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## Homework Statement

For an even function, the Fourier series takes the form

[itex]^{\infty}_{n=0}\Sigma A_n cos(\frac{2\pi n x}{\lambda})[/itex]

where [itex]\lambda[/itex] is the wavelength of the function. In this problem you will see how to find the Fourier coefficients [itex]A_n[/itex].

**a.)**Prove that

[itex]A_0 = \frac{1}{\lambda} \int^{\lambda}_0 F(x)dx[/itex]

(Hint: Integrate F(x) from x = 0 to x = [itex]\lambda[/itex].)

**b.)**Prove that, for m > 0,

[itex]A_m = \frac{2}{\lambda} \int^{\lambda}_0 F(x) cos(\frac{2\pi m x}{\lambda}) dx[/itex]

where A

_{m}(rather than A

_{n}) is the Fourier coefficient for reasons that will become apparent in your proof. [Hint: Multiply both sides of equation (6.57) by [itex]cos (\frac{2\pi m x}{\lambda})[/itex], and integrate from 0 to λ.]

## Homework Equations

Equation (6.57): [itex]F(x) = ^\infty _{n=0} \Sigma A_n cos(\frac{2\pi n x}{\lambda})[/itex]

Final result (a) should be: [itex]A_0 = \frac{1}{\lambda} \int^{\lambda}_0 F(x) dx[/itex]

Final result (b) should be: [itex]A_m = \frac{2}{\lambda} \int^{\lambda}_0 F(x) cos(\frac{2\pi m x}{\lambda}) dx[/itex]

For a square wave: [itex]A_n = \frac{2}{\pi n} cos(\frac{\pi a n}{\lambda})[/itex] (Eq. 6.26)

## The Attempt at a Solution

For part (a) I tried integrating as suggested. I ended up with

[itex]\int^{\lambda}_0 A_n cos(\frac{2\pi n x}{\lambda})dx = \frac{\lambda A_{n} sin(2\pi n)}{2\pi n}[/itex]

...which doesn't do anything for me. Substituting 0 for n sets the whole sine argument equal to 0 (which is problematic in its own right, given the known expected outcome) but even worse puts a 0 in the denominator. If I substitute the function for A

_{n}for an even square wave, I get

[itex]\frac{2}{\pi n} * cos(\frac{\pi a n}{\lambda}) * \frac{sin({2\pi n})}{2\pi n}[/itex]

...which still doesn't do a thing for me, for the same reasons. I'm pretty badly lost here, and I'm not sure where to take the problem.