Chapter V: Titchmarsh's book “The theory of the Riemann Zeta function”
up vote
0
down vote
favorite
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 at 19:39
Dubglass
1098
add a comment |
up vote
0
down vote
favorite
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 at 19:39
Dubglass
1098
Here's a link to the book if anyone wants it.
– Mason
Dec 4 at 19:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 at 19:39
Dubglass
1098
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 at 19:39
Dubglass
1098
asked Dec 4 at 19:39
Dubglass
1098
asked Dec 4 at 19:39
Dubglass
1098
asked Dec 4 at 19:39
Dubglass
1098
asked Dec 4 at 19:39
Dubglass
1098
1098
Here's a link to the book if anyone wants it.
– Mason
Dec 4 at 19:53
add a comment |
Here's a link to the book if anyone wants it.
– Mason
Dec 4 at 19:53
Here's a link to the book if anyone wants it.
– Mason
Dec 4 at 19:53
Here's a link to the book if anyone wants it.
– Mason
Dec 4 at 19:53
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 at 20:46
p4sch
4,800217
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "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',
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%2f3026046%2fchapter-v-titchmarshs-book-the-theory-of-the-riemann-zeta-function%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
up vote
1
down vote
accepted
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 at 20:46
p4sch
4,800217
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
add a comment |
up vote
1
down vote
accepted
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 at 20:46
p4sch
4,800217
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 at 20:46
p4sch
4,800217
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 at 20:46
p4sch
4,800217
answered Dec 4 at 20:46
p4sch
4,800217
answered Dec 4 at 20:46
p4sch
4,800217
answered Dec 4 at 20:46
p4sch
4,800217
4,800217
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
add a comment |
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
Thanks! I think I got it.
– Dubglass
Dec 4 at 20:59
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3026046%2fchapter-v-titchmarshs-book-the-theory-of-the-riemann-zeta-function%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
Here's a link to the book if anyone wants it.
– Mason
Dec 4 at 19:53