Numerical Methods and Computational Techniques

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Numerical Methods and Computational Techniques
Solution of Transcendental and Polynomial
Equations
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Dictionary meaning of Transcendent More than
ordinary, Supernatural, Superlative Transcendental
Number A number which is not ordinary. For
example Pi (p) 3.14159 approximately
Or e
Hence f (x) involving ex or ln (x) are
transcendental functions. Further Sin (x) Cos
(x)
Hence f (x) involving sin (x), cos (x) etc are
also transcendental functions.
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Root of transcendental or polynomial function is
the value of x for which f (x) 0 In graphical
form it is point of intersection of graph of f
(x) with X axis.
f (x)
roots
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• Numerical Methods to determine the roots
• Birge Vieta Method
• Bairstow Method
• Bisection Method
• False Position Method
• Simple Fixed Point Iteration Method
• Newton Raphson (Tangent) Method
• Secant Method

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• Birge Vieta Method
• Used for finding roots of polynomial functions.
• Uses synthetic division of polynomial to
  extract factor of the given polynomial in the
  form of (x p).

Problem Find roots of f (x) 2x³ 5x 1 using
Birge Vieta Method.
Solution Assume that x 1 is root of the
equation. Hence initial approximation of the
solution is p0 1.
Synthetic Division will be performed as
below Let f (x) a0x3 a1x2 a2x a3
p0
a0
a1
a2
a3
p0b0
p1b1
p2b2
p0
b1a1p0b0
b0
b1
b2
b3
p1 p0 b3/c2
s i m i l a r l y
Repeat synthetic division using p1
c0
c1
c2
c3
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Iteration No. 1
1
2
0
-5
1
2
2
-3
1
2
2
-3
-2
2
4
1
2
4
1
-1
p1 p0 b3/c2 1 (-2)/1 3
Iteration No. 2
3
2
0
-5
1
6
18
39
Not required
3
2
6
13
40
6
36
147
2
12
49
187
p2 p1 b3/c2 3 40/49 2.1837
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Iteration No. 3
2.1837
2
0
-5
1
4.3674
9.5371
9.9076
2.1837
2
4.3674
4.5371
10.9076
4.3674
19.0742
2
8.7348
23.6113
p3 p2 b3/c2 2.1837 9.9076/23.6113 1.7217
Iteration No. 4
1.7217
2
0
-5
1
3.4434
5.9285
1.5986
1.7217
2
3.4434
0.9285
2.5986
3.4434
11.857
2
6.8868
12.785
p4 p3 b3/c2 1.7217 2.5986/12.785 1.5185
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Iteration No. 5
1.5185
2
0
-5
1
3.037
4.6117
-0.5896
1.5185
2
3.037
-0.3883
0.4104
3.037
9.2234
2
6.074
8.8351
p5 p4 b3/c2 1.5185 0.4104/8.8351 1.4721
Iteration No. 6
1.4721
2
0
-5
1
2.9442
4.3342
-0.9801
1.4721
2
2.9442
-0.6658
0.01986
2.9442
8.6683
2
5.8884
8.0025
p6 p5 b3/c2 1.4721 0.01986/8.0025
1.469624
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Click here to download Excel sheet for performing
synthetic division and find pn.
Thus one of the roots of given f (x) is x
1.47114
To verify f (1.47114) 2 1.471143 5 1.47114
1 0.012138021435088 0
Performing further division by x 1.47114, we
get the deflated polynomial as
1.47114
2
0
-5
1
2.94228
4.3285
-0.9879
2
2.94228
-0.6715
0.01213 0
Therefore f (x) (2x2 2.94228x 0.6715) (x
1.47114)
If (2x2 2.94228x 0.6715) 0 then x 0.2008
or x -1.67195
To verify f (0.2008) 0.012192769 0 and f
(-1.67195) 0.012155754 0
Thus f (x) has three roots as x 1.47114, x
0.2008 and x -1.67195
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3. Bisection Method Used for finding roots of
polynomial functions or transcendental
functions. Uses two guess points which are on
either sides of the root.
Problem Find roots of f (x) 2.5x³ 17x² 22x
11 using Bisection Method.
Solution Let xa 2 and xb 4 so that f(xa)
f(xb) is negative. Hence the required root lies
between the initial two guesses.
Now xr (xa xb) / 2
If f(xr)f(xa) is negative then xbxr else xaxr
R e p e a t
Graphically
xb
xr
xb
xa
xr
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Iteration xa xb Xr
1 2 3 3
2 2 4 2.5
3 2 2.5 2.25
4 2.25 2.5 2.375
5 2.375 2.5 2.4375
6 2.375 2.4375 2.40625
7 2.40625 2.4375 2.421875
8 2.421875 2.4375 2.429688
9 2.421875 2.429688 2.429688
10 2.425782 2.429688 2.427735
Hence x 2.427735 is root of the
equation. f(2.427735) -0.0140314332742740625
0
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4. False Position Method Used for finding roots
of polynomial functions or transcendental
functions. Uses two guess points which are on
either sides of the root. Normally, Faster than
Bisection Method.
Problem Find roots of f (x) 3x³ 20x² 25x
15 using False Position Method.
Solution Let xa 3 and xb 6 so that f(xa)
f(xb) is negative. Hence the required root lies
between the initial two guesses.
If f(xr)f(xa) is negative then xbxr else xaxr
R e p e a t
Graphically
xb
xr
xb
xa
xr
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Iteration xa xb xr
1 3 6 3.320755
2 3.320755 6 3.704273
3 3.704273 6 4.075649
4 4.075649 6 4.363149
5 4.363149 6 4.547058
6 4.547058 6 4.650068
7 4.650068 6 4.703378
8 4.703378 6 4.729822
9 4.729822 6 4.742659
10 4.742659 6 4.748827
Hence x 4.748827 is root of the
equation. f(4.748827) 0