Calculating the numerical value of the regularized generalized hypergeometric function

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

HypergeometricPFQRegularized({1},{0,0},0)({−1.5},{−1.,−0.5},3600.)\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?

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1

 

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.♦
Oct 27 ’15 at 23:43

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2 Answers
2

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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 -> 3600.0`60}

(* 4.070133*10^52 *)

Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1,
b2}, z] /. {a -> -3/2, b1 -> -1, b2 -> -0.5`60, z -> 3600}

(* 4.070133*10^52 *)

Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1,
b2}, z] /. {a -> -3/2, b1 -> -1.0`60, b2 -> -1/2, z -> 3600}

(* 4.070133*10^52 *)

Derivative[{1}, {0, 0}, 0][HypergeometricPFQRegularized][{a}, {b1,
b2}, z] /. {a -> -1.5`60, 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.♦
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.♦
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