How to convert recursive to closed form

Question

For example if $f(x+1)=af(x)+b$ is there a way to find the closed form?

Like what is $f(x+1)=\frac67f(x)+\frac67$ is closed form?

Answer 1

You can first solve for $n$ integer with $u_n=f(n)$

• if $a=1$ then $u_{n+1}-u_n=b$ is telescopic and $u_n=u_0+nb$
• if $a\neq 1$ then $u_{n+1}=au_n+b$ has general solution of homogeneous equation $u_n=Ua^n$ and particular solution $c=ac+b\iff c=\frac b{1-a}$ therefore $u_n=Ua^n+\frac b{1-a}$ and $u_0=U+\frac b{1-a}$ therefore $u_n=a^n(u_0-\frac b{1-a})+\frac b{1-a}$

If you do not impose any more condition of regularity on $f$ then you'll have a piecewise function for all initial seeds $\{x\}\in[0,1)$, i.e $u_0=f(\{x\})$ and $f(x)$ the expression given for $u_n$ where $n$ is replaced by $x$.

Note: for $a<0$, the function $f$ is not well defined for $x<0$ (because of $a^x$)