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Fixed-point iteration

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Through Numerical analysis, Fixed point (mathematics) and Real number

More specifically, given a function f defined on the real numbers with real values and given a point x0 in the domain of f, the fixed-point iteration is xn+1=f(xn),n=0,1,2,… which gives rise to the sequence x0,x1,x2,… of iterated function applications x0,f(x0),f(f(x0)),… which is hoped to converge to a point xfix. If f is continuous, then one can prove that the obtained xfix is a fixed point of f, i.e., f(xfix)=xfix.

More generally, the function f can be defined on any metric space with values in that same space.

Examples

  • A first simple and useful example is the Babylonian method for computing the square root of a > 0, which consists in taking f(x)=12(ax+x), i.e. the mean value of x and a/x, to approach the limit x=a (from whatever starting point x0≫0). This is a special case of Newton's method quoted below.
  • The fixed-point iteration xn+1=cos⁡xn converges to the unique fixed point of the function f(x)=cos⁡x for any starting point x0. This example does satisfy (at the latest after the first iteration step) the assumptions of the Banach fixed-point theorem. Hence, the error after n steps satisfies |xn−x|≤qn1−q|x1−x0|=Cqn (where we can take q=0.85, if we start from x0=1.) When the error is less than a multiple of qn for some constant q, we say that we have linear convergence. The Banach fixed-point theorem allows one to obtain fixed-point iterations with linear convergence.
  • The requirement that f is continuous is important, as the following example shows. The iteration xn+1={xn2,xn≠01,xn=0 converges to 0 for all values of x0. However, 0 is not a fixed point of the function f(x)={x2,x≠01,x=0 as this function is not continuous at x=0, and in fact has no fixed points.

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