Mixing
Limit of series of Heaviside step functions
$begingroup$
I couldn't find a better title, because it is a very specific limit that I want to show:
$$f_b(y)=frac{2}{b}sumlimits_{k=1}^inftythetaleft(y-frac{kpi}{b}right) rightarrow f(y)=frac{2}{pi}y,;brightarrowinfty$$
I thought, that it would be helpful to use this definition of the Heaviside step function:
$$theta=limlimits_{epsilonrightarrow0}frac{1}{pi}left[arctan(x/epsilon)+frac{pi}{2}right]$$
Could it be some kind of Fourier series I have to work with?
Thanks for reading!
real-analysis functional-analysis limits analysis
asked Jan 13 at 10:39
ZapatistaZapatista
82
$endgroup$
add a comment |
$begingroup$
I couldn't find a better title, because it is a very specific limit that I want to show:
$$f_b(y)=frac{2}{b}sumlimits_{k=1}^inftythetaleft(y-frac{kpi}{b}right) rightarrow f(y)=frac{2}{pi}y,;brightarrowinfty$$
I thought, that it would be helpful to use this definition of the Heaviside step function:
$$theta=limlimits_{epsilonrightarrow0}frac{1}{pi}left[arctan(x/epsilon)+frac{pi}{2}right]$$
Could it be some kind of Fourier series I have to work with?
Thanks for reading!
real-analysis functional-analysis limits analysis
asked Jan 13 at 10:39
ZapatistaZapatista
82
$endgroup$
add a comment |
$begingroup$
I couldn't find a better title, because it is a very specific limit that I want to show:
$$f_b(y)=frac{2}{b}sumlimits_{k=1}^inftythetaleft(y-frac{kpi}{b}right) rightarrow f(y)=frac{2}{pi}y,;brightarrowinfty$$
I thought, that it would be helpful to use this definition of the Heaviside step function:
$$theta=limlimits_{epsilonrightarrow0}frac{1}{pi}left[arctan(x/epsilon)+frac{pi}{2}right]$$
Could it be some kind of Fourier series I have to work with?
Thanks for reading!
real-analysis functional-analysis limits analysis
asked Jan 13 at 10:39
ZapatistaZapatista
82
$endgroup$
I couldn't find a better title, because it is a very specific limit that I want to show:
$$f_b(y)=frac{2}{b}sumlimits_{k=1}^inftythetaleft(y-frac{kpi}{b}right) rightarrow f(y)=frac{2}{pi}y,;brightarrowinfty$$
I thought, that it would be helpful to use this definition of the Heaviside step function:
$$theta=limlimits_{epsilonrightarrow0}frac{1}{pi}left[arctan(x/epsilon)+frac{pi}{2}right]$$
Could it be some kind of Fourier series I have to work with?
Thanks for reading!
real-analysis functional-analysis limits analysis
real-analysis functional-analysis limits analysis
asked Jan 13 at 10:39
ZapatistaZapatista
82
asked Jan 13 at 10:39
ZapatistaZapatista
82
asked Jan 13 at 10:39
ZapatistaZapatista
82
asked Jan 13 at 10:39
ZapatistaZapatista
82
asked Jan 13 at 10:39
ZapatistaZapatista
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $y le 0$ the limes is zero and for $y > 0$ we get $f(z) = 2 y/pi$. If $y > 0$ is fixed, we only have to consider the sum up to $n = lceil b y/pi rceil$, because the other terms are zero. All summands up to $n-1$ are $1$. Note that $2/b rightarrow 0$, thus the $n$-th summand is negligible, i.e.
$$left|frac{2}{b} sum_{k=1}^n Theta(y-kpi/b)-frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b)right| le frac{2}{b} rightarrow 0.$$
Since $n/b rightarrow y/pi$ we get that
$$frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b) = frac{2n}{b} rightarrow frac{2y}{pi}.$$
answered Jan 13 at 11:31
p4schp4sch
5,295317
$endgroup$
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
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%2f3071872%2flimit-of-series-of-heaviside-step-functions%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
$begingroup$
For $y le 0$ the limes is zero and for $y > 0$ we get $f(z) = 2 y/pi$. If $y > 0$ is fixed, we only have to consider the sum up to $n = lceil b y/pi rceil$, because the other terms are zero. All summands up to $n-1$ are $1$. Note that $2/b rightarrow 0$, thus the $n$-th summand is negligible, i.e.
$$left|frac{2}{b} sum_{k=1}^n Theta(y-kpi/b)-frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b)right| le frac{2}{b} rightarrow 0.$$
Since $n/b rightarrow y/pi$ we get that
$$frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b) = frac{2n}{b} rightarrow frac{2y}{pi}.$$
answered Jan 13 at 11:31
p4schp4sch
5,295317
$endgroup$
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
add a comment |
$begingroup$
For $y le 0$ the limes is zero and for $y > 0$ we get $f(z) = 2 y/pi$. If $y > 0$ is fixed, we only have to consider the sum up to $n = lceil b y/pi rceil$, because the other terms are zero. All summands up to $n-1$ are $1$. Note that $2/b rightarrow 0$, thus the $n$-th summand is negligible, i.e.
$$left|frac{2}{b} sum_{k=1}^n Theta(y-kpi/b)-frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b)right| le frac{2}{b} rightarrow 0.$$
Since $n/b rightarrow y/pi$ we get that
$$frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b) = frac{2n}{b} rightarrow frac{2y}{pi}.$$
answered Jan 13 at 11:31
p4schp4sch
5,295317
$endgroup$
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
add a comment |
$begingroup$
For $y le 0$ the limes is zero and for $y > 0$ we get $f(z) = 2 y/pi$. If $y > 0$ is fixed, we only have to consider the sum up to $n = lceil b y/pi rceil$, because the other terms are zero. All summands up to $n-1$ are $1$. Note that $2/b rightarrow 0$, thus the $n$-th summand is negligible, i.e.
$$left|frac{2}{b} sum_{k=1}^n Theta(y-kpi/b)-frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b)right| le frac{2}{b} rightarrow 0.$$
Since $n/b rightarrow y/pi$ we get that
$$frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b) = frac{2n}{b} rightarrow frac{2y}{pi}.$$
answered Jan 13 at 11:31
p4schp4sch
5,295317
$endgroup$
For $y le 0$ the limes is zero and for $y > 0$ we get $f(z) = 2 y/pi$. If $y > 0$ is fixed, we only have to consider the sum up to $n = lceil b y/pi rceil$, because the other terms are zero. All summands up to $n-1$ are $1$. Note that $2/b rightarrow 0$, thus the $n$-th summand is negligible, i.e.
$$left|frac{2}{b} sum_{k=1}^n Theta(y-kpi/b)-frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b)right| le frac{2}{b} rightarrow 0.$$
Since $n/b rightarrow y/pi$ we get that
$$frac{2}{b} sum_{k=1}^{n-1} Theta(y-kpi/b) = frac{2n}{b} rightarrow frac{2y}{pi}.$$
answered Jan 13 at 11:31
p4schp4sch
5,295317
edited Jan 13 at 15:52
answered Jan 13 at 11:31
p4schp4sch
5,295317
answered Jan 13 at 11:31
p4schp4sch
5,295317
answered Jan 13 at 11:31
p4schp4sch
5,295317
5,295317
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
add a comment |
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Wow, thank you so much. But I don't really understand how you did the first estimation where you get k(y/n-pi/b).
$endgroup$
– Zapatista
Jan 13 at 15:11
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
$begingroup$
Sorry, there was a mistake in my argumentation. I have corrected the answer and now the solution is also much simpler. :-)
$endgroup$
– p4sch
Jan 13 at 15:53
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%2f3071872%2flimit-of-series-of-heaviside-step-functions%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
