Owen's T statistical function —> Taylor development? - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: Owen's T statistical function —> Taylor development? (/thread-21048.html) |
Owen's T statistical function —> Taylor development? - Gil - 12-24-2023 10:08 PM I have got the Owen's T statistical function '1/(2*pi)*Integral (from 0,to a,e^(-h^2*((1+t^2)/2))/(1+t^2),t) '. Normally, h & a are real numbers. Wolfram Alpha gives an example with a complex argument. I imagine that, to achieve that, the original function with the integral was transformed with a Taylor's development or some alike idea. For real h=4 and a=1, the numerical integral on HP50 gives correctly 1.58351193827E-5 (as in Wolfram). I tried the Taylor development with the following program, but I get alternatively a positive and a negative value, but no sign of converging value toward 1.58351193827E-5: \<< 4 1 \-> h a \<< '1/(2*\pi)*e^(-h^2*((1+t^2)/2))/(1+t^2)' EVAL DUP 'f' STO 'last.deriv' STO f 't=0' SUBST EVAL a * \->NUM 2 40 FOR i last.deriv 't' \.d EVAL DUP 'last.deriv' STO 't=0' SUBST EVAL a i ^ * i ! / \->NUM + NEXT \>> \>> Could somebody try and help me give the searched approached "solution" ? Apparently not an easy task, after discovering: https://www.researchgate.net/publication/5142787_Fast_and_Accurate_Calculation_of_Owen's_T_Function And, supposing that the found program or Taylor's algorithm is OK for real number, can we use it with complex numbers? Many thanks in advance. |