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Numerical Analysis
Chapter-2
Prepared by-
Md. Harun Or Rashid
Lecturer
Dept. of CSE, BAUET
False Position Method
This is the oldest method for finding the real root of a
nonlinear equation 𝑓 𝑥 = 0. It is also known as Regula-
Falsi or method of Chords. In this method, we choose two
points a and b such that 𝑓 𝑎 and 𝑓 𝑏 are of opposite signs,
then a root must lie in between a and b. Now the equation of
the chord joining the two points [𝑎, 𝑓 𝑎 ] and [𝑏, 𝑓 𝑏 ] is
given by-
𝑦−𝑓(𝑎)
𝑥−𝑎
=
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
By taking the point of intersection on x-axis [y=0] as an
approximation to the root we obtain-
2
False Position Method (Cont’d)
0−𝑓(𝑎)
𝑥−𝑎
=
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝑥 − 𝑎 = −
(𝑏−𝑎) 𝑓(𝑎)
𝑓(𝑏)−𝑓(𝑎)
𝑥 = 𝑎 −
(𝑏−𝑎) 𝑓(𝑎)
𝑓(𝑏)−𝑓(𝑎)
𝑥 =
𝑎𝑓(𝑏)−𝑏𝑓(𝑎)
𝑓(𝑏)−𝑓(𝑎)
Which is the first approximation to the root of 𝑓 𝑥 = 0.
3
False Position Method (Cont’d)
4
Fig: Graphical representation of the False Position method.
False Position Method (Cont’d)
It can be easily programmed using the following computational
steps:
1. Choose two real numbers a and b such that 𝑓 𝑎 𝑓 𝑏 < 0. [N.B.: a and b must
be closer as much as possible]
2. Set 𝑥𝑟 =
𝑎𝑓(𝑏)−𝑏𝑓(𝑎)
𝑓(𝑏)−𝑓(𝑎)
.
3. (a) If 𝑓 𝑎 𝑓 𝑥𝑟 < 0, the root lies in the interval (𝑎, 𝑥𝑟). Then, set 𝑏 = 𝑥𝑟 and
go to step 2 above.
(b) If 𝑓 𝑎 𝑓 𝑥𝑟 > 0, the root lies in the interval (𝑥𝑟, 𝑏). Then, set
𝑎 = 𝑥𝑟 and go to step 2 above.
(c) If 𝑓 𝑎 𝑓 𝑥𝑟 = 0, it means that 𝑥𝑟 is a root of the equation 𝑓 𝑥 = 0 and
the computation may be terminated.
5
False Position Method (Cont’d)
Example: Find a real root of the equation 𝑥3
−2𝑥 − 5 = 0 correct
to three decimal places.
Solution: Given that,
𝑥3
− 2𝑥 − 5 = 0
Let, 𝑓 𝑥 = 𝑥3
−2𝑥 − 5
Taking a=2 and b=3
∴ 𝑓 2 = 23
−2 × 2 − 5 = −1 (< 0)
∴ 𝑓 3 = 33
−2 × 3 − 5 = 16 (> 0)
Hence a root lies between 2 and 3.
6
7
False Position Method (Cont’d)
Step No. a b 𝒙𝒓=
𝒂𝒇(𝒃)−𝒃𝒇(𝒂)
𝒇(𝒃)−𝒇(𝒂)
𝒇 𝒙 = 𝒙𝟑−𝟐𝒙 − 𝟓
1 2 3 2.058823529 −0.390799918
2 2.058823529 3 2.08126366 −0.147204064
3 2.08126366 3 2.089639211 −0.054676498
4 2.089639211 3 2.092739575 −0.020202859
5 2.092739575 3 2.093883709 −0.007450502
6 2.093883709 3 2.094305451 −0.002745669
7 2.094305451 3 2.094460845 −0.001011578
The difference between 𝒙𝟔 and 𝒙𝟕 is 0.000155394. Hence a
root, correct to three decimal places is 2.094.
8
False Position Method(Cont’d)
Self Practise
Example- 2.7, 2.8, 2.9
N.B.: While computing trigonometric ratio, the specified
value must be converted into degree using,
Degree =
180 × 𝑋
3.14
Where, X is in radian unit.
Task related to False position Method
1. Write some advantages of false position method.
2. Write some disadvantages of false position method.
9
Any Question???
10

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NA_Lecture-7.pdf

  • 1. Numerical Analysis Chapter-2 Prepared by- Md. Harun Or Rashid Lecturer Dept. of CSE, BAUET
  • 2. False Position Method This is the oldest method for finding the real root of a nonlinear equation 𝑓 𝑥 = 0. It is also known as Regula- Falsi or method of Chords. In this method, we choose two points a and b such that 𝑓 𝑎 and 𝑓 𝑏 are of opposite signs, then a root must lie in between a and b. Now the equation of the chord joining the two points [𝑎, 𝑓 𝑎 ] and [𝑏, 𝑓 𝑏 ] is given by- 𝑦−𝑓(𝑎) 𝑥−𝑎 = 𝑓(𝑏)−𝑓(𝑎) 𝑏−𝑎 By taking the point of intersection on x-axis [y=0] as an approximation to the root we obtain- 2
  • 3. False Position Method (Cont’d) 0−𝑓(𝑎) 𝑥−𝑎 = 𝑓(𝑏)−𝑓(𝑎) 𝑏−𝑎 𝑥 − 𝑎 = − (𝑏−𝑎) 𝑓(𝑎) 𝑓(𝑏)−𝑓(𝑎) 𝑥 = 𝑎 − (𝑏−𝑎) 𝑓(𝑎) 𝑓(𝑏)−𝑓(𝑎) 𝑥 = 𝑎𝑓(𝑏)−𝑏𝑓(𝑎) 𝑓(𝑏)−𝑓(𝑎) Which is the first approximation to the root of 𝑓 𝑥 = 0. 3
  • 4. False Position Method (Cont’d) 4 Fig: Graphical representation of the False Position method.
  • 5. False Position Method (Cont’d) It can be easily programmed using the following computational steps: 1. Choose two real numbers a and b such that 𝑓 𝑎 𝑓 𝑏 < 0. [N.B.: a and b must be closer as much as possible] 2. Set 𝑥𝑟 = 𝑎𝑓(𝑏)−𝑏𝑓(𝑎) 𝑓(𝑏)−𝑓(𝑎) . 3. (a) If 𝑓 𝑎 𝑓 𝑥𝑟 < 0, the root lies in the interval (𝑎, 𝑥𝑟). Then, set 𝑏 = 𝑥𝑟 and go to step 2 above. (b) If 𝑓 𝑎 𝑓 𝑥𝑟 > 0, the root lies in the interval (𝑥𝑟, 𝑏). Then, set 𝑎 = 𝑥𝑟 and go to step 2 above. (c) If 𝑓 𝑎 𝑓 𝑥𝑟 = 0, it means that 𝑥𝑟 is a root of the equation 𝑓 𝑥 = 0 and the computation may be terminated. 5
  • 6. False Position Method (Cont’d) Example: Find a real root of the equation 𝑥3 −2𝑥 − 5 = 0 correct to three decimal places. Solution: Given that, 𝑥3 − 2𝑥 − 5 = 0 Let, 𝑓 𝑥 = 𝑥3 −2𝑥 − 5 Taking a=2 and b=3 ∴ 𝑓 2 = 23 −2 × 2 − 5 = −1 (< 0) ∴ 𝑓 3 = 33 −2 × 3 − 5 = 16 (> 0) Hence a root lies between 2 and 3. 6
  • 7. 7 False Position Method (Cont’d) Step No. a b 𝒙𝒓= 𝒂𝒇(𝒃)−𝒃𝒇(𝒂) 𝒇(𝒃)−𝒇(𝒂) 𝒇 𝒙 = 𝒙𝟑−𝟐𝒙 − 𝟓 1 2 3 2.058823529 −0.390799918 2 2.058823529 3 2.08126366 −0.147204064 3 2.08126366 3 2.089639211 −0.054676498 4 2.089639211 3 2.092739575 −0.020202859 5 2.092739575 3 2.093883709 −0.007450502 6 2.093883709 3 2.094305451 −0.002745669 7 2.094305451 3 2.094460845 −0.001011578 The difference between 𝒙𝟔 and 𝒙𝟕 is 0.000155394. Hence a root, correct to three decimal places is 2.094.
  • 8. 8 False Position Method(Cont’d) Self Practise Example- 2.7, 2.8, 2.9 N.B.: While computing trigonometric ratio, the specified value must be converted into degree using, Degree = 180 × 𝑋 3.14 Where, X is in radian unit.
  • 9. Task related to False position Method 1. Write some advantages of false position method. 2. Write some disadvantages of false position method. 9