After finishing some BP(back propagation) problem, I ended up by these equations: $$W_1(0) = 1.5$$$$W_2(0) =0$$ $$W_1(x+1) = W_1(x) -2B(x)+\frac{W_2(x)-W_1(x)-1}{5}$$ $$W_2(x+1) = W_2(x) -2B(x) +\frac{W_1(x)-W_2(x)+1}{5}$$ where $$B(odd) = -0.1$$ and $$B(even) = 0.1$$ I want a closed formula for $W_1(x)$ and $W_2(x)$, and I googled to find a way to approach, but I didn't find any solution.
Could someone give me a way to solve it?
Write it as
$$v(x+1) = v(x) -2b(x)+c(w(x)-v(x)-1)$$ $$w(x+1) = w(x) -2b(x) +c(v(x)-w(x)+1)$$
$c = \frac15$.
Subtracting,
$v(x+1)-w(x+1) =v(x)-w(x)+c(2w(x)-2v(x)-2)\\ =(v(x)-w(x))(1-2c)-2c $
Let $u(x)=v(x)-w(x) $. Then $u(x+1)=du(x)-2c$ where $d=1-2c $.
From this, we can get $u(x)$.
Adding,
$v(x+1)+w(x+1) =v(x)+w(x)-4b(x) $.
Let $t(x)=v(x)+w(x)$. Then $t(x+1)=t(x)-4b(x) $.
From this we can easily get $t(x) $.
Then $v(x)=(t(x)+u(x))/2, w(x)=(t(x)-u(x))/2 $.