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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.2012-06-12
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    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.2012-06-12
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    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 much2012-06-12
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    Any comments on the answer I posted?2012-06-15
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    @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.2012-06-18

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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 $$A={\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-B\sin(2\pi fk))\over\sum_{k=0}^{n-1}\cos^2(2\pi fk)}$$