Root finding using the Bisection Method - execution of the program

In a previous post I provided a program for finding the root of a function f(x) using the bisection method. To show how this works, consider the problem of calculating where the lines y=x and y=cos x intersect on a graph of y against x (see the above plot from Wolfram Alpha). There is a single point of intersection somewhere between x=0 and x=π/2 and to find it we need to obtain the solution of x=cos x, i.e. x-cos x=0. This trigonometric equation has no obvious solutions and so we can use the bisection method to find a numerical solution (where f(x)=x-cos x).

The program given is already set up for this. On running the program the display shows A? This is the lower limit of the interval that is being bisected. Enter [0] and press [EXE]. The display then shows B? This is the upper limit of the interval. Enter [SHIFT] [π] [/] [2] and press [EXE]. The display then shows D? This the error tolerance. Enter [1] [EXP] [-] [5] and press [EXE].

After a moment or two the display shows that the calculated root C is 0.739085126 and pressing [EXE] once more, the number of iterations X used in the calculations was 18. The maximum error in the root is  ±0.0000060, smaller than the tolerance of ±0.00001 requested, as expected (the maximum error can be calculated from C-A). So we can probably quote the root as being 0.7391 to four decimal places and be confident that this is correct (other maths software shows that the root is 0.739085133 to 9 decimal places).

Repeating the bisection with different intervals but the same level of tolerance in the error gives the following results:-

A=0.5, B=1, C=0.739082336, X=16
A=0.25, B=1.5, C=0.739091873, X=17
A=0.7, B=0.8, C=0.73908081, X=14

and this confirms that quoting the root as 0.7391 to 4 d.p. is correct.