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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