HP Prime Numerical Integration

08032022, 06:15 PM
Post: #1




HP Prime Numerical Integration
Hello,
I'd like to know the method of numerical integration the hp prime uses, because it's great. For example, I tried to compute the incomplete elliptic integral of the first kind: EllipticF(x,m) = integrate 1/sqrt(1  msin^2(t)) dt from 0 to x For parameters x=2, m=3, so: EllipticF(2,3) = integrate 1/sqrt(1  3sin^2(t)) dt from 0 to 2 Wolfram Alpha tells me that the result is: EllipticF(2,3) = 1.0010771.490278*i On HP Prime the command would be: int(1/SQRT(13*(SIN(T)^2)),T,0,2) When approximated in HP Prime this results in 1.00107718634  1.49027809758*i When I tried the same thing in Wolfram Alpha: N[integrate (1/sqrt(13sin^2(t)) dt from 0 to 2)] the result is 0.9996911316703074  1.4848975811276912*i Then I tried other methods, such as programming my own methods in python. I tried the Romberg integration, Gauss Legandre Quadrature, Simpson Rule, Gauss Kronrod. None of which worked as good. Even romberg in HP Prime can't calculate it well enough. romberg(1/sqrt(13*(sin(t)^2)),t,0,2) returns the last approximation 0.9911988698181.47660568998*i The scipy.integrate.quad in python works well enough, but it's not as good as HP Prime int. Since 1/sqrt(13sin^2(x)) has poles from 0 to 2, I suspect that some pole finding algorithm might be used. 

08042022, 12:18 PM
Post: #2




RE: HP Prime Numerical Integration
Look at the thread just below yours:
https://www.hpmuseum.org/forum/thread18566.html 

08042022, 04:50 PM
Post: #3




RE: HP Prime Numerical Integration
(08032022 06:15 PM)Ruda975 Wrote: int(1/SQRT(13*(SIN(T)^2)),T,0,2) XCAS int(...) is using the same algorithm, with full 53 bits precision (HP Prime use 48 bits) XCAS> int(1/sqrt(13*sin(x)^2), x, 0., 2.) Low accuracy, error estimate 3.0628046608e06. Error might be underestimated if initial boundary was +/infinity 1.00107391075  1.49027871636*i XCAS only get 6 digits accuracy, HP Prime getting 7 is just lucky. With a pole within integral limits, even XCAS getting 6 correct digits is luck. Error estimate seems to be in the right ballpark ... but, don't bet on it.  To accurately calculate EllipticF by integration, we locate the pole, and not touch it. pole = ± asin(sqrt(1/3)) + n*pi ≈ ± 0.6155 + n*pi For pi/2 .. 2, there is no pole. N[integrate (1/sqrt(13sin^2(t)) dt from pi/2 to 2)] = 0.31885796059775706 i Using AGM, complete elliptic integral of the first kind does not care where pole is. EllipticK(3) = pi/2 / AGM(1, sqrt(13)) = 1.0010773804561062  1.1714200841467699 i Sum 2 terms, we have EllipticF(2,3) = 1.0010773804561062  1.490278044744527 i 

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