Floating problems??

Anon

Here is the code...

<?php
function filter($L, $H, $Av, $x) {
//preconditions:
// 0<=$x<=1
// $L<=$Av<$H

$b = pow($L-$Av, 1/3);
$a = pow($H-$Av, 1/3)-$b;
$f = pow($a*$x + $b, 3)+$Av;
return $f;
}
echo "<table>";
$y=0.5;
$x=filter(0,100,75,$y);
echo $x . "<br>\n";
?>

And this is the output (no matter what I put in):

-1.#IND

What exactly am I doing wrong? Am I supposed to explicitly define floats or something?

nashirak

look at this $x=filter(0,100,75,$y);

What is wrong with this is not your code but your algorithm... if you have:

0 - 75 then take the 1/3 (cubed root ) or this number.... well from math we know that the cubed root of any negative number is an imaginary number ie.

the answer is : sq of -75 which is 75i . The computer has no idea how to handle imaginary numbers (let me clarify PHP has no idea how to do imaginary numbers. You can fudge imaginary numbers with complicated math programs).

Try to keep all your code in the real world and not the imaginary one 🙂.

Anon

That shouldn't be.. the cubic root of any negative number should be -abs(n)^(1/3) (eg: (-3)³ = -27) I'll try focing it to use positives, and just handle everything by grabbing the sign before hand, I suppose.. it'd make everything work nice and smoothly.

nashirak

Look in a math book the cube root of any negative number is always going to come out imaginary for the fact that:
No number times itself will equal a negative number: if you want proof you could say something like:

-27^(1/3/) factor out 27^1/3 and put it outside:

so 3*-1^1/3 ( and you know from any math book that the nth root of negative one is always going to equal i (you can look that one up I have forgoten the proof)
So therefore it can be shown that

-27^1/3 == 3*-1^1/3 where -1^1/3 is equal to i (imaginary) so therefore -27^1/3 == 3i .
However you are right when it come to doing powers. But that is different. Using the same logic:

-3³ factor out a -1 so -1(3^3), so 3³ is a real number and so is -1. So the answer is -27. -1^1/n (where n is any real rational number) can never be a real number.

I hope that clarifies stuff. Your solution sounds ok, but just know that the positive answer you are getting is not the "real" answer to the problem. It still lies in the imaginary plane. 🙂

If -3i is equal to anything its:
x(t) = c1 cos 3t + c2 sin 3t (where its a solution for a differential equation)

P.S. to any math majors or profs out there forgive me I am shure I havent done the best job of proving this.