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.

$H(x)$ could be defined as anything on $[0,1)$, then the relation
$$
D(x) = H(x) + H(x-1)\tag{1}
$$
would define $H(x)$ on the rest of $\mathbb{R}$. For example, for $x\ge0$,
$$
H(x)=(-1)^{\lfloor x\rfloor}H(x-\lfloor x\rfloor)+\sum_{k=0}^{\lfloor x\rfloor-1}(-1)^kD(x-k)\tag{2}
$$
and for $x<0$,
$$
H(x)=(-1)^{\lfloor x\rfloor}H(x-\lfloor x\rfloor)+\sum_{k=\lfloor x\rfloor}^{-1}(-1)^kD(x-k)\tag{3}
$$
Polynomial solutions:
Some interest was shown for polynomial solutions when $D(x)$ is a polynomial. As was done in another recent answer, we can use Taylor's Theorem to show that
$$
e^{-\mathcal{D}}P(x)=P(x-1)\tag{4}
$$
where $\mathcal{D}=\dfrac{\mathrm{d}}{\mathrm{d}x}$, at least for polynomial $P$. Using $(4)$, $(1)$ becomes
$$
D(x)=\left(I+e^{-\mathcal{D}}\right)H(x)\tag{5}
$$
Noting that
$$
(1+e^{-x})^{-1}=\tfrac{1}{2}+\tfrac{1}{4}x-\tfrac{1}{48}x^3+\tfrac{1}{480}x^5+\dots\tag{6}
$$
we can get a polynomial solution of $(1)$ with
$$
H(x)=\left(\tfrac{1}{2}I+\tfrac{1}{4}\mathcal{D}-\tfrac{1}{48}\mathcal{D}^3+\tfrac{1}{480}\mathcal{D}^5+\dots\right)D(x)\tag{7}
$$
Polynomial example:
For example, if $D(x)=x^2$, then using $(7)$, $H(x)=\tfrac{1}{2}x^2+\tfrac{1}{2}x$. Check: $$\left(\tfrac{1}{2}x^2+\tfrac{1}{2}x\right)+\left(\tfrac{1}{2}(x-1)^2+\tfrac{1}{2}(x-1)\right)=x^2$$
