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I am confused about using expm1 function in java The Oracle java doc for Math.expm1 says:

Returns exp(x) -1. Note that for values of x near 0, the exact sum of expm1(x) + 1 is much closer to the true result of ex than exp(x).

but this page says:

However, for negative values of x, roughly -4 and lower, the algorithm used to calculate Math.exp() is relatively ill-behaved and subject to round-off error. It's more accurate to calculate ex - 1 with a different algorithm and then add 1 to the final result.

should we use expm1(x) for negative x values or near 0 values?

Masood_mj
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  • You can always pull some other algorithm off the net and use it. No need to confine yourself to Java's library. – Hot Licks Aug 17 '12 at 19:37

2 Answers2

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The implementation of double at the bit level means that you can store doubles near 0 with much more precision than doubles near 1. That's why expm1 can give you much more accuracy for near-zero powers than exp can, because double doesn't have enough precision to store very accurate numbers very close to 1.

I don't believe the article you're citing is correct, as far as the accuracy of Math.exp goes (modulo the limitations of double). The Math.exp specification guarantees that the result is within 1 ulp of the exact value, which means -- to oversimplify a bit -- a relative error of at most 2^-52, ish.

Louis Wasserman
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You use expm1(x) for anything close to 0. Positive or negative.

The reason is because exp(x) of anything close to 0 will be very close to 1. Therefore exp(x) - 1 will suffer from destructive cancellation when x is close to 0.

expm1(x) is properly optimized to avoid this destructive cancellation.

From the mathematical side: If exp is implemented using its Taylor Series, then expm1(x) can be done by simply omitting the first +1.

Mysticial
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  • But that page says: "roughly -4 and lower", does lower means here negative values close to 0 or smaller negative values? – Masood_mj Aug 17 '12 at 19:40
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    @Masood_mj I actually disagree with that statement. It seems that the author is confusing two different sources of numerical instability in the taylor expansion of `exp(x)`. If you apply the series directly to large negative numbers, then yes the it will be unstable. So in practice you flip `x` positive and take the reciprocal at the end. But I can't imagine that `Math.exp(x)` doesn't already do that. – Mysticial Aug 17 '12 at 19:43