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 Kommentare
Walter Roberson
am 3 Nov. 2011
Will the coefficients always be either 0 or 1 ? A binary polynomial?
faa nad
am 5 Nov. 2011
Antworten (4)
Daniel Baboiu
am 3 Nov. 2011
0 Stimmen
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 Kommentare
faa nad
am 5 Nov. 2011
Walter Roberson
am 5 Nov. 2011
In mathematics, we mentally extract the coefficients in order to do the division.
Andrei Bobrov
am 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 Kommentar
faa nad
am 5 Nov. 2011
Walter Roberson
am 5 Nov. 2011
0 Stimmen
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 Kommentare
faa nad
am 6 Nov. 2011
Walter Roberson
am 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
am 6 Nov. 2011
Walter Roberson
am 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
am 21 Feb. 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
am 18 Nov. 2022
0 Stimmen
syms x y
p = x^3 - x*y^2 + 1;
d = x + y;
[r,q] = polynomialReduce(p,d)
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