Solving Fourier Coefficients for A_m & B_m

In summary, the problem involves finding the coefficients ##A_m## and ##B_m## in a Fourier series expression, with the given result for ##B_1## and ##B_3## and all other coefficients being 0. The proposed solution involves multiplying both sides by cosine and integrating, but this method may not work since ##m## is a dummy index. It is instead suggested to use trig identities to simplify the expression before solving for the coefficients.
  • #1
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Homework Statement


I've reached a relation but then I need to obtain the coefficients ##A_m## and ##B_m##'s, those are the only unknowns.
Here's the expression: ##\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=T_0\sin ^3 \theta##.

Homework Equations


Fourier coefficients.
Result is probably ##B_1=\frac{3T_0}{4a}## and ##B_3=-\frac{T_0}{4a^3}##.
All the other coefficients being 0.

The Attempt at a Solution


So I am not really sure. I have something of the form ##\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=f(\theta) ##.
So in order to get let's say the ##A_m## first, I guess I must multiplicate both sides by ##cos (m \theta )## and integrate from ... 0 to ##2 \pi##?
I'd get ##\sum _{m=0}^\infty a^m A_m \int _0^{2\pi } \cos ^2 (m \theta ) d \theta + \sum _{m=0}^\infty a^m B_m \int _0^{2\pi } \sin (m \theta ) \cos (m\theta ) d \theta=\int _0 ^{2\pi } f(\theta ) d \theta##. Where an integral vanishes.
I'd be left with ##\sum _{m=0}^\infty a^m A_m \int _0^{2\pi } \cos ^2 (m \theta ) d \theta =\int _0 ^{2\pi } f(\theta ) d \theta##.
I don't really know how to evaluate those integrals nor how to isolate ##A_m##.
 
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  • #2
You can't multiply by ##\cos m\theta## because ##m## is your dummy index. Also, you forgot to multiply the righthand side. You can multiply by ##\cos n\theta## and then do the integrals. Only the term with n=m will survive.

But for this problem, it's probably better to use trig identities to express ##\sin^3 \theta## in terms of sines and cosines without exponents.
 

Related to Solving Fourier Coefficients for A_m & B_m

1. What is the purpose of solving Fourier coefficients for A_m and B_m?

The purpose of solving Fourier coefficients for A_m and B_m is to find a way to represent a periodic function as a combination of simple sine and cosine functions. This allows us to break down a complex function into simpler components, making it easier to analyze and understand.

2. How do you calculate A_m and B_m for a given function?

To calculate A_m and B_m for a given function, we use the Fourier series formula, which involves integrating the function multiplied by sine and cosine functions over one period. The resulting coefficients, A_m and B_m, represent the amplitudes of the sine and cosine components, respectively.

3. What does the value of A_m and B_m tell us about the original function?

The value of A_m and B_m tells us about the amplitude of the sine and cosine components in the original function. A higher value for A_m or B_m indicates a larger contribution of that component to the overall function, while a lower value indicates a smaller contribution. Additionally, the phase of the sine and cosine components can be determined from the values of A_m and B_m.

4. Can A_m and B_m be negative?

Yes, A_m and B_m can be negative. The negative sign indicates that the corresponding sine or cosine component in the Fourier series has a phase shift of 180 degrees or pi radians. This means that the component is shifted in the opposite direction compared to a positive value.

5. What is the significance of higher order coefficients, such as A_2 and B_3?

Higher order coefficients, such as A_2 and B_3, represent higher frequency components in the Fourier series. These components have a shorter period and oscillate more rapidly compared to lower order components. The presence of higher order coefficients can indicate the presence of sharp changes or discontinuities in the original function.

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