CODEWITHSUNDEEP
algorithmmathnumericexponent
Simplexity
Simplexity

Reputation: 1

High power and double precision

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

e.g.

(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.

Upvotes: 0

Views: 215

Answers (2)

Vitaly Olegovitch
Vitaly Olegovitch

Reputation: 3547

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());

}

Upvotes: 0

Daniel Fischer
Daniel Fischer

Reputation: 183898

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.

Upvotes: 4

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