Power method help(recursive)

[Steve Shipman]

Ok I have this recursive method which evalutes the value of positive exponents fine, but I need it to work with negative exponents, and im not sure how to do this. This is the code I have now

    
public static double power(double x, int n)
{
   if (n == 0)
      return 1.0;                   // anchor case
   else 
      return power(x, n - 1) * x;   // inductive step (n > 0)
}
 

Thanks in advance!

[Keith Lynn]

I think the easiest thing to do would be to simply evaluate the power and then divide 1 by the power. If you're going to have large powers I would recommend using BigInteger or BigDecimal.

[Henry Wong]

It is actually pretty straightforward. Here is a hint...

Is there a way to calculate a value to a negative exponent based on knowing the value to a positive exponent? If there is, can you recursively calculate the value to a positive exponent, and use that to calculate the value to the negative exponent?

Henry

[Steve Shipman]

Doh, didnt even think of the 1 divided by the power thing. But unfortunatly that wont work either, I had this

public static double power(double x, int n)
{
   if (n == 0)
      return 1.0;                   // anchor case
   else 
      return power(x, n - 1) * x;   // inductive step (n > 0)
}

 
public static double negativePower(double x, int n)
{
if (n == 0)
return 1.0;
else
return (1 / (power(x, n -1) * x)); 
}
 

but it gives me the same problem as the original, just goes into a stack over flow exception.

[Henry Wong]

Originally posted by Ikara Lott:
Doh, didnt even think of the 1 divided by the power thing. But unfortunatly that wont work either, I had this

but it gives me the same problem as the original, just goes into a stack over flow exception.

Close... two more hints...

1. You don't need to have a special negative power method, you can do the whole thing recusively within the power method.

2. When you "1 divided by the power thing", don't you also have to change the sign?

Henry

[Steve Shipman]

ohhhhh change the sign! Haha yeah that might help.

Ok well I tried this:

public static double power(double x, int n)
{
  Math.abs(n);
  

   if (n == 0)
      return 1.0;                   // anchor case
   else 
      return (1 / (power(x, n - 1) * x));   // inductive step (n > 0)
}
 
 

but that was a no go, same error
[ March 28, 2006: Message edited by: Ikara Lott ]

[Henry Wong]

When I said that you can do the whole thing with one power() method, I meant that you have to do it without breaking the method. Try it with positive numbers, you'll see that it no longer works...

Henry

[Steve Shipman]

hmmm ok I tried this for the last return statement, isntead of what I had above

return (power(x, (Math.abs(n)) - 1) * x); 

And it will work with finding the regular exponents, like if I do 2^-2, ill get 4. However when I add the 1 divded by the power part

return (1 / (power(x, (Math.abs(n)) - 1) * x)); 

I dont get the error, but I dont get the correct answer either 2^-3 gives me .5

[Keith Lynn]

The second method should not be recursive. It should look more like this

public static double negativePower(double x, int n)
{
if (n == 0)
return 1.0;
else
return (1.0 / (power(x, Math.abs(n))); 
}

Note that when you have a negative power you don't change the sign of the number.

[ March 28, 2006: Message edited by: Keith Lynn ]
[ March 28, 2006: Message edited by: Keith Lynn ]

[Steve Shipman]

"When I said that you can do the whole thing with one power() method, I meant that you have to do it without breaking the method. Try it with positive numbers, you'll see that it no longer works..."

Oh yeah, I didnt break it into one method yet, im just trying to get the negative ones to work. Hehe I did see it didnt work with positive ones anymore, but im just getting the negative exponent thing worked out first

[Steve Shipman]

"The second method should not be recursive. It should look more like this"

Thats part of the way I have to do it though, I have to have it evaluate negative exponents recursivly

[Henry Wong]

Oh yeah, I didnt break it into one method yet, im just trying to get the negative ones to work. Hehe I did see it didnt work with positive ones anymore, but im just getting the negative exponent thing worked out first

It's a recursive algorithm, the negative calls depends on the positive calls. The negative exponents will not work, if the positive exponents do not work.

Henry

[Steve Shipman]

ahh I see, yeah when I split it back up into 2 methods it works fine, now ill just go back and make it one method. Thanks for your help guys

[Keith Lynn]

Here is one way of looking at it.

Whether it's a negative or positive power, the stopping point will be 0.

Note the following

x^3 = x * x^2

x^-3 = x^-1 * x^-2

if power is 0
   return 1
otherwise
   if the power is positive
      the value returned is x * the result of calling the method on power - 1
   otherwise
      the value returned is 1/x * the result of calling the method on power + 1

[ March 28, 2006: Message edited by: Keith Lynn ]

[Ryan McGuire]

Originally posted by Ikara Lott:
...

By the way, nice name.