# Trig Series Sum

I have equation that I try to solve for one of the values

$$\sum_{k=0}^{n-1}\cos(2 \pi fk)(x_{k}- \mu-A\cos(2 \pi fk)-B\sin(2 \pi fk))$$ I know to set equal $0$ I try to solve for A but how to take sum ?

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What are you really trying to get? It is not clear what your question is. –  Pedro Tamaroff Jun 12 '12 at 0:03
Are you trying to find a closed form for the sum? Part of that sum will be $\sum x_k\cos(2\pi fk)$. There's not much you can do with that, if you don't know what the $x_k$ are. –  Gerry Myerson Jun 12 '12 at 0:35
Hi,I assume I know values for x. I try to solve for A so I set that summ equal to 0 and get A by itself. thank you very much –  nanme Jun 12 '12 at 0:45
Any comments on the answer I posted? –  Gerry Myerson Jun 15 '12 at 12:51
@nname: By removing the equations from your question, you made it impossible to understand. That constitutes vandalism. I have undone your edits, and if you do something like that again, you may be suspended. –  Zev Chonoles Jun 18 '12 at 22:41

## 1 Answer

I think OP wants to set the sum to zero and solve for $A$. So, $$\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-A\cos(2\pi fk)-B\sin(2\pi fk))=0$$ becomes $$\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-B\sin(2\pi fk))-A\sum_{k=0}^{n-1}\cos^2(2\pi fk)=0$$ which gives us $$A={\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-B\sin(2\pi fk))\over\sum_{k=0}^{n-1}\cos^2(2\pi fk)}$$

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