How to get precise Math.exp() in j2me?

Question

I am using j2me and I need to get quite precise exp() for values up to 4. Problem with the j2me is that it's Math library doesn't have pow() and exp() method. To solve this, I just used this method to implement pow():

public static double pow(double num1, double num2) {
  double result = 1;
  for (int i = 0; i < num2; i++)
    result *= num1;
  return result;
}

This enabled me to have exp functionality by using setting e as constant (2.718281828459045) and calling pow:

double calculation =  (20.386 - (5132.000 / (t + 273.15)));
System.out.println("calc: " + pow(2.71,calculation));
calculation = pow(2.7182818284590452,calculation) * 1.33;

My problem is that result is quite inaccurate, for example if I compare math.exp and my pow method for number 3,75, results are like this:

Pow function returns: 54.5980031309658

Math function returns: 42.52108200006278

So I would need advice, how to implement exp functionality in j2me environment with highest precision possible.

Your function names no sense whatsoever for exponents that are not fractions. Are you surprised that this isn't correct? — duffymo, Feb 01 '16 at 13:41
Possible duplicate of [J2ME power(double, double) math function implementation](http://stackoverflow.com/questions/2076913/j2me-powerdouble-double-math-function-implementation) — agold, Feb 01 '16 at 13:43
@Ferrybig: Yes you are right, but the [accepted answer](http://stackoverflow.com/a/2076930/1771479) (link only answer) has a good [link](https://community.oracle.com/docs/DOC-983720) which gives algorithms to implement a `pow` function. — agold, Feb 01 '16 at 13:47
@duffymo Only name for function that you can see is pow function? What is wrong with that name? Function works correct for values 0,1, butt I have problemes with higher values. — Jure, Feb 01 '16 at 13:48
No, your implementation is laughably wrong. Please tell me how you should calculate it for an easy one: exponent of 0.5. (aka square root). You need Newton iterative method. — duffymo, Feb 01 '16 at 13:49
@duffymo I know what you mean know. I first tryed some library from link, that agold gave. But it wasn't any good, so tried this method and I didn't even notice that it uses integer values. — Jure, Feb 01 '16 at 13:52
No, it uses doubles. Your algorithm is all wrong from beginning to end. Read up on Newton and iterative method. — duffymo, Feb 01 '16 at 13:57
see related **Q/A**'s: [How Math.Pow (and so on) actualy works](http://stackoverflow.com/a/19072451/2521214) , [Power by squaring for negative exponents](http://stackoverflow.com/a/30962495/2521214) and [fixed point exp2,log2,pow](http://stackoverflow.com/a/18169727/2521214). Especially the first two will lead to solution ... the last is just implementation from one of my bignum class and without reading the first two links will not help — Spektre, Feb 02 '16 at 08:10
@Spektre, thank you for you help. I managed to find answer in similar question. — Jure, Feb 02 '16 at 10:10

score 1 · Accepted Answer · edited May 23 '17 at 12:30

I helped my self with bharath answer in this question: How to get the power of a number in J2ME

Since exp method is just pow, where we use Euler's number for the first argument, I used bharath method:

public double powSqrt(double x, double y)
    {
        int den = 1024, num = (int)(y*den), iterations = 10;
        double n = Double.MAX_VALUE;

        while( n >= Double.MAX_VALUE && iterations > 1)
        {
            n = x;

            for( int i=1; i < num; i++ )n*=x;

            if( n >= Double.MAX_VALUE ) 
            {
                iterations--;
                den = (int)(den / 2);
                num = (int)(y*den);
            }
        }   

        for( int i = 0; i <iterations; i++ )n = Math.sqrt(n);

        return n;
    }

Method call:

calculation = powSqrt(2.7182818284590452,calculation) * 1.33;

Result is almost as good as Math.pow() method.

PS: I don't know if this is duplicated thread, if so you can delete it.

How to get precise Math.exp() in j2me?

1 Answers1