- #1
MartynaJ
- 18
- 1
- Homework Statement
- If ##f(x)=-f(x+L/2)##, where L is the period of the periodic function ##f(x)##, then the coefficient of the even term of its Fourier series is zero. Hint: we can use the shifting property of the Fourier transform.
- Relevant Equations
- See above please
Summary:: If ##f(x)=-f(x+L/2)##, where L is the period of the periodic function ##f(x)##, then the coefficient of the even term of its Fourier series is zero. Hint: we can use the shifting property of the Fourier transform.
So here's my attempt to this problem so far:
##f(x)=-f(x+\frac{L}{2})## Then using the shifting property of the Fourier Transform we get: ##F(u)=-F(u)e^{2\pi i u\frac{L}{2}}##
And a periodic function is a function in the form ##f(x)=f(x+L)##. Now using the shifting property of the Fourier Transform we get: ##F(u)=F(u)e^{2\pi i u L}##
Making these two functions equal, I get:
##-e^{2\pi i u\frac{L}{2}}=e^{2\pi i u L}##
Now I don't know what else to do to prove the question. Did I go wrong anywhere?
I also know that if the even terms of Fourier series is zero, this means that function is odd, i.e. ##f(-x)=-f(x)##
So here's my attempt to this problem so far:
##f(x)=-f(x+\frac{L}{2})## Then using the shifting property of the Fourier Transform we get: ##F(u)=-F(u)e^{2\pi i u\frac{L}{2}}##
And a periodic function is a function in the form ##f(x)=f(x+L)##. Now using the shifting property of the Fourier Transform we get: ##F(u)=F(u)e^{2\pi i u L}##
Making these two functions equal, I get:
##-e^{2\pi i u\frac{L}{2}}=e^{2\pi i u L}##
Now I don't know what else to do to prove the question. Did I go wrong anywhere?
I also know that if the even terms of Fourier series is zero, this means that function is odd, i.e. ##f(-x)=-f(x)##
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