Baby Rudin exercise 7.12
1
$begingroup$
There is a standard solution posted for this problem(here is the link: Rudin's 'Principle of Mathematical Analysis' Problem 7.12). I understand everything except the last step(the inequality). I don't see why there exists an interval $[c,d]$ that the first term is less than $varepsilon/3$ for all $n?$
real-analysis
edited Dec 28 '18 at 18:53
AugSB
3,38921733
asked Dec 28 '18 at 18:21
aud999aud999
944
$endgroup$
add a comment |
1
$begingroup$
There is a standard solution posted for this problem(here is the link: Rudin's 'Principle of Mathematical Analysis' Problem 7.12). I understand everything except the last step(the inequality). I don't see why there exists an interval $[c,d]$ that the first term is less than $varepsilon/3$ for all $n?$
real-analysis
edited Dec 28 '18 at 18:53
AugSB
3,38921733
asked Dec 28 '18 at 18:21
aud999aud999
944
$endgroup$
1
$begingroup$
If you understood the argument right above that wherein it discuss bounding the difference of integrals of $phi$ for various choices of $phi$, then you just take $epsilon/3$ in that line rather than $epsilon$.
$endgroup$
– DudeMan
Dec 28 '18 at 18:28
$begingroup$
I mean, there exists [c,d] for every n, but how do we guarantee that there is an interval [c,d] that exists for all n.( n approaches infinity. )
$endgroup$
– aud999
Dec 29 '18 at 8:35
$begingroup$
I think I got it now.. It is because of the dominance of g
$endgroup$
– aud999
Dec 29 '18 at 9:15
add a comment |
1
1
1
$begingroup$
There is a standard solution posted for this problem(here is the link: Rudin's 'Principle of Mathematical Analysis' Problem 7.12). I understand everything except the last step(the inequality). I don't see why there exists an interval $[c,d]$ that the first term is less than $varepsilon/3$ for all $n?$
real-analysis
edited Dec 28 '18 at 18:53
AugSB
3,38921733
asked Dec 28 '18 at 18:21
aud999aud999
944
$endgroup$
There is a standard solution posted for this problem(here is the link: Rudin's 'Principle of Mathematical Analysis' Problem 7.12). I understand everything except the last step(the inequality). I don't see why there exists an interval $[c,d]$ that the first term is less than $varepsilon/3$ for all $n?$
real-analysis
real-analysis
edited Dec 28 '18 at 18:53
AugSB
3,38921733
asked Dec 28 '18 at 18:21
aud999aud999
944
edited Dec 28 '18 at 18:53
AugSB
3,38921733
asked Dec 28 '18 at 18:21
aud999aud999
944
edited Dec 28 '18 at 18:53
AugSB
3,38921733
edited Dec 28 '18 at 18:53
AugSB
3,38921733
edited Dec 28 '18 at 18:53
AugSB
3,38921733
3,38921733
asked Dec 28 '18 at 18:21
aud999aud999
944
asked Dec 28 '18 at 18:21
aud999aud999
944
asked Dec 28 '18 at 18:21
aud999aud999
944
944
1
$begingroup$
If you understood the argument right above that wherein it discuss bounding the difference of integrals of $phi$ for various choices of $phi$, then you just take $epsilon/3$ in that line rather than $epsilon$.
$endgroup$
– DudeMan
Dec 28 '18 at 18:28
$begingroup$
I mean, there exists [c,d] for every n, but how do we guarantee that there is an interval [c,d] that exists for all n.( n approaches infinity. )
$endgroup$
– aud999
Dec 29 '18 at 8:35
$begingroup$
I think I got it now.. It is because of the dominance of g
$endgroup$
– aud999
Dec 29 '18 at 9:15
add a comment |
1
$begingroup$
If you understood the argument right above that wherein it discuss bounding the difference of integrals of $phi$ for various choices of $phi$, then you just take $epsilon/3$ in that line rather than $epsilon$.
$endgroup$
– DudeMan
Dec 28 '18 at 18:28
$begingroup$
I mean, there exists [c,d] for every n, but how do we guarantee that there is an interval [c,d] that exists for all n.( n approaches infinity. )
$endgroup$
– aud999
Dec 29 '18 at 8:35
$begingroup$
I think I got it now.. It is because of the dominance of g
$endgroup$
– aud999
Dec 29 '18 at 9:15
1
1
$begingroup$
If you understood the argument right above that wherein it discuss bounding the difference of integrals of $phi$ for various choices of $phi$, then you just take $epsilon/3$ in that line rather than $epsilon$.
$endgroup$
– DudeMan
Dec 28 '18 at 18:28
$begingroup$
If you understood the argument right above that wherein it discuss bounding the difference of integrals of $phi$ for various choices of $phi$, then you just take $epsilon/3$ in that line rather than $epsilon$.
$endgroup$
– DudeMan
Dec 28 '18 at 18:28
$begingroup$
I mean, there exists [c,d] for every n, but how do we guarantee that there is an interval [c,d] that exists for all n.( n approaches infinity. )
$endgroup$
– aud999
Dec 29 '18 at 8:35
$begingroup$
I mean, there exists [c,d] for every n, but how do we guarantee that there is an interval [c,d] that exists for all n.( n approaches infinity. )
$endgroup$
– aud999
Dec 29 '18 at 8:35
$begingroup$
I think I got it now.. It is because of the dominance of g
$endgroup$
– aud999
Dec 29 '18 at 9:15
$begingroup$
I think I got it now.. It is because of the dominance of g
$endgroup$
– aud999
Dec 29 '18 at 9:15
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
});
}
});
draft saved
draft discarded
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%2f3055162%2fbaby-rudin-exercise-7-12%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
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3055162%2fbaby-rudin-exercise-7-12%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
$begingroup$
If you understood the argument right above that wherein it discuss bounding the difference of integrals of $phi$ for various choices of $phi$, then you just take $epsilon/3$ in that line rather than $epsilon$.
$endgroup$
– DudeMan
Dec 28 '18 at 18:28
$begingroup$
I mean, there exists [c,d] for every n, but how do we guarantee that there is an interval [c,d] that exists for all n.( n approaches infinity. )
$endgroup$
– aud999
Dec 29 '18 at 8:35
$begingroup$
I think I got it now.. It is because of the dominance of g
$endgroup$
– aud999
Dec 29 '18 at 9:15