Math.pow Question

[Michael B. Williams]

Please help

The Math.pow call is retuning Infinity? What am I doing wrong - be
gental I java newbie. Thanks in advance

import java.math.*;


double
interest = .0575,
J = 0,
N = 0,
ftemp;

J = interest / 1200 ;
ftemp = ( 1-(1+J) );
N = N * (-1);

System.out.println( Math.pow( ftemp , N ) );

 

[Andrew Thompson]

Please help

Uuuggh!
<http://www.physci.org/codes/javafaq.jsp#cljh>
[ The link was still on the clipboard
from the last poster! ]

...and post compileable code rather than
the snippets you had, please..
<http://www.physci.org/codes/sscce.jsp>

 

[Woebegone]

Please help

The Math.pow call is retuning Infinity? What am I doing wrong - be
gental I java newbie. Thanks in advance

import java.math.*;


double
interest = .0575,
J = 0,
N = 0,
ftemp;

J = interest / 1200 ;
ftemp = ( 1-(1+J) );
N = N * (-1);

System.out.println( Math.pow( ftemp , N ) );

My result is 1.0, which seems reasonable to me since your exponent is -0.0:
is this the exact code that produces Infinity? By my reckoning, x^(-y) ==
1/(x^y), so you have ftemp^(-0) == 1/(ftemp^0) == 1/1 == 1. The expected
results are laid out fairly clearly in the JavaDoc:

"Returns the value of the first argument raised to the power of the second
argument. Special cases:
- If the second argument is positive or negative zero, then the result is
1.0. "

http://java.sun.com/j2se/1.4.2/docs/api/java/lang/Math.html#pow(double,
double)

 

[Chris Smith]

The Math.pow call is retuning Infinity? What am I doing wrong - be
gental I java newbie. Thanks in advance

When I compile and run your code, I get 1.0 as the answer, not Infinity.
That's the result I'd expect. If you're still getting 1.0, then I
suspect that's not the code that you're compiling or running. Are you
sure you aren't running an older version of the code by mistake?

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