Challenge Math Challenge - October 2018

[nuuskur]

@fresh_42
I haven't looked too much into it, yet, but at first glance it seems trivialising. The extended topology gains virtually every set in $2^{\overline{\mathbb C}}$. I'm probably crazy for saying that :D

 

[fresh_42]

@fresh_42
I haven't looked too much into it, yet, but at first glance it seems trivialising. The extended topology gains virtually every set in $2^{\overline{\mathbb C}}$. I'm probably crazy for saying that :D

I'm not sure I understood your remark. What is it that's trivial? And what do you mean by gain sets? Since the original topology is still intact, yourself noted $\tau \subseteq \bar{\tau}$, we still have the original sets in $\tau$ resp. $\tau^C$.

 

[Infrared]

It can be done with any topological space, at least as long as closed subsets of compact sets are compact again.

Fortunately, closed subsets of compact sets are always compact (in any topological space).

 

[fresh_42]

Fortunately, closed subsets of compact sets are always compact (in any topological space).

Thanks, I was too lazy to check. I thought at least some separation property might be necessary.

 

[member 587159]

Solution for (2)

Spoiler

(a) We claim that $$\sigma (\mathcal{F}) = \{A \subseteq X \mid |A| \leq |\mathbb{N}| \lor |A^c| \leq |\mathbb{N}|\}$$

Clearly the right side is contained in the left side, as we can write countable sets as countable union of singeltons. It is also clear that the right set contains $\mathcal{F}$, since singeltons are finite. It remains to check that the right side is a sigma-algebra, which is evident. Just note that the countable union of countable sets remains countable. The other inclusion then follows by minimality.

(b) Define $$\mathcal{S}:=\{B \in \sigma(A) \mid \exists S_0: |S_0| \leq |\mathbb{N|: B \in \sigma(S_0)}\}$$ If we believe the claim the exercise asks for, we have to prove that $\sigma(A)= \mathcal{S}$. Clearly, the right side is contained in the left side. The other inclusion follows again because the right side is a sigma algebra containing $A$ (because $A \in \sigma(\{A\})$).

 

[fresh_42]

Solution for (2)

Spoiler

(a) We claim that $$\sigma (\mathcal{F}) = \{A \subseteq X \mid |A| \leq |\mathbb{N}| \lor |A^c| \leq |\mathbb{N}|\}$$

Clearly the right side is contained in the left side, as we can write countable sets as countable union of singeltons. It is also clear that the right set contains $\mathcal{F}$, since singeltons are finite. It remains to check that the right side is a sigma-algebra, which is evident. Just note that the countable union of countable sets remains countable. The other inclusion then follows by minimality.

(b) Define $$\mathcal{S}:=\{B \in \sigma(A) \mid \exists S_0: |S_0| \leq |\mathbb{N|: B \in \sigma(S_0)}\}$$ If we believe the claim the exercise asks for, we have to prove that $\sigma(A)= \mathcal{S}$. Clearly, the right side is contained in the left side. The other inclusion follows again because the right side is a sigma algebra containing $A$ (because $A \in \sigma(\{A\})$).

The first part is correct, although very short. I'll give a more complete solution for the sake of readability.

A $\sigma-$algebra $\sigma(\mathcal{B})$ over a set $\mathcal{B}$ is a subset of the power set $P(\mathcal{B})$, which contains $\mathcal{B}$ as an element or equivalently $\emptyset$, is closed under complements to $\mathcal{B}$, and closed under the union of countable many sets. For an arbitrary set $X$ and a family of subsets $B \subseteq \mathcal{P}(X)$, $\sigma(B)$ denotes the intersection of all $\sigma-$algebras of subsets of $X$ that contain $B$, i.e. $\sigma(B)=\cap \{\,\sigma(C)\,|\,C\subseteq B\,\}$ with complements are taken to $X$.

We define $\mathcal{A}:=\{\,A\subseteq X\,|\,A \text{ is countable or }X-A \text{ is countable }\,\}$ and show $\mathcal{A}=\sigma(\mathcal{F})$.
Every countable set $A=\{\,x_i\,|\,i\in \mathbb{N}\,\}=\bigcup_{i\in\mathbb{N}}\{\,x_i\,\}\in \sigma(\mathcal{F})$ and also each set with a countable complement. Thus $\mathcal{A}\subseteq \sigma(\mathcal{F})\,.$ In order to show $\sigma(\mathcal{F})\subseteq \mathcal{A}$, we show that $\mathcal{A}$ is a $\sigma-$algebra which contains $\mathcal{F}\,.$

1. $\emptyset \in \mathcal{A}$ since it is countable.
2. If $A\in\mathcal{A}$ then $X-A\in \mathcal{A}$ since $A=X-(X-A)\,.$
3. Let $A_i\in \mathcal{A}\; , \;i\in \mathbb{N}.$ Then either all $A_i$ are countable, and then $\cup_{i\in \mathbb{N}}A_i$ is countable, too, and in $\mathcal{A}$, or there is an index $j$ with an uncountable set $A_j$. In this case is $X-A_j$ countable. Then we have that $X-\cup_{i\in \mathbb{N}}A_i = \cap_{i\in \mathbb{N}} X-A_i \subseteq X-A_j$ is countable, and again $\cup_{i\in \mathbb{N}}A_i \in \mathcal{A}\,.$
4. If $A=\{\,x\,\}\in \mathcal{F}$, then it is countable and so is $A \in \mathcal{A}$, i.e. $\mathcal{F}\subseteq \mathcal{A}\,. \square$

Your solution to the second part confuses me. $S \subseteq \mathcal{P}(X)$ is a given set of sets, but you define an (other?) $S$ in the first line? Regarding your question about what has to be shown: Given any set $A$ from the $\sigma-$algebra of $S$, which are both given, prove that there is a countable subset $S_0$ of the given set $S$, such that $A \in \sigma(S_0)$.

Your definition looks a bit right, as one should consider $\mathcal{A}:=\bigcup\{\,\sigma(C)\,|\,C\subseteq S \text{ countable }\,\}$. But this is at prior simply a set of sets. It should be shown that this is already a $\sigma-$algebra, and that it contains $S$. If this has been shown, the proof is a final small conclusion from that.

 

[member 587159]

The first part is correct, although very short. I'll give a more complete solution for the sake of readability.

A $\sigma-$algebra $\sigma(\mathcal{B})$ over a set $\mathcal{B}$ is a subset of the power set $P(\mathcal{B})$, which contains $\mathcal{B}$ as an element or equivalently $\emptyset$, is closed under complements to $\mathcal{B}$, and closed under the union of countable many sets. For an arbitrary set $X$ and a family of subsets $B \subseteq \mathcal{P}(X)$, $\sigma(B)$ denotes the intersection of all $\sigma-$algebras of subsets of $X$ that contain $B$, i.e. $\sigma(B)=\cap \{\,\sigma(C)\,|\,C\subseteq B\,\}$ with complements are taken to $X$.

We define $\mathcal{A}:=\{\,A\subseteq X\,|\,A \text{ is countable or }X-A \text{ is countable }\,\}$ and show $\mathcal{A}=\sigma(\mathcal{F})$.
Every countable set $A=\{\,x_i\,|\,i\in \mathbb{N}\,\}=\bigcup_{i\in\mathbb{N}}\{\,x_i\,\}\in \sigma(\mathcal{F})$ and also each set with a countable complement. Thus $\mathcal{A}\subseteq \sigma(\mathcal{F})\,.$ In order to show $\sigma(\mathcal{F})\subseteq \mathcal{A}$, we show that $\mathcal{A}$ is a $\sigma-$algebra which contains $\mathcal{F}\,.$

1. $\emptyset \in \mathcal{A}$ since it is countable.
2. If $A\in\mathcal{A}$ then $X-A\in \mathcal{A}$ since $A=X-(X-A)\,.$
3. Let $A_i\in \mathcal{A}\; , \;i\in \mathbb{N}.$ Then either all $A_i$ are countable, and then $\cup_{i\in \mathbb{N}}A_i$ is countable, too, and in $\mathcal{A}$, or there is an index $j$ with an uncountable set $A_j$. In this case is $X-A_j$ countable. Then we have that $X-\cup_{i\in \mathbb{N}}A_i = \cap_{i\in \mathbb{N}} X-A_i \subseteq X-A_j$ is countable, and again $\cup_{i\in \mathbb{N}}A_i \in \mathcal{A}\,.$
4. If $A=\{\,x\,\}\in \mathcal{F}$, then it is countable and so is $A \in \mathcal{A}$, i.e. $\mathcal{F}\subseteq \mathcal{A}\,. \square$

Your solution to the second part confuses me. $S \subseteq \mathcal{P}(X)$ is a given set of sets, but you define an (other?) $S$ in the first line? Regarding your question about what has to be shown: Given any set $A$ from the $\sigma-$algebra of $S$, which are both given, prove that there is a countable subset $S_0$ of the given set $S$, such that $A \in \sigma(S_0)$.

Your definition looks a bit right, as one should consider $\mathcal{A}:=\bigcup\{\,\sigma(C)\,|\,C\subseteq S \text{ countable }\,\}$. But this is at prior simply a set of sets. It should be shown that this is already a $\sigma-$algebra, and that it contains $S$. If this has been shown, the proof is a final small conclusion from that.

Sorry for the second part. I'll try to be more explicit.

Define $$\mathcal{B} =\{C \in \sigma(S)\mid \exists S_0 \subseteq S \mathrm{\ at \ most \ countable \ with\ } C \in \sigma(S_0)\}$$

One then shows that $\mathcal{B}$ is a sigma-algebra containing $S$ and we are done.

This $\mathcal{B}$ looks different than your $\mathcal{A}$, but works . I just called it curled S but that was confusing as something was already called S. For some strange reason I must have thought that our sigma algebra was generated by a certain set A. Apologies for that. Maybe too much mathematics for today...

 

[fresh_42]

Sorry for the second part. I'll try to be more explicit.

Define $$\mathcal{B} =\{C \in \sigma(S)\mid \exists S_0 \subseteq S \mathrm{\ at \ most \ countable \ with\ } C \in \sigma(S_0)\}$$

One then shows that $\mathcal{B}$ is a sigma-algebra containing $S$ and we are done.

This $\mathcal{B}$ looks different than your $\mathcal{A}$, but works . I just called it curled S but that was confusing as something was already called S. For some strange reason I must have thought that our sigma algebra was generated by a certain set A. Apologies for that. Maybe too much mathematics for today...

O.k., but still very short. Let me post my proof for all who are less familiar with the constructions.

Let $\mathcal{A}=\bigcup \{\,\sigma(C)\,|\,C\subseteq S \text{ countable}\,\}\,.$ We show that $\mathcal{A}$ is a $\sigma-$algebra over $X$ which contains $S.$

1. $\emptyset \in \sigma(\emptyset)=\{\emptyset,X\}$ and $\emptyset$ is countable with $\emptyset \subseteq S$, so $\emptyset \in \mathcal{A}\,.$
2. With $A\in \mathcal{A}$ we have a countable set $C\subseteq S$ with $A\in \sigma(C)$, and so is $X-A \in \mathcal{A}$.
3. Let $A_i \in \mathcal{A}$ for $i \in \mathbb{N}$. Then there are countable sets $C_i \subseteq S$ with $A_i \in \sigma(C_i)$. The set $C=\bigcup_{i\in \mathbb{N}} C_i \subseteq S$ is also countable. Since $C_i \subseteq C$ and $\sigma$ is monotone, we have $A_i\in \sigma(C)$ and so $\bigcup_{i\in \mathbb{N}} A_i \in \sigma(C) \in \mathcal{A}$.

If $A \in S$, then $\{\,A\,\}\subseteq S$ is countable and $A \in \sigma(\{A\})=\{\,\emptyset, A,X-A,X\,\}$, i.e. $A\in \mathcal{A}$ and thus $S\subseteq \mathcal{A}$.

Since $\mathcal{A}$ is a $\sigma-$algebra which contains $S$, we have $\sigma(S)\subseteq \mathcal{A}$. This means that for all $A\in \sigma(S)$ there is a countable set $S_0\subseteq S$ with $A\in \sigma(S_0)$.

 

[YoungPhysicist]

It’s November 9th now.Will there be a November math challenge?

 

[member 587159]

It’s November 9th now.Will there be a November math challenge?

I think there are still too much open problems of the previous editions.

 

[YoungPhysicist]

I think there are still too much open problems of the previous editions.

Oh~~all right.It is just that since I joined PF, there is a new math challenge every month.

 

[fresh_42]

Oh~~all right.It is just that since I joined PF, there is a new math challenge every month.

We currently have 25 open problems of all possible degrees of difficulty:

https://www.physicsforums.com/threads/math-challenge-october-2018.956631/
https://www.physicsforums.com/threads/basic-math-challenge-september-2018.954490/
https://www.physicsforums.com/threads/intermediate-math-challenge-september-2018.954495/

As nobody even has asked about hints or at least attempted to solve them, we concluded a lack of interest. And to provide a collection of questions à la 'Solve this integral / series!' didn't appear worthwhile to us.

 

[member 587159]

Here is another solution for (11), using complex analysis. I hope this is the elegant solution you were looking for. The method I use is the same as in this video:

Spoiler

Note that the roots of $x^2 - x +1$ over $\mathbb{C}$ are given by $\frac{1\pm \sqrt{3}i}{2}$. Only the one with the positive square root sign lies in the upper half plane.

These are poles of order 1. Using the technique in the video, one obtains:

$$\int_{-\infty}^{+\infty} \frac{4}{x^2 - x +1} = 2 \pi i Res\left(\frac{4}{x^2-x+1}, \frac{1+\sqrt{3}i}{2}\right) = 2\pi i \lim_{x \to \frac{1+\sqrt{3}i}{2}} \frac{4}{x-1/2 + \sqrt{3}/2i} = \frac{8 \pi}{\sqrt{3}}$$

 

[Fred Wright]

Dear Dr. @wrobel ,
The quadrature solution of problem #1 results in an intractable integral:

Spoiler

$$\int \frac {x^2dx}{\sqrt {x^4-\frac {1}{4}} }=\sqrt {C_2}(\frac{2^{\frac {7}{2}}}{3}t^{\frac {3}{2}} +C_1)$$

I note that the integrand is complex when $\left | x \right | < 1$ and real otherwise. How am I wrong?

 

[wrobel]

The quadrature solution of problem #1 results in an intractable integral

There is no need to compute such integrals explicitly. For example, if you obtained an equation $\dot x=\frac{1}{\sin(x^2)}$ then you
can write
$$\int\sin(x^2)dx=t+const$$ and that is all. Integration in quadrature does not imply that you will present all the integrals and inverse functions as combination of elementary functions.

 

[nuuskur]

Not lack of interest, rather lack of time

 

[member 587159]

Not lack of interest, rather lack of time

Indeed, same for me. Over time, I think the questions will get solved though.

 

[Greg Bernhardt]

We currently have 25 open problems of all possible degrees of difficulty:

Maybe compile these into a new competition for better organization?

 

[lpetrich]

Good idea. I'd like to see that.

 

[fresh_42]

With additional hints or without?

 

[Periwinkle]

8. Let $f$, $g$: $\mathbb{R} \rightarrow \mathbb{R}$ be two functions with $f\,''(x) + f\,'(x)g(x) - f(x) = 0$. Show that if $f(a) = f(b) = 0$ then $f(x) = 0$ for all $x\in [a,b]$. (by @QuantumQuest )

According to Weierstrass's maximum value theorem, if a function is continuous on $[a,b]$, then it is bounded on this interval and takes its maximum and minimum. If function $f(x)$ is not constant, then its maximum or minimum is different from $f(a) = f(b)=0$. Let this be the maximum that the function takes at an internal $x_0$ point of the interval. According to one of Fermat's theorems, in this case - provided that the function $f (x)$ can be derived at this point - the derivative $f'(x_0)$ here is $0$. Therefore, the following equation is correct:
$$f\,''(x_0) + 0 \cdot g(x_0) - f(x_0) = 0$$ $f(x_0)$ is positive, so $f\,''(x_0)$ is also positive. Therefore, the function $f(x)$ has a local minimum at $x_0$, which cannot be a local maximum.

 

[benorin]

4) b) $$\int_A \frac{1}{x^2+y}\, d\lambda(x,y) =\int_0^1\int_0^1 \frac{dydx}{x^2+y} =\int_0^1\left[\log (x^2+y) \right|_{y=0}^1\, dx$$
$$=\int_0^1\left[ \log (1+x^2) -2\log(x)\right] dx $$
$$=\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\left[ \tfrac{x^2}{1+x^2} -1\right] dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\tfrac{1}{1+x^2} dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)-2\tan ^{-1}(x)\right|_{x=0}^{1} = \tfrac{\pi}{2}+\log 2$$

 

[member 587159]

4) b) $$\int_A \frac{1}{x^2+y}\, d\lambda(x,y) =\int_0^1\int_0^1 \frac{dydx}{x^2+y} =\int_0^1\left[\log (x^2+y) \right|_{y=0}^1\, dx$$
$$=\int_0^1\left[ \log (1+x^2) -2\log(x)\right] dx $$
$$=\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\left[ \tfrac{x^2}{1+x^2} -1\right] dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\tfrac{1}{1+x^2} dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)-2\tan ^{-1}(x)\right|_{x=0}^{1} = \tfrac{\pi}{2}+\log 2$$

Hi. I think you should justify the first equality. You are using a theorem for this. Which theorem and why can you apply it?

 

[benorin]

If the Improper Riemann and Lebesgue integrals both exist, they are equal?

 

[fresh_42]

If the Improper Riemann and Lebesgue integrals both exist, they are equal?

No.

 

[benorin]

I thumbed through baby Rudin, and alas, no joy. Rudin doesn't distinguish betwixt $d\lambda (x,y)$ and $d\lambda (x)\, d\lambda (y)$, everything is just $d\mu$ no matter the dimension.

 

[fresh_42]

It is neither about the Lebesgue measure nor the dimension. It has to do with the order of integration, something you always have to consider, if you integrate over an area or a volume, because you can't do it in a single step.

 

[member 587159]

If the Improper Riemann and Lebesgue integrals both exist, they are equal?

The problem is that in order to calculate the exact value, like you did, you need to somehow show that the integral exists, which is the original question.

So basically, you assumed what the exercise asks you to prove to get your solution, which seems problematic to me.

I thumbed through baby Rudin, and alas, no joy. Rudin doesn't distinguish betwixt $d\lambda (x,y)$ and $d\lambda (x)\, d\lambda (y)$, everything is just $d\mu$ no matter the dimension.

Baby Rudin does not treat product measures, so you won't get the relevant stuff in that book. See Papa Rudin for a treatment of this instead.

 

[benorin]

7. a) Determine $\int_1^\infty \frac{\log(x)}{x^3}\,dx\,.$
7. b) Determine for which $\alpha$ the integral $\int_0^\infty x^2\exp(-\alpha x)\,dx$ converges.
7. c) Find a sequence of functions $f_n\, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,n\in \mathbb{N}$ such that $$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$
7. d) Find a family of functions $f_r\, : \,\mathbb{R}^+\longrightarrow \mathbb{R}\, , \,r\in \mathbb{R}$ such that
$$
\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx \neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx
$$
7. e) Find an example for which
$$
\dfrac{d}{dx}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy
$$
(by @fresh_42 )

Note: I'm going to post this incomplete answer so you can get started, then come back and do the rest later...

7. a) Let $I_a:=\int_1^\infty \frac{\log(x)}{x^3}\,dx$. Let $x=e^u\Rightarrow dx=e^u du$ hence

$$I_a=\int_0^\infty ue^{-3u}e^u \, du=\int_0^\infty ue^{-2u}\, du=\int_0^\infty \tfrac{v}{2}e^{-v}\, \tfrac{dv}{2}=\tfrac{1}{4}\Gamma (2)=\boxed{\tfrac{1}{4}}$$

7. b) Determine for which $\alpha$ the integral $\int_0^\infty x^2\exp(-\alpha x)\,dx$ converges.
Let $I_{\alpha}:=\int_0^\infty x^2e^{-\alpha x}\,dx$. For $\alpha$ positive, let $u=\alpha x$, then

$$I_{\alpha}=\int_0^\infty \left( \tfrac{u}{\alpha}\right) ^2e^{-u}\,\tfrac{du}{\alpha}=\tfrac{1}{\alpha ^3}\int_0^\infty u^2e^{-u}\,du=\tfrac{1}{\alpha ^3}\Gamma (3)=\tfrac{2}{\alpha ^3}$$

Hence $I_{\alpha}$ converges for positive $\alpha$. $I_{\alpha}$ obviously diverges for $\alpha$ non-positive. Let $\alpha :=a+b i$ then $\left| I_{a+b i}\right| \leq\tfrac{2}{a^3}$ whenever $a>0$. Hence $I_{\alpha}$ converges for $\Re \left[\alpha \right] >0$.

7. c) Find a sequence of functions $f_n\, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,n\in \mathbb{N}$ such that $$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$

Define $f_n (x):=\begin{cases}xne^{-\tfrac{1}{2}nx^2} & \text{if } x \geq 0 \\0 & \text{if } x \leq 0\end{cases}$

Aside: Let $|a|<1$. Consider the sequence $\left\{ na^n\right\}$: this sequence trivially converges to zero if $a=0$. For $0<|a|<1$, we may write, with $\theta >0$

$$|a|=\tfrac{1}{1+\theta}\Rightarrow |a^n|=\tfrac{1}{1+\binom{n}{1}\theta +\cdots +\binom{n}{n}\theta ^n}$$

so we have for $n=1, 2, 3,\ldots$,

$$|a^n|<\tfrac{1}{1+\binom{n}{2}\theta ^2}\Rightarrow |na^n|<\tfrac{1\cdot 2}{(n-1)\theta ^2}$$

Thus we have $|na^n|<\epsilon ,$ as soon as, $\tfrac{1\cdot 2}{(n-1)\theta ^2}< \epsilon$
i.e. for every $$n>1+\tfrac{2}{\epsilon \cdot \theta ^2}$$
Hence $na^n\rightarrow 0$ for $|a|<1$. End Aside.

Back to our sequence of functions $\left\{ f_n\right\}$: Let

$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx=\sum_{n=1}^\infty \int_{0}^\infty xne^{-\tfrac{1}{2}nx^2}\,dx=\sum_{n=1}^\infty \int_{0}^\infty u^{0}e^{-u}\,du=\sum_{n=1}^\infty \Gamma (1) \rightarrow +\infty$

By the aside with $a=e^{-\tfrac{1}{2}x^2}<1\forall x>0$, we have

$\int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx=\int_\mathbb{R} 0 \, dx=0$

hence

$$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$.

7. d) Find a family of functions $f_r\, : \,\mathbb{R}^+\longrightarrow \mathbb{R}\, , \,r\in \mathbb{R}$ such that
$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx\neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx$$

Define $f_r(x):=\begin{cases}\left\|\tfrac{1}{r}\right\| x(1-x^2)^{\left\|\tfrac{1}{r}\right\| }& \text{if } 0\leq x \leq 1 \\0 & \text{ otherwise } \end{cases}$

where $\left\| y\right\|:=\text{nint}(y)$. Let n be the integer $n:=\left\| \tfrac{1}{r}\right\|$,

$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx=\lim_{r \to 0}\int_{0}^{1}\left\|\tfrac{1}{r}\right\| x(1-x^2)^{\left\|\tfrac{1}{r}\right\| }\, dx$$
$$ =\lim_{n \to \infty}\tfrac{n}{2}\int_{0}^{1}u^{n}\, du =\lim_{n \to \infty}\tfrac{n}{2(n+1)}=\tfrac{1}{2} $$
but
$$ \int_{\mathbb{R}} \lim_{r \to 0} f_r(x) \, dx = \int_{0}^{1} \lim_{r \to 0} \left\| \tfrac{1}{r} \right\| x(1-x^2)^{ \left\| \tfrac{1}{r} \right\|} \, dx $$
$$ = \int_{0}^{1} \lim_{n \to\infty} n x(1-x^2)^{n}\,dx =0 $$
where the limit was evaluated by the aside to problem 7 c). Hence
$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx\neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx$$

 

[fresh_42]

I don't see why you can swap limit and sum in 7.c), i.e. why $\sum_{\mathbb{N}}f_n(x) = 0$. This wasn't what you have shown.

... you know that you can look up the solutions?