Answer each question: Section XXXXXXXXXXUse the three methods they are; Newton’s Method, Secant Method, Method of False Position, to find solutions to within 10-7 for the following problems. a. x2 -4x...

12)
Use the three methods they are; Newton’s Method, Secant Method, Method of False Position, to find
solutions to within 10-7 for the following problems.
a. x2 -4x +4 – lnx = 0 for 1< x < 2 and for 2 < x < 4
b. x+ 1- 2 sin? x = 0 for 0 < x < ½ and for ½ < x < 1
Solution:
a) Newton’s method
??+1 = ?? –
? ??
?′ ??

? ? = ?2 − 4? + 4 − ln ?
?′ ? = 2? − 4 −
1
?


For 1< x < 2
x1 = 1
x2 = x1 – [f(x1)/ f’(x1)]
f(x1) = f(1) = 1-4+4-ln(1) = 1
f’(x1)=f’(1) = 2-4-1 = -3
x2 =1-[1/-3] = 1+(1/3) = 4/3= 1.3333333
f(x2) = 0.1567624
f’(x2) = -2.0833333
x3 = 1.4085793
f(x3) = 0.0071968
f’(x3) = -1.8927766
x4 = 1.4123816
f(x4) = 0.0000810
f’(x4) = -1.8832608
x5 = 1.4123912
ε = x5 – x4 = 0.000104
so the solution for the equation is x = 1.4123912
For 2< x < 4
n xn f(xn) f’(xn)
1, x1 4 2.6137056 3.7500000
2, x2 3.3030118 0.5030050 2.3032696
3, x3 3.0846244 0.0499802 1.8450602
4, x4 3.0575358 0.0007727 1.7880108
5, x5 3.0571037 0.0000003 1.7871004
6, x6 3.0571035
So x = 3.0571035
Ans: x =1.4123912 and x = 3.0571035
Secant method:
)()(
))((
1
1
1






ii
iii
ii
xfxf
xxxf
xx
For 1< x < 2
x0 = 2 and x-1 = 1
Iteration
Number,
i
1ix i
x
1i
x

)( 1ixf )( ixf  1ixf
? =
??+1 − ??
??
∗ 100
1 1 2 1.5906161 1.0000000 -0.6931472 -0.2965262 25.737442%
2 2 1.5906161 1.2845479 -0.6931472 -0.2965262 0.2614649 23.82692%
3 1.5906161 1.2845479 1.4279661 -0.2965262 0.2614649 -0.0290283 10.04353%
4 1.2845479 1.4279661 1.4136346 0.2614649 -0.0290283 -0.0023397 1.0036303%
5 1.4279661 1.4136346 1.4123782 0.0290283 -0.0023397 0.0000244 0.0889563%
6 1.4136346 1.4123782 1.4123912 -0.0023397 0.0000244 -0.0000001 0.0009204%
7 1.4123782 1.4123912 1.4123912 0.0000244 -0.0000001 -0.0000001 0.0000000%
X = 1.4123912
For 2< x < 4
x0 = 4 and x-1 = 2
Iteration
Number,
i
1ix i
x
1i
x

)( 1ixf )( ixf  1ixf
? =
??+1 − ??
??
∗ 100
1 2 4 2.4192186 -0.6931472 2.6137056 -0.7077004 65.342644
2 4 2.4192186 2.7560397 2.6137056 -0.7077004 -0.4421987 13.922723
3 2.4192186 2.7560397 3.3170226 -0.7077004 -0.4421987 0.5354810 16.912242
4 2.7560397 3.3170226 3.0097689 -0.4421987 0.5354810 -0.0822301 9.2629366
5 3.3170226 3.0097689 3.0506707 0.5354810 -0.0822301 -0.0114525 1.3407478
6 3.0097689 3.0506707 3.0572890 -0.0822301 -0.0114525 0.0003315 0.2169457
7 3.0506707 3.0572890 3.0570982 -0.0114525 0.0003315 -0.0000096 0.0062412

X = 3.0570982
Ans: x =1.4123912 and x = 3.0570982
False position method:
   
   ii
iiii
i
xfxf
xfxxfx
x






1
11
1
For 1< x < 2
x0 = 2 and x-1 = 1
Iteration
Number,
i
1ix i
x
1i
x

)( 1ixf )( ixf  1ixf
? =
??+1 − ??
??
∗ 100
1 1 2 1.5906161 1.0000000 -0.6931472 -0.2965262 20.469195
2 1 1.5906161 1.4555373 1.0000000 -0.2965262 -0.0789355 8.4922298
3 1 1.4555373 1.4222100 1.0000000 -0.0789355 -0.0183707 2.2896936
4 1 1.4222100 1.4145936 1.0000000 -0.0183707 -0.0041416 0.5355314
5 1 1.4145936 1.4128836 1.0000000 -0.0041416 -0.0009271 0.1208835
6 1 1.4128836 1.4125012 1.0000000 -0.0009271 -0.0002072 0.027066
7 1 1.4125012 1.4124158 1.0000000 -0.0002072 0.0060495
x = 1.4124158
For 2< x < 4
x0 = 4 and x-1 = 2
Iteration
Number,
i
1ix i
x
1i
x

)( 1ixf )( ixf  1ixf
? =
??−1 − ??+1
??−1
∗ 100
1 2 4 2.4192186 -0.6931472 2.6137056 -0.7077004 20.960932
2 2.4192186 4 2.7560397 -0.7077004 2.6137056 -0.4421987 13.922722
3 2.7560397 4 2.9360446 -0.4421987 2.6137056 -0.2008838 6.5312872
4 2.9360446 4 3.0119816 -0.2008838 2.6137056 -0.0784914 2.58637
5 3.0119816 4 3.0407874 -0.0784914 2.6137056 -0.0288781 0.9563749
6 3.0407874 4 3.0512696 -0.0288781 2.6137056 -0.0103900 0.3447215
7 3.0512696 4 3.0550261 -0.0103900 2.6137056 -0.0037081 0.1231113
8 3.0550261 4 3.0563648 -0.0037081 2.6137056 -0.0013196 0.0438207
9 3.0563648 4 3.0568410 -0.0013196 2.6137056 -0.0004691 0.0155805
10 3.0568410 4 3.0570103 -0.0004691 2.6137056 -0.005537
x = 3.0570103
Ans: x =1.4124158 and x = 3.0570103
b) x+ 1- 2 sin? x = 0 for 0 < x < ½ and for ½ < x < 1
f(x) = x+ 1- 2 sin? x
f’(x) = 1 - 2?cos? ?
Newton’s method:...