05/25/2015, 11:29 PM
I noticed a possible pattern.
\( \sum_{k=0}^{z-1} e^{nk} = \frac{e^{nz} - 1}{e^{n} - 1} \)
next
\( \sum_{k=0}^{z-1} e^{e^k} = \sum_{k=0}^{z-1} f(k) = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\sum_{k=0}^{z-1} e^{nk}\right) = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{z-1} e^{nk} = \sum_{n=0}^{\infty} \frac{e^{nz} - 1}{n! \left(e^{n} - 1\right)} \)
next
define : \( e^{e^{e^x}} = \sum_{n=0}^{\infty} a_n e^{nx} \)
then
\( \begin{align}\sum_{n=0}^{x-1} e^{e^{e^n}} &= a_0 x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} \left(e^{nx} - 1\right) \\ &= \left(a_0 - \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1}\right) x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} e^{nx}\end{align} \)
Now it appears
The continuum sum up to z-1 of exp^[n](x) is given by
CS [ exp^[n](x) , z-1] = P_n(x) + F_n(x)
where F_n is 2pi i periodic and P_n is a polynomial of degree (at most) n.
I guess there is a simple reason for it.
Right ??
What if n is not an integer ? say n = 3/2 ?
Does that imply P_{3/2}(x) =< O( x^{3/2} ) ??
regards
tommy1729
\( \sum_{k=0}^{z-1} e^{nk} = \frac{e^{nz} - 1}{e^{n} - 1} \)
next
\( \sum_{k=0}^{z-1} e^{e^k} = \sum_{k=0}^{z-1} f(k) = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\sum_{k=0}^{z-1} e^{nk}\right) = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{z-1} e^{nk} = \sum_{n=0}^{\infty} \frac{e^{nz} - 1}{n! \left(e^{n} - 1\right)} \)
next
define : \( e^{e^{e^x}} = \sum_{n=0}^{\infty} a_n e^{nx} \)
then
\( \begin{align}\sum_{n=0}^{x-1} e^{e^{e^n}} &= a_0 x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} \left(e^{nx} - 1\right) \\ &= \left(a_0 - \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1}\right) x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} e^{nx}\end{align} \)
Now it appears
The continuum sum up to z-1 of exp^[n](x) is given by
CS [ exp^[n](x) , z-1] = P_n(x) + F_n(x)
where F_n is 2pi i periodic and P_n is a polynomial of degree (at most) n.
I guess there is a simple reason for it.
Right ??
What if n is not an integer ? say n = 3/2 ?
Does that imply P_{3/2}(x) =< O( x^{3/2} ) ??
regards
tommy1729