Fourier series of $(-1)^{[x]}$
$begingroup$
I've been trying to find the Fourier series of $f(x)=(-1)^{[x]}$ ($[x]$ is the floor function) on $[-2,2]$.
I've drawn the graph and noticed that $f$ is odd and therefore $a_n = 0, forall n in mathbb{N}$. Calculating the coefficient $b_n$ resulted in $b_n = frac{2}{npi}left(-2cos(frac{npi}{2}) + 1 + (-1)^nright)$. However when plotting the series, I does not look like $f(x)=(-1)^{[x]} $. How can I best approach finding Fourier series for this type of functions?
real-analysis
asked Jan 13 at 13:25
ZacharyZachary
1799
$endgroup$
add a comment |
$begingroup$
I've been trying to find the Fourier series of $f(x)=(-1)^{[x]}$ ($[x]$ is the floor function) on $[-2,2]$.
I've drawn the graph and noticed that $f$ is odd and therefore $a_n = 0, forall n in mathbb{N}$. Calculating the coefficient $b_n$ resulted in $b_n = frac{2}{npi}left(-2cos(frac{npi}{2}) + 1 + (-1)^nright)$. However when plotting the series, I does not look like $f(x)=(-1)^{[x]} $. How can I best approach finding Fourier series for this type of functions?
real-analysis
asked Jan 13 at 13:25
ZacharyZachary
1799
$endgroup$
$begingroup$
$f$ is odd $2$-periodic so $a_n=0$, $b_n = frac{1}{2} int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$
$endgroup$
– reuns
Jan 13 at 13:49
$begingroup$
Why do you have $sin(2pi nx/2)$? I don't see where the first $2$ comes from.
$endgroup$
– Zachary
Jan 13 at 14:12
$begingroup$
$b_n = int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$ then $f(x) = sum_{n ge 1} b_n sin(2pi n x/2)$
$endgroup$
– reuns
Jan 13 at 14:41
$begingroup$
But why do you use $2pi n x/2$ for the argument of $sin$, instead of $pi n x/2$. (Formula: $b_ n = frac1L int_{-L}^L f(x) sin (n pi x/L) $).
$endgroup$
– Zachary
Jan 13 at 14:45
$begingroup$
Has it something to do with the period of $f$?
$endgroup$
– Zachary
Jan 13 at 14:46
add a comment |
$begingroup$
I've been trying to find the Fourier series of $f(x)=(-1)^{[x]}$ ($[x]$ is the floor function) on $[-2,2]$.
I've drawn the graph and noticed that $f$ is odd and therefore $a_n = 0, forall n in mathbb{N}$. Calculating the coefficient $b_n$ resulted in $b_n = frac{2}{npi}left(-2cos(frac{npi}{2}) + 1 + (-1)^nright)$. However when plotting the series, I does not look like $f(x)=(-1)^{[x]} $. How can I best approach finding Fourier series for this type of functions?
real-analysis
asked Jan 13 at 13:25
ZacharyZachary
1799
$endgroup$
I've been trying to find the Fourier series of $f(x)=(-1)^{[x]}$ ($[x]$ is the floor function) on $[-2,2]$.
I've drawn the graph and noticed that $f$ is odd and therefore $a_n = 0, forall n in mathbb{N}$. Calculating the coefficient $b_n$ resulted in $b_n = frac{2}{npi}left(-2cos(frac{npi}{2}) + 1 + (-1)^nright)$. However when plotting the series, I does not look like $f(x)=(-1)^{[x]} $. How can I best approach finding Fourier series for this type of functions?
real-analysis
real-analysis
asked Jan 13 at 13:25
ZacharyZachary
1799
asked Jan 13 at 13:25
ZacharyZachary
1799
asked Jan 13 at 13:25
ZacharyZachary
1799
asked Jan 13 at 13:25
ZacharyZachary
1799
asked Jan 13 at 13:25
ZacharyZachary
1799
1799
$begingroup$
$f$ is odd $2$-periodic so $a_n=0$, $b_n = frac{1}{2} int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$
$endgroup$
– reuns
Jan 13 at 13:49
$begingroup$
Why do you have $sin(2pi nx/2)$? I don't see where the first $2$ comes from.
$endgroup$
– Zachary
Jan 13 at 14:12
$begingroup$
$b_n = int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$ then $f(x) = sum_{n ge 1} b_n sin(2pi n x/2)$
$endgroup$
– reuns
Jan 13 at 14:41
$begingroup$
But why do you use $2pi n x/2$ for the argument of $sin$, instead of $pi n x/2$. (Formula: $b_ n = frac1L int_{-L}^L f(x) sin (n pi x/L) $).
$endgroup$
– Zachary
Jan 13 at 14:45
$begingroup$
Has it something to do with the period of $f$?
$endgroup$
– Zachary
Jan 13 at 14:46
add a comment |
$begingroup$
$f$ is odd $2$-periodic so $a_n=0$, $b_n = frac{1}{2} int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$
$endgroup$
– reuns
Jan 13 at 13:49
$begingroup$
Why do you have $sin(2pi nx/2)$? I don't see where the first $2$ comes from.
$endgroup$
– Zachary
Jan 13 at 14:12
$begingroup$
$b_n = int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$ then $f(x) = sum_{n ge 1} b_n sin(2pi n x/2)$
$endgroup$
– reuns
Jan 13 at 14:41
$begingroup$
But why do you use $2pi n x/2$ for the argument of $sin$, instead of $pi n x/2$. (Formula: $b_ n = frac1L int_{-L}^L f(x) sin (n pi x/L) $).
$endgroup$
– Zachary
Jan 13 at 14:45
$begingroup$
Has it something to do with the period of $f$?
$endgroup$
– Zachary
Jan 13 at 14:46
$begingroup$
$f$ is odd $2$-periodic so $a_n=0$, $b_n = frac{1}{2} int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$
$endgroup$
– reuns
Jan 13 at 13:49
$begingroup$
$f$ is odd $2$-periodic so $a_n=0$, $b_n = frac{1}{2} int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$
$endgroup$
– reuns
Jan 13 at 13:49
$begingroup$
Why do you have $sin(2pi nx/2)$? I don't see where the first $2$ comes from.
$endgroup$
– Zachary
Jan 13 at 14:12
$begingroup$
Why do you have $sin(2pi nx/2)$? I don't see where the first $2$ comes from.
$endgroup$
– Zachary
Jan 13 at 14:12
$begingroup$
$b_n = int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$ then $f(x) = sum_{n ge 1} b_n sin(2pi n x/2)$
$endgroup$
– reuns
Jan 13 at 14:41
$begingroup$
$b_n = int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$ then $f(x) = sum_{n ge 1} b_n sin(2pi n x/2)$
$endgroup$
– reuns
Jan 13 at 14:41
$begingroup$
But why do you use $2pi n x/2$ for the argument of $sin$, instead of $pi n x/2$. (Formula: $b_ n = frac1L int_{-L}^L f(x) sin (n pi x/L) $).
$endgroup$
– Zachary
Jan 13 at 14:45
$begingroup$
But why do you use $2pi n x/2$ for the argument of $sin$, instead of $pi n x/2$. (Formula: $b_ n = frac1L int_{-L}^L f(x) sin (n pi x/L) $).
$endgroup$
– Zachary
Jan 13 at 14:45
$begingroup$
Has it something to do with the period of $f$?
$endgroup$
– Zachary
Jan 13 at 14:46
$begingroup$
Has it something to do with the period of $f$?
$endgroup$
– Zachary
Jan 13 at 14:46
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%2f3072004%2ffourier-series-of-1x%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%2f3072004%2ffourier-series-of-1x%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
$begingroup$
$f$ is odd $2$-periodic so $a_n=0$, $b_n = frac{1}{2} int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$
$endgroup$
– reuns
Jan 13 at 13:49
$begingroup$
Why do you have $sin(2pi nx/2)$? I don't see where the first $2$ comes from.
$endgroup$
– Zachary
Jan 13 at 14:12
$begingroup$
$b_n = int_{-1}^1 f(x) sin(2 pi n x/2)dx= int_0^1 sin(2 pi n x/2)dx = ?$ then $f(x) = sum_{n ge 1} b_n sin(2pi n x/2)$
$endgroup$
– reuns
Jan 13 at 14:41
$begingroup$
But why do you use $2pi n x/2$ for the argument of $sin$, instead of $pi n x/2$. (Formula: $b_ n = frac1L int_{-L}^L f(x) sin (n pi x/L) $).
$endgroup$
– Zachary
Jan 13 at 14:45
$begingroup$
Has it something to do with the period of $f$?
$endgroup$
– Zachary
Jan 13 at 14:46