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Bisection Method

Mt

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0% found this document useful (0 votes)
20 views7 pages

Bisection Method

Mt

Uploaded by

mulokoziavitha9
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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MT171 – One Variable Calculus

and Differential Equatios for


Non-Majors – 2020/2021

Idrissa S. A.

Bisection Method
Department of Mathematics – UDSM
May 31, 2022
The Bisection Method
Suppose that f is a real-valued function defined and con-
tinuous on a bounded closed interval [a, b] such that
f (ξ) = 0 for some ξ ∈ [a, b].
A very simple iterative method for the solution of the
nonlinear equation f (x) = 0 can be constructed by be-
ginning with an interval [a0 , b0 ] which is known to contain
the required solution, simply choose a0 = a and b0 = b,
and successively halving its size. This method is mainly
based on the IVT which promise the existence of a
root ξ in [a, b] if f (a)f (b) < 0.
Let k ≥ 0, and that f (ak ) and f (bk ) have opposite signs;
we then conclude from the IVT that the interval (ak , bk )
contains a solution of f (x) = 0.
ISA MT171 – Bisection Method 2/7
The Bisection Method
y

f (b)

y  f (x)

f (p1) p3

a  a1 p2 pp b  b1 x
1
f (p2)
f (a)

a1 p1 b1

a2 p2 b2

a3 p
ISA 3 bMT171
3 – Bisection Method 3/7
The Bisection Method
Consider the midpoint pk of the interval (ak , bk ) defined
by
1
pk = (ak + bk )
2
If f (pk ) = 0, then we have located a solution ξ of f (x) =
0, and the iteration stops. Else, we define the new inter-
val (ak+1 , bk+1 ) by
(
(ak , pk ) if f (pk )f (bk ) > 0
(ak+1 , bk+1 ) =
(pk , bk ) if f (pk )f (bk ) < 0

and repeat this procedure until you meet the stoping rule.
ISA MT171 – Bisection Method 4/7
The Bisection Method

Example 1
The equation f (x) = x 3 + 4x 2 − 10 = 0 has a root in
[1, 2]. Estimate the root by the bisection method.
From the solution table, we see that after 13 iterations,
c13 = 1.365112 approximates the root ξ. The BM has
significant drawbacks, it is very slow. However, it always
converges to a solution and for that reason is often used
as a starter.

ISA MT171 – Bisection Method 5/7


The Bisection Method
k ak bk pk f (ak ) f (pk ) f (ak )f (p
1 1.000000 2.000000 1.500000 < 0 > 0 <0
2 1.000000 1.500000 1.250000 < 0 < 0 >0
3 1.250000 1.500000 1.375000 < 0 > 0 <0
4 1.250000 1.375000 1.312500 < 0 < 0 >0
5 1.312500 1.375000 1.343750 < 0 < 0 >0
6 1.343750 1.375000 1.359375 < 0 < 0 >0
7 1.359375 1.375000 1.367188 < 0 > 0 >0
8 1.359375 1.367188 1.363281 < 0 > 0 >0
9 1.363281 1.367188 1.365234 > 0 > 0 >0
10 1.363281 1.365234 1.364258 > 0 > 0 >0
11 1.364258 1.365234 1.364746 > 0 > 0 >0
12 1.364746 1.365234 1.364990 > 0 > 0 >0
13 1.364990 1.365234 1.365112
ISA MT171 – Bisection Method 6/7
The Bisection Method Algorithm
inputs; k = 0, Niteration , a, b
ak = a, bk = b.
WHILE k < Niteration DO
compute pk ; pk = 12 (ak + bk ),
compute f (ak )f (pk ),
IF f (ak )f (pk ) < 0, THEN
ak+1 = min{ak , pk } , bk+1 = max{ak , pk },
ELSE
ak+1 = min{bk , pk } , bk+1 = max{bk , pk }.
END IF
k = k + 1.
END WHILE
print pk .
ISA MT171 – Bisection Method 7/7

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