Let $f_n (x) = n^2 x (1-x^2)^n$ on $[0,1]$. Explain why convergence is not uniform on $[0,1]$
I'm not sure what to do here. I tried $x = frac{1}{sqrt n}$ and then $f_n = n^frac{3}2 (1-frac{1}n)^n$ which looks eerily close to $n^frac{3}2 e$, which obviously goes to infinity and therefore would not converge uniformly. Also its a 2 parter- are there intervals such that $f_n$ does uniformly converge? I was thinking it converges for any interval $[0,a]$ where $0 < a < 1$. Since $$|f_n - 0| = |f_n| = |n^2 x (1-x^2)^n| le |n^2 a(1-a^2)^n|$$
and for any $epsilon > 0$ we can find an $N in N$ s.t. $|n^2 x (1-x^2)^n| < epsilon$ And this would satisfy the definition of uniform convergence right?
analysis proof-verification
asked Nov 28 at 7:53
zodross
1496
add a comment |
I'm not sure what to do here. I tried $x = frac{1}{sqrt n}$ and then $f_n = n^frac{3}2 (1-frac{1}n)^n$ which looks eerily close to $n^frac{3}2 e$, which obviously goes to infinity and therefore would not converge uniformly. Also its a 2 parter- are there intervals such that $f_n$ does uniformly converge? I was thinking it converges for any interval $[0,a]$ where $0 < a < 1$. Since $$|f_n - 0| = |f_n| = |n^2 x (1-x^2)^n| le |n^2 a(1-a^2)^n|$$
and for any $epsilon > 0$ we can find an $N in N$ s.t. $|n^2 x (1-x^2)^n| < epsilon$ And this would satisfy the definition of uniform convergence right?
analysis proof-verification
asked Nov 28 at 7:53
zodross
1496
add a comment |
I'm not sure what to do here. I tried $x = frac{1}{sqrt n}$ and then $f_n = n^frac{3}2 (1-frac{1}n)^n$ which looks eerily close to $n^frac{3}2 e$, which obviously goes to infinity and therefore would not converge uniformly. Also its a 2 parter- are there intervals such that $f_n$ does uniformly converge? I was thinking it converges for any interval $[0,a]$ where $0 < a < 1$. Since $$|f_n - 0| = |f_n| = |n^2 x (1-x^2)^n| le |n^2 a(1-a^2)^n|$$
and for any $epsilon > 0$ we can find an $N in N$ s.t. $|n^2 x (1-x^2)^n| < epsilon$ And this would satisfy the definition of uniform convergence right?
analysis proof-verification
asked Nov 28 at 7:53
zodross
1496
I'm not sure what to do here. I tried $x = frac{1}{sqrt n}$ and then $f_n = n^frac{3}2 (1-frac{1}n)^n$ which looks eerily close to $n^frac{3}2 e$, which obviously goes to infinity and therefore would not converge uniformly. Also its a 2 parter- are there intervals such that $f_n$ does uniformly converge? I was thinking it converges for any interval $[0,a]$ where $0 < a < 1$. Since $$|f_n - 0| = |f_n| = |n^2 x (1-x^2)^n| le |n^2 a(1-a^2)^n|$$
and for any $epsilon > 0$ we can find an $N in N$ s.t. $|n^2 x (1-x^2)^n| < epsilon$ And this would satisfy the definition of uniform convergence right?
analysis proof-verification
analysis proof-verification
asked Nov 28 at 7:53
zodross
1496
asked Nov 28 at 7:53
zodross
1496
asked Nov 28 at 7:53
zodross
1496
asked Nov 28 at 7:53
zodross
1496
asked Nov 28 at 7:53
zodross
1496
1496
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
For each fixed $x$, $f_n(x) to 0$. You have already proved that the convergence is not uniform: $f_n(frac 1 {sqrt n}) to infty$. [ Use the fact that $(1-frac 1 n)^{n} >frac 1 {2e}$ for $n$ sufficiently large]. This also proves that convergence is not uniform on $[0,a]$ for any $a>0$. However, $f_n to 0$ uniformly on $[a,1]$ for any $a >0$ [because $n^{2}(1-a^{2})^{n} to 0$].
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
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%2f3016869%2flet-f-n-x-n2-x-1-x2n-on-0-1-explain-why-convergence-is-not-unifo%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
For each fixed $x$, $f_n(x) to 0$. You have already proved that the convergence is not uniform: $f_n(frac 1 {sqrt n}) to infty$. [ Use the fact that $(1-frac 1 n)^{n} >frac 1 {2e}$ for $n$ sufficiently large]. This also proves that convergence is not uniform on $[0,a]$ for any $a>0$. However, $f_n to 0$ uniformly on $[a,1]$ for any $a >0$ [because $n^{2}(1-a^{2})^{n} to 0$].
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
add a comment |
For each fixed $x$, $f_n(x) to 0$. You have already proved that the convergence is not uniform: $f_n(frac 1 {sqrt n}) to infty$. [ Use the fact that $(1-frac 1 n)^{n} >frac 1 {2e}$ for $n$ sufficiently large]. This also proves that convergence is not uniform on $[0,a]$ for any $a>0$. However, $f_n to 0$ uniformly on $[a,1]$ for any $a >0$ [because $n^{2}(1-a^{2})^{n} to 0$].
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
add a comment |
For each fixed $x$, $f_n(x) to 0$. You have already proved that the convergence is not uniform: $f_n(frac 1 {sqrt n}) to infty$. [ Use the fact that $(1-frac 1 n)^{n} >frac 1 {2e}$ for $n$ sufficiently large]. This also proves that convergence is not uniform on $[0,a]$ for any $a>0$. However, $f_n to 0$ uniformly on $[a,1]$ for any $a >0$ [because $n^{2}(1-a^{2})^{n} to 0$].
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
For each fixed $x$, $f_n(x) to 0$. You have already proved that the convergence is not uniform: $f_n(frac 1 {sqrt n}) to infty$. [ Use the fact that $(1-frac 1 n)^{n} >frac 1 {2e}$ for $n$ sufficiently large]. This also proves that convergence is not uniform on $[0,a]$ for any $a>0$. However, $f_n to 0$ uniformly on $[a,1]$ for any $a >0$ [because $n^{2}(1-a^{2})^{n} to 0$].
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
answered Nov 28 at 8:00
Kavi Rama Murthy
48.4k31854
48.4k31854
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.
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%2f3016869%2flet-f-n-x-n2-x-1-x2n-on-0-1-explain-why-convergence-is-not-unifo%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
