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

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On Mon, 9 Nov 2015 04:21:14 -0800 (PST), Salvatore DI DIO 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  !!!!

Python floats have close to 16 decimal digits of precision.  When you
compute 1+1/p with large p, the result will be close to 1, so digits of
1/p beyond the 16th place will be damaged by rounding.  For p of
900000000, the first nearly-9 digits of 1/p are zero, so the first
"significant" digit is the 10th, and beyond the 16th digit -- the 7th
significant digit -- they're damaged, so you can only expect about 
16-9 = 7 significant digits to be accurate.  And as it turns out,
2.7182820518605446 is good to about 7 significant figures.

It takes longer to explain than to see: roundoff limits you to ...

   (digits of p) + (good digits in the result) = 16  (approximately)

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Calulation in  lim  (1  + 1 /n) ^n  when  n -> infinite Salvatore DI DIO <artyprog@gmail.com> - 2015-11-09 04:21 -0800
  Re: Calulation in lim (1 + 1 /n) ^n when n -> infinite Chris Angelico <rosuav@gmail.com> - 2015-11-09 23:45 +1100
    Re: Calulation in lim (1 + 1 /n) ^n when n -> infinite Salvatore DI DIO <artyprog@gmail.com> - 2015-11-09 07:13 -0800
  Re: Calulation in lim (1 + 1 /n) ^n when n -> infinite Oscar Benjamin <oscar.j.benjamin@gmail.com> - 2015-11-09 12:50 +0000
    Re: Calulation in lim (1 + 1 /n) ^n when n -> infinite Salvatore DI DIO <artyprog@gmail.com> - 2015-11-09 07:13 -0800
    Re: Calulation in lim (1 + 1 /n) ^n when n -> infinite Salvatore DI DIO <artyprog@gmail.com> - 2015-11-09 07:15 -0800
  Re: Calulation in  lim  (1  + 1 /n) ^n  when  n -> infinite Peter Pearson <pkpearson@nowhere.invalid> - 2015-11-09 17:51 +0000
    Re: Calulation in  lim  (1  + 1 /n) ^n  when  n -> infinite Salvatore DI DIO <artyprog@gmail.com> - 2015-11-10 00:26 -0800

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