Mixing
In what sense does integration “raise the power by 1”
$begingroup$
I am working on a problem that requires me to compare the value of a particular, but generically defined function, with the definite integral of that function.
Naturally, if the function is a polynomial then I can compute the definite integral, and the result is explicitly a polynomial of degree 1 higher than the original function.
I am wondering in what precise sense (if any) one can say this in general. That is, that integrating raises the degree by 1. Perhaps using Taylor series or big-$O$ notation.
To be concrete, I want to compare $f(x^p)$ with $$int_x^1 f(t^p) dt $$ as $p$ gets large, where $f(0)=0$ and $f(1)=1$ say.
calculus integration definite-integrals
asked Jan 21 at 12:34
user434180user434180
1108
$endgroup$
add a comment |
$begingroup$
I am working on a problem that requires me to compare the value of a particular, but generically defined function, with the definite integral of that function.
Naturally, if the function is a polynomial then I can compute the definite integral, and the result is explicitly a polynomial of degree 1 higher than the original function.
I am wondering in what precise sense (if any) one can say this in general. That is, that integrating raises the degree by 1. Perhaps using Taylor series or big-$O$ notation.
To be concrete, I want to compare $f(x^p)$ with $$int_x^1 f(t^p) dt $$ as $p$ gets large, where $f(0)=0$ and $f(1)=1$ say.
calculus integration definite-integrals
asked Jan 21 at 12:34
user434180user434180
1108
$endgroup$
$begingroup$
you don't mean "genetically", you mean "generically" :P
$endgroup$
– terrace
Jan 21 at 12:37
1
$begingroup$
Indeed I don’t. Thanks for picking it up!
$endgroup$
– user434180
Jan 21 at 12:37
$begingroup$
Integration term by term of taylor series is indeed allowed, see dpmms.cam.ac.uk/~agk22/uniform.pdf since Taylor series converge locally uniformly
$endgroup$
– Calvin Khor
Jan 21 at 12:42
add a comment |
$begingroup$
I am working on a problem that requires me to compare the value of a particular, but generically defined function, with the definite integral of that function.
Naturally, if the function is a polynomial then I can compute the definite integral, and the result is explicitly a polynomial of degree 1 higher than the original function.
I am wondering in what precise sense (if any) one can say this in general. That is, that integrating raises the degree by 1. Perhaps using Taylor series or big-$O$ notation.
To be concrete, I want to compare $f(x^p)$ with $$int_x^1 f(t^p) dt $$ as $p$ gets large, where $f(0)=0$ and $f(1)=1$ say.
calculus integration definite-integrals
asked Jan 21 at 12:34
user434180user434180
1108
$endgroup$
I am working on a problem that requires me to compare the value of a particular, but generically defined function, with the definite integral of that function.
Naturally, if the function is a polynomial then I can compute the definite integral, and the result is explicitly a polynomial of degree 1 higher than the original function.
I am wondering in what precise sense (if any) one can say this in general. That is, that integrating raises the degree by 1. Perhaps using Taylor series or big-$O$ notation.
To be concrete, I want to compare $f(x^p)$ with $$int_x^1 f(t^p) dt $$ as $p$ gets large, where $f(0)=0$ and $f(1)=1$ say.
calculus integration definite-integrals
calculus integration definite-integrals
asked Jan 21 at 12:34
user434180user434180
1108
asked Jan 21 at 12:34
user434180user434180
1108
edited Jan 21 at 12:37
user434180
asked Jan 21 at 12:34
user434180user434180
1108
asked Jan 21 at 12:34
user434180user434180
1108
asked Jan 21 at 12:34
user434180user434180
1108
1108
$begingroup$
you don't mean "genetically", you mean "generically" :P
$endgroup$
– terrace
Jan 21 at 12:37
1
$begingroup$
Indeed I don’t. Thanks for picking it up!
$endgroup$
– user434180
Jan 21 at 12:37
$begingroup$
Integration term by term of taylor series is indeed allowed, see dpmms.cam.ac.uk/~agk22/uniform.pdf since Taylor series converge locally uniformly
$endgroup$
– Calvin Khor
Jan 21 at 12:42
add a comment |
$begingroup$
you don't mean "genetically", you mean "generically" :P
$endgroup$
– terrace
Jan 21 at 12:37
1
$begingroup$
Indeed I don’t. Thanks for picking it up!
$endgroup$
– user434180
Jan 21 at 12:37
$begingroup$
Integration term by term of taylor series is indeed allowed, see dpmms.cam.ac.uk/~agk22/uniform.pdf since Taylor series converge locally uniformly
$endgroup$
– Calvin Khor
Jan 21 at 12:42
$begingroup$
you don't mean "genetically", you mean "generically" :P
$endgroup$
– terrace
Jan 21 at 12:37
$begingroup$
you don't mean "genetically", you mean "generically" :P
$endgroup$
– terrace
Jan 21 at 12:37
1
1
$begingroup$
Indeed I don’t. Thanks for picking it up!
$endgroup$
– user434180
Jan 21 at 12:37
$begingroup$
Indeed I don’t. Thanks for picking it up!
$endgroup$
– user434180
Jan 21 at 12:37
$begingroup$
Integration term by term of taylor series is indeed allowed, see dpmms.cam.ac.uk/~agk22/uniform.pdf since Taylor series converge locally uniformly
$endgroup$
– Calvin Khor
Jan 21 at 12:42
$begingroup$
Integration term by term of taylor series is indeed allowed, see dpmms.cam.ac.uk/~agk22/uniform.pdf since Taylor series converge locally uniformly
$endgroup$
– Calvin Khor
Jan 21 at 12:42
add a comment |
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
});
}
});
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%2f3081827%2fin-what-sense-does-integration-raise-the-power-by-1%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
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%2f3081827%2fin-what-sense-does-integration-raise-the-power-by-1%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 don't mean "genetically", you mean "generically" :P
$endgroup$
– terrace
Jan 21 at 12:37
1
$begingroup$
Indeed I don’t. Thanks for picking it up!
$endgroup$
– user434180
Jan 21 at 12:37
$begingroup$
Integration term by term of taylor series is indeed allowed, see dpmms.cam.ac.uk/~agk22/uniform.pdf since Taylor series converge locally uniformly
$endgroup$
– Calvin Khor
Jan 21 at 12:42