Dirac delta function divided by Dirac delta function
$begingroup$
Is the following defined: (Dirac delta function divided by Dirac delta function)
$$f = frac{delta}{delta} = ?$$
distribution-theory dirac-delta
edited Jan 15 at 21:10
Qmechanic
5,17711858
asked Sep 6 '13 at 22:09
OSEOSE
95110
$endgroup$
add a comment |
$begingroup$
Is the following defined: (Dirac delta function divided by Dirac delta function)
$$f = frac{delta}{delta} = ?$$
distribution-theory dirac-delta
edited Jan 15 at 21:10
Qmechanic
5,17711858
asked Sep 6 '13 at 22:09
OSEOSE
95110
$endgroup$
2
$begingroup$
I have purged all the comments from this post. Please remember to keep comments on topic, and try not to berate other users.
$endgroup$
– Alex Becker
Sep 6 '13 at 22:57
$begingroup$
@Flybynight Please join me in chat so that we may discuss this further.
$endgroup$
– Potato
Sep 6 '13 at 22:58
add a comment |
$begingroup$
Is the following defined: (Dirac delta function divided by Dirac delta function)
$$f = frac{delta}{delta} = ?$$
distribution-theory dirac-delta
edited Jan 15 at 21:10
Qmechanic
5,17711858
asked Sep 6 '13 at 22:09
OSEOSE
95110
$endgroup$
Is the following defined: (Dirac delta function divided by Dirac delta function)
$$f = frac{delta}{delta} = ?$$
distribution-theory dirac-delta
distribution-theory dirac-delta
edited Jan 15 at 21:10
Qmechanic
5,17711858
asked Sep 6 '13 at 22:09
OSEOSE
95110
edited Jan 15 at 21:10
Qmechanic
5,17711858
asked Sep 6 '13 at 22:09
OSEOSE
95110
edited Jan 15 at 21:10
Qmechanic
5,17711858
edited Jan 15 at 21:10
Qmechanic
5,17711858
edited Jan 15 at 21:10
Qmechanic
5,17711858
5,17711858
asked Sep 6 '13 at 22:09
OSEOSE
95110
asked Sep 6 '13 at 22:09
OSEOSE
95110
asked Sep 6 '13 at 22:09
OSEOSE
95110
95110
2
$begingroup$
I have purged all the comments from this post. Please remember to keep comments on topic, and try not to berate other users.
$endgroup$
– Alex Becker
Sep 6 '13 at 22:57
$begingroup$
@Flybynight Please join me in chat so that we may discuss this further.
$endgroup$
– Potato
Sep 6 '13 at 22:58
add a comment |
2
$begingroup$
I have purged all the comments from this post. Please remember to keep comments on topic, and try not to berate other users.
$endgroup$
– Alex Becker
Sep 6 '13 at 22:57
$begingroup$
@Flybynight Please join me in chat so that we may discuss this further.
$endgroup$
– Potato
Sep 6 '13 at 22:58
2
2
$begingroup$
I have purged all the comments from this post. Please remember to keep comments on topic, and try not to berate other users.
$endgroup$
– Alex Becker
Sep 6 '13 at 22:57
$begingroup$
I have purged all the comments from this post. Please remember to keep comments on topic, and try not to berate other users.
$endgroup$
– Alex Becker
Sep 6 '13 at 22:57
$begingroup$
@Flybynight Please join me in chat so that we may discuss this further.
$endgroup$
– Potato
Sep 6 '13 at 22:58
$begingroup$
@Flybynight Please join me in chat so that we may discuss this further.
$endgroup$
– Potato
Sep 6 '13 at 22:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't know if the following is what you are looking for, but: To give the division a sense, what you can do is look for functions $phi in mathcal E^0(mathbb R)$ (that is continuous functions $mathbb R to mathbb R$ that fulfill $phi delta = delta$. As for any $psi in mathcal D(mathbb R)$ we have $$(phi delta)(psi) =delta(phipsi) = phi(0)psi(0) = phi(0)delta(psi), $$
that is $phi delta = phi(0)delta$, so we have $$phi delta = delta iff phi(0) = 1. $$
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
$endgroup$
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
add a comment |
$begingroup$
Remember $delta$ is 'operationally' defined by its effect under integration. So $delta$/$delta$ needs more context, for example, like
($delta$(x-y),f(x))/($delta$(x-y),g(x)),
where (.,.) is the inner product, before the question can be answered. In this case $delta$/$delta$ = 1 only if f(y)=g(y). Otherwise the results could be one of many other things.
answered May 4 '14 at 15:41
user45664user45664
266213
$endgroup$
add a comment |
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%2f486100%2fdirac-delta-function-divided-by-dirac-delta-function%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
$begingroup$
I don't know if the following is what you are looking for, but: To give the division a sense, what you can do is look for functions $phi in mathcal E^0(mathbb R)$ (that is continuous functions $mathbb R to mathbb R$ that fulfill $phi delta = delta$. As for any $psi in mathcal D(mathbb R)$ we have $$(phi delta)(psi) =delta(phipsi) = phi(0)psi(0) = phi(0)delta(psi), $$
that is $phi delta = phi(0)delta$, so we have $$phi delta = delta iff phi(0) = 1. $$
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
$endgroup$
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
add a comment |
$begingroup$
I don't know if the following is what you are looking for, but: To give the division a sense, what you can do is look for functions $phi in mathcal E^0(mathbb R)$ (that is continuous functions $mathbb R to mathbb R$ that fulfill $phi delta = delta$. As for any $psi in mathcal D(mathbb R)$ we have $$(phi delta)(psi) =delta(phipsi) = phi(0)psi(0) = phi(0)delta(psi), $$
that is $phi delta = phi(0)delta$, so we have $$phi delta = delta iff phi(0) = 1. $$
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
$endgroup$
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
add a comment |
$begingroup$
I don't know if the following is what you are looking for, but: To give the division a sense, what you can do is look for functions $phi in mathcal E^0(mathbb R)$ (that is continuous functions $mathbb R to mathbb R$ that fulfill $phi delta = delta$. As for any $psi in mathcal D(mathbb R)$ we have $$(phi delta)(psi) =delta(phipsi) = phi(0)psi(0) = phi(0)delta(psi), $$
that is $phi delta = phi(0)delta$, so we have $$phi delta = delta iff phi(0) = 1. $$
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
$endgroup$
I don't know if the following is what you are looking for, but: To give the division a sense, what you can do is look for functions $phi in mathcal E^0(mathbb R)$ (that is continuous functions $mathbb R to mathbb R$ that fulfill $phi delta = delta$. As for any $psi in mathcal D(mathbb R)$ we have $$(phi delta)(psi) =delta(phipsi) = phi(0)psi(0) = phi(0)delta(psi), $$
that is $phi delta = phi(0)delta$, so we have $$phi delta = delta iff phi(0) = 1. $$
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
answered Sep 6 '13 at 22:39
martinimartini
70.8k45991
70.8k45991
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
add a comment |
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
$begingroup$
what is D(R)?
$endgroup$
– user114766
Oct 17 '17 at 14:52
add a comment |
$begingroup$
Remember $delta$ is 'operationally' defined by its effect under integration. So $delta$/$delta$ needs more context, for example, like
($delta$(x-y),f(x))/($delta$(x-y),g(x)),
where (.,.) is the inner product, before the question can be answered. In this case $delta$/$delta$ = 1 only if f(y)=g(y). Otherwise the results could be one of many other things.
answered May 4 '14 at 15:41
user45664user45664
266213
$endgroup$
add a comment |
$begingroup$
Remember $delta$ is 'operationally' defined by its effect under integration. So $delta$/$delta$ needs more context, for example, like
($delta$(x-y),f(x))/($delta$(x-y),g(x)),
where (.,.) is the inner product, before the question can be answered. In this case $delta$/$delta$ = 1 only if f(y)=g(y). Otherwise the results could be one of many other things.
answered May 4 '14 at 15:41
user45664user45664
266213
$endgroup$
add a comment |
$begingroup$
Remember $delta$ is 'operationally' defined by its effect under integration. So $delta$/$delta$ needs more context, for example, like
($delta$(x-y),f(x))/($delta$(x-y),g(x)),
where (.,.) is the inner product, before the question can be answered. In this case $delta$/$delta$ = 1 only if f(y)=g(y). Otherwise the results could be one of many other things.
answered May 4 '14 at 15:41
user45664user45664
266213
$endgroup$
Remember $delta$ is 'operationally' defined by its effect under integration. So $delta$/$delta$ needs more context, for example, like
($delta$(x-y),f(x))/($delta$(x-y),g(x)),
where (.,.) is the inner product, before the question can be answered. In this case $delta$/$delta$ = 1 only if f(y)=g(y). Otherwise the results could be one of many other things.
answered May 4 '14 at 15:41
user45664user45664
266213
answered May 4 '14 at 15:41
user45664user45664
266213
answered May 4 '14 at 15:41
user45664user45664
266213
answered May 4 '14 at 15:41
user45664user45664
266213
266213
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%2f486100%2fdirac-delta-function-divided-by-dirac-delta-function%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
2
$begingroup$
I have purged all the comments from this post. Please remember to keep comments on topic, and try not to berate other users.
$endgroup$
– Alex Becker
Sep 6 '13 at 22:57
$begingroup$
@Flybynight Please join me in chat so that we may discuss this further.
$endgroup$
– Potato
Sep 6 '13 at 22:58