My Math Forum Computation of L(4)

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 August 29th, 2012, 10:01 PM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Computation of L(4) Define $L(z)= \sum_{n=1}^{\infty} \frac{1}{^z n}$ I am able to calculate it only for z = 2,3. Mathematica overflows if I try to calculate L(4). Any idea how to compute L(4) ? Any help will be appreciated, Balarka .
 August 30th, 2012, 08:00 AM #2 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Computation of L(4) What does the subscript of n mean? Pochhammer symbol?
August 30th, 2012, 08:12 AM   #3
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Re: Computation of L(4)

Quote:
 Originally Posted by The_Fool What does the subscript of n mean? Pochhammer symbol?
Its not a subscript. its a superscript on the left side. This is the tetration.

 August 30th, 2012, 09:15 AM #4 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Computation of L(4) Ah, OK. That's easier to write than the double up arrow notation. I doubt you'll be able to. The number $^{\small 4} 3$ has about $3\cdot 10^{\small 12}+1$ digits. You'd need nearly 3 terabytes of memory just to store that number. Supercomputers can do it, but ours can't. Therefore I must stop calculating after two terms. $L(4)\approx 1.0000152587890625...$ with a LOT of zeroes following. I'd say that it's practically equal to 65537/65536. With this you could make almost integers.
 August 30th, 2012, 10:28 PM #5 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Computation of L(4) Ok, another question related to L(z) : Mathematica says $L(\infty)= \sum_{n=1}^{\infty} \frac{-\ln(n)}{W(-\ln(n))}$ diverges. How can we prove this? Thanks in advance, Balarka .
 August 31st, 2012, 05:02 AM #6 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Computation of L(4) I don't see why it would say that. It converges substantially faster than the series for e. Clearly $\lim_{z\rightarrow\infty}L(z)=1$ from the definition in your original post. The second term goes to zero, and the subsequent terms are smaller and go to zero quicker. The First term is always one.
August 31st, 2012, 05:23 AM   #7
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Re: Computation of L(4)

Quote:
 Originally Posted by The_Fool The second term goes to zero
Why the second term goes to zero? I don't see any reason why it goes to zero.

 August 31st, 2012, 05:25 AM #8 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Computation of L(4) $\lim_{z\rightarrow\infty}$$\frac{1}{2\uparrow\upar row z}$$=0$. The limit of the remaining terms goes go zero faster. This is because $\lim_{z\rightarrow\infty}$$2\uparrow\uparrow z$$=\infty$.
August 31st, 2012, 05:35 AM   #9
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Re: Computation of L(4)

Quote:
 Originally Posted by The_Fool This is because $\lim_{z\rightarrow\infty}$$2\uparrow\uparrow z$$=\infty$.
Oh, you're right. For a minute, I thought that it is a solution to $2^L= L$ which is approximately $0.82 - i 1.56$ but I forgot that it is divergent since
2 > e^(1/e)

Balarka

.

August 31st, 2012, 05:40 AM   #10
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Re: Computation of L(4)

Quote:
 Originally Posted by The_Fool I don't see why it would say that.
Then I guess I implemented it in Mathematica wrongly. I totally forgot that $^{\infty} z$ is convergent if and only if $e^{-e} < z < e^{\frac{1}{e}}$

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