Precision Tail-off?

[Stephen Tucker]

Thanks, one and all, for your reponses.

This is a hugely controversial claim, I know, but I would consider this
behaviour to be a serious deficiency in the IEEE standard.

Consider an integer N consisting of a finitely-long string of digits in
base 10.

Consider the infinitely-precise cube root of N (yes I know that it could
never be computed unless N is the cube of an integer, but this is a
mathematical argument, not a computational one), also in base 10. Let's
call it RootN.

Now consider appending three zeroes to the right-hand end of N (let's call
it NZZZ) and NZZZ's infinitely-precise cube root (RootNZZZ).

The *only *difference between RootN and RootNZZZ is that the decimal point
in RootNZZZ is one place further to the right than the decimal point in
RootN.

None of the digits in RootNZZZ's string should be different from the
corresponding digits in RootN.

I rest my case.

Perhaps this observation should be brought to the attention of the IEEE. I
would like to know their response to it.

Stephen Tucker.


On Thu, Feb 16, 2023 at 6:49 PM Peter Pearson <pkpearson at nowhere.invalid>
wrote:

> On Tue, 14 Feb 2023 11:17:20 +0000, Oscar Benjamin wrote:
> > On Tue, 14 Feb 2023 at 07:12, Stephen Tucker <stephen_tucker at sil.org>
> wrote:
> [snip]
> >> I have just produced the following log in IDLE (admittedly, in Python
> >> 2.7.10 and, yes I know that it has been superseded).
> >>
> >> It appears to show a precision tail-off as the supplied float gets
> bigger.
> [snip]
> >>
> >> For your information, the first 20 significant figures of the cube root
> in
> >> question are:
> >>    49793385921817447440
> >>
> >> Stephen Tucker.
> >> ----------------------------------------------
> >> >>> 123.456789 ** (1.0 / 3.0)
> >> 4.979338592181744
> >> >>> 123456789000000000000000000000000000000000. ** (1.0 / 3.0)
> >> 49793385921817.36
> >
> > You need to be aware that 1.0/3.0 is a float that is not exactly equal
> > to 1/3 ...
> [snip]
> > SymPy again:
> >
> > In [37]: a, x = symbols('a, x')
> >
> > In [38]: print(series(a**x, x, Rational(1, 3), 2))
> > a**(1/3) + a**(1/3)*(x - 1/3)*log(a) + O((x - 1/3)**2, (x, 1/3))
> >
> > You can see that the leading relative error term from x being not
> > quite equal to 1/3 is proportional to the log of the base. You should
> > expect this difference to grow approximately linearly as you keep
> > adding more zeros in the base.
>
> Marvelous.  Thank you.
>
>
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