Finding Fourier Expansion of f(t): Part b & c

zyphriss2
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Find the Fourier expansion of one period of f(t)=1+t absolute value of t<1

I found this to be 1+2/pi Sigma(0 to infinity) ((-1^(n+1))/n)sinnpit by just the standard methods of the a0 an and bn formuals, which I know is correct

Now the parts I am having problems with is part b and c which our teacher has not covered much at all and I cannot find any help online.

Part b. Use the Fourier expansion of f to find the sum of the Series
Sigma(0toinfinity)(-1^n)/(2n+1)

Part C. If f denotes the function defined on (-infinity to infinity) by the Fourier series of f, find F(1)+F(-5)-3F(0)
 
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Just to make things a bit more clear on the notation side, I take it your expansion is:

1+\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n} \sin{n\pi t}

As for question b:

Notice that if you set t equal to t_0 \equiv \frac{1}{2}, then

\sin{n\pi t_0} = \sin{\frac{n\pi}{2}}

which is zero whenever n is even and (-1)^{(n-1)/2 whenever n is odd... Do you see where I'm going with this?

part C:
I take it by F(0) you just mean f(0)? Have you tried filling in these numbers into the Fourier series?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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