A smooth function whose derivatives satisfy smallness condition near a point
1
$begingroup$
Let $fin C^{infty}left(mathbb{R}rightarrow mathbb{R}right)$. Suppose that, for $ngeq 1$, we have
$$lim_{xrightarrow 0}frac{f^{prime}(x)+a_{1}xf^{primeprime}(x)+...+a_{n-1}x^{n-1}f^{(n)}(x)}{x^n}$$
exists, for a given sequence of reals $(a_{n})$.
I want to show that there does not exist a function $f$ that satisfies the countably many conditions above except for a function constant near $0$, and functions of the form $f(x):=g(x^{n+1})$.
Is this true ? How prove it if true?
real-analysis calculus analysis
asked Jan 23 at 16:21
MedoMedo
642214
$endgroup$
|
show 4 more comments
1
$begingroup$
Let $fin C^{infty}left(mathbb{R}rightarrow mathbb{R}right)$. Suppose that, for $ngeq 1$, we have
$$lim_{xrightarrow 0}frac{f^{prime}(x)+a_{1}xf^{primeprime}(x)+...+a_{n-1}x^{n-1}f^{(n)}(x)}{x^n}$$
exists, for a given sequence of reals $(a_{n})$.
I want to show that there does not exist a function $f$ that satisfies the countably many conditions above except for a function constant near $0$, and functions of the form $f(x):=g(x^{n+1})$.
Is this true ? How prove it if true?
real-analysis calculus analysis
asked Jan 23 at 16:21
MedoMedo
642214
$endgroup$
$begingroup$
Is there supposed to be some sort of condition on $(a_n)$ or $g$? $-cos$ would satisfy this relation for some particular sequence that you could recursively construct with the analytic extension of $g=cos(sqrt{x})$.
$endgroup$
– Robert Wolfe
Jan 23 at 18:19
$begingroup$
Thanks a lot for your comment. I am afraid I don't follow though. $lim_{xrightarrow {0}^{+}} frac{g^{prime}(x)}{x}=frac{1}{2}lim_{xrightarrow {0}^{+}} frac{sin{sqrt{x}}}{xsqrt{x}}=+infty$.
$endgroup$
– Medo
Jan 24 at 19:11
$begingroup$
In fact, if $a_{n}=0$, then $f$ must be constant near $0$ because of the condition that $lim_{xrightarrow 0^{+}}frac{f^{prime}(x)}{x^n}$ exists for all $ngeq 1$, which means that $f^{prime}$ must be identically zero near the origin.
$endgroup$
– Medo
Jan 24 at 19:15
1
$begingroup$
If you define $f=-cos$ and take $a_1=-1, a_2=0, a_3=1/3, a_4=0, a_5=-2/15, a_6=0, a_7=17/315, cdots$ and continue to recursively define this sequence, you'll get a nonzero $f$.
$endgroup$
– Robert Wolfe
Jan 24 at 20:35
$begingroup$
Could you further explain, please ? I do not understand the sequence (cannot predict the next term(s)) and I do not know what is going to happen to the numerator either. Thanks.
$endgroup$
– Medo
Jan 24 at 23:25
|
show 4 more comments
1
1
1
$begingroup$
Let $fin C^{infty}left(mathbb{R}rightarrow mathbb{R}right)$. Suppose that, for $ngeq 1$, we have
$$lim_{xrightarrow 0}frac{f^{prime}(x)+a_{1}xf^{primeprime}(x)+...+a_{n-1}x^{n-1}f^{(n)}(x)}{x^n}$$
exists, for a given sequence of reals $(a_{n})$.
I want to show that there does not exist a function $f$ that satisfies the countably many conditions above except for a function constant near $0$, and functions of the form $f(x):=g(x^{n+1})$.
Is this true ? How prove it if true?
real-analysis calculus analysis
asked Jan 23 at 16:21
MedoMedo
642214
$endgroup$
Let $fin C^{infty}left(mathbb{R}rightarrow mathbb{R}right)$. Suppose that, for $ngeq 1$, we have
$$lim_{xrightarrow 0}frac{f^{prime}(x)+a_{1}xf^{primeprime}(x)+...+a_{n-1}x^{n-1}f^{(n)}(x)}{x^n}$$
exists, for a given sequence of reals $(a_{n})$.
I want to show that there does not exist a function $f$ that satisfies the countably many conditions above except for a function constant near $0$, and functions of the form $f(x):=g(x^{n+1})$.
Is this true ? How prove it if true?
real-analysis calculus analysis
real-analysis calculus analysis
asked Jan 23 at 16:21
MedoMedo
642214
asked Jan 23 at 16:21
MedoMedo
642214
asked Jan 23 at 16:21
MedoMedo
642214
asked Jan 23 at 16:21
MedoMedo
642214
asked Jan 23 at 16:21
MedoMedo
642214
642214
$begingroup$
Is there supposed to be some sort of condition on $(a_n)$ or $g$? $-cos$ would satisfy this relation for some particular sequence that you could recursively construct with the analytic extension of $g=cos(sqrt{x})$.
$endgroup$
– Robert Wolfe
Jan 23 at 18:19
$begingroup$
Thanks a lot for your comment. I am afraid I don't follow though. $lim_{xrightarrow {0}^{+}} frac{g^{prime}(x)}{x}=frac{1}{2}lim_{xrightarrow {0}^{+}} frac{sin{sqrt{x}}}{xsqrt{x}}=+infty$.
$endgroup$
– Medo
Jan 24 at 19:11
$begingroup$
In fact, if $a_{n}=0$, then $f$ must be constant near $0$ because of the condition that $lim_{xrightarrow 0^{+}}frac{f^{prime}(x)}{x^n}$ exists for all $ngeq 1$, which means that $f^{prime}$ must be identically zero near the origin.
$endgroup$
– Medo
Jan 24 at 19:15
1
$begingroup$
If you define $f=-cos$ and take $a_1=-1, a_2=0, a_3=1/3, a_4=0, a_5=-2/15, a_6=0, a_7=17/315, cdots$ and continue to recursively define this sequence, you'll get a nonzero $f$.
$endgroup$
– Robert Wolfe
Jan 24 at 20:35
$begingroup$
Could you further explain, please ? I do not understand the sequence (cannot predict the next term(s)) and I do not know what is going to happen to the numerator either. Thanks.
$endgroup$
– Medo
Jan 24 at 23:25
|
show 4 more comments
$begingroup$
Is there supposed to be some sort of condition on $(a_n)$ or $g$? $-cos$ would satisfy this relation for some particular sequence that you could recursively construct with the analytic extension of $g=cos(sqrt{x})$.
$endgroup$
– Robert Wolfe
Jan 23 at 18:19
$begingroup$
Thanks a lot for your comment. I am afraid I don't follow though. $lim_{xrightarrow {0}^{+}} frac{g^{prime}(x)}{x}=frac{1}{2}lim_{xrightarrow {0}^{+}} frac{sin{sqrt{x}}}{xsqrt{x}}=+infty$.
$endgroup$
– Medo
Jan 24 at 19:11
$begingroup$
In fact, if $a_{n}=0$, then $f$ must be constant near $0$ because of the condition that $lim_{xrightarrow 0^{+}}frac{f^{prime}(x)}{x^n}$ exists for all $ngeq 1$, which means that $f^{prime}$ must be identically zero near the origin.
$endgroup$
– Medo
Jan 24 at 19:15
1
$begingroup$
If you define $f=-cos$ and take $a_1=-1, a_2=0, a_3=1/3, a_4=0, a_5=-2/15, a_6=0, a_7=17/315, cdots$ and continue to recursively define this sequence, you'll get a nonzero $f$.
$endgroup$
– Robert Wolfe
Jan 24 at 20:35
$begingroup$
Could you further explain, please ? I do not understand the sequence (cannot predict the next term(s)) and I do not know what is going to happen to the numerator either. Thanks.
$endgroup$
– Medo
Jan 24 at 23:25
$begingroup$
Is there supposed to be some sort of condition on $(a_n)$ or $g$? $-cos$ would satisfy this relation for some particular sequence that you could recursively construct with the analytic extension of $g=cos(sqrt{x})$.
$endgroup$
– Robert Wolfe
Jan 23 at 18:19
$begingroup$
Is there supposed to be some sort of condition on $(a_n)$ or $g$? $-cos$ would satisfy this relation for some particular sequence that you could recursively construct with the analytic extension of $g=cos(sqrt{x})$.
$endgroup$
– Robert Wolfe
Jan 23 at 18:19
$begingroup$
Thanks a lot for your comment. I am afraid I don't follow though. $lim_{xrightarrow {0}^{+}} frac{g^{prime}(x)}{x}=frac{1}{2}lim_{xrightarrow {0}^{+}} frac{sin{sqrt{x}}}{xsqrt{x}}=+infty$.
$endgroup$
– Medo
Jan 24 at 19:11
$begingroup$
Thanks a lot for your comment. I am afraid I don't follow though. $lim_{xrightarrow {0}^{+}} frac{g^{prime}(x)}{x}=frac{1}{2}lim_{xrightarrow {0}^{+}} frac{sin{sqrt{x}}}{xsqrt{x}}=+infty$.
$endgroup$
– Medo
Jan 24 at 19:11
$begingroup$
In fact, if $a_{n}=0$, then $f$ must be constant near $0$ because of the condition that $lim_{xrightarrow 0^{+}}frac{f^{prime}(x)}{x^n}$ exists for all $ngeq 1$, which means that $f^{prime}$ must be identically zero near the origin.
$endgroup$
– Medo
Jan 24 at 19:15
$begingroup$
In fact, if $a_{n}=0$, then $f$ must be constant near $0$ because of the condition that $lim_{xrightarrow 0^{+}}frac{f^{prime}(x)}{x^n}$ exists for all $ngeq 1$, which means that $f^{prime}$ must be identically zero near the origin.
$endgroup$
– Medo
Jan 24 at 19:15
1
1
$begingroup$
If you define $f=-cos$ and take $a_1=-1, a_2=0, a_3=1/3, a_4=0, a_5=-2/15, a_6=0, a_7=17/315, cdots$ and continue to recursively define this sequence, you'll get a nonzero $f$.
$endgroup$
– Robert Wolfe
Jan 24 at 20:35
$begingroup$
If you define $f=-cos$ and take $a_1=-1, a_2=0, a_3=1/3, a_4=0, a_5=-2/15, a_6=0, a_7=17/315, cdots$ and continue to recursively define this sequence, you'll get a nonzero $f$.
$endgroup$
– Robert Wolfe
Jan 24 at 20:35
$begingroup$
Could you further explain, please ? I do not understand the sequence (cannot predict the next term(s)) and I do not know what is going to happen to the numerator either. Thanks.
$endgroup$
– Medo
Jan 24 at 23:25
$begingroup$
Could you further explain, please ? I do not understand the sequence (cannot predict the next term(s)) and I do not know what is going to happen to the numerator either. Thanks.
$endgroup$
– Medo
Jan 24 at 23:25
|
show 4 more comments
0
active
oldest
votes
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
});
}
});
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%2f3084701%2fa-smooth-function-whose-derivatives-satisfy-smallness-condition-near-a-point%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3084701%2fa-smooth-function-whose-derivatives-satisfy-smallness-condition-near-a-point%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
$begingroup$
Is there supposed to be some sort of condition on $(a_n)$ or $g$? $-cos$ would satisfy this relation for some particular sequence that you could recursively construct with the analytic extension of $g=cos(sqrt{x})$.
$endgroup$
– Robert Wolfe
Jan 23 at 18:19
$begingroup$
Thanks a lot for your comment. I am afraid I don't follow though. $lim_{xrightarrow {0}^{+}} frac{g^{prime}(x)}{x}=frac{1}{2}lim_{xrightarrow {0}^{+}} frac{sin{sqrt{x}}}{xsqrt{x}}=+infty$.
$endgroup$
– Medo
Jan 24 at 19:11
$begingroup$
In fact, if $a_{n}=0$, then $f$ must be constant near $0$ because of the condition that $lim_{xrightarrow 0^{+}}frac{f^{prime}(x)}{x^n}$ exists for all $ngeq 1$, which means that $f^{prime}$ must be identically zero near the origin.
$endgroup$
– Medo
Jan 24 at 19:15
1
$begingroup$
If you define $f=-cos$ and take $a_1=-1, a_2=0, a_3=1/3, a_4=0, a_5=-2/15, a_6=0, a_7=17/315, cdots$ and continue to recursively define this sequence, you'll get a nonzero $f$.
$endgroup$
– Robert Wolfe
Jan 24 at 20:35
$begingroup$
Could you further explain, please ? I do not understand the sequence (cannot predict the next term(s)) and I do not know what is going to happen to the numerator either. Thanks.
$endgroup$
– Medo
Jan 24 at 23:25