# Exercises for March 23rd 2012 entry

Here are the solutions to the exercises of March 23rd entry, A Motivational Problem for the Study of Fourier Series.

Exercise 1:Show that in polar coordinates, the Laplacian is given by.

**Solution:**

In terms of and , and are functions given by and . We thus have :

Therefore, by the chain rule,

- .

It is important at this point to realize that 1. and 2. are *differential operators* while their evaluation at , for example , are *functions*. So that is the evaluation of 1. at the function . With this clarified, we can write explicitly what we have to calculate :

Applying distributivity of the differential operators on sums of functions and then the Leibniz rule (product rule) when the operator acts on a product of functions, we get, after simplifying,

In a similar way, we obtain (using the commutativity of and )

.

Thus, summing these two expressions and using the identity , we get

as wanted.

QED

Exercise 2:Show that if is a twice differentiable function satisfying for some , then there exists such that .

**Solution:**

We proceed in two steps. First, we show that can be written as for some . We’ll then show how this gives us the desired result.

Let be a twice differentiable function satisfying . But then we have so is also twice differentiable. It’s first and second differentials are given respectively by and By induction, is then infinitely differentiable and we have for even values of and for odd values of . The Taylor series expansion of at is thus given by

But, summing the Taylor series representation of and , we get

Taking and , we get

as wanted.

To conclude, let and . We get, with Euler’s formula,

But since is an even function and an odd function, this can be rewritten as

We also have

and

So putting all this together, we get

as desired.

QED

Exercise 3:Show that if the only solutions of the differential equationwhich are twice differentiable when , are given by linear combinations of and when , and 1 and when .

**Solution:**

If , then we have so that This means that so . We conclude that for some .

In case , we make the substitution . We then have

So we get . By the chain rule, this can be rewritten as which, by the product rule, is equal to But since , this gives us which is equal to by our original hypothesis.

The problem thus reduces to solving the resulting differential equation . But this is very similar to what we have done in exercise 2! In fact, it can be shown that there exists such that (the argument is the same as in exercise 2 but we consider and instead of and ) so, since , we have

which is

Hence

for some as wanted.

QED

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