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 |
.
.
. |
|
.
.
. |
|
.
.
. |
|
.
.
. |
| ai |
= |
bi - pbi-1 |
Þ |
bi |
= |
ai + pbk-1 |
.
.
. |
|
.
.
. |
|
.
.
. |
|
.
.
. |
| 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 :
|
