I would be grateful if someone please rewrite or expand this please. I have problem multiplying two sigmas ($\sum $)
$$ (d(n)-\sum_{k=-\infty}^{\infty} h_k x(n-k)) \times (d(n)-\sum_{l=-\infty}^{\infty} h_l x(n-l))^{*} $$
By the way $^*$ is the conjugate.
Many thanks
$$\begin{align} & \left(d(n)-\sum_{k=-\infty}^{\infty} h_k x(n-k)\right) \times \left(d(n)-\sum_{l=-\infty}^{\infty} h_l x(n-l)\right)^{*} \tag{1}\\ & = \left(d(n)-\sum_{k=-\infty}^{\infty} h_k x(n-k)\right) \times \left(d(n)^* -\left(\sum_{l=-\infty}^{\infty} h_l x(n-l)\right)^*\right) \tag{2}\\ & = d(n)d(n)^* - d(n)^*\sum_{k=-\infty}^{\infty} h_k x(n-k) - d(n) \left(\sum_{l=-\infty}^{\infty} h_l x(n-l)\right)^* + \left( \sum_{k=-\infty}^{\infty} h_k x(n-k) \right)\left(\sum_{l=-\infty}^{\infty} h_l x(n-l)\right)^* \tag{3}\\ & = d(n)d(n)^* - d(n)^*\sum_{k=-\infty}^{\infty} h_k x(n-k) - d(n) \left(\sum_{k=-\infty}^{\infty} h_k x(n-k)\right)^* + \left( \sum_{k=-\infty}^{\infty} h_k x(n-k) \right)\left(\sum_{k=-\infty}^{\infty} h_k x(n-k)\right)^* \tag{4}\\ & = |d(n)|^2 - 2\,\mathbb{R}\left\{d(n)^*\sum_{k=-\infty}^{\infty} h_k x(n-k)\right\} + \left| \sum_{k=-\infty}^{\infty} h_k x(n-k) \right|^2\tag{5}\\ \end{align}$$
where $\mathbb{R}$ is the real part of a complex number.
From $(1)$ to $(2)$ we use the fact that $(a \pm b)^*$ = $a^* \pm b^*$.
From $(2)$ to $(3)$ we use the fact that $(a + b)(c + d)$ = $ac + ad + bc + bd$.
From $(3)$ to $(4)$ we use the fact that $\sum_{l=-\infty}^{\infty} f(l) = \sum_{k=-\infty}^{\infty} f(k)$ (and that is how the $l$ disappears, it has no meaning outside of the sum).
From $(4)$ to $(5)$ we use the fact that $a \times a^* = |a|^2$ and $a^*b + ab^* = 2\mathbb{R}\left\{a^*b\right\} = 2\mathbb{R}\left\{ab^*\right\}$.