If $f(x)$ and $g(\alpha)$ is a pair of Fourier transforms, then how can we show that $df/dx$ and $i\alpha g(\alpha)$ is a pair of Fourier transforms?
Conjugate pairs in Fourier transforms but with Fourier coefficients
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real-analysis
fourier-analysis
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2If you'd like others to take the time to help you, it only makes sense to take the time to expand on what you have tried and to typeset your question so it's a bit more readable. You've asked 31 question here; by now you should know how to ask a question on this site. – 2012-11-13
1 Answers
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While I agree that you should type your question more carefully, here's the solution anyway:
- Write down the definition for the inverse Fourier transform
- Take the derivative.
$$ f(x)=\int_{-\infty}^\infty g(\alpha)e^{i\alpha x}d\alpha \\ \frac{d}{dx}f(x)=\int_{-\infty}^\infty g(\alpha)\frac{d}{dx}e^{i\alpha x}d\alpha\\ =\int_{-\infty}^\infty i\alpha g(\alpha)e^{i\alpha x}d\alpha $$
Since we now have the function $i\alpha g(\alpha)$ in the inverse Fourier transform formula, it must be the Fourier transform of $\frac{df}{dx}$.
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0I am actually having some trouble with the differentiation. Can you please edit your solution to help? – 2012-11-14
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0You're learning the Fourier transform and you don't know how to differentiate an exponential function? I find this strange. Well, anyway, $\frac{d}{dx}e^{i\alpha x}=i\alpha e^{i\alpha x}$. You just use the chain rule. – 2012-11-14
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0Can you show how the inverse FT connects to taking the derivative? they are 2 disjoint steps and I dont see how to proceed with this – 2012-11-14