Application of Lebesgue's dominated convergence theorem, exercise 4 p.17, vi) from “Intégration”, T....
0
$begingroup$
As a new year's resolution, I'm going through basic exercises and here is one that causes me some trouble. One needs to study the following limit using the dominated convergence theorem:
$$lim_{nrightarrow infty} n^2 int_0^1 (1-x)^n sin(pi x), dx$$
What I can say:
- By integration by parts, and solving a difference equation, one can certainly obtain the answer but that does not seem to be the goal of the exercise.
- The integral alone (without the $n^2$ factor) tends to 0: one can simply dominate by the constant function 1, and since the integrand converges to 0. Hence we are with an indetermined limit, something that goes to infinity times something that goes to 0.
- Doing the change of variable $y=nx$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = n int_0^n left(1- frac{y}{n}right)^n sinleft(pi frac{y}{n}right), dy$$
while $y=n^2 x$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = int_0^{n^2} left(1- frac{y}{n^2}right)^n sinleft(pi frac{y}{n^2}right), dy$$
In the first case the integral alone goes to 0 while n goes to infinity. The last formula seems promising, unfortunately I'm not able to dominate the integrand by an integrable function. - I also thought about cutting the integral in two parts but at the moment it leads nowhere.
calculus measure-theory lebesgue-integral
asked Jan 8 at 17:26
Noix07Noix07
1,241922
$endgroup$
add a comment |
0
$begingroup$
As a new year's resolution, I'm going through basic exercises and here is one that causes me some trouble. One needs to study the following limit using the dominated convergence theorem:
$$lim_{nrightarrow infty} n^2 int_0^1 (1-x)^n sin(pi x), dx$$
What I can say:
- By integration by parts, and solving a difference equation, one can certainly obtain the answer but that does not seem to be the goal of the exercise.
- The integral alone (without the $n^2$ factor) tends to 0: one can simply dominate by the constant function 1, and since the integrand converges to 0. Hence we are with an indetermined limit, something that goes to infinity times something that goes to 0.
- Doing the change of variable $y=nx$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = n int_0^n left(1- frac{y}{n}right)^n sinleft(pi frac{y}{n}right), dy$$
while $y=n^2 x$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = int_0^{n^2} left(1- frac{y}{n^2}right)^n sinleft(pi frac{y}{n^2}right), dy$$
In the first case the integral alone goes to 0 while n goes to infinity. The last formula seems promising, unfortunately I'm not able to dominate the integrand by an integrable function. - I also thought about cutting the integral in two parts but at the moment it leads nowhere.
calculus measure-theory lebesgue-integral
asked Jan 8 at 17:26
Noix07Noix07
1,241922
$endgroup$
$begingroup$
$n^2 int_0^1 (1-x)^n sin(pi x), dx== int_0^1 n^2(1-x)^n sin(pi x), dx$
$endgroup$
– David C. Ullrich
Jan 8 at 17:33
$begingroup$
How do you dominate the integrand by sthg independent of n?
$endgroup$
– Noix07
Jan 8 at 17:52
$begingroup$
I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2lambda^t$ for real $t>>0$ (given $0<lambda<1$).
$endgroup$
– David C. Ullrich
Jan 8 at 18:01
$begingroup$
Indeed by the change of variable $lambda=1-x$ the integral is equal to $int_0^1 n^2 lambda^n sin(pi lambda), dlambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $lambda$ fixed. I find $t=-frac{2}{ln lambda}$ and plugging this back to $t^2 lambda^t sin (pi lambda)$ yields $ frac{4 sin (pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha...
$endgroup$
– Noix07
Jan 9 at 11:00
1
$begingroup$
$-2/log(lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential.
$endgroup$
– David C. Ullrich
Jan 9 at 14:30
add a comment |
0
0
0
$begingroup$
As a new year's resolution, I'm going through basic exercises and here is one that causes me some trouble. One needs to study the following limit using the dominated convergence theorem:
$$lim_{nrightarrow infty} n^2 int_0^1 (1-x)^n sin(pi x), dx$$
What I can say:
- By integration by parts, and solving a difference equation, one can certainly obtain the answer but that does not seem to be the goal of the exercise.
- The integral alone (without the $n^2$ factor) tends to 0: one can simply dominate by the constant function 1, and since the integrand converges to 0. Hence we are with an indetermined limit, something that goes to infinity times something that goes to 0.
- Doing the change of variable $y=nx$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = n int_0^n left(1- frac{y}{n}right)^n sinleft(pi frac{y}{n}right), dy$$
while $y=n^2 x$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = int_0^{n^2} left(1- frac{y}{n^2}right)^n sinleft(pi frac{y}{n^2}right), dy$$
In the first case the integral alone goes to 0 while n goes to infinity. The last formula seems promising, unfortunately I'm not able to dominate the integrand by an integrable function. - I also thought about cutting the integral in two parts but at the moment it leads nowhere.
calculus measure-theory lebesgue-integral
asked Jan 8 at 17:26
Noix07Noix07
1,241922
$endgroup$
As a new year's resolution, I'm going through basic exercises and here is one that causes me some trouble. One needs to study the following limit using the dominated convergence theorem:
$$lim_{nrightarrow infty} n^2 int_0^1 (1-x)^n sin(pi x), dx$$
What I can say:
- By integration by parts, and solving a difference equation, one can certainly obtain the answer but that does not seem to be the goal of the exercise.
- The integral alone (without the $n^2$ factor) tends to 0: one can simply dominate by the constant function 1, and since the integrand converges to 0. Hence we are with an indetermined limit, something that goes to infinity times something that goes to 0.
- Doing the change of variable $y=nx$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = n int_0^n left(1- frac{y}{n}right)^n sinleft(pi frac{y}{n}right), dy$$
while $y=n^2 x$ leads to
$$ n^2 int_0^1 (1-x)^n sin(pi x), dx = int_0^{n^2} left(1- frac{y}{n^2}right)^n sinleft(pi frac{y}{n^2}right), dy$$
In the first case the integral alone goes to 0 while n goes to infinity. The last formula seems promising, unfortunately I'm not able to dominate the integrand by an integrable function. - I also thought about cutting the integral in two parts but at the moment it leads nowhere.
calculus measure-theory lebesgue-integral
calculus measure-theory lebesgue-integral
asked Jan 8 at 17:26
Noix07Noix07
1,241922
asked Jan 8 at 17:26
Noix07Noix07
1,241922
asked Jan 8 at 17:26
Noix07Noix07
1,241922
asked Jan 8 at 17:26
Noix07Noix07
1,241922
asked Jan 8 at 17:26
Noix07Noix07
1,241922
1,241922
$begingroup$
$n^2 int_0^1 (1-x)^n sin(pi x), dx== int_0^1 n^2(1-x)^n sin(pi x), dx$
$endgroup$
– David C. Ullrich
Jan 8 at 17:33
$begingroup$
How do you dominate the integrand by sthg independent of n?
$endgroup$
– Noix07
Jan 8 at 17:52
$begingroup$
I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2lambda^t$ for real $t>>0$ (given $0<lambda<1$).
$endgroup$
– David C. Ullrich
Jan 8 at 18:01
$begingroup$
Indeed by the change of variable $lambda=1-x$ the integral is equal to $int_0^1 n^2 lambda^n sin(pi lambda), dlambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $lambda$ fixed. I find $t=-frac{2}{ln lambda}$ and plugging this back to $t^2 lambda^t sin (pi lambda)$ yields $ frac{4 sin (pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha...
$endgroup$
– Noix07
Jan 9 at 11:00
1
$begingroup$
$-2/log(lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential.
$endgroup$
– David C. Ullrich
Jan 9 at 14:30
add a comment |
$begingroup$
$n^2 int_0^1 (1-x)^n sin(pi x), dx== int_0^1 n^2(1-x)^n sin(pi x), dx$
$endgroup$
– David C. Ullrich
Jan 8 at 17:33
$begingroup$
How do you dominate the integrand by sthg independent of n?
$endgroup$
– Noix07
Jan 8 at 17:52
$begingroup$
I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2lambda^t$ for real $t>>0$ (given $0<lambda<1$).
$endgroup$
– David C. Ullrich
Jan 8 at 18:01
$begingroup$
Indeed by the change of variable $lambda=1-x$ the integral is equal to $int_0^1 n^2 lambda^n sin(pi lambda), dlambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $lambda$ fixed. I find $t=-frac{2}{ln lambda}$ and plugging this back to $t^2 lambda^t sin (pi lambda)$ yields $ frac{4 sin (pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha...
$endgroup$
– Noix07
Jan 9 at 11:00
1
$begingroup$
$-2/log(lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential.
$endgroup$
– David C. Ullrich
Jan 9 at 14:30
$begingroup$
$n^2 int_0^1 (1-x)^n sin(pi x), dx== int_0^1 n^2(1-x)^n sin(pi x), dx$
$endgroup$
– David C. Ullrich
Jan 8 at 17:33
$begingroup$
$n^2 int_0^1 (1-x)^n sin(pi x), dx== int_0^1 n^2(1-x)^n sin(pi x), dx$
$endgroup$
– David C. Ullrich
Jan 8 at 17:33
$begingroup$
How do you dominate the integrand by sthg independent of n?
$endgroup$
– Noix07
Jan 8 at 17:52
$begingroup$
How do you dominate the integrand by sthg independent of n?
$endgroup$
– Noix07
Jan 8 at 17:52
$begingroup$
I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2lambda^t$ for real $t>>0$ (given $0<lambda<1$).
$endgroup$
– David C. Ullrich
Jan 8 at 18:01
$begingroup$
I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2lambda^t$ for real $t>>0$ (given $0<lambda<1$).
$endgroup$
– David C. Ullrich
Jan 8 at 18:01
$begingroup$
Indeed by the change of variable $lambda=1-x$ the integral is equal to $int_0^1 n^2 lambda^n sin(pi lambda), dlambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $lambda$ fixed. I find $t=-frac{2}{ln lambda}$ and plugging this back to $t^2 lambda^t sin (pi lambda)$ yields $ frac{4 sin (pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha...
$endgroup$
– Noix07
Jan 9 at 11:00
$begingroup$
Indeed by the change of variable $lambda=1-x$ the integral is equal to $int_0^1 n^2 lambda^n sin(pi lambda), dlambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $lambda$ fixed. I find $t=-frac{2}{ln lambda}$ and plugging this back to $t^2 lambda^t sin (pi lambda)$ yields $ frac{4 sin (pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha...
$endgroup$
– Noix07
Jan 9 at 11:00
1
1
$begingroup$
$-2/log(lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential.
$endgroup$
– David C. Ullrich
Jan 9 at 14:30
$begingroup$
$-2/log(lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential.
$endgroup$
– David C. Ullrich
Jan 9 at 14:30
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Ok, do this :$$n^2int_0^1(1-x)^nsin(pi x)=int_0^nleft(1-frac xnright)^n nsinleft(frac{pi x}{n}right).$$Of course here
$$left|nsinleft(frac{pi x}{n}right)right|lepi x;$$a little calculus shows that $$log(1-x/n)le-x/nquad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|le g=lim f_n=pi xe^{-x}$.)
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
add a comment |
1
$begingroup$
First, compute the limit when you replaced $sin$ by the identity function.
Then, prove that if $x in [0,1]$, $0 geq sin(pi x)-pi xgeq -Cx^2$ for some positive constant $C$.
answered Jan 8 at 17:35
MindlackMindlack
4,900211
$endgroup$
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
add a comment |
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%2f3066469%2fapplication-of-lebesgues-dominated-convergence-theorem-exercise-4-p-17-vi-fr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Ok, do this :$$n^2int_0^1(1-x)^nsin(pi x)=int_0^nleft(1-frac xnright)^n nsinleft(frac{pi x}{n}right).$$Of course here
$$left|nsinleft(frac{pi x}{n}right)right|lepi x;$$a little calculus shows that $$log(1-x/n)le-x/nquad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|le g=lim f_n=pi xe^{-x}$.)
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
add a comment |
2
$begingroup$
Ok, do this :$$n^2int_0^1(1-x)^nsin(pi x)=int_0^nleft(1-frac xnright)^n nsinleft(frac{pi x}{n}right).$$Of course here
$$left|nsinleft(frac{pi x}{n}right)right|lepi x;$$a little calculus shows that $$log(1-x/n)le-x/nquad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|le g=lim f_n=pi xe^{-x}$.)
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
add a comment |
2
2
2
$begingroup$
Ok, do this :$$n^2int_0^1(1-x)^nsin(pi x)=int_0^nleft(1-frac xnright)^n nsinleft(frac{pi x}{n}right).$$Of course here
$$left|nsinleft(frac{pi x}{n}right)right|lepi x;$$a little calculus shows that $$log(1-x/n)le-x/nquad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|le g=lim f_n=pi xe^{-x}$.)
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
Ok, do this :$$n^2int_0^1(1-x)^nsin(pi x)=int_0^nleft(1-frac xnright)^n nsinleft(frac{pi x}{n}right).$$Of course here
$$left|nsinleft(frac{pi x}{n}right)right|lepi x;$$a little calculus shows that $$log(1-x/n)le-x/nquad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|le g=lim f_n=pi xe^{-x}$.)
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
edited Jan 10 at 1:37
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 9 at 14:21
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
add a comment |
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
@Noix07 I think I got it - considering the record you should look carefully at the details.
$endgroup$
– David C. Ullrich
Jan 10 at 0:29
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
$begingroup$
Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $displaystyle n^2 int_0^1 (1-x)^n sin (pi x), dx = left[ frac{n^2 (1-x)^{n+1}}{-(n+1)} sin (pi x)right]_ 0^1 + int_0^1 frac{ pi n^2 (1-x)^{n+1}}{n+1} cos (pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $pi frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} times 1$...
$endgroup$
– Noix07
Jan 10 at 9:42
add a comment |
1
$begingroup$
First, compute the limit when you replaced $sin$ by the identity function.
Then, prove that if $x in [0,1]$, $0 geq sin(pi x)-pi xgeq -Cx^2$ for some positive constant $C$.
answered Jan 8 at 17:35
MindlackMindlack
4,900211
$endgroup$
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
add a comment |
1
$begingroup$
First, compute the limit when you replaced $sin$ by the identity function.
Then, prove that if $x in [0,1]$, $0 geq sin(pi x)-pi xgeq -Cx^2$ for some positive constant $C$.
answered Jan 8 at 17:35
MindlackMindlack
4,900211
$endgroup$
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
add a comment |
1
1
1
$begingroup$
First, compute the limit when you replaced $sin$ by the identity function.
Then, prove that if $x in [0,1]$, $0 geq sin(pi x)-pi xgeq -Cx^2$ for some positive constant $C$.
answered Jan 8 at 17:35
MindlackMindlack
4,900211
$endgroup$
First, compute the limit when you replaced $sin$ by the identity function.
Then, prove that if $x in [0,1]$, $0 geq sin(pi x)-pi xgeq -Cx^2$ for some positive constant $C$.
answered Jan 8 at 17:35
MindlackMindlack
4,900211
answered Jan 8 at 17:35
MindlackMindlack
4,900211
answered Jan 8 at 17:35
MindlackMindlack
4,900211
answered Jan 8 at 17:35
MindlackMindlack
4,900211
4,900211
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
add a comment |
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
$begingroup$
Ok it works: $int_0^1 (1-x)^n pi x, dx= frac{pi}{(n+1)(n+2)}$ and $int_0^1 (1-x)^n C x^2, dx= frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $xin [0,1/ pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $sin(pi x)= pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm
$endgroup$
– Noix07
Jan 8 at 17:50
add a comment |
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%2f3066469%2fapplication-of-lebesgues-dominated-convergence-theorem-exercise-4-p-17-vi-fr%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
$begingroup$
$n^2 int_0^1 (1-x)^n sin(pi x), dx== int_0^1 n^2(1-x)^n sin(pi x), dx$
$endgroup$
– David C. Ullrich
Jan 8 at 17:33
$begingroup$
How do you dominate the integrand by sthg independent of n?
$endgroup$
– Noix07
Jan 8 at 17:52
$begingroup$
I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2lambda^t$ for real $t>>0$ (given $0<lambda<1$).
$endgroup$
– David C. Ullrich
Jan 8 at 18:01
$begingroup$
Indeed by the change of variable $lambda=1-x$ the integral is equal to $int_0^1 n^2 lambda^n sin(pi lambda), dlambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $lambda$ fixed. I find $t=-frac{2}{ln lambda}$ and plugging this back to $t^2 lambda^t sin (pi lambda)$ yields $ frac{4 sin (pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha...
$endgroup$
– Noix07
Jan 9 at 11:00
1
$begingroup$
$-2/log(lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential.
$endgroup$
– David C. Ullrich
Jan 9 at 14:30