# Tricky demonstration

this is what I'm trying to demonstrate

With the following system of $$n \times n$$ equations

$$G_1B_1+G_2B_2+\cdots+G_nB_n=G_1A_1+G_2A_2+\cdots+G_nA_n$$

$$B_1-B_2=A_2-A_1$$

$$B_2-B_3=A_3-A_2$$

$$\vdots \hspace{1cm} \vdots \hspace{1cm} \vdots \hspace{1cm} \vdots$$

$$B_{n-1}-B_{n}=A_n-A_{n-1}$$

Prove that in general any $$B_{V}$$ with $$V \in \{1,2,\cdots,n\}$$ is equal to:

$$B_V=\left(\Gamma_1A_1+\Gamma_2A_2+\cdots+\Gamma_nA_n\right)-A_V$$

With $$\Gamma_V=\dfrac{2G_V}{G1+G_2+\cdots+G_n}$$

This is my first time solving a problem like that, so the first thing that I tried, was to write the system in a matrix form

$$\left( \begin{array}{cccccc} G_1 & G_2 & G_3 & G_4 & \cdots & G_{n-1} & G_n \\ 1 & -1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & -1 \\ \end{array} \right)\left( \begin{array}{c} B_1 \\ B_2 \\ B_3 \\ B_4 \\ \vdots \\ B_{n-1}\\ B_n \end{array} \right)$$=$$\left( \begin{array}{cccccc} G_1 & G_2 & G_3 & G_4 & \cdots & G_{n-1} & G_n \\ -1 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & -1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & -1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & -1 & 1 \\ \end{array} \right)\left( \begin{array}{c} A_1 \\ A_2 \\ A_3 \\ A_4 \\ \vdots \\ A_{n-1}\\ A_n \end{array} \right)$$

But it gets me nowhere and I think that it isn't the right approach to achieve that demonstration, This is not a book exercise, is just a formula that I found in a paper and that I don't understant how was obtained.

This is the paper https://ccrma.stanford.edu/~jingjiez/portfolio/gtr-amp-sim/pdfs/Wave%20Digital%20Filters%20Theory%20and%20Practice.pdf , equation (29)

First you should note that $$A_1+B_1= A_2+B_2= \dots = A_n + B_n.$$
Now starting from the first given equation we have $$0= \sum_i G_i (A_i -B_i)= -\sum_i G_i (A_i +B_i) +\sum_i G_i (A_i +B_i) +\sum_i G_i (A_i -B_i)$$ so that $$0= -\sum_i G_i (A_i +B_i) +2\sum_i G_i A_i,$$ or $$\sum_i G_i (A_i +B_i)= 2\sum_i G_i A_i.$$
Now recall that $$A_i+B_i$$ does not depend on $$i$$; we can replace them all by $$A_j+B_j$$ for a fixed $$j$$ and get: $$\left(\sum_i G_i\right) (A_j +B_j)= 2\sum_i G_i A_i$$ which is what you wanted to prove.