Solving Finite Series with Real Y

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The discussion focuses on finding the sum of the finite series |ysin(x)| + |y2sin(2x)| + ... + |ynsin(nx)|, where y is a real number. Participants suggest using complex exponentials to transform the sine function into a geometric series, although this approach is complicated by the presence of absolute values. A proposed method involves determining the signs of the terms before applying the modulus, allowing for separate summation of positive and negative terms. The challenge lies in predicting the signs based on the value of x, especially when x is a rational multiple of pi. The consensus is that while the task is complex, breaking it down logically can lead to a closed-form solution.
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I cannot figure out the sum of this finite series:
|ysin(x)|+|y2sin(2x)|+...+|ynsin(nx)|
where y is real.
so I want to listen any opinion may help me>
 
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Maybe try to express Sine with a complex exponential. Then you get a geometric series. This should work.
 
I had tried but it doesnot work it works just when there are no absolute values
 
Be creative ;) It works if you work through the cases, i.e. use logic to split positive and negative terms. It's messy, but gives you a closed form equation in the end.
 
thanks but I am not sure what you mean
 
I realize it's harder than I thought first.
I meant first determine which terms are going to be negative and which ones are positive (before the modulus).
Then sum both sets separately, because knowing the sign you can actually remove the modulus and replace it by *(-1) wherever you determined the sign to be negative. Depending on x you will have some runs of positive only terms and some negative only terms.
If x is a rational part of pi then it should be predictable. I haven't completely the calculation though...
 
I think it is so hard to do that in this manner because it is so hard to predict which terms should be positive or negative
 
At least for x=rational*pi it's fairly easy. But haven't put more thought in it.
 

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