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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

Element118
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Radhakrishnamurty P
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1 Answers1

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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.

Mohsen Shahriari
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  • Thanks.When does each form hold? – Radhakrishnamurty P Dec 01 '15 at 14:18
  • What is the significance of each of the forms? Please give intuitive examples. I would appreciate the derivation, if possible, with explanation of what is happening at each step. – Radhakrishnamurty P Dec 01 '15 at 14:23
  • See for example chapter 10 of the book "Introduction to Functional Equations" by Sahoo and Kannappan. – Mohsen Shahriari Dec 01 '15 at 18:17
  • https://books.google.com/books?id=f2rRBQAAQBAJ&pg=PA143&lpg=PA143&dq=introduction+functional+equations+sahoo&source=bl&ots=enAi2wWfUH&sig=W0HIx4P3pgugFqqbUpZxgY7M60U&hl=en&sa=X&ved=0ahUKEwj1krTdorvJAhVCoQ4KHX4-AkMQ6AEIMDAC#v=onepage&q&f=false – Mohsen Shahriari Dec 01 '15 at 18:23
  • @RadhakrishnamurtyP Did you find the link useful? If not, I can add extra information. – Mohsen Shahriari Dec 03 '15 at 16:48
  • As I cannot download books, I could not see the content you referred to. As such, additional information that you could provide would be useful to me. If possible, you may send a pdf file containing that information from that book. Please don't misunderstand me for having asked for the pdf of the relevant content. If you have no difficulty in sending, only then you please send it. – Radhakrishnamurty P Dec 05 '15 at 04:19
  • I added a question and answer and gave you the link at the end of my post. I also want to mention that the link I gave you in my comment, leads to a page in which you can read the book online and there was no need of downloading. – Mohsen Shahriari Dec 11 '15 at 15:47
  • Thank you, @Mohsen Shahriari, I read through few pages there but I am not enlightened. Suppose we choose Cosh(x) as the solution instead of Cos(x). What difference does it make? Can we use the 'cosh' solution to addition of vectors, if not why not? What would the 'Cos' solution give in the case: x+y=90 degrees? I am getting Sin 2x = Sin 2x. What does it mean in the context of addition of two vectors? Please help me with these questions. – Radhakrishnamurty P Dec 18 '15 at 15:51
  • I'm not sure about what you're asking for, but I can say that the solution of the form $\cos(ax)$ is related to the Euclidean geometry and the solution of the form $\cosh(ax)$ is related to the hyperbolic geometry. I think you need to study about hyperbolic geometry first, to understand the relevance. – Mohsen Shahriari Dec 18 '15 at 16:31