High power and double precision

Question

How do you solve the below equation in some programming language of your choice?

(1-1/X)^Y

Easy!

But how about when X & Y are very big and X>>Y

eg

(1-1/X)^Y
where 
X = 10^40
Y = 10^12

Looks like it should be a simple enough problem, but getting around the double precision problem before applying the power is something I was not able to figure out.

Answer 1

Well, (1 - 1/X)^Y = exp(Y*log(1 - 1/X)) . If X is very large, and much larger than Y , you can approximate the logarithm with

log(1 - 1/x) = -1/x -1/(2*X^2) + O(1/X^3)

and calculate

exp(-(Y/X+ Y/(2*X*X)))

If X isn't that much larger than Y , using a third or even fourth term of the Taylor series of the logarithm may be necessary.

Answer 2

Using GNU Octave the calculations are approximate:

octave:1> x = 10^40
x =  1.0000e+40
octave:2> y = 10^12
y =  1.0000e+12
octave:3> (1-1/x)^y
ans =  1

octave:8> exp(-(y/x + y /(2*x*x)))
ans =  1

Provided that the calculation made by Daniel Fischer is correct, the code to calculate exp(-(Y/X+ Y/(2*X*X))) in Java using BigDecimal is:

public static void main(String[] args) {
    BigDecimal x = new BigDecimal(10,MathContext.UNLIMITED).pow(40);
    BigDecimal y = new BigDecimal(10,MathContext.UNLIMITED).pow(12);

    BigDecimal twoXSquared = new BigDecimal(2,MathContext.UNLIMITED).multiply(x).multiply(x);
    BigDecimal yDividedByTwoXSquared = y.divide(twoXSquared);

    BigDecimal yDividedByX = y.divide(x);


    BigDecimal exponent = new BigDecimal(-1,MathContext.UNLIMITED).multiply(yDividedByX.add(yDividedByTwoXSquared));
    System.out.println(exponent.toEngineeringString());

    BigDecimal result = new BigDecimal(Math.E,MathContext.UNLIMITED).pow(exponent.intValue());

    System.out.println(result.toEngineeringString());

}