Calulation in lim (1 + 1 /n) ^n when n -> infinite

[Chris Angelico]

On Mon, Nov 9, 2015 at 11:21 PM, Salvatore DI DIO <artyprog at gmail.com> wrote:
> I was trying to show that this limit was 'e'
> But when I try large numbers I get errors
>
> def lim(p):
>     return math.pow(1 + 1.0 / p , p)
>
>>>> lim(500000000)
> 2.718281748862504
>>>> lim(900000000)
> 2.7182820518605446  !!!!
>
>
> What am i doing wrong ?
>

Floating point error is going to start becoming a major problem here.
Eventually, 1.0/p will underflow to zero, and you'll simply get a
result of 1.0:

>>> lim(900000000000000000000)
1.0

You could try using decimal.Decimal instead; that can give you rather
more precision. Or if you have a lot of patience, fractions.Fraction.

>>> def lim(p):
    return (1 + decimal.Decimal(1) / p) ** p

>>> lim(500000000)
Decimal('2.718281825740763411884758912')
>>> lim(900000000)
Decimal('2.718281826948888665260445479')
>>> lim(900000000000000000000)
Decimal('2.718281556630875980862943027')

Definitely not perfect yet... but let's crank up the precision.

>>> decimal.getcontext().prec=200
>>> lim(500000000)
Decimal('2.7182818257407634118847589119866382434352189174535941028176465917058364010909850212743201574998148959449308935216863472501568773422911827403060031110458945666576132256505421665510943751730347310393611')
>>> lim(900000000)
Decimal('2.7182818269488886655322736616533366198705073189604924113513719666470682138645212144697520852594221964992741852547403862053248620085931699038380869918694830598723560639286551799819233225160142059050076')
>>> lim(900000000000000000000)
Decimal('2.7182818284590452353587773147812963615153818054494196724217932791045775211554493949916225628968841872263472976692247791433837657091684393783360159727396877123886624050885921242672885287748600658242797')

Now we're starting to get pretty close to the actual value of e. You
can push the precision further if you like; it'll take longer to
calculate, and dump a bigger pile of digits onto your screen, but
it'll be more accurate.

I don't recommend using fractions.Fraction unless you have a really
fast computer and a LOT of patience. It'll take three parts of forever
to get a result... but that result _will_ be perfectly accurate :)

ChrisA