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1. Solutions to Non-linear Equations
False Position Method
1
Dr. Umer Farooq Ahmed
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2. False Position Method
 Frequently, as in the case shown in Fig., the function in the
interval [a, b] is either concave up or concave down.
 In this case, one of the endpoints of the interval stays the
same in all the iterations, while the other endpoint advances
toward the root.
 In other words, the numerical solution advances toward the
root only from one side.
 The convergence toward the solution could be faster if the
other endpoint would also "move" toward the root.
 Food for thought: How we move other point?
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3. False Position Method
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4. False Position Method
 Consider an equation f(x) =
0, which contains only one
variable, i.e. x.
 To find the real root of the
equation f(x) = 0, we
consider a sufficiently small
interval (a, b) where a < b
such that f(a) and f(b) will
have opposite signs.
According to the
intermediate value theorem,
this implies a root lies
between a and b.
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5. False Position Method
 Also, the curve y = f(x) will
meet the x-axis at a certain
point between A[a, f(a)] and
B[b, f(b)].
 Now, the equation of the
chord joining A[a, f(a)] and
B[b, f(b)] is given by:
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6. False Position Method
 Let y = 0 be the point of
intersection of the chord
equation (given above) with
the x-axis. Then,
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7. False Position Method
 This can be simplified as
 Thus, the first approximation
is
x1 = [af(b) – bf(a)]/ [f(b) – f(a)]
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8. False Position Method
 Also, x1 is the root of f(x) if
f(x1) = 0.
 If f(x1) ≠ 0 and if f(x1) and
f(a) have opposite signs, then
we can write the second
approximation as:
 x2 = [af(x1) – x1f(a)]/ [f(x1)
– f(a)]
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9. False Position Method
 Similarly, we can estimate x3,
x4, x5, and so on.
 Geometrical representation of
the roots of the equation f(x)
= 0 can be shown as:
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10. 10
Method of False Position
Find a root of an equation f(x)=2x3-2x-5 using False Position method (Regula
Falsi method)
Solution
Here 2x3-2x-5=0
Let f(x)=2x3-2x-5
x 0 1 2
f(x) -5 -5 7
Approximate root of
the equation 2x3-2x-
5=0 using
False Position
method is 1.60056
False Position Method: Example

11. 11
False Position Method: Algorithm

12. Regula- Falsi Method
• Oldest method ( dates back to ancient Egyptians)
• Also known as method of linear interpolation or False
position method.
• Effective alternate to Bisection method.
• Roots convergence is faster than Bisection method.
• Computational efforts are less.
Advantages

13. Home Tasks
 Apply bisection Method and regula falsi method and find p3
for f (x)
= x - cos x on [0, 1]. Also compare results
√
 Use the Bisection method and Regula Falsi to find solutions
accurate to within 10-2
for x3 - 7x2 + 14x - 6 = 0 on
each interval.
– a. [0, 1]
– b. [1, 3.2]
 Also compare results
 Use the Bisection method to find solutions, accurate to within 10-5
for the following problems.
 3x - ex = 0 for 1 x 2
≤ ≤
 2x + 3 cos x - ex = 0 for 0 x 1
≤ ≤
 x2 - 4x + 4 - ln x = 0 for 1 x 2 and 2 x 4
≤ ≤ ≤ ≤
13

14. Error Analysis of Bisection Method
14
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15.  What is the maximum error after n iterations of
bisection method?
 How many iterations will be required to obtained the root
to the desired accuracy.
Bisection Method
Bisection Method

16. 12 16
-34.8 17.6
16
-12.6 17.6
12
14 15
-12.6 1.5
14.5
15
-5.
8
1.5
-34.8
Change
of sign
Change
of sign
17.6
Change
of sign
-12
.6
Change
of
sign
Iter1
Iter2
Iter3
14
16
14
Error Estimate
At the n-th iteration:
endpoints of the inteval
Length of the interval
Iter4

17. Error Estimates for
Bisection
At the iter1:
True root
live inside
this interval
At the iter2:
At the nth iteration:
the absolute
error in the
n-th iteration
Theorem 2.1
16
-12.6 17.6
12
-34.8
Iter1
14
14 15
-12.6 1.5 17.6
Iter2
16

18. Theorem 2.1
It is important to realize that Theorem
2.1 gives only a bound for
approximation error and that this
bound might be quite conservative.
For example,
Remark
1 14.0000 9.0000e-01
2 15.0000 1.0000e-01
3 14.5000 4.0000e-01
4 14.7500 1.5000e-01
5 14.8750 2.5000e-02
6 14.9375 3.7500e-02
7 14.9063 6.2500e-03
8 14.8906 9.3750e-03
9 14.8984 1.5625e-03
Example:

19. Iterations till desired error

20. Theorem 2.1
Determine the number of iterations
necessary to solve f (x) = 0 with
accuracy 10−2
using a1 = 12 and b1 =
16.
Example
Example:
the desired error
solve for n:
1 14.0000 9.0000e-01
2 15.0000 1.0000e-01
3 14.5000 4.0000e-01
4 14.7500 1.5000e-01
5 14.8750 2.5000e-02
6 14.9375 3.7500e-02
7 14.9063 6.2500e-03
8 14.8906 9.3750e-03
9 14.8984 1.5625e-03
10 14.9023 2.3437e-03
11 14.9004 3.9062e-04
12 14.8994 5.8594e-04
It is important to keep in mind that the error analysis
gives only a bound for the number of iterations. In
many cases this bound is much larger than the actual
number required.
Remark
2
2n