Suppose we have differentiable functions $F$, $f_1, \dots, f_n$, and $g_1, \dots, g_n$ satisfy the following relation $$ F(x+y) = \sum_{i=1}^n f_i(x) g_i(y).$$ What are the possible forms of $F$?
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Differentiate with respect to $x$ and then with respect to $y$, you get $$F'(x+y)=\sum_i f_i^\prime(x)g_i(y)=\sum_i f_i(x)g_i^\prime(y).$$ Substitute to the last equality some values $y_j$ then $$\sum_i f_i^\prime(x)g_i(y_j)=\sum_i f_i(x)g_i^\prime(y_j),\quad j=1,\ldots,n.$$ Assuming that $\det g_i(y_j)\neq 0$, solve this, and obtain $$f_i^\prime(x)=\sum_{j} c_{i,j}f_j(x).$$ This is a system of linear equation with constant coefficients from which it follows that $f_i$ are (generalized) exponential sums. For similar reasons, $g_i$ are also generalized exponential sums. The rest is algebra.
answered Dec 12, 2014 at 23:13
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$\begingroup$ To be more concrete, $F$ is a solution if and only if it can be written $$F(x)=x^{k_1}e^{\lambda_1x}+\dots +x^{k_m}e^{\lambda_m x},$$ where $\sum k_j=n$. You prove the "only if" part and the "if" part is obvious. Of course, trigonometric solutions are included by choosing $\lambda_j$ in complex conjugate pairs. $\endgroup$ Commented Dec 13, 2014 at 7:18
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$\begingroup$ I meant of course a linear combination $C_1x^{k_1}e^{\lambda_1}+\dots$. $\endgroup$ Commented Dec 13, 2014 at 7:48
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$\begingroup$ Or even more precisely, $C_1x^{k_1}e^{\lambda_1 x}+\ldots$ :-) $\endgroup$ Commented Dec 13, 2014 at 14:22