Bisection Method Fortran

The Bisection Method

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The Bisection Method at the same time gives a proof of theIntermediate Value Theoremandprovides a practical method to find roots of equations. If yourcalculator can solve equations numerically, it most likely uses acombination of the Bisection Method and the Newton-RaphsonMethod.

Recall the statement of the Intermediate Value Theorem: Letf (x) be a continuous function on the interval [a, b]. Ifd [f (a), f (b)], then there is a c [a, b] such thatf (c) = d.

By replacing f (x) by f (x) - d, we may assume that d = 0; itthen suffices to obtain the following version: Let f (x) be acontinuous function on the interval [a, b]. If f (a) and f (b)have opposite signs, then there is a c [a, b] such thatf (c) = 0.

Here is an outline of its proof: Let's assume that f (a) < 0, whilef (b) > 0, the other case being handled similarly. Set a₀ = a andb₀ = b.

Now consider the midpoint m₀ = ,and evaluate f (m₀). If f (m₀) < 0, set a₁ = m₀ and b₁ = b₀.If f (m₀) > 0, set a₁ = a₀ and b₁ = m₀. (If f (m₀)=0, you'realready done.) What situation are we in now? f (a₁) andf (b₁) still have opposite signs, but the length of the interval[a₁, b₁] is only half of the length of the original interval[a₀, b₀]. Note also that a₀a₁ and that b₀b₁.

You probably guess this by now: we will do this procedure againand again.

Here is the second step: Consider the midpoint m₁ = , and evaluate f (m₁). If f (m₁) < 0, seta₂ = m₁ and b₂ = b₁. If f (m₁) > 0, set a₂ = a₁ andb₂ = m₁. (If f (m₁)=0, you're already done.) What situationare we in now? f (a₂) and f (b₂) still have opposite signs,but the length of the interval [a₂, b₂] is only a quarter ofthe length of the original interval [a₀, b₀]. Note also thata₀a₁a₂ and that b₀b₁b₂.

The red line shows the interval [a_n, b_n].

Continuing in this fashion we construct by induction two sequences

with the following properties:

(a_n) is an increasing sequence, (b_n) is a decreasingsequence.
a_nb_n for all n.
f (a_n) < 0 for all n, f (b_n) > 0 for all n.
b_n - a_n = 2^-n(b - a) for all n.

It follows from the first two properties that the sequences(a_n) and (b_n) converge; set

a_n = a, b_n = b.

The third property and the continuity of the function f (x) implythat f ( Bisection Method Fortran

Bisection Method Using Fortran

a) 0 and that f

Bisection Method In Fortran

(b) 0.

The crucial observation is the fact that the fourth propertyimplies that a = b. Consequently, f (a) = f (b) = 0, and we are done.

Example. Let's compute numerical approximations for the with the help of the bisection method. We setf (x) = x² - 2. Let us start with an interval of length one: a₀ = 1and b₁ = 2. Note that f (a₀) = f (1) = - 1 < 0, and f (b₀) = f (2) = 2 > 0.Here are the first 20 applications of the bisection algorithm:

Bisection Method Fortran Definition

A comparison of the Bisection Method and theNewton-Raphson Method. TheNewton-Raphson Method is often much faster than the BisectionMethod. In the last example, we started with an interval oflength 1. After 10 steps, the interval [a₁₀, b₁₀] haslength 1/1024. Consequently every 10 steps of the BisectionMethod will give us about 3 digits more accuracy - that is ratherslow. (On the Newton-Raphson Method page, we did the sameexample, compare the speeds of convergence!)

The Newton-Raphson Method can be unreliable: If the algorithmencounters a point x where f '(x) = 0, it crashes; if itencounters points where the derivative is very close to 0, itwill become very unreliable.

Bisection Method Fortran Formula

The Bisection Method on the other hand will always work, once youhave found starting points a and b where the function takesopposite signs.

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