Second approximation: The root will lies between 3.6106 and 4... Find the root of the equation tan x + tan h x = 0 which lies in the interval 1.6, 3.0 correct to four significant digits
Trang 1First approximation:
2 0 1 0 0
1) 0
( )
−
−
= 1 – 1.5 1 ( 0.2817) 3.7225 0.2817
+
= 1 + 0.14085 1.0352 4.0042 = Now, f(x2) = f(1.0352)
= 1.0352e1.0352 – 3
= 2.9148 – 3 = – 0.0852
Thus, the root lies between 1.0352 and 1.5 Then taking x0 = 1.0352, x1 = 1.5, f(x0) = – 0.0852
and f(x1) = 3.7225
Second approximation: The next approximation to the root is
x3 = x0 – 1 0 0
1) 0
( )
f x
−
−
= 1.0352 – 1.5 1.0352 ( 0.0852)
3.7225 0.0852
+
= 1.0352 + 0.0396
3.8077 = 1.0456 Now, f(x3) = f(1.0456)
= 1.0456e1.0456 – 3
= 2.9748 – 3 = –0.0252
Thus, the root lies between 1.0456 and 1.5 Then taking x0 = 1.0456, x1 = 1.5, f(x0) = –0.0252
and f(x1) = 3.7225
Third approximation: The next approximation to the root is
4 0 1 0 0
1) 0
( )
−
−
= 1.0456 – 1.5 1.0456 ( 0.0252)
3.7225 0.0252
+
= 1.0456 + 0.01153.7477 =1.0487 Now, f(x4) = f(1.0487)
= (1.0487)e1.0487 – 3
= 2.9929 – 3 = – 0.0071 Thus, the root lies between 1.0487 and 1.5
Fourth approximation: Taking x0 = 1.0487, x1 = 1.5, f(x0) = – 0.0071 and f(x1) = 3.7225 Then the root becomes
5 0 1 0 0
1) 0
( )
−
−
Trang 2= 1.0487 – 1.5 1.0487 ( 0.0071)
3.7225 0.0071
+
= 1.0487 + 0.0032
3.7296 = 1.0496 Now, f(x5) = f(1.0496)
= (1.0496) e1.0496 – 3
= 2.9982 – 3 = – 0.0018
Thus the root lies between 1.0496, and 1.5
Fifth approximation: Taking x0 = 1.0496, x1 = 1.5, f(x0) = – 0.0018 and f(x1) = 3.7225 Then the next approximation to the root is given by
6 0 1 0 0
1) 0
( )
−
−
= 1.0496 – 1.5 1.0496 ( )
0 0018 3.7225 0.0018
+
= 1.0496 + 0.00018 1.0498
Hence, the root is approximately 1.0498 correct to three decimal places
Example 6 Find a real root of the equation x 2 – log e x – 12 = 0 using Regula-Falsi method correct
to three places of decimals.
Sol Let f(x) = x2 – loge x – 12 = 0
So that f(3) = 32 – loge 3 – 12 = – 4.0986 and f(4) = 42 – loge 4 – 12 and 2.6137
Therefore, f(3) and f(4) are of opposite signs Therefore, a real root lies between 3 and 4 For
the approximation to the root, taking
x0 = 3, x1 = 4, f(x0) = – 4.0986 and f(x1) = 2.6137
First approximation: By Regula-Falsi method, the root is
2 0 ( )1 0( ) ( )0
−
−
4 3
2.6137 4.0986
−
+
= 3 + 4.0986 3.6106 6.7123 = Now, f(x2) = f(3.6106)
= (3.6106)2 – loge (3.6106) – 12
= 13.0364 – 13.2839 = – 0.2475
Second approximation: The root will lies between 3.6106 and 4 Therefore for next approximation, taking
x0 = 3.6106, x1 = 4, f (x0) = – 0.2475 and f(x1) = 2.6137 Then the root is
3 0 ( )1 0( ) ( )0
−
−
Trang 3= 3.6106 – 4 3.6106 ( 0.2475)
2.6137 0.2475
+
= 3.6106 + 0.0964 3.6443
2.8612= Now, f(x3) = f(3.6443)
= (3.6443)2 – loge (3.6443) – 12
= 13.2809 – 13.2932 = – 0.0123
Third approximation: The root lies between 3.6443 and 4 Therefore, taking x0 = 3.6443,
x1 = 4, f(x0) = – 0.0123 and f(x1) = 2.6137 Then the root is given by
x4 = x0 –
( )1 0( ) ( )0
f x
−
−
= 3.6443 – 4 3.6443 ( 0.0123)
2.6137 0.0123
+
= 3.6443 + 0.0044
2.626 = 3.6459 Now, f(x4) = f(3.6459)
= (3.6459)2 – loge (3.6459) – 12
= 13.2926 – 13.2936 = – 0.001
Fourth approximation: The root lies between 3.6459 and 4 Therefore, taking x0 = 3.6459,
x1 = 4, f(x0) = – 0.001 and f(x1) = 2.6137 Then the root is
5 0 ( )1 0( ) ( )0
−
−
= 3.6459 – 4 3.6459 ( 0.001)
2.6137 0.001
−
− +
= 3 6459 + 0.000352.6147 =3.6460 Now, f(x5) = f (3.6460)
= (3.6460)2 – loge (3.6460) – 12
= 13.2933 – 13.2936 = – 0.0003 Fifth approximation: The root lies between 3.6460 and 4 Then for next approximation,
taking x0 = 3.6460, x1 = 4, f(x0) = – 0.0003 and f(x1) = 2.6137 Then the root is
x6 = 0 1 0 0
( ) ( ) ( )
−
−
−
= 3.6460 – 4 3.6460 ( 0.0003)
2.6137 0.0003
+
= 3.6460 + 0.00011
2.614 = 3.6461 Hence the root is approximated by 3.646 correct to three decimal places
Trang 4Example 7 (1) Solve x 3 – 5x + 3 = 0 by using Regula-Falsi method.
(2) Use the method of Falsi Position to solve x 3 – x – 4 = 0
Sol
(1) Let f(x) = x3 – 5x + 3
Since f(0.65) = 0.024625
and f(0.66) = – 0.012504
Hence root lies between 0.65 and 0.66
Let x0 = 0.65 and x1 = 0.66
Using method of Falsi Position,
2 0 ( )1 0( ) ( )0
−
−
= 0.65 – 0.66 0.65 (0.024625)
0.012504 0.024625
= 0.656632282 Now, f(x2) = – 0.00004392
Hence root lies between 0.65 and 0.656632282
Using method of Falsi Position,
3 0 ( )1 0( ) ( )0
−
−
= 0.65 – 0.656632282 0.65 (0.024625)
0.00004392 0.024625
= 0.656620474
Since x2 and x3 are same up to 4 decimal places hence the required root is 0.6566 correct up
to four decimal places Similarly the other roots of this equation are 1.8342 and –2.4909 (2) Let f(x) = x3 – x – 4
Since f(1.79) = – 0.054661
and f(1.80) = 0.032
Hence root lies between 1.79 and 1.80
Let x0 = 1.79 and x1 = 1.80
Using method of Falsi Position,
2 0 ( )1 0( ) ( )0
−
−
= 1.79 – 1.80 1.79 ( 0.054661)
0.032 0.054661
+
= 1.796307
Trang 5Now, f(x2) = – 0.00012936
Hence root lies between 1.796307 and 1.80
Using method of Falsi Position,
3 0 ( )1 0( ) ( )0
−
−
= 1.796307 – 1.8 1.796307 ( 0.00012936)
0.032 0.00012936
= 1.796321
Since x2 and x3 are same up to 4 decimal places hence the required root is 1.7963 correct up
to four decimal places
Example 8 Find the root of the equation tan x + tan h x = 0 which lies in the interval (1.6, 3.0)
correct to four significant digits using of Falsi Position.
Sol Let f(x) = tan x + tan h x = 0
Since f(2.35) = – 0.03
and f(2.37) = 0.009
Hence the root lies between 2.35 and 2.37
Let x0 = 2.35 and x1 = 2.37
Using method of Falsi Position,
x2 = 0 ( )1 0( ) ( )0
−
−
−
= 2.35 – 2.37 2.35 ( 0.03)
0.009 0.03
+
= 2.35 + 0.02 (0.03) 2.365
Hence the root lies between 2.365 and 2.37
Using method of Falsi Position,
( )1 0( ) ( )
−
−
= 2.365 – 2.37 2.365 ( 0.00004)
0.009 0.00004
+
= 2.365 + 0.005 (0.00004)
0.00904
= 2.365 Hence the required root is 2.365 correct to four significant digits
Trang 6PROBLEM SET 2.2
1 Find the real root of the equation x3 – 2x – 5 = 0 by the method of Falsi Position correct
2 Find the real root of the equation x log10 x = 1.2 by Regula-Falsi method correct to four
3 Find the positive root of xe x = 2 by the method of Falsi Position [Ans 0.852605]
4 Apply Falsi Position method to find smallest positive root of the equation x – e –x = 0 correct
5 Find the real root of the equations:
6 Find the rate of convergence of Regula-Falsi method
7 Find real cube root of 18 by Regula-Falsi method [Ans 2.62074]
8 Discuss method of Falsi Position
2.6 ITERATION METHOD (METHOD OF SUCCESSIVE APPROXIMATION)
This method is also known as the direct substitution method or method of fixed iterations
To find the root of the equation f(x) = 0 by successive approximations, we rewrite the given
equation in the form
Now, first we assume the approximate value of root (let x0) , then substitute it in g(x) to have
a first approximation x1 given by
Similarly, the second approximation x2 is given by
2.6.1 Procedure For Iteration Method To Find The Root of The Equation f(x) = 0
Step 1: Take an initial approximation as x0.
Step 2: Find the next (first) approximation x1 by using x1 = g(x0)
Step 3: Follow the above procedure to find the successive approximations x i+1 by using
x i+1 = g(x i ), i = 1, 2, 3
Step 4: Stop the evaluation where relative error ≤ ε, where ε is the prescribed accuracy
Note 1: The iteration method x = g(x) is convergent if 1( )
g x < 1
Note 2: When g1( )x > ⇒1 g1( )x >1 org1( )x < −1, the iterative process is divergent
Trang 72.6.2 Rate of Convergence of Iteration Method
Let f(x) = 0 be the equation which is being expressed as x = g(x) The iterative formula for solving
the equation is
x i +1 = g(x i)
If a is the root of the equation x = g(x) lying in the interval ]a,b[, α = g (α)
The iterative formula may also be written as
x i+1=g x( +x i − α)
Then by mean value theorem
x i + 1 = g(α) + (x1 – α)g’ (c i) Where α < c i < b
⇒ x i+1 = α + (x i – α) g’ (c i)
Now, if e i+1 , e i are the error for the approximation x i + 1 and x i
Therefore, e i+1 = x i+1–α, e i = x i – α
Using this in (1), we get
e i +1 = e i g’ (c i)
Here g(x) is a continuous function, therefore, it is bounded
∴ g c'( )i ≤k, where k ∈ ]a,b[ is a constant.
∴ e i+1 ≤ e i k
i
e e
+ ≤ k
Hence, by definition, the rate of convergence of iteration method is 1 In other words, iteration method converges linearly
Example 1 Find a real root correct upto four decimal places of the equation 2x – log 10 x – 7 = 0 using iteration method.
Sol Here, we have f(x) = 2x – log10 x – 7 = 0
Now, we find that f(3) = – 1.447 = – ve and f (4) = 0.398 = +ve
Therefore, at least one real root of f(x) = 0 lies between x = 3 and x = 4.
Now, the given equation can be re-written as
x = 1
2[log10 x + 7] = g(x), say.
Now, g’ (x) = 1,
2x from which we clearly note that
1( )
g x < 1 for all x ∈ (3, 4) Again since f( )4 < f( )3 , therefore, root is nearer to x = 4 Let the initial approximation be
x0 = 3.6 because f(3.6) tends to zero Then from the iterative formula x i+1 = g(x i), we obtain
x1 = g (x0) = 1
2[log10 x0 + 7] =
1
2 [log10 3.6 + 7] = 3.77815
x2 = g (x1) = g (3.77815) = 3.78863
Trang 8x3 = g (x2) = g (3.78863) = 3.78924
x4 = g (x3) = g (3.78924) = 3.78927
Hence, the root of the equation correct to the four places of decimal is 3.7892
Example 2 Solve x = 0.21 sin (0.5 + x) by iteration method starting with x = 0.12.
Sol Here, x = 0.21 sin (0.05 + x)
Here we observe that f x( ) < 1
⇒ Method of iteration can be applied
Now, first approximation of x is given by
x(1) = 0.21 sin (0.5 + 0.12) = 0.21 sin (0.62)
= 0.21(0.58104) = 0.1220
The second approximation of x is given by
x(2) = 0.21 sin (0.5 + 0.122) = 0.21 sin (0.622)
= 0.21(0.58267) = 0.1224
The third approximation of x is given by
x(3) = 0.21 sin (0.5+0.1224) = 0.21 sin (0.6224)
= 0.21(0.58299) = 0.12243
The fourth approximation of x is given by
x(4) = 0.21 sin (0.5 + 0.12243) = 0.21 sin (0.62243)
= 0.21(0.58301) = 0.12243
Here, we observe that x(3) = x(4)
Hence, the required root is given by x = 0.12243.
Example 3 The equation sin x = 5x – 2 can be put as x = sin –1 (5x – 2) and also as x = 1
5 (sin x + 2)
suggesting two iterating procedures for its solution Which of these, if any, would succeed and which would fall to give root in the neighbourhood of 0.5.
Sol In First case, φ(x) = sin–1 (5x – 2)
5
1− 5x−2 Hence, φ′( )x > 1 for all x for which (5x – 2)2 < 1 or x < 3/5 or x < 0.6 in neighbourhood
of 0.5 Thus the method would not give convergent sequence
In Second case, φ(x) = 1
5 (sin x + 2)
5 cos x
Trang 9Hence, φ′( ) ≤1
5
x for all x because cosx≤1
∴ φ’(x) will succeed.
Hence taking x = φ(x) = 1
5 (sin x + 2) and initial value x0 = 0.5, we have the first approximation
x1 given by
x1 = 1
5 (sin 0.5 + 2) = 0.4017
x2 = 1
5 (sin 0.4017 + 2) = 0.4014
x3 = 1
5 (sin 0.4014 + 2) = 0.4014 Hence up to four places of decimal, the value of required root is 0.4014
Example 4 Find the real root of the equation cos x = 3x – 1 correct to three decimal places, using
iteration method.
We observe that f(0) = 2 = +ve and f(π/2) = – 3 (π/2) + 1 = –ve
⇒ Roots lies between 0 and
2
π
Now, the given equation can be re-written as x = 1
3 (cos x + 1) = g(x) (say)
Then, we have ′( )= −sin = ′( ) <1
3
x
Hence iteration method can be applied
Take the first approximation x0 = 0
Then we can find the successive approximation as:
x1 = g(x0) = 1
3 [cos 0 + 1] = 0.667
x2 = g(x1) = 1
3 [cos (0.667) + 1] = 0.5953
x3 = g(x2) = 13 [cos (0.5953) + 1] = 0.6093
x4 = g(x3) = 1
3 [cos (0.6093) + 1] = 0.6067
x5 = g(x4) = 13 [cos (0.6067) + 1] = 0.6072
x6 = g(x5) = 1
3 [cos (0.6072) + 1] = 0.6071
Now, x5 and x6 being almost same Hence the required root is given by 0.607
Trang 10Example 5 Find the real root of equation f(x) = x 3 + x 2 – 1 = 0 by using iteration method.
Sol Here, f(0)= – 1 and f(1) = 1 so a root lies between 0 and 1 Now, x = 1 x1+ so that,
φ(x) = 1
1 x+
∴ φ′ (x) = –
( )3/2
1
2 1 x+
We have, φ′( )x < 1 for x < 1
Hence iterative method can be applied
x1 = φ (x0) = 1 0.81649
1.5=
x2 = φ (x1) = 1.816491 =0.74196
x8 = 0.75487
Example 6 Find the cube root of 15 correct to four significant figures by iterative method.
Sol Let x = (15)1/3 therefore x3 – 15 = 0
Real root of the equation lies in (2,3) The equation may be written as
x =
3
15 20
( ) 20
x
= φ
2 3 20
x
therefore φ′( )x < 1 (for x ~ 2.5)
Iterative formula is x i + 1 =
3
15 20 20
(1) Put i = 0, x0 = 2.5, we get x1 = 2.47
Put i = 1 in (1), x2 = 2.466 (where x1 = 2.47)
Similarly, x3 = 2.4661
Therefore 320 correct to 3 decimal places is 2.466.
Example 7 Find the reciprocal of 41 correct to 4 decimal places by iterative formula
x i+1 = x i (2 – 41x i ).
Sol Iterative formula is x i+1 = x i (2 – 41x i) (1) Putting i = 0, x1 = x0 (2 – 41x0)
Let x0 = 0.02 then x1 = (0.02) (2 – 0.82) = 0.024
Putting i = 1 in (1) x2 = (0.024) {2 – (41 × 0.024)} = 0.0244