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Consider the difference equation $c(x+h)-c(x)=f(x)$, where $f(x)=\sum\limits_{i=1}^{s}f_ix^i$ is a polynomial. What is the solution of this equation?

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If $f$ is a nonzero polynomial of degree $s$, there exists a unique polynomial function $g$ of degree $s+1$ and with no constant term, such that $f(x)=g(x+h)-g(x)$ identically.

Thus, $c(x+h)-c(x)=f(x)$ if and only if $c(x+h)-c(x)=g(x+h)-g(x)$, hence:

The solutions are exactly the functions $c=g+a$ where $a$ is periodic with period $h$.

To find such a polynomial function $g$, write $f(x)=\sum\limits_{k=0}^sf_kx^k$ with $f_s\ne 0$ and look for $g(x)=\sum\limits_{k=0}^{s}g_kx^{k+1}$ solving the equation $f(x)=g(x+h)-g(x)$. One gets, for every $0\leqslant k\leqslant s$, $ f_k=\sum\limits_{i=k}^{s}{i+1\choose k}h^{i+1-k}g_i. $ This is a triangular linear system of size $s+1$, which gives $(f_k)_{0\leqslant k\leqslant s}$ as a function of $(g_k)_{0\leqslant k\leqslant s}$. If $h\ne0$, the diagonal coefficients of this Cramér system are nonzero, hence it has a unique solution $(g_k)_{0\leqslant k\leqslant s}$, which yields the desired polynomial function $g$.

Note: More systematic approaches are based on the Calculus of finite differences and on Newton's series.