Uniform convergence of Binomial series on $Bbb{R}$
$begingroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
$endgroup$
add a comment |
$begingroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
$endgroup$
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
add a comment |
$begingroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
$endgroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
edited Jan 8 at 7:24
Omojola Micheal
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,999424
1,999424
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
add a comment |
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
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%2f3065878%2funiform-convergence-of-binomial-series-on-bbbr%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
$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
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.
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%2f3065878%2funiform-convergence-of-binomial-series-on-bbbr%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
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24