read this if you’re new to the Putnam series

### Due:

13/6/2018 Two problems.

### Problems:

##### 1996 Putnam | A6

Let $$c > 0$$ be a constant. Give a complete description, with proof, of the set of all continuous functions $$f: \mathcal{R} \rightarrow \mathcal{R}$$ such that $$f(x) = f(x^2 + c) \forall x \in \mathcal{R}$$.

Note: $$\mathcal{R}$$ is set of real numbers.

_ I was super late in doing this. Like, a week late. Not that I wasn’t trying but, I am still trying to work on finding time._

I failed to do it when I did the next problem. Lol

This problem is beautiful. If $$f(x) = f(x^2 + c)$$, then let’s look at all the roots of $$x^2 -x + c$$, the idea is so simple yet so beautiful and that I am dumbfounded that such an ideology exists and that I didn’t think like that before.

$$x^2 + c = x$$ for some $$x$$ which is a root.

Writing discriminant $$D^2 = 1 - 4c$$ and noting that there are two cases, when $$c > \frac{1}{4}$$ and otherwise.

For $$c > \frac{1}{4}$$, the equation has no real roots. Let $$x_n = x_{n-1}^2 + c$$, then, $$f(x)$$ defined on $$[x_{n-1}, x_{n}]$$ can be transformed to be defined on $$[x_{n}, x_{n+1}]$$ and so on.

In the absense of real roots, this is the extent to which we can say. This is a generic statement. And, if in case there are real roots,

when $$c \leq \frac{1}{4}$$, let the real roots be $$\alpha, \beta$$ with $$\alpha \leq \beta$$.

Looking at the $$f(x) = f(x^2 + c)$$, one captures that motion on x-axis is increaing. Let’s suppose we have a very large value and we want to see all the previous values where the function value is the same. The way we define the sequence then is $$x_{n+1} = \sqrt{x_n + c}$$. Let’s find the limit of this sequence, noting that this is a decreasing sequence.

##### 1996 Putnam | B4

For any square matrix $$A$$, we can define $$sinA$$ by the usual power series:

$sinA = \sum_{n=0}^\infty \frac{ (-1)^n} {(2n+1)!} A^{2n+1}$

Prove or disprove: there exists a $$2\times 2$$ matrix $$A$$ with real entries such that

$sinA = \begin{bmatrix} 1 & 1996 \\ 0 & 1 \end{bmatrix}$

_ I was super late in doing this. Like, a week late. Not that I wasn’t trying but, I am still trying to work on finding time._

I didn’t get the answer and since I was already two weeks late, I looked up the answer. I know. *sigh*

In my defense, I think I was on the right track but didn’t get intuition on how to proceed forward; but now that I have seen this, I’m intuition-ed to think like this more.

Suppose, $$\exists A \in \mathcal{R}_{2\times 2}$$

Further suppose that $$A$$ is not defective, meaning, $$A$$ has $$n (=2)$$ independent eigen values, which would mean that $$A$$ has an eigenvalue decomposition, and thus, diagonalizable.

$A = Q^{-1} \Lambda Q$
 $$Lambda$$being the diagonal matrix made up of the eigenvalues of$$A$$.

Re-expressing $$A$$, we get,

$\Lambda = Q A Q^{-1}$ $\implies sin \Lambda = Q sin A Q^{-1}$

Since, $$Q$$ is orthogonal, $$n$$-the power of matrix is easier to compute. Also, we note that, $$sin \Lambda$$ is diagonal since $$\Lambda$$ is a diagonal matrix.

$$sin \Lambda$$ is diagonal $$\implies$$ $$sin A$$ is diagonal.

This is a contradiction, implying $$A$$ is defective and eigenvalues of $$A$$ are same (since $$A$$ is $$2\times 2$$).

Given any two dimensional matrix, one can write it’s characteristics equation as follows:

$\lambda^2 - tr(A) \lambda + det(A) = 0$

If such a matrix has equal eigenvalues, it implies that

$tr(A)^2 = 4 det(A)$

Assuming, $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$,

The above equation transforms to be,

$(a-d)^2 + 4bc = 0$

Setting, $$a = d$$ implies one of $$b,c$$ is zero.

Therefore, $$A = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$$

Which means,

$sin A = \begin{bmatrix} sin(a) & b cos(a) \\ 0 & sin(a) \end{bmatrix}$

Comparing that with question, we get obvious contradiction. Hence, no such $$A$$ can exist.

QED

My thoughts: I came to like making use of the eigenvalue decomposition but that was the end of my train of thought. Me not happy with this. I should sit and think more about it.