RE: Problem with hypergeometric function

• To: mathgroup at smc.vnet.net
• Subject: [mg34818] RE: [mg34796] Problem with hypergeometric function
• From: "DrBob" <majort at cox-internet.com>
• Date: Sat, 8 Jun 2002 05:21:32 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```>>switch to arbitrary precision arithmetic in all those cases in which
they do not know for sure if the algorithm that is being used will give
reliable results in case of using machine size arithmetic

WHAT arbitrary-precision numbers would be used in place of the original
machine-precision numbers?  The process of converting an arbitrary
precision quantity to machine precision is well-defined; the reverse is
not.

For instance, it's easy to see that Pi is approximately
3.141592653589793` --- that's the closest 16-digit approximation to Pi,
and there's only ONE such number within the defined distance from Pi.

However, if you have a machine-number that displays as
9.86960440108936`, what arbitrary precision number would you replace it
with?  There are infinitely many algebraic numbers (for instance) within
10^-16 (for instance) of 9.86960440108936`.

Quick, now... which one do you choose?

Maybe you'd use:

Rationalize[9.86960440108936`, 10^-20]

404289898/40963131

Or... you might somehow notice:

NumberForm[N[Pi^2], 20]

9.86960440108936

Now... just because Pi^2 is an interesting choice, that's no reason in
itself not to choose a nice rational fraction instead, and there's no
way to discover the approximation Pi^2... but that's how I got
9.86960440108936` in the first place, so I knew it would work nicely.
But, if you knew where the number came from, you'd be using an arbitrary

For a problem like the one you describe below, you MAY not get the same
answer with two replacements for the same machine-precision number, even
though they're both very close.

Bobby

-----Original Message-----
From: Ignacio Rodriguez [mailto:ignacio at sgirmn.pluri.ucm.es]
To: mathgroup at smc.vnet.net
Subject: [mg34818] [mg34796] Problem with hypergeometric function

\$Version

Microsoft Windows 3.0 (April 25, 1997)

Hi all,
I have noticed some problems when trying to evaluate numerically certain

hypergeometric functions.
For example:

f=HypergeometricPFQ[{1/2},{1,3/2},-8000]

N[f]

-1.34969 x 10^57

a bit big, isn't it?

\$MaxExtraPrecision=200
N[N[f,30]]

0.00586605

This seems more reasonable. The reason for this odd behaviour is related

to how this expressions are evaluated. Essencially, N applies itself to
any subexpression of f, as if MapAll were used.
So, in the first case, HypergeometricPFQ finds machine precission
numbers as its arguments, and evaluates itself in the same way. The
algorithm is obviously not very fortunate (a series expansion, I
guess?), and so is not the result. In the second case, their arguments
are arbitrary precision numbers, and even though the same problems are
present, using extremely high precision numbers for the intermediate
calculation does the trick.

My version of Mathematica is a bit old, and I would like to know if this

I would also like to recommend to Mathematica developers to switch to
arbitrary precision arithmetic in all those cases in which they do not
know for sure if the algorithm that is being used will give reliable
results in case of using machine size arithmetic.

```

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