Checking approximation of E

MathWorld page gives a simple numeric formula for e that's allegedly correct for first 10^25 digits. It states that e is approximately

(1 + 9^-4^(7*6))^3^2^85

Any idea how to check whether this formula is correct even for the first 10 digits? Here's another way of writing the right hand side

Power[Plus[1, Power[9, 开发者_JAVA技巧Times[-1, Power[4, Times[7, 6]]]]], Power[3, Power[2, 85]]]

This problem does not need Mathematica at all. First, it is easy to show that 9^(4^(7*6)) is exactly equal to 3^2^85, since

 9^(4^(7*6)) = 3^(2*4^(7*6)) = 3^(2^(1+2*(7*6))) = 3^2^85

Then, we know that one of the ways to represent e is as a limit

e = lim (1+1/n)^n, n->infinity

The only question is what is the error given that n is very large but finite. We have

(1+1/n)^n = e^log((1+1/n)^n) = e^(n*log(1+1/n)) = e^(1-1/(2n)+O(1/n^2)) = e + O(1/n),

Given the n = 3^2^85, i we take the log(10,n) = 2^85 log(10,3) ~ 1.85 *10^25, we get an estimate similar to the quoted one

Repeatedly taking logs is a nice (usually) generally-applicable solution to problems of this sort. Here's a more special-case approach to this problem: recall that e = lim(n->infinity, (1+1/n)^n). So to be a good approximation to e, all we need is for 9^(4^(42)) (the denominator of the fractional part) to be sufficiently close to 3^(2^85) and big.

In this case, they're identical, so we have n=3^(2^85), and it's going to be a very good approximation to e. These are big numbers, but not unworkably so:

>>> from mpmath import *
>>> iv.dps = 50 # let's use interval arithmetic, just for fun
>>> x = mpi(9)**(-(4**(42)))
>>> up = (mpi(3)**(2**85))
>>> x
mpi('1.4846305545498656772753385085652043615636250118238876e-18457734525360901453873570', 
'1.4846305545498656772753385085652043615636250118238899e-18457734525360901453873570')
>>> 1/x
mpi('6.7356824695231749871315222528985858700759934154677854e+18457734525360901453873569', 
'6.7356824695231749871315222528985858700759934154678156e+18457734525360901453873569')
>>> up
mpi('6.7356824695231749871315222528985858700759934154678005e+18457734525360901453873569', 
'6.7356824695231749871315222528985858700759934154678156e+18457734525360901453873569')
>>> 0 in (1/x-up)
True

Working out the exact error bounds on e is left as an exercise for the reader ;-) -- hint: compare the number of digits of accuracy the mathworld page claims and the above numbers, and ask why that might be, thinking of the series of approximations (1+1/1)^1, (1+1/2)^2, etc.