Question for php pros

DomMike · 02-25-2005, 12:02 PM

Hi,
I was wondering if any of you pros could help me, translate this function so I can use it in a php script.
Thanks
=========
To calculate it, you would first calculate 1 + J then take that to the -N (minus N) power, subtract that from the number 1. Now take the inverse of that (if you have a 1/X button on your calculator push that). Then multiply the result times J and then times P.

The one-liner for a program would be (adjust for your favorite language):

M = P * ( J / (1 - (1 + J) ** -N))

axilant · 02-25-2005, 01:04 PM

Quote:

M = P * ( J / (1 - (1 + J) ** -N))

PHP Code:

  $m = $p*($j/(1-(1+$j)**-$n)); 

????: NamePros.com http://www.namepros.com/programming/72040-question-for-php-pros.html

i think thats what u need?

i didnt try this... but i think its what u want. http://php.net/ and look at there math functions im sure you can figure it out....

Peter · 02-26-2005, 06:37 AM

Originally Posted by axilant

PHP Code:

  $m = $p*($j/(1-(1+$j)**-$n));

i think thats what u need?

i didnt try this... but i think its what u want. http://php.net/ and look at there math functions im sure you can figure it out....

all you have done is put a $ before the letters. somehow i do not think doing that will suddenly solve his problem.

eagle12 · 02-26-2005, 07:25 AM

This is from the function you showed:

PHP Code:

  $M = $P*($J/(1-(pow((1+$J), (-1*$N)))))

This is from the word description:

PHP Code:

  $M = $P*($J*(1/(1-(pow((1+$J), (-1*$N)))))) 

????: NamePros.com http://www.namepros.com/showthread.php?t=72040

There is a slight difference between the two, because you say J * the inverse of (1-((1+J)^-N)). But the function you put at the bottom just says J / (1-((1+J)^-N)).

Let me know if this helps.

majinbuu1023 · 02-26-2005, 03:32 PM

ok im no php pro but it looks like one of you guys solved it lol

DomMike · 02-26-2005, 08:38 PM

Thanks a lot eagle12.
I'll try and let you know.
Mike

02-25-2005, 12:02 PM	THREAD STARTER #1 (permalink)
DomMike NamePros Member Join Date: Apr 2003 Posts: 78	Question for php pros Hi, I was wondering if any of you pros could help me, translate this function so I can use it in a php script. Thanks ========= To calculate it, you would first calculate 1 + J then take that to the -N (minus N) power, subtract that from the number 1. Now take the inverse of that (if you have a 1/X button on your calculator push that). Then multiply the result times J and then times P. The one-liner for a program would be (adjust for your favorite language): M = P * ( J / (1 - (1 + J) ** -N)) __________________ Mike

02-26-2005, 07:25 AM	#4 (permalink)
eagle12 NamePros Member Join Date: Nov 2003 Location: Ontario, Canada Posts: 127	This is from the function you showed: PHP Code: `$M = $P($J/(1-(pow((1+$J), (-1$N)))))` This is from the word description: PHP Code: `$M = $P($J(1/(1-(pow((1+$J), (-1$N)))))) ????: NamePros.com http://www.namepros.com/showthread.php?t=72040` There is a slight difference between the two, because you say J the inverse of (1-((1+J)^-N)). But the function you put at the bottom just says J / (1-((1+J)^-N)). Let me know if this helps. Last edited by eagle12; 02-26-2005 at 07:30 AM.

02-26-2005, 03:32 PM	#5 (permalink)
majinbuu1023 Senior Member Join Date: Jan 2005 Location: New Zealand Posts: 3,747	ok im no php pro but it looks like one of you guys solved it lol __________________ WowHumor.net - Funny World of Warcraft Pictures

02-26-2005, 08:38 PM	THREAD STARTER #6 (permalink)
DomMike NamePros Member Join Date: Apr 2003 Posts: 78	Thanks a lot eagle12. I'll try and let you know. Mike __________________ Mike

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