4
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

$$f(2^y)-f(2^{2y})=1$$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.