Computing limits in PHP

Question

I want to compute n^(2/3) as n approaches infinity in PHP.

What I tried:

$n = "INF";
echo pow($n, (2/3));

Desired result:

INF

Actual result:

0

Any suggestions? How to compute limits in PHP?

UPDATE:

OK, first problem solved - I just needed to remove the quotation mark.
But is this actually the way to calculate a limit? Is it interpreted as a limit?

I don't think it is a duplicate - that's another question. – user3419936 Mar 14 '14 at 19:35 — user3419936, Mar 14 '14 at 19:35

score 1 · Accepted Answer · answered Mar 14 '14 at 19:30

1

You're passing in the string "INF" into the exponential expression, which isn't valid. Try passing in just POW:

$n = INF;
echo pow($n, (2/3));
// INF

answered Mar 14 '14 at 19:30

Chris Forrence

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That actually solved the problem! Is that interpreted as a limit in PHP? – user3419936 Mar 14 '14 at 19:33
2

INF is the constant used when a number is too large for PHP. You can, though check `pow(INF, 0)` and see that it resolves to 1 correctly. In addition, `pow(INF, -1) == 0` is true – Chris Forrence Mar 14 '14 at 20:02

Computing limits in PHP

1 Answers1