## Putnam Calculus Problem: Sine and Cosine

Supposedly, the following is a former Putnam Exam, which I came across here on Math.StackExchange.

Suppose $f,g: \mathbb{R}\rightarrow\mathbb{R}$ are nonconstant, differentiable functions satisfying

$\displaystyle f(x+y)=f(x)f(y)-g(x)g(y),\indent g(x+y)=f(x)g(y)+f(y)g(x)$

and $f'(0)=0$ for all $x,y\in\mathbb{R}$. Prove that

$\displaystyle f(x)^{2}+g(x)^{2}=1$

Here is my attempt at a solution. I believe it’s correct–otherwise, I would not be posting it. If it’s not, please drop me a comment highlighting any errors.

Proof. Observe that

$\begin{array}{lcl}\displaystyle f(x)^{2}+g(x)^{2}&=&\displaystyle\left(f\left(\frac{x}{2}\right)^{2}-g\left(\frac{x}{2}\right)^{2}\right)^{2}+\left(f\left(\frac{x}{2}\right)g\left(\frac{x}{2}\right)+g\left(\frac{x}{2}\right)f\left(\frac{x}{2}\right)\right)^{2}\\[.7 em]&=&\displaystyle f\left(\frac{x}{2}\right)^{4}-2f\left(\frac{x}{2}\right)^{2}g\left(\frac{x}{2}\right)^{2}+g\left(\frac{x}{2}\right)^{4}\\[.7 em]&+&\displaystyle f\left(\frac{x}{2}\right)^{2}g\left(\frac{x}{2}\right)^{2}+2f\left(\frac{x}{2}\right)^{2}g\left(\frac{x}{2}\right)^{2}+g\left(\frac{x}{2}\right)^{2}f\left(\frac{x}{2}\right)^{2}\\[.7 em]&=&\displaystyle\left(f\left(\frac{x}{2}\right)^{2}+g\left(\frac{x}{2}\right)^{2}\right)^{2}\end{array}$

By induction, we see that

$\displaystyle\left(f(x)^{2}+g(x)^{2}\right)^{2^{-n}}=f\left(\frac{x}{2^{n}}\right)^{2}+g\left(\frac{x}{2^{n}}\right)^{2}$

Letting $n\rightarrow\infty$ and using the continuity of $f$ and $g$, we see that

$\displaystyle1=\lim_{n\rightarrow\infty}\left(f(x)^{2}+g(x)^{2}\right)^{2^{-n}}=\lim_{n\rightarrow\infty}f\left(\frac{x}{2^{n}}\right)^{2}+g\left(\frac{x}{2^{n}}\right)^{2}=f(0)^{2}+g(0)^{2}$

I claim that $g(0)=0$. Suppose not. Then

$\displaystyle g(0)=g(0+0)=2g(0)f(0)\Longrightarrow f(0)=\frac{1}{2}$,

which is a contradiction since $f(0)^{2}+g(0)^{2}=1$. Moreover,

$\displaystyle f(0)=f(0+0)=f(0)^{2}-g(0)^{2}=f(0)^{2}\Longrightarrow f(0)=1$

Using the limit definition of the derivative, we obtain the identities

$\begin{array}{lcl}\displaystyle f'(x)=\lim_{h\rightarrow0}\dfrac{f(x+h)-f(x)}{h}&=&\displaystyle\lim_{h\rightarrow 0}\dfrac{f(x)f(h)-g(x)g(h)-f(x)}{h}\\[.7 em]&=&\displaystyle\lim_{h\rightarrow0}\dfrac{f(x)[f(h)-1]-g(x)g(h)}{h}\\[.7 em]&=&\displaystyle f(x)f'(0)-g(x)g'(0)\\[.7 em]&=&\displaystyle-g(x)g'(0),\end{array}$

$\begin{array}{lcl}\displaystyle g'(x)=\lim_{h\rightarrow0}\dfrac{g(x+h)-g(x)}{h}&=&\displaystyle\lim_{h\rightarrow0}\dfrac{f(x)g(h)+g(x)f(h)-g(x)}{h}\\[.7 em]&=&\displaystyle\lim_{h\rightarrow0}\dfrac{f(x)g(h)+g(x)[f(h)-1]}{h}\\[.7 em]&=&\displaystyle f(x)g'(0)+g(x)f'(0)\\[.7 em]&=&\displaystyle f(x)g'(0)\end{array}$

Hence,

$\begin{array}{lcl}\displaystyle \frac{d}{dx}\left[f(x)^{2}+g(x)^{2}\right]=2\left[f(x)f'(x)+g(x)g'(x)\right]&=&\displaystyle2\left[-f(x)g(x)g'(0)+g(x)f(x)g'(0)\right]\\&=&\displaystyle0,\end{array}$

which implies that the function $x\mapsto f(x)^{2}+g(x)^{2}$ is constant. Since $f(0)^{2}+g(0)^{2}=1$, we conclude that

$\displaystyle f(x)^{2}+g(x)^{2}=1,\indent\forall x\in\mathbb{R}$

$\Box$

Additionally, we can ask what are the only nonconstant differentiable functions $f,g: \mathbb{R}\rightarrow\mathbb{R}$ that satisfy the above hypotheses. I claim that $f(x)$ and $g(x)$ must be, respectively, of the forms

$\displaystyle f(x)=\cos(\alpha x),\indent g(x)=\sin(\alpha x)$,

where $\alpha\in\mathbb{R}$. Our formulas for $f'(x)$ and $g'(x)$ obtained above show that $f$ and $g$ are infinitely differentiable. Since $\max\left\{\left|f(x)\right|,\left|g(x)\right|\right\}\leq1$, by induction, we see that

$\displaystyle \max\left\{\left|g^{(n)}(x)\right|,\left|f^{(n)}(x)\right|\right\}\leq\left|g'(0)\right|^{n},\indent\forall x\in\mathbb{R},\forall n\in\mathbb{Z}^{\geq0}$

I claim that $f$ and $g$ are analytic and $x=0$ with infinite radius of convergence. Fix $x\in\mathbb{R}$. For each $n\in\mathbb{Z}^{\geq 0}$, Taylor’s theorem with remainder (the Lagrange form) tells us that there exists $\xi_{n}\in[0,x]$ such that

$\displaystyle f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}+\frac{f^{(n+1)}(\xi_{n})}{(n+1)!}x^{n+1}$

But

$\displaystyle\left|f^{(n+1)}(\xi_{n})\right|\frac{\left|x\right|^{n+1}}{(n+1)!}\leq\frac{\left|x\right|^{n+1}}{(n+1)!}\longrightarrow 0, n \longrightarrow\infty$

We conclude that $f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^{k}$, where the series converges uniformly on compact subsets. A completely analogous argument shows that $g(x)=\sum_{k=0}^{\infty}\frac{g^{(k)}(0)}{k!}x^{k}$. We can now use the power series method to find series expressions for $f$ and $g$. An induction argument together with the result $f(0)=1$ shows that

$\displaystyle f^{(n)}(0)=\begin{cases}{(-1)^{k}(g'(0))^{n}}&{n=2k,k\in\mathbb{Z}^{\geq0}} \\ {0}&{n=2k+1,k\in\mathbb{Z}^{\geq0}}\end{cases}$

Similarly,

$\displaystyle g^{(n)}(0)=\begin{cases}{(-1)^{k}(g'(0))^{n}}&{n=2k+1,k\in\mathbb{Z}^{\geq0}} \\ {0}&{n=2k,k\in\mathbb{Z}^{\geq0}}\end{cases}$

We conclude that

$\displaystyle f(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}(g'(0)x)^{2k}}{(2k)!}x^{2k}=\cos(g'(0)x)$

and

$\displaystyle g(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}(g'(0)x)^{2k+1}}{(2k+1)!}=\sin(g'(0)x)$

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### 2 Responses to Putnam Calculus Problem: Sine and Cosine

1. NoOne says:

Your statement of the problem is wrong — you have an incorrect sign in the second equation and you have omitted the condition $f'(0)=0$.