numerical solutions of algebraic and transcendental equationsย ยท intermediate value theorem: if is...
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Chapter 5
Numerical Solutions of Algebraic and Transcendental Equations
5.1 Introduction
An expression of the form ๐ ๐ฅ = ๐0๐ฅ๐+๐1๐ฅ
๐โ1 + โฏ + ๐๐โ1๐ฅ + ๐๐ , ๐0 โ 0 is
called a polynomial of degree โ๐โand the polynomial ๐ ๐ฅ = 0 is called an algebraic
equation of ๐๐กโ degree.If ๐ ๐ฅ contains trigonometric, logarithmic or exponential
functions, then ๐ ๐ฅ = 0 is called a transcendental equation. For example ๐ฅ2 +2 sin ๐ฅ + ๐๐ฅ = 0 is a transcendental equation.
If ๐ ๐ฅ is an algebraic polynomial of degree less than or equal to 4, direct methods for
finding the roots of such equation are available. But if ๐ ๐ฅ is of higher degree or it
involves transcendental functions, direct methods do not exist and we need to apply
numerical methods to find the roots of the equation ๐ ๐ฅ = 0.
Some useful results
If ๐ผ is root of the equation ๐ ๐ฅ = 0, then ๐ ๐ผ = 0 Every equation of ๐๐กโdegree has exactly ๐ roots (real or imaginary)
Intermediate Value Theorem: If ๐ ๐ฅ is a continuous function in a closed
interval ๐, ๐ and ๐ ๐ &๐ ๐ are having opposite signs, then the equation
๐ ๐ฅ = 0 has atleast one real root or odd number of roots between ๐and ๐.
If ๐ ๐ฅ is a continuous function in the closed interval ๐, ๐ and ๐ ๐ & ๐ ๐
are of same signs, then the equation ๐ ๐ฅ = 0 has no root or even number of
roots between ๐and ๐.
5.2Numerical methods to find roots of algebraic and transcendental equations
Most numerical methods use iterative procedures to find an approximate root of an
equation ๐ ๐ฅ = 0. They require an initial guess of the root as starting value and each
subsequent iteration leads closer to the actual root.
Order of convergence: For any iterative numerical method, each successive iteration
gives an approximation that moves progressively closer to actual solution. This is known
as convergence. Any numerical method is said have order of convergence ๐, if ๐ is the
largest positive number such that ๐๐+1 โค ๐ ๐๐ ๐ , where ๐๐ and ๐๐+1 are errors in
๐๐กโand (๐ + 1)๐กโ iterations, ๐ is a finite positive constant.
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5.2.1 Bisection Method (or Bolzano Method) Bisection method is used to find an
approximate root in an interval by repeatedly
bisecting into subintervals. It is a very simple
and robust method but it is also relatively
slow. Because of this it is often used to obtain
a rough approximation to a solution which is
then used as a starting point for more rapidly
converging methods. This method is based on
the intermediate value theorem for continuous
functions.
Algorithm:
Let ๐ ๐ฅ be a continuous function in the interval ๐, ๐ , such that ๐ ๐ and ๐ ๐ are of
opposite signs, i.e. ๐ ๐ . ๐ ๐ < 0.
Step1. Take the initial approximation given by ๐ฅ0 =๐+๐
2, one of the three conditions
arises for finding the 1st approximation ๐ฅ1
i. ๐(๐ฅ0) = 0, we have a root at ๐ฅ0.
ii. If ๐ ๐ . ๐ ๐ฅ0 < 0, the root lies between ๐ and ๐ฅ0 โด ๐ฅ1 =๐+ ๐ฅ0
2 and repeat the
procedure by halving the interval again.
iii. I๐ ๐ ๐ . ๐ ๐ฅ0 < 0, the root lies between ๐ฅ0 and ๐ โด ๐ฅ1 = ๐ฅ0+๐
2 and repeat the
procedure by halving the interval again.
iv. Continue the process until root is found to be of desired accuracy.
Remarks:
Convergence is not unidirectional as none of the end points is fixed. As a result
convergence of Bisection method is very slow.
Repeating the procedure ๐ times, the new interval will be exactly half the length of
the previous one, until the root is found of desired accuracy (error less than โ). โด
and at the end of๐๐กโ iteration, the interval containing the root will be of length ๐โ๐
2๐, such that
๐โ๐
2๐<โ
โ log ๐โ๐
2๐< log โ
โ log ๐ โ ๐ โ log 2๐ < log โ
โ log ๐ โ ๐ โ log โ < ๐ log 2
โ ๐ >log ๐โ๐ โlog โ
log 2
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โดIn bisection method, the minimum number of iterations required to achieve the
desired accuracy (error less than โ)are log
๐โ๐
โ
log 2.
Example1Apply bisection method to find a root of the equation ๐ฅ4 + 2๐ฅ3 โ ๐ฅ โ 1 = 0
Solution:๐ ๐ฅ = ๐ฅ4 + 2๐ฅ3 โ ๐ฅ โ 1
Here ๐ 0 = โ1and ๐ 1 = 1 โ ๐ 0 . ๐ 1 < 0
Also ๐ ๐ฅ is continuous in 0,1 , โด atleast one root exists in 0,1
Initial approximation: ๐ = 0, ๐ = 1
๐ฅ0 =0+1
2= .5, ๐ 0.5 = โ1.1875, ๐ 0.5 . ๐ 1 < 0
First approximation: ๐ = 0.5, ๐ = 1
๐ฅ1 =0.5+1
2= 0.75, ๐ 0.75 = โ0.5898, ๐ 0.75 . ๐ 1 < 0
Second approximation: ๐ = 0.75, ๐ = 1
๐ฅ2 =0.75+1
2= 0.875, ๐ 0.875 = 0.051, ๐ 0.75 . ๐ 0.875 < 0
Third approximation: ๐ = 0.75, ๐ = 0.875
๐ฅ3 =0.75+0.875
2= 0.8125, ๐ 0.8125 = โ0.30394, ๐ 0.8125 . ๐ 0.875 < 0
Fourth approximation: ๐ = 0.8125, ๐ = 0.875
๐ฅ4 =0.8125+0.875
2= 0.84375,๐ 0.84375 = โ0.135, ๐ 0.84375 . ๐ 0.875 < 0
Fifth approximation: ๐ = 0.84375, ๐ = 0.875
๐ฅ5 =0.84375+0.875
2= 0.8594, ๐ 0.8594 = โ0.0445, . ๐ 0.8594 . ๐ 0.875 < 0
Sixth approximation: ๐ = 0.8594, ๐ = 0.875
๐ฅ6 =0.8594 + 0.875
2= 0.8672, ๐ 0.8672 = 0.0027, ๐ 0.8594 . ๐ 0.8672 . < 0
Seventh approximation: ๐ = 0.8594, ๐ = 0.8672
๐ฅ7 =0.8594+0.8672
2= 0.8633
First 2 decimal places have been stabilized; hence 0.8633 is the real root correct to two
decimal places.
Example2 Apply bisection method to find a root of the equation ๐ฅ3 โ 2๐ฅ2 โ 4 = 0
correct to three decimal places.
Solution:๐ ๐ฅ = ๐ฅ3 โ 2๐ฅ2 โ 4
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Here ๐ 2 = โ4 and ๐ 3 = 5 โ ๐ 2 . ๐ 3 < 0
Also ๐ ๐ฅ is continuous in 2,3 , โด atleast one root exists in 2,3
Initial approximation: ๐ = 2, ๐ = 3
๐ฅ0 =2+3
2= 2.5, ๐ 2.5 = โ1.8750, ๐ 2.5 . ๐ 3 < 0
First approximation: ๐ = 2.5, ๐ = 3
๐ฅ1 =2.5+3
2= 2.75, ๐ 2.75 = 1.6719, ๐ 2.5 . ๐ 2.75 < 0
Second approximation: ๐ = 2.5, ๐ = 2.75
๐ฅ2 =2.5+2.75
2= 2.625, ๐ 2.625 = 0.3066, ๐ 2.5 . ๐ 2.625 < 0
Third approximation: ๐ = 2.5, ๐ = 2.625
๐ฅ3 =2.5+2.625
2= 2.5625, ๐ 2.5625 = โ.3640, ๐ 2.5625 . ๐ 2.625 < 0
Fourth approximation: ๐ = 2.5625, ๐ = 2.625
๐ฅ4 =2.5625+2.625
2= 2.59375, ๐ 2.59375 = โ.0055, ๐ 2.59375 . ๐ 2.625 < 0
Fifth approximation: ๐ = 2.59375, ๐ = 2.625
๐ฅ5 =2.59375+2.625
2= 2.60938, ๐ 2.60938 = .1488, ๐ 2.59375 . ๐ 2.60938 < 0
Sixth approximation: ๐ = 2.59375, ๐ = 2.60938
๐ฅ6 =2.59375+2.60938
2= 2.60157,๐ 2.60157 = .0719, ๐ 2.59375 . ๐ 2.60157 < 0
Seventh approximation: ๐ = 2.59375, ๐ = 2.60157
๐ฅ7 =2.59375+2.60157
2= 2.59765, ๐ 2.59765 = .0329, ๐ 2.59375 . ๐ 2.59765 < 0
Eighth approximation: ๐ = 2.59375, ๐ = 2.59765
๐ฅ8 =2.59375+2.59765
2= 2.5957, ๐ 2.5957 = .0136, ๐ 2.59375 . ๐ 2.5957 < 0
Ninth approximation: ๐ = 2.59375, ๐ = 2.5957
๐ฅ9 =2.59375+2.5957
2= 2.5947, ๐ 2.5947 = โ.004, ๐ 2.5947 . ๐ 2.5957 < 0
Tenth approximation: ๐ = 2.5947, ๐ = 2.5957
๐ฅ10 =2.5947+2.5957
2= 2.5952
Hence 2.5952 is the real root correct to three decimal places.
Example3 Apply bisection method to find a root of the equation ๐ฅ๐๐ฅ = 1 correct to
three decimal places.
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Solution:๐ ๐ฅ = ๐ฅ๐๐ฅ โ 1
Here ๐ 0 = โ1 and ๐ 1 = ๐ โ 1 = 1.718 โ ๐ 0 . ๐ 1 < 0
Also ๐ ๐ฅ is continuous in 0,1 , โด atleast one root exists in 0,1
Initial approximation: ๐ = 0, ๐ = 1
๐ฅ0 =0+1
2= 0.5, ๐ 0.5 = โ0.1756, ๐ 0.5 . ๐ 1 < 0
First approximation: ๐ = 0.5, ๐ = 1
๐ฅ1 =0.5+1
2= 0.75, ๐ 0.75 = 0.5877, ๐ 0.5 . ๐ 0.75 < 0
Second approximation: ๐ = 0.5, ๐ = 0.625
๐ฅ2 =0.5+0.75
2= 0.625, ๐ 0.625 = 0.8682, ๐ 0.5 . ๐ 0.625 < 0
Third approximation: ๐ = 0.5, ๐ = 0.625
๐ฅ3 =0.5+0.625
2= 0.5625, ๐ 0.5625 = โ0.0128, ๐ 0.5625 . ๐ 0.625 < 0
Fourth approximation: ๐ = 0.5625, ๐ = 0.625
๐ฅ4 =0.5625+0.625
2= 0.59375, ๐ 0.59375 = 0.0751, ๐ 0.5625 . ๐ 0.59375 < 0
Fifth approximation: ๐ = 0.5625, ๐ = 0.59375
๐ฅ5 =0.5625+0.59375
2= 0.5781, ๐ 0.5781 = 0.0305, ๐ 0.5625 . ๐ 0.5781 < 0
Sixth approximation: ๐ = 0.5625, ๐ = 0.5781
๐ฅ6 =0.5625+0.5781
2= 0.5703,๐ 0.5703 = .0087, ๐ 0.5625 . ๐ 0.5703 < 0
Seventh approximation: ๐ = 0.5625, ๐ = 0.5703
๐ฅ7 =0.5625+0.5703
2= 0.5664, ๐ 0.5664 = โ.002, ๐ 0.5664 . ๐ 0.5703 < 0
Eighth approximation: ๐ = 0.5664, ๐ = 0.5703
๐ฅ8 =0.5664+0.5703
2= 0.5684, ๐ 0.5684 = 0.0035, ๐ 0.5664 . ๐ 0.5684 < 0
Ninth approximation: ๐ = 0.5664, ๐ = 0.5684
๐ฅ9 =0.5664+0.5684
2= 0.5674, ๐ 0.5674 = .0007, ๐ 0.5664 . ๐ 0.5674 < 0
Tenth approximation: ๐ = 0.5664, ๐ = 0.5674
๐ฅ10 =0.5664+0.5674
2= 0.5669, ๐ 0.5669 = โ.0007, ๐ 0.5669 . ๐ 0.5674 < 0
Eleventh approximation: ๐ = 0.5669, ๐ = 0.5674
๐ฅ11 =0.5669+0.5674
2= 0.56715, ๐ 0.56715 = .00001~0
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Hence 0.56715 is the real root correct to three decimal places.
Example4 Using bisection method find an approximate root of the equation sin ๐ฅ =1
๐ฅ
correct to two decimal places.
Solution:๐ ๐ฅ = ๐ฅsin ๐ฅ โ 1
Here ๐ 1 = sin 1 โ 1 = โ0.1585 and ๐ 2 = 2sin 2 โ 1 = 0.8186
Also ๐ ๐ฅ is continuous in 1,2 , โด atleast one root exists in 1,2
Initial approximation: ๐ = 1, ๐ = 2
๐ฅ0 =1+2
2= 1.5, ๐ 1.5 = 0.4963, ๐ 1 . ๐ 1.5 < 0
First approximation: ๐ = 1, ๐ = 1.5
๐ฅ1 =1+1.5
2= 1.25, ๐ 1.25 = 0.1862, ๐ 1 . ๐ 1.25 < 0
Second approximation: ๐ = 1, ๐ = 1.25
๐ฅ2 =1+1.25
2= 1.125, ๐ 1.125 = 0.0151, ๐ 1 . ๐ 1.125 < 0
Third approximation: ๐ = 1, ๐ = 1.125
๐ฅ3 =1+1.125
2= 1.0625, ๐ 1.0625 = โ0.0718, ๐ 1.0625 . ๐ 1.125 < 0
Fourth approximation: ๐ = 1.0625, ๐ = 1.125
๐ฅ4 =1.0625+1.125
2= 1.09375, ๐ 1.09375 = โ0.0284, ๐ 1.09375 . ๐ 1.125 < 0
Fifth approximation: ๐ = 1.09375, ๐ = 1.125
๐ฅ5 =1.09375+1.125
2= 1.10937, ๐ 1.10937 = โ0.0066, ๐ 1.10937 . ๐ 1.125 < 0
Sixth approximation: ๐ = 1.10937, ๐ = 1.125
๐ฅ6 =1.10937+1.125
2= 1.11719,๐ 1.11719 = .0042, ๐ 1.10937 . ๐ 1.11719 < 0
Seventh approximation: ๐ = 1.10937, ๐ = 1.11719
๐ฅ7 =1.10937+1.11719
2= 1.11328, ๐ 1.11328 = โ.0012~0
Hence 1.11328 is the real root correct to two decimal places.
5.2.2 Regula- Falsi Method (Geometrical Interpretation) Regula-Falsi method is also known as method of false position as false position of curve
is taken as initial approximation. Let ๐ฆ = ๐(๐ฅ) be represented by the curve ๐ด๐ต.The real
root of equation ๐ ๐ฅ = 0 is ๐ผ as shown in adjoining figure. The false position of curve
๐ด๐ต is taken as chord ๐ด๐ต and initial approximation ๐ฅ0 is the point of intersection of chord
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๐ด๐ต with ๐ฅ-axis. Successive approximations ๐ฅ1, ๐ฅ2 , โฆare given by point of intersection of
chord ๐ดโฒ๐ต, ๐ดโฒโฒ ๐ต, โฆwith ๐ฅ โ axis, until the root is found to be of desired accuracy.
Now equation of chord AB in two-point form is given by:
๐ฆ โ ๐ ๐ =๐ ๐ โ๐ ๐
๐โ๐(๐ฅ โ ๐)
To find ๐ฅ0 (point of intersection of chord AB with ๐ฅ -
axis), put ๐ฆ = 0
โ โ๐ ๐ =๐ ๐ โ๐ ๐
๐โ๐ ๐ฅ0 โ ๐
โ (๐ฅ0 โ ๐) = โ ๐โ๐ ๐ ๐
๐(๐)โ๐(๐)
โ ๐ฅ0 = ๐ โ ๐โ๐
๐ ๐ โ๐ ๐ ๐(๐)
Repeat the procedure until the root is found to the desired
accuracy.
Remarks:
Rate of convergence is much faster than that of bisection method.
Unlike bisection method, one end point will converge to the actual root ๐ , whereas the other end point always remains fixed. As a result Regula- Falsi
method has linear convergence.
Example5 Apply Regula-Falsi method to find a root of the equation ๐ฅ3 + ๐ฅ โ 1 =0 correct to two decimal places.
Solution:๐ ๐ฅ = ๐ฅ3 + ๐ฅ โ 1
Here ๐ 0 = โ1 and ๐ 1 = 1 โ ๐ 0 . ๐ 1 < 0
Also ๐ ๐ฅ is continuous in 0,1 , โด atleast one root exists in 0,1
Initial approximation:๐ฅ0 = ๐ โ ๐โ๐
๐ ๐ โ๐ ๐ ๐(๐) ; ๐ = 0, ๐ = 1
โ ๐ฅ0 = 0 โ 1โ0
๐ 1 โ๐ 0 ๐ 0 = 0 โ
1
1โ โ1 (โ1) = 0.5
๐ 0.5 = โ0.375, ๐ 0.5 . ๐ 1 < 0
First approximation: ๐ = 0.5, ๐ = 1
๐ฅ1 = 0.5 โ 1โ0.5
๐ 1 โ๐ 0.5 ๐ 0.5 = 0 โ
0.5
1โ โ0.375 (โ0.375) = 0.636
๐ 0.636 = โ0.107, ๐ 0.636 . ๐ 1 < 0
Second approximation: ๐ = 0.636, ๐ = 1
๐ฅ2 = 0.636 โ 1โ0.636
๐ 1 โ๐ 0.636 ๐ 0.636 = 0.636 โ
0.364
1โ โ0.107 โ0.107 = 0.6711
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๐ 0.6711 = โ0.0267, ๐ 0.6711 . ๐ 1 < 0
Third approximation: ๐ = 0.6711, ๐ = 1
๐ฅ3 = .6711 โ 1โ0.6711
๐ 1 โ๐ 0.6711 ๐ . 6711 = .6711 โ
0.3289
1โ โ0.0267 โ.0267 = 0.6796
First 2 decimal places have been stabilized; hence 0.6796 is the real root correct to
two decimal places.
Example6 Use Regula-Falsi method to find a root of the equation ๐ฅ log10 ๐ฅ โ 1.2 = 0 correct to two decimal places.
Solution:๐ ๐ฅ = ๐ฅ log10 ๐ฅ โ 1.2
Here ๐ 2 = โ0.5979 and ๐ 3 = 0.2314 โ ๐ 2 . ๐ 3 < 0
Also ๐ ๐ฅ is continuous in 2,3 , โด atleast one root exists in 2,3
Initial approximation: ๐ฅ0 = ๐ โ ๐โ๐
๐ ๐ โ๐ ๐ ๐(๐) ; ๐ = 2, ๐ = 3
โ ๐ฅ0 = 2 โ 3โ2
๐ 3 โ๐ 2 ๐ 2 = 2 โ
1
0.2314โ โ0.5979 (โ0.5979) = 2.721
๐ 2.721 = โ0.0171, ๐ 2.721 . ๐ 3 < 0
First approximation: ๐ = 2.721, ๐ = 3
๐ฅ1 = 2.721 โ 3โ2.721
๐ 3 โ๐ 2.721 ๐ 2.721 = 2.721 โ
0.279
.2314โ โ0.0171 โ0.0171 = 2.7402
๐ 2.7402 = โ0.0004, ๐ 2.7402 . ๐ 3 < 0
Second approximation: ๐ = 2.7402, ๐ = 3
๐ฅ2 = 2.7402 โ 3โ2.7402
๐ 3 โ๐ 2.7402 ๐ 2.7402 = 2.7402 โ
0.2598
.2314โ โ.0004 โ.0004 = 2.7407
First two decimal places have been stabilized; hence 2.7407 is the real root correct to
two decimal places.
Example7 Use Regula-Falsi method to find a root of the equation tan ๐ฅ + tanh ๐ฅ =0 upto three iterations only.
Solution:๐ ๐ฅ = tan ๐ฅ + tanh ๐ฅ
Here ๐ 2 = โ1.2210 and ๐ 3 = 0.8525 โ ๐ 2 . ๐ 3 < 0
Also ๐ ๐ฅ is continuous in 2,3 , โด atleast one root exists in 2,3
Initial approximation:๐ฅ0 = ๐ โ ๐โ๐
๐ ๐ โ๐ ๐ ๐(๐) ; ๐ = 2, ๐ = 3
โ ๐ฅ0 = 2 โ 3โ2
๐ 3 โ๐ 2 ๐ 2 = 2 โ
1
0.8525โ โ1.221 (โ1.221) = 2.5889
๐ 2.5889 = 0.3720, ๐ 2 . ๐ 2.5889 < 0
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First approximation: ๐ = 2, ๐ = 2.5889
๐ฅ1 = 2 โ 2.5889โ2
๐ 2.5889 โ๐ 2 ๐ 2 = 2 โ
0.5889
0.3720โ โ1.2210 (โ1.2210) = 2.4514
๐ 2.4514 = 0.1596, ๐ 2 . ๐ 2.4514 < 0
Second approximation: ๐ = 2, ๐ = 2.4514
๐ฅ2 = 2 โ 2.4514โ2
๐ 2.4514 โ๐ 2 ๐ 2 = 2 โ
0.4514
0.1596โ โ1.2210 (โ1.2210) = 2.3992
๐ 2.3992 = 0.0662, ๐ 2 . ๐ 2.3992 < 0
Third approximation: ๐ = 2, ๐ = 2.3992
๐ฅ2 = 2 โ 2.3992โ2
๐ 2.3992 โ๐ 2 ๐ 2 = 2 โ
0.3992
0.0662โ โ1.2210 (โ1.2210) = 2.3787
โดReal root of the equationtan ๐ฅ + tanh ๐ฅ = 0 after three iterations is 2.3787
Example8 Use Regula-Falsi method to find a root of the equation ๐ฅ๐๐ฅ โ 2 = 0 correct to
three decimal places.
Solution: ๐ ๐ฅ = ๐ฅ๐๐ฅ โ 2
Here ๐ 0 = โ2 and ๐ 1 = 0.7183 โ ๐ 0 . ๐ 1 < 0
Also ๐ ๐ฅ is continuous in 0,1 , โด atleast one root exists in 0,1
Initial approximation: ๐ฅ0 = ๐ โ ๐โ๐
๐ ๐ โ๐ ๐ ๐(๐) ; ๐ = 0, ๐ = 1
โ ๐ฅ0 = 0 โ 1โ0
๐ 1 โ๐ 0 ๐ 0 = 0 โ
1
0.7183โ โ2 (โ2) = 0.7358
๐ 0.7358 = โ0.4643, ๐ 0.7358 . ๐ 1 < 0
First approximation: ๐ = 0.7358, ๐ = 1
๐ฅ1 = 0.7358 โ 1โ0.7358
๐ 1 โ๐ 0.7358 ๐ 0.7358 = 0.7358 โ
0.2642
0.7183โ โ0.4643 (โ0.4643) =
0.8395
๐ 0.8395 = โ0.0564, ๐ 0.8395 . ๐ 1 < 0
Second approximation: ๐ = 0.8395, ๐ = 1
๐ฅ2 = 0.8395 โ 1โ0.8395
๐ 1 โ๐ 0.8395 ๐ 0.8395 = 0.8395 โ
0.1605
0.7183โ โ0.0564 (โ0.0564) =
0.8512
๐ 0.8512 = โ0.006, ๐ 0.8512 . ๐ 1 < 0
Third approximation: ๐ = 0.8512, ๐ = 1
๐ฅ2 = 0.8512 โ 1โ0.8512
๐ 1 โ๐ 0.8512 ๐ 0.8512 = 0.8512 โ
0.1488
0.7183โ โ0.006 (โ0.006) = 0.8524
๐ 0.8524 = โ0.009, ๐ 0.8524 . ๐ 1 < 0
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Fourth approximation: ๐ = 0.8474 ๐ = 1
๐ฅ4 = 0.8524 โ 1โ0.8524
๐ 1 โ๐ 0.8524 ๐ 0.8524 = 0.8524 โ
0.1476
0.7183โ โ0.0009 (โ0.0009) =
0.8526
๐ 0.8526 = โ0.00002~0,
โด Real root of the equation ๐ฅ๐๐ฅ โ 2 = 0 correct to three decimal places is 0.8526
5.2.3 Newton-Raphson Method (Geometrical Interpretation) Newton-Raphson method named after Isaac Newton and Joseph Raphson is a powerful
technique for solving equations numerically. The Newton-Raphson method in one
variable is implemented as follows:
Let ๐ผ be an exact root and ๐ฅ0 be the initial approximate
root of the equation ๐ ๐ฅ = 0. First approximation ๐ฅ1 is
taken by drawing a tangent to curve ๐ฆ = ๐ ๐ฅ at the
point ๐ฅ0, ๐ ๐ฅ0 . If ๐ is the angle which tangent
through the point ๐ฅ0, ๐ ๐ฅ0 makes with ๐ฅ- axis, then
slope of the tangent is given by:
tan ๐ = ๐ ๐ฅ0
๐ฅ0โ๐ฅ1= ๐โฒ ๐ฅ0
โ ๐ฅ1 = ๐ฅ0 โ๐ ๐ฅ0
๐โฒ ๐ฅ0
Similarly ๐ฅ2 = ๐ฅ1 โ๐ ๐ฅ1
๐โฒ ๐ฅ1
โฎ The required root to desired accuracy is obtained by drawing tangents to the curve
at points ๐ฅ๐ , ๐ ๐ฅ๐ successively.
โด ๐ฅ๐+1 = ๐ฅ๐ โ๐ ๐ฅ๐
๐โฒ ๐ฅ๐
Newton-Raphson method works very fast but sometimes it fails to converge as shown
below:
Case I:
If any of the approximations encounters a zero
derivative (extreme point), then the tangent at that
point goes parallel to ๐ฅ-axis, resulting in no further
approximations as shown in given figure where third
approximation tends to infinity.
Case II:
Sometimes Newton-Raphson method may run into an
infinite cycle or loop as shown in adjoining figure.
Change in initial approximation may untangle the
problem.
๐ฝ
๐๐,๐ ๐๐
Page | 11
Case III:
In case of a point of discontinuity, as shown in given
figure, subsequent roots may diverge instead of
converging.
Remarks:
Newton-Raphson method can be used for solving both algebraic and
transcendental equations and it can also be used when roots are complex.
Initial approximation ๐ฅ0 can be taken randomly in the interval ๐, ๐ , such
that ๐ ๐ . ๐ ๐ < 0
Newton-Raphson method has quadratic convergence, but in case of bad choice
of ๐ฅ0 (the initial guess), Newton- Raphson method may fail to converge
This method is useful in case of large value of ๐โฒ ๐ฅ๐ i.e. when graph of ๐ ๐ฅ
while crossing ๐ฅ -axis is nearly vertical
Example 9 Use Newton-Raphson method to find a root of the equation ๐ฅ3 โ 5๐ฅ + 3 = 0 correct to three decimal places.
Solution: ๐ ๐ฅ = ๐ฅ3 โ 5๐ฅ + 3
โ ๐โฒ ๐ฅ = 3๐ฅ2 โ 5
Here ๐ 0 = 3 and ๐ 1 = โ1 โ ๐ 0 . ๐ 1 < 0
Also ๐ ๐ฅ is continuous in 0,1 , โด atleast one root exists in 0,1
Initial approximation: Let initial approximation ๐ฅ0 in the interval 0,1 be 0.8
By Newton-Raphson method ๐ฅ๐+1 = ๐ฅ๐ โ๐ ๐ฅ๐
๐โฒ ๐ฅ๐
First approximation:
๐ฅ1 = ๐ฅ0 โ๐ ๐ฅ0
๐โฒ ๐ฅ0 , where ๐ฅ0 = 0.8, ๐ 0.8 = โ0.488, ๐โฒ 0.8 = โ3.08
โ ๐ฅ1 = 0.8 โโ0.488
โ3.08= 0.6416
Second approximation:
๐ฅ2 = ๐ฅ1 โ๐ ๐ฅ1
๐โฒ ๐ฅ1 , where ๐ฅ1 = 0.6415 , ๐ 0.6416 = 0.0561 , ๐โฒ 0.6416 = โ3.7650
โ ๐ฅ2 = 0.6416 โ0.05611
โ3.7650= 0.6565
Third approximation:
๐ฅ3 = ๐ฅ2 โ๐ ๐ฅ2
๐โฒ ๐ฅ2 , where ๐ฅ2 = 0.6565 , ๐ 0.6565 = 0.0004 , ๐โฒ 0.6565 = โ3.7070
โ ๐ฅ3 = 0.6565 โ0.0004
โ3.7070= 0.6566
Hence a root of the equation ๐ฅ3 โ 5๐ฅ + 3 = 0 correct to three decimal places is 0.6566
Page | 12
Example 10 Find the approximate value of 28 correct to 3 decimal places using
Newton Raphson method.
Solution: ๐ฅ = 28 โ ๐ฅ2 โ 28 = 0
โด ๐ ๐ฅ = ๐ฅ2 โ 28
โ ๐โฒ ๐ฅ = 2๐ฅ
Here ๐ 5 = โ3 and ๐ 6 = 8 โ ๐ 5 . ๐ 6 < 0
Also ๐ ๐ฅ is continuous in 5,6 , โด atleast one root exists in 5,6
Initial approximation: Let initial approximation ๐ฅ0 in the interval 5,6 be 5.5
By Newton-Raphson method ๐ฅ๐+1 = ๐ฅ๐ โ๐ ๐ฅ๐
๐โฒ ๐ฅ๐
First approximation:
๐ฅ1 = ๐ฅ0 โ๐ ๐ฅ0
๐โฒ ๐ฅ0 , where ๐ฅ0 = 5.5, ๐ 5.5 = 2.25, ๐โฒ 5.5 = 11
โ ๐ฅ1 = 5.5 โ๐.๐๐
11= 5.2955
Second approximation:
๐ฅ2 = ๐ฅ1 โ๐ ๐ฅ1
๐โฒ ๐ฅ1 , where ๐ฅ1 = 5.2955, ๐ 5.2955 = 0.0423, ๐โฒ 5.2955 = 10.591
โ ๐ฅ2 = 5.2955 โ0.0423
10.591= 5.2915
Third approximation:
๐ฅ3 = ๐ฅ2 โ๐ ๐ฅ2
๐โฒ ๐ฅ2 , where ๐ฅ2 = 5.2915, ๐ 5.2915 = โ0.00003, ๐โฒ 5.2915 = 10.583
โ ๐ฅ3 = 5.2915 โโ0.00003
10.583= 5.2915
Hence value of 28 correct to three decimal places is 5.2915
Example 11 Use Newton-Raphson method to find a root of the equation ๐ฅ sin ๐ฅ +cos ๐ฅ = 0 correct to three decimal places.
Solution: ๐ ๐ฅ = ๐ฅ sin ๐ฅ + cos ๐ฅ
โ ๐โฒ ๐ฅ = ๐ฅ cos ๐ฅ + sin ๐ฅ โ sin ๐ฅ = ๐ฅ cos ๐ฅ
Here ๐ ๐
2 = 1.5708 and ๐ ๐ = โ1 โ ๐
๐
2 . ๐ ๐ < 0
Also ๐ ๐ฅ is continuous in ๐
2, ๐ โด atleast one root exists in
๐
2, ๐
Initial approximation: Let initial approximation ๐ฅ0 in the interval ๐
2, ๐ be ๐
Page | 13
By Newton-Raphson method ๐ฅ๐+1 = ๐ฅ๐ โ๐ ๐ฅ๐
๐โฒ ๐ฅ๐
First approximation:
๐ฅ1 = ๐ฅ0 โ๐ ๐ฅ0
๐โฒ ๐ฅ0 , where ๐ฅ0 = ๐, ๐ ๐ = โ1, ๐โฒ ๐ = โ3.1416
โ ๐ฅ1 = 3.1416 โโ1
โ3.1416= 2.8233
Second approximation:
๐ฅ2 = ๐ฅ1 โ๐ ๐ฅ1
๐โฒ ๐ฅ1 , where ๐ฅ1 = 2.8233, ๐ 2.8233 = โ0.0662, ๐โฒ 2.8233 = โ2.6815
โ ๐ฅ2 = 2.8233 โโ0.0662
โ2.6815= 2.7986
Third approximation:
๐ฅ3 = ๐ฅ2 โ๐ ๐ฅ2
๐โฒ ๐ฅ2 , where ๐ฅ2 = 2.798, ๐ 2.7986 = โ0.0006, ๐โฒ 2.7986 = โ2.6356
โ ๐ฅ3 = 2.7986 โโ0.0006
โ2.6356= 2.7984
Hence a root of the equation ๐ฅ sin ๐ฅ + cos ๐ฅ = 0 correct to three decimal places is
2.7984
Example 12 Use Newton Raphson method to derive a formula to find ๐5
, ๐๐๐ .
Hence evaluate 435
correct to 3 decimal places.
Solution: ๐ฅ = ๐5
โ ๐ฅ5 โ ๐ = 0
๐ ๐ฅ = ๐ฅ5 โ ๐
โ ๐โฒ ๐ฅ = 5๐ฅ4
By Newton-Raphson method ๐ฅ๐+1 = ๐ฅ๐ โ๐ ๐ฅ๐
๐โฒ ๐ฅ๐
โ ๐ฅ๐+1 = ๐ฅ๐ โ๐ฅ๐
5โ๐
5๐ฅ๐4 =
4
5๐ฅ๐ +
๐
5๐ฅ๐4
To evaluate 435
, putting ๐ = 43 , โดNewton-Raphson formula is given by
๐ฅ๐+1 =4
5๐ฅ๐ +
43
5๐ฅ๐4
Let initial approximation ๐ฅ0 be 2
First approximation:
๐ฅ1 =4
5๐ฅ0 +
43
5๐ฅ04 , where ๐ฅ0 = 2
โ ๐ฅ1 =8
5+
43
80= 2.1375
Page | 14
Second approximation:
๐ฅ2 =4
5๐ฅ1 +
43
5๐ฅ14 , where ๐ฅ1 = 2.1375
โ ๐ฅ2 =4(2.1375)
5+
43
5(2.1375)4= 2.1220
Third approximation:
๐ฅ3 =4
5๐ฅ2 +
43
5๐ฅ24 , where ๐ฅ2 = 2.1220
โ ๐ฅ3 =4(2.1220)
5+
43
5(2.1220)4= 2.1217
Fourth approximation:
๐ฅ4 =4
5๐ฅ3 +
43
5๐ฅ34 , where ๐ฅ3 = 2.1217
โ ๐ฅ4 =4(2.1217)
5+
43
5(2.1217)4= 2.1217
Hence value of 435
correct to four decimal places is 2.1217
5.2.3.1 Generalized Newtonโs Method for Multiple Roots
Result: If ๐ผ is a root of equation ๐ ๐ฅ = 0 with multiplicity ๐, then it is also a root of
equation ๐โฒ ๐ฅ = 0 with multiplicity (๐ โ 1) and also of the equation ๐โฒโฒ ๐ฅ = 0 with
multiplicity (๐ โ 1) and so on.
For example (๐ฅ โ 1)3 = 0 has โ1โ as a root with multiplicity 3
3(๐ฅ โ 1)2 = 0 has โ1โ as the root with multiplicity 2
6(๐ฅ โ 1) = 0 has โ1โ as the root with multiplicity 1
โด The expressions ๐ฅ๐ โ ๐๐ ๐ฅ๐
๐โฒ ๐ฅ๐ , ๐ฅ๐ โ (๐ โ 1)
๐โฒ ๐ฅ๐
๐โฒโฒ ๐ฅ๐ , ๐ฅ๐ โ (๐ โ 2)
๐โฒโฒ ๐ฅ๐
๐โฒโฒโฒ ๐ฅ๐ are
equivalent
Generalized Newtonโs method is used to find repeated roots of an equation as is given as:
If ๐ผ be a root of equation ๐ ๐ฅ = 0 which is repeated ๐ times,
Then ๐ฅ๐+1 = ๐ฅ๐ โ ๐๐ ๐ฅ๐
๐โฒ ๐ฅ๐ ~ ๐ฅ๐ โ (๐ โ 1)
๐โฒ ๐ฅ๐
๐โฒโฒ ๐ฅ๐
Example 13 Use Newton-Raphson method to find a double root of the equation
๐ฅ3 โ ๐ฅ2 โ ๐ฅ + 1 = 0 upto three iterations.
Solution: ๐ ๐ฅ = ๐ฅ3 โ ๐ฅ2 โ ๐ฅ + 1
๐โฒ ๐ฅ = 3๐ฅ2 โ 2๐ฅ โ 1
๐ โฒโฒ (๐ฅ) = 6๐ฅ โ 2
Page | 15
Let the initial approximation ๐ฅ0 = 0.7
First approximation:
๐ฅ1 = ๐ฅ0 โ2๐ ๐ฅ0
๐โฒ ๐ฅ0 Also ๐ฅ1 = ๐ฅ0 โ
๐โฒ ๐ฅ0
๐โฒโฒ ๐ฅ0
โ ๐ฅ1 = 0.7 โ0.306
โ0.93= 1.0290 And ๐ฅ1 = 0.7 โ
โ0.93
2.2= 1.1227
โด ๐ฅ1 =1.029+1.1227
2= 1.0759, ๐(๐ฅ1) = .012
Second approximation:
๐ฅ2 = ๐ฅ1 โ2๐ ๐ฅ1
๐โฒ ๐ฅ1 Also ๐ฅ2 = ๐ฅ1 โ
๐โฒ ๐ฅ1
๐โฒโฒ ๐ฅ1
โ ๐ฅ2 = 1.0759 โ0.0239
0.3209= 1.001 And ๐ฅ2 = 1.0759 โ
0.3209
4.4554= 1.004
โด ๐ฅ2 =1.001+1.004
2= 1.0025 , ๐(๐ฅ2) = .00001
Third approximation:
๐ฅ3 = ๐ฅ2 โ2๐ ๐ฅ2
๐โฒ ๐ฅ2 Also ๐ฅ3 = ๐ฅ2 โ
๐โฒ ๐ฅ2
๐โฒโฒ ๐ฅ2
โ ๐ฅ3 = 1.0025 โ0.00003
0.0100= 0.995 And ๐ฅ3 = 1.0025 โ
0.0100
4.015= 1.0000
โด ๐ฅ3 =0.995+1.000
2= 0.9975 , ๐(๐ฅ3) = .00001
The double root of the equation ๐ฅ3 โ ๐ฅ2 โ ๐ฅ + 1 = 0 after three iterations is
0.9975.
5.2.3.1 Convergence of Newton Raphson Method
Let ๐ผ be an exact root of the equation ๐ ๐ฅ = 0
โ ๐ ๐ผ = 0
Also let ๐ฅ๐ and ๐ฅ๐+1 be two successive approximations to the root ๐ผ.
If โ๐ and โ๐+1 are the corresponding errors in the approximations ๐ฅ๐ and ๐ฅ๐+1
Then ๐ฅ๐ = ๐ผ +โ๐ โฆ โ
and ๐ฅ๐+1 = ๐ผ +โ๐+1 โฆ โก
Now by Newton Raphson method
๐ฅ๐+1 = ๐ฅ๐ โ๐ ๐ฅ๐
๐โฒ ๐ฅ๐ โฆ โข
Using โ and โก in โข
โ ๐ผ +โ๐+1= ๐ผ +โ๐โ๐ ๐ผ+โ๐
๐โฒ ๐ผ+โ๐
Page | 16
โโ๐+1=โ๐๐ โฒ ๐ผ+โ๐ โ๐(๐ผ+โ๐ )
๐ โฒ (๐ผ+โ๐ )
โโ๐+1=โ๐ ๐ โฒ ๐ผ +โ๐๐ โฒโฒ ๐ผ + โฆ โ ๐ ๐ผ +โ๐๐ โฒ ๐ผ +
โ๐2
2!๐ โฒโฒ ๐ผ +โฏ
๐ โฒ ๐ผ +โ๐๐ โฒโฒ ๐ผ + โฆ By Taylorโs expansion
โโ๐+1=โ๐
2๐ โฒโฒ ๐ผ โ โ๐
2
2!๐ โฒโฒ ๐ผ +โฏ
๐ โฒ ๐ผ 1+โ๐๐โฒโฒ ๐ผ
๐โฒ ๐ผ +โฏ
โต ๐ ๐ผ = 0
โโ๐+1= โ๐
2
2
๐ โฒโฒ (๐ผ)
๐ โฒ (๐ผ)+ โฏ 1 +
โ๐๐ โฒโฒ ๐ผ
๐ โฒ ๐ผ + โฏ
โ1
โโ๐+1= โ๐
2
2
๐ โฒโฒ (๐ผ)
๐ โฒ (๐ผ)+ โฏ 1 โ
โ๐๐ โฒโฒ ๐ผ
๐ โฒ ๐ผ + โฏ
โโ๐+1=โ๐
2
2
๐ โฒโฒ (๐ผ)
๐ โฒ (๐ผ) Neglecting higher order terms
โโ๐+1= ๐พ โ๐2 Where ๐ =
1
2
๐ โฒโฒ (๐ผ)
๐ โฒ (๐ผ)
โด Newton Raphson method has convergence of order 2 or quadratic convergence.
5.3 Iterative Methods for Solving Simultaneous Linear Equations
Consider a system of linear equations:
๐1๐ฅ + ๐1๐ฆ + ๐1๐ง = ๐1
๐2๐ฅ + ๐2๐ฆ + ๐2๐ง = ๐2
๐3๐ฅ + ๐3๐ฆ + ๐3๐ง = ๐3
โฆโ
We have been using direct methods for solving a system of linear equations. Direct
methods produce exact solution after a finite number of steps whereas iterative
methods give a sequence of approximate solutions until solution is obtained up to
desired accuracy. Common iterative methods for solving a system of linear
equations are:
1. Gauss-Jacobiโs iteration method
2. Gauss-Seidalโs iteration method
These methods require partial pivoting before application.
Partial pivoting: It is about changing rows of a system of linear equations given by
โ such that ๐1 โฅ ๐2 , ๐3 ; ๐2 โฅ ๐3.
Complete pivoting: It is the process of selecting the largest element in the magnitude
as the pivot element by interchanging row as well as columns of the system. Order
of variables is also changed in the procedure. In particular for the system given by โ ,
complete pivoting would require ๐1 โฅ ๐2, ๐3 ; ๐2 โฅ ๐1, ๐3 , if ๐1 and ๐2 are to be
taken as pivots.
Page | 17
5.3.1 Gauss-Jacobiโs Iteration Method
The concept of the Gauss- Jacobiโs iteration scheme is extremely simple with the
assumptions that the system has unique solution and diagonal elements are non-zeros.
Algorithm: Gauss-Jacobiโs iteration method
1. Take the system of linear equations given by โ after partial pivoting and solve
each equation in the system for the diagonal value of variables such that
๐ฅ =1
๐1(๐1 โ ๐1๐ฆ โ ๐1๐ง)
๐ฆ =1
๐2(๐2 โ ๐2๐ฅ โ ๐2๐ง)
๐ง =1
๐3(๐3 โ ๐3๐ฅ โ ๐3๐ฆ)
โฆโก
2. Rewrite โก in generalized form given by:
๐ฅ๐+1 =1
๐1(๐1 โ ๐1๐ฆ๐ โ ๐1๐ง๐)
๐ฆ๐+1 =1
๐2(๐2 โ ๐2๐ฅ๐ โ ๐2๐ง๐)
๐ง๐+1 =1
๐3(๐3 โ ๐3๐ฅ๐ โ ๐3๐ฆ๐ )
โฆโข
3. Take ๐ฅ0 = ๐ฆ0 = ๐ง0 = 0 as initial approximation (in general if a better
approximation can not be judged) and substitute in the system given by โข
โด ๐ฅ1 =๐1
๐1 , ๐ฆ1 =
๐2
๐2 , ๐ง1 =
๐3
๐3
4. Putting ๐ = 1 , substitute the values of ๐ฅ1 , ๐ฆ1 and ๐ง1 in โข to get next
approximations of ๐ฅ2, ๐ฆ2 and ๐ง2. Continue the procedure until the difference
between two consecutive approximations is negligible.
Example 14 Solve the following system of equations using Gauss Jacobi's method
5๐ฅ โ 2๐ฆ + 3๐ง = โ1
โ3๐ฅ + 9๐ฆ + ๐ง = 2
2๐ฅ โ ๐ฆ โ 7๐ง = 3
Solution: The given system of equations is satisfying rules of partial pivoting.
Rewriting in general form as given in โข
๐ฅ๐+1 =1
5(โ1 + 2๐ฆ๐ โ 3๐ง๐)
๐ฆ๐+1 =1
9 2 + 3๐ฅ๐ โ ๐ง๐
๐ง๐+1 =1
7 โ3 + 2๐ฅ๐ โ ๐ฆ๐
Taking ๐ฅ0 = ๐ฆ0 = ๐ง0 = 0 as initial approximation
First Approximation:
๐ฅ1 = โ1
5= โ0.2 ๐ฆ1 =
2
9= 0.222 , ๐ง1 = โ
3
7= โ0.429
Page | 18
Second Approximation:
๐ฅ2 =1
5 โ1 + 2๐ฆ1 โ 3๐ง1 , ๐ฆ2 =
1
9 2 + 3๐ฅ1 โ ๐ง1 , ๐ง2 =
1
7(โ3 + 2๐ฅ1 โ ๐ฆ1)
โ ๐ฅ2 =1
5 โ1 + 2 0.222 โ 3(โ0.429 ) = 0.146
๐ฆ2 =1
9 2 + 3(โ0.2 + 0.429) = 0.203
๐ง2 =1
7 โ3 + 2 โ0.2 โ 0.222 = โ0.517
Third Approximation:
๐ฅ3 =1
5 โ1 + 2๐ฆ2 โ 3๐ง2 , ๐ฆ3 =
1
9 2 + 3๐ฅ2 โ ๐ง2 , ๐ง3 =
1
7(โ3 + 2๐ฅ2 โ ๐ฆ2)
โ ๐ฅ3 =1
5 โ1 + 2 0.203 โ 3(โ0.517 ) = 0.191
๐ฆ3 =1
9 2 + 3(0.146 + 0.517) = 0.328
๐ง3 =1
7 โ3 + 2 0.146 โ 0.203 = โ0.416
Fourth Approximation:
๐ฅ4 =1
5 โ1 + 2๐ฆ3 โ 3๐ง3 , ๐ฆ4 =
1
9 2 + 3๐ฅ3 โ ๐ง3 , ๐ง4 =
1
7(โ3 + 2๐ฅ3 โ ๐ฆ3)
โ ๐ฅ4 =1
5 โ1 + 2 0.328 โ 3(โ0.416 ) = 0.181
๐ฆ4 =1
9 2 + 3(0.191 + 0.416) = 0.332
๐ง4 =1
7 โ3 + 2 0.191 โ 0.328 = โ0.421
Fifth Approximation:
๐ฅ5 =1
5 โ1 + 2๐ฆ4 โ 3๐ง4 , ๐ฆ5 =
1
9 2 + 3๐ฅ4 โ ๐ง4 , ๐ง5 =
1
7(โ3 + 2๐ฅ4 โ ๐ฆ4)
โ ๐ฅ5 =1
5 โ1 + 2 0.332 โ 3(โ0.421 ) = 0.185
๐ฆ5 =1
9 2 + 3(0.181 + 0.421) = 0.329
๐ง5 =1
7 โ3 + 2 0.181 โ 0.332 = โ0.424
Sixth Approximation:
๐ฅ6 =1
5 โ1 + 2๐ฆ5 โ 3๐ง5 , ๐ฆ6 =
1
9 2 + 3๐ฅ5 โ ๐ง5 , ๐ง6 =
1
7(โ3 + 2๐ฅ5 โ ๐ฆ5)
โ ๐ฅ6 =1
5 โ1 + 2 0.329 โ 3(โ0.424 ) = 0.186
๐ฆ6 =1
9 2 + 3(0.185 + 0.424) = 0.331
๐ง6 =1
7 โ3 + 2 0.185 โ 0.329 = โ0.423
Page | 19
Values of variables have been stabilized, โด approximate solution is given by
๐ฅ = 0.186, ๐ฆ = 0.331and ๐ง = โ0.423
Example 15 Compute 4 iterations to find an approximate solution of the given system of
equations using Gauss Jacobi's method.
๐ฅ + ๐ฆ + 5๐ง = โ1
5๐ฅ โ ๐ฆ + ๐ง = 10
2๐ฅ + 4๐ฆ = 12
Solution: Rearranging the given equations by partial pivoting
5๐ฅ โ ๐ฆ + ๐ง = 10
2๐ฅ + 4๐ฆ = 12
๐ฅ + ๐ฆ + 5๐ง = โ1
Rewriting in general form as given in โข
๐ฅ๐+1 =1
5(10 + ๐ฆ๐ โ ๐ง๐ )
๐ฆ๐+1 =1
4 12 โ 2๐ฅ๐
๐ง๐+1 =1
5(โ1 โ ๐ฅ๐ โ ๐ฆ๐)
Taking ๐ฅ0 = ๐ฆ0 = ๐ง0 = 0 as initial approximation
First Approximation:
๐ฅ1 =10
5= 2 , ๐ฆ1 =
12
4= 3 , ๐ง1 = โ
1
5
Second Approximation:
๐ฅ2 =1
5 10 + ๐ฆ1 โ ๐ง1 , ๐ฆ2 =
1
4 12 โ 2๐ฅ1 , ๐ง2 =
1
5(โ1 โ ๐ฅ1 โ ๐ฆ1)
โ ๐ฅ2 =1
5 10 + 3 +
1
5 , ๐ฆ2 =
1
4 12 โ 4 , ๐ง2 =
1
5(โ1 โ 2 โ 3)
โด ๐ฅ2 = 2.64 , ๐ฆ2 = 2 , ๐ง2 = โ1.2
Third Approximation:
๐ฅ3 =1
5 10 + ๐ฆ2 โ ๐ง2 , ๐ฆ3 =
1
4 12 โ 2๐ฅ2 , ๐ง3 =
1
5(โ1 โ ๐ฅ2 โ ๐ฆ2)
โ ๐ฅ3 =1
5 10 + 2 + 1.2 , ๐ฆ3 =
1
4 12 โ 2(2.64) , ๐ง3 =
1
5(โ1 โ 2.64 โ 2)
โด ๐ฅ3 = 2.64 , ๐ฆ3 = 1.68 , ๐ง3 = โ0.928
Fourth Approximation:
๐ฅ4 =1
5 10 + ๐ฆ3 โ ๐ง3 , ๐ฆ3 =
1
4 12 โ 2๐ฅ3 , ๐ง3 =
1
5(โ1 โ ๐ฅ3 โ ๐ฆ3)
โ ๐ฅ4 =1
5 10 + 1.68 + 0.928 , ๐ฆ4 =
1
4 12 โ 2(2.64) , ๐ง4 =
1
5(โ1 โ 2.64 โ 1.68)
Page | 20
โด ๐ฅ4 = 2.52 , ๐ฆ4 = 1.68 , ๐ง4 = โ1.064
Approximate solution after 4 iterations is given by ๐ฅ = 2.52, ๐ฆ = 1.68, ๐ง = โ1.064
5.3.2Gauss-Seidalโs Iteration Method
Gauss-Seidel method is an improvement of the basic Gauss-Jordan method. Here the
improved values of variables are utilized as soon as they are obtained.
โด System of equations given in โข is improved by taking latest values of the variables
as
๐ฅ๐+1 =1
๐1 ๐1 โ ๐1๐ฆ๐ โ ๐1๐ง๐
๐ฆ๐+1 =1
๐2 ๐2 โ ๐2๐ฅ๐+1 โ ๐2๐ง๐
๐ง๐+1 =1
๐3 ๐3 โ ๐3๐ฅ๐+1 โ ๐3๐ฆ๐+1
Gauss-Seidel scheme usually converges faster than Jacobiโs iteration method.
Example 16 Solve the system of equations given in Example 14 using Gauss Seidal's
method
5๐ฅ โ 2๐ฆ + 3๐ง = โ1
โ3๐ฅ + 9๐ฆ + ๐ง = 2
2๐ฅ โ ๐ฆ โ 7๐ง = 3
Also compare the results obtained in two methods.
Solution: The given system of equations is satisfying rules of partial pivoting.
Using Gauss Seidal's approximations, system can be rewritten as
๐ฅ๐+1 =1
5 โ1 + 2๐ฆ๐ โ 3๐ง๐
๐ฆ๐+1 =1
9 2 + 3๐ฅ๐+1 โ ๐ง๐
๐ง๐+1 =1
7 โ3 + 2๐ฅ๐+1 โ ๐ฆ๐+1
Taking ๐ฅ0 = ๐ฆ0 = ๐ง0 = 0 as initial approximation
First Approximation:
๐ฅ1 = โ1
5= โ0.2 ๐ฆ1 =
2
9= 0.222 , ๐ง1 = โ
3
7= โ0.429
Second Approximation:
๐ฅ2 =1
5 โ1 + 2๐ฆ1 โ 3๐ง1 , ๐ฆ2 =
1
9 2 + 3๐ฅ2 โ ๐ง1 , ๐ง2 =
1
7(โ3 + 2๐ฅ2 โ ๐ฆ2)
โ ๐ฅ2 =1
5 โ1 + 2 0.222 โ 3(โ0.429 ) = 0.146
๐ฆ2 =1
9 2 + 3(0.146 + 0.429) = 0.319
Page | 21
๐ง2 =1
7 โ3 + 2 0.146 โ 0.319 = โ0.432
Third Approximation:
๐ฅ3 =1
5 โ1 + 2๐ฆ2 โ 3๐ง2 , ๐ฆ3 =
1
9 2 + 3๐ฅ3 โ ๐ง2 , ๐ง3 =
1
7(โ3 + 2๐ฅ3 โ ๐ฆ3)
โ ๐ฅ3 =1
5 โ1 + 2 0.319 โ 3(โ0.432 ) = 0.187
๐ฆ3 =1
9 2 + 3(0.187 + 0.432) = 0.333
๐ง3 =1
7 โ3 + 2 0.187 โ 0.333 = โ0.423
Fourth Approximation:
๐ฅ4 =1
5 โ1 + 2๐ฆ3 โ 3๐ง3 , ๐ฆ4 =
1
9 2 + 3๐ฅ4 โ ๐ง3 , ๐ง4 =
1
7(โ3 + 2๐ฅ4 โ ๐ฆ4)
โ ๐ฅ4 =1
5 โ1 + 2 0.333 โ 3(โ0.423 ) = 0.187
๐ฆ4 =1
9 2 + 3(0.187 + 0.423) = 0.332
๐ง4 =1
7 โ3 + 2 0.187 โ 0.332 = โ0.423
Values of variables have been stabilized, โด approximate solution is given by
๐ฅ = 0.187, ๐ฆ = 0.332 and ๐ง = โ0.423
Clearly numbers of iterations for the solution to converge in Gauss Seidalโs
method are much less than Gauss Jacobiโs method.
Exercise 5A
1. Find the real root of the equation ๐ฅ3 โ 2๐ฅ โ 5 = 0 correct to three decimal
places using Bisection method.
2. Perform three iterations to find root of the equation ๐ฅ๐๐ฅ โ cos ๐ฅ = 0 using
Regula-Falsi method.
3. Find the real root of the equation ๐ฅ3 โ 3๐ฅ โ 5 = 0 correct to three decimal
places using Newton- Raphson method.
4. Solve the system of equations:
10๐ฅ1 โ 2๐ฅ2 โ ๐ฅ3 โ ๐ฅ4 = 3
2๐ฅ1 โ 10๐ฅ2 + ๐ฅ3 + ๐ฅ4 = โ15
๐ฅ1 + ๐ฅ2 โ 10๐ฅ3 + 2๐ฅ4 = โ27
๐ฅ1 + ๐ฅ2 + 2๐ฅ3 โ 10๐ฅ4 = 9 using Gauss- Jacobi method. Compute results for 2 iterations.
5. Solve the system of equations given in Q4 upto 2 iterations, using Gauss-
Seidal method.