3
$\begingroup$

I am looking for a value $a \approx 14$ with some nice property. So I am going to define some things with this value $a$ and then ask what $a$ does the trick I want (If there is some $a$ that does the trick at all).

Definitions and Intro

Let $f(x,t) = \frac{\ln(t+a)^x}{t}$ and note that $f_x(x,t)=\frac{\ln(t+a)^{x}\ln(\ln(t+a))}{t}$ where $f_x$ refers to $\frac{d}{dx} f(x,t)$

Now define $$g(x) = \lim_{m\to\infty} \sum_{t=1}^m f(x,t)-\int_1^m f(x,t)dt $$

Note that $g(0) =\gamma$ the Euler Mascheroni constant and the generalization above can be found under the generalization section of that wiki (So I am not conjuring this idea from thin air). In fact, when $a=0$ it seems that $g(x)$ is connected with what is referred to as Stieljes Constants.

It looks to me that there may exist some $a$ value that $g(x)=g'(x)$. Which would be kind of interesting. Because this would mean that $g(x)= \gamma e^x$.

Here's a graph which led me to these suspicions. I won't reproduce the image of the graph because it just looks like $y=\gamma e^x$. The interesting thing is that the numerical derivative nearly overlays the function.

The Question Does there exist some $a$ that does this? And what is it?

Some preliminary notes/ attempts to make progress

We should note that $$g'(x) = \lim_{m\to\infty} \sum_{t=1}^m f_x(x,t)-\int_1^m f_x(x,t)dt $$

Which allows for a little algebraic manipulations after we take the assumption $g'(x) =g(x)$. These manipulations haven't really helped me find out what $a$ is...

Motivations David Hilbert referred to the puzzle of proving the irrationality of $\gamma$ as "unapproachable." Which explains my title... I am just looking for some approaches to $\gamma$ which may communicate some information about this constant.

$\endgroup$
  • $\begingroup$ I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$ $\endgroup$ – Mason Nov 20 '18 at 18:11
  • $\begingroup$ Another way to think about this is that the $a$ value just changes the index of the summation and the integral. $\endgroup$ – Mason Nov 22 '18 at 14:32
  • $\begingroup$ I would think that we can prove that there isn't one. Or there is one. I think there may be because of the graph which I've linked. $\endgroup$ – Mason Nov 22 '18 at 21:44
  • 1
    $\begingroup$ Why do you think that for some $a$ then (for every $x$ in some interval) $g_a(x) = g_a'(x)$ ? The Stieltjes constant are the derivatives of $F_0(s) = (s-1) \zeta(s)$ at $s=1$. So try finding the analytic function $F_a(s)$ whose derivatives at $s=1$ are related to $g_a(n)$ $\endgroup$ – reuns Nov 22 '18 at 21:46
  • $\begingroup$ A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=\gamma$ . Then choose $x_0\neq 0$ and test, if $g’(x_0)=g(x_0)$ (e.g. up to 8 digits behind the decimal point). Then you know, whether it makes sense to expect $g'(x)=g(x)$ or not. $\endgroup$ – user90369 Nov 29 '18 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.