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Graeffe's Root SquaringMethod

This is a direct method to find the roots of any polynomial equation with real coefficients. The basic idea behind this method is to separate the roots of the equations by squaring the roots. This can be done by separating even and odd powers of x in 

 Pn(x) = xn + a1 xn-1 + a2 xn-2 + .  .  . + a n-1x + an = 0 

and squaring on both sides. Thus we get,


 (xn + a2 xn-2 + a 4 xn-4 + .  .  . )2 = (x n + a1 xn-1 + a3 xn-3 + .  .  . )2

x2nn - (a12 - 2a2)x2n-2 + (a22 - 2a1a3 + 2a4)x2n-4 +.  .  . +(-1)nan2 = 0

substituting y for -x2 we have 


yn +b1yn-1 + . . . + b n-1y + bn = 0

where 

b1 = a12 - 2a
b2 = a22 - 2a1a3 + 2a4



bn = an2

Thus all the bi's ( i = 0, 1, 2, . . . n) are known in terms of ai's. The roots of this equation are -s12,. -s22, . . ., sn2 where s1, s2 ,  .  .  ., sn are the roots of Pn (x) = 0.

A typical coefficient bk of  bi , i = 1, 2, . . .n  is obtained by following 

The terms alternate in sign starting with a +ve sign. The first term is the square of the coefficient ak . The second term is twice the product of the nearest nebhouring coefficients ai-1 and ai+1. The third is twice the product of the next nebhouring coefficients ai-2 and ai+2. This procedure is continued until there are no available coefficients to form the cross products.

This procedure can be repeated many times so that the final equation


xn + B1 xn-1 +  .  .  . + Bn-1x + Bn= 0

has the roots R1, R2 , .  .  ., Rn such that

Ri = -si (2^m) ,                    i = 1, 2, . . ., m

if we repeat the process for m times.

If we assume |s1| > |s2| > . . . |sn|     then    |R1| >> |R2| >> . . .>> |R n|

that is the roots Ri are very widely separated for large m.

Now we have 

-B1 = åRi @ R
 B2 = åRiRj@ R 1R
-B3 = å RiRjRk@ R 1R2R3
  .
  .
  .
(-1)nBn = R1 R2 . . .R

which gives        Ri = -Bi / B i-1 ,              i = 1, 2, . . . n

where B0 = 1 .

since                 | si |2^m = | Ri |          i = 1, 2, . . . n
Þ     | si | = | Ri |2^-m          i = 1, 2, . . . n

This determines the absolute values of the roots and substitution in the original equation will give the sign of the roots.

Example

Find the roots of   x3 -7x2 +14x - 8 = 0

a[]                1          -7               14                  -8 

b[]                1          21               84                  64 

roots =         4.583             2               0.873

b[]                1         273              4368            4096 

roots =         4.065             2               0.984

b[]                1         65793          1.68E7         1.68E7 

roots =         4.002             2               0.9995

Thus the absolute values of the roots are 4, 2, 1

Since f(1) = 0, f(2) = 0 and f(4) = 0, the signs of the roots 1, 2 and  4 are all positive.


 
Worked out problems
  Find the roots of the following polynomial equations
 Exapmple 1   x2 - 10x + 23 = 0  Solution
 Exapmple 2  x3 + 3x2 - 4 = 0  Solution
 Exapmple 3  x3 - 4x2  + 5x  - 2 = 0    Solution
 Exapmple 4  x3- 6x2 + 11x - 6 = 0  Solution
 Exapmple 5  x3 - x2- x + 1 = 0   Solution
 Exapmple 6  x2+ 6x - 3 = 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