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

Mon Nov 9 07:50:34 EST 2015

On 9 November 2015 at 12:21, 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 ?

You're performing a floating point calculation and expecting exact results.
Try this:
 >>> lim(10 ** 17)
 1.0
Why does this happen? Well in this case that number is 10**17 and it
turns out that
 >>> 1 + 1.0 / 10**17
 1.0
This is because there aren't enough digits in double precision
floating point to represent the difference between 1 and 1+1e-17. As p
gets larger the addition 1+1.0/p because less and less accurate. The
error in computing that is amplified by raising to a large power p.
You can use more digits by using the decimal module:
 >>> from decimal import Decimal, localcontext
 >>> def lim(p):
 ... return (1 + 1 / Decimal(p)) ** p
 ...
 >>> with localcontext() as ctx:
 ... ctx.prec = 100
 ... lim(10**17)
 ...
 Decimal('2.718281828459045221768878329057436445543726874642885850945607978722364313911964199165598158907225076')
You can also install sympy and find this result symbolically:
 >>> from sympy import Symbol, limit, oo
 >>> p = Symbol('p', integer=True)
 >>> limit((1 + 1/p)**p, p, oo)
 E
--
Oscar