Numerical values of integrals
6 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Qian Feng
am 7 Nov. 2016
Bearbeitet: Qian Feng
am 1 Sep. 2021
I am dealing with a problem of finding accurate numerical values of integrals. Specifically, the integral is introduced by using the best approximation scheme (Legendre Polynomials) to approximate a vector valued function whose indefinite integral is not easy to be explicitly written down. The code is provided as follows:
r1 = 0.7; r2 = 1; r3 = r2 - r1; d = 8;
syms y real
le = [];
for i = 0:d
l = [coeffs(legendreP(i,(2/r1)*y+1)) zeros(1,d-i)];
le = [le; l];
end
leg = [];
for i = 0:d
l = [coeffs(legendreP(i,(2*y + r1 + r2)/r3)) zeros(1,d-i)];
leg = [leg; l];
end
syms x
t = [];
for i = 0 : d
t = [t ; x^i];
end
xp = t;
lp = le*xp; la = leg*xp;
fi1 = [exp(sin(x)); exp(cos(x))]; fi2 = [sin(x^2); cos(x^2)];
ny1 = size(fi1,1); ny2 = size(fi2,1); ny = ny1 + ny2;
ga1 = fi1*lp'; ga2 = fi2*la';
Ga1 = double(int(ga1,x,-r1,0)*diag(2.*(0:d)+1)*(1/r1));
Ga2 = double(int(ga2,x,-r2,-r1)*diag(2.*(0:d)+1)*(1/r3));
ep1 = fi1 - Ga1*lp; ep2 = fi2 - Ga2*la;
E1 = double(int(ep1*ep1',x,-r1,0)); E2 = double(int(ep2*ep2',x,-r2,-r1));
The code works fine until d = 8 when an error is returned to state that DOUBLE cannot convert the input expression into a double array. If the input expression contains a symbolic variable, use VPA.
I have tried vpa function but the same problem happens still.
One may suggest to use integral numerical integration instead of int. However, the numerical integration seems produce inaccurate result compared to the symbolic representation. Note that the error matrix E1 and E2 become extremely small when the approximation degree d becomes large.
To summarize, the problem here is how to extract, or accurately calculate if anyone has suggestions, the numerical values of E1 and E2.
Thanks a lot!
2 Kommentare
John D'Errico
am 7 Nov. 2016
You say that we can try it ourselves, but since there are undefined variables, we cannot do so.
Undefined function or variable 'n'.
So trying it ourselves stops at line 1.
Akzeptierte Antwort
Walter Roberson
am 7 Nov. 2016
If you change your first line to
syms n
r1 = 0.7; r2 = 1; r3 = r2 - r1; d = 8; di = n*(d+1);
then you can complete down through E1. After that, unfortunately MATLAB does not know how to do the integral, even though there is a closed-form solution for it. You will need to switch to numeric:
FF = ep2*ep2';
FF1 = matlabFunction(simplify(FF(1));
FF2 = matlabFunciton(simplify(FF(2));
E2(1) = integral(FF1, -r2, -r1);
E2(2) = integral(FF2, -r2, -r1);
26 Kommentare
Walter Roberson
am 30 Dez. 2016
Bearbeitet: Walter Roberson
am 24 Sep. 2017
No, you need the number of digits of vpa to remain high so that vpa is able to converge. But then you can transform those higher number of symbolic digits into a numeric floating point value with double().
... or just use vpaintegral() directly if you have a new enough MATLAB.
Weitere Antworten (1)
Karan Gill
am 30 Nov. 2016
Bearbeitet: Karan Gill
am 17 Okt. 2017
F2 = vpaintegral(ep2*ep2',x,-r2,-r1)
F2 =
[ 1.53919e-23, 2.0475e-23]
[ 2.0475e-23, 2.73446e-23]
4 Kommentare
Walter Roberson
am 24 Sep. 2017
My tests with a different package suggested that 20 or so digits of precision was needed to get a decent result, so 1e-20 like you used for Ga1 might be enough.
Siehe auch
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!