Calculate e^x in Java

ckoniecny · 07-23-2007, 11:45 PM

I'm trying to calculate e^x, without any built-in functions such as Math.exp. This is what I have so far:

Code:

    public static double math_ex(int iter, int x) {
       double ex = 0;
       int i = 1;
       
       while(i <= iter) {
          if (i == 1) {
             ex += x;
          } else {
             ex += Math.pow(x,(double)i/factorial(i));
          }
          i++;
       }
       ex += 1;
       return ex;       
    }

My input is:

Code:

Enter iterations: 
6
Enter x: 
3
e^3 = 11.989024

But when I use the calculator to figure it out, it is 20.0855........

Can you see anything thats wrong with my function?

Thanks.

Wim Sturkenboom · 07-24-2007, 06:06 AM

Not a java programmer, but if Math.exp is a built=in function, I assume that Math.pow is as well.

Don't know what is wrong with your code, but is simple multiplying in a loop not easier.

something like

Code:

result=1;
while(iter--)
  result *= x;
return result;

This is not bug free.

[edit]
OK, it's still vague but the factorial is probably to calculate e; so calculate e first and fill it in in place of x. Will be a bit faster as well as you only have to calculate e once.[/edit]

oami · 07-24-2007, 06:29 AM

Quote:

Originally Posted by ckoniecny

Code:

             ex += Math.pow(x,(double)i/factorial(i));

This line should be

Code:

             ex += Math.pow(x,(double)i)/factorial(i);

according to the series e^x=1+x+x^2/2!+...+x^n/n!
(not x^(n/n!) )

pixellany · 07-24-2007, 06:31 AM

What is the algorithm that you intended to code? (I just ran the numbers by hand and it clearly is not converging on the right answer)

Dan04 · 07-24-2007, 10:22 PM

Quote:

Originally Posted by pixellany

What is the algorithm that you intended to code? (I just ran the numbers by hand and it clearly is not converging on the right answer)

It looks like it's supposed to be the MacLaurin series sum(x**n / n! for n in {0, 1, 2, ...}).

Dan04 · 07-24-2007, 10:27 PM

BTW, what type does your factorial function use? Because of overflow, "int" is only accurate up to n=12, and even "long" only gets you to n=20.

jiml8 · 07-25-2007, 10:28 AM

In addition to changing it as oami specified, you have to either start i at 0 or add one to the result.