SOLVING EQUATIONS BY ITERATIVE METHODS 79A
Bisecting this interval gives
1. 50390625 + 1. 5078125
2i.e. 1. 505859375Hence
f(1.505859375)= 1. 505859375 + 3 −e^1.^505859375
=−0.0021666...Sincef(1.50589375) is negative andf(1.50390625)
is positive, a root lies betweenx= 1 .50589375 and
x= 1 .50390625.
Bisecting this interval gives
1. 505859375 + 1. 50390625
2i.e. 1. 504882813Hence
f(1.504882813)= 1. 504882813 + 3 −e^1.^504882813
=+0.001256...Since f(1.504882813) is positive and
f(1.505859375) is negative,a root lies between x= 1 .504882813 and x=
1 .505859375.
Bisecting this interval gives
1. 504882813 + 1. 50589375
2i.e.1.505388282The last two values of xare 1.504882813 and
1.505388282, i.e. both are equal to 1.505, correct
to 3 decimal places.
Hence the root ofx+ 3 =exisx=1.505, correct
to 3 decimal places.
The above is a lengthy procedure and it is proba-
bly easier to present the data in a table as shown in
the table.
Problem 3. Solve, correct to 2 decimal places,
the equation 2 lnx+x=2 using the method of
bisection.Let f(x)=2lnx+x− 2
f(0.1)=2ln(0.1)+ 0. 1 − 2 =− 6. 5051 ...
(Note that ln 0 is infinite that
is whyx=0 was not chosen)x 1 x 2 x 3 =x 1 +x 2
2f(x 3 )0 + 2
1 +1.2817...
2 −2.3890...
1 2 1.5 +0.0183...
1.5 2 1.75 −1.0046...
1.5 1.75 1.625 −0.4534...
1.5 1.625 1.5625 −0.2082...
1.5 1.5625 1.53125 −0.0927...
1.5 1.53125 1.515625 −0.0366...
1.5 1.515625 1.5078125 −0.0090...
1.5 1.5078125 1.50390625 +0.0046...
1.50390625 1.5078125 1.505859375 −0.0021...
1.50390625 1.505859375 1.504882813 +0.0012...
1.504882813 1.505859375 1.505388282f(1)=2ln1+ 1 − 2 =− 1
f(2)=2ln2+ 2 − 2 =+ 1. 3862 ...A change of sign indicates a root lies betweenx= 1
andx=2.
Since 2 lnx+x=2 then 2 lnx=−x+2; sketches
of 2 lnxand−x+2 are shown in Fig. 9.4.f(x)212 x− 2f(x) = −x+ 2f(x) = 2 ln x0Figure 9.4