Hello Oscar,
why is the float approximation of Rational('2.4') more precise of
Float('2.4')?
Davide.
Il giorno lun 12 ott 2020 alle ore 14:14 Oscar Benjamin <
[email protected]> ha scritto:
> Hi,
>
> The evalf function is intended to compute the result to any given
> specified precision and return a result such that it is correct to
> almost every digit. However evalf needs access to the full expression
> to be able to do that. If you first substitute a Float in because the
> expression will partially or fully evaluate before evalf gets a
> chance:
>
> In [5]: expr = cos(2*x)
>
> In [6]: expr.evalf(50, subs={x:2.4})
> Out[6]: 0.087498983439446392365833650247410931894111629665389
>
> In [7]: expr.subs(x, 2.4).evalf(50)
> Out[7]: 0.087498983439446398335803678492084145545959472656250
>
> Note that eval isn't doing anything in the second example because the
> subs already evaluated to a Float (ignoring our request for 50
> digits):
>
> In [10]: expr.subs(x, 2.4)
> Out[10]: 0.0874989834394464
>
> The first example gives an accurate result for cos(2x) based on the
> true value of the float 2.4 (which is not exactly equal to 12/5). If
> we really want to calculate cos(2x) for x=12/5 with minimal rounding
> error we should use:
>
> In [9]: expr.subs(x, Rational('2.4')).evalf(50)
> Out[9]: 0.087498983439446569320215257649487633957449890596100
>
> --
> Oscar
>
> On Mon, 12 Oct 2020 at 12:44, s5s <[email protected]> wrote:
> >
> > Hi,
> >
> > First, apologies if this has been discussed before. I'm browsing through
> the tutorial and encountered what appears to be an inaccuracy in the
> documentation. In particular, I am reading Basic Operations section which
> states that
> >
> > To numerically evaluate an expression with a Symbol at a point, we might
> use subs followed by evalf, but it is more efficient and numerically stable
> to pass the substitution to evalf using the subs flag, which takes a
> dictionary of Symbol: point pairs.
> >
> > However, when I test this in jupyter lab using "%timeit" it appears the
> reverse is true. I'll avoid posting images, but I got the following results:
> >
> > expr = cos(x) - 2*sin(exp(x))
> > %timeit expr.subs(x, 0).evalf()
> > 92.9 µs ± 628 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
> >
> > %timeit expr.evalf(subs=dict(x=0))
> > 264 µs ± 7.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
> >
> > expr = cos(2*x)
> > %timeit expr.evalf(subs={x: 2.4})
> > 73.8 µs ± 3.28 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
> >
> > %timeit expr.subs(x, 2.4).evalf()
> > 35.1 µs ± 434 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
> >
> > Perhaps the documentation needs to elaborate more what it means by
> "stable" and "efficient" and under what conditions using one is preferable
> over the other because from the tests I performed, it appears using
> subs().evalf() is the way to go. The docs are either incorrect or out of
> date.
> >
> > Best regards
> >
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