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I am reading the chapter titled Fourier Analysis from Kreyszing's book "Advanced Engineering Mathematics".

There is a section which talks about changing the period from $$2\pi$$ to $$2L$$ and I am trying to understand how this "change of scale" takes place. I think this is a change of variables but I am unable to derive it by myself. Here is a snapshot of that section: https://pasteboard.co/oubtbmgJsS0r.png

I understand the need behind why you would want to do something like this but unable to do it myself.

For additional context, I have tried to do some manipulation myself on this function having period $p$ but without success : $$f(x+p) = f(x)$$

Can someone please help me with this change of variable?

Also, why is the value of $v$ in the textbook set to $\frac{\pi}{L}x$? What's the idea behind this?

ishan_ae
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1 Answers1

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Given that $f(x)$ has a period of $p$:

$$f(x+p) = f(x)$$

then $f\left(\frac{p}{2\pi}v\right)$, as a function of $v$, has a period of $2\pi$:

$$f\left(\frac{p}{2\pi}(v+2\pi)\right) = f\left(\frac{p}{2\pi}v+p\right) = f\left(\frac{p}{2\pi}v\right)$$

peterwhy
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  • Thank you. This manipulation makes sense. The two functions clearly have a period $p$ and that's what we want. – ishan_ae Apr 09 '23 at 16:48
  • Sorry, I could not edit the previous answer. I quote from the text "Then we can introduce a new variable $v$ such that $f(x)$, as a function of $v$, has period $2\pi$". In your answer if we compare the two functions then $x = \frac{p}{2\pi}v $. This will give us $x$ in terms of $v$. Then the value of $p$ is set to $2L$. I am confused as to why would something like this need to be done. – ishan_ae Apr 09 '23 at 16:57
  • @ishan_ae Your image says setting $p=2L$ is for practical reason, maybe a following Sec in your source would explain more. – peterwhy Apr 09 '23 at 19:33