (a) (in which a is rational and a\$lt\$\$alpha\$) and an upper class (A) (in which A is rational and A\$ge\$\$alpha\$).

I’d like to prove that there are infinitely many rational numbers between any given two real numbers definied in this way by using this idea of sets.

I’m not sure if it’s possible, but I believe so, because Hardy says “All these results are immediate consequences of our definitions” after a series of examples I managed to prove except for this one.

It might be useful to tell you that, according to thoses definitions, the relations of magnitude between two real numbers are defined in this way:

\$alpha\$\$lt\$\$beta\$ if, and only if, (a)\$subset\$(b) and (A)\$supset\$(B)

\$alpha\$\$gt\$\$beta\$ if, and only if, (a)\$supset\$(b) and (A)\$subset\$(B)

(where (a), (A) and (b), (B) are the lower and upper classes of \$alpha\$ and \$beta\$).

Evidently, if \$alpha\$\$lt\$\$beta\$ and x is a rational number between \$alpha\$ and \$beta\$, then x\$in\$(A)\$cap\$(b). Showing that there are infinitely many elements in (A)\$cap\$(b) would show that there are infinitely many rational numbers between \$alpha\$ and \$beta\$, but I’m stuck at this point and cannot find a way to develop the proof from this approach.

Both hints and solutions are welcome.

Greetings!

Let \$a,binmathbb R\$, WLOG \$a< b\$ then \$b-a>0\$. Let us assume that \$x=b-a\$ and \$y=5\$ then there exist (by Archimedean property) \$ninmathbb N\$ s.t \$\$nx>y\$\$
\$\$n(b-a)>5\$\$ \$\$nb-na>5\$\$. Since difference between \$nb\$ and \$na\$ is greater than \$5\$,then there exists atleast one \$minmathbb Z\$ s.t \$\$na<m<nb\$\$ \$\$a<frac mn<b\$\$
\$\$a<r<b\$\$,where \$r=frac mninmathbb Q\$ i.e there exist one rational number \$r\$ between \$a\$ and \$b\$.Again \$a<r\$,there exist \$r_1inmathbb Q\$ s.t \$\$a<r_1<r\$\$.Proceeding this way we can generate as many as rational number as we wish.

Hope this will help!!!

This is really an immediate consequence of the definition of real numbers as given by Hardy. To clarify I repeat the definition here:

A real number $$alpha$$ is a pair of sets $$A, B$$ such that

• $$Acap B=emptyset, Acup B=mathbb{Q}, Aneq emptyset neq B$$.
• if $$xin A, yin B$$ then $$x.
• if $$xin A$$ then there is a $$yin A$$ with $$x. Note that repeated use of this property shows that there are infinitely many $$yin A$$ with $$x.

If $$alpha=(A, B), beta=(C, D)$$ are two real numbers defined in this manner then we say that $$alpha$$ is less than $$beta$$ and write $$alpha if $$Asubset C$$.

Also if $$r$$ is rational number then there is a corresponding real number $$alpha_r$$ defined as $$(A_r, B_r)$$ where $$A_r={xmid xinmathbb {Q}, x and $$B_r={xmid xinmathbb {Q}, xgeq r}$$.

Let’s now suppose we are given two real numbers $$alpha=(A, B), beta=(C, D)$$ such that $$alpha . By definition this means that $$Asubset C$$ and hence $$S=Csetminus Aneq emptyset$$. It is easy to show that the set $$S$$ contains infinitely many rationals.

Since $$Sneq emptyset$$ there is a member $$xin S$$ and then $$xin C, xnotin A$$. Since $$beta=(C, D)$$ is a real number it follows that there are infinitely many members $$yin C$$ with $$y>x$$. And none of these members $$y$$ are in $$A$$ because if $$yin A$$ then $$x implies $$xin A$$ which is not the case. So all these members $$yin S=Csetminus A$$.

And if $$r$$ is one such rational number lying in this set $$S=Csetminus A$$ then the corresponding real number $$alpha_r=(A_r, B_r)$$ as defined earlier is such that $$alpha. It is in this sense that there are infinitely many rational numbers between any two given real numbers.

## Math Genius: Proof of continuity in metric space

My first attempt:

Let $$U subset Y$$ be an open set. Then since $$f$$ is continuous, $$f^{-1}(U)$$ is open in $$X$$. Now, if $$xin f^{-1}(U)$$ then $$x_nin f^{-1}(U)$$ for all $$ngeq N_1$$

Furthermore, $$f(x_n)in U$$ for all $$ngeq N_2$$

We can choose $$N=text{ max }{N_1,N_2}$$ such that $$x_ninf^{-1}(U)$$ and $$f(x_n)in U$$. So $$x_nto x$$ implies $$f(x_n)to f(x)$$

Would this be correct? I’m not sure whether the thing where I introduced $$N_1,N_2$$ is neccessary.

Proving that $$lim_{ntoinfty}f(x_n)=f(x)$$ means proving that, for each neighborhood $$V$$ of $$f(x)$$, there is some $$NinBbb N$$ such that $$ngeqslant Nimplies f(x_n)in V$$. I did not see you proving this.

Let $$U$$ be an open subset of $$V$$ such that $$f(x)in U$$. Since $$f$$ is continuous, $$f^{-1}(U)$$ is an open subset of $$X$$. Besides, since $$f(x)in U$$, $$xin f^{-1}(U)$$. So, there is some $$NinBbb N$$ such that $$ngeqslant Nimplies x_nin f^{-1}(U)$$. And therefore$$ngeqslant Nimplies f(x_n)in Uimplies f(x_n)in V.$$

## Math Genius: Asymptotic Expansion of an integral in the method of Steepest Descent

This is an exercise 2.4 in Grigis and Sjostrand’s Microlocal Analysis for Differential Operators

The following problem appears in the method of steepest descent. Show that for $$u in C_c^{infty}left( mathbb{R}^n right)$$, $$lambda geq 1$$:
$$int_{mathbb{R}^n}^{}e^{- frac{lambda x^2}{2}}u(x) dx = sum_{k=0}^{N-1}frac{left( 2 pi right)^{n/2}}{k! lambda^{k + n/2}}left( left( frac{1}{2} Delta right)^{k}u right)left( 0 right) + S_Nleft( u, lambda right)$$
where $$|S_{N, lambda}| leq C_{N, n}lambda^{ – N – frac{n}{2}} sum_{alpha = 2N} sup| partial^{alpha}uleft( x right)|$$

By Plancherel’s theorem and the formula for the Fourier Transform of a Gaussian we can write that
$$int_{mathbb{R}^n}^{}e^{ – lambda frac{x^2}{2}} u(x) dx = left(frac{2 pi }{lambda}right)^{frac{n}{2}}int_{mathbb{R}^n}^{}e^{ – frac{|xi|^2}{2 lambda}}hat{u} left( xi right) d xi$$
After using Taylors theorem with Remainder to expand the exponential and using Fourier Inversion it is straightforward to see that
$$left(frac{2 pi }{lambda}right)^{frac{n}{2}}int_{mathbb{R}^n}^{}e^{ – frac{|xi|^2}{2 lambda}}hat{u} left( xi right) d xi = sum_{k=0}^{N-1}frac{left( 2 pi right)^{n/2}}{k! lambda^{k + n/2}}left( left( frac{1}{2} Delta right)^{k}u right)left( 0 right) + S_Nleft( u, lambda right)$$
where $$|S_N(u,lambda)| leq C_{N, n} lambda^{ – N – frac{n}{2}} sum_{alpha = 2N}int_{mathbb{R}^{n}}^{}| xi^{alpha}hat{u} left( xi right)| dxi$$
However, I do not see how to prove the remainder bound that the problem asks for from this step. Can it be done or should there be more care taken with estimating the remainder from Taylor’s theorem?

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## Math Genius: \$v(Q)lesum_{i=1}^k v(Q_i)\$ where \$Q_1,…,Q_k\$ are rectangles that cover the rectangle \$Q\$

What shown below is a reference from “Analysis on manifolds” by James R. Munkres.

Well I don’t formally understand why each $$Rsubseteq Q$$ is contained in at least one of the rectangles $$Q_1,…,Q_k$$ so I ask to prove this formally. Could someone help me, please?

As stated in the beginning, “… each of the rectangles $$Q,Q_1,ldots, Q_k$$ is a union of subrectangles determined by $$P$$“.

Thus, $$Q_j = bigcup_{l=1}^{m_j} R_{jl}$$ for each $$j=1,ldots,k$$ and since $$Q_1,ldots, Q_k$$ cover $$Q$$, we have

$$Q subset bigcup_{j=1}^k Q_j = bigcup_{j=1}^kbigcup_{l=1}^{m_j}R_{jl}$$

If $$R subset Q$$, then as a member of the partition $$P$$ it must belong to the set $${R_{jl}}$$ and so is contained in at least one of the rectangles $$Q_1, ldots , Q_k$$.

Let $$a = (a_x, a_y)$$ be any interior point of $$R$$ (say the center of $$R$$ for sake of concreteness), and let $$Q_j$$ be any rectangle of $${Q_1, ldots, Q_k}$$ that contains $$a$$. We will show that $$Q_j$$ contains $$R$$. Let $$Q_j$$ be defined by the equations $$x=l$$, $$x=r$$, $$y=d$$, and $$y=u$$. Suppose for contradiction that $$b = (b_x, b_y)in R$$ is not in $$Q_j$$. Without loss of generality suppose that $$b_x > a_x$$, that is $$b$$ is to the right of $$a$$ (the cases $$b_x < a_x$$, $$b_y > a_y$$, and $$b_y < a_y$$ are similar). Then the right vertical edge of $$Q_j$$ must pass between $$c$$ and $$d$$, i.e. $$a_x leq r < b_x$$. But this means that $$R$$ would not be a subrectangle of $$P$$ (since the line $$x=r$$ would further partition $$R$$).

## Math Genius: \$v(Q)lesum_{i=1}^k v(Q_i)\$ where \$Q_1,…,Q_k\$ are rectangles that cover the rectangle \$Q\$

What shown below is a reference from “Analysis on manifolds” by James R. Munkres.

Well I don’t formally understand why each $$Rsubseteq Q$$ is contained in at least one of the rectangles $$Q_1,…,Q_k$$ so I ask to prove this formally. Could someone help me, please?

As stated in the beginning, “… each of the rectangles $$Q,Q_1,ldots, Q_k$$ is a union of subrectangles determined by $$P$$“.

Thus, $$Q_j = bigcup_{l=1}^{m_j} R_{jl}$$ for each $$j=1,ldots,k$$ and since $$Q_1,ldots, Q_k$$ cover $$Q$$, we have

$$Q subset bigcup_{j=1}^k Q_j = bigcup_{j=1}^kbigcup_{l=1}^{m_j}R_{jl}$$

If $$R subset Q$$, then as a member of the partition $$P$$ it must belong to the set $${R_{jl}}$$ and so is contained in at least one of the rectangles $$Q_1, ldots , Q_k$$.

Let $$a = (a_x, a_y)$$ be any interior point of $$R$$ (say the center of $$R$$ for sake of concreteness), and let $$Q_j$$ be any rectangle of $${Q_1, ldots, Q_k}$$ that contains $$a$$. We will show that $$Q_j$$ contains $$R$$. Let $$Q_j$$ be defined by the equations $$x=l$$, $$x=r$$, $$y=d$$, and $$y=u$$. Suppose for contradiction that $$b = (b_x, b_y)in R$$ is not in $$Q_j$$. Without loss of generality suppose that $$b_x > a_x$$, that is $$b$$ is to the right of $$a$$ (the cases $$b_x < a_x$$, $$b_y > a_y$$, and $$b_y < a_y$$ are similar). Then the right vertical edge of $$Q_j$$ must pass between $$c$$ and $$d$$, i.e. $$a_x leq r < b_x$$. But this means that $$R$$ would not be a subrectangle of $$P$$ (since the line $$x=r$$ would further partition $$R$$).

## Math Genius: Does \$lim_{Mtoinfty}int_{0}^infty (1+x^2)^{-s}frac{sin Mx}{x}dx\$ exist?

Does the limit \$displaystyle limlimits_{Mtoinfty}int_{0}^infty (1+x^2)^{-s}frac{sin Mx}{x}dx\$ exist? Where \$s>0\$ be a fix real number.

i.e. does the integral \$displaystyle int_{0}^infty dyint_{0}^infty (1+x^2)^{-s}cos xydx\$ converage in some sense?

\$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}\$

begin{align}
&color{#f00}{lim_{M to infty}int_{0}^{infty}pars{1 + x^{2}}^{-s},
{sinpars{Mx} over x},dd x} =
half,lim_{M to infty}
bracks{Mint_{-infty}^{infty}pars{1 + x^{2}}^{-s},
{sinpars{Mx} over Mx},dd x}
\[3mm] = &
half,lim_{M to infty}
bracks{Mint_{-infty}^{infty}pars{1 + x^{2}}^{-s}, overbrace{%
halfint_{-1}^{1}expo{ic kMx},dd k}
^{ds{= {sinpars{Mx} over Mx}}},dd x}
\[3mm] = &
{1 over 4},lim_{M to infty}bracks{Mint_{-1}^{1}
int_{-infty}^{infty}pars{1 + x^{2}}^{-s}expo{ic kMx},dd x,dd k}
\[3mm] = &
{1 over 4},lim_{M to infty}bracks{Mint_{-1}^{1}
int_{-infty}^{infty}exppars{ic kMx – slnpars{1 + x^{2}}},dd x,dd k}
end{align}

For ‘large \$M\$’, the asymptotic result for the \$x\$-integral is given by
\$ds{root{pi over s},exppars{-,{M^{2} over 4s},k^{2}}}\$ such that
begin{align}
&color{#f00}{lim_{M to infty}int_{0}^{infty}pars{1 + x^{2}}^{-s},
{sinpars{Mx} over x},dd x} =
{1 over 4},root{pi over s},lim_{M to infty}bracks{Mint_{-1}^{1}
exppars{-,{M^{2} over 4s},k^{2}},dd k}
\[3mm] = &
{1 over 4},root{pi over s},lim_{M to infty}braces{%
Mbracks{{2root{pi s} over M},mathrm{erf}pars{M over 2root{s}}}}
= {pi over 2},
underbrace{lim_{M to infty}mathrm{erf}pars{M over 2root{s}}}
_{ds{= 1}} = color{#f00}{pi over 2}
end{align}

\$ds{mathrm{erf}}\$ is the Error Function.

By switching to Fourier transforms or through other techniques it is not difficult to prove that
\$\$ lim_{Mto +infty}frac{sin(Mx)}{pi x}=delta(x) \$\$
hence:
\$\$ lim_{Mto +infty}int_{0}^{+infty}frac{1}{(1+x^2)^s}cdotfrac{sin(Mx)}{x},dx = frac{1}{2}lim_{Mto +infty}int_{-infty}^{+infty}frac{1}{(1+x^2)^s}cdotfrac{sin(Mx)}{x},dx = color{red}{frac{pi}{2}}.\$\$

Tagged :

## Math Genius: Does \$lim_{Mtoinfty}int_{0}^infty (1+x^2)^{-s}frac{sin Mx}{x}dx\$ exist?

Does the limit \$displaystyle limlimits_{Mtoinfty}int_{0}^infty (1+x^2)^{-s}frac{sin Mx}{x}dx\$ exist? Where \$s>0\$ be a fix real number.

i.e. does the integral \$displaystyle int_{0}^infty dyint_{0}^infty (1+x^2)^{-s}cos xydx\$ converage in some sense?

\$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}\$

begin{align}
&color{#f00}{lim_{M to infty}int_{0}^{infty}pars{1 + x^{2}}^{-s},
{sinpars{Mx} over x},dd x} =
half,lim_{M to infty}
bracks{Mint_{-infty}^{infty}pars{1 + x^{2}}^{-s},
{sinpars{Mx} over Mx},dd x}
\[3mm] = &
half,lim_{M to infty}
bracks{Mint_{-infty}^{infty}pars{1 + x^{2}}^{-s}, overbrace{%
halfint_{-1}^{1}expo{ic kMx},dd k}
^{ds{= {sinpars{Mx} over Mx}}},dd x}
\[3mm] = &
{1 over 4},lim_{M to infty}bracks{Mint_{-1}^{1}
int_{-infty}^{infty}pars{1 + x^{2}}^{-s}expo{ic kMx},dd x,dd k}
\[3mm] = &
{1 over 4},lim_{M to infty}bracks{Mint_{-1}^{1}
int_{-infty}^{infty}exppars{ic kMx – slnpars{1 + x^{2}}},dd x,dd k}
end{align}

For ‘large \$M\$’, the asymptotic result for the \$x\$-integral is given by
\$ds{root{pi over s},exppars{-,{M^{2} over 4s},k^{2}}}\$ such that
begin{align}
&color{#f00}{lim_{M to infty}int_{0}^{infty}pars{1 + x^{2}}^{-s},
{sinpars{Mx} over x},dd x} =
{1 over 4},root{pi over s},lim_{M to infty}bracks{Mint_{-1}^{1}
exppars{-,{M^{2} over 4s},k^{2}},dd k}
\[3mm] = &
{1 over 4},root{pi over s},lim_{M to infty}braces{%
Mbracks{{2root{pi s} over M},mathrm{erf}pars{M over 2root{s}}}}
= {pi over 2},
underbrace{lim_{M to infty}mathrm{erf}pars{M over 2root{s}}}
_{ds{= 1}} = color{#f00}{pi over 2}
end{align}

\$ds{mathrm{erf}}\$ is the Error Function.

By switching to Fourier transforms or through other techniques it is not difficult to prove that
\$\$ lim_{Mto +infty}frac{sin(Mx)}{pi x}=delta(x) \$\$
hence:
\$\$ lim_{Mto +infty}int_{0}^{+infty}frac{1}{(1+x^2)^s}cdotfrac{sin(Mx)}{x},dx = frac{1}{2}lim_{Mto +infty}int_{-infty}^{+infty}frac{1}{(1+x^2)^s}cdotfrac{sin(Mx)}{x},dx = color{red}{frac{pi}{2}}.\$\$

Tagged :