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Birge-Vieta Method
This is an iterative method to find a real root of the nth degree polynomial equation f(x) = Pn(x) = 0 of the form 


an xn + an-1 xn-1 + .  .  . + a1 x + a0 = 0

The theory can be understood better if we consider the above nth degree polynomial in the form


xn + a1 xn-1 + a2 xn-2 + .  .  . + an-1x + an = 0

If s is a real root of Pn(x) = 0 then Pn(x) = (x-s)Qn-1(x) where Qn-1(x) is an (n-1)th degree polynomial of the form


Qn-1(x) = xn-1 + b1 xn-2 +  .  .  . + bn-2x + bn-1.

If p is any approximation to s then Pn(x) = (x-p)Qn-1(x) + R where R is the residue which depends on p.
Now starting with p, we can use some iterative method to improve the value of p such that 
R(p) = 0.
If we apply the Newton-Raphson method with a starting value p0, the iterative scheme can be written as

 
pi+1= pi -   Pn(pi)
                     i = 0,1,2...
 P'(pi)
Now by comparing the coefficients of  Pn and (x-p)Qn-1(x) + R we get
a1 = b1 - p Þ b1 = a1 + p
a2 = b2 - pb1 Þ b2 = a2 + pb1
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ai = bi - pbi-1 Þ bi = ai + pbk-1
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an = R - pbn-1 Þ R = bn = an + pbn-1
(or)
bi = ai + pbi-1 i=1,2,...n
with b0=1 and R = bn=Pn(p)
To find P'n(p), let us differentiate the equation


   bi = ai + pbi-1

with respect to p


Þ                 dbi / dp = bi-1 + p (dbi-1 / dp)

if we substitute (dbi / dp) = ci-1  then


Þ            ci-1 = bi-1 + pci-2              (or)

                                      ci     = bi + pci-1            i=1, 2, . . ., n-1

Then the  cn-1 obtained from the last equation is nothing but 


ci-1 = dbn / dp = dR / dp = P'n(p)

That is the Newton's method now can be written as
pi+1 = pi - bn / cn-1
On convergence this iterative process will give one root p of the polynomial equation Pn(x) = 0. Now the deflated polynomial equation Qn-1(x) = 0 can be used to find the other real roots. 
This method is often called as Birge-Vieta method.

 
Example :
Find the real root of  x3 - x2 - x + 1 = 0
In this problem the coefficients are a0 = 1, a1 = -1, a2 = -1, a3 = 1
Let the initial approximation to p be p0 = 0.5
a0
a1
a2
a3
p0 = 5
+1
-1
-1
+1
+0.5
-0.25
-0.625
+1
-0.5
-1.25
+0.375
 (b4 = R) 
+0.5
  0
-0.625
+1
  0
-1.25
   (c2 = R')

 
 
 
p1 = p0 - b4 / c3  = 0.5 -  0.375   = 0.5 +  0.375   = 0.5 + 0.3 = 0.8
-1.25  1.25

 
 
 
a0
a1
a2
a3
p1 = 0.8
+1
-1
-1
+1
+0.8
-0.16
-0.928
+1
-0.2
-1.16
+0.072
 (b4 = R)
+0.8
+0.48
+1
+0.6
-0.68
 (c2 = R')

 
 
p2   = 0.8 -  0.072  = 0.8 +  0.072   = 0.8 + 0.1059 = 0.9059
-0.68  0.68

 
 
a0
a1
a2
a3
p2 = 0.9059
+1
-1
-1
+1
+0.9059
-0.0852
-0.9831
+1
-0.0941
-1.0852
+0.0169
 (b4 = R)
+0.9059
+0.7354
+1
+0.8118
-0.3498
 (c2 = R')

 
 
p3  = 0.905 -  0.0169  = 0.905 +  0.0169   = 0.905 + 0.0483 = 0.9533
-0.3498  0.3498
The exact root is 1.0
Worked out problems
  Find a root and the corrponding polynomial factor for the following polynomial equations
 Exapmple 1   x4 - 3x3 + 3x2 - 3x + 2 = 0  Solution
 Exapmple 2  x4-x-10 = 0  Solution
 Exapmple 3  x3 - 6x2  + 11x  - 6 = 0  Solution
 Exapmple 4  x3 - 4x2 + 5x - 2 = 0  Solution
 Exapmple 5  x4 - x3 + 3x2 + x - 4 = 0  Solution
 Exapmple 6  x3 - x - 4 = 0  Solution
Problems to workout

 
Work out here :
Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations