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I'm given that

$$ \sum_{t=1}^N \epsilon^2(t) = \sum_{t=1}^N\left[y(t) - \sum_{n=0}^{N/2}\left\{a_n\cos\left(\frac{2\pi nt}{N}\right) + b_n\sin\left(\frac{2\pi nt}{N}\right)\right\}\right]^2 $$

and am required to derive $\hat{a}_0$, $\hat{a}_{N/2}$, $\hat{b}_0$, $\hat{b}_{N/2}$ and $\hat{a}_n$, $\hat{b}_n$ for $n = 1, \cdots (N/2)-1$.

So far, I've derived these expressions for $a$

$$ 0 = \sum_{t=1}^N\left[-2\sum_{n=0}^{N/2}\cos\left(\frac{2\pi nt}{N}\right)\left(y(t) - \sum_{n=0}^{N/2}a_n\cos\left(\frac{2\pi nt}{N}\right)-\sum_{n=0}^{N/2}b_n\sin\left(\frac{2\pi nt}{N}\right)\right)\right] $$

Which gives $\hat{a}_0 = \bar{y}$ for $n=0$ which is correct, but for $n = N/2$ it gives $\hat{a}_{N/2} = \sum_{t=1}^N y(t)\cos(\pi t)$ but it should be $\hat{a}_{N/2} = \sum_{t=1}^N (-1)^t/N$ which I'm not arriving at.

For $b$

$$ 0 = \sum_{t=1}^N\left[-2\sum_{n=0}^{N/2}\sin\left(\frac{2\pi nt}{N}\right)\left(y(t) - \sum_{n=0}^{N/2}a_n\cos\left(\frac{2\pi nt}{N}\right)-\sum_{n=0}^{N/2}b_n\sin\left(\frac{2\pi nt}{N}\right)\right)\right] $$

Which, for $n=0$ and $n=N/2$ caused the whole expression to go to $0$, which I guess it okay since $\hat{b}_0$ and $\hat{b}_{N/2}$ are $0$.

The expressions for the parameters are $\hat{a}_0 = \bar{y}$, $\hat{a}_{N/2} = \sum_{t=1}^N (-1)^t/N$, $\hat{b}_0 = 0 = \hat{b}_{N/2}$ and

$$ \hat{a}_n = \frac{2}{N}\sum_{t=0}^Ny(t)\cos(2\pi nt/N) $$ $$ \hat{b}_n = \frac{2}{N}\sum_{t=0}^Ny(t)\sin(2\pi nt/N) $$

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  • $\begingroup$ Use $\cos(\pi t)=(-1)^t$ for integral $t.$ $\endgroup$ – whuber Nov 21 '18 at 19:48
  • $\begingroup$ but it still leaves $\sum_{t=1}^N y(t)$ in $\hat{a}_{N/2}$ expression $\endgroup$ – Syed Ali Nov 23 '18 at 16:51
  • $\begingroup$ Since you haven't made any assumptions about $y,$ we must presume that all coefficients really do depend on $y.$ You should therefore suspect typographical errors in the "correct" formula. $\endgroup$ – whuber Nov 23 '18 at 16:53

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