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 June 24th, 2010, 06:28 AM #1 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Does this equal e? It is the sum of the inverse of a modification of the factorial function, where exponentiation is used instead of multiplication. $1 + 1 + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{4^{3^2}} + \frac{1}{5^{4^{3^2}}} ...= e$ 1. Is the above statement true? 2. Any tips on how I should prove it? PS. Sorry if I've chosen the wrong category, I'm not sure where this belongs.
 June 24th, 2010, 06:36 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 930 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Does this equal e? e is larger. The factorial grows much more slowly than the tetration-like function you use in the denominator. The value of your expression is almost exactly 6160393/2359296 = 2.611114925808376736111111.... The approximation is correct to over a hundred thousand decimal places.
 June 24th, 2010, 06:43 AM #3 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: Does this equal e? Oh, I agree that was obvious.
 June 24th, 2010, 06:57 AM #4 Math Team   Joined: Apr 2010 Posts: 2,765 Thanks: 351 Re: Does this equal e? I don't know whether it helps you, but I assume, you know: $\sum_{k=0}^{\infty}\frac{1}{k!}=e=1+\sum_{k=1}^{\i nfty}\frac{1}{k!}$ Your function is basicly $1+\sum_{k=1}^{\infty}\frac{1}{k^{(k-1)!}$. Hoempa
June 24th, 2010, 07:21 AM   #5
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Re: Does this equal e?

Quote:
 Originally Posted by Hoempa Your function is basicly $1+\sum_{k=1}^{\infty}\frac{1}{k^{(k-1)!}$.
That wasn't what I figured at all. That's double exponential (the k-th term is about $k^{-k^k}$) where what I was looking at was tetrational (the k-th term was about $k^{-k^{k^{k^k}}}$ with k ks.)

 June 24th, 2010, 07:30 AM #6 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: Does this equal e? We have Sigma for addition, Pi for multiplication, but we don't have anything for exponentiation. Is there a notation for integer-repetition where one could specify the operation itself?
 June 24th, 2010, 07:35 AM #7 Math Team   Joined: Apr 2010 Posts: 2,765 Thanks: 351 Re: Does this equal e? Can't you use $a^{b^{c^{d}}}=a^{bcd}$ like I did, or is it false? I tried to describe brangelito's function, and I do agree e is larger. Brangelito, do you mean this Hoempa
June 24th, 2010, 07:39 AM   #8
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Re: Does this equal e?

Quote:
 Originally Posted by brangelito We have Sigma for addition, Pi for multiplication, but we don't have anything for exponentiation. Is there a notation for integer-repetition where one could specify the operation itself?
No, not really. The normal way of representing this would be with recursion: f(1) = 1, f(n) = n^f(n-1).

Quote:
 Originally Posted by Hoempa Brangelito, do you mean this
That's what I use to represent tetration when all the numbers are the same. In this case they aren't, though...

June 24th, 2010, 07:41 AM   #9
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Re: Does this equal e?

Quote:
 Originally Posted by Hoempa Can't you use $a^{b^{c^{d}}}=a^{bcd}$ like I did, or is it false?
That's false. $a^{b^{c^d}}=$a^(b^(c^d)) which is in general not equal to ((a^b)^c)^d$=a^{bcd}$ since exponentiation is not associative. 3^(3^3) = 7625597484987, while (3^3)^3 = 729.

 June 24th, 2010, 07:50 AM #10 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: Does this equal e? I actually meant the tetration way, or at least that's what I calculated.

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