sexp by continuum product ?

[tommy1729]

We all know

\( integral \) \( dx/x = ln(x) \)
\( integral \) \( dx/(x ln(x)) = ln(ln(x)) \)
\( integral \) \( dx/( x ln(x) ln(ln(x)) ) = ln(ln(ln(x))) \)

So now the idea occurs : let \( ctp_k (f(x,k)) \) denote continuum product with respect to k ,

\( integral \) \( dx / ctp_k (sexp(slog(x) - k)) = sexp(slog(x) - k) \)

And the continuum product is defined as e^(continuum sum ( log (a_k ))

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.

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.

So we ( try to ... ) use this idea to compute a (probably unique and analytic ) sexp for bases > sqrt(e).

regards

tommy1729

Moderator's note: If you quote things, please mark it accordingly best with link to its origin. Otherwise I would consider it as wilful deceit. I modified your post accordingly.