46
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
PROBLEM SET 2.1
1. Find the smallest root lying in the interval (1, 2) up to four decimal places for the equation
x
6
– x
4
– x
3
– 1 = 0 by Bisection Method [Ans. 1.4036]
2. Find the smallest root of x
3
– 9x + 1 = 0, using Bisection Method correct to three decimal
places. [Ans. 0.111]
3. Find the real root of e
x
= 3x by Bisection Method. [Ans. 1.5121375]
4. Find the positive real root of x – cos x = 0 by Bisection Method, correct to four decimal
places between 0 and 1. [Ans. 0.7393]
5. Find a root of x
3
– x – 11 = 0 using Bisection Method correct to three decimal places which
lies between 2 and 3. [Ans. 2.374]
6. Find the positive root of the equation xe
x
= 1 which lies between 0 and 1.
[Ans. 0.5671433]
7. Solve x
3
– 9x + 1 = 0 for the root between x = 2 and x = 4 by the method of Bisection.

, f(x
0
)} and {x
1
, f (x
1
)} is

()
() ()
()
10
00
10
fx fx
xxx
xx
−
−= −
−
yf
The method consists in replacing the curve
AB by means of the chord AB and taking the point
of intersection of the chord with X-axis as an
approximation to the root.
So the abscissa of the point where chord cuts y = 0 is given by

()
()
()

can also be put in the following form:

()
()
()
()
01 10
2
10
xf x xf x
x
fx fx
−
=
−
In general, the (i + 1)th approximation to the root is given by

() ( )
() ( )
11
1
1
iiii
i
ii
xfx xfx
x
fx fx
−−
+

()
01 10
2
10
xf x xf x
x
fx fx
−
=
−
and also evaluate f(x
2
).
Step 3: If f(x
2
) f(x
1
) < 0, then go to the next step. If not, rename x
0
as x
1
and then go to
the next step.
Step 4: Evaluate successive approximations using the formula

() ( )
() ( )
11
1
1

ii
xx
where
ε
is the prescribed accuracy.
2.5.2 Order (or Rate) of Convergence of False Position Method
The general iterative formula for False Position Method is given by

() ( )
() ( )
11
1
1
iiii
i
ii
xfxxfx
x
fx f x
−−
+
−
−
=
−
(1)
where x
i–1
, x
i

, e
i
and e
i+1
are the successive
errors in (i – 1)th, ith

and (i + 1)th iterations respectively, then
e
i–1
= x
i–1
–
α
, e
i
= x
i
–
α
, e
i + 1
= x
i + 1
–
α
or x
i–1
= α + e
i–1

−α+
α+ =α+ −
α+ − α+
or
()
()
()
()
1
1
1
ii i
ii
ii
ee f e
ee
fefe
−
+
−
−α+
=−
α+ − α+
(3)
Expanding f(
α
+ e
i
) and f(α + e
i –1

−

′′′
−α+α+α+



=−
 
′′′ ′′
α+ α+ α+ − α+ ′α+ α+
 
 
 
i.e.,
()
() () ()
()
() ()
−
+
−
−

′′′
−α+α+α



=−

ee f f
, [on ignoring the higher order terms]
i.e.
() () ()
() ()
+
−

′′′
α+ α+ α


==
+


′′
α+ α




2
1
1
2

'
2
i

1
2
=
2
i
i
ii
ii
e
ef f
ee
ee
ff
[since f(
α
) = 0]
i.e.
()
()
()
()
2
1
1
2
1
2
i
i
ii

[on dividing numerator and denominator by f ′(
α
)
i.e.
()
()
()
()
1
2
1
1
1
22
iii
iii
ff
eee
eee
ff
−
−
+

′′ ′′

αα
+

=− + +

αα
+

=− + −



′′
αα



ALGEBRAIC AND TRANSCENDENTAL EQUATION
49
i.e.,
()
()
()
()
2
22
11
1
()
() ()
22()4
ii i i i i i
ii
ff
f

α
1
2
1
0
2
iii i
f
eee e
f
If e
i–1
and e
i
are very small, then ignoring 0(e
2
i
), we get

()
()
11
2
iii
f
eee
f
+−
′′
α

e
i
= Ae
1
k
i−
or e
i–1
= (e
i
/A)
1/k
Now, substituting the value of e
i+1
and e
i–1
in (5), we get
Ae
k
i
= e
i
.
1/
.
k
i
e
M
A

eAe A
e
+
==
Comparing this with
1
lim
i
k
i
i
e
A
e
+
→∞

≤



, we see that order (or rate) of convergence of false
position method is 1.618.
Example 1. Find a real root of the equation f(x) = x
3
– 2x – 5 = 0 by the method of false position
up to three places of decimal.
Sol. Given that f(x)= x
3
– 2x – 5 = 0

=−
−
()
32 1
2 1 2 2.0588
16 1 17
−
=− −=+ =
+
Now, f(x
2
)= f(2.0588)
= (2.0588)
3
– 2 (2.0588) – 5 = – 0.3911
Therefore, root lies between 2.0588 and 3.
Second approximation: Now, taking x
0
= 2.0588, x
1
= 3, f(x
0
) = – 0.3911, f(x
1
) = 16, then by
Regula-Falsi method, we get

()
()
()

) = 16. Then by
Regula-Falsi method, we get

()
()
()
10
40 0
10
xx
xx fx
fx fx
−
=−
−
= 2.0813 –
3 2.0813
16 0.1468
−
+
(– 0.1468) = 2.0813 + 0.0084 = 2.0897
Now, f(x
4
)= f(2.0897)
= (2.0897)
3
– 2 (2.0897) – 5
= 9.1254 – 9.1794 = – 0.054
Therefore, root lies between 2.0897 and 3.
Fourth approximation: Now, taking x

= 2.0897 + 0.0031 = 2.0928
ALGEBRAIC AND TRANSCENDENTAL EQUATION
51
Now, f(x
5
)= f(2.0928)
= (2.0928)
3
– 2(2.0928) – 5
= 9.1661 – 9.1856 = – 0.0195
Therefore, root lies between 2.0928 and 3.
Fifth approximation: Now, taking x
0
= 2.0928, x
1
= 3, f(x
0
) = – 0.0195, f(x
1
) = 16, then we
get

()
()
()
10
60 0
10
xx
xx fx

get

()
()
()
10
70 0
10
xx
xx fx
fx fx
−
=−
−
= 2.0939 –
()
32.0939
0.0074
16 0.0074
−
−
+
= 2.0939 + 0.00042 = 2.0943
Now, f(x
7
)= f(2.0943)
= (2.0943)
3
– 2(2.0943) – 5
= 9.1858 – 9.1886 = – 0.0028

−
+
= 2.0943 + 0.00016 = 2.0945
Hence, the root is 2.094 correct to three decimal places.
Example 2. Find the real root of the equation f(x) = x
3
– 9x + 1 = 0 by Regula-Falsi method.
Sol. Let f(x)= x
3
– 9x + 1 = 0 (1)
So that f(2) = (2)
3
– 9(2) + 1 = – 9
f(3) = (3)
3
– 9(3) + 1 = 1
52
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Since f(2) and f(3) are of opposite signs, therefore the root lies between 2 and 3, so taking
x
0
= 2, x
1
= 3, f(x
0
) = – 9, f(x
1
) = 1, then by Regula-Falsi method, we get
First approximation:
()

f(x
0
) = – 0.711, f(x
1
) = 1. Then

()
()
()
10
30 0
10
xx
xx fx
fx fx
−
=−
−
= 2.9 –
()
32.9
0.711
1 0.711
−
−
+
= 2.9 + 0.0416 = 2.9416
Now, f(x
3
)= f(2.9416)

0.0207
10.0207
−
−
+
= 2.9416 + 0.0012 = 2.9428
Now, f(x
4
)= f (2.9428)
= (2.9428)
3
– 9(2.9428) + 1
= 25.4849 – 25. 4852 = – 0.0003
Fourth approximation: The root lies between 2.9428 and 3. Therefore, taking
x
0
= 2.9428, x
1
= 3, f (x
0
) = – 0.0003, f (x
1
) = 1. Then by False Position method, we have

()
()
()
10
50 0
10

– x
3
– 1
f(1.4) = (1.4)
6
– (1.4)
4
– (1.4)
3
– 1 = – 0.056
f(1.41) = (1.41)
6
– (1.41)
4
– (1.41)
3
– 1 = 0.102
Hence the root lies between 1.4 and 1.41.
Using the method of False Position,

()
()
()
10
20 0
10
xx
xx fx
fx fx
−

Using the method of False Position,

() ()
()
12
32 2
12
xx
xx fx
fx fx
−
=−
−
= 1.4035 –
−
−
+
1.41 1.4035
( 0.0016)
0.102 0.0016
= 1.4035 +
()
0.0065
0.0016 1.4036
0.1036

=


Now, f(1.4036) = (1.4036)

−
+
= 1.4036 +
()
0.0064
0.00003 1.4036
0.10203

=


Since, x
3
and x
4
are approximately the same upto four places of decimal, hence the required
root of the given equation is 1.4036.
54
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Example 4. Find a real root of the equation f(x) = x
3
– x
2
– 2 = 0 by Regula-Falsi method.
Sol. Let f(x)= x
3
– x
2
– 2 = 0
Then, f(0) = – 2, f(1) = – 2 and f(2) = 2

=− − =+ =
+
Now,
2
() (1.5)
fx f=
= (1.5)
3
– (1.5)
2
– 2
= 3.375 – 4.25 = – 0.875
Thus, the root lies between 1.5 and 2.
Second approximation: Taking
01 0
1.5, 2, ( ) 0.875
xxfx== =−
and
1
()2.
fx =
Then the next
approximation to the root is given by

10
30 0
1) 0
()
(()
xx

) = – 0.2197 and f(x
1
) = 2. Then the next
appoximation to the root is given by

10
40 0
1) 0
()
(()
xx
xx fx
fx fx
−
=−
−
= 1.6522 –
()
2 1.6522
0.2197
2 0.2197
−
−
+
= 1.6522 + 0.0344 = 1.6866
Now, f(x
4
)= f(1.6866)
= (1.6866)
3

()
1.6866
0.0469
2 0.0469
−
−
+
= 1.6866 + 0.0072 = 1.6938
Now, f(x
5
)= f(1.6938)
= (1.6938)
3
– (1.6938)
2
– 2
= 4.8594 – 4.8690 = – 0.0096
Thus, the root lies between 1.6938 and 2.
Fifth approximation: Taking x
0
= 1.6938, x
1
= 2, f(x
0
) = – 0.0096 and f(x
1
) = 2. Then the next
approximation to the root is given by

10

0
= 1.6953, x
1
= 2, f(x
0
) = – 0.0016 and f(x
1
) = 2. Then the next
approximation to the root is

10
70 0
1) 0
()
(()
xx
xx fx
fx fx
−
=−
−
= 1.6953 –
()
2 1.6953
0.0016
2 0.0016
−
−
+
= 1.6953 + 0.0002 = 1.6955