Simplifying the Fourier Series Function: Tips & Tricks

darkmagic
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Homework Statement



f(x) = 1 0<t<1
= -1 1<t<2

How can I simplify this given that function(on the attachment).

What I mean is that how can I write the function in any other way?

In addition, How can I know if the function can be written in other form?
How can I write the function in other form?



Homework Equations





The Attempt at a Solution

 

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If n is even, 1- (-1)n= 1- 1= 0! If n is odd, 1- (-1)n= 1- (-1)= 2.

So
\frac{2}{\pi}\sum_{n=1}^\infty \frac{[1- (-1)^n] sin(n\pi t)}{n}
is just
\frac{2}{\pi}\sum \frac{2 sin(n\pi t)}{n}
where now the sum runs only over odd n. One way to show that is to use 2n+1 rather than n in the body of the sum. That way, as n goes over all non-negative integers, 2n+1 goes over all positive odd integers:
\frac{4}{\pi}\sum_{n=0}^\infty \frac{sin((2n+1)\pi t)}{2n+1}
 
Can 2n+1 be 2n-1 provided that n=1 to infinity?
How can I know if the function can be converted in some form?
 
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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