Fourier Transform of f'(ax): Discrepancy in Results?

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SUMMARY

The Fourier transform of \(f'(ax)\) is derived using the scaling theorem and the derivative rule. The correct formulation is \(H(\omega) = \frac{i \omega}{a^2} F\left(\frac{\omega}{a}\right)\), where \(H(\omega)\) is the Fourier transform of \(h(x) = f'(ax)\). The discrepancy arises from the application of the scaling theorem and the derivative rule, which must be carefully combined to avoid additional factors. The key takeaway is that the scaling factor is \( \frac{1}{a^2} \) when considering both transformations.

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Poirot1
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What is the Fourier transform of $f'(ax)$, where a>0 is a constant? Firstly, I reasoned that (lets say $F[f]$ is the Fourier transform of f) $F[f('x)]=\frac{1}{a}F[f](\frac{k}{a})$ by scaling theorem, then using the derivative rule we get $F[f'(ax)]=\frac{ik}{a}F[f(x)](\frac{k}{a})$. But when I did it manually (substitution and integration by parts), I got an extra factor of $\frac{1}{a}$. So which is correct and why the discrepency?
 
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Poirot said:
What is the Fourier transform of $f'(ax)$, where a>0 is a constant? Firstly, I reasoned that (lets say $F[f]$ is the Fourier transform of f) $F[f('x)]=\frac{1}{a}F[f](\frac{k}{a})$ by scaling theorem, then using the derivative rule we get $F[f'(ax)]=\frac{ik}{a}F[f(x)](\frac{k}{a})$. But when I did it manually (substitution and integration by parts), I got an extra factor of $\frac{1}{a}$. So which is correct and why the discrepency?

Let: \(g(x)=f'(x)\) and \(h(x)=g(a x)\), and \(H(\omega)\), \(G(\omega)\) and \(F(\omega)\) be the FT of \(h(x)\), \(g(x)\) and \(h(x)\) respectively.

then:

\[ H(\omega)=\frac{1}{a}G\left(\frac{\omega}{a}\right) \]

and:

\[ G(\omega) = i \omega F(\omega)\]

Now substitute the second into the first to get:

\[ H(\omega)=\frac{1}{a} \frac{ i \omega}{a}F\left(\frac{\omega}{a}\right) = \; \frac{i \omega}{a^2} F\left( \frac{\omega}{a}\right) \]

CB
 
Last edited:

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