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Suppose $f(x)$ and $g(x)$ are non-constant real-valued differentiable functions $\in\Bbb R$

Furthermore, suppose that $f(x + y) = f(x)f(y) − g(x)g(y)$ and $g(x + y) = f(x)g(y) + g(x)f(y)\ \forall\ x,y.$

If $f′(0) = 0$, prove that $(f(x))^2 + (g(x))^2 = 1\ \forall\ x.$

Firstly apologies for the format. Unfortunately this proof was meant to be the focus of a group project, but now with the deadline looming the group has let me down and quite frankly I have no idea how to prove this. I only need the proof to present in the form of a poster. If anyone could provide a proof I would be extremely grateful.

Thanks,

Callum

Mauro ALLEGRANZA
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Callum Perkins
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1 Answers1

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With $h(x)=f(x)^2+g(x)^2$, some elementary algebra gives you $$h(x+y)=[f(x)f(y) − g(x)g(y)]^2+[f(x)g(y) + g(x)f(y)]^2=[f(x)^2+g(x)^2][f(y)^2+g(y)^2]=h(x)h(y).$$ Under your assumptions, that means $h(x)=c^x$. Your equations with $x=y=0$ give $g(0)=0$ and $f(0)=1$ (since the functions are not constant, otherwise $f=g=0$ would be a solution, too), so $h'(0)=2f'(0)f(0)+2g'(0)g(0)=0$. Thus, $c=1$, and $h(x)=1$ for all $x$.