$ int_{0}^{1} sum_{m=1}^{infty} a_m sin(mpi x)sin(l pi x)dx = frac{a_l}{2}$
$begingroup$
Why is this true?
$$int_{0}^{1}u^0(x)sin(lpi x)dx = int_{0}^{1}u(0,x)sin(lpi x)dx$$ $$ = int_{0}^{1} sum_{m=1}^{infty} a_m sin(mpi x)sin(lpi x)dx$$ $$ = frac{a_l}{2}$$
where coefficients ${a_m}$ are determined by $u(0,x) = u^0(x)$ which are the initial conditions to $u_t(t,x) = u_{xx}(t,x)$
ordinary-differential-equations pde fourier-series heat-equation
edited Dec 7 '18 at 13:27
amWhy
1
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
$endgroup$
add a comment |
$begingroup$
Why is this true?
$$int_{0}^{1}u^0(x)sin(lpi x)dx = int_{0}^{1}u(0,x)sin(lpi x)dx$$ $$ = int_{0}^{1} sum_{m=1}^{infty} a_m sin(mpi x)sin(lpi x)dx$$ $$ = frac{a_l}{2}$$
where coefficients ${a_m}$ are determined by $u(0,x) = u^0(x)$ which are the initial conditions to $u_t(t,x) = u_{xx}(t,x)$
ordinary-differential-equations pde fourier-series heat-equation
edited Dec 7 '18 at 13:27
amWhy
1
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
$endgroup$
add a comment |
$begingroup$
Why is this true?
$$int_{0}^{1}u^0(x)sin(lpi x)dx = int_{0}^{1}u(0,x)sin(lpi x)dx$$ $$ = int_{0}^{1} sum_{m=1}^{infty} a_m sin(mpi x)sin(lpi x)dx$$ $$ = frac{a_l}{2}$$
where coefficients ${a_m}$ are determined by $u(0,x) = u^0(x)$ which are the initial conditions to $u_t(t,x) = u_{xx}(t,x)$
ordinary-differential-equations pde fourier-series heat-equation
edited Dec 7 '18 at 13:27
amWhy
1
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
$endgroup$
Why is this true?
$$int_{0}^{1}u^0(x)sin(lpi x)dx = int_{0}^{1}u(0,x)sin(lpi x)dx$$ $$ = int_{0}^{1} sum_{m=1}^{infty} a_m sin(mpi x)sin(lpi x)dx$$ $$ = frac{a_l}{2}$$
where coefficients ${a_m}$ are determined by $u(0,x) = u^0(x)$ which are the initial conditions to $u_t(t,x) = u_{xx}(t,x)$
ordinary-differential-equations pde fourier-series heat-equation
ordinary-differential-equations pde fourier-series heat-equation
edited Dec 7 '18 at 13:27
amWhy
1
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
edited Dec 7 '18 at 13:27
amWhy
1
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
edited Dec 7 '18 at 13:27
amWhy
1
edited Dec 7 '18 at 13:27
amWhy
1
edited Dec 7 '18 at 13:27
amWhy
1
1
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
asked Dec 7 '18 at 13:08
pablo_mathscobarpablo_mathscobar
856
856
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is true because of the trigonometric identity
$$
sin(A)sin(B)=frac12left[cosfrac{A-B}2-cosfrac{A+B}2right]
$$
Series and integral can be switched in position if $sum_{m=0}^infty a_m^2<infty$, which is the case if $u^0in L^2([0,1])$.
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
$endgroup$
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%2f3029877%2fint-01-sum-m-1-infty-a-m-sinm-pi-x-sinl-pi-xdx-fraca-l%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
$begingroup$
It is true because of the trigonometric identity
$$
sin(A)sin(B)=frac12left[cosfrac{A-B}2-cosfrac{A+B}2right]
$$
Series and integral can be switched in position if $sum_{m=0}^infty a_m^2<infty$, which is the case if $u^0in L^2([0,1])$.
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
It is true because of the trigonometric identity
$$
sin(A)sin(B)=frac12left[cosfrac{A-B}2-cosfrac{A+B}2right]
$$
Series and integral can be switched in position if $sum_{m=0}^infty a_m^2<infty$, which is the case if $u^0in L^2([0,1])$.
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
It is true because of the trigonometric identity
$$
sin(A)sin(B)=frac12left[cosfrac{A-B}2-cosfrac{A+B}2right]
$$
Series and integral can be switched in position if $sum_{m=0}^infty a_m^2<infty$, which is the case if $u^0in L^2([0,1])$.
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
$endgroup$
It is true because of the trigonometric identity
$$
sin(A)sin(B)=frac12left[cosfrac{A-B}2-cosfrac{A+B}2right]
$$
Series and integral can be switched in position if $sum_{m=0}^infty a_m^2<infty$, which is the case if $u^0in L^2([0,1])$.
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
answered Dec 7 '18 at 14:18
LutzLLutzL
57.2k42054
57.2k42054
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%2f3029877%2fint-01-sum-m-1-infty-a-m-sinm-pi-x-sinl-pi-xdx-fraca-l%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
