Re: The next exercise

[Michael Heerdegen <michael_heerdegen@web.de>]

<tomas@tuxteam.de> writes:

> My choice was school maths. But I cheated, because I
> studied physics later, so those maths were kept warm
> for some longer while ;-)

Ok, I owe the mathematical background, at least a bit.  The general
background is that

  /n + 1\ n
  |- - -|   --> e
  \  n  /

and if you set n := 1000 you get that 1001^1000 is approximately
e*1000^1000 (*).

Regards,

Michael.



(*) But is this enough to be sure that the first three digits are those
of e?  To be really sure you either check with some sort of calculator
which can calculate such large numbers - there are several ways in Emacs
as we saw!

Or you remember how e can be introduced: with

         /    1\ n               /   1\ n+1
  a_n := |1 + -|   and   b_n := |1 + -|  
         \    n/                 \   n/  

you got that a_n is strictly monotonously increasing, b_n strictly
monotonously decreasing, and b_n - a_n --> 0, so they both converge and
the limit is called e.  From a_n < e < b_n you then also see that

  e    b_n   n+1       b_n   b_n   n+1
 --- < --- = ---  and  --- < --- = ---
 a_n   a_n    n         e    a_n    n

and from that you can even conclude how good the approximation is.  For
n=1000 the exact value must fulfill

   1
 ----- *e*1000^1000 < 1001^1000 < 1.001*e*1000^1000
 1.001

which is good enough for three digits 271 from e.

This is a quite surprising thing, I mean, here are no functions and
derivatives, no fractions, no combinatorics etc involved.  Still,
(1+10^n)^(10^n), a simple product of natural numbers, always starts with
the first n digits of e for every n.  More and more the larger n gets.

Ok - I really apologize for the little mathematical excursion.  Have a
look at M-x calc (it's built in) if you haven't yet.  It's worth it!