Functional Equations

If a relation links $f(x)$ and $f(g(x))$ where $g(g(x))=x$ (e.g. $g(x)=\tfrac1x$ or $1-x$), replace $x$ by $g(x)$ to obtain a second relation, then eliminate $f(g(x))$ algebraically to isolate $f(x)$.

• $f(xy)=f(x)f(y)$: power-type, $f(x)=x^k$; useful values come from $f(1)=1$ and $f(1/a)=1/f(a)$.
• $f(x+y)=f(x)f(y)$: exponential, $f(x)=a^x$; so $f(x)f(y)f(z)=f(x+y+z)$.
• $f(x+y)=f(x)+f(y)$: additive (Cauchy), $f(x)=cx$.
• $f(x-y)=f(x)f(y)$: put $x=y$ to get $f(0)=f(x)^2$, so $f$ is constant $0$ or $1$ on the values it takes; with $f(0)=1$ and $f(1)=a$ the useful line is $f(n)=a^n$ — treat it like the exponential form after fixing $f(0)$.
• $f(xy)=f(x+y)$ for all $x,y$: a fourth signature the three above actively mislead on — put $y=0$: $f(0)=f(x)$ for every $x$, so $f$ is constant. Then $f(2020)=f(0)=f(1)=\dots$, and any sum $\sum f(k)$ is just $n$ times that constant.
• Finishing an exponential-type equation usually needs a GP sum: if $f(x)=2^x$ then $\sum_{x=1}^{n}f(x)=2+4+\dots+2^n=2^{n+1}-2$, and $\sum_{x=1}^{n}4^x=\dfrac{4(4^n-1)}{3}$ — the formula is $a\dfrac{r^n-1}{r-1}$ (Sequence & Series). If a stem says $f(a),f(b),f(c)$ are in GP with $f(x)=k^x$, that is $b^2=ac$ on the exponents' images, i.e. $f(b)^2=f(a)f(c)\Rightarrow 2b=a+c$.

Given $f(g(x))=h(x)$, set $t=g(x)$, solve $x$ in terms of $t$, and substitute: $f(t)=h(\,x(t)\,)$. Renaming $t\to x$ gives the explicit rule. (E.g. from $f(x+1)$ put $t=x+1\Rightarrow x=t-1$.)