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katkatkat

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Show G(k)=[itex]\sqrt{2π}[/itex]g

Given that G(k) is the Fourier transform of F(x), g

F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf

SO FAR

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf

mult by e

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf

=g

Problem starts here... I have arg of x-y in the remaining part, with dx...

Thought was that sub dx=dx-dy+dx=d(x-y)+dx <-Pretty sure I'm allowed to do this

=g

=[itex]\sqrt{2π}[/itex]g

I'm definitely on the right track up to the problems part, do I say that ∫arg*dy is equal to zero because F(x) is a function of x... or a different reasoning... not sure... If I can make the right half of the equation = 0, I'll have the desired result

=[itex]\sqrt{2π}[/itex]g

=G(k)

_{1}(k)g_{2}(k)Given that G(k) is the Fourier transform of F(x), g

_{1}(k) is Fourier trans of f_{1}(x), g_{2}(k) is Fourier trans of f_{2}(X) andF(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf

_{1}(y)f_{2}(x-y)SO FAR

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e

^{-ikx}dx <-def'n of Fourier transform=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf

_{1}(y)f_{2}(x-y)e^{-ikx}dxmult by e

^{-iky}e^{iky}=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf

_{1}(y)e^{-iky}[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}dx=g

_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}dxProblem starts here... I have arg of x-y in the remaining part, with dx...

Thought was that sub dx=dx-dy+dx=d(x-y)+dx <-Pretty sure I'm allowed to do this

=g

_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}d(x-y)+g_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}d(y)=[itex]\sqrt{2π}[/itex]g

_{1}(k)g_{2}(k)+g_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}d(y)I'm definitely on the right track up to the problems part, do I say that ∫arg*dy is equal to zero because F(x) is a function of x... or a different reasoning... not sure... If I can make the right half of the equation = 0, I'll have the desired result

=[itex]\sqrt{2π}[/itex]g

_{1}(k)g_{2}(k)=G(k)

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