Mixing
How to use Lebesgue Dominated Convergence theorem in this example?
$begingroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
$endgroup$
add a comment |
$begingroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
$endgroup$
add a comment |
$begingroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
$endgroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
$endgroup$
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
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%2f3063733%2fhow-to-use-lebesgue-dominated-convergence-theorem-in-this-example%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
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
$endgroup$
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
$endgroup$
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
$endgroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
answered Jan 6 at 11:26
mathworker21mathworker21
8,9171928
8,9171928
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
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%2f3063733%2fhow-to-use-lebesgue-dominated-convergence-theorem-in-this-example%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
