Applying sine orthogonality
$begingroup$
I have a confusion as to why this is a viable procedure in the following image:
$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$
To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.
Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.
fourier-analysis orthonormal
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
$endgroup$
add a comment |
$begingroup$
I have a confusion as to why this is a viable procedure in the following image:
$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$
To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.
Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.
fourier-analysis orthonormal
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
$endgroup$
3
$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17
$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19
1
$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20
$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26
$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37
add a comment |
$begingroup$
I have a confusion as to why this is a viable procedure in the following image:
$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$
To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.
Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.
fourier-analysis orthonormal
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
$endgroup$
I have a confusion as to why this is a viable procedure in the following image:
$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$
To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.
Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.
fourier-analysis orthonormal
fourier-analysis orthonormal
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
edited Dec 14 '18 at 21:49
Viktor Glombik
1,0931527
1,0931527
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
asked Dec 14 '18 at 21:14
sangstarsangstar
853215
853215
3
$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17
$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19
1
$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20
$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26
$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37
add a comment |
3
$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17
$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19
1
$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20
$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26
$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37
3
3
$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17
$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17
$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19
$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19
1
1
$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20
$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20
$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26
$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26
$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37
$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your eigenfunctions are solutions of
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0) = 0,;; y(a)=0.
$$
That is,
$y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.
Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
$$
y''+lambda y= 0,;;; 0 le x le a \
y'(0)=0,;; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.
There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0)=0,;;; Ay(a)+By'(a)=0.
$$
The functions $sin(alpha_n x)$ are solutions where
$$
Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
$$
The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.
Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
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%2f3039893%2fapplying-sine-orthogonality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your eigenfunctions are solutions of
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0) = 0,;; y(a)=0.
$$
That is,
$y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.
Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
$$
y''+lambda y= 0,;;; 0 le x le a \
y'(0)=0,;; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.
There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0)=0,;;; Ay(a)+By'(a)=0.
$$
The functions $sin(alpha_n x)$ are solutions where
$$
Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
$$
The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
Your eigenfunctions are solutions of
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0) = 0,;; y(a)=0.
$$
That is,
$y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.
Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
$$
y''+lambda y= 0,;;; 0 le x le a \
y'(0)=0,;; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.
There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0)=0,;;; Ay(a)+By'(a)=0.
$$
The functions $sin(alpha_n x)$ are solutions where
$$
Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
$$
The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
Your eigenfunctions are solutions of
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0) = 0,;; y(a)=0.
$$
That is,
$y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.
Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
$$
y''+lambda y= 0,;;; 0 le x le a \
y'(0)=0,;; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.
There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0)=0,;;; Ay(a)+By'(a)=0.
$$
The functions $sin(alpha_n x)$ are solutions where
$$
Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
$$
The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
Your eigenfunctions are solutions of
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0) = 0,;; y(a)=0.
$$
That is,
$y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.
Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
$$
y''+lambda y= 0,;;; 0 le x le a \
y'(0)=0,;; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.
There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0)=0,;;; Ay(a)+By'(a)=0.
$$
The functions $sin(alpha_n x)$ are solutions where
$$
Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
$$
The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 15 '18 at 17:51
DisintegratingByPartsDisintegratingByParts
59.5k42580
59.5k42580
add a comment |
add a comment |
$begingroup$
If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.
Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.
Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
add a comment |
$begingroup$
If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.
Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.
Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 20 '18 at 17:07
DisintegratingByPartsDisintegratingByParts
59.5k42580
59.5k42580
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%2f3039893%2fapplying-sine-orthogonality%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
3
$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17
$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19
1
$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20
$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26
$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37