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tiagotorres
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I tried to find the cosine series of the function [tex]f(x) = \sin x[/tex], using the equation below:
[tex]S(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)[/tex]
where: [tex]a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx) dx[/tex]
I found:
[tex]a_0 = \frac{4}{\pi}[/tex]
[tex]a_n = \frac{2 }{\pi (1 - n^2)} (\cos(n \pi) + 1)[/tex]
Therefore:
[tex]S(x) = \frac{2}{\pi} (1 - 2 \sum_{n=1}^{\infty} \frac{\cos(2nx)}{4n^2 - 1})[/tex]
Making the graph of the first terms of the function above on my calculator, I noticed that this is actually [tex]| \sin x |[/tex], rather than just [tex]\sin x[/tex]. Why does this happen? Is there a way of figuring out the sum not using a calculator?
Thanks
[tex]S(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)[/tex]
where: [tex]a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx) dx[/tex]
I found:
[tex]a_0 = \frac{4}{\pi}[/tex]
[tex]a_n = \frac{2 }{\pi (1 - n^2)} (\cos(n \pi) + 1)[/tex]
Therefore:
[tex]S(x) = \frac{2}{\pi} (1 - 2 \sum_{n=1}^{\infty} \frac{\cos(2nx)}{4n^2 - 1})[/tex]
Making the graph of the first terms of the function above on my calculator, I noticed that this is actually [tex]| \sin x |[/tex], rather than just [tex]\sin x[/tex]. Why does this happen? Is there a way of figuring out the sum not using a calculator?
Thanks
Last edited: