2.4 - Convergence of the Newton Method and Modified Newton Method
Consider the problem of finding x ∗ , the solution of the equation: fx  0 for x in a, b . Assume that
′
f x is continuous and f x ≠ 0 for x in a, b .
′
1. Newton’s Method:
Suppose that x ∗ is a simple zero of fx. Then we know fx  x − x ∗ Qx where
lim∗ Qx ≠ 0.
x→x
In Section 2.2, the Newton Method is introduced for solving fx  0. Recall
Algorithm Newton Method: newton.m
Given fx, x 0 in a, b and , for n  1, 2, . . . ,
(1) compute x n  x n−1 −
fx n−1 
′
f x n−1 
;
(2) if x n − x n−1   or fx n   , then the algorithm is terminated and
x ∗ ≈ x n ; otherwise goto Step (1).
Convergence and Rate of Convergence:
In this section, we study the convergence of Newton’s Method and will answer the following questions:
(i) Does the Newton Method converge for any given starting approximation x 0 ?
(ii) If Newton Method converges, what is the rate of convergence and what is the order of convergence?
(iii) Does the Newton Method still converge when fx has multiple zeros at x ∗ ? Note that in this case,
′
f x ∗   0.
From Section 2.3, we know that the sequence of iterations x n  n1 generated by Newton’s Method can
also be considered as iterations generated by the Fixed-Point Algorithm with function
fx
.
g newton x  x − ′
f x
By the Fixed-Point Theorem, the sequence of iterations x n n1 converges to x ∗ if
g ′ x
≤ K  1 for all x in a, b
where x 0 in a, b , and the rate of convergence is OK n . We now investigate conditions on
fx under which the property g ′newton x  1 for x ∈ a, b is satisfied.
Observe that
′
g ′newton
x  1 −
′
′′
f xf x − fxf x
′
f x
′
2
′
′′
 1−1
fxf x
′
f x
2
′′

fxf x
′
f x
′′
2
.
We know that f x ∗   0. If f x ∗  ≠ 0, f x ≤ M 1 for all x in a, b and f x ≤ M 2 for all x
in a, b , then there exists an interval ā, b̄ ⊆ a, b such that
1
′′
fxf x
′
≤ K  1 and for x 0 in ā, b̄ , lim
x  x∗.
n→ n
2
f x
Local Convergence Property of Newton’s Method:
′
′′
′
Theorem 1 Let f , f and f be continuous in a, b . If x ∗ is in a, b and f x ∗  ≠ 0, then there
exists a   0 such that x n  generated by Newton’s method converges to x ∗ for any x 0 in
x ∗ − , x ∗   .
Note that:
a. Since the convergence of Newton method depends on , lim n→ x n  x ∗ only locally.
′′
∗
n
b. By the Fixed-Point Theorem, x n  x  OK  where
fxf x
′
f x
≤ K  1 for all x in
2
x ∗ − , x ∗   .
Example Determine if the iterations generated by Newton’s Method for solving x 2 − 2  0 converges.
′
fx  x 2 − 2, f x  2x, f ′′ x  2. Consider a, b  1, 2.
′′
fxf x
′
g Newton x 
y
′
f x
2
2x 2 − 2
2x 2

 1 1 − 22
2
x
≤ g ′newton 1  1  K
2
0.5
0.4
(1) lim n→ x n 
0.3
2 and
0.2
(2) x n 
0.1
2  O0. 5 n 
0.0
1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
x
′
g Newton x
Example Approximate 2 by finding the solution of x 2 − 2  0 for x ≥ 0 within 10 −8 .
2
Let x 0  1. Then g1  1 − 1 − 2  3 . By the Fixed-Point Theorem, we can find an upper bound
2
21
N for the number of iterations needed:
ln10 −8 1 − 0. 5/ 32 − 1 
 26. 575 425
N
ln0. 5
Let N  27.
Use the MatLab program newton.m to find an approximation.
fun@(x) x.^2-2;
2
fpfun@(x) 2*x;
[xsol, flag, ct]newton(fun, fpfun,1,10^(-8),100)
xsol 
1.00000000000000
1.50000000000000
1.41666666666667
1.41421568627451
1.41421356237469
flag  1
ct  5
Example Find an approximation of the solution of the equation x 3  x  1  0 for x in −1, 0 within
10 −5 .
(1) Compute x n  by (a) Newton’s Method with x 0  −1, (b) the Bisection Method and (c) the
Fixed point Iteration with gx  −1 and p 0  −1.
x2  1
(2) Compare numerical results to determine which algorithm has the best performance for
this problem.
(3) Can we reach algebraically the same conclusion?
3
′
(1) (a) Newton’s Method: fx  x 3  x  1, f x  3x 2  1, x n  x n−1 − x n−1 2 x n−1  1 .
3x n−1  1
fun@(x) x.^3x1;
fpfun@(x) 3*x.^21;
[xsol,flag,ct]newton(fun,fpfun,-1,10^(-5),100)
xsol 
-1.00000000000000
-0.75000000000000
-0.68604651162791
-0.68233958259731
-0.68232780394651
flag 1
ct 5
(b) Bisection method:
bisect
the leftend point a  -1
the rightend point b  0
epsilon  10^(-5)
the approximation to the solution - cN  -0.68233489990234
function value at the solution - fcN  -1.700736159326866e-005
number of iterations - Nit 17
3
(c) Fixed-point Algorithm:
fixpt
input initial point p0  -1
input the tolerance for stopping criterion, ex:.0001,10^(-8) 10^(-5)
input the maximum number of iterations 100
Algorithm converges with number of iterations  25
fixed point p  -0.68232442571947
(2) Compare the numbers of approximations generate by all three algorithms:
algorithm
Newton Method Bisection Method Fixed-Point Iteration
Numerical of Approximations 5
17
25
The Newton Method has the best performance.
(3) (a) Newton Method:
3
′′
g newton x  x − x 2 x  1 , f x  6x,
3x  1
x 3  x  16x graphically
g ′newton x 

0. 8
3x 2  1 2
0.8
x n  x ∗  O0. 8 n 
0.4
0.6
x 0  −1, x 1  g newton −1  − 34
N
−5
ln10 1 − 0. 8/ −
ln0. 8
3
4
1 
0.2
 52. 59
Note that N  5 by the algorithm for   10 −5 .
0.0
-1.0
-0.8
-0.6
-0.4
′
g x
(b) Bisection Method:
x n  x ∗  O 21n   x ∗  O0. 5 n 
ln1 − 0 − ln10 −5 
 16. 609 640 5, N  17
N
ln2
(c) Fixed-Point Iteration:
4
-0.2
0.0
x
y
Check the maximum value of |g ′ x| :
2x
g ′ x  −
2
2
x  1
2|x|
≤ 0. 7  K
|g ′ x| 
2
x 2  1
y=0.7
0.6
y
0.4
0.2
n
x n  x ∗  O0. 7 
ln 10 −5  ln1 − 0. 7
N
 35. 654
ln0. 7
0.0
-1.0
Note that N  25 by the algorithm for   10 −5 .
-0.5
0.0
x
2|x|
2
x  1
Algebraically, we find the Bisection Method has the best performance so we cannot reach the same
conclusion as we have obtained from numerical results.
y  |g ′ x| 
2
Order of convergence:
Theorem 2 Modified Fixed Point Theorem:
Let p be a fixed point of gx. If g ′ p  0 and g ′′ p ≠ 0, then p n → p quadratically (  2)
with an asymptotic error constant   1 |g ′′ p|.
2
′
′′
Proof: Assume that g p  0 but g p ≠ 0. By the Taylor Theorem, we have for x ≠ p,
g ′′ cx
g ′ p
gx  gp 
x − p 
x − p 2 , where cx is between x and p.
1!
2!
Then for x  p n ,
g ′′ cx n 
g ′′ cp n 
g ′ p
gp n   gp 
p n − p 
p n − p 2  gp 
p n − p 2
1!
2!
2!
where cp n−1  is between x and p n−1 .
g ′′ cp n−1 
p n − p  gp n−1  − gp  gp 
p n−1 − p 2 − gp  1 g ′′ cp n−1  p n−1 − p
2
2!
and
p n1 − p
1 g ′′ cp n   1 g ′′ lim cp n 
 lim
 1 g ′′ p .
lim
2
n→
n→ 2
n→
2
2
pn − p
So, the order of convergence is 2 and the asymptotic error constant 1 |g ′′ p|.
2
Theorem 3 Let x n  be generated by the Newton Method. If x ∗ is a simple zero of fx, then x n → x ∗
′′
quadratically with an asymptotic error constant
f x ∗ 
.
′
2f x ∗ 
′′
Proof: By the Modified Fixed-Point Theorem, we have g ′ x 
5
fxf x
′
f x
2
. Then
2
′
g ′′ x 
′′
f xf x  fxf
x
′
2
′
4
f x
f x
′

′′′
′′
f xf x  fxf
′′′
′
′
′
′′
∗
g x 
f x ∗ 

′′
2
3
2
′′
f x ∗ 
′
f x ∗ 
the asymptotic error constant is 1 |g ′′ x ∗ | 
2
′
f x − 2fx f x
x
f x
fx ∗ 0
′′
− fxf x 2f xf ′′ x
3
′′

f x ∗ 
′
f x ∗ 
,
′′
f x ∗ 
′
2f x ∗ 
.
2. Zero of Multiplicity m:
Definition A solution x ∗ of fx  0 is a zeros of multiplicity m of f if for x ≠ x ∗ , f can be expressed as
fx  x − x ∗  m qx
where lim x→x ∗ qx ≠ 0.
Theorem 4 Let f be continuously differentiable on a, b . f has a simple zero at x ∗ in a, b if and
′
only if fx ∗   0 but f x ∗  ≠ 0.
Proof Let f have a simple zero at x ∗ in a, b . By the definition, fx  x − x ∗ qx where
lim x→x ∗ qx ≠ 0. Since
′
f ′ x  qx  x − x ∗ q ′ x, f x ∗   qx ∗  ≠ 0.
′
′
Now let f x ∗  ≠ 0. Since f x exists on a, b , for x ≠ x ∗ by the Taylor Theorem
′
f cx
′
fx  fx  
x − x ∗   f xx − x ∗  where cx is between x and x ∗ .
1!
Since
∗
′
′
′
lim f cx  f x ∗  ≠ 0, qx  f cx exists and qx ∗  ≠ 0.
x→x ∗
So, fx  x − x ∗ qx where lim x→x ∗ qx ≠ 0.
Theorem 5 Let f have n continuous derivatives on a, b . f has a zero of multiplicity m at x ∗ if and
′
only if fx ∗   0, f x ∗   0, . . . , f m−1 x ∗   0 but f m x ∗  ≠ 0.
Example Let fx  e x − x − 1. f has a zero at x  0. Determine the multiplicity m of this zero.
Three ways to determine m.
′
′′
a. Use Theorem 5: find m such that f 0  f 0 . . .  f m−1 0  0 but f m 0 ≠ 0. Check:
′
′
′′
′′
f x  e x − 1, f 0  1 − 1  0. f x  e x , f 0  1 ≠ 0. So, m  2.
b. Use the definition: find m such that fx  x m qx where lim x→0 qx ≠ 0. That is
6
lim
x→0
fx
≠ 0.
xm
x
x
fx
lim x  lim e − xx − 1  lim e − 1  0.
x→0
x→0
x→0
1
x
x
x
fx
lim 2  lim e − x2 − 1  lim e − 1  lim e  1 ≠ 0
x→0 x
x→0
x→0
x→0 2
2
2x
x
So, m  2.
c. Use the definition and Taylor series of fx.
lim
x→0
2
3
1  x  x  x . . .
2!
3!
xm
x
fx
 lim e − xm − 1  lim
x
xm
x→0
x→0
2
3
x  x . . .
3!
 lim 2!
 1 if m  2.
xm
x→0
2
−x−1
So, m  2.
Modified Newton’s Methods:
′
′
Let x ∗ be a zero of multiplicity 2 of f on a, b . Then we know f x ∗   0. Though f x n  ≠ 0, as
′
′
′
x n  x ∗ , f x n  → f x ∗   0. So, f x n  ≈ 0 for large n. In this case, does x n still converge to
x ∗ (theoretically)? If so, what is the order of convergence? Let
fx
.
gx  x − ′
f x
To answer the first question, we need to find out if
lim∗ |g ′ x|  1.
x→x
Observe that
′
′
g x  1 −
f x
2
′′
− fxf x
′
f x
′′
′
lim g x  lim∗
xx ∗
xx
fxf x
′
f x
2

L’Hopital
′

′
 lim∗
xx
2
f x
′
0
0
2
′′
fxf x
′′
f xf x  fxf
′
′′
2f xf x
′′′
x
′′′
fxf x
 1  lim∗
2 xx 2f ′ xf ′′ x
′′′
f xf x  fxf 4 x
 1  1 lim∗
 1 0  1  1
2
′′
′
′′′
2
2 xx
2
2
f x  f xf x
So, x n  still converges to x ∗ . However, since lim xx ∗ g ′ x  1 ≠ 0, x n → x ∗ linearly instead of
2
′
′′
′′′
quadratically. If f x ∗   0, f x ∗   0 but f x ∗  ≠ 0, then
7
′
′′′
f xf x  fxf 4 x
lim∗ g x  1  1 lim∗
2
′′
′
′′′
x→x
2 xx
2
f x  f xf x
′
′′
′′′
f xf x  2f ′ xf 4 x  f xf 5 x
 1  lim∗
′′
′′′
′
4
2 x→x
2 3f xf x  f xf x
′′′
 1  1 lim∗
2
2 x→x
′′
2
f x
 3f ′′ xf 4 x  3f ′ xf 5 x  f xf
′′′
3 f x
2
′′
 4f xf
4
′
x  f xf
5
6
x
x
 1  1  2
2
3
23
′
′′
′′′
4
If f x ∗   0, f x ∗   0, f x ∗   0 but f
′′′
f x
lim g ′ x  1  1 lim∗
x→x ∗
2
2 x→x
 1  1 lim∗
2
2 x→x
2
′′
 3f ′′ xf 4 x  3f ′ xf 5 x  f xf
′′′
3 f x
2f
′′′
x ∗  ≠ 0, then
2
′′
 4f xf
4
′
x  f xf
5
6
x
x
xf 4 x3f ′′′ xf 4 x3f ′′ xf 5 xf ′′′ xf 6 x3f ′′ xf 5 x3f ′ xf 6 xf ′′ xf 7 x
′′′
6f
xf 4 x4f ′′′ xf 4 xf ′′ xf 5 x4f ′′ xf 5 xf ′ xf 6 x
′′′
′′′
′′
6
5f xf 4 x  6f ′′ xf 5 x  f xf x  3f ′ xf 6 x  f xf
 1  1 lim∗
′′′
′′
′
5
6
2
2 x→x
10f xf 4 x  5f xf x  f xf x
2
′′
x
′′′
5f 4 x  5f xf 5 x . . .
1
1

 lim∗
 1  1
2
2 x→x 10f 4 x 2  10f ′′′ xf 5 x . . .
2
2
′
7
Theorem If f x ∗   0, f x ∗   0, … , f
m
1
2
 3
4
m1
x ∗   0 but f
x ∗  ≠ 0, then
1.
lim g ′ x  m m
x→x ∗
Example Consider solving the equation: lnx  1 − x  0 using Newton’s Method with x 0  1.
1.
8
fun@(x) log(x1)-x;
fpfun@(x) 1./(x1)-1;
[xsol,flag,ct]newton(fun,fpfun,1,10^(-5),100)
xsol 
0.5
0.48
1.00000000000000
0.38629436111989
0.46
0.17219218899218
0.44
0.08154040269002
0.03970514407998
0.42
0.01959491748522
0.4
0.00973408497441
0.00485132683348
0.38
1
0.00242175034192
2
3
4
flag 1
y
ct 9
5
6
7
8
x n1 −x ∗
x n −x ∗
 orderconv(xsol,0,9,1)
′
1 − 1  0 when x  0 and f ′′ x  −
1
≠0
1x
1  x 2
when x  0, x  0 is a zero of multiplicity 2 of f. Can we improve the convergence of Newton’s
Method for the problems where m  1?
Clearly, x n  → 0 linearly. Since f x 
Modified Newton’s Methods: newtonM.m
fx 
(1) x n  x n−1 − m ′ n−1
f x n−1 
x n−1 
fx
(2) x n  x n−1 −
where x  ′
.
′
 x n−1 
f x
Let x ∗ be a solution of fx  0. For each method, we will check if
(a) x n → x ∗ ; and
(b) x n → x ∗ quadratically.
′
′′
Let us just check out the case where m  2. By the definition, f x ∗   0 but f x ∗  ≠ 0.
fx
fx
, and g Newton x  x − ′
. Then
(1) Let gx  x − 2 ′
f x
f x
gx  x − m
fx
′
f x
 x2 x−
fx
′
f x
−x
 x  2g Newton x − 2x  2g Newton x − x.
gx ∗   2g Newton x ∗  − x ∗  2x ∗ − x ∗  x ∗ .
So, x ∗ is a fixed point of gx. Since
9
g ′ x  d 2g Newton x − x
dx
 2g ′Newton x − 1
g ′ x ∗   2g ′Newton x ∗  − 1  2 1 − 1  0,
2
x n  converges to x ∗ and converges to x ∗ quadratically.
(2) This method can also be considered as solving x  0 using Newton’s Method where  has a
simple zero at x ∗ .
fx
∗
x  lim
x   lim
x→p
x→p
 x  d
dx
′

′
f x
L’Hopital
′
fx
2
f x

′
f x
′
0
0
′
f x
lim
x→p

′′
f x
′
2
x→x
′
0
0

L’Hopital
′
 1−
′
′
2
f x
fxf x
′
x→x
′′′
2
f x
x
′′
2f xf x
′
1  lim∗
x→x
2
′′
x→x
 1− 1
2
 1 − lim∗
f xf x  fxf
1 − lim∗
fxf x
′′
fxf x
f x
0
′′
′′
x→x
′′
f x ∗ 
′′
− fxf x
f x
 ′ x ∗   lim∗  ′ x  lim∗ 1 −
f x ∗ 
′′
f xf x  fxf
′′
f x
2
′
 f xf
′′′
x
′′′
x
 1 −0  1 ≠ 0
2
2
Since  ′ x ∗  ≠ 0,  has a simple zero at x ∗ and x n  → x ∗ quadratically.
Example Consider solving the equation: lnx  1 − x  0 using both modified Newton’s Methods with
x 0  1.
fun@(x) log(x1)-x;
fpfun@(x) 1./(x1)-1;
0.38
[xsol,flag,ct]newtonM(fun,fpfun,1,10^(-5),100,2)
xsol 
0.36
0.34
1.00000000000000
(1)
0.32
-0.22741127776022
0.3
-0.01951471341458
0.28
-0.00012819465259
0.26
flag  1
0.24
ct  4
0.22
1
orderconv(xsol,0,4,2)
1.2
1.4
1.6
1.8
y
(2) fx  lnx  1 − x,
′
f x 
10
1 − 1  1 − x − 1  −x
x1
x1
x1
2
2.2
x n1 −x ∗
x n −x ∗
2
2.4
2.6
2.8
3
and
′′
1
f x  − x  1 − 2x  −
.
x  1 2
x  1
Define
x 
fx
′
f x

lnx  1 − x
lnx  1 − xx  1
−
−x
x
x1
and
′′
 ′ x  1 −
fxf x
′
f x
2
 1−
′
′
f x
 1
lnx  1 − xx  1
x
 1
lnx  1 − x
x2
fun@(x) -(log(x1)-x).*(x1)./x;
f ′′ x
fx
f x
−
x1
−x
1
x  1 2
0.35
fpfun@(x) 1(log(x1)-x).*(x1)./x.^2;
[xsol,flag,ct]newton(fun,fpfun,1,10^(-5),100)
xsol 
0.3
0.25
1.00000000000000
0.11460991822207
0.2
0.00366562036799
0.15
0.00000445171110
flag 1
ct 4
 orderconv(xsol,0,4,2)
0.1
1
1.2
1.4
1.6
1.8
y
2
x n1
2.2
2.4
2.6
2.8
−x ∗
x n −x ∗
2
Example Use the Newton Method to solve fx  e 3x − 27x 6  27x 4 e x − 9x 2 e 2x for x in 3, 5.
Determine the multiplicity of the zero of fx.
fx  e 3x − 27x 6  27x 4 e x − 9x 2 e 2x
′
f x  3e 3x − 162x 5  274x 3 e x  x 4 e x  − 92xe 2x  2x 2 e 2x 
 3e 3x − 162x 5  27x 3 4  xe x − 18x1  xe 2x
fun@(x) exp(3*x)-27*x.^627*x.^4.*exp(x)-9*x.^2.*exp(2*x);
fpfun@(x) 3*exp(3*x)-162*x.^527*x.^3*(4x).*exp(x)-18*x.*(1x).*exp(2*x);
[xsol,flag,ct]newton(fun,fpfun,3,10^(-8),200)
ct24
xsol(ct)3.73315956142332
for k1:ct-2;
r1(k)abs(xsol(k1)-xsol(ct))/abs(xsol(k)-xsol(ct));
end
plot(r1)
11
3
title(’|x_{n1}-x*|/x_n-x*|’)
for k1:ct-2;
r2(k)abs(xsol(k1)-xsol(ct))/(abs(xsol(k)-xsol(ct)))^2;
end
plot(r2)
title(’|x_{n1}-x*|/x_n-x*|^2’)
|x
|x
-x*|/|x -x*|
k+1
0.75
4000
0.7
3500
0.65
3000
0.6
2500
0.55
2000
0.5
1500
0.45
1000
0.4
0.35
0
k
500
5
10
15
20
25
0
0
So, the Newton Method converges linearly and m  1.
Now let m  2 and use the Modified Newton Method:
 [xsol,flag,ct]newtonM(fun,fpfun,3,10^(-8),200,2);
ct14
xsol(14)3.733124800815943
for k1:ct-2;
r3(k)abs(xsol(k1)-xsol(ct))/abs(xsol(k)-xsol(ct));
end
subplot(1,2,1),plot(r3)
title(’|x_{n1}-x*|/x_n-x*|’)
for k1:ct-2;
r4(k)abs(xsol(k1)-xsol(ct))/(abs(xsol(k)-xsol(ct)))^2;
end
subplot(1,2,2),plot(r4)
title(’|x_{n1}-x*|/x_n-x*|^2’)
12
2
-x*|/|x -x*|
k+1
k
5
10
15
20
25
|x
n+1
-x*|/x -x*|
|x
n
2.2
n+1
-x*|/x -x*|
2
n
700
2
600
1.8
500
1.6
1.4
400
1.2
300
1
0.8
200
0.6
100
0.4
0.2
0
5
10
15
0
0
5
So, the Modified Newton Method converges linearly with m  2.
Now let m  3 and use the Modified Newton Method again:
 [xsol,flag,ct]newtonM(fun,fpfun,3,10^(-8),200,3);
ct9
xsol(9) 3.733078834748490
for k1:ct-2;
r5(k)abs(xsol(k1)-xsol(ct))/abs(xsol(k)-xsol(ct));
end
subplot(1,2,1),plot(r5)
title(’|x_{n1}-x*|/x_n-x*|’)
for k1:ct-2;
r6(k)abs(xsol(k1)-xsol(ct))/(abs(xsol(k)-xsol(ct)))^2;
end
subplot(1,2,2),plot(r6)
title(’|x_{n1}-x*|/x_n-x*|^2’)
13
10
15
|x
n+1
-x*|/x -x*|
|x
n
4
n+1
-x*|/x -x*|
2
n
5
4.5
3.5
4
3
3.5
2.5
3
2
2.5
2
1.5
1.5
1
1
0.5
0
0.5
0
2
4
6
8
0
0
2
4
6
Hence, the Modified Newton Method convergence quadratically and m  3.
14
8
Exercises:
1. Consider solving the equation cosx − x  0 for x in 0,  .
2
(1) Compare numerical results you generated in Problem 4 of Section 2.3 to determine which algorithm
among the Newton Method, the Bisection Method and the Fixed-Point Iteration has the best performance
for this problem.
(2) What can we say algebraically about the performances of these three methods for this problem?
Show your work in detail to support your answer.
2. Let fx  27x 4  162x 3 − 180x 2  62x − 7.
(1) Show first that fx has a zero of multiplicity 3 at x  1 .
3
(2) Use Newton Method to solve this equation with x 0  0 within 10 −8 .
|x n1 − x ∗ |
|x n1 − x ∗ |
(3) Verify using the graphs of the sequences
and
that the iterations
∗
|x n − x |
|x n − x ∗ | 2
obtained in (2) converge linearly. Estimate the asymptotic error constant.
(4) Choose one modified Newton Method to solve this equation with x 0  0 within 10 −8 .
|x n1 − x ∗ |
|x n1 − x ∗ |
(5) Verify using the graphs of the sequences
and
that the iterations
∗
|x n − x |
|x n − x ∗ | 2
obtained in (4) converge quadratically.
3. Let fx  x1 − cosx.
(1) Show first that fx has a zero of multiplicity 3 at x  0.
(2) Use Newton Method to solve this equation with x 0  1 within 10 −8 .
|x n1 − x ∗ |
|x n1 − x ∗ |
(3) Verify using the graphs of the sequences
and
that the iterations
|x n − x ∗ |
|x n − x ∗ | 2
obtained in (2) converge linearly. Estimate the asymptotic error constant.
(4) Choose one modified Newton Method to solve this equation with x 0  0 within 10 −8 .
|x n1 − x ∗ |
|x n1 − x ∗ |
(5) Verify using the graphs of the sequences
and
that the iterations
|x n − x ∗ |
|x n − x ∗ | 2
obtained in (4) converge quadratically.
4. Let fx  1  1 x 2 − x sinx − 12 cos2x.
2
4
a. Verify that x  0 is a zero of fx. Find the multiplicity m of the zero x  0.
b. Use Newton Method to find the zero x  0 of fx within 10 −10 with x 0  −0. 5. Use a proper
Modified Newton Method to find the zero x  0 (quadratically) within 10 −10 with x 0  −0. 5.
c. Use Newton Method to find another zero of fx within 10 −10 with x 0  3. What is the
zero? Estimate the order of convergence  and asymptotic error constant . Is this zero simple? If
not, find a proper Modified Newton Method to find this zero (quadratically) within 10 −10 with
x 0  3. What is the multiplicity of this zero?
5. The sum of two real numbers is 20. If each number is added to its square root, the product of the two
sums is 155.55. Determine the two numbers to within 10 −8 by the Newton Method.
15