i know that equation for d’Alembert’s equation. is looking so $g(x+y)+g(x-y)=2*g(x)*g(y)$ so am trying to find actual solution for this equation,first i took $x=y=0$ and i got $2*g(0)$=$2*g(0)^2$ from here $g(0)=0$ or $g(0)=1$ if i take $x=y$ ,then $g(2*x)$+$g(0)$=$2*g(x)^2$,or if we take $y=-x$,we will get $g(0)$+$g(2*x)$=$(2*g(x)*g(-x))$ ,so i think that $g(x)=g(-x)$ ,am i right?so $g(x)$ should be of the form $g(x)$=$(x^n)$ where $n$ is even,or $g(x)=cos(x)$ right?

EDITED: but because $x^n$ for x=0 never equal 1,so it should be $cos(x)$ width additional additive constant a or b as you would like so final form should be like this $g(x)$=$b+cos(x)$ correct?

  • $\begingroup$ A note on LaTeX: enclose entire formulas in dollar signs, not pieces thereof: write $g(x+y)+g(x-y)=2g(x)g(y)$, not $g(x+y)$+$g(x-y)$=2*$g(x)$*$g(y)$ $\endgroup$ – t.b. Apr 13 '12 at 9:37
  • $\begingroup$ ok thanks @t.b i will fix up it right now $\endgroup$ – dato datuashvili Apr 13 '12 at 9:39
  • $\begingroup$ @datodatuashvili Note that $g(x)=\cosh x$ and $g(x)=2^{-x-1}((3-\sqrt{5})^x+(3+\sqrt{5})^x)$ are solutions to the equation. $\endgroup$ – Mohsen Shahriari Jun 12 '15 at 11:11

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