M Sc I Practical Lab I

Exp# 1: Bisection Method

Manish Sharad Hiray

September 14, 2023

0.1 Aim
To use the Bisection Method to find the roots of given function

0.2 Apparatus
Scientific Calculator, Graph, Scale

0.3 Theory
Bisection Method

This is the simple method based on the successive application of intermediate value The-
orem.

Theorem:

If a function f (x) is continuous in [a, b] and if f (a) and f (b) have opposite signs, then
the equation f (x) = 0 has at least one real root say α in the interval [a, b]

The theorem can also be stated as that if f (a) · f (b) < 0, then f (x) = 0 has atleast one real
root say c in [a, b].
The bisection method can be applied as following steps:

1. Let f (x) be continuous in [a, b] and f (a) · f (b) is negative then f (x) = 0 has a root in
[a, b].

2. Find the middle point of [a, b] say c and construct the intervals [a, c] and [c, b]. Find
whether f (a) · f (c) < 0 or f (c) · f (b) < 0.
If f (a) · f (c) < 0 then the root lies in [a, c] and if f (c) · f (b) < 0, then the root lies in
[c, b].
This way we can have reduced the interval [a, b] to half the interval, let’s denote it now
by [a1 , b1 ].

3. Again repeat the step no 2 to [a1 , b1 ] and the interval [a1 , b1 ] is halved to [a2 , b2 ] and
repeat the process.

1
 4. At some stage, you will either get an exact root or an infinite sequence of interval
[a1 , b1 ],[a2 , b2 ],.......[an , bn ]..... and you have

f (an ).f (bn ) < 0

and
1
b n − an = (b − a)
2n
.

5. Observe that a1 , a2 , ........, an , .. is a monotonic non-decreasing sequence and b1 , b2 , ........., bn

is a monotonic non-increasing sequence and thses exists common limit.

α = lim an = lim bn ; and α is the root.

n→∞ n→∞

6. If we allow the error ϵ, then the number of iteration can be found as follows:
∵ peermissible error is ϵ

∴ b n − an ≤ ϵ

b−a
i.e. ≤ϵ
2n

∴ b − a ≤ ϵ · 2n

∴ log(b − a) − n log 2 ≤ logϵ

log(a − b) − logϵ
∴ n≥
log2

0.4 Solved Example

Carry out four steps of the bisection method for the root of (2x + 1)2 = 4cosπx
lying in the interval 14 , 13 .


Solutin:

1 1



(i) Let f (x) = (2x + 1)2 − 4cosπx. Now check f 4
and f 3
.

1 1


1 1

f 4
< 0 and f
> 0 ∴ Root lies in
3
,
4 3
.

(ii) Now, let c1 = the mid point of 41 , 31 .

 

1
+ 13
c1 = 4
2
= 0.2917 and f (c1 ) > 0

∴ the root lies in [0.25, 0.2917]

2
 (iii) Now, let c2 = mid point of [0.25, 0.2917]

0.25+0.2917
c2 = 2
= 0.2709 Now, f (c2 ) < 0

∴ The root lies in [0.2709, 0.2917].

(iv) Now, let c3 = mid point of [0.2709, 0.2917]

0.2709+0.2917
c3 = 2
= 0.2813 Now, f (c3 ) < 0

∴ The root lies in [0.2813, 0.2917].

(v) Now, let c4 = mid point of [0.2813, 0.2917]

0.2813+0.2917
c4 = 2
= 0.2865 Now, f (43 ) < 0

∴ The root lies in [0.2865, 0.2917].

Here, we have only two-decimal place accuracy, i.e., α = 0.29.

Sr No Iteration No Middle point ci f (ci ) Old Interval New Interval

1 0 - - [0.25 , 0.33] -
2 I 0.2917 0.0724 [0.25 , 0.33] [0.25 , 0.2917]
3 II 0.2709 -0.2597 [0.25 , 0.2917] [0.2709 , 0.2917]
4 III 0.2813 -0.0953 [0.2709 , 0.2917] [0.2813 , 0.2917]
5 IV 0.2865 -0.0119 [0.2813 , 0.2917] [0.2856 , 0.2917]

0.5 Problems
Prob# 1 Obtain the real root of x3 − x2 − x − 1 = 0 lying between 1.7 and 1.9, correct
upto 4 places of dcimal by bisection method.

Iteration No Middle point ci f (ci ) Old Interval New Interval

Prob# 2 Use the method of bisector to find a real root of the equation x4 +2x3 −x−1 =
0 in the interval [0, 1] . How many iterations are required if thepermissible
error is 0.0005.

Iteration No Middle point ci f (ci ) Old Interval New Interval

Prob# 3 Using method of besection, find the cube-root of 100, using six iterations.
Determine whether the number of iterations are enough for three significant
digit accuracy.