Use this with 2 as starting point to find a fraction that is

Use this, with 2 as starting point, to find a fraction that is within 10^-14 of Squareroot 2 . In this problem, you find a root of the following equation within the given interval using the Bisection method. x^3 + 4x^2= 10, [1, 2] 2) e^x = x + 2, [1, 2] 5x +In x= 10000, [1900, 2100] Write a program Bisection. m that has the following syntax for calling: [sol, -1] = Bisection ( \'f\', a, b, tol, max_iter) ; Determine an approximation to the root that is accurate with the relative error less than 10^- 4. Produce a table that looks like the following:

Solution

Given that we an initial bound on the problem [a, b], then the maximum error of using either a or b as our approximation is h = b a. Because we halve the width of the interval with each iteration, the error is reduced by a factor of 2, and thus, the error after n iterations will be h/2ⁿ.

Therefore, thus, if _step is fixed, then we may immediately know how many steps are required, after which we are assured that the absolute error is less than _step. The inequality

may be solved for an integer value of n by finding:

For example, suppose that our initial interval is [0.7, 1.5]. If we have an _step value of 1e-5, then we require a minimum of log₂( 0.8/1e-5 ) = 17 steps.

Examples

Example 1

Consider finding the root of f(x) = x² - 3. Let _step = 0.01, _abs = 0.01 and start with the interval [1, 2].

Table 1. Bisection method applied to f(x) = x² - 3.

Thus, with the seventh iteration, we note that the final interval, [1.7266, 1.7344], has a width less than 0.01 and |f(1.7344)| < 0.01, and therefore we chose b = 1.7344 to be our approximation of the root.

Example 2

Consider finding the root of f(x) = e^-x(3.2 sin(x) - 0.5 cos(x)) on the interval [3, 4], this time with _step = 0.001, _abs = 0.001.

Table 1. Bisection method applied to f(x) = e^-x(3.2 sin(x) - 0.5 cos(x)).

Thus, after the 11th iteration, we note that the final interval,

a	b	f(a)	f(b)	c = (a + b)/2	f(c)	Update	new b a
1.0	2.0	-2.0	1.0	1.5	-0.75	a = c	0.5
1.5	2.0	-0.75	1.0	1.75	0.062	b = c	0.25
1.5	1.75	-0.75	0.0625	1.625	-0.359	a = c	0.125
1.625	1.75	-0.3594	0.0625	1.6875	-0.1523	a = c	0.0625
1.6875	1.75	-0.1523	0.0625	1.7188	-0.0457	a = c	0.0313
1.7188	1.75	-0.0457	0.0625	1.7344	0.0081	b = c	0.0156
1.71988/td>	1.7344	-0.0457	0.0081	1.7266	-0.0189	a = c	0.0078