- #1
Charles49
- 87
- 0
If we want to solve $$f(x)=0$$ we can re-write the equation as
$$g(x)=x$$ and use the fixed point method, i.e, $$x_{n+1}=g(x_n)$$ starting with a guess $$x_0.$$ I was wondering if something similar can be done with
$$\Lambda(x,y)=h(x,y).$$
$$g(x)=x$$ and use the fixed point method, i.e, $$x_{n+1}=g(x_n)$$ starting with a guess $$x_0.$$ I was wondering if something similar can be done with
$$\Lambda(x,y)=h(x,y).$$