Evaluating \sum_{k=1}^{n-1} k \cos(kx)

Danitat6

Danitat6

Answered question

2022-01-23

Evaluating k=1n1kcos(kx)

Answer & Explanation

izumrledk

izumrledk

Beginner2022-01-24Added 15 answers

Using the above suggestion from User Simply Beautiful Art, that is, by writing the expression as a derivative of a sine function, we finally get:
k=1n1kcos(kx)=k=1n1ddxsinkx
=ddx(sin(kx2)sin(x2)sinn12x)
=12(kcos(kx2)sinx2sin(kx2)sin2x2cos(x2))sin(n12x)+n12sin(kx2)sinx2cos(n12x)
where in the second row we use again the formula from the reference given in the opening post.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?