# bisection method on $f(x) = \sqrt{x} − 1.1$

the question goes something like this :

Apply the bisection method using the code in slides to find the root of the function $$f(x) = \sqrt{x} − 1.1$$ starting from the interval $$[0, 2]$$ with a tolerance of $$\epsilon = 1.e-8$$

• How many iterations are required ? Does the iteration count match the expectations, based on our convergence analysis ?
• What is the resulting absolute error ? Could this absolute error be predicted by our convergence analysis ?

the first part of the question is pretty easy to code out and here is my code for the same :

def func(x):
return np.sqrt(x)-1.1

def bisection(a, b, tol):

if (func(a) * func(b) >= 0):
return

c = a
itercount=0
while ((b-a) >= tol):

c = (a+b)/2

if (func(c) == 0.0):
return c

if (func(c)*func(a) < 0):
b = c
else:
a = c
itercount+=1
return c, itercount


we get the theoretical number of iterations using the relation $$n_{_{theoretical}} = \log_{_{2}} \left( \frac{b-a}{\epsilon}\right)$$ which is very close to that counted by the programme ( $$n_{_{calculated}} = 28$$).

the absolute error = $$8.94 \cdot 10^{-10}$$ , but I have a problem understanding how this error could have been predicted by the convergence analysis, any hints or solutions are welcome.

Let $$S_n$$ be the $$nth$$ root approximation generated by the Bisection method, and $$a_n,b_n$$ denote the $$nth$$ $$a,b$$ for every iteration. We also have the exact root $$S$$ between the interval $$(a,b)$$ and $$f(a)f(b)<0$$ on the interval $$[a,b] = [0, 2]$$
For $$n=1$$, it is easy to see that $$b_1-a_1=b-a$$ For $$n>1$$, iterations of the Bisection method has been made so: $$b_n-a_n=\frac{b-a}{2^{n-1}}$$
From the Biscetion method: $$S_n=a_n+\frac{b_n-a_n}{2}$$ $$S_n-a_n=\frac{b_n-a_n}{2}$$ Obviously $$S\in(a_n,b_n)$$ so $$|S_n-S|\leqslant \frac{b_n-a_n}{2}$$ $$|S_n-S|\leqslant\frac{b-a}{2^n}$$
Let $$\epsilon = |S_n-S|$$ $$\epsilon \leqslant\frac{b-a}{2^n}$$ $$n\leqslant \log_2(\frac{b-a}{\epsilon})$$
We now see that the $$nth$$ iteration corresponds to the distance between $$S_n$$ and $$S$$, which is bounded by $$\frac{b-a}{2^n}$$