# Roots of Equations ```Approximating Roots of Equations
Roots of single equations can always be written as solutions of the form f ( x)  0 . For example,
if f ( x)  ax 2  bx  c  0 , there will be two possible roots given by
 b  b 2  4ac
.
2a
But only polynomials to the fourth degree can be solved using algebraic operations. As an
example of a more complicated case consider f ( x)  tan x  x  0 .
x
There are no algebraic solutions. The roots must be found numerically
Roots by Bisection: For f ( x)  0
Bisection Algorithm
Begin
1. Set n=0
2. Find a0 &amp; b0
3. Set m=(an+bn)/2
4. If f(a0)f(m)&lt;0 then
an+1=an, bn+1=m
Else
an+1=m, bn+1= bn
5. End if
6. Set n=n+1
7. If |bn-an|&gt;then
Go to step 3
Else
Root=(an+bn)/2
End if
End
Linear Interpolation
By similar triangles
x3  x1
0  f ( x1 )

x2  x1 f ( x2 )  f ( x1 )
or
x3  x1  ( x2  x1 )
f ( x1 )
x f ( x2 )  x2 f ( x1 )
 1
f ( x2 )  f ( x1 )
f ( x2 )  f ( x1 )
Continuing
x4 
x2 f ( x3 )  x3 f ( x2 )
f ( x3 )  f ( x 2 )
Then
xn 1 
xn1 f ( xn )  xn f ( xn1 )
f ( xn )  f ( xn 1 )
Newton’s Method
Using a Taylor series
f ( x)  f ( x0 )  ( x  x0 ) f ' ( x0 )
Setting
f ( x1 )  0 ,
and using the Taylor series
f ( x0 )
x1  x0 
.
f ' ( x0 )
Similarly
f ( x1 )
x2  x1 
f ' ( x1 )
and in general
f ( xn )
xn1  xn 
f ' ( xn )
For example to find
2 use f ( x)  x 2  2  0 . Then f ' ( x)  2 x and
f ( xn )
x  2 2 xn  xn  2 xn  2 xn 1
xn1  xn 
 xn  n




f ' ( xn )
2 xn
2 xn
2 xn
2 xn
2
2
2
2
Take x0  0 then
x1 
1
 1  1.5
2
x2  0.75  0.666...  1.41666...
x3  1.4142157
x4  1.412136
x5  1.412136
As a second example consider f ( x)  e x  2 x  0 then f ' ( x)  e x  2 .
Take x0  0, f(x0 )  e 0 -2  0  1, f '(x0 )  e 0 -2  1 then
1
x1  0 
 1, f ( x1 )  e  2, f '( x1)  e  2 then
1
x2  1  1  0  x3  1, x4  0, x5  1, ...
As a last example consider f ( x)  tan x  x, f ' ( x)  sec 2 x  1  tan 2 x .
1
3 
Take x0    
  3.927 then x1  6.854, x2  21.92, x3  5,551, x4  18,700 ...
2
2 
Convergence of Newton’s Method: We can write
f ( xN )
xN 1  g ( xN )  xN 
.
f ' ( xN )
If x  r is a root, then f (r )  0 and r  g (r ) . The error can be found by subtracting, then
x N 1  r  g ( x N )  g (r ) ,
is the error. Performing a Taylor series expansion about x N  r ,
1
g ( x N )  g (r )  g ' (r )( x N  r )  g ' ' ( )( x N  r ) 2
2
f (r )
where r    x N . We can find the terms using g (r )  r 
then
f ' (r )
f ' (r ) f (r ) f ' ' (r ) f (r ) f ' ' (r )
g ' (r )  1 


0
f ' (r )
 f ' (r )2
 f ' (r )2
since f (r )  0 . Hence
1
g ( x N )  g (r )  g ' ' ( )( x N  r ) 2
2
Gives the error to second order as
E N  x N 1  r  g ( x N )  g (r ) 
1
g ' ' ( )( x N  r ) 2
2
Fixed Point Iteration
Sometimes is referred to as back substitution. Recall from Newton’s method that the root used
xN 1  g ( xN ) and converged at the root r when r  g (r ) , or at f (r )  r  g (r )  0 .
As an example consider f ( x)  x 2  2 x  3  0
1) The exact solution can be found by factoring.
f ( x)  ( x  3)( x  1) =0.
Then the solutions are: x  1, 3 .
2) Choose x2  2x  3 or xN 1  2 xN  3  g ( xN ) .
For x0  4 then
x1  3.316, x2  3.104
x3  3.034, x4  3.011
x5  3.004, x6  3.001, ...
3) Writing
f ( x)  x( x  2)  3  0 then we can set xN 1 
3
.
xN  2
For x0  4 then
x1  1.500, x2  6.000
x3  0.375, x4  1.263
x5  0.919, x6  1.028
x7  0.991, x8  1.003
4) Now try
x N 1
2
xN
3

2
Again take x0  4 then
x1  6.5
x2  19.635
x3  191.0
Convergence Criteria for Fixed Point Iteration
For a root at r  g (r ) and x N 1  g ( x N ) the error is
x N 1  r  g ( x N )  g (r ) 
By the mean value theorem g ' ( ) 
g ( x N )  g (r )
x N  r 
xN  r
g ( x N )  g (r )
, where x N    r (for x N  r ).
xN  r
Let eN  x N  r then eN 1  x N 1  r  g ' ( )eN or eN 1  g ' ( ) eN .
Hence
e N 1
eN
 g ' ( ) and at the root
eN 1
 g ' (r )
eN
Hence if g (x ) and g ' ( x ) are continuous in the interval about the root x  r of x  g (x ) , then
x  g (x) for N  1, 2, 3,... will converge to the root x  r provided that x1 is chosen in an
interval where g ' ( x)  1 .
The condition is sufficient but not necessary.
Convergence Criteria for Newton’s Method
Recall that Newton’s method can be written as
x N 1  x N 
f (xN )
 g(xN ) .
f ( x N )
Since
g ' ( x) 
f ( x) f ' ' ( x)
 f ' ( x)2
convergence will occur when
f ( x) f ' ' ( x)
 f ' ( x)2
when x  r .
1
,
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