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