Roots of Functions and Polynomials:
The transfer function obtained by using polynomial approximation will be presented in the symbolic form by the rational function
F ( s )
 i n 

0 m  i

0 a i s i b i s i

N ( s )
D ( s )
In many applications we need to know zeros and roots of this function, so the following form is desired
F ( s )

K b
 i

1
 s
 z i
 i m


1
 s
 p i
 where z i
are roots (zeros) of N(s) and p i
are roots of D(s). The best known general method for finding roots is the Newton-Raphson method. This is an iterative method which uses function and its derivative values. We start with an initial point x
0
and expand the function f(x) around this point: f
   
1 0
 f '
 
0 x
1
 x
0


...
If we neglect higher order terms and want to find x, such that f(x
1
) = 0 then
 x
1
 x
0
  
0
  f
 
0 and x
1
 x
0
 f f
'
  x
0
 
0
Because this is an approximated solution we have to iterate. The iterations can be illustrated as shown below on the following fig.

The Newton-Raphson method is defined by the following algorithm.
1.
Set k = 0 and select x
0
.
2.
Calculate
 x k
  f
   
3.
Calculate x k+1
= x k
+ k
 x k k
4.
If
 x
  k
stop, else set k = k + 1 and go to step ____
The method averages quickly to the solution, provided that initial point x
0
was correctly chosen.
The method which always averages to a root of polynomial F
1
is called Laguerre's method and is defined by
X k

1 
X k 
P
1 n k nP n
( x

)
 
H k where
H k 
 n

1

 n

1
 
P
1 n
  
2  nP n
    n
"
 and P n
(x) is a polynomial of degree n. It is advantageous to first compute roots with small absolute values, thus initial estimates should start with x 0 = 0.
Example (Laguerre's method)
Find one root of the polynomial
P
5
(x) = 8x 5 + 12x 4 + 14x 3 + 12x 2 + 6x + 1 we have P
5
1
( x )

40 x
4 
48 x
3 
42 x
2 
26

16 and P
5
11 ( x )

160 x 3 start at x
0
= 0.

144 x 2 
84 x

26
Calculate P
5
(x 0 ) = 1
H x
1
0

 x
4
0

4


6
P
1
5
2 
5
5 P
5  

1 x

0
26

H
0



56
6


P
1
5
5 .
1
7 .
48
H
0


6

7 .
48

0 .
37
 selected to have x
1
P
11
5
 x
0

26 small

Second iteration: P
5

0 .
02 P
5
1
H
1 x
2 

4

4 x
1 


0 .
45

2
P
1
5
5 P
5

 x
1
5
 

0 .
02


H
1
6 .
53

0 .
37



0 .
628
5

0 .
02
0 .
45

0 .
79

0 .
45
H
1 
0 .
79
 
0 .
45
P
5
11
Third iteration: P
5

0 .
0012 P
5
1
H
2 x
3 
 x
2
4

4



0 .
071

2
5 P
5 x

2
5
P
5
1 

0 .
0012

H
2
 
0 .
45
2 .
78



0 .
014
5

0 .
0012
0 .
071

0 .
118
Solution is: x
1
 x
2
 x
3
 
0 .
5 x
4
  j

0 .
071
H
2
 j
P
5
11

0 .
118
 
0 .
48 x
5

6 .
5

2 .
78
I.
Solution of a nonlinear equ. by Taylow expansion and local inverse
Assume that we are solving f(x) = 0.
We know that a local inverse x(f) can be determined if f
' 
0 .
Then x
' f 0

1 f
' x
0
If we differentiate this equation we can get higher order derivatives. So
'' x f
0
 
  x
0

2
'' f x
0
 dx df f
0
 
  x
0

3 f x
''
0 also x
' '' f 0

3
      x 0 x 0 x 0

4 f
' '' x 0 and x


' v  
15 f
0
4
10 f
  f x
0
  x
0

6

6 f f
      x f
'' x
0 x
''
0
0 f f x
' ''
0 x
' ''
0

 x f
0
  x
0

5 f iv x
0 f
  x
0

5 f x iv
0 x
0


6 f
' x

15
''
0 f f x
''
0
    z
0 x
0
3 etc.


1

Using local inverse we can approximate x
   
0
 x
1 f
0
 f

1
2
'' x f
0
 f
2 
1
6
' '' x f
0
 f
3 
1
24 x iv f
0
 f
4
Now, assume that at a given x
0
we evaluated f
 
0
 f x
0

...
and derivatives f
' x
0
, f
'' x
0
, f
' '' x
0
, f iv x
0
...
In order to solve f(x) = 0 which gives xˆ we will simply find value x f

0

* x
*
If the x(f) is a true inverse of f(x) then x

 x and we get solution without iterations.
Since all derivatives of x(f) are calculated at the nominal point f
0
, then to change argument of to zero we use
 f
  f
0
.
So for instance if we limit expansion to 1st derivative we get Newton-Raphon method x
 x
0
 x
'  f
 x
0

 f f
'
  x
 x
 x
0

 f f
1
  f f
0
0
'
For the first two derivatives we have x
 x
0
 x '
 f
 x
2
''
 f
2  x
0
 f f
0
0
'




'' f f
 
3

0
0 so
 x
  f
0 f
0
'


1
 f
0
2
'' f
 
0
0
2


For the first three derivatives we have x
 x
0
 x
'  f
 x
''
2
 f
7  x
' ''
6
 f
3  x
0
 f
0 f
0
'


1
 f
0
2 f

   
0
0
'' f
3 f
0
0
''
2
 f
0
' '' f
0
' f
0
''



