how to do polynomial division
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hi...
i wanna do polynomial divison given numerator=x^5+x^4+x^3 and
denominator=x^3+x+1 ;remainder should be x...how to implement it in matlab..
2 commentaires
Walter Roberson
le 3 Nov 2011
Will the coefficients always be either 0 or 1 ? A binary polynomial?
faa nad
le 5 Nov 2011
Réponses (4)
Daniel Baboiu
le 3 Nov 2011
0 votes
You have two choices: 1. Use the Symbolic Math Toolbox 2. Store all coefficients as a vector (including the coefficients which are 0), then use this representation to implement division steps as described below: http://en.wikipedia.org/wiki/Polynomial_long_division
2 commentaires
faa nad
le 5 Nov 2011
Walter Roberson
le 5 Nov 2011
In mathematics, we mentally extract the coefficients in order to do the division.
Andrei Bobrov
le 5 Nov 2011
[a,b]=deconv([1 1 1 0 0 0],[1 0 1 1])
add
p1=[1 1 1 0 0 0]
p2=[1 0 1 1]
[a b] = deconv(p1,p2)
syms x
k = cellfun(@(y) y*x.^(numel(y)-1:-1:0).',{a b p2},'un',0)
k = [k{:}]
out = k(1) + k(2)/k(3)
1 commentaire
faa nad
le 5 Nov 2011
Walter Roberson
le 5 Nov 2011
0 votes
As you have restricted this to symbolic expressions without ever extracting the coefficients (at least not in code you write, even if it gets done "under the hood"), then the solution is to use the MuPAD Standard Library function pdivide
I could offer a very nice and efficient calculation for polynomials up to order 52 where the coefficients are all 0 or 1, if we are allowed to extract the coefficients in the code (which you could stuff in to a subroutine and never look at again), but I gather that efficiency and simplicity are not important for your purposes.
5 commentaires
faa nad
le 6 Nov 2011
Walter Roberson
le 6 Nov 2011
pdivide is a member of the MuPAD Standard Library, and must be invoked within MuPAD. There is no direct interface between it and MATLAB. You can access it using evalin(symengine) or feval(symengine)
feval(symengine, 'pdivide', x^5+x^4+x^3, x^3+x+1)
faa nad
le 6 Nov 2011
Walter Roberson
le 6 Nov 2011
Do you have the symbolic toolbox installed and licensed? If not, then you cannot do what you are asking for, as only the symbolic toolbox hides extracting the coefficients of polynomials.
Maria Maximina
le 21 Fév 2014
hi! o have one question for you! i know it is long time ago.. but anyway.. jajaja if u do that operation that you suggested, and you get a vector like:
[a,b,polinom]
what do actually a and b mean??? thanks!
Ahmed J. Abougarair
le 18 Nov 2022
0 votes
syms x y
p = x^3 - x*y^2 + 1;
d = x + y;
[r,q] = polynomialReduce(p,d)
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