Way to solve basic functional equations 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...
 A: 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.

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