# 2012 Putnam Exam A3

This question mostly eluded me during the exam itself.

Problem: Suppose that $f: [-1, 1] \rightarrow \mathbb{R}$ continuously, and that

\begin{align} \text{(i)}\qquad &f(x) = \frac{2 - x^2}{2} f\left(\frac{x^2}{2-x^2}\right)\\ \text{(ii)}\qquad &f(0) = 1\\[6pt] \text{(iii)}\qquad &\lim_{x\to 1^-}\frac{f(x)}{\sqrt{1-x}}\ \text{exists} \end{align}

Determine the closed form of $f$, and prove that it is unique. ---

All that I managed to do was - using the eventual heuristic of "powers of $1 - x^2$ will cancel neatly in (iii) and work in (ii), so let's see if they work in (i)" - discover that

$$f(x) = \sqrt{1 - x^2}$$

Now, this $f$ is an involution, so it suffices to prove that if $g$ is a function satisfying (i) to (iii), then $g(f(x)) = x$. This, however, was more than I could do.

How do you prove uniqueness? (I ran across one solution I did not understand; please, be gentle.)

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Have you seen the discussion at artofproblemsolving.com/Forum/… ? – Gerry Myerson Dec 4 '12 at 4:56
@Gerry I had not. echinodermata's answer there is really clever. Thanks for the recommendation! – Chris Dec 4 '12 at 5:16
@user1296727: perhaps you can post a solution now, to keep this question from going unanswered? – user641 Mar 21 '13 at 23:05

Disclaimer: I am borrowing much from the Art of Problem Solving link alluded to in the comments.

Since you have guessed a solution, namely $x \mapsto \sqrt{1-x^2}$, so let us exploit it. Denote $s(x) := \sqrt{1-x^2}$, and note, as you did, that $s(s(x)) = x$. It is a good idea to express $f$ in an alternative way so that for $f(x) = s(x)$ things become very simple. This is vague, but a way to proceed is to write $f(x) = h(s(x))$ for some continuous $h:[0,1] \to \mathbb{R}$; since this is equivalent to saying that $h(y) = f(s(y))$, there is a 1 to 1 correspondence between possible $f$'s and $h$'s.

Let us see what the conditions say about $h$:

(i) $$h(y) = f(s(y)) = \frac{1 + y^2}{2} f\left(\frac{1-y^2}{1+y^2}\right) = \frac{1 + y^2}{2}h\left(s\left(\frac{1-y^2}{1+y^2}\right)\right) = \frac{1 + y^2}{2} h\left( \frac{2y}{1+y^2} \right)$$ (ii) $h(1) = 1$

(iii) $\lim_{y \to 0+} \frac{h(y)}{y}$ exists.

There are several ways to proceed at this stage. To keep things elegant, let us notice the striking resemblance between the expression $\frac{2y}{1+y^2}$ and the formula for $\tanh$ of doubled angle. In fact, if $y = \tanh \alpha$, then $\frac{2y}{1+y^2} = \tanh 2 \alpha$. Writing additionally $\frac{1+y^2}{2}$ as $\frac{y}{ \frac{2y}{1+y^2}}$ we conclude that (i) can be re-expressed in a nicer form: $$h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} h(\tanh 2\alpha)$$ Iterating this as many times as we like, we conclude: $$h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} \frac{\tanh 2 \alpha}{\tanh 4 \alpha} \dots \frac{\tanh 2^{k-1} \alpha}{\tanh 2^{k} \alpha} h(\tanh 2^k\alpha) = \tanh \alpha \frac{h(\tanh 2^k\alpha)}{\tanh 2^k \alpha}$$ Passing to the limit $k \to \infty$ (and remembering that $\tanh \beta \to 1$ as $\beta \to \infty$ we conclude that: $$h(\tanh \alpha) = \tanh \alpha$$ This means that $h(y) = y$. Translating this back to $f$ we conclude that: $$f(x) = h(s(x)) = s(x) = \sqrt{1-x^2}$$

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