Contraction iteration error bound ($d(x_n,x_{\ast })\le L^{n}d(x_1,x_0)/(1-L)$)

Successive steps satisfy $d(x_{n+1},x_n)\le L^{n}d(x_1,x_0)$. Summing the remaining steps with the triangle inequality gives a bound available before the iteration starts:After computing a new step, the contraction mapping gives the often sharper boundEquivalently, the residual at any $z\in X$ satisfies $d(z,x_{\ast })\le d(z,Tz)/(1-L)$. These estimates convert visible step sizes into guaranteed distances to an unknown fixed point. When $L$ is close to one, the factor $1/(1-L)$ can be large: a small step alone needs interpretation through the contraction constant.