deduce the derivative of Fourier series when the series is not continuous.












0












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    What does it mean to say $f'$ is "soft piecewise"?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:36


















0












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    What does it mean to say $f'$ is "soft piecewise"?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:36
















0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 23:06









Bernard

119k639112




119k639112










asked Dec 30 '18 at 22:54









Bvss12Bvss12

1,785618




1,785618












  • $begingroup$
    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






$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36












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


















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
















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














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





















































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







Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

張江高科駅