## calculus and analysis – Integrate just spits out the input

Hi new to Mathematica here. I am trying to integrate this function. I am not so sure if it can be integrated or I have maybe messed up the syntax somewhere. I would be very happy for any help. The function is the first line after simplification. Thanks.

``````(-R^2 + r1^2 + d^2 t^2 + 2 d r1 t theta)/(R^4 -
2 R^2 r1 (r1 + d t theta) + r1^2 (r1^2 + d^2 t^2 + 2 d r1 t theta))

Integrate(%4, {t, 0, 1},
Assumptions -> { R, r1, d, theta} (Element) Reals && {R, r1, d} >
0 && -1 < theta < 1 )
Integrate((-R^2 + r1^2 + d^2 t^2 + 2 d r1 t theta)/(
R^4 - 2 R^2 r1 (r1 + d t theta) +
r1^2 (r1^2 + d^2 t^2 + 2 d r1 t theta)), {t, 0, 1},
Assumptions -> (R | r1 | d | theta) (Element) Reals && {R, r1, d} >
0 && -1 < theta < 1)
``````

## real analysis – Prove the measure is a premeasure

Let $$mathcal{A}={S subset (0,1) mid S text{or} S^c text{is finite}}$$. Let $$rho_0(S)=0$$ if $$S$$ is finite or $$rho_0(S)= infty$$ if $$S^c$$ is finite, prove $$rho_0$$ is a premeasure.

Attempt:Since $$varnothing$$ is finite, $$rho_0(varnothing)=0$$ by definition of $$rho_0$$. Let $${A_j}_{j=1}^{infty}$$ be a sequence of disjoint sets in $$mathcal{A}$$ such that $$bigcuplimits_{j=1}^{infty}A_j in mathcal{A}$$. Since $$|bigcuplimits_{j=1}^{infty}A_j| geq infty$$, we must have $$(bigcuplimits_{j=1}^{infty}A_j)^c=bigcaplimits_{j=1}^{infty}A_j^c$$ is finite. So that $$A_j^c$$ is finite for some $$j$$ and $$|A_j| geq infty$$. So that $$infty=rho_o(bigcuplimits_{j=1}^{infty}A_j)=rho_0(A_j) leq sumlimits_{j=1}^{infty}rho_0(A_j)$$
and $$sumlimits_{j=1}^{infty}sigma_0(A_j)=infty$$. Thus, $$rho_0(bigcuplimits_{j=1}^{infty}A_j)=sumlimits_{j=1}^{infty}rho_0(A_j)$$ and $$rho_0$$ is a premeasure.

My question is, is it safe to assume that $$A^c_j$$ is finite for some $$j$$? Or do I have to consider a case where all the $$A^c_j$$ are infinite?

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## real analysis – Power series end up with \$ln(-infty)\$

I am trying to find the radius of convergence of the power series $$a_n=frac{n^2}{4^n+3n}$$ $$sum_{n=1}^infty a_nz^n$$.

Using hadamard’s formula, we have that $$R=frac{1}{limsup|frac{n^2}{4^n+3n}|^frac{1}{n}}$$

So to compute this, we take the logarithm and raise everything to the $$e$$ then apply l’hopital’s rule giving

$$lim_{ntoinfty}left(frac{n^2}{4^n+3n}right)^frac{1}{n} =e^{lim _{n rightarrow infty} frac{ln left(n^{2}right)-ln left(4^{n}+3 nright)}{n}}=e^{ln (0-infty)} =e^{ln(-infty)} =0$$

However I have ended up with $$ln(-infty)$$ which is undefined. Where did my resasoning go wrong here?

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## fa.functional analysis – Is there a notion of „flatness” in point-set topology?

In algebraic geometry, flat morphisms are usually associated with the intuition that they have „continuously varying fibers”. Is there a notion in topology formalizing the same intuition? Consider for example a map $$p colon P to X$$ of good (locally compact Hausdorff, say) topological spaces and assume for the moment that it is proper (i.e. has compact fibers). If the fibers vary continuously, then the following property should be true for example:

(*) If $$f colon P to mathbf{R}$$ is a continuous function, then the induced function $$widetilde{f} colon X to mathbf{R}, qquad widetilde{f}(x) := sup_{q in p^{-1}({x})}f(q)$$
is also continuous.

This is true for example if the fibers are constant, i.e. if $$p$$ is a trivial bundle $$p colon X times F to X$$. In that case, the map $$f colon X times F to mathbf{R}$$ induces a continuous map $$X to C(F, mathbf{R})$$, the space of continuous functions equipped with the compact-open topology (which is the topology of uniform convergence) and the map $$widetilde{f}$$ factors as $$X to C(F, mathbf{R}) xrightarrow{sup} mathbf{R}$$.

But there are also examples where the fibers are not (locally) constant: For example the map $$mathbf{C} to mathbf{C}$$, $$z mapsto z^2$$ has continuously varying fibers in the intuitive sense (as $$z$$ approaches the origin, the two points in the fiber get closer to each other). This is also a flat map in algebraic geometry.

For an example which is not flat in the intuitive sense and where also the above property is clearly wrong, consider the first projection
$$p colon {(x,y) in mathbf{R} times (0,1) mid x cdot y = 0 } to mathbf{R}.$$
(This example is of course also inspered by algebraic geometry.)

I should perhaps add that I am aware of the related questions The topological analog of flatness? and Flatness in Algebraic Geometry vs. Fibration in Topology. But I am interested in the purely point-set-topological setting and in fact I care specifically about the property (*) (I have a specific example of a continuous map $$p$$ in mind for which I hope (*) to be true and I’m trying to gain some intuition.)

## fa.functional analysis – How to show that \$left|F_n^* – Fright|={{O}_{p}}left({{n}^{-frac{1}{2}}}right)\$

Let $$X_1,ldots,X_n$$ be iid from a cdf $$F$$ on $$R^d$$, $$X_1^*,ldots,X_n^*$$ iid from empirical cdf $$F_n$$. Let $$F_n^*$$ be empirical cdf based on $$X_i^*$$‘s. Using DKW inequality, Let $$rholeft(F_1,F_2right)=left|F_1-F_2right|$$ How to show that

(a)$${{rho }_{infty }}left(F_{n}^{*},Fright)xrightarrow{a.s}0$$

(b)$${{rho }_{infty }}left(F_{n}^{*},Fright)={{O}_{p}}left({{n}^{-frac{1}{2}}}right)$$

(c)$${{rho }_{L_p}}(F_{n}^{*},F)={{O}_{p}}({{n}^{-frac{1}{2}}})$$

My thought is that

(a)
begin{equation} begin{aligned} P({{rho }_{infty }}(F_{n}^{*},F)>z)& < P({{rho }_{infty }}(F_{n}^{*},{{F}_{n}})+{{rho }_{infty}}left({{F}_{n}},Fright)>z) \ & < int_{0}^{z}{P({{rho }_{infty }}({{F}_{n}},F)>{{z}_{1}})}P({{rho }_{infty }}(F_{n}^{*},{{F}_{n}})>z-{{z}_{1}})mathrm d{{z}_{1}} quad text{(Correct?)}\ & le int_{0}^{z}{{{C}_{varepsilon ,d}}{{e}^{-(2-varepsilon )nz_{1}^{2}}}{{{{C}’}}_{varepsilon ,d}}{{e}^{-(2-varepsilon )n{{(z-{{z}_{1}})}^{2}}}}d{{z}_{1}}} (DKW)\ & ={{C}_{varepsilon ,d}}{{{{C}’}}_{varepsilon ,d}}int_{0}^{z}{{{e}^{-(2-varepsilon )nz_{1}^{2}}}{{e}^{-(2-varepsilon )n{{(z-{{z}_{1}})}^{2}}}}d{{z}_{1}}} \ & ={{C}_{varepsilon ,d}}{{{{C}’}}_{varepsilon ,d}}int_{0}^{z}{{{e}^{-(2-varepsilon )n({{z}^{2}}-2{{z}_{1}}z+2z_{1}^{2})}}d{{z}_{1}}} \ & ={{C}_{varepsilon ,d}}{{{{C}’}}_{varepsilon ,d}}{{e}^{frac{-(2-varepsilon )n{{z}^{2}}}{2}}}int_{0}^{z}{{{e}^{-(2-varepsilon )n2{{({{z}_{1}}-frac{z}{2})}^{2}}}}d{{z}_{1}}propto {{{{C}”}}_{varepsilon ,d}}{{e}^{-{{C}_{varepsilon ,d}}^{prime prime prime }n{{z}^{2}}}}} end{aligned} end{equation}

$$begin{equation} sumlimits_{n=1}^{infty }{P({{rho }_{infty }}(F_{n}^{*},F)>z)le sumlimits_{n=1}^{infty }{{{{{C}”}}_{varepsilon ,d}}{{e}^{-{{C}_{varepsilon ,d}}^{prime prime prime }n{{z}^{2}}}}}}
$$begin{equation} {{rho }_{infty }}(F_{n}^{*},F)xrightarrow{a.s}0 end{equation}$$

(b)

From part a $$P({{rho }_{infty }}(F_{n}^{*},F)>z)le C{{e}^{an{{z}^{2}}}}$$, where $$a,C$$ are constant. Let $$C{{e}^{an{{z}^{2}}}}=epsilon$$, then we can derive $$z=sqrt{frac{ln varepsilon -ln C}{an}}$$. So $$exists$$ constant $${{D}_{varepsilon }}=frac{sqrt{ln varepsilon -ln C}}{sqrt{a}}$$ s.t. $$forallepsilon>0$$, $$P({{rho }_{infty }}(F_{n}^{*},F)>frac{{{D}_{varepsilon }}}{sqrt{n}})le varepsilon$$. In other words, $${{rho }_{infty }}(F_{n}^{*},F)={{O}_{p}}({{n}^{-frac{1}{2}}})$$.

(c)$$rho_{L_p}=|F_n^*-F|_{L_p}le|F_n^*-F_n|_{L_p}+|F_n-F|_{L_p}le(|F_n^*-F_n|_{infty}|F_n^*-F_n|_{L_1})^{frac{1}{p}}+(|F_n-F|_{infty}|F_n-F|_{L_1})^{frac{1}{p}}$$

Then I think we can also use DKW inequality, but I don’t know how to deal with this $$||_{L_1}$$, do you have some ideas?

## complex analysis – Imaginary differential into fourier transform

I’ m looking for a way to relate the Fourier trasform in the form
$$tilde F[vec q] = int d^2x , e^{i vec q cdot vec x} fleft[ sqrt{x^2 + d^2} right],$$

and the fourier transform

$$F'[vec q] = int d^2x , e^{i vec q cdot vec x} fleft[ vert x vert right],$$

by trasnforming the first integral with the change of variables $$vec x rightarrow vec x – i vec d$$, with $$vec d parallel vec q$$.

As a result, I expect to see :

$$tilde F[vec q] = e ^ {- q d} F'[vec q],$$

but the procedure would require to go into the complex plane and doesn’ t seem intuitive how to get the expected result. Is there any hint to get there?

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## real analysis – Proof of general delta method

I have found proof of the “delta method”, but I cannot understand some steps in this proof.

Theorem :
Let $$X_1, X_2,…$$ and $$Y$$ be random k-vectors satisfying $$a_n(X_n-c)to_dY$$ where $$cinmathcal{R^k}$$ and $${a_n}$$ is a sequence of positive numbers with $$lim_{ntoinfty}a_n=infty$$. Let $$g$$ be a function from $$mathcal{R}^ktomathcal{R}$$.

If $$g$$ is differentiable at $$c$$,then$$a_n(g(X_n)-g(c))to_d(bigtriangledown g(c))^TY$$
where $$bigtriangledown g(c)$$ denotes the $$k$$-vector of partial derivatives of $$g$$ at $$x$$.

Proof:

Let$$Z_n=a_n(g(X_n)-g(c))-a_n(bigtriangledown g(c))^T(X_n-c)$$

The differentiability of $$g$$ at $$c$$ implies that for any $$epsilon>0,exists$$ a $$delta_{epsilon}>0$$ s.t.$$|(g(X_n)-g(c))-(bigtriangledown g(c))^T(X_n-c)|leepsilon|x-c|$$
whenever $$|x-c|. Let $$eta>0$$ be fixed,$$P(|Z_n|geeta)le P(|X_n-c|gedelta_{epsilon})+P(a_n|X_n-c|geeta/epsilon)$$
Since $$a_ntoinfty$$, $$a_n(g(X_n)-g(c))to_d(bigtriangledown g(c))^TY$$, Slutsky theorem implies $$X_nto_pc$$(Why?)

Since $$|.|$$ is continuous, $$a_n(X_n-c)to_dY$$ implies $$a_n|X_n-c|to_d|Y|$$. WLOG, assume $$eta/epsilon$$ is a continuous point of $$F_{|Y|}$$. Then,$$limsup_nP(|Z_n|geeta)le lim_{ntoinfty}P(|X_n-c|gedelta_{epsilon})+lim_{ntoinfty}P(a_n|X_n-c|geeta/epsilon)=P(|Y|geeta/epsilon)$$

The proof is complete since $$epsilon$$ can be arbitrary.(Why?)