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> This is a different case, concerning functions that are expensive to compute to a high degree of precision (e.g. due to high sensitivity to rounding of interm

> This is a different case, concerning functions that are expensive to compute to a high degree of precision (e.g. due to high sensitivity to rounding of intermediate results) but "reputable" libraries will still try really hard to give good results. This is what I was referring to. See where it says "Why can't y^w be rounded within half an ulp like SQRT? Because nobody knows how much computation it would cost."... that's why I was confused how a arbitrary-precision library would handle something like this. User asks to raise some long decimal to another long decimal power and it seems to me you might sometimes either have to return a very wrong result or spend so much time computing that it won't be worth it.

yorwba 6y ago
> User asks to raise some long decimal to another long decimal power and it seems to me you might sometimes either have to return a very wrong result or spend so much time computing that it won't be worth it. y^w isn't that expensive. Usually it gets expressed roughly as exp(log(y^w)) = exp(w * log(y)), where exp and log are implemented using polynomial series. So if you want more precision, you just need to evaluate a few more polynomial terms.
- dataflow 6y ago
  If just a few more terms would suffice then why does he say "nobody knows how much computation it would cost"? Are you saying he's just referring to the half-ulp + epsilon case here too? If so then I've definitely mixed these up!
  - yorwba 6y ago
    Yeah, the question he asks is "Why can't y^w be rounded within half an ulp like SQRT?" not "Why can't y^w be rounded within slightly more than half an ulp like SQRT?" It's really just that tiny epsilon difference that turns "just a few more terms" into "nobody knows how much it'd cost". The reason SQRT can actually be rounded within half an espilon in a predictable amount of time is that SQRT(x) >= y + 0.5 precisely when x >= (y + 0.5)^2 = y^2 + y + 0.25, which means you can determine whether to round up or not with a simple comparison.
    - dataflow 6y ago
      Gotcha. Thanks for clarifying this!

yorwba 6y ago

> User asks to raise some long decimal to another long decimal power and it seems to me you might sometimes either have to return a very wrong result or spend so much time computing that it won't be worth it. y^w isn't that expensive. Usually it gets expressed roughly as exp(log(y^w)) = exp(w * log(y)), where exp and log are implemented using polynomial series. So if you want more precision, you just need to evaluate a few more polynomial terms.

dataflow 6y ago
If just a few more terms would suffice then why does he say "nobody knows how much computation it would cost"? Are you saying he's just referring to the half-ulp + epsilon case here too? If so then I've definitely mixed these up!
- yorwba 6y ago
  Yeah, the question he asks is "Why can't y^w be rounded within half an ulp like SQRT?" not "Why can't y^w be rounded within slightly more than half an ulp like SQRT?" It's really just that tiny epsilon difference that turns "just a few more terms" into "nobody knows how much it'd cost". The reason SQRT can actually be rounded within half an espilon in a predictable amount of time is that SQRT(x) >= y + 0.5 precisely when x >= (y + 0.5)^2 = y^2 + y + 0.25, which means you can determine whether to round up or not with a simple comparison.
  - dataflow 6y ago
    Gotcha. Thanks for clarifying this!