If you start with x1=1 and define the subsequent terms of {xn} by the rule xn=xn-1+cosxn-1, you

step 1 All right, so here, we're given that x sub 1 is equal to 1, and then we have that x sub n is equ

If you start with x1=1 and define the subsequent terms of...

If you start with $x_{1}=1$ and define the subsequent terms of $\left\{x_{n}\right\}$ by the rule $x_{n}=x_{n-1}+\cos x_{n-1},$ you generate a sequence that converges rapidly to $\pi / 2 .$ (a) Try it. (b) Use the accompanying figure to explain why the convergence is so rapid.

Key Concepts

Recurrence Relation

A recurrence relation defines each term of a sequence in terms of one or more of its previous terms. In this context, the sequence is generated by an iterative process, where each new value is calculated by adding a function of the previous term to that previous term.

Fixed Point Iteration

Fixed point iteration is a method for finding a point x* such that f(x*) = x*. For the sequence defined by x? = x??? + cos(x???), the fixed point is reached when cos(x*) = 0, which occurs at x* = ?/2. This means that once the sequence gets sufficiently close to ?/2, it remains near that value since the update becomes negligibly small.

Local Convergence Analysis

The rapid convergence is explained by analyzing the behavior of the iterative function near the fixed point. When the derivative of the function at the fixed point is zero, perturbations from the fixed point decrease very quickly. In this sequence, the derivative f'(x) = 1 - sin(x), and at x = ?/2 the derivative becomes 1 - 1 = 0, resulting in a locally flat function which leads to a rapid reduction in the error and thus a fast convergence.

Graphical Interpretation

A graphical explanation involves plotting the function f(x) = x + cos(x) along with the line y = x. The point where these curves intersect represents the fixed point. Near this intersection, because f(x) is almost tangent to the line y = x due to the zero derivative, the successive iterates jump directly to the fixed point with minimal deviation, thereby visually explaining the rapid approach to ?/2.

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