Fix exponential() #226
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Fix exponential() #226
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robpike
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Jun 6, 2025
Fix various issues in exponential() like: - ** 1000 needing over 1600 iterations - ** -1000 yielding a large positive number - ** 2000 not converging after 2570 iterations The fix involves scaling down x to [0.5, 1) and expanding the Taylor series with increased precision. The iteration count is now <= 17 with up to 47 multiplications to scale the result in the worst case.
- Optimize bounds check on input value by only checking sign & exponents - Transform negative arguments using exp(x) = 1/exp(-x) - Dynamically set target exponent for argument reduction - Dynamically Set extra precision - Add comments documenting the above points - Optimize inner loop with one less multiplication - Optimize inner loop and result squaring by managing temp values manually - Panic on infinite result
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Update with f460a05. Sorry about the substantial changes, but this is however much cleaner and addresses all the points you've raised. |
db47h
commented
Jun 7, 2025
| // term < 1 ulp of sum ⇒ term < 0.5 × 2^(sum.exp-sum.prec+1) | ||
| // 0 ≤ term < 1 × 2^term.exp ⇒ 2^term.exp ≤ 2^(sum.exp-sum.prec) | ||
| // Because of argument reduction, 1 ≤ sum < 1+2^(-k+1) ⇒ sum.exp == 1 | ||
| if term.Sign() == 0 || term.MantExp(nil) <= sum.MantExp(nil) /* ==1 */ -int(sum.Prec()) { |
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Although sum.MantExp(nil) is known to be 1, I've left it just in case I messed up somewhere.
robpike
reviewed
Jun 7, 2025
and fix typos.
db47h
commented
Jun 8, 2025
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Just a quick note regarding the exit condition of the loop where we check that |
robpike
approved these changes
Jun 11, 2025
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Fix various issues in exponential() like:
The fix involves scaling down x to [0.5, 1) and expanding the Taylor series with increased precision.
The iteration count is now <= 17 with up to 47 multiplications to scale the result in the worst case.
The results are almost always properly rounded, and off by 1 ulp in the worst case.
I'd like to add a test for this, but because of #225, there is no really clean way to do it.