- #1
Koldstream
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Homework Statement
This is an example provided by my lecturer in his notes. He puts practically zero working in.
When i work the problem through i do not get the same answer as he does.
In this section i have copied the exact text from the problem:
Find the Fourier transform of cos(x). Your answer will include delta functions
We shall use the definition [tex]\delta(k)=\int\limits_\infty^\infty dx \ e^{ikx}[/tex]
and
[tex]\delta(k)=\delta(-k)[/tex]
he then simply writes:
[tex]Fourier[cos(x)]=\frac{1}{\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\ \frac{e^{ix}+e^{-ix}}{2}=\sqrt{\frac{\pi}{2}}\left[\delta(k-1)+\delta(k+1)\right][/tex]
As you can imagine this makes me very irritated so I start to solve it myself
Homework Equations
[tex]Fourier[f(x)]=\frac{1}{\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\ f(x)[/tex]
[tex]\delta(k)=\int\limits_\infty^\infty dx \ e^{ikx}[/tex]
The Attempt at a Solution
So i work it through line by line:
[tex]Fourier[cos(x)]=\frac{1}{\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\ \frac{e^{ix}+e^{-ix}}{2}[/tex]
[tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\left(\ e^{ix}+e^{-ix}\right)[/tex]
[tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx+ix}+e^{-ikx-ix}[/tex]
[tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{ix(-k+1)}+e^{ix(-k-1)}[/tex]
[tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{ix(-k+1)}+\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \e^{ix(-k-1)}[/tex]
now recognise
[tex]\delta(k)=\int\limits_\infty^\infty dx \ e^{ikx}[/tex]
can be used. However the definition for the Fourier transform shows that the constant should be:
[tex]\frac{1}{2\pi}[/tex]
therefore we need a constant such that [tex]\frac{1}{2\sqrt{2\pi}}*x=\frac{1}{2\pi}[/tex]
this implies x = [tex]\frac{2}{\sqrt{2\pi}}[/tex]
this leads to:
[tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}x\frac{2}{\sqrt{2\pi}}\ \delta(-k+1)+\frac{1}{2\sqrt{2\pi}}x\frac{2}{\sqrt{2\pi}}\ \delta(-k-1)[/tex]
[tex]Fourier[cos(x)]=\frac{1}{2\pi}\ \delta(-k+1)+\frac{1}{2\pi}}\ \delta(-k-1)[/tex]
then recognise [tex]\delta(k)=\delta(-k)[/tex]
[tex]Fourier[cos(x)]=\frac{1}{2\pi}\ \delta(k-1)+\frac{1}{2\pi}}\ \delta(k+1)[/tex]
This is of course not:
[tex]
Fourier[cos(x)]=\sqrt{\frac{\pi}{2}}\left[\delta(k-1)+\delta(k+1)\right]
[/tex]
The constant is incorrect.
I am certain that he is correct and that I am wrong.
Why have I got it wrong?
Thanks very much for your time.