Dirac delta function + constant [closed]
$begingroup$
If we sum the Dirac delta function with a constant, what is the result?
I.e., $k+delta(x)$, where $k$ is a constant.
dirac-delta
edited Jan 7 at 13:44
amWhy
192k28225439
asked Jan 7 at 13:28
ARFARF
31
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ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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closed as unclear what you're asking by Arthur, Hans Lundmark, Abcd, Glorfindel, amWhy Jan 7 at 19:25
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 8 more comments
$begingroup$
If we sum the Dirac delta function with a constant, what is the result?
I.e., $k+delta(x)$, where $k$ is a constant.
dirac-delta
edited Jan 7 at 13:44
amWhy
192k28225439
asked Jan 7 at 13:28
ARFARF
31
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as unclear what you're asking by Arthur, Hans Lundmark, Abcd, Glorfindel, amWhy Jan 7 at 19:25
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Think about the values that the Delta function takes on over the real line and at $x=0$.
$endgroup$
– aleden
Jan 7 at 13:30
$begingroup$
something like: $+infty$ when $x=0$ and $0$ in everywhere else?
$endgroup$
– ARF
Jan 7 at 13:40
3
$begingroup$
The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $delta(x) + k$ is just $delta(x) + k$, and that's it.
$endgroup$
– Arthur
Jan 7 at 13:42
1
$begingroup$
@PeterMelech No, $<delta +k,phi>$ should be $phi(0)=kintphi$. (It certainly can't be $phi(0)+k$; that's not linear in $phi$.)
$endgroup$
– David C. Ullrich
Jan 7 at 14:23
1
$begingroup$
@PeterMelech The pairing is linear : $<kdelta,phi>=kphi(0)$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:54
|
show 8 more comments
$begingroup$
If we sum the Dirac delta function with a constant, what is the result?
I.e., $k+delta(x)$, where $k$ is a constant.
dirac-delta
edited Jan 7 at 13:44
amWhy
192k28225439
asked Jan 7 at 13:28
ARFARF
31
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If we sum the Dirac delta function with a constant, what is the result?
I.e., $k+delta(x)$, where $k$ is a constant.
dirac-delta
dirac-delta
edited Jan 7 at 13:44
amWhy
192k28225439
asked Jan 7 at 13:28
ARFARF
31
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 13:44
amWhy
192k28225439
asked Jan 7 at 13:28
ARFARF
31
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 13:44
amWhy
192k28225439
edited Jan 7 at 13:44
amWhy
192k28225439
edited Jan 7 at 13:44
amWhy
192k28225439
192k28225439
asked Jan 7 at 13:28
ARFARF
31
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 13:28
ARFARF
31
asked Jan 7 at 13:28
ARFARF
31
31
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ARF is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as unclear what you're asking by Arthur, Hans Lundmark, Abcd, Glorfindel, amWhy Jan 7 at 19:25
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Arthur, Hans Lundmark, Abcd, Glorfindel, amWhy Jan 7 at 19:25
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Think about the values that the Delta function takes on over the real line and at $x=0$.
$endgroup$
– aleden
Jan 7 at 13:30
$begingroup$
something like: $+infty$ when $x=0$ and $0$ in everywhere else?
$endgroup$
– ARF
Jan 7 at 13:40
3
$begingroup$
The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $delta(x) + k$ is just $delta(x) + k$, and that's it.
$endgroup$
– Arthur
Jan 7 at 13:42
1
$begingroup$
@PeterMelech No, $<delta +k,phi>$ should be $phi(0)=kintphi$. (It certainly can't be $phi(0)+k$; that's not linear in $phi$.)
$endgroup$
– David C. Ullrich
Jan 7 at 14:23
1
$begingroup$
@PeterMelech The pairing is linear : $<kdelta,phi>=kphi(0)$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:54
|
show 8 more comments
$begingroup$
Think about the values that the Delta function takes on over the real line and at $x=0$.
$endgroup$
– aleden
Jan 7 at 13:30
$begingroup$
something like: $+infty$ when $x=0$ and $0$ in everywhere else?
$endgroup$
– ARF
Jan 7 at 13:40
3
$begingroup$
The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $delta(x) + k$ is just $delta(x) + k$, and that's it.
$endgroup$
– Arthur
Jan 7 at 13:42
1
$begingroup$
@PeterMelech No, $<delta +k,phi>$ should be $phi(0)=kintphi$. (It certainly can't be $phi(0)+k$; that's not linear in $phi$.)
$endgroup$
– David C. Ullrich
Jan 7 at 14:23
1
$begingroup$
@PeterMelech The pairing is linear : $<kdelta,phi>=kphi(0)$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:54
$begingroup$
Think about the values that the Delta function takes on over the real line and at $x=0$.
$endgroup$
– aleden
Jan 7 at 13:30
$begingroup$
Think about the values that the Delta function takes on over the real line and at $x=0$.
$endgroup$
– aleden
Jan 7 at 13:30
$begingroup$
something like: $+infty$ when $x=0$ and $0$ in everywhere else?
$endgroup$
– ARF
Jan 7 at 13:40
$begingroup$
something like: $+infty$ when $x=0$ and $0$ in everywhere else?
$endgroup$
– ARF
Jan 7 at 13:40
3
3
$begingroup$
The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $delta(x) + k$ is just $delta(x) + k$, and that's it.
$endgroup$
– Arthur
Jan 7 at 13:42
$begingroup$
The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $delta(x) + k$ is just $delta(x) + k$, and that's it.
$endgroup$
– Arthur
Jan 7 at 13:42
1
1
$begingroup$
@PeterMelech No, $<delta +k,phi>$ should be $phi(0)=kintphi$. (It certainly can't be $phi(0)+k$; that's not linear in $phi$.)
$endgroup$
– David C. Ullrich
Jan 7 at 14:23
$begingroup$
@PeterMelech No, $<delta +k,phi>$ should be $phi(0)=kintphi$. (It certainly can't be $phi(0)+k$; that's not linear in $phi$.)
$endgroup$
– David C. Ullrich
Jan 7 at 14:23
1
1
$begingroup$
@PeterMelech The pairing is linear : $<kdelta,phi>=kphi(0)$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:54
$begingroup$
@PeterMelech The pairing is linear : $<kdelta,phi>=kphi(0)$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:54
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There's a lot of confusion here. The "delta function" is sometimes treated "informally", as though it were a function with the property that $$int f(x)delta(x)=f(0)$$for all $f$. There may be some merit to such an "informal" presentation in some contexts, but in fact it's wrong; there is no such function.
$newcommandD{mathcal D}$
If we want to get this straight we alas need to start with a more formally correct definition. First, $D$ is the space of all infinitely differentiable functions with compact support. A distribution is a map $u:DtoBbb R$ such that (i) $u$ is linear (ii) $u$ is continuous.
Explaining exactly what it means to say $u$ is continuous gets a little complicated. We don't need to get that straight here, we just do need to keep in mind that a distribution is not a function (ie not a function defined on the line), rather a distribution is a certain sort of linear mapping from $D$ to $Bbb R$.
The space of distributions is denoted $D'$. So if $uinD'$ and $phiinD$ then $$u(phi)inBbb R.$$The notation $$<u,phi>:=u(phi)$$is often convenient.
If $f$ is a (locally integrable) function on $Bbb R$ then $f$ may be regarded as a distribution; here the distribution corresponding to the function $f$ is defined by $$<f,phi>=int fphi.$$(Where $int$ means $int_{-infty}^infty$.)
If $u$and $v$ are distributions and $k$ is a constant then the distributions $u+v$ and $ku$ are defined by $$<u+v,phi>=<u,phi>+<v,phi>$$and $$<ku,phi>=k<u,phi>.$$
And the distibution $delta$ is defined by $$<delta,phi>=phi(0).$$
And now we can answer the question of what $delta+k$ is if $k$ is a constant. Here we regard $k$ as a constant function; so applying the definitions above gives $$<delta+k,phi>=<delta,phi>+<k,phi>=phi(0)+int kphi=phi(0)+kintphiquad(phiinD).$$
That really is all there is to be said; $<delta+k,phi>=phi(0)=kintphi$ really is the definition of $delta+k$. Honest.
Not the answer you wanted? Sorry, nothing to be done about that. How would we plot $delta+k$? We don't "plot" $delta+k$; that makes no sense, because $delta+k$ is not a function.
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There's a lot of confusion here. The "delta function" is sometimes treated "informally", as though it were a function with the property that $$int f(x)delta(x)=f(0)$$for all $f$. There may be some merit to such an "informal" presentation in some contexts, but in fact it's wrong; there is no such function.
$newcommandD{mathcal D}$
If we want to get this straight we alas need to start with a more formally correct definition. First, $D$ is the space of all infinitely differentiable functions with compact support. A distribution is a map $u:DtoBbb R$ such that (i) $u$ is linear (ii) $u$ is continuous.
Explaining exactly what it means to say $u$ is continuous gets a little complicated. We don't need to get that straight here, we just do need to keep in mind that a distribution is not a function (ie not a function defined on the line), rather a distribution is a certain sort of linear mapping from $D$ to $Bbb R$.
The space of distributions is denoted $D'$. So if $uinD'$ and $phiinD$ then $$u(phi)inBbb R.$$The notation $$<u,phi>:=u(phi)$$is often convenient.
If $f$ is a (locally integrable) function on $Bbb R$ then $f$ may be regarded as a distribution; here the distribution corresponding to the function $f$ is defined by $$<f,phi>=int fphi.$$(Where $int$ means $int_{-infty}^infty$.)
If $u$and $v$ are distributions and $k$ is a constant then the distributions $u+v$ and $ku$ are defined by $$<u+v,phi>=<u,phi>+<v,phi>$$and $$<ku,phi>=k<u,phi>.$$
And the distibution $delta$ is defined by $$<delta,phi>=phi(0).$$
And now we can answer the question of what $delta+k$ is if $k$ is a constant. Here we regard $k$ as a constant function; so applying the definitions above gives $$<delta+k,phi>=<delta,phi>+<k,phi>=phi(0)+int kphi=phi(0)+kintphiquad(phiinD).$$
That really is all there is to be said; $<delta+k,phi>=phi(0)=kintphi$ really is the definition of $delta+k$. Honest.
Not the answer you wanted? Sorry, nothing to be done about that. How would we plot $delta+k$? We don't "plot" $delta+k$; that makes no sense, because $delta+k$ is not a function.
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
$endgroup$
add a comment |
$begingroup$
There's a lot of confusion here. The "delta function" is sometimes treated "informally", as though it were a function with the property that $$int f(x)delta(x)=f(0)$$for all $f$. There may be some merit to such an "informal" presentation in some contexts, but in fact it's wrong; there is no such function.
$newcommandD{mathcal D}$
If we want to get this straight we alas need to start with a more formally correct definition. First, $D$ is the space of all infinitely differentiable functions with compact support. A distribution is a map $u:DtoBbb R$ such that (i) $u$ is linear (ii) $u$ is continuous.
Explaining exactly what it means to say $u$ is continuous gets a little complicated. We don't need to get that straight here, we just do need to keep in mind that a distribution is not a function (ie not a function defined on the line), rather a distribution is a certain sort of linear mapping from $D$ to $Bbb R$.
The space of distributions is denoted $D'$. So if $uinD'$ and $phiinD$ then $$u(phi)inBbb R.$$The notation $$<u,phi>:=u(phi)$$is often convenient.
If $f$ is a (locally integrable) function on $Bbb R$ then $f$ may be regarded as a distribution; here the distribution corresponding to the function $f$ is defined by $$<f,phi>=int fphi.$$(Where $int$ means $int_{-infty}^infty$.)
If $u$and $v$ are distributions and $k$ is a constant then the distributions $u+v$ and $ku$ are defined by $$<u+v,phi>=<u,phi>+<v,phi>$$and $$<ku,phi>=k<u,phi>.$$
And the distibution $delta$ is defined by $$<delta,phi>=phi(0).$$
And now we can answer the question of what $delta+k$ is if $k$ is a constant. Here we regard $k$ as a constant function; so applying the definitions above gives $$<delta+k,phi>=<delta,phi>+<k,phi>=phi(0)+int kphi=phi(0)+kintphiquad(phiinD).$$
That really is all there is to be said; $<delta+k,phi>=phi(0)=kintphi$ really is the definition of $delta+k$. Honest.
Not the answer you wanted? Sorry, nothing to be done about that. How would we plot $delta+k$? We don't "plot" $delta+k$; that makes no sense, because $delta+k$ is not a function.
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
$endgroup$
add a comment |
$begingroup$
There's a lot of confusion here. The "delta function" is sometimes treated "informally", as though it were a function with the property that $$int f(x)delta(x)=f(0)$$for all $f$. There may be some merit to such an "informal" presentation in some contexts, but in fact it's wrong; there is no such function.
$newcommandD{mathcal D}$
If we want to get this straight we alas need to start with a more formally correct definition. First, $D$ is the space of all infinitely differentiable functions with compact support. A distribution is a map $u:DtoBbb R$ such that (i) $u$ is linear (ii) $u$ is continuous.
Explaining exactly what it means to say $u$ is continuous gets a little complicated. We don't need to get that straight here, we just do need to keep in mind that a distribution is not a function (ie not a function defined on the line), rather a distribution is a certain sort of linear mapping from $D$ to $Bbb R$.
The space of distributions is denoted $D'$. So if $uinD'$ and $phiinD$ then $$u(phi)inBbb R.$$The notation $$<u,phi>:=u(phi)$$is often convenient.
If $f$ is a (locally integrable) function on $Bbb R$ then $f$ may be regarded as a distribution; here the distribution corresponding to the function $f$ is defined by $$<f,phi>=int fphi.$$(Where $int$ means $int_{-infty}^infty$.)
If $u$and $v$ are distributions and $k$ is a constant then the distributions $u+v$ and $ku$ are defined by $$<u+v,phi>=<u,phi>+<v,phi>$$and $$<ku,phi>=k<u,phi>.$$
And the distibution $delta$ is defined by $$<delta,phi>=phi(0).$$
And now we can answer the question of what $delta+k$ is if $k$ is a constant. Here we regard $k$ as a constant function; so applying the definitions above gives $$<delta+k,phi>=<delta,phi>+<k,phi>=phi(0)+int kphi=phi(0)+kintphiquad(phiinD).$$
That really is all there is to be said; $<delta+k,phi>=phi(0)=kintphi$ really is the definition of $delta+k$. Honest.
Not the answer you wanted? Sorry, nothing to be done about that. How would we plot $delta+k$? We don't "plot" $delta+k$; that makes no sense, because $delta+k$ is not a function.
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
$endgroup$
There's a lot of confusion here. The "delta function" is sometimes treated "informally", as though it were a function with the property that $$int f(x)delta(x)=f(0)$$for all $f$. There may be some merit to such an "informal" presentation in some contexts, but in fact it's wrong; there is no such function.
$newcommandD{mathcal D}$
If we want to get this straight we alas need to start with a more formally correct definition. First, $D$ is the space of all infinitely differentiable functions with compact support. A distribution is a map $u:DtoBbb R$ such that (i) $u$ is linear (ii) $u$ is continuous.
Explaining exactly what it means to say $u$ is continuous gets a little complicated. We don't need to get that straight here, we just do need to keep in mind that a distribution is not a function (ie not a function defined on the line), rather a distribution is a certain sort of linear mapping from $D$ to $Bbb R$.
The space of distributions is denoted $D'$. So if $uinD'$ and $phiinD$ then $$u(phi)inBbb R.$$The notation $$<u,phi>:=u(phi)$$is often convenient.
If $f$ is a (locally integrable) function on $Bbb R$ then $f$ may be regarded as a distribution; here the distribution corresponding to the function $f$ is defined by $$<f,phi>=int fphi.$$(Where $int$ means $int_{-infty}^infty$.)
If $u$and $v$ are distributions and $k$ is a constant then the distributions $u+v$ and $ku$ are defined by $$<u+v,phi>=<u,phi>+<v,phi>$$and $$<ku,phi>=k<u,phi>.$$
And the distibution $delta$ is defined by $$<delta,phi>=phi(0).$$
And now we can answer the question of what $delta+k$ is if $k$ is a constant. Here we regard $k$ as a constant function; so applying the definitions above gives $$<delta+k,phi>=<delta,phi>+<k,phi>=phi(0)+int kphi=phi(0)+kintphiquad(phiinD).$$
That really is all there is to be said; $<delta+k,phi>=phi(0)=kintphi$ really is the definition of $delta+k$. Honest.
Not the answer you wanted? Sorry, nothing to be done about that. How would we plot $delta+k$? We don't "plot" $delta+k$; that makes no sense, because $delta+k$ is not a function.
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
edited Jan 8 at 13:08
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
answered Jan 7 at 17:10
David C. UllrichDavid C. Ullrich
59.4k43893
59.4k43893
add a comment |
add a comment |
$begingroup$
Think about the values that the Delta function takes on over the real line and at $x=0$.
$endgroup$
– aleden
Jan 7 at 13:30
$begingroup$
something like: $+infty$ when $x=0$ and $0$ in everywhere else?
$endgroup$
– ARF
Jan 7 at 13:40
3
$begingroup$
The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $delta(x) + k$ is just $delta(x) + k$, and that's it.
$endgroup$
– Arthur
Jan 7 at 13:42
1
$begingroup$
@PeterMelech No, $<delta +k,phi>$ should be $phi(0)=kintphi$. (It certainly can't be $phi(0)+k$; that's not linear in $phi$.)
$endgroup$
– David C. Ullrich
Jan 7 at 14:23
1
$begingroup$
@PeterMelech The pairing is linear : $<kdelta,phi>=kphi(0)$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:54