(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This isn't really homework, just research I'm doing for myself...

we start by defining:

[tex]\sin_\ell(x) = (1 - \frac{1}{1+\tan^\ell(x)})^{\frac{1}{\ell}}\,\,\,for\, \,\,x \in [0, \pi][/tex]

for [itex]\ell[/itex] divisible by two, or [itex]\frac{1}{\ell}[/itex] divisible by two.

[tex]\sin_\ell(x + \pi) = -\sin_\ell(x) = \sin_\ell(-x)[/tex]

and then for:

[tex]a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin_\ell(t) \cdot \cos(n\cdot t) dt[/tex]

[tex]b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin_\ell(t) \cdot \sin(n\cdot t) dt[/tex]

[tex]\sin_\ell(x) = \sum_{n=1}^{\infty} [a_n \cdot \cos(n\cdot x) + b_n \cdot \sin(n \cdot x)][/tex]

2. Relevant equations

what is [tex] b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin_\ell(t) \cdot \sin(n\cdot t) dt[/tex]

3. The attempt at a solution

Well, all I've really managed to deduce is that [tex]a_n[/tex] is always zero, this was found through the odd property of [tex]\sin_\ell(x)[/tex]

so we get

[tex]\sin_\ell(x) = \sum_{n=1}^{\infty}b_n \cdot \sin(n \cdot x)[/tex]

plugging in [itex]x = \frac{\pi}{2}[/itex] we get:

[tex]\sum_{n=0}^{\infty} b_{2n + 1} \cdot (-1)^{n} = 1[/tex]

otherwise, I have no clue how to tackle this integral. To me it seems insurpassably complicated. Perhaps someone can simply point me towards the correct integration technique. Or perhaps there is an alternative way to find the fourier coefficients that don't involve evaluating this clunky integral.

This is only a small part of the much bigger problem I'm analysing, so please don't think I'm just trying to find the easy way out. This part of the problem just goes beyond my schooling. Thanks, any help or suggestions would be greatly appreciated.

edit:

well I managed to reformulate:

[tex]\int_{-\pi}^{\pi} \sin_\ell(t) \cdot \sin(n\cdot t) dt = \int_{-\pi}^{0} \sin_\ell(t) \cdot \sin(n\cdot t) dt + \int_{0}^{\pi} \sin_\ell(t) \cdot \sin(n\cdot t) dt[/tex]

[tex]\int_{0}^{\pi} \sin_\ell(t-\pi) \cdot \sin(n\cdot t - n \cdot \pi) dt + \int_{0}^{\pi} \sin_\ell(t) \cdot \sin(n\cdot t) dt = ((-1)^{n+1} + 1) \cdot \int_{0}^{\pi} (1 - \frac{1}{1+\tan^\ell(t)})^{\frac{1}{\ell}} \cdot \sin(n\cdot t) dt[/tex]

which means we can rewrite the equation as:

[tex]\sin_\ell(x) = \sum_{n=0}^{\infty} b_{2n+1} \cdot \sin((2n+1)\cdot x)[/tex]

or

[tex]\sin_\ell(x) = \sum_{n=0}^{\infty} \sin((2n + 1)\cdot x) \frac{2}{\pi} \int_{0}^{\pi} \sin_\ell(t) \cdot \sin((2n + 1) \cdot t) dt[/tex]

Actually, after plugging this into mathematica it does not return a algebraic result, which leaves me to believe this integral is NOT elementary. I'm a little inadequate at how to give the proof of this, but it seems sufficiently complex to not be elementary (save when [itex]\ell = 2[/itex])

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# Homework Help: Help with the coefficients of a fourier series

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