Uniform convergence of a sequence on limited t range
$begingroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
$endgroup$
add a comment |
$begingroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
$endgroup$
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
add a comment |
$begingroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
$endgroup$
Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t in [0,1]$ and $n≥1$
$$f_n(t)= frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$frac {-t^2}{n+t}$$
But I am not sure how to get to uniform convergence from here.
I believe it converges to $t$ uniformly, but could someone please clarify?
sequences-and-series convergence uniform-convergence
sequences-and-series convergence uniform-convergence
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
edited Dec 12 '18 at 13:45
Arnaud D.
15.9k52443
15.9k52443
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
asked Dec 12 '18 at 13:19
MathExchange123MathExchange123
61
61
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
add a comment |
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
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%2f3036676%2funiform-convergence-of-a-sequence-on-limited-t-range%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
$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
$begingroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
$endgroup$
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals
$$left|frac{1cdot 1}{1+1} - 1right| = left|frac12right|=frac12$$ while the right side equals $$frac{-1^2}{1+1}=-frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $epsilon$, there exists some $N$ such that $sup_{tin[0,1]} |f_n(t)-f(t)| < epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$frac{t^2}{n+t} leq frac{1}{n+t}leq frac1n$$ should help you through the final steps.
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
edited Dec 12 '18 at 13:29
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:24
5xum5xum
91.1k394161
91.1k394161
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
add a comment |
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
Does $ frac {-t^2}{n+t}$ being negative affect this? Also I'm not quite sure how to find the supremum on [0,1] or pick a suitable n. Could you please help me with this?
$endgroup$
– MathExchange123
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
@MathExchange123 Please, carefully read the first part of my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:28
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
$begingroup$
Thank you, I missed that the first time.
$endgroup$
– MathExchange123
Dec 12 '18 at 13:44
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%2f3036676%2funiform-convergence-of-a-sequence-on-limited-t-range%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
$begingroup$
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
$endgroup$
– 5xum
Jan 15 at 8:50