Solve for equation algebraically Is it possible to write the following function as $H(x)=$'some expresssion` ?
$$D(x) = H(x) + H(x-1)$$
Edit:
Hey everyone, thanks for all the great responses, and just to clarify H(x) and D(x) are always going to be polynomials, I wasn't sure if that made too big of a difference so I didn't mention it before.
Edit 2: 
I'm sincerely sorry if I was too general, I've only really used polynomial equations in my education thus far, so I didn't realize they might not seam just regular to everyone else. Once again I'm very sorry If I didn't give everyone an opportunity to answer the question because of my vagueness. I sincerely appreciate all of your answers and time.

I'll use lower case, so $d(x) = h(x) + h(x-1)$.
Let $D(x) = \sum \limits_{n=0}^{\infty} d(n) x^n/n!$
and $H(x) = \sum \limits_{n=0}^{\infty} h(n) x^n/n!$.
$H'(x) = \sum \limits_{n=1}^{\infty} h(n) x^{n-1}/(n-1)!

 = \sum \limits_{n=0}^{\infty} h(n+1) x^n/n!$
so $H(x) + H'(x) = \sum \limits_{n=0}^{\infty} (h(n)+h(n+1)) x^n/n!
 = \sum \limits_{n=0}^{\infty} d(n+1) x^n/n!$.
Since $D'(x) = \sum \limits_{n=0}^{\infty} d(n+1) x^n/n!$,
$H(x) + H'(x) = D'(x)$.
The usual integrating factor is $e^x$:
$e^x D'(x) = e^x(H(x) + H'(x)) 
=(e^x H(x))'$,
so
$e^x H(x) = \int e^x D'(x) dx
$
or
$ H(x) = e^{-x} \int e^x D'(x) dx
$.
If you can do the integral
(and I haven't made too many mistakes)
,
you can get $H$.
You can integrate by parts, but you still have to deal with
$\int e^x D(x) dx$.
Now I'll see what was entered while I was entering this ...
