GCD of Polynomials

Find polynomial GCD by "Leading-coefficient Elinimation"

Description

In the longhand polynomial division given as
      P(k) = P(k-2) - P(k-1)*Q(k)
The quotient Q(k) and the remainder P(k) are obtained from dividing the dividend P(k-2) by the divisor P(k-1). If we can make Q(k) = 1, by converting P(k-2) and P(k-1) into equal degree and monic, then the longhand polynomial division becomes simply the "monic polynomial subtraction" (MPS):
      P(k) = P(k-2) - P(k-1)
For a pair of given polynomials p(x) and q(x) of degree n and m, n>m, we set
      P(1) = p(x)/p_0
      P(2) = q(x)*x^(n-m)/q_0
Applying the MPS repeatedly starting from k=3, until k=K+1, such that
      P(K+1) = P(k-1) - P(k) = 0
then we get our desired polynomial GCD as
      gcd(p,q) = P(K).
The source code uses only basic MATLAB built-in functions. Its listing is only 17 lines total !
Amazingly, this simple routine gives the expected results for the test polynomials and their derivatives of very high degree, such as
      p(x) = (x + 1)^1000
      p(x) = (x + 123456789)^30
      p(x) = (1234x + 56789)^60
      p(x) = (x^4-2x^3+3x^2-4x +5)^50
      p(x) = (x^4 - 1)^25

  *************** UPDATE (10/05/09): **************
The approach "Leading-coefficient Elinimation" is revised from the original "Monic Polynomial Subtraction".
It also reduces almost half of the total arithematic operations.
The total source code listing is only 12 lines!

Acknowledgements

This file inspired Polynomials With Multiple Roots Solved.

• MATLAB release: MATLAB 6.5 (R13)

[Feng Cheng Chang]

Pozos:
To answer your question, first find r the GCD of two given polynomials p and q, then, u=p/r, v=q/r and w=u*v.
The partial fraction expansion (PFE) of 1/w will give 1/w=y/u+x/v,
where the two desired polynomials x and y can thus be determined after performing reverse PFE. Thus,
1 = x*u+y*v, or
r=x*p+y*q.
In addition, we may also find the two polynomials t and s, so that
1=u/t+v/s, or
r=p/t+q/s.
It is interesting to note that x and y are expected to be, respectively, equal to 1/t and 1/s. However, it isn't so !
Please let me know if you want find out more detail derivation about these relations.
FC Chang

[bullpoz Pozos]

FC Chang:
I would like a favor.I've made a gui that represents the gcd
it is like this: r=poly_gcd(p,q)
i've been asked to make this:
r=xp+ψq
if you understand can you tell me how to show this expression out of the gcd code
Thanks

[Feng Cheng Chang]

Pozos:
I do not understand your question. But if you means that you want make a gui from my article, you may go ahead to do so.
Please see my other articles, such as "Solve multiple-root polynomials' that may refer computation of polynomial GCD.
FC Chang

[bullpoz Pozos]

Who can i make a gui of that?please help!