Is there any general way to solve basic functional equations?

For example we have algebraic ways to solve algebraic equations (find $x$)!

But for functional equations like :

$$f(x) + f(x-1) = 0$$

or, $$f(x)-f(x^2)=1$$

How does one find $f(x)$?

I can solve one or two using trial and error but when i used WolframAlpha it was able to solve both [Links : 1 and 2] correctly!

Solutions were : 1. $f(x)=-((-1)^x)$ and 2. $f(x)=\dfrac{\log{\log{x}}}{\log{2}}$

How to solve it? I am not asking the about the complicated ones having $f(f(x))$ or $f(x)f(y)$...Just the basic ones having just $x^n$ or $2x-1$ or something like that...

P.S. - I have seen a lot of questions similar to this but none of theme were general or answered my question...

  • $\begingroup$ For 1. $f_a(x)= a(-1)^x$ for any $a$ will work. This are the only functions when working form $\mathbb Z$ to $\mathbb Z$. $\endgroup$ – wythagoras Jun 12 '15 at 8:27
  • $\begingroup$ @NeilRoy please correct your post yourself then if you reject my edits. Please notice that both W|A hyperlinks are identical. $\endgroup$ – dbanet Sep 27 '15 at 11:39
  • $\begingroup$ @dbanet Oh...I'm sorry! Please re-edit it... And Thanks! $\endgroup$ – NeilRoy Sep 28 '15 at 5:51

The first is a basic linear recurrence, solved from its characteristic equation: the solution of such equations are known to have an exponential form, $ar^x$, and the equation turns to

$$ar^x+ar^{x-1}=0,$$simplifying to$$r+1=0$$ and from the root $r=-1$, $$f(x)=a\cdot(-1)^x.$$ The value of $a$ is arbitrary. By the way, if you don't give more conditions on $f$, only values one unit apart are related to each other, so that any function $g(x)$ defined in range $[0,1)$ can serve to extend $f$ on all $\mathbb R$.

$$f(x)=g(\rfloor x\lfloor)\cdot(-1)^{\lfloor x\rfloor}$$

($\rfloor x\lfloor:=x-\lfloor x \rfloor$ denotes the fractional part of $x$.)

For instance, taking $g(x)=x(1-x)$, we get an alternation of parabolas.

enter image description here

The second equation can be transformed to the first form.

Indeed, $x=\ln_2(y)$ makes


and $y=\ln_2(z)$ makes

$$f(2^{2^z})-f(2^{2^{z+1}})=1,$$ i.e. $$h(z)-h(z+1)=1.$$

By a process similar to above, the solution is

$$h(z)=-z+a,$$ and $$f(x)=-\ln_2(\ln_2(x))+a.$$ And the general solution $$f(x)=-\ln_2(\ln_2(x))+g(\rfloor\ln_2(\ln_2(x))\lfloor).$$

With these methods, you can address the linear recurrences of the form

$$\sum_k c_kf(x+k)=RHS(x),$$

$$\sum_k c_kf(kx)=RHS(x)$$or

$$\sum_k c_kf(x^k)=RHS(x).$$

  • $\begingroup$ Hi @ Yves Daoust, What did you use to draw the graph? Thank you. $\endgroup$ – Jack Jun 12 '15 at 11:03
  • 1
    $\begingroup$ Microsoft Mathematics. Free and rich-featured. $\endgroup$ – Yves Daoust Jun 12 '15 at 11:06

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