Proving a sequence of functions does not converge uniformly. Can it be assumed that $Ngeq 1$ an epsilon N...
$begingroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
$endgroup$
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
|
show 1 more comment
$begingroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
$endgroup$
I'm trying to prove that the function $f_n(x) = frac{n}{nx+1}$ for $x in(0,1)$
, where $n$ is in the natural numbers, does not converge uniformly to $frac{1}{x}$ by proving $exists epsilon > 0, forall N, exists x, exists n> N$ such that $|f_n(x) - frac{1}{x}| geq epsilon$. I know it says for all $N$, but it doesn't make much sense to me that we'd be given an $N$ less than $1$ since the function requires that $n$ is in the natural numbers. I don't need help with the proof, just clarification on whether $N$ could be negative.
real-analysis
real-analysis
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
edited Dec 5 '18 at 3:54
Trigginometric
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
asked Dec 5 '18 at 3:50
TrigginometricTrigginometric
114
114
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
|
show 1 more comment
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
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%2f3026587%2fproving-a-sequence-of-functions-does-not-converge-uniformly-can-it-be-assumed-t%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
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
$begingroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
$endgroup$
I agree that this is getting way too complicated, but if you really want...
Theorem
Let $p(epsilon,x,n)$ be some proposition. Suppose it is proven that
$$existsepsilon,,forall Ngeq1,,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Then it is also true that
$$existsepsilon,,forall Ninmathbb{R},,exists x,,exists n>N,|,p(epsilon,x,n)text{.}$$
Proof
By hypothesis, some $epsilon$ exists. Pick the same $epsilon$. We are now given $Ninmathbb{R}$. Suppose $Ngeq1$, then obviously use the hypothesis to find $x$ and $n$.
Now suppose that $N<1$. By hypothesis we can find $x$, $n>1$ such that $p(epsilon,x,n)$ is true. Then $n>1>N$ and therefore we have found $x$, $n>N$ such that $p(epsilon,x,n)$ is true.
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
edited Dec 5 '18 at 4:16
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
answered Dec 5 '18 at 3:55
obscuransobscurans
1,152311
1,152311
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
add a comment |
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
My only issue is that in my proof I picked $x=frac{1}{N+1}$ where $xin (0,1)$ and then on the last inequality right before I say everything is greather than or equal to epsilon, I have an $frac{N}{2}$ (It looks like $frac{N}{2} geq frac{1}{2} geq epsilon$) So if $N$ could be negative, my proof falls apart in multiple ways.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:59
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
The argument above proves $existsepsilonforall Ngeq1exists xexists n>N|text{statement},Rightarrow,existsepsilonforall Nexists xexists n>N|text{statement}$ by copying. So prove the statement assuming $Ngeq1$.
$endgroup$
– obscurans
Dec 5 '18 at 4:02
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
I'm don't think I understand what you mean.
$endgroup$
– Trigginometric
Dec 5 '18 at 4:06
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
$begingroup$
See rewrite. Just prove using $Ngeq1$, as asked.
$endgroup$
– obscurans
Dec 5 '18 at 4:18
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%2f3026587%2fproving-a-sequence-of-functions-does-not-converge-uniformly-can-it-be-assumed-t%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
$begingroup$
We have not covered simple convergence as of yet, only uniform and pointwise.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:53
$begingroup$
Simple = "pointwise."
$endgroup$
– Will M.
Dec 5 '18 at 3:53
$begingroup$
I totally missed inserting that the domain was for $xin (0,1)$. The question actually does ask us to prove pointwise convergence for all $x$ in the domain $(0,1)$ and I've done that. My ONLY question here is if it can be assumed the $N$ is non-negative.
$endgroup$
– Trigginometric
Dec 5 '18 at 3:55
$begingroup$
Perhaps I should have shown my proof from the start. Let $epsilon =frac{1}{4}$ let $n = N+1$, let $x=frac{1}{n}$. Then $|frac{n}{nx+1} - frac{1}{x}| = frac{1}{x(nx+1)}=frac{1}{frac{1}{N}(frac{N+1}{N+1} + 1)} = frac{1}{frac{2}{N}} = frac{N}{2} geq frac{1}{2} geq epsilon$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:03
$begingroup$
But it does convergence simply, for any given $x$ in the specified domain (which I stupidly forgot to mention earlier so you may have missed), $f_n(x)$ converges to $frac{1}{x}$
$endgroup$
– Trigginometric
Dec 5 '18 at 4:07