Mixing
Approaching The Euler-Mascheroni Constant
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 $$g(x) = lim_{mtoinfty} sum_{t=1}^m f(x,t)-int_1^m f(x,t)dt $$
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 $$g'(x) = lim_{mtoinfty} sum_{t=1}^m f_x(x,t)-int_1^m f_x(x,t)dt $$
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.
real-analysis sequences-and-series eulers-constant
asked Nov 20 '18 at 15:16
Mason
1,9641530
|
show 2 more comments
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 $$g(x) = lim_{mtoinfty} sum_{t=1}^m f(x,t)-int_1^m f(x,t)dt $$
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 $$g'(x) = lim_{mtoinfty} sum_{t=1}^m f_x(x,t)-int_1^m f_x(x,t)dt $$
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.
real-analysis sequences-and-series eulers-constant
asked Nov 20 '18 at 15:16
Mason
1,9641530
I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$
– Mason
Nov 20 '18 at 18:11
Another way to think about this is that the $a$ value just changes the index of the summation and the integral.
– Mason
Nov 22 '18 at 14:32
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.
– Mason
Nov 22 '18 at 21:44
1
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 $g_a(n)$
– reuns
Nov 22 '18 at 21:46
A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=gamma$ . Then choose $x_0neq 0$ 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.
– user90369
Nov 29 '18 at 13:34
|
show 2 more comments
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 $$g(x) = lim_{mtoinfty} sum_{t=1}^m f(x,t)-int_1^m f(x,t)dt $$
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 $$g'(x) = lim_{mtoinfty} sum_{t=1}^m f_x(x,t)-int_1^m f_x(x,t)dt $$
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.
real-analysis sequences-and-series eulers-constant
asked Nov 20 '18 at 15:16
Mason
1,9641530
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 $$g(x) = lim_{mtoinfty} sum_{t=1}^m f(x,t)-int_1^m f(x,t)dt $$
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 $$g'(x) = lim_{mtoinfty} sum_{t=1}^m f_x(x,t)-int_1^m f_x(x,t)dt $$
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.
real-analysis sequences-and-series eulers-constant
real-analysis sequences-and-series eulers-constant
asked Nov 20 '18 at 15:16
Mason
1,9641530
asked Nov 20 '18 at 15:16
Mason
1,9641530
edited Nov 20 '18 at 17:03
asked Nov 20 '18 at 15:16
Mason
1,9641530
asked Nov 20 '18 at 15:16
Mason
1,9641530
asked Nov 20 '18 at 15:16
Mason
1,9641530
1,9641530
I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$
– Mason
Nov 20 '18 at 18:11
Another way to think about this is that the $a$ value just changes the index of the summation and the integral.
– Mason
Nov 22 '18 at 14:32
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.
– Mason
Nov 22 '18 at 21:44
1
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 $g_a(n)$
– reuns
Nov 22 '18 at 21:46
A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=gamma$ . Then choose $x_0neq 0$ 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.
– user90369
Nov 29 '18 at 13:34
|
show 2 more comments
I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$
– Mason
Nov 20 '18 at 18:11
Another way to think about this is that the $a$ value just changes the index of the summation and the integral.
– Mason
Nov 22 '18 at 14:32
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.
– Mason
Nov 22 '18 at 21:44
1
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 $g_a(n)$
– reuns
Nov 22 '18 at 21:46
A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=gamma$ . Then choose $x_0neq 0$ 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.
– user90369
Nov 29 '18 at 13:34
I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$
– Mason
Nov 20 '18 at 18:11
I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$
– Mason
Nov 20 '18 at 18:11
Another way to think about this is that the $a$ value just changes the index of the summation and the integral.
– Mason
Nov 22 '18 at 14:32
Another way to think about this is that the $a$ value just changes the index of the summation and the integral.
– Mason
Nov 22 '18 at 14:32
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.
– Mason
Nov 22 '18 at 21:44
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.
– Mason
Nov 22 '18 at 21:44
1
1
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 $g_a(n)$
– reuns
Nov 22 '18 at 21:46
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 $g_a(n)$
– reuns
Nov 22 '18 at 21:46
A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=gamma$ . Then choose $x_0neq 0$ 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.
– user90369
Nov 29 '18 at 13:34
A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=gamma$ . Then choose $x_0neq 0$ 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.
– user90369
Nov 29 '18 at 13:34
|
show 2 more comments
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I suppose I could just ask more broadly about the class of functions $f$ such that $g(x)=g'(x)$
– Mason
Nov 20 '18 at 18:11
Another way to think about this is that the $a$ value just changes the index of the summation and the integral.
– Mason
Nov 22 '18 at 14:32
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.
– Mason
Nov 22 '18 at 21:44
1
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 $g_a(n)$
– reuns
Nov 22 '18 at 21:46
A suggestion (rough method): Find out numerically, for which $a$ is $g’(0)=gamma$ . Then choose $x_0neq 0$ 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.
– user90369
Nov 29 '18 at 13:34