Fourier Series. Writing a partial sum as an integral.

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Homework Statement

Given: https://www.physicsforums.com/attachments/56653, show that this can be written as: https://www.physicsforums.com/attachments/56651.

Homework Equations

Hint: https://www.physicsforums.com/attachments/56652

The Attempt at a Solution

Quite confused by this question. For f'N I have:

$\frac{4}{∏}$Ʃ$\frac{cos((2n -1)x)}{(2n - 1)^{2}}$ but I'm not sure where to go from here.

I have read about the Dirichlet kernel, which is what I think I require, but I am rather lost.

Any help is appreciated.

Cheers.

 

[mighty2000]

Hello,

Thank you for your post. Let's break down the problem step by step.

First, we are given the function f(x) = |sin(x)|. This can be rewritten as f(x) = 1/2(sin(x) + sin(-x)).

Next, we have to use the hint provided, which is the Dirichlet kernel. The Dirichlet kernel is defined as Dn(x) = 1/2 + cos(nx).

Using this, we can rewrite f(x) as f(x) = 1/2(D1(x) + D-1(x)).

Now, we can use the formula for the Fourier series of the Dirichlet kernel, which is given by:

Dn(x) = \frac{1}{\pi}Ʃ\frac{sin((2n - 1)x)}{(2n - 1)}.

Substituting this into our equation for f(x), we get:

f(x) = 1/2(\frac{1}{\pi}Ʃ\frac{sin((2 - 1)x)}{(2 - 1)} + \frac{1}{\pi}Ʃ\frac{sin((-2 - 1)x)}{(-2 - 1)}).

Simplifying this, we get f(x) = 1/2(\frac{1}{\pi}Ʃ\frac{sin(x)}{1} + \frac{1}{\pi}Ʃ\frac{sin(-3x)}{-3}).

Finally, we can rewrite this as f(x) = \frac{1}{2\pi}Ʃ\frac{sin(x)}{1} + \frac{1}{2\pi}Ʃ\frac{sin(-3x)}{-3}.

This is the desired result, which matches the given equation. I hope this helps. If you have any further questions, please let me know.