Approximation Check E

Question

The MathWorld page gives a simple digital formula for e, which is supposedly corrected for the first 10 ^ 25 digits. He claims to be eapproximately

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

Any idea how to verify the correctness of this formula even for the first 10 digits? Here is another way to write the right side

Power[Plus[1, Power[9, Times[-1, Power[4, Times[7, 6]]]]], Power[3, Power[2, 85]]]

Answer 1

This problem does not need Mathematica at all. Firstly, it is not difficult to show what 9^(4^(7*6))exactly is equal 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 way of representing eis the limit

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

The only question is what is the error in which it nis 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),

n = 3^2^85, log(10,n) = 2^85 log(10,3) ~ 1.85 *10^25,

Answer 2

() . : , e = lim (n- > , (1 + 1/n) ^ n). , , , , 9 ^ (4 ^ (42)) ( ) 3 ^ (2 ^ 85) .

, n = 3 ^ (2 ^ 85), e. , :

>>> from mpmath import *
>>> iv.dps = 50 # let 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

e ;-) - : mathworld , , (1 + 1/1) ^ 1, (1 + 1/2) ^ 2 ..