deduce the derivative of Fourier series when the series is not continuous.
$begingroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
edited Dec 30 '18 at 23:06
Bernard
121k740116
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
$endgroup$
add a comment |
$begingroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
edited Dec 30 '18 at 23:06
Bernard
121k740116
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
$endgroup$
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
add a comment |
$begingroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
edited Dec 30 '18 at 23:06
Bernard
121k740116
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
$endgroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
fourier-series
edited Dec 30 '18 at 23:06
Bernard
121k740116
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
edited Dec 30 '18 at 23:06
Bernard
121k740116
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
edited Dec 30 '18 at 23:06
Bernard
121k740116
edited Dec 30 '18 at 23:06
Bernard
121k740116
edited Dec 30 '18 at 23:06
Bernard
121k740116
121k740116
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,811619
1,811619
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
add a comment |
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
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%2f3057261%2fdeduce-the-derivative-of-fourier-series-when-the-series-is-not-continuous%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%2f3057261%2fdeduce-the-derivative-of-fourier-series-when-the-series-is-not-continuous%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
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36