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What can one say about functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the condition $f(x+1)-f(x-1)=2f'(x)$? Is is possible to find all such functions, or is this defining equation the best characterisation that one is likely to get?

I know that all polynomials with degree at most 2 satisfy the conditions. This is most easily seen by noting that constants, $f(x)=x$ and $f(x)=x^2$ satisfy the conditions and then noticing that if two functions $f$ and $g$ satisfy the condition then so does any linear combination of $f$ and $g$.

One can also see that no polynomial with degree more than 2 will satisfy the condition by noting that $f(x)=x^3$ does not satisfy the condition and then noticing that if $f$ satisfies the condition then so does its derivative (Provided that $f'$ is also differentiable, which is obviously the case for polynomials). Thus the existence of any polynomial $f$ of degree more than 2 which satisfies the condition would imply that there exists such a cubic by repeatedly differentiating $f$, and the existence of such a cubic implies that $f(x)=x^3$ satisfies the condition because $f(x)=x^3$ is a linear combination of the obtained cubic and some quadratic polynomial.

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    $\begingroup$ Such functions are infinitely differentiable. Did you try a Taylor series? $\endgroup$ – Pedro Tamaroff Apr 30 '14 at 22:23
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    $\begingroup$ This is a differential-difference equation. There are methods to solve such equations using Laplace transform described in a book in Schaum's series. $\endgroup$ – Jean-Claude Arbaut Apr 30 '14 at 22:24
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Let $f(x)$ be any differentiable function defined on $[0,1]$ and let $g(x)$ be any integrable function defined on $[1,2]$. Extend $f$ for $x\in (1,2]$ by defining $$f(x) = \int_1^x g(t)\,dt + c$$ where $c$ is chosen to make $f$ continuous at $1$. Also require that $g(1) = f'(1)$, so that $f$ will be also differentiable at $1$. Extend $f$ to $x \in (2,3]$ by $$f(x) = f(x-2) + g(x-1).$$ Now $f$ is defined on $[0,3]$ and satisfies the formula given in the question.

Repeat the above starting with $f$ so far defined on $[0,3]$ to extend the definition of $f$ to $[0,4]$, etc. to extend the definition to $[0, \infty)$.

One can work this procedure backwards to extend the definition of $f$ to all of $\mathbb{R}$. Thus any such $f$ satisfying the original formula can be created from starting functions $f$ and $g$ as described.

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  • $\begingroup$ I assume you mean to define $f(x)=f(x-2)+2g(x-1)$ instead. However, you need stronger hypotheses to make this work. For instance, as $x$ approaches $2$ from below, we get $f'(2)=g(2)$, but as $x$ approaches $2$ from above, we get $f'(2)=f'(0)+2g'(1)$. So $f$ will only be differentiable at $2$ if $g(2)=f'(0)+2g'(1)$. In general, you need an infinite sequence of similar equations involving higher derivatives to make sure $f$ is continuous and differentiable at all integers. $\endgroup$ – Eric Wofsey Dec 10 '17 at 17:18
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All such functions can be obtained in the following way. Start with any smooth ($C^\infty$) function $f:[-1,1]\to\mathbb{R}$ such that $$f^{(n)}(1)-f^{(n)}(-1)=2f^{(n+1)}(0)$$ for all $n\geq 0$. We can then extend $f$ smoothly to $[-1,2]$ by defining $f(x)=f(x-2)+2f'(x-1)$ for $x\in[1,2]$. The equations above guarantee that all the higher derivatives of $f$ at $1$ from the left and right will agree, so $f$ is smooth at $1$. But now, $f$ satisfies $f^{(n)}(x+1)-f^{(n)}(x-1)=2f^{(n+1)}(x)$ for all $n$ and all $x\in [0,1]$. In particular, these equations for $x=1$ allow us to extend $f$ to $[2,3]$ in the same way. We can similarly extend to $[3,4]$ and $[4,5]$ and so on, and also to $[-2,-1]$, $[-3,-2]$, and so on, and obtain a smooth function $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x+1)-f(x-1)=2f'(x)$ for all $x$.

Conversely, given any $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+1)-f(x-1)=2f'(x)$ for all $x$, note that $f$ must be smooth: it is differentiable by assumption, but then its derivative can be written in terms of $f$ itself so its derivative is differentiable, and so on. So restricting $f$ to $[-1,1]$ we get an initial function as above, and the values of $f$ on all of $\mathbb{R}$ can be recovered by the process above.

Let me finally remark that there are indeed many functions $f:[-1,1]\to\mathbb{R}$ satisfying the conditions above. Indeed, those conditions only involve the derivatives of $f$ at $-1$, $0$, and $1$. So we can first prescribe $f$'s behavior near those three points, and then fill in the values between them however we want, as long as we do so smoothly. Moreover, given any two sequences $(a_n)$ and $(b_n)$ of real numbers, if we define $c_n=a_n+2b_{n+1}$, then it is possible to construct $f$ such that $f^{(n)}(-1)=a_n$, $f^{(n)}(0)=b_n$, and $f^{(n)}(1)=c_n$ (by Borel's lemma).

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let $a=x-1$ then we have

$$ \frac{f(a+2)-f(a)}{2} =f'(a+1)$$

which is the means that the the derivative in the middle of the interval is equal to the average rate of change of the interval, so we can see that all functions of the form $f(x)=mx+b$ are solutions.

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When derivative of a function and function are linear in each other, chances are there that it is of exponential nature. Take, $f(x)=\lambda a^{\alpha x} +\mu x+c$ and solving it we get,

$$ a^{\alpha}- a^{-\alpha}=\alpha \Rightarrow a=(\frac{\alpha \pm \sqrt{\alpha^2+4}}{2}) $$

Choose $\alpha$ (or $a$) sutiably.

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