Integrating from polyint(p)

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Tatte Berklee
Tatte Berklee am 3 Aug. 2020
Kommentiert: Tatte Berklee am 4 Aug. 2020
Hi folks!
I have a question regarding univariate integration from the coefficients produced (e.g. 1x8 double) from polynomial integration.
An example code I have is:
[p,S] = polyfit(x,y,N);
q=polyint(p);
At this point, the code produces something like 1x8 double with polynomial coefficients.
But the ultimate integral I would like to compute is an indefinite integral with lower and upper limit being 1 and inf respectively in the form of a fraction:
x/(the polynomial I got for q with x being the variable).
How would I do this?

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John D'Errico
John D'Errico am 4 Aug. 2020
Bearbeitet: John D'Errico am 4 Aug. 2020
Sorry, but polyint cannot integrate a rational polynomial, and certainly not over an infinite region. Depending on the roots of the polynomial, it may or may not have an integral.
The simplest way to do your integration is to use the symbolic toolbox.
syms x
P = x^4 + 8*x^3 + 23*x^2 + 28*x + 12;
int(x/P,x,[1,inf])
ans =
log((3*2^(1/2))/8) + 2/3
If you have only the coefficients of the polynomial, you can still do it.
pcoeff = [1 8 23 28 12];
roots(pcoeff)
ans =
-3
-2
-2
-1
So a happy list of roots.
int(x/poly2sym(pcoeff,x),[1,inf])
ans =
log((3*2^(1/2))/8) + 2/3
  1 Kommentar
Tatte Berklee
Tatte Berklee am 4 Aug. 2020
John, thank you. Since I know exactly to which order my polynomial goes, I did something like: (e.g. n=6)
fun = @(x) x/(p(1,1)*x.^6+p(1,2)*x.^5+p(1,3)*x.^4+p(1,4)*x.^3+p(1,5)*x.^2+p(1,6)*x+p(1,7)).^2;
exp = integral(fun,0,Inf,'ArrayValued',true);
Does my code look valid?

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