Closed Forms of Certain Zeta constants?
$begingroup$
The Riemann Zeta function $zeta(s)=sum_{n=1}^infty frac{1}{n^s}$ converges for $operatorname{Re}(s)>1$. I am interested in some particular "irrational " values of it such as:
$zeta(pi)=1.176241738ldots$,
$zeta(e)=1.2690096043ldots$,
$zeta(sqrt2)=3.020737679ldots$,- $ldots$
Are there closed form representations for these and constants? Are there formulas which consists of these constants?
riemann-zeta closed-form constants
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
$endgroup$
add a comment |
$begingroup$
The Riemann Zeta function $zeta(s)=sum_{n=1}^infty frac{1}{n^s}$ converges for $operatorname{Re}(s)>1$. I am interested in some particular "irrational " values of it such as:
$zeta(pi)=1.176241738ldots$,
$zeta(e)=1.2690096043ldots$,
$zeta(sqrt2)=3.020737679ldots$,- $ldots$
Are there closed form representations for these and constants? Are there formulas which consists of these constants?
riemann-zeta closed-form constants
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
$endgroup$
2
$begingroup$
The Riemann Zeta Function exists for all complex numbers except for $;z=1;$ ....Perhaps you meant the summatory form $$sum_{n=1}^inftyfrac1{n^s};,;;text{which converges for Re},(s)>1; ?$$
$endgroup$
– DonAntonio
Sep 28 '13 at 16:32
$begingroup$
Yes that. Specially !and @DonAntonio does the Riemann zeta function exists for $0$ too?
$endgroup$
– Shivam Patel
Sep 28 '13 at 16:36
$begingroup$
Yes Shivam: for any complex value except $;1;$ .
$endgroup$
– DonAntonio
Sep 28 '13 at 16:42
2
$begingroup$
In some sense, $zeta(sqrt{2})$ is already a closed form.
$endgroup$
– Hurkyl
Sep 28 '13 at 17:32
add a comment |
$begingroup$
The Riemann Zeta function $zeta(s)=sum_{n=1}^infty frac{1}{n^s}$ converges for $operatorname{Re}(s)>1$. I am interested in some particular "irrational " values of it such as:
$zeta(pi)=1.176241738ldots$,
$zeta(e)=1.2690096043ldots$,
$zeta(sqrt2)=3.020737679ldots$,- $ldots$
Are there closed form representations for these and constants? Are there formulas which consists of these constants?
riemann-zeta closed-form constants
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
$endgroup$
The Riemann Zeta function $zeta(s)=sum_{n=1}^infty frac{1}{n^s}$ converges for $operatorname{Re}(s)>1$. I am interested in some particular "irrational " values of it such as:
$zeta(pi)=1.176241738ldots$,
$zeta(e)=1.2690096043ldots$,
$zeta(sqrt2)=3.020737679ldots$,- $ldots$
Are there closed form representations for these and constants? Are there formulas which consists of these constants?
riemann-zeta closed-form constants
riemann-zeta closed-form constants
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
edited Dec 7 '18 at 10:07
José Carlos Santos
166k22132235
166k22132235
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
asked Sep 28 '13 at 16:24
Shivam PatelShivam Patel
2,11111725
2,11111725
2
$begingroup$
The Riemann Zeta Function exists for all complex numbers except for $;z=1;$ ....Perhaps you meant the summatory form $$sum_{n=1}^inftyfrac1{n^s};,;;text{which converges for Re},(s)>1; ?$$
$endgroup$
– DonAntonio
Sep 28 '13 at 16:32
$begingroup$
Yes that. Specially !and @DonAntonio does the Riemann zeta function exists for $0$ too?
$endgroup$
– Shivam Patel
Sep 28 '13 at 16:36
$begingroup$
Yes Shivam: for any complex value except $;1;$ .
$endgroup$
– DonAntonio
Sep 28 '13 at 16:42
2
$begingroup$
In some sense, $zeta(sqrt{2})$ is already a closed form.
$endgroup$
– Hurkyl
Sep 28 '13 at 17:32
add a comment |
2
$begingroup$
The Riemann Zeta Function exists for all complex numbers except for $;z=1;$ ....Perhaps you meant the summatory form $$sum_{n=1}^inftyfrac1{n^s};,;;text{which converges for Re},(s)>1; ?$$
$endgroup$
– DonAntonio
Sep 28 '13 at 16:32
$begingroup$
Yes that. Specially !and @DonAntonio does the Riemann zeta function exists for $0$ too?
$endgroup$
– Shivam Patel
Sep 28 '13 at 16:36
$begingroup$
Yes Shivam: for any complex value except $;1;$ .
$endgroup$
– DonAntonio
Sep 28 '13 at 16:42
2
$begingroup$
In some sense, $zeta(sqrt{2})$ is already a closed form.
$endgroup$
– Hurkyl
Sep 28 '13 at 17:32
2
2
$begingroup$
The Riemann Zeta Function exists for all complex numbers except for $;z=1;$ ....Perhaps you meant the summatory form $$sum_{n=1}^inftyfrac1{n^s};,;;text{which converges for Re},(s)>1; ?$$
$endgroup$
– DonAntonio
Sep 28 '13 at 16:32
$begingroup$
The Riemann Zeta Function exists for all complex numbers except for $;z=1;$ ....Perhaps you meant the summatory form $$sum_{n=1}^inftyfrac1{n^s};,;;text{which converges for Re},(s)>1; ?$$
$endgroup$
– DonAntonio
Sep 28 '13 at 16:32
$begingroup$
Yes that. Specially !and @DonAntonio does the Riemann zeta function exists for $0$ too?
$endgroup$
– Shivam Patel
Sep 28 '13 at 16:36
$begingroup$
Yes that. Specially !and @DonAntonio does the Riemann zeta function exists for $0$ too?
$endgroup$
– Shivam Patel
Sep 28 '13 at 16:36
$begingroup$
Yes Shivam: for any complex value except $;1;$ .
$endgroup$
– DonAntonio
Sep 28 '13 at 16:42
$begingroup$
Yes Shivam: for any complex value except $;1;$ .
$endgroup$
– DonAntonio
Sep 28 '13 at 16:42
2
2
$begingroup$
In some sense, $zeta(sqrt{2})$ is already a closed form.
$endgroup$
– Hurkyl
Sep 28 '13 at 17:32
$begingroup$
In some sense, $zeta(sqrt{2})$ is already a closed form.
$endgroup$
– Hurkyl
Sep 28 '13 at 17:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no reason to suspect that these have a "closed form". There isn't even a known closed form for $zeta(3)$...
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
$endgroup$
add a comment |
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%2f507957%2fclosed-forms-of-certain-zeta-constants%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no reason to suspect that these have a "closed form". There isn't even a known closed form for $zeta(3)$...
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
$endgroup$
add a comment |
$begingroup$
There is no reason to suspect that these have a "closed form". There isn't even a known closed form for $zeta(3)$...
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
$endgroup$
add a comment |
$begingroup$
There is no reason to suspect that these have a "closed form". There isn't even a known closed form for $zeta(3)$...
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
$endgroup$
There is no reason to suspect that these have a "closed form". There isn't even a known closed form for $zeta(3)$...
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
answered Sep 28 '13 at 17:15
Bruno JoyalBruno Joyal
42.9k695186
42.9k695186
add a comment |
add a comment |
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%2f507957%2fclosed-forms-of-certain-zeta-constants%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
2
$begingroup$
The Riemann Zeta Function exists for all complex numbers except for $;z=1;$ ....Perhaps you meant the summatory form $$sum_{n=1}^inftyfrac1{n^s};,;;text{which converges for Re},(s)>1; ?$$
$endgroup$
– DonAntonio
Sep 28 '13 at 16:32
$begingroup$
Yes that. Specially !and @DonAntonio does the Riemann zeta function exists for $0$ too?
$endgroup$
– Shivam Patel
Sep 28 '13 at 16:36
$begingroup$
Yes Shivam: for any complex value except $;1;$ .
$endgroup$
– DonAntonio
Sep 28 '13 at 16:42
2
$begingroup$
In some sense, $zeta(sqrt{2})$ is already a closed form.
$endgroup$
– Hurkyl
Sep 28 '13 at 17:32