Simplify logic for Math.tanh #412
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| if (b === Infinity) { return -1; } | ||
| return (a - b) / (Math.exp(x) + Math.exp(-x)); | ||
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| return (Math.expm1(x) - Math.expm1(-x)) / (Math.exp(x) + Math.exp(-x)); |
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Note: I not sure 4 math function calls really should be needed here, but I'll trust it has some accuracy benefit or something so not changing.
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You're correct about neglibile performance difference for Math.sign changes, reverted. |
benjoffe
changed the title
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Thanks, this looks great - if it's not too much to ask, could you rebase this on latest master? |
What about this one? |
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@Yaffle i'd be happy to review a new PR that changed the implementation, with assorted tests (altho i'd prefer the edge cases be explicit) |
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Previous PR for tanh left some code that was no longer needed (the checks against Infinity are redundant with the early exit).