Numerical values of integrals

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Qian Feng
Qian Feng on 7 Nov 2016
Edited: Qian Feng on 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 Comments
Qian Feng
Qian Feng on 17 Nov 2016
Sorry, now the code should be correct

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Accepted Answer

Walter Roberson
Walter Roberson on 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 Comments
Qian Feng
Qian Feng on 4 Jan 2017
Right, so I need larger valued of digits for vpa function.

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More Answers (1)

Karan Gill
Karan Gill on 30 Nov 2016
Edited: Karan Gill on 17 Oct 2017
Use "vpaintegral" introduced in 16b: https://www.mathworks.com/help/symbolic/vpaintegral.html.
F2 = vpaintegral(ep2*ep2',x,-r2,-r1)
F2 =
[ 1.53919e-23, 2.0475e-23]
[ 2.0475e-23, 2.73446e-23]
  4 Comments
Walter Roberson
Walter Roberson on 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.

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