Why should the series representation of the zeta function know about its analytic continuation?
$begingroup$
In physics, when we calculate the renormalized sum of $S=sum_{n=1}^infty n$, we usually use an exponential regulator and instead first calculate
$$S_epsilon = sum_{n=1}^infty ne^{-epsilon n} = frac{e^epsilon}{(e^epsilon -1)^2}. $$
There's nothing controversial about this, but an example by Lubos Motl can be found here.
Now, if we expand this around small $epsilon>0$ we obtain that
begin{align*}
S_epsilon &= frac{1}{epsilon^2} - frac{1}{12} + {cal O}(epsilon^2)\
&=zeta(-1) + frac{1}{epsilon^2} + {cal O}(epsilon^2).
end{align*}
So that when we take $epsilon to 0$ (in which limit $S_epsilon to S)$, we obtain that
$$S to zeta(-1) + text{divergent piece.}$$
Now, I know that this "decomposition", if you will, into a convergent and divergent part is independent of the sum regulator, and hence always yields the same result. Furthermore, the domain of validity of the series representation of the $zeta$ function is only when $mathbb{R}(s) >0$. Therefore, we ought to expect it to be of no use to us when naively applying it in regions where $mathbb{R}(s)<1$.
However, it is useful. Namely, its renormalized sum gives us the correct value for the analytic continuation of the $zeta$ function. Furthermore, since the analytic continuation of a complex function is unique I feel there must be a connection between these two things.
Question: How does the series representation "know" about its analytic continuation? That is, what exactly is the connection between the naive summation and its analytic continuation?
riemann-zeta analytic-continuation
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
$endgroup$
add a comment |
$begingroup$
In physics, when we calculate the renormalized sum of $S=sum_{n=1}^infty n$, we usually use an exponential regulator and instead first calculate
$$S_epsilon = sum_{n=1}^infty ne^{-epsilon n} = frac{e^epsilon}{(e^epsilon -1)^2}. $$
There's nothing controversial about this, but an example by Lubos Motl can be found here.
Now, if we expand this around small $epsilon>0$ we obtain that
begin{align*}
S_epsilon &= frac{1}{epsilon^2} - frac{1}{12} + {cal O}(epsilon^2)\
&=zeta(-1) + frac{1}{epsilon^2} + {cal O}(epsilon^2).
end{align*}
So that when we take $epsilon to 0$ (in which limit $S_epsilon to S)$, we obtain that
$$S to zeta(-1) + text{divergent piece.}$$
Now, I know that this "decomposition", if you will, into a convergent and divergent part is independent of the sum regulator, and hence always yields the same result. Furthermore, the domain of validity of the series representation of the $zeta$ function is only when $mathbb{R}(s) >0$. Therefore, we ought to expect it to be of no use to us when naively applying it in regions where $mathbb{R}(s)<1$.
However, it is useful. Namely, its renormalized sum gives us the correct value for the analytic continuation of the $zeta$ function. Furthermore, since the analytic continuation of a complex function is unique I feel there must be a connection between these two things.
Question: How does the series representation "know" about its analytic continuation? That is, what exactly is the connection between the naive summation and its analytic continuation?
riemann-zeta analytic-continuation
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
$endgroup$
1
$begingroup$
To see what is really happening you should look at $Gamma(s) zeta(s) = int_0^infty t^{s-1} f(t)dt$ where $f(t) = sum_{n=1}^infty e^{-nt}$. That $g(t)=t f(t)$ is $C^infty$ at $t=0$ implies that $int_0^1 t^{s-1} (f(t)- t^{-1}sum_{k=0}^K frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $Re(s) > -K$, whence the poles of $Gamma(s) zeta(s) $ are at the negative integers, of order $1$ and residue $frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $zeta(-k) = (-1)^{k} g^{(k-1)}(0)$
$endgroup$
– reuns
Jan 10 at 6:01
add a comment |
$begingroup$
In physics, when we calculate the renormalized sum of $S=sum_{n=1}^infty n$, we usually use an exponential regulator and instead first calculate
$$S_epsilon = sum_{n=1}^infty ne^{-epsilon n} = frac{e^epsilon}{(e^epsilon -1)^2}. $$
There's nothing controversial about this, but an example by Lubos Motl can be found here.
Now, if we expand this around small $epsilon>0$ we obtain that
begin{align*}
S_epsilon &= frac{1}{epsilon^2} - frac{1}{12} + {cal O}(epsilon^2)\
&=zeta(-1) + frac{1}{epsilon^2} + {cal O}(epsilon^2).
end{align*}
So that when we take $epsilon to 0$ (in which limit $S_epsilon to S)$, we obtain that
$$S to zeta(-1) + text{divergent piece.}$$
Now, I know that this "decomposition", if you will, into a convergent and divergent part is independent of the sum regulator, and hence always yields the same result. Furthermore, the domain of validity of the series representation of the $zeta$ function is only when $mathbb{R}(s) >0$. Therefore, we ought to expect it to be of no use to us when naively applying it in regions where $mathbb{R}(s)<1$.
However, it is useful. Namely, its renormalized sum gives us the correct value for the analytic continuation of the $zeta$ function. Furthermore, since the analytic continuation of a complex function is unique I feel there must be a connection between these two things.
Question: How does the series representation "know" about its analytic continuation? That is, what exactly is the connection between the naive summation and its analytic continuation?
riemann-zeta analytic-continuation
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
$endgroup$
In physics, when we calculate the renormalized sum of $S=sum_{n=1}^infty n$, we usually use an exponential regulator and instead first calculate
$$S_epsilon = sum_{n=1}^infty ne^{-epsilon n} = frac{e^epsilon}{(e^epsilon -1)^2}. $$
There's nothing controversial about this, but an example by Lubos Motl can be found here.
Now, if we expand this around small $epsilon>0$ we obtain that
begin{align*}
S_epsilon &= frac{1}{epsilon^2} - frac{1}{12} + {cal O}(epsilon^2)\
&=zeta(-1) + frac{1}{epsilon^2} + {cal O}(epsilon^2).
end{align*}
So that when we take $epsilon to 0$ (in which limit $S_epsilon to S)$, we obtain that
$$S to zeta(-1) + text{divergent piece.}$$
Now, I know that this "decomposition", if you will, into a convergent and divergent part is independent of the sum regulator, and hence always yields the same result. Furthermore, the domain of validity of the series representation of the $zeta$ function is only when $mathbb{R}(s) >0$. Therefore, we ought to expect it to be of no use to us when naively applying it in regions where $mathbb{R}(s)<1$.
However, it is useful. Namely, its renormalized sum gives us the correct value for the analytic continuation of the $zeta$ function. Furthermore, since the analytic continuation of a complex function is unique I feel there must be a connection between these two things.
Question: How does the series representation "know" about its analytic continuation? That is, what exactly is the connection between the naive summation and its analytic continuation?
riemann-zeta analytic-continuation
riemann-zeta analytic-continuation
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
asked Jan 10 at 5:53
InertialObserverInertialObserver
387110
387110
1
$begingroup$
To see what is really happening you should look at $Gamma(s) zeta(s) = int_0^infty t^{s-1} f(t)dt$ where $f(t) = sum_{n=1}^infty e^{-nt}$. That $g(t)=t f(t)$ is $C^infty$ at $t=0$ implies that $int_0^1 t^{s-1} (f(t)- t^{-1}sum_{k=0}^K frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $Re(s) > -K$, whence the poles of $Gamma(s) zeta(s) $ are at the negative integers, of order $1$ and residue $frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $zeta(-k) = (-1)^{k} g^{(k-1)}(0)$
$endgroup$
– reuns
Jan 10 at 6:01
add a comment |
1
$begingroup$
To see what is really happening you should look at $Gamma(s) zeta(s) = int_0^infty t^{s-1} f(t)dt$ where $f(t) = sum_{n=1}^infty e^{-nt}$. That $g(t)=t f(t)$ is $C^infty$ at $t=0$ implies that $int_0^1 t^{s-1} (f(t)- t^{-1}sum_{k=0}^K frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $Re(s) > -K$, whence the poles of $Gamma(s) zeta(s) $ are at the negative integers, of order $1$ and residue $frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $zeta(-k) = (-1)^{k} g^{(k-1)}(0)$
$endgroup$
– reuns
Jan 10 at 6:01
1
1
$begingroup$
To see what is really happening you should look at $Gamma(s) zeta(s) = int_0^infty t^{s-1} f(t)dt$ where $f(t) = sum_{n=1}^infty e^{-nt}$. That $g(t)=t f(t)$ is $C^infty$ at $t=0$ implies that $int_0^1 t^{s-1} (f(t)- t^{-1}sum_{k=0}^K frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $Re(s) > -K$, whence the poles of $Gamma(s) zeta(s) $ are at the negative integers, of order $1$ and residue $frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $zeta(-k) = (-1)^{k} g^{(k-1)}(0)$
$endgroup$
– reuns
Jan 10 at 6:01
$begingroup$
To see what is really happening you should look at $Gamma(s) zeta(s) = int_0^infty t^{s-1} f(t)dt$ where $f(t) = sum_{n=1}^infty e^{-nt}$. That $g(t)=t f(t)$ is $C^infty$ at $t=0$ implies that $int_0^1 t^{s-1} (f(t)- t^{-1}sum_{k=0}^K frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $Re(s) > -K$, whence the poles of $Gamma(s) zeta(s) $ are at the negative integers, of order $1$ and residue $frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $zeta(-k) = (-1)^{k} g^{(k-1)}(0)$
$endgroup$
– reuns
Jan 10 at 6:01
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068286%2fwhy-should-the-series-representation-of-the-zeta-function-know-about-its-analyti%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068286%2fwhy-should-the-series-representation-of-the-zeta-function-know-about-its-analyti%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
To see what is really happening you should look at $Gamma(s) zeta(s) = int_0^infty t^{s-1} f(t)dt$ where $f(t) = sum_{n=1}^infty e^{-nt}$. That $g(t)=t f(t)$ is $C^infty$ at $t=0$ implies that $int_0^1 t^{s-1} (f(t)- t^{-1}sum_{k=0}^K frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $Re(s) > -K$, whence the poles of $Gamma(s) zeta(s) $ are at the negative integers, of order $1$ and residue $frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $zeta(-k) = (-1)^{k} g^{(k-1)}(0)$
$endgroup$
– reuns
Jan 10 at 6:01