What is the solution for the following functional equation?
$g(x)g(z) = g(x+z)+g(x-z)$
The solution given is: $g(z) = 2\cos(z)$.
In the derivation of the result (using Taylor expansion), there is a step that is like this:
$g''(x) = bg(x)$, where $g''(x)$ is the second derivative of $g(x)$ and $b$ is a constant. Differentiating this equation twice $g^{(4)}(x) = b^2g(x)$.
To me the last equation appears incorrect as I get $b^3$ in place of $b^2$. Even if $b^3$ is correct, I don't know how to get the final solution.
I seek help in this connection. If some one could help, I would like to get the solution of the functional equation and the method used.
P. Radhakrishnamurty
As @Element118 has noticed, it's not obvious that your question about $g^{(4)}(x)$ is relevant to the solution of the functional equation. But it's easy to see that $g^{(4)}(x)=b^2g(x)$. Differentiate the sides of $g^{\prime\prime}(x)=bg(x)$ twice and get $g^{(4)}(x)=bg^{\prime\prime}(x)$. Note that since $b$ is a constant, the derivative of $bg(x)$ with respect to $x$ is equal to $bg^\prime(x)$. Now substitue $g^{\prime\prime}(x)$ from the first equation and you're done.
For a full explanation of the solutions of the functional equation, note that letting $f(x)=\frac{g(x)}2$, we have $f(x+y)+f(x-y)=2f(x)f(y)$ which is the well known d'Alembert functional equation. Every continuous solution to this equation is of the form $f(x)=0$, $f(x)=\cos(ax)$ or $f(x)=\cosh(ax)=\frac{e^{ax}+e^{-ax}}2$. See the answer to this question for example.