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# Solution Set I

In our first blog post Problem set 1 we presented three questions. We received many emails regarding the blog and were happy to see the interest it generated. There were also (different) solutions posted on our blog site. In this blog we present our set of solutions, we look forward to feedback on the same. Our next set of questions will be posted in middle of May.

(High School Students) Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1\times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?

Solution : The only possible areas in the answer are among the perfect squares 1, 4, 9, 16, 25, 36, 49. You can experiment with different arrangements. Graph paper would be helpful, and you might enjoy cutting the possible squares out of paper and shifting them around physically.

(Undergraduate Students) Given $f\in C([0,\infty ))$ such that $f(x)\rightarrow 0$ as $x\rightarrow \infty$ show that for any $\epsilon >0$ there is a polynomial $p$ such that

$\left\vert f(x)-e^{-x}p(x)\right\vert <\epsilon \,\,\, \forall x\in \lbrack 0,\infty ).$

Solution : Initial hunch is to hammer this problem to with the Weierstrass Approximation Theorem (see Terrence Tao’s blog for a more general version of the same). However, one needs to be careful in getting uniformity in approximation. Any simplification of the below proof is welcome.

Define a function $g:[0.1] \rightarrow \mathbb{R}$ by

$g(x)= \left \{ \begin{array}{ll} f(-\log (x)) & \mbox{ if } 0

Let $\epsilon >0$ be given. By Weierstrass Approximation Theorem ($g$ is continuous ?) we can find a polynomial $q : [0,1] \rightarrow \mathbb{R}$ such that

$\left\vert g(x)-q(x)\right\vert <\epsilon /2 \hspace{1.5in} (1)$

for $0\leq x\leq 1.$ Suppose $q(x) = \sum_{j=0}^{N}c_{j}x^{j}$ then from the above it is easy to note that $\left\vert c_{0}\right\vert <\epsilon /2$ (why ?). Translating the inequality in (1) for $f$ we have

$\left\vert f(x)-\sum_{j=1}^{N}c_{j}e^{-jx}\right\vert <\epsilon \hspace{1.5in} (2)$

for all $x\geq 0.$ Note that $N$ is now fixed. To finish the proof, using (2), we need to show that for $j \in \mathbb{N}$ and any $\eta >0$ there is a polynomial $p_j : [0,1] \rightarrow\mathbb{R}$ such that

$\left\vert e^{-jx}- e^{-x}p_j(x)\right\vert <\eta \hspace{1.5in} (3)$

for $0 \leq x.$ We prove the above by the method of mathematical induction on $j.$ Let $\eta >0$ be given.

• Verification of base case: For $j=1$ we take $p=1.$
• Inductive step assumption: For $j= k$ assume that there exists a polynomial $p_k : [0,1] \rightarrow \mathbb{R}$ such that

$\left\vert e^{-kx}-e^{-x}p_k(x)\right\vert <\eta \hspace{1.5in} (4)$

for all $x \geq 0.$

• Proof for $j =k+1$, : Let $x \geq 0.$ Using (4) with $x$ replaced by $\frac{k+1}{k}x$ we have

$\left\vert e^{-(k+1)x}-e^{-(\frac{k+1}{k})x}p_k(\frac{k+1}{k}x)\right\vert <\eta\,\, \mbox{ for all } x \geq 0 \hspace{1.5in}(5)$

We will now to find a polynomial $h$ such that $e^{-x}h(x)$ approximates $e^{-(\frac{k+1}{k})x}p_k(\frac{k+1}{k}x).$ Let $n \geq 1, \mbox{ and } \phi_n(x)=\sum_{m=0}^{2n}\frac{(-x/k)^{m}}{m!}.$ Then

$\left\vert e^{-x}\phi_n(x) p_k(\frac{1+k}{k}x)-e^{-(1+\frac{1}{k})x}p_k(\frac{1+k}{k}x)\right\vert = e^{-x(1-\frac{1}{k})} p_k(\frac{1+k}{k}x) e^{-\frac{x}{k}} \left\vert \phi_n(x) -e^{-\frac{x}{k}}\right\vert \hspace{.5in}(6)$

We estimate the right hand side by observing:-
(a)

$M\equiv M(k,p_k,\eta):= \sup \{\left\vert e^{-x(1-\frac{1}{k})}p_k(\frac{1+k}{k}x)\right\vert :x\geq 0\} < \infty \hspace{0.5in}$ (why ?)

and
(b) for any $n \geq 1, x \geq 0$

$e^{-\frac{x}{k}} \left\vert e^{-x/j}-\phi (x)\right\vert \leq e^{-\frac{x}{k}} \left\vert \frac{(-x/j)^{2n}}{(2n)!}\right\vert \leq e^{-2n}\frac{(2n)^{2n}}{(2n)!},$

where we have used the inequality

$\sum_{m=0}^{2n-1}\frac{ (-x/k)^{m}}{m!}\leq e^{-x/k}\leq \sum_{m=0}^{2n}\frac{(-x/k)^{m}}{m!}, \hspace{1in}$ (Taylor’s Theorem)

and the fact that $e^{-\frac{x}{k}}$ attains its maximum at
the point $x=2kn.$

(c) Stirling’s formula, implies that there exists $c>0, L \in \mathbb{N}$ such that if $n \geq L$ then

$e^{-2n}\frac{(2n)^{2n}}{(2n)!} \leq \frac{c}{\sqrt{n}}.$

Therefore if $n_0 \in \mathbb{N}$ and $n_0> \max\{L, \frac{Mc}{\eta}\}$

$e^{-x(1-\frac{1}{k})} p_k(\frac{1+k}{k}x) e^{-\frac{x}{k}} \left\vert \phi_{n_0}(x) -e^{-\frac{x}{k}}\right\vert \leq \frac{M c}{\sqrt{n_0}} < \eta \hspace{1.5in}(7)$

If we define the polynomial $h$ by $h(x) = \phi_{n_0}(x) p_k(\frac{1+k}{k}x)$ for $x \geq 0$, from (5), (6), and (7) we have

$\left\vert e^{-(k+1)x} - e^{-x}h(x) \right\vert < 2\eta.$

As $\eta$ was arbitrary we have proved the case $j=k+1$..

This completes the proof.

(Masters Students and above) If $K$ is a compact subset of $\mathbb{R}^{n}$ show that the set

$A=\{x\in \mathbb{R}^{n}: d(x,K)=1\}$

has Lebesgue measure $0.$

Solution :
Suppose $K\subset B(0,\frac{1}{3}), x \in \mathbb{R}^n \mbox{ with } d(x,K)=1.$ Let $0 Let $y\in K$ with $\left\Vert x-y\right\Vert =1.$ Then

$\begin{array}{ll}\left\Vert tx-y\right\Vert ^{2}-\left\Vert x-y\right\Vert ^{2}&=\left\Vert tx-x\right\Vert ^{2}+2 \cr &=\left\Vert tx-x\right\Vert ^{2}+2(t-1)\left\Vert x\right\Vert ^{2}-2\cr &=(t^{2}-1)\left\Vert x\right\Vert ^{2}+2(1-t)\left\Vert x\right\Vert \left\Vert y\right\Vert. \end{array}$

Now note that $\left\Vert y\right\Vert \leq \frac{1}{3}$ and $\left\Vert x\right\Vert \geq \frac{2}{3}.$Hence

$\left\Vert tx-y\right\Vert ^{2}-\left\Vert x-y\right\Vert ^{2}<0$

proving that $tx\notin A.$

Similarly if $t>1$ then

$\left\Vert tx-y\right\Vert ^{2}-\left\Vert x-y\right\Vert ^{2}=(t^{2}-1)\left\Vert x\right\Vert ^{2}+2(1-t)\left\Vert x\right\Vert \left\Vert y\right\Vert >\frac{2}{3}(t^{2}-1)-2(t-1)\frac{1}{3}>0$

so $tx\notin A.$

Therefore we have shown that $K\subset B(0,\frac{1}{3})$ then

$x\in A\Rightarrow tx\notin A$

for any $t\in (0,\infty )\backslash \{1\}$.

Using polar coordinates (See Real and Complex Analysis by Walter Rudin, 3rd Edition
Problem 6, Chapter 8) we conclude from that $A$ has measure $0.$

By translation the same conclusion holds if $K$ is contained in some
open ball of radius $\frac{1}{3}.$ The general case is handled by noting
that $K$ is the union of a finite number of compact subsets of diameter not
exceeding $\frac{1}{3}$ and any point in $A$ has distance $1$ from one of
these subsets.

This completes the proof.

## 7 thoughts on “Solution Set I”

1. Please tell me if there is some problem in this proof of the undergraduate problem, which I feel is much simpler than the given proof. Thank you!

It is clear that for any polynomial $p(x)$, $\lim_{x \to \infty} e^{-x}p(x) = 0$.

We are given that $\lim_{x \to \infty}f(x) = 0$, i.e. for given $\epsilon > 0$, $\exists M$ such that $\forall x > M$, $|f(x)-0|=|f(x)| M, |f(x)| N$, $|e^{-x}p(x)| L$,

$|f(x)-e^{-x}p(x)| < |f(x)|+|e^{-x}p(x)| < \epsilon /2+ \epsilon /2 = \epsilon$

Hence for given $\epsilon$ we now need to find a polynomial $p(x)$ such that $\forall x \in [0, L]$, $|f(x) - e^{-x}p(x)| < \epsilon.$ Equivalently, $|e^{x}f(x) - p(x)|< \epsilon \cdot e^x < \epsilon \cdot e^L=\epsilon '$. This is given by the Weierstrauss Approximation Theorem applied on $h(x) = e^xf(x)$ on $[0, L]$.

Hence proved.

(Apologies for multiple posts, I am new to both LaTeX and WordPress)

• The argument is circular. The interval $[0,L]$ and the polynomial $P$ are interdependent.

2. Please tell me if there is some problem in this proof of the undergraduate problem, which I feel is much simpler than the given proof. Thank you!

It is clear that for any polynomial $p(x)$, $\lim_{x \to \infty} e^{-x}p(x) = 0$.

We are given that $\lim_{x \to \infty}f(x) = 0$, i.e. for given $\epsilon>0$, $\exists M$ such that $\forall x>M$, $|f(x)-0|=|f(x)|M, |f(x)|N$, $|e^{-x}p(x)|L$,

$|f(x)-e^{-x}p(x)|<|f(x)|+|e^{-x}p(x)|<\epsilon /2+ \epsilon /2 = \epsilon$

Hence for given $\epsilon$ we now need to find a polynomial $p(x)$ such that $\forall x\in[0, L]$, $|f(x) - e^{-x}p(x)|< \epsilon.$ Equivalently, $|e^xf(x)-p(x)|<\epsilon \cdot e^x<\epsilon \cdot e^L=\epsilon '$. This is given by the Weirstrauss Approximation Theorem applied on $h(x) = e^xf(x)$ on $[0, L]$.

Hence proved.

(Apologies for multiple posts, I was not aware of how to use LaTeX on WordPress.)

3. Please tell me if there is some problem in this proof of the undergraduate problem, which I feel is much simpler than the given proof. Thank you!

It is clear that for any polynomial $p(x)$, $\lim_{x \to \infty} e^{-x}p(x) = 0$.

We are given that $\lim_{x \to \infty}f(x) = 0$, i.e. for given $\epsilon>0$, $\exists M$ such that $\forall x>M$, $|f(x)-0|=|f(x)|M, |f(x)|N$, $|e^{-x}p(x)|L$,
$$|f(x)-e^{-x}p(x)|<|f(x)|+|e^{-x}p(x)|<\epsilon /2+ \epsilon /2 = \epsilon$$

Hence for given $\epsilon$ we now need to find a polynomial $p(x)$ such that $\forall x\in[0, L]$, $|f(x) - e^{-x}p(x)|< \epsilon.$ Equivalently, $|e^xf(x)-p(x)|<\epsilon \cdot e^x<\epsilon \cdot e^L=\epsilon '$. This is given by the Weierstrauss Approximation Theorem applied on $h(x) = e^xf(x)$ on $[0, L]$.

Hence proved.

4. Please tell me if there is some problem in this proof of the undergraduate problem, which I feel is much simpler than the given proof. Thank you!

It is clear that for any polynomial $p(x)$, $\lim_{x \to \infty} e^{-x}p(x) = 0$.

We are given that $\lim_{x \to \infty}f(x) = 0$, i.e. for given $\epsilon>0$, $\exists M$ such that $\forall x>M$, $|f(x)-0|=|f(x)|M, |f(x)|N$, $|e^{-x}p(x)|L$,
$$|f(x)-e^{-x}p(x)|<|f(x)|+|e^{-x}p(x)|<\epsilon /2+ \epsilon /2 = \epsilon$$

Hence for given $\epsilon$ we now need to find a polynomial $p(x)$ such that $\forall x\in[0, L]$, $|f(x) – e^{-x}p(x)|< \epsilon.$ Equivalently, $|e^xf(x)-p(x)|<\epsilon \cdot e^x<\epsilon \cdot e^L=\epsilon '$. This is given by the Weirstrauss Approximation Theorem applied on $h(x) = e^xf(x)$ on $[0, L]$.

Hence proved.

5. Kavi Rama Murthy says:

Well, the idea is to use the standard approximation of continuous functions by polynomials on a closed interval. Since such an approximation is not valid for an infinite interval we had to ‘transform’ the given infinite interval to a finite closed interval. The first choice was the transformatiom y=1/x, but this did not solve the problem. So as a next trial we looked at -log(x) and it worked (with some effort, of course). Incidentally, we do not intend to post routine exrecises because there are plenty of them in text books. The problems are not intended to have one step solutions.

6. Hmmm. Questions are quite hard. I have one question to ask regarding the undergraduate problem: What made you to define $g(x)$ in the manner which you have defined?