Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.