We know that that the difference equation formula for computing an output sample at time $ n$ is based upon its past/present input samples and past output samples in time domain:
$\displaystyle y(n)$ $\displaystyle =$ $\displaystyle b_0 \,x(n) + b_1 \,x(n - 1) + \cdots + b_M \,x(n - M)$
$\displaystyle \qquad\quad\; - a_1 \,y(n - 1) - \cdots - a_N \,y(n - N)$
$\displaystyle =$ $\displaystyle \sum_{i=0}^M b_i \,x(n-i) - \sum_{j=1}^N a_j \,y(n-j)$
My question is why 'j' begins from 1 rather than from value 0 like for 'i' and also how can we write the below output equation in difference equation form:
$\displaystyle y(n) = 0.01\, x(n - 5) + 0.002\, x(n - 1) + 0.99\, y(n - 1) $
$\displaystyle y(n) = 0.01\, x(n) + 0.002\, x(n - 1) $
$\displaystyle y(5) = 0.01\, x(3) + x(1)$
Secondly why y(n) sequence values are calculated as a sum of a series of input / output, cause I have been studying that $y(n) = \{2,4,6,8\}$ when $x(n) = \{1,2,3,4\}$ and $n =\{1,2,3,4\}$ (example per say) i.e. $y(n) = 2\cdot x(n)$. There was no mention that for $y(n)$ ($y(2)$ say) value we need to add values of $x(n)$ and $y(n)$ i.e. (take values of $x(1)$ or $y(1)$) or $y(n-1)$)?