# Tetration Forum

Full Version: using the sum , hoping for convergence
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without going into details , the idea is simple.

ignoring some details we consider :

we have a continuum sum operator.

tet'(x) = tet'(0)* continuum product till tet(x).

we can use the continuum sum to compute the continuum product.

once we have tet'(x) , we can find tet(x) by integration.

the problem with the above is that its an equation with selfreference.

the proposed solution is to do the same but with iteration.

consider any coo solution to tet(x).

thats our starting function tet_0(x).

tet_1 ' (x) = tet_0 ' (0) * continuum product till tet_0 (x).

tet_1 (x) = integral tet_1 ' (x).

we continue :

tet_2 ' (x) = tet_1 ' (0) * continuum product till tet_1 (x).

tet_2 (x) = integral tet_2 ' (x).

etc

till we get our converging limit.

basicly ...

tommy1729
This is a bit problematic because we don't know how to define the continuum sum for all functions in the sequence.
actually we do.
How?

If we start with the constant function 1, we get

1 -> Gamma(x + 1) -> ?

If we start with the constant function e, we get

e -> e^x -> integral( exp( (x^2 + x) / 2) ) -> ?
(08/28/2010, 10:04 AM)kobi_78 Wrote: [ -> ]How?

If we start with the constant function 1, we get

1 -> Gamma(x + 1) -> ?

If we start with the constant function e, we get

e -> e^x -> integral( exp( (x^2 + x) / 2) ) -> ?

for starters , we start with a coo solution to tetration mapping the positive reals to the reals.

secondly , we use the q-method for the continuum sum.

this method is explained by mike 3 :

We have

$f(z) = \sum_{n=0}^{\infty} a_n b^{nz}$

or, even better

$f(z) = \sum_{n=-\infty}^{\infty} a_n b^{nz}$.

Then,

$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} a_n \frac{b^{nz} - 1}{b^n - 1}$

with the term at $n = 0$ interpreted as $a_0 z$, so,

$\sum_{n=0}^{z-1} f(n) = \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1}\right) + a_0 z + \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1} b^{nz}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1} b^{nz}\right)$.

However, when viewed in the complex plane, we see that exp-series are just Fourier series

$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{i \frac{2\pi}{P} n z}$.

which represent a periodic function with period $P$.

Thus it would seem that only periodic functions can be continuum-summed this way. Tetration is not periodic, so how could this help?

Well, we could consider the possibility of continuum-summing an aperiodic function by taking a limit of a sequence of periodic functions that converge to it. The hypothesis I have is that if $f_0, f_1, f_2, ...$ is a sequence of periodic analytic functions converging to a given $f$, then their continuum sums, if they converge to anything, converge to the same thing, regardless of the sequence of functions.

Would you have any ideas to prove or refute this hypothesis?

An example. Let $f(z) = z$, the identity function. We can't continuum-sum it with the exp-series directly. But now let $f_u(z) = u \sinh\left(\frac{z}{u}\right)$, so that $\lim_{u \rightarrow \infty} f_u(z) = f(z)$, a sequence of periodic (with imaginary period $2 \pi i u$) entire functions converging to $f(z)$. The continuum sum is, by using the exponential expansion of sinh giving $f_u(z) = -\frac{u}{2} e^{-\frac{z}{u}} + \frac{u}{2} e^{\frac{z}{u}}$,

$\sum_{n=0}^{z-1} f_u(n) = \frac{u}{2} \left(\frac{e^{\frac{z}{u}} - 1}{e^{\frac{1}{u}} - 1} - \frac{e^{-\frac{z}{u}} - 1}{e^{-\frac{1}{u}} - 1}\right)$.

Though I didn't bother to try to work it out by hand, instead using a computer math package, the limit is $\frac{x(x-1)}{2}$ as $u \rightarrow \infty$, agreeing with the result from Faulhaber's formula.

Another example is the function $f(z) = \frac{1}{z}$, or better, $f(z) = \frac{1}{z+1}$. We can construct periodic approximations like $f_u(z) = \frac{1}{u \sinh\left(\frac{z}{u}\right) + 1}$, and take the limit at infinity. The terms in the Fourier series are even worse. If we use a numerical approximation of the series with period , I get the continuum sum from 0 to -1/2 as ~0.6137 (rounded, act. more like something over 0.61369) suggesting the process is recovering the digamma function, as can be seen by setting 1/2 in the canonical formula $-\gamma + \Psi(x + 1) = \sum_{n=0}^{x-1} \frac{1}{x+1}$ yielding 0.61370563888011...

Trying it with $log(1 + z)$, so as to attempt to evaluate $z! = \exp\left(\sum_{n=0}^{z-1} \log(1+z)\right)$ yields values that agree with the gamma function, providing more evidence that gamma is the natural extension of the factorial function to the complex plane.

Thus it seems this continuum sum is recovering all the expected sums and extensions.

--

this should get you started.

regards

tommy1729