How Do You Prove a Specific Fourier Transform Property?

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Discussion Overview

The discussion centers around proving a specific property of the Fourier transform, specifically the relationship between the Fourier transform of a function and its Fourier transform evaluated at a negated argument. Participants explore various approaches to derive the property and express concerns about the correctness of their methods.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a method to prove that \( F[\widetilde{f}(x)] = \frac{f(-k)}{2 \pi} \) using the definition of the Fourier transform, but expresses confusion about the presence of \( 2 \pi \) in the numerator.
  • Another participant agrees with the initial approach but suggests that the problem statement may be incorrect, referencing a wiki table for clarification.
  • Some participants discuss the use of the Dirac delta function to continue the proof, leading to the conclusion that \( F[\widetilde{f}(x)] = 2\pi f(-k) \), but there is uncertainty about the steps leading to this conclusion.
  • There is a correction regarding the formulation of the Fourier transform, indicating that the initial assumptions may not align with standard definitions.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial problem statement and the methods used to prove the property. While some agree on the use of the Dirac delta function, there is no consensus on the validity of the initial approach or the problem statement itself.

Contextual Notes

Participants note that the discussion relies on specific definitions of the Fourier transform, which may vary. There are unresolved mathematical steps and assumptions that could affect the validity of the proposed proofs.

mathmari
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Hey! :o

Could you give me a hint how to prove the following property of the Fourier transform, when $F[f(x)]=\widetilde{f}(x)$, where $F[f(x)]$ is the Fourier transform of $f(x)$?

$$F[ \widetilde{f}(x) ]= \frac{f(-k)}{2 \pi}$$

We know that: $ \widetilde{f}(k)=\int_{- \infty}^{+ \infty}{ {f}(x) e^{-i k x}}dx$ and $f(x)=\frac{1}{2 \pi}\int_{- \infty}^{+ \infty}{ \widetilde{f}(k) e^{i k x}}dx$.

I thought the following:

$F[\widetilde{f}(x)]=\int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{-ikx}}dx=2 \pi \frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{ix(-k)}}dx=2 \pi f(-k)$

But then $2 \pi$ would be at the numerator..
Have I done something wrong or can we not prove this in this way?

Then an other idea is:
$\widetilde{f}(x)=\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk \Rightarrow F[\widetilde{f}(x)]=F[\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk]=\int_{-\infty}^{+\infty}{\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-2ikx}}}dkdx$
How could we continue?

Or is there an other way to prove the property?? (Worried)
 
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mathmari said:
Hey! :o

Could you give me a hint how to prove the following property of the Fourier transform, when $F[f(x)]=\widetilde{f}(x)$, where $F[f(x)]$ is the Fourier transform of $f(x)$?

$$F[ \widetilde{f}(x) ]= \frac{f(-k)}{2 \pi}$$

We know that: $ \widetilde{f}(k)=\int_{- \infty}^{+ \infty}{ {f}(x) e^{-i k x}}dx$ and $f(x)=\frac{1}{2 \pi}\int_{- \infty}^{+ \infty}{ \widetilde{f}(k) e^{i k x}}dx$.

I thought the following:

$F[\widetilde{f}(x)]=\int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{-ikx}}dx=2 \pi \frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{ix(-k)}}dx=2 \pi f(-k)$

But then $2 \pi$ would be at the numerator..
Have I done something wrong or can we not prove this in this way?

Hi! :)

That is the correct way to prove it and your result is the right result.
None of the other common versions of the Fourier transform will give the result in your problem statement.
So it appears that your problem statement is wrong.

You can see the same result in this wiki table in the row labeled 105.
Your version of the Fourier transform is the last Fourier column in the table.
Then an other idea is:
$\widetilde{f}(x)=\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk \Rightarrow F[\widetilde{f}(x)]=F[\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk]=\int_{-\infty}^{+\infty}{\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-2ikx}}}dkdx$
How could we continue?

This is not correct.
It should be:
\begin{aligned}\widetilde{f}(x)&=\int_{-\infty}^{+\infty}{f(k')e^{-ik'x}}dk' \\
\Rightarrow F[\widetilde{f}(x)]
&=F[\int_{-\infty}^{+\infty} f(k')e^{-ik'x} dk'] \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-ik'x} e^{-ikx} dk'dx \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx
\end{aligned}

Anyway, I do not see how you can continue with that.
 
I like Serena said:
Hi! :)

That is the correct way to prove it and your result is the right result.
None of the other common versions of the Fourier transform will give the result in your problem statement.
So it appears that your problem statement is wrong.

You can see the same result in this wiki table in the row labeled 105.
Your version of the Fourier transform is the last Fourier column in the table.

This is not correct.
It should be:
\begin{aligned}\widetilde{f}(x)&=\int_{-\infty}^{+\infty}{f(k')e^{-ik'x}}dk' \\
\Rightarrow F[\widetilde{f}(x)]
&=F[\int_{-\infty}^{+\infty} f(k')e^{-ik'x} dk'] \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-ik'x} e^{-ikx} dk'dx \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx
\end{aligned}

Anyway, I do not see how you can continue with that.

Ok! (Smile)
Thank you very much! (Smirk)
 
I realized that actually we can continue from here using the Dirac delta function:
\begin{aligned}
F[\widetilde{f}(x)]
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx \\
&=\int_{-\infty}^{+\infty} dk' f(k') \int_{-\infty}^{+\infty} e^{-i(k'+k)x} dx \\
&=\int_{-\infty}^{+\infty} dk' f(k')\ 2\pi\ \delta(k'+k) \\
&= 2\pi f(-k)
\end{aligned}
 
I like Serena said:
I realized that actually we can continue from here using the Dirac delta function:
\begin{aligned}
F[\widetilde{f}(x)]
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx \\
&=\int_{-\infty}^{+\infty} dk' f(k') \int_{-\infty}^{+\infty} e^{-i(k'+k)x} dx \\
&=\int_{-\infty}^{+\infty} dk' f(k')\ 2\pi\ \delta(k'+k) \\
&= 2\pi f(-k)
\end{aligned}

Ok! (flower)
Thanks a lot! (Smirk)
 

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