I'm basically a total novice with functional equations and have some questions regarding the solving technuiqes of them. Although, i'm adware of the lack of general solving methods, I have noticed that transformation of variables is one of the most common approaches to solve a functional equation.
Example: If we have a functional equation (polynomial with real coefficients):

$f(x+1)-f(x)=f(x)-f(x-1)+2$ (I). Then it might be a good idea to let:

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

so that our equation can be written:


But I have noticed that (I) actually can be written as:

$h(x+1)=h(x)+2$ and I can't see why. If we let $t=x-1$ we can write the right hand side of our equation as $f(t+1)-f(t)+2=h(t)+2$ but I can't see why the left hand side becomes $h(x+1)$ instead of $h(x)$.

My question more specifically:

How may $f(x+1)-f(x)=f(x)-f(x-1)+2$ be written as $h(x+1)=h(x)+2$


1 Answer 1


In terms of finite differences, your first idea has $h(x)$ as the forward difference of $f$ at $x$, i.e., $h(x)=f(x+1)-f(x)$. Whatever source you are looking at suggests it is better if $h(x)$ is the backward difference of $f$ at $x$. Then $h(x+1)=f(x+1)-f(x)$ and $h(x)=f(x)-f(x-1)$ are simply consequences of how $h$ is defined, and we can even get one more as $g(x)=h(x+1)-h(x)=2$.

We could also take the forward difference and see that $h(x)=f(x+1)-f(x)$ means that $h(x-1)=f(x)-f(x-1)$ and we could still get a similar final equation of $h(x)=h(x-1)+2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .