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Let $\lambda > 0$ and let $f(x)$ be a periodic function that has period $a$. How to show that function $g(x)=f'(\lambda x)$ is periodic and determine its period. Just some hints, please. I have achieved this far( I have used following function $g(x)=f(\lambda x)$): $\lambda > 0$ $g(x+a)=f(\lambda(x+a))=f(\lambda x+ \lambda a) = ...$

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2 Answers 2

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Assume $f(t)=f(t+a)$. Then $f'(t)=f'(t+a)$.

Now $f'(\lambda x)=f'(\lambda x + a)$, taking $t=\lambda x$ from above.

I leave the last step for you. Since $g(x)=f'(\lambda x)$, what value can we add to $x$ in both the left and right expressions and still yield a true statement?

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If $a$ is a period of $f$ (and $f$ is differentiable), then $a$ is clearly also a period of $f'$. Finally $\frac a\lambda$ is a period of $g$ as seen by plugging it in: $g(x+\frac a\lambda)=f'(\lambda x+a)=f'(\lambda x)=g(x)$.

Do you also need to show that $g$ has no period smaller than $\frac a\lambda$ if $a$ is the smallest period of $f$?

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hi, why don't you use $g(x+a)=f'(\lambda(x+a))...$? and how would you show that finally $...f(x)$ if first is $g(x+a)$? –  laovultai Sep 9 '12 at 12:19
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I don't want to calculate $g(x+a)$ because I don't expect $a$ to be a period of $g$. I don't understand your second question. –  Hagen von Eitzen Sep 9 '12 at 12:49
    
I mean would it be possible to show that $g(x+a)=...=g(x)$. And if that is possible, how would you show that g is also periodical in this way? –  laovultai Sep 9 '12 at 12:50
    
How did you came up period of g to be $\frac{a}{\lambda}$? just by guessing? –  laovultai Sep 9 '12 at 12:51
    
@alvoutila: Not guessing, more like thinking about a concrete example. Look at $\sin x$ and $\sin(2x)$. The second wiggles twice as fast, has period $1/2$. –  André Nicolas Sep 9 '12 at 15:13

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