# Help on a Putnam Problem from the 90s

Let $$f$$ be an infinitely differentiable real-valued function defined on the real numbers. If

$$f\left(\frac{1}{n}\right) = \frac{n^2}{n^2+1}, \quad n = 1,2,3,...,$$

compute the values of the derivatives $$f^{(k)}(0)$$ for $$k = 1,2,3,\dots$$.

I had a friend who gave me the recommendations to consider instances where we will have $$f\left(\frac{1}{n}\right) = \frac{1}{n}$$ $$\implies \frac{1}{n} = \frac{n^2}{n^2 + 1}$$ $$\implies n + \frac{1}{n} = n^2$$ From what I got on Wolfram, there are no integer solutions to this problem, and thus I was back square one on this problem. I guess my real problem is that I have no way of inferring what $$f(x)$$ looks like for all $$x \in \mathbb{R}$$ based on my current information of $$f$$. Any recommendations on this problem?

• So... $f(x)=\frac1{1+x^2}$? Oct 6, 2016 at 0:23
• Not quite: $1/(1+x^2) + e^{-1/x^2}\sin(\pi/x)$ has the same property.
– zhw.
Oct 6, 2016 at 1:41
• Reference: This is problem A4 from the 1992 competition.
– user940
Oct 6, 2016 at 9:53

Lemma: If $g\in C^\infty(\mathbb R),$ $a_1 > a_2 > \cdots \to 0,$ and $g(a_n) = 0$ for all $n,$ then $g^{(k)}(0) = 0$ for all $k.$

Proof: We have $g(0)=0$ by continuity. By Rolle's theorem, for each $n,$ $g'(c_n) = 0$ for some $c_n\in (a_{n+1},a_n).$ Note that $c_1>c_2>\ \cdots \to 0.$ This implies $g'(0)=0$ by continuity of $g'.$ Now $g'\in C^\infty$ and vanishes along the sequence $c_n,$ so we can apply the same argument to it to conclude $g''(d_n) = 0, d_n\in (c_{n+1},c_n).$ We again get $d_1>d_2 > \cdots \to 0,$ so $g''(0)=0.$ This process can be repeated indefinitely, giving the conclusion.

Back to our problem: Let $g(x) = f(x)-1/(1+x^2).$ Then $g\in C^\infty$ and $g(1/n) = 0$ for all $n.$ From the lemma, $g^{(k)}(0) = 0$ for all $k.$ Thus the derivatives of $f$ at $0$ equal the corresponding derivatives of $1/(1+x^2).$ Since $1/(1+x^2) = 1-x^2 + x^4 - \cdots$ for $|x|<1,$ the derivatives of $f$ can be read off from the power series as $f^{(2k)}(0) = (-1)^k(2k)!, f^{(2k+1)}(0)=0$ for all $k.$

If $f(x)$ is analytic (around $0$), we would then have $f(x)=\frac1{1+x^2}$

See then the geometric series expansion of $f(x)$.

$$f(x)=\frac1{1+x^2}=\sum_{n=0}^\infty(-1)^nx^{2n}$$

By Taylor's theorem,

$$f^{(k)}(0)=\begin{cases}0;&k\text{ is odd}\\k!;&(k\mod4)=0\\-(k!);&{(k\mod4)=2}\end{cases}$$

• $f(x)=\frac{1}{1+x^2}\;\;$ maybe, but you need to prove that, first.
– dxiv
Oct 6, 2016 at 0:35
• If we assume that $f$ is analytic then $f(x) = \frac{1}{1+x^2}$ since they agree on a set with an accumulation point. Oct 6, 2016 at 0:41
• @Vik78 Good point. Oct 6, 2016 at 0:49
• The problem doesn't assume that $f$ is analytic, but it's likely that was intended. Oct 6, 2016 at 0:52
• @TobiasKienzler It doesn't. Oct 6, 2016 at 11:39

Partial Answer, way too long for a comment:

$f'(0)$ is easy to calculate:

By the mean value Theorem, there exists $c_n \in (\frac{1}{2n}, \frac{1}{n})$ such that

$$f'(c_n)=\frac{f(\frac{1}{n})-f(\frac{1}{2n})}{\frac{1}{n}-\frac{1}{2n}}=2n \left(\frac{n^2}{n^2+1}-\frac{4n^2}{4n^2+1}\right)=-\frac{6n^3}{(n^2+1)(4n^2+1)}$$

Now, since $f$ is twice differentiable, $f'$ is continuous at $x=0$.

By the squeeze Theorem, $\lim_n c_n=0$.

Therefore, as $f'$ is continuous at $x=0$, we have

$$f'(0)=\lim_n f'(c_n) =0$$

For $f''(0)$ Use the following version of MVT:

If $f$ is twice differentiable on $[a,b]$ and $f'$ is continuous on $(a,b)$, then there exists some $c \in (a,b)$ such that $$f(b)-2 f\left(\frac{a+b}{2}\right)+f(a)=\frac{1}{4} (b-a)^2 f''(c)$$

Now, the tricky part is to pick $a=\frac{1}{n}$ and $b=\frac{1}{m}$ such that $\frac{a+b}{2}=\frac{1}{k}$.

One way to do this is to start from $$\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \Rightarrow \\ \frac{1}{3n}+\frac{1}{6n}=\frac{1}{2n} =\frac{2}{4n} \Rightarrow \\ \frac{\frac{1}{3n}+\frac{1}{6n}}{2} =\frac{1}{4n}$$

Use the generalized version of MVT for $a=\frac{1}{6n}$ and $b=\frac{1}{3n}$ to get a formula for $f''(c_n)$, and then use that by the same argument as above we have $$f''(0)=\lim_n f''(c_n)$$

Unfortunately, this method cannot most likely be extended too much,but who knows. For each order of derivatives there should be a version of MVT, but it will most likely be tricky to make sure that all terms are of the given form. And the most likely replacement for the generalized MVT, the Lagrange remainder estimate for the Taylor remainder, includes derivatives at $\frac{1}{n}$ in the formula, which you cannot most likely not get. (you could apply this formula at $x=\frac{1}{n}$ or $x=0$, but each leads to the same issue).

But maybe someone can get an idea from here and find a solution.

• I like this method... for its purity. Oct 6, 2016 at 1:02
• @SimpleArt Unfortunately, while I can most likely extend the solution to few more derivatives, I don't see this leading to a solution which would work for all $k's$... But maybe someone smarter can find one :) Oct 6, 2016 at 1:07
• Hm, taking a quick glance, you are considering we approach this with the Grünwald–Letnikov derivative, upon which we want to solve systems of equations that have variables of the form $\frac1n$... I'll think about this. Oct 6, 2016 at 1:10
• Try $\frac1{n!},\frac1{(n+1)!}$ for the $n$th derivative. Oct 6, 2016 at 1:15
• @SimpleArt Thank you for the link, I was not aware of this. And yes this seem to be the approach, the problem I see is making all $x+mh$ of the form $\frac{1}{n}$.... Oct 6, 2016 at 1:15

For variety... recall the example that if $f$ is twice differentiable at $0$, then we can derive

$$f''(0) = \lim_{h\to 0} \frac{f'(h) - f'(0)}{h} = \lim_{h \to 0} \frac{f(2h) - 2 f(h) + f(0)}{h^2}$$

Or more generally,

$$f^{(k)}(0) = \lim_{h \to 0} \frac{1}{h^k} \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} f(ih)$$

In particular, if $g(x) = \frac{1}{1+x^2}$, then these limits prove $f^{(k)}(0) = g^{(k)}(0)$, by considering the subsequence of $h$ of the form $\frac{1}{k! n}$.

First a simple reformulation: since $\frac{n^2}{n^2+1}=\frac1{1+(1/n)^2}$ for all $n>0$, the hypothesis is that $f$ takes the same values as $g:x\mapsto\frac1{1+x^2}$ for $x\in S=\{\,\frac1n\mid n\in\Bbb N_{>0}\,\}$ and is smooth. It is not given that $f$ is analytic, in which case it would have to be equal to $g$ (two analytic functions that agree on a set with a limit point must be equal), and one could just compute the derivatives of$~g$ at$~0$, which is easy. For the record, being an even function the odd derivatives at$~0$ are all zero, and the $2n$-th derivative of$~g$ at$~0$ equals $(-1)^n(2n)!$.

So the main point of the question is whether we can know these derivatives at$~0$ for certain at all, given only the values of the $f$ on the set $S$ and its smoothness. The answer is "Yes, we can!", thanks to Taylor's theorem. In fact we need only the simplest form of that theorem, which states that if $f$ is $k$ times differentiable at$~0$ then $$f(x)=\sum_{i=0}^k\frac{f^{(k)}(0)}{i!}x^i+o(x^k)\quad\text{as x\to0}.$$

Since $f$ is smooth, at has all its derivatives at$~0$, and therefore Taylor's theorem applies to$~f$ for every$~k\in\Bbb N$, as it does to$~g$ (which is analytic). Now if the derivatives of $f$ at$~0$ did not all coincide with those of$~g$, then one could take $k$ to be the first order for which they differ, and by subtracting the Taylor estimates for $f$ and$~g$ would get $$f(x)-g(x)=cx^k+o(x^k)\quad\text{as x\to0}$$ for some constant $c\neq0$. This would force $f-g$ to be nonzero in some neighbourhood of$~0$ except at$~0$ itself, contradicting the fact that it is zero on$~S$. Hence all derivatives of $f$ at$~0$ coincide with those of$~g$.

Note that not only do we not need that $f$ is analytic, we don't even need that it is smooth (on$~\Bbb R$ or in a neighbourhood of$~0$), just that all its derivatives at$~0$ exist. One couldn't possibly weaken that hypothesis on$~f$ even further, given that the question is to determine those derivatives.