2015 Continuum sum conjecture
#1
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
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#2
The answer appears to be simple.

That is considering we work with the same continuum sum.

Within an analytic region :
Let r > 0.
exp^[1+r](x) = exp^[r](0) + F(exp(x))

where F(0) = 0 and is analytic.

Therefore the continuum sum CS is given by

T_r(x) = CS exp^[1+r](x) = exp^[r](0) x + G(exp(x)).

Therefore T(x) - T(x - 2pi i) = exp^[r](0) * 2 pi i.

regards

tommy1729
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#3
One wonders what happens with r < 0.

I will post a related thread.
I have some ideas but little time.

regards

tommy1729
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#4
Notice the counterintuitive nature

sum [ f(x) - f(x - period f) ] =/= sum [ f(x) ] - sum[ f(x - period f) =/= 0.
even if f(x) - f(x - period f) = 0 and sum [ 0 ] = 0.

I have to think more about this.

Any ideas ?

I considered using tommysum[f(x)] := continuum sum [f(x) - f(0)].

regards

tommy1729
Reply


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