Find all functions $f∶ \Bbb R \to \Bbb R$ such that, for all $n \in \Bbb N, \left[\frac{{f(x+n) - f(x-n)}}{{2n}} = f'(x)\right]$

Question

There was a similar question in the Putnam ($2010$ A$2$) but I can't wrap my head around this one. I've tried attacking it in some standard ways of eliminating arbitrary functions but it doesn't seem to be an easy manipulation. Also noticed that this is the expression for the average rate change of a differentiable function in the interval $[x-n,x+n]$ but again don't know how I can use that information. Linear and quadratic functions seem to fit the bill but how do you find the other ones or prove that these are the only kinds?

Answer 1

Start with $$f'(x) = \frac12 \Big(f(x+1)-f(x-1)\Big).$$ As $f$ is differentiable, it follows that $\frac12 (f(x+1)-f(x-1))$ is differentiable, so $f'$ is differentiable. Let's differentiate it! $$\begin{align*} f''(x) &= \frac12 \Big(f'(x+1)-f'(x-1)\Big) \\ &= \frac12 \left(\frac12\Big(f(x+2)-f(x)\Big) - \frac12\Big(f(x)-f(x-2)\Big)\right) \\ &= \frac14\Big(f(x+2)-2f(x)+f(x-2)\Big). \end{align*}$$

Once again, we note that since $f$ is differentiable, the expression on the right side of this equation is differentiable, so $f''$ is differentiable. Differentiating again yields $$\begin{align*} f'''(x) &= \frac14\Big(f'(x+2)-2f'(x)+f'(x-2)\Big) \\ &= \frac14\left(\frac12\Big(f(x+3)-f(x+1)\Big) - \frac22\Big(f(x+1)-f(x-1)\Big) + \frac12\Big(f(x-1)-f(x-3)\Big)\right) \\ &= \frac18\Big(f(x+3)-3f(x+1) + 3f(x-1)-f(x-3)\Big) \\ &= \frac18\Big(\big(f(x+3)-f(x-3)\big) - 3\big(f(x+1)-f(x-1)\big)\Big) \\ &= \frac18(6f'(x)-6f'(x)) = 0. \end{align*}$$

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In particular, this shows that if $f$ is differentiable and $$f'(x) = \frac{f(x+n)-f(x-n)}{2n}$$ for every real $x$ and for $n \in \{1,3\},$ then $f$ is thrice-differentiable and $f'''(x) = 0$.

By integrating (or equivalently, applying the mean value theorem to the derivatives of $f(x)-f(0)-f'(0)x-\frac12f''(0)x^2$), we see that $f(x) = ax^2 + bx + c$.

To finish, it suffices to check that every quadratic function satisfies the given condition for any natural number $n$.

Answer 2

Continuing the answer from Brian Moehring, we have $f'''(x) = 0$. Assuming $f(x)$ has a convergent Taylor series at $x$, we have,

$$f(x) = a+bx+cx^2$$ $$f'(x) = b+2cx$$ $$f(x+n)-f(x-n) = b2n + c((x+n)^2-(x-n)^2) = b2n + c(4nx)$$

Hence $\frac{f(x+n)-f(x-n)}{2n} = b+2cx = f'(x)$.

So any function of the form $f(x) = a + bx + cx^2$ will work.