# Calculating the numerical value of the regularized generalized hypergeometric function

I'm trying to calculate the numerical value of the regularized generalized hypergeometric functions:

$\qquad \sf{HypergeometricPFQRegularized}^{(\{1\},\{0,0\},0)}(\{-1.5\},\{-1.,-0.5\},3600.)$

I tried

HypergeometricPFQRegularized^({0}, {0, 1} , 0)[{-1.5}, {-1., -0.5}, 3600.]


but I got the expression returned unevaluated

HypergeometricPFQRegularized^({1}, {0, 0}, 0))[{-1.5}, {-1., -0.5}, 3600.]

Mathematica didn't calculate it numerically. Is there a way to get numerical values?

• In Mathematica functions are called with [square brackets] not (parens). Additionally, indices on functions are not written with ^ but are passed as parameters to the function. Take a look at reference.wolfram.com/language/ref/… for more details on the syntax. – evanb Oct 27 '15 at 23:39
• Unfortunately, even after entering the correct N[Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{-3/2}, {-1, -1/2}, 3600]], it still doesn't work. Let me see what I can do... – J. M. will be back soon Oct 27 '15 at 23:43

Set at least one parameter or variable to a high precision numeric value or numerically evaluate the exact expression using arbitrary precision.

$Version (* "10.3.0 for Mac OS X x86 (64-bit) (October 9, 2015)" *) Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1, b2}, z] /. {a -> -3/2, b1 -> -1, b2 -> -1/2, z -> 3600.060} (* 4.070133*10^52 *) Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1, b2}, z] /. {a -> -3/2, b1 -> -1, b2 -> -0.560, z -> 3600} (* 4.070133*10^52 *) Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1, b2}, z] /. {a -> -3/2, b1 -> -1.060, b2 -> -1/2, z -> 3600} (* 4.070133*10^52 *) Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1, b2}, z] /. {a -> -1.560, b1 -> -1, b2 -> -1/2, z -> 3600} (* 4.070133*10^52 *)  EDIT: After restart, last entry did not evaluate until arbitrary precision was increased and then only with warning message. However, result agrees with results above. N[Derivative[{1}, {0, 0}, 0][ HypergeometricPFQRegularized][{-3/2}, {-1, -1/2}, 3600], 75] (* N::meprec: Internal precision limit$MaxExtraPrecision = 50. reached while evaluating (HypergeometricPFQRegularized^({1},{0,0},0))[{-(3/2)},{-1,-(1/2)},3600]. >>

4.07013311778051868807295020202207635064552896440315759211918737043238\
348*10^52  *)

• What version are you on? For some reason, this doesn't work on 10.3. – J. M. will be back soon Oct 28 '15 at 2:42
• @J.M. - I am using 10.3 on a Mac. Note edit above. – Bob Hanlon Oct 28 '15 at 2:54
• Huh, it worked after restarting. But, I've also noticed that if you try to evaluate at machine precision first before trying arbitrary precision, the thing remains unevaluated. Bizarre... – J. M. will be back soon Oct 28 '15 at 2:57

The trick is to use high precision:

SetPrecision[
(HypergeometricPFQRegularized^({0}, {1, 0}, 0))[{-1.5}, {-1., -0.5}, 3600.],
100]


it gives:

-2.8734042033205156581184947784*10^34`