Numerical values of integrals

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Qian Feng
Qian Feng il 7 Nov 2016
Modificato: Qian Feng il 1 Set 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 Commenti
John D'Errico
John D'Errico il 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.
Qian Feng
Qian Feng il 17 Nov 2016
Sorry, now the code should be correct

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Risposta accettata

Walter Roberson
Walter Roberson il 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 Commenti
Walter Roberson
Walter Roberson il 30 Dic 2016
Modificato: Walter Roberson il 24 Set 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.
Qian Feng
Qian Feng il 4 Gen 2017
Right, so I need larger valued of digits for vpa function.

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Più risposte (1)

Karan Gill
Karan Gill il 30 Nov 2016
Modificato: Karan Gill il 17 Ott 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 Commenti
Qian Feng
Qian Feng il 24 Set 2017
Thanks Walter, so what number do you suggest that I should choose for the AbsTol and RelTol in this case?
Walter Roberson
Walter Roberson il 24 Set 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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