matlab,numerical-methods,numerical-integration

You are changing the parameters of your your ODEs discontinuously in time. This results in a very stiff system and less accurate, or even completely wrong, results. In this case, because the your ODE is so simple when I = 0, an adaptive solver like ode45 will take very large...

matlab,plot,numerical-integration

Yes, you can use trapz to integrate over set of discrete data: http://www.mathworks.com/help/matlab/math/integration-of-numeric-data.html Assuming your x-axes data is saved to vector t, and your best fit line data is saved to fx, then the area is easily calculated like: area = trapz(t, fx); ...

algorithm,numerical-methods,numerical,numerical-integration

numerical integration always return just a number if you do not want the number but function instead then you can not use numerical integration for this task directly Polynomial approach you can use any approximation/interpolation technique to obtain a polynomial representing f(x) then integrate as standard polynomial (just change...

octave,numerical-methods,numerical-integration

I have tried your function in Matlab. Your code is not stalling. It is rather that the size of trap_1_midpoints increases exponentionaly. With that the computation time of trap_1 increases also exponentionaly. This is what you experience as stalling. I also found a possible bug in your code. I guess...

python,algorithm,recursion,numerical-methods,numerical-integration

Your code is refining to meet an error tolerance in each individual subinterval. It's also using a low-order integration rule. Improvements in both of these can significantly reduce the number of function evaluations. Rather than considering the error in each subinterval separately, more advanced codes compute the total error over...

I figured out a wrap around function that will force the correct output, where funMatch and funNoMatch are 4d functions and my limits were stored in Limit. f = @(r1, t1, r2, t2) selective(funMatch, funNoMatch, r1 == r2, r1, t1, r2, t2); res = integralN(f, Limit(1,1), Limit(1,2), Limit(2,1), Limit(2,2), Limit(3,1),...

c,loops,for-loop,integration,numerical-integration

I suggest you move the variable initialisation to within the for loop like this: int main(void) { int N; double upper, lower, total; for(N=2;N<20000;N+=100) { upper = DBL_EPSILON*2; lower = 0; total = 0; while(upper<FLT_MAX) { total=total+tra(upper, lower, N); lower=upper; upper=upper*2; } printf("Integral is %.10f\n", total); } return 0; }...

matlab,interpolation,triangulation,numerical-integration

You could assume your function to be piecewise linear on your integration area and then integrate it using a midpoint quadrature rule: For every triangle you compute the midpoint value as the mean of the nodal values and multiply by the triangle's area. You sum it all up to get...

python,scipy,constraints,ode,numerical-integration

if algebraic manipulation fails, you can go for a numerical solution of your constraint, running for example fsolve at each timestep: import sys from numpy import linspace from scipy.integrate import odeint from scipy.optimize import fsolve y0 = [0, 5] time = linspace(0., 10., 1000) F_lon = 10. mass = 1000....

X = eps:0.01:1; Y = (X.^3)./(exp(X)-1.0); Z = trapz(X,Y); ...

matlab,ode,numerical-integration

The error is pretty clear: arrays of function handles aren't allowed in Matlab. You may be able to use cell arrays in some cases. See this question. I'm not sure that the ODE suite functions support the evaluation of separate events functions in a cell array (R2013b gives me an...

matlab,ode,differential-equations,numerical-integration

Summary: There are problems in integrating Hamiltonian systems with normal numerical integrators, and your special initial conditions aggravate this to the point where the numerical solution has no resemblance with the correct one. There's nothing wrong with your implementation per se. However, the initial conditions you use are not the...

c++,integral,numerical-integration

Unfortunately Simpson's rule can't be applied directly to multiple integrals. What you need to do is derive interpolant surfaces or hypersurfaces for double or triple integrals, respectively. For a double integral you end up evaluating the function on a grid of nine points instead of the three points you use...