Expanding Dirac delta function with Hermite polynomial
$begingroup$
My question is related to a formula in this paper
In that paper, they try to expand Dirac delta function $delta(x)$, which has the property
$$
int delta(x)f(x) , dx = f(0),
$$
using Hermite polynomial. So they write
$$
delta(x) = sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2}
$$
and get the coefficient $A_n$ by
$$
begin{align}
int H_{2m}(x) delta(x) , dx &= int H_{2m}(x) sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2} \
Rightarrow H_{2m}(0) &= A_m sqrt {pi}4^m (2m)! \
Rightarrow A_m &= frac{(-1)^m}{m! 4^m sqrt{pi}} ~~~~~~~~(H_{2n}(0)=frac{(2n)!(-1)^n}{n!})
end{align}
$$
Usual $delta(x)$ function has property that it equals to zero for $xneq 0$, but $delta(x) rightarrow infty $ for $x=0$
Now following above expansion, if we plug $x=0$ to the formula, we get
$$
begin{align}
delta(0) & = sum_{n=0}^{infty}A_n H_{2n}(0) \
& = sum_{n=0}^{infty} frac{(2n)!}{n!n!4^nsqrt{pi}}
end{align}
$$
But this series converges, so the usual property of $delta(x)$ is not recovered. So my question is, is this expansion for $delta(x)$ valid?
dirac-delta
edited May 31 '18 at 3:48
Michael Hardy
1
asked May 31 '18 at 2:36
user42298user42298
18116
$endgroup$
add a comment |
$begingroup$
My question is related to a formula in this paper
In that paper, they try to expand Dirac delta function $delta(x)$, which has the property
$$
int delta(x)f(x) , dx = f(0),
$$
using Hermite polynomial. So they write
$$
delta(x) = sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2}
$$
and get the coefficient $A_n$ by
$$
begin{align}
int H_{2m}(x) delta(x) , dx &= int H_{2m}(x) sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2} \
Rightarrow H_{2m}(0) &= A_m sqrt {pi}4^m (2m)! \
Rightarrow A_m &= frac{(-1)^m}{m! 4^m sqrt{pi}} ~~~~~~~~(H_{2n}(0)=frac{(2n)!(-1)^n}{n!})
end{align}
$$
Usual $delta(x)$ function has property that it equals to zero for $xneq 0$, but $delta(x) rightarrow infty $ for $x=0$
Now following above expansion, if we plug $x=0$ to the formula, we get
$$
begin{align}
delta(0) & = sum_{n=0}^{infty}A_n H_{2n}(0) \
& = sum_{n=0}^{infty} frac{(2n)!}{n!n!4^nsqrt{pi}}
end{align}
$$
But this series converges, so the usual property of $delta(x)$ is not recovered. So my question is, is this expansion for $delta(x)$ valid?
dirac-delta
edited May 31 '18 at 3:48
Michael Hardy
1
asked May 31 '18 at 2:36
user42298user42298
18116
$endgroup$
$begingroup$
The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $ge C/sqrt{n},$ at which point divergence to $+infty$ is immediate.
$endgroup$
– stochasticboy321
May 31 '18 at 3:06
add a comment |
$begingroup$
My question is related to a formula in this paper
In that paper, they try to expand Dirac delta function $delta(x)$, which has the property
$$
int delta(x)f(x) , dx = f(0),
$$
using Hermite polynomial. So they write
$$
delta(x) = sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2}
$$
and get the coefficient $A_n$ by
$$
begin{align}
int H_{2m}(x) delta(x) , dx &= int H_{2m}(x) sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2} \
Rightarrow H_{2m}(0) &= A_m sqrt {pi}4^m (2m)! \
Rightarrow A_m &= frac{(-1)^m}{m! 4^m sqrt{pi}} ~~~~~~~~(H_{2n}(0)=frac{(2n)!(-1)^n}{n!})
end{align}
$$
Usual $delta(x)$ function has property that it equals to zero for $xneq 0$, but $delta(x) rightarrow infty $ for $x=0$
Now following above expansion, if we plug $x=0$ to the formula, we get
$$
begin{align}
delta(0) & = sum_{n=0}^{infty}A_n H_{2n}(0) \
& = sum_{n=0}^{infty} frac{(2n)!}{n!n!4^nsqrt{pi}}
end{align}
$$
But this series converges, so the usual property of $delta(x)$ is not recovered. So my question is, is this expansion for $delta(x)$ valid?
dirac-delta
edited May 31 '18 at 3:48
Michael Hardy
1
asked May 31 '18 at 2:36
user42298user42298
18116
$endgroup$
My question is related to a formula in this paper
In that paper, they try to expand Dirac delta function $delta(x)$, which has the property
$$
int delta(x)f(x) , dx = f(0),
$$
using Hermite polynomial. So they write
$$
delta(x) = sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2}
$$
and get the coefficient $A_n$ by
$$
begin{align}
int H_{2m}(x) delta(x) , dx &= int H_{2m}(x) sum_{n=0}^{infty}A_n H_{2n}(x)e^{-x^2} \
Rightarrow H_{2m}(0) &= A_m sqrt {pi}4^m (2m)! \
Rightarrow A_m &= frac{(-1)^m}{m! 4^m sqrt{pi}} ~~~~~~~~(H_{2n}(0)=frac{(2n)!(-1)^n}{n!})
end{align}
$$
Usual $delta(x)$ function has property that it equals to zero for $xneq 0$, but $delta(x) rightarrow infty $ for $x=0$
Now following above expansion, if we plug $x=0$ to the formula, we get
$$
begin{align}
delta(0) & = sum_{n=0}^{infty}A_n H_{2n}(0) \
& = sum_{n=0}^{infty} frac{(2n)!}{n!n!4^nsqrt{pi}}
end{align}
$$
But this series converges, so the usual property of $delta(x)$ is not recovered. So my question is, is this expansion for $delta(x)$ valid?
dirac-delta
dirac-delta
edited May 31 '18 at 3:48
Michael Hardy
1
asked May 31 '18 at 2:36
user42298user42298
18116
edited May 31 '18 at 3:48
Michael Hardy
1
asked May 31 '18 at 2:36
user42298user42298
18116
edited May 31 '18 at 3:48
Michael Hardy
1
edited May 31 '18 at 3:48
Michael Hardy
1
edited May 31 '18 at 3:48
Michael Hardy
1
1
asked May 31 '18 at 2:36
user42298user42298
18116
asked May 31 '18 at 2:36
user42298user42298
18116
asked May 31 '18 at 2:36
user42298user42298
18116
18116
$begingroup$
The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $ge C/sqrt{n},$ at which point divergence to $+infty$ is immediate.
$endgroup$
– stochasticboy321
May 31 '18 at 3:06
add a comment |
$begingroup$
The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $ge C/sqrt{n},$ at which point divergence to $+infty$ is immediate.
$endgroup$
– stochasticboy321
May 31 '18 at 3:06
$begingroup$
The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $ge C/sqrt{n},$ at which point divergence to $+infty$ is immediate.
$endgroup$
– stochasticboy321
May 31 '18 at 3:06
$begingroup$
The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $ge C/sqrt{n},$ at which point divergence to $+infty$ is immediate.
$endgroup$
– stochasticboy321
May 31 '18 at 3:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Plugging the series into Wolfram Alpha gives us that it diverges, see here. So the expansions seems to be valid. Just a minor thing: it is usually a trap to actually evaluate distributions like the Dirac function, as they are not actually functions and only really make sense under an integral sign. Viewing it as a functional with the property you listed at the top is much more clear and avoids function having infinite values and other weird stuff.
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
$endgroup$
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
add a comment |
$begingroup$
The value for $N$-th approximation to the delta function at zero turns out to be
$$
frac{1}{sqrt{pi}} frac{(2N+1)!!}{2^N N!}
$$
This can be verified by induction over $N$. The values are quite small, climbing up in a leasurely fashion. For $N=50$ the value is $approx 4.5$.
answered Jan 22 at 14:46
MSMMSM
111
$endgroup$
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%2f2802591%2fexpanding-dirac-delta-function-with-hermite-polynomial%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
$begingroup$
Plugging the series into Wolfram Alpha gives us that it diverges, see here. So the expansions seems to be valid. Just a minor thing: it is usually a trap to actually evaluate distributions like the Dirac function, as they are not actually functions and only really make sense under an integral sign. Viewing it as a functional with the property you listed at the top is much more clear and avoids function having infinite values and other weird stuff.
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
$endgroup$
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
add a comment |
$begingroup$
Plugging the series into Wolfram Alpha gives us that it diverges, see here. So the expansions seems to be valid. Just a minor thing: it is usually a trap to actually evaluate distributions like the Dirac function, as they are not actually functions and only really make sense under an integral sign. Viewing it as a functional with the property you listed at the top is much more clear and avoids function having infinite values and other weird stuff.
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
$endgroup$
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
add a comment |
$begingroup$
Plugging the series into Wolfram Alpha gives us that it diverges, see here. So the expansions seems to be valid. Just a minor thing: it is usually a trap to actually evaluate distributions like the Dirac function, as they are not actually functions and only really make sense under an integral sign. Viewing it as a functional with the property you listed at the top is much more clear and avoids function having infinite values and other weird stuff.
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
$endgroup$
Plugging the series into Wolfram Alpha gives us that it diverges, see here. So the expansions seems to be valid. Just a minor thing: it is usually a trap to actually evaluate distributions like the Dirac function, as they are not actually functions and only really make sense under an integral sign. Viewing it as a functional with the property you listed at the top is much more clear and avoids function having infinite values and other weird stuff.
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
answered May 31 '18 at 2:55
whpowell96whpowell96
59115
59115
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
add a comment |
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
$begingroup$
I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks!
$endgroup$
– user42298
May 31 '18 at 3:10
add a comment |
$begingroup$
The value for $N$-th approximation to the delta function at zero turns out to be
$$
frac{1}{sqrt{pi}} frac{(2N+1)!!}{2^N N!}
$$
This can be verified by induction over $N$. The values are quite small, climbing up in a leasurely fashion. For $N=50$ the value is $approx 4.5$.
answered Jan 22 at 14:46
MSMMSM
111
$endgroup$
add a comment |
$begingroup$
The value for $N$-th approximation to the delta function at zero turns out to be
$$
frac{1}{sqrt{pi}} frac{(2N+1)!!}{2^N N!}
$$
This can be verified by induction over $N$. The values are quite small, climbing up in a leasurely fashion. For $N=50$ the value is $approx 4.5$.
answered Jan 22 at 14:46
MSMMSM
111
$endgroup$
add a comment |
$begingroup$
The value for $N$-th approximation to the delta function at zero turns out to be
$$
frac{1}{sqrt{pi}} frac{(2N+1)!!}{2^N N!}
$$
This can be verified by induction over $N$. The values are quite small, climbing up in a leasurely fashion. For $N=50$ the value is $approx 4.5$.
answered Jan 22 at 14:46
MSMMSM
111
$endgroup$
The value for $N$-th approximation to the delta function at zero turns out to be
$$
frac{1}{sqrt{pi}} frac{(2N+1)!!}{2^N N!}
$$
This can be verified by induction over $N$. The values are quite small, climbing up in a leasurely fashion. For $N=50$ the value is $approx 4.5$.
answered Jan 22 at 14:46
MSMMSM
111
answered Jan 22 at 14:46
MSMMSM
111
answered Jan 22 at 14:46
MSMMSM
111
answered Jan 22 at 14:46
MSMMSM
111
111
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%2f2802591%2fexpanding-dirac-delta-function-with-hermite-polynomial%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
$begingroup$
The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $ge C/sqrt{n},$ at which point divergence to $+infty$ is immediate.
$endgroup$
– stochasticboy321
May 31 '18 at 3:06