Tim Peters added the comment:
Steven, you certainly _can_ ;-) check first whether `r**n == x`, but can you
prove `r` is the best possible result when it's true? Offhand, I can't. I
question it because it rarely seems to _be_ true (in well less than 1% of the
random-ish test cases I tried). An expensive test that's rarely true tends to
make things slower overall rather than faster.
As to whether a Newton step won't make things worse, that requires seeing the
exact code you're using. There are many mathematically equivalent ways to code
"a Newton step" that have different numeric behavior. If for some unfathomable
(to me) reason you're determined to stick to native precision, then - generally
speaking - the numerically best way to do "a correction step" is to code it in
the form:
r += correction # where `correction` is small compared to `r`
Coding a Newton step here as, e.g., `r = ((n-1)*r + x/r**(n-1))/n` in native
precision would be essentially useless: multiple rounding errors show up in
the result's last few bits, but the last few bits are the only ones we're
_trying_ to correct.
When `correction` is small compared to `r` in `r += correction`, the rounding
errors in the computation of `correction` show up in correction's last few
bits, which are far removed from `r`'s last few bits (_because_ `correction` is
small compared to `r`). So that way of writing it _may_ be helpful.
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue27761>
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