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sage integration differes from Mathematica

asked 2012-01-30 00:22:54 -0500

anonymous user


Hi, I'm trying to integrate

E(x) = ae^(-0.5(x/s)^2)cos(fx+phi)


A = Integrate[a Exp[-0.5 x^2/[Sigma]^2] Cos[f x + [Phi]], x]

gives me

0.626657 a E^(-0.5 f^2 [Sigma]^2) [Sigma] (Erf[(0.707107 x)/[Sigma]+(0.+0.707107 I) f [Sigma]] (Cos[[Phi]]-(0.+1. I) Sin[[Phi]])+Erfi[((0.+0.707107 I) x)/[Sigma]+0.707107 f [Sigma]] ((0.-1. I) Cos[[Phi]]+Sin[[Phi]]))


E = ae^(-0.5(x/s)^2)cos(fx+phi)

A = integrate(E, x)

Sages result is a different one (sorry, cannot copy output, but you can easily reproduce it yourself). The difference isn't obvious at first glance (at least not to me), but you can see it doing e.g.

A /. {a -> 100, f -> 1, [Sigma] -> 8, [Phi] -> 0, x -> 0}

gives 3.11179*10^-15 + 0. I

and Sage

show(A(a=100, f=1, s=8, phi=0, x=0).n())

results in 1.26977421759869e10-11+101.642197289814i

The result differs at least at zero. Anyone any idea what I'm doing wrong?



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answered 2012-01-30 05:08:56 -0500

achrzesz gravatar image

The indefinite integral of a given function is only defined up to an additive constant

(and your results differ by a constant)

The definte integral, for example

sage: ii=integrate(100exp(-1/2x^2/8^2)*cos(x),x,0,+oo);

sage: ii


is the same in Mathematica (WolframAlpha)

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Agree. Thanks haven't thought about the constant, which is of course right, however I'm not sure if that's actually the problem: Mathematica (for the above function): A(0) = 3.11179*10^-15 + 0. I A(1) = 83.9607 + 0. I A(20) = 3.15709 + 0. I Sage: A(0) = 1.26977421759869e-11+101.642197289814i A(1) = 83.9606807305601 A(20) = 3.15709049993006 To me that doesn't look like a constant... Furthermore I cannot plot the stuff ("plot(A(a=100, f=1, s=8, phi=0), (x, 10, 16))"), with this error: verbose 0 (4190:, generate_plot_points) WARNING: When plotting, failed to evaluate function at 200 points. verbose 0 (4190:, generate_plot_points) Last error message: 'unable to simplify to float approximation' Thanks Ben

BenMoon gravatar imageBenMoon ( 2012-01-31 01:40:57 -0500 )edit

answered 2012-01-31 12:21:28 -0500

achrzesz gravatar image

Hi Ben

For special values of the constants there are no differences in the integral and the plot:


L=[[l,numerical_integral(100exp(-1/2x^2/8^2)*cos(x),0,l)[0]] for l in srange(-6,6,0.1)];


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Cheers man! So I've got to evaluate the stuff numerically before plotting. Usually I only do numpy and rarely sympy and recently had a look into sage again after some time. I start liking it ;) Thanks Ben

BenMoon gravatar imageBenMoon ( 2012-01-31 20:20:48 -0500 )edit

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Asked: 2012-01-30 00:22:54 -0500

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Last updated: Jan 31 '12