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Given a function $g(x)$, is it possible to find a function $f(x)$ that satisfies

$$ f(x+1) - f(x-1) = g(x) $$

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    $\begingroup$ What kind of functions are we talking about here? $\endgroup$ Jan 13, 2011 at 2:00

4 Answers 4

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Pick any choice of $f$ on the interval $[-1,1)$. By using the property $f(x) = g(x-1) + f(x-2)$, you can extend this definition to $[1,3)$, and so on to the rest of the positive real line. Similarly, you can extend $f$ to the negative real line using $f(x) = f(x+2) - g(x+1)$. This is one solution to your equation.

Of course, a different choice of $f$ on $[-1,1)$ will give you a different result. This corresponds to the fact that adding any periodic function $h$ with period 2 to $f$ still satisfies the original equation. In other words, there are infinitely many solutions.

For particular restricted classes of functions, as Qiaochu alludes to, there may be a single canonical solution. For example, if $g$ is a polynomial of degree $n$, there is a unique polynomial solution for $f$. This is a polynomial of degree $n+1$, and you can find it by expanding out the original equation and comparing powers starting with the highest. A similar fact holds for exponentials and trigonometric functions (except of course when the trigonometric functions have period 2, in which case you would have to include multiples of $x \sin \pi x$ and $x \cos \pi x$ in $f$ to get a solution).

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Let $f(x)=\frac{1}{2}x g(0)$ for $-1\le x \le 1$, $f(x)=g(x-1) + f(x-2)$ for $x>1$, and $f(x)=f(x+2)-g(x+1)$ for $x<-1$.

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There might be something to gain by interpreting $g(x)$ as being an approximation to the derivative of $f(x)$, (barring a factor of 2). It's not really the derivative, but the centered finite-difference.

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I don't know whether this procedure is correct or not:

$f(x+1)-f(x-1)=g(x)$

$x\to x+1$:

$f(x+2)-f(x)=g(x+1)$

$x\to2x$:

$f(2x+2)-f(2x)=g(2x+1)$

$f(2(x+1))-f(2x)=g(2x+1)$

$f(2x)=\sum_x g(2x+1)+\Theta_1(x)$, where $\Theta_1(x)$ is an arbitrary periodic function with unit period

$f(x)=h\left(\dfrac{x}{2}\right)+\Theta(x)$, where $h(x)=\sum_x g(2x+1)$ and $\Theta(x)$ is an arbitrary periodic function with period $2$

If $\lim_{x\to+\infty}g(x)=0$, according to http://en.wikipedia.org/wiki/Indefinite_sum#Mueller.27s_formula, the result can be further simplified to $f(x)=\sum_{n=0}^\infty(g(2n+1)-g(x+2n+1))+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with period $2$, $\sum_{n=0}^\infty(g(2n+1)-g(x+2n+1))$ i.e. kernelisation of $\sum_{n=0}^{\frac{x}{2}-1}g(2n+1)$

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  • $\begingroup$ In other words: solutions to this problem are equivalent to solutions of the "indefinite summation" problem. $\endgroup$
    – GEdgar
    Jul 1, 2012 at 22:24

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