Find $|f|^2$ in two ways
$begingroup$
Question: Consider the Hilbert space $L^2(mathbb R/mathbb Z)$ of
periodic measurable functions. Recall that it has a Hilbert space
basis consisting of functions $chi_n(x)=e^{2pi i n x}$ where $nin
> mathbb Z$. Let $fin L^2(mathbb R/mathbb Z)$ be the periodic
function defined by $f(x)= x$, for $xin [0,1)$. Compute $||f||^2$ in
two ways, directly, and then using the Parseval's identity, i.e. using
the Fourier series expansion of $f$.
My attempt:
We have that $$|f(x)|^2 = langle f, f rangle = int_0^1 |f(x)|^2 dx = int_0^1 |x|^2 dx = frac{|x|^3}{3}dx = frac{1}{3}$$ since $x in [0,1)$.
I am not sure if this is right though. I feel like the norm should be
1...
Now, I will find the norm using Parseval's identity using the Fourier series expansion of $f$.
By Parseval's, given the basis $chi_n$ such that $langle chi_n, chi_m rangle=delta_{n,m}$ like in our problem, we have that for every $xin L^2(mathbb{R}/mathbb{Z})$ $$sum_n |langle x,chi_n rangle|=|x|^2$$
For our problem, we have $f(x)=x$ for $xin [0,1)$, thus we have that $$f(x)=x=sum_n langle x,chi_n rangle chi_n$$
$$Rightarrow |f(x)|^2=|x|^2=bigg(|sum_n langle x,chi_n rangle chi_n|bigg)^2$$
I am not sure if this is right so far or how to continue from here.
Any help would be much appreciated.
real-analysis functional-analysis norm fourier-series
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
$endgroup$
add a comment |
$begingroup$
Question: Consider the Hilbert space $L^2(mathbb R/mathbb Z)$ of
periodic measurable functions. Recall that it has a Hilbert space
basis consisting of functions $chi_n(x)=e^{2pi i n x}$ where $nin
> mathbb Z$. Let $fin L^2(mathbb R/mathbb Z)$ be the periodic
function defined by $f(x)= x$, for $xin [0,1)$. Compute $||f||^2$ in
two ways, directly, and then using the Parseval's identity, i.e. using
the Fourier series expansion of $f$.
My attempt:
We have that $$|f(x)|^2 = langle f, f rangle = int_0^1 |f(x)|^2 dx = int_0^1 |x|^2 dx = frac{|x|^3}{3}dx = frac{1}{3}$$ since $x in [0,1)$.
I am not sure if this is right though. I feel like the norm should be
1...
Now, I will find the norm using Parseval's identity using the Fourier series expansion of $f$.
By Parseval's, given the basis $chi_n$ such that $langle chi_n, chi_m rangle=delta_{n,m}$ like in our problem, we have that for every $xin L^2(mathbb{R}/mathbb{Z})$ $$sum_n |langle x,chi_n rangle|=|x|^2$$
For our problem, we have $f(x)=x$ for $xin [0,1)$, thus we have that $$f(x)=x=sum_n langle x,chi_n rangle chi_n$$
$$Rightarrow |f(x)|^2=|x|^2=bigg(|sum_n langle x,chi_n rangle chi_n|bigg)^2$$
I am not sure if this is right so far or how to continue from here.
Any help would be much appreciated.
real-analysis functional-analysis norm fourier-series
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
$endgroup$
$begingroup$
In general, it is not a good practice to keep both $f(x)$ and $x$ to represent the identity function on $[0, 1)$. This might be causing some confusion. Try to use $f$ only to functions on your Hilbert space and $x$ as a dummy variable. Also, I think there are some typos above, like a square missing in the sum of Parseval identity
$endgroup$
– Daniel
Dec 2 '18 at 0:24
1
$begingroup$
Your evaluation of $lVert f rVert^2$ is correct. For the second part, you should use a very similar reasoning, computing for each fixed $n$ the value of $langle f, chi_n rangle$ by the definition of inner product on this space, that is, by computing an integral.
$endgroup$
– Daniel
Dec 2 '18 at 0:26
$begingroup$
@Daniel Thank you for your comments. I am getting The Fourier coefficients for are given by $$a_n = int_0^{1}f(x)e^{2pi inx}dx=int_0^{1}xe^{2pi inx}dx $$ which gives $a_0=1/2$ and $a_n=1/(ipi n)$. Now, from Parseval's, $$|f|^2=sum_{n=-infty}^{infty}|a_n|^2= 1/4 $$ But shouldn't it be the same norm for both methods?
$endgroup$
– MathIsHard
Dec 2 '18 at 3:52
1
$begingroup$
It is the same norm. Notice the sum is not $frac14$, this is only $|a_0|^2$. Since $|a_n| = |a_{-n}|$, what you actually get is $$ sum_{n in mathbb{Z}} |a_n|^2 = frac14 + sum_{m geq 1} frac{2}{(2pi m)^2} = frac14 + frac1{2pi^2} sum_{m geq 1} frac{1}{m^2}. $$ Since $lVert f rVert^2 = frac13$, this is equivalent to showing the famous result that $sum_{m geq 1} frac{1}{m^2} = frac{pi^2}{6}$.
$endgroup$
– Daniel
Dec 2 '18 at 13:41
$begingroup$
Oh thank you so much. I appreciate it. I forgot I was taking the absolute value of all of them so that they don’t all cancel except the zeroth one.
$endgroup$
– MathIsHard
Dec 2 '18 at 18:03
add a comment |
$begingroup$
Question: Consider the Hilbert space $L^2(mathbb R/mathbb Z)$ of
periodic measurable functions. Recall that it has a Hilbert space
basis consisting of functions $chi_n(x)=e^{2pi i n x}$ where $nin
> mathbb Z$. Let $fin L^2(mathbb R/mathbb Z)$ be the periodic
function defined by $f(x)= x$, for $xin [0,1)$. Compute $||f||^2$ in
two ways, directly, and then using the Parseval's identity, i.e. using
the Fourier series expansion of $f$.
My attempt:
We have that $$|f(x)|^2 = langle f, f rangle = int_0^1 |f(x)|^2 dx = int_0^1 |x|^2 dx = frac{|x|^3}{3}dx = frac{1}{3}$$ since $x in [0,1)$.
I am not sure if this is right though. I feel like the norm should be
1...
Now, I will find the norm using Parseval's identity using the Fourier series expansion of $f$.
By Parseval's, given the basis $chi_n$ such that $langle chi_n, chi_m rangle=delta_{n,m}$ like in our problem, we have that for every $xin L^2(mathbb{R}/mathbb{Z})$ $$sum_n |langle x,chi_n rangle|=|x|^2$$
For our problem, we have $f(x)=x$ for $xin [0,1)$, thus we have that $$f(x)=x=sum_n langle x,chi_n rangle chi_n$$
$$Rightarrow |f(x)|^2=|x|^2=bigg(|sum_n langle x,chi_n rangle chi_n|bigg)^2$$
I am not sure if this is right so far or how to continue from here.
Any help would be much appreciated.
real-analysis functional-analysis norm fourier-series
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
$endgroup$
Question: Consider the Hilbert space $L^2(mathbb R/mathbb Z)$ of
periodic measurable functions. Recall that it has a Hilbert space
basis consisting of functions $chi_n(x)=e^{2pi i n x}$ where $nin
> mathbb Z$. Let $fin L^2(mathbb R/mathbb Z)$ be the periodic
function defined by $f(x)= x$, for $xin [0,1)$. Compute $||f||^2$ in
two ways, directly, and then using the Parseval's identity, i.e. using
the Fourier series expansion of $f$.
My attempt:
We have that $$|f(x)|^2 = langle f, f rangle = int_0^1 |f(x)|^2 dx = int_0^1 |x|^2 dx = frac{|x|^3}{3}dx = frac{1}{3}$$ since $x in [0,1)$.
I am not sure if this is right though. I feel like the norm should be
1...
Now, I will find the norm using Parseval's identity using the Fourier series expansion of $f$.
By Parseval's, given the basis $chi_n$ such that $langle chi_n, chi_m rangle=delta_{n,m}$ like in our problem, we have that for every $xin L^2(mathbb{R}/mathbb{Z})$ $$sum_n |langle x,chi_n rangle|=|x|^2$$
For our problem, we have $f(x)=x$ for $xin [0,1)$, thus we have that $$f(x)=x=sum_n langle x,chi_n rangle chi_n$$
$$Rightarrow |f(x)|^2=|x|^2=bigg(|sum_n langle x,chi_n rangle chi_n|bigg)^2$$
I am not sure if this is right so far or how to continue from here.
Any help would be much appreciated.
real-analysis functional-analysis norm fourier-series
real-analysis functional-analysis norm fourier-series
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
asked Dec 2 '18 at 0:16
MathIsHardMathIsHard
1,242416
1,242416
$begingroup$
In general, it is not a good practice to keep both $f(x)$ and $x$ to represent the identity function on $[0, 1)$. This might be causing some confusion. Try to use $f$ only to functions on your Hilbert space and $x$ as a dummy variable. Also, I think there are some typos above, like a square missing in the sum of Parseval identity
$endgroup$
– Daniel
Dec 2 '18 at 0:24
1
$begingroup$
Your evaluation of $lVert f rVert^2$ is correct. For the second part, you should use a very similar reasoning, computing for each fixed $n$ the value of $langle f, chi_n rangle$ by the definition of inner product on this space, that is, by computing an integral.
$endgroup$
– Daniel
Dec 2 '18 at 0:26
$begingroup$
@Daniel Thank you for your comments. I am getting The Fourier coefficients for are given by $$a_n = int_0^{1}f(x)e^{2pi inx}dx=int_0^{1}xe^{2pi inx}dx $$ which gives $a_0=1/2$ and $a_n=1/(ipi n)$. Now, from Parseval's, $$|f|^2=sum_{n=-infty}^{infty}|a_n|^2= 1/4 $$ But shouldn't it be the same norm for both methods?
$endgroup$
– MathIsHard
Dec 2 '18 at 3:52
1
$begingroup$
It is the same norm. Notice the sum is not $frac14$, this is only $|a_0|^2$. Since $|a_n| = |a_{-n}|$, what you actually get is $$ sum_{n in mathbb{Z}} |a_n|^2 = frac14 + sum_{m geq 1} frac{2}{(2pi m)^2} = frac14 + frac1{2pi^2} sum_{m geq 1} frac{1}{m^2}. $$ Since $lVert f rVert^2 = frac13$, this is equivalent to showing the famous result that $sum_{m geq 1} frac{1}{m^2} = frac{pi^2}{6}$.
$endgroup$
– Daniel
Dec 2 '18 at 13:41
$begingroup$
Oh thank you so much. I appreciate it. I forgot I was taking the absolute value of all of them so that they don’t all cancel except the zeroth one.
$endgroup$
– MathIsHard
Dec 2 '18 at 18:03
add a comment |
$begingroup$
In general, it is not a good practice to keep both $f(x)$ and $x$ to represent the identity function on $[0, 1)$. This might be causing some confusion. Try to use $f$ only to functions on your Hilbert space and $x$ as a dummy variable. Also, I think there are some typos above, like a square missing in the sum of Parseval identity
$endgroup$
– Daniel
Dec 2 '18 at 0:24
1
$begingroup$
Your evaluation of $lVert f rVert^2$ is correct. For the second part, you should use a very similar reasoning, computing for each fixed $n$ the value of $langle f, chi_n rangle$ by the definition of inner product on this space, that is, by computing an integral.
$endgroup$
– Daniel
Dec 2 '18 at 0:26
$begingroup$
@Daniel Thank you for your comments. I am getting The Fourier coefficients for are given by $$a_n = int_0^{1}f(x)e^{2pi inx}dx=int_0^{1}xe^{2pi inx}dx $$ which gives $a_0=1/2$ and $a_n=1/(ipi n)$. Now, from Parseval's, $$|f|^2=sum_{n=-infty}^{infty}|a_n|^2= 1/4 $$ But shouldn't it be the same norm for both methods?
$endgroup$
– MathIsHard
Dec 2 '18 at 3:52
1
$begingroup$
It is the same norm. Notice the sum is not $frac14$, this is only $|a_0|^2$. Since $|a_n| = |a_{-n}|$, what you actually get is $$ sum_{n in mathbb{Z}} |a_n|^2 = frac14 + sum_{m geq 1} frac{2}{(2pi m)^2} = frac14 + frac1{2pi^2} sum_{m geq 1} frac{1}{m^2}. $$ Since $lVert f rVert^2 = frac13$, this is equivalent to showing the famous result that $sum_{m geq 1} frac{1}{m^2} = frac{pi^2}{6}$.
$endgroup$
– Daniel
Dec 2 '18 at 13:41
$begingroup$
Oh thank you so much. I appreciate it. I forgot I was taking the absolute value of all of them so that they don’t all cancel except the zeroth one.
$endgroup$
– MathIsHard
Dec 2 '18 at 18:03
$begingroup$
In general, it is not a good practice to keep both $f(x)$ and $x$ to represent the identity function on $[0, 1)$. This might be causing some confusion. Try to use $f$ only to functions on your Hilbert space and $x$ as a dummy variable. Also, I think there are some typos above, like a square missing in the sum of Parseval identity
$endgroup$
– Daniel
Dec 2 '18 at 0:24
$begingroup$
In general, it is not a good practice to keep both $f(x)$ and $x$ to represent the identity function on $[0, 1)$. This might be causing some confusion. Try to use $f$ only to functions on your Hilbert space and $x$ as a dummy variable. Also, I think there are some typos above, like a square missing in the sum of Parseval identity
$endgroup$
– Daniel
Dec 2 '18 at 0:24
1
1
$begingroup$
Your evaluation of $lVert f rVert^2$ is correct. For the second part, you should use a very similar reasoning, computing for each fixed $n$ the value of $langle f, chi_n rangle$ by the definition of inner product on this space, that is, by computing an integral.
$endgroup$
– Daniel
Dec 2 '18 at 0:26
$begingroup$
Your evaluation of $lVert f rVert^2$ is correct. For the second part, you should use a very similar reasoning, computing for each fixed $n$ the value of $langle f, chi_n rangle$ by the definition of inner product on this space, that is, by computing an integral.
$endgroup$
– Daniel
Dec 2 '18 at 0:26
$begingroup$
@Daniel Thank you for your comments. I am getting The Fourier coefficients for are given by $$a_n = int_0^{1}f(x)e^{2pi inx}dx=int_0^{1}xe^{2pi inx}dx $$ which gives $a_0=1/2$ and $a_n=1/(ipi n)$. Now, from Parseval's, $$|f|^2=sum_{n=-infty}^{infty}|a_n|^2= 1/4 $$ But shouldn't it be the same norm for both methods?
$endgroup$
– MathIsHard
Dec 2 '18 at 3:52
$begingroup$
@Daniel Thank you for your comments. I am getting The Fourier coefficients for are given by $$a_n = int_0^{1}f(x)e^{2pi inx}dx=int_0^{1}xe^{2pi inx}dx $$ which gives $a_0=1/2$ and $a_n=1/(ipi n)$. Now, from Parseval's, $$|f|^2=sum_{n=-infty}^{infty}|a_n|^2= 1/4 $$ But shouldn't it be the same norm for both methods?
$endgroup$
– MathIsHard
Dec 2 '18 at 3:52
1
1
$begingroup$
It is the same norm. Notice the sum is not $frac14$, this is only $|a_0|^2$. Since $|a_n| = |a_{-n}|$, what you actually get is $$ sum_{n in mathbb{Z}} |a_n|^2 = frac14 + sum_{m geq 1} frac{2}{(2pi m)^2} = frac14 + frac1{2pi^2} sum_{m geq 1} frac{1}{m^2}. $$ Since $lVert f rVert^2 = frac13$, this is equivalent to showing the famous result that $sum_{m geq 1} frac{1}{m^2} = frac{pi^2}{6}$.
$endgroup$
– Daniel
Dec 2 '18 at 13:41
$begingroup$
It is the same norm. Notice the sum is not $frac14$, this is only $|a_0|^2$. Since $|a_n| = |a_{-n}|$, what you actually get is $$ sum_{n in mathbb{Z}} |a_n|^2 = frac14 + sum_{m geq 1} frac{2}{(2pi m)^2} = frac14 + frac1{2pi^2} sum_{m geq 1} frac{1}{m^2}. $$ Since $lVert f rVert^2 = frac13$, this is equivalent to showing the famous result that $sum_{m geq 1} frac{1}{m^2} = frac{pi^2}{6}$.
$endgroup$
– Daniel
Dec 2 '18 at 13:41
$begingroup$
Oh thank you so much. I appreciate it. I forgot I was taking the absolute value of all of them so that they don’t all cancel except the zeroth one.
$endgroup$
– MathIsHard
Dec 2 '18 at 18:03
$begingroup$
Oh thank you so much. I appreciate it. I forgot I was taking the absolute value of all of them so that they don’t all cancel except the zeroth one.
$endgroup$
– MathIsHard
Dec 2 '18 at 18:03
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%2f3022048%2ffind-f-2-in-two-ways%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%2f3022048%2ffind-f-2-in-two-ways%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
$begingroup$
In general, it is not a good practice to keep both $f(x)$ and $x$ to represent the identity function on $[0, 1)$. This might be causing some confusion. Try to use $f$ only to functions on your Hilbert space and $x$ as a dummy variable. Also, I think there are some typos above, like a square missing in the sum of Parseval identity
$endgroup$
– Daniel
Dec 2 '18 at 0:24
1
$begingroup$
Your evaluation of $lVert f rVert^2$ is correct. For the second part, you should use a very similar reasoning, computing for each fixed $n$ the value of $langle f, chi_n rangle$ by the definition of inner product on this space, that is, by computing an integral.
$endgroup$
– Daniel
Dec 2 '18 at 0:26
$begingroup$
@Daniel Thank you for your comments. I am getting The Fourier coefficients for are given by $$a_n = int_0^{1}f(x)e^{2pi inx}dx=int_0^{1}xe^{2pi inx}dx $$ which gives $a_0=1/2$ and $a_n=1/(ipi n)$. Now, from Parseval's, $$|f|^2=sum_{n=-infty}^{infty}|a_n|^2= 1/4 $$ But shouldn't it be the same norm for both methods?
$endgroup$
– MathIsHard
Dec 2 '18 at 3:52
1
$begingroup$
It is the same norm. Notice the sum is not $frac14$, this is only $|a_0|^2$. Since $|a_n| = |a_{-n}|$, what you actually get is $$ sum_{n in mathbb{Z}} |a_n|^2 = frac14 + sum_{m geq 1} frac{2}{(2pi m)^2} = frac14 + frac1{2pi^2} sum_{m geq 1} frac{1}{m^2}. $$ Since $lVert f rVert^2 = frac13$, this is equivalent to showing the famous result that $sum_{m geq 1} frac{1}{m^2} = frac{pi^2}{6}$.
$endgroup$
– Daniel
Dec 2 '18 at 13:41
$begingroup$
Oh thank you so much. I appreciate it. I forgot I was taking the absolute value of all of them so that they don’t all cancel except the zeroth one.
$endgroup$
– MathIsHard
Dec 2 '18 at 18:03