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 . 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 such that as show that for any there is a polynomial such that

**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 by

Let be given. By Weierstrass Approximation Theorem ( is continuous ?) we can find a polynomial such that

for Suppose then from the above it is easy to note that (why ?). Translating the inequality in (1) for we have

for all Note that is now fixed. To finish the proof, using (2), we need to show that for and any $\eta >0$ there is a polynomial such that

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

**Verification of base case:**For we take**Inductive step assumption:**For assume that there exists a polynomial such thatfor all

**Proof for , :**Let Using (4) with replaced by we haveWe will now to find a polynomial such that approximates Let Then

We estimate the right hand side by observing:-

(a)(why ?)

and

(b) for anywhere we have used the inequality

and the fact that attains its maximum at

the point(c) Stirling’s formula, implies that there exists such that if then

Therefore if and

If we define the polynomial by for , from (5), (6), and (7) we have

As was arbitrary we have proved the case ..

This completes the proof.

**(Masters Students and above)** If is a compact subset of show that the set

has Lebesgue measure

**Solution : **

Suppose Let Let with Then

Now note that and Hence

proving that

Similarly if then

so

Therefore we have shown that then

for any .

Using polar coordinates (See Real and Complex Analysis by Walter Rudin, 3rd Edition

Problem 6, Chapter 8) we conclude from that has measure

By translation the same conclusion holds if is contained in *some*

open ball of radius The general case is handled by noting

that is the union of a finite number of compact subsets of diameter not

exceeding and any point in has distance from one of

these subsets.

This completes the proof.

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 , .

We are given that , i.e. for given , such that , , ,

Hence for given we now need to find a polynomial such that , Equivalently, . This is given by the Weierstrauss Approximation Theorem applied on on .

Hence proved.

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

The argument is circular. The interval and the polynomial are interdependent.

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 , .

We are given that , i.e. for given , such that , , ,

Hence for given we now need to find a polynomial such that , Equivalently, . This is given by the Weirstrauss Approximation Theorem applied on on .

Hence proved.

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

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 , .

We are given that , i.e. for given , such that , , ,

$$

Hence for given we now need to find a polynomial such that , Equivalently, . This is given by the Weierstrauss Approximation Theorem applied on on .

Hence proved.

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.

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.

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