Higher order derivatives of Composition of Dirac delta distributions
$begingroup$
There are two equations showed in Gel'fand and Shilov's book (Generalized Functions I Properties and Operations) on page 183 and 185:
$${delta}^{(k-1)}(1-x^2)=frac{(-1)^{k-1}}{2^kx^{k-1}}[{delta}^{(k-1)}(x-1)-{delta}^{(k-1)}(x+1)];tag{1}$$
$${delta}^{(k)}(f(x))=sum_n frac{1}{|f'(x_n)|}(frac{1}{f'(x)}frac{d}{dx})^k{delta}(x-x_n),tag{2}$$
where $x_n$ are $n$ simple roots of $f(x).$
I think the first equation should satisfy the second equation. But it seems not. Why? And I would like to know the right equation.
analysis distribution-theory dirac-delta
$endgroup$
add a comment |
$begingroup$
There are two equations showed in Gel'fand and Shilov's book (Generalized Functions I Properties and Operations) on page 183 and 185:
$${delta}^{(k-1)}(1-x^2)=frac{(-1)^{k-1}}{2^kx^{k-1}}[{delta}^{(k-1)}(x-1)-{delta}^{(k-1)}(x+1)];tag{1}$$
$${delta}^{(k)}(f(x))=sum_n frac{1}{|f'(x_n)|}(frac{1}{f'(x)}frac{d}{dx})^k{delta}(x-x_n),tag{2}$$
where $x_n$ are $n$ simple roots of $f(x).$
I think the first equation should satisfy the second equation. But it seems not. Why? And I would like to know the right equation.
analysis distribution-theory dirac-delta
$endgroup$
add a comment |
$begingroup$
There are two equations showed in Gel'fand and Shilov's book (Generalized Functions I Properties and Operations) on page 183 and 185:
$${delta}^{(k-1)}(1-x^2)=frac{(-1)^{k-1}}{2^kx^{k-1}}[{delta}^{(k-1)}(x-1)-{delta}^{(k-1)}(x+1)];tag{1}$$
$${delta}^{(k)}(f(x))=sum_n frac{1}{|f'(x_n)|}(frac{1}{f'(x)}frac{d}{dx})^k{delta}(x-x_n),tag{2}$$
where $x_n$ are $n$ simple roots of $f(x).$
I think the first equation should satisfy the second equation. But it seems not. Why? And I would like to know the right equation.
analysis distribution-theory dirac-delta
$endgroup$
There are two equations showed in Gel'fand and Shilov's book (Generalized Functions I Properties and Operations) on page 183 and 185:
$${delta}^{(k-1)}(1-x^2)=frac{(-1)^{k-1}}{2^kx^{k-1}}[{delta}^{(k-1)}(x-1)-{delta}^{(k-1)}(x+1)];tag{1}$$
$${delta}^{(k)}(f(x))=sum_n frac{1}{|f'(x_n)|}(frac{1}{f'(x)}frac{d}{dx})^k{delta}(x-x_n),tag{2}$$
where $x_n$ are $n$ simple roots of $f(x).$
I think the first equation should satisfy the second equation. But it seems not. Why? And I would like to know the right equation.
analysis distribution-theory dirac-delta
analysis distribution-theory dirac-delta
edited Jan 5 at 16:24
Qmechanic
5,17711858
5,17711858
asked Jan 3 at 7:08
MathwuMathwu
113
113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.
We calculate$^2$
$$ u_k(x)~:=~(-1)^kdelta^{(k)}(1!-!x^2)
~=~delta^{(k)}(x^2!-!1)~=~left.delta^{(k)}(y) right|_{y=x^2-1}$$
$$~=~left.left(frac{d}{dy}right)^kdelta(y)right|_{y=x^2-1}
~=~left(frac{1}{2x}frac{d}{dx}right)^kdelta(x^2!-!1)
~=~left(frac{1}{2x}frac{d}{dx}right)^kfrac{1}{2}sum_{pm}delta(x!mp!1). tag{A}$$
By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d}{dx}frac{1}{2x}+frac{1}{2x^2}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4}sum_{pm}pm delta^{prime}(x!mp!1)+frac{1}{4}sum_{pm}delta(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d^2}{dx^2}frac{1}{4x^2}+frac{d}{dx}frac{3}{4x^3}+frac{3}{4x^4}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8}sum_{pm}delta^{primeprime}(x!mp!1)+frac{3}{8}sum_{pm}pm delta^{prime}(x!mp!1)+frac{3}{8}sum_{pm}delta(x!mp!1),cr
&~~~vdotsend{align}tag{B}$$
By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4x}sum_{pm}delta^{prime}(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{1}{4x^2}frac{d^2}{dx^2}-frac{1}{4x^3}frac{d}{dx}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8x^2}sum_{pm}delta^{primeprime}(x!mp!1)
-frac{1}{8x^3}sum_{pm} delta^{prime}(x!mp!1),cr
&~~~vdotsend{align}tag{C}$$
which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.
--
$^1$ In this answer, we make repeated use of the following distribution identities:
$$delta(f(x))~=~sum_{i}^{f(x_i)=0}frac{1}{|f'(x_i)|}delta(x!-!x_i), tag{D}$$
$$ {f(x)-f(y)}delta(x!-!y)~=~0, tag{E}$$
and derivatives thereof:
$$ left(frac{d}{dx}right)^k[{f(x)-f(y)}delta(x!-!y)]~=~0. tag{F}$$
$^2$ We have for convenience shifted OP's definition $k-1to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.
$endgroup$
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%2f3060318%2fhigher-order-derivatives-of-composition-of-dirac-delta-distributions%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
$begingroup$
TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.
We calculate$^2$
$$ u_k(x)~:=~(-1)^kdelta^{(k)}(1!-!x^2)
~=~delta^{(k)}(x^2!-!1)~=~left.delta^{(k)}(y) right|_{y=x^2-1}$$
$$~=~left.left(frac{d}{dy}right)^kdelta(y)right|_{y=x^2-1}
~=~left(frac{1}{2x}frac{d}{dx}right)^kdelta(x^2!-!1)
~=~left(frac{1}{2x}frac{d}{dx}right)^kfrac{1}{2}sum_{pm}delta(x!mp!1). tag{A}$$
By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d}{dx}frac{1}{2x}+frac{1}{2x^2}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4}sum_{pm}pm delta^{prime}(x!mp!1)+frac{1}{4}sum_{pm}delta(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d^2}{dx^2}frac{1}{4x^2}+frac{d}{dx}frac{3}{4x^3}+frac{3}{4x^4}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8}sum_{pm}delta^{primeprime}(x!mp!1)+frac{3}{8}sum_{pm}pm delta^{prime}(x!mp!1)+frac{3}{8}sum_{pm}delta(x!mp!1),cr
&~~~vdotsend{align}tag{B}$$
By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4x}sum_{pm}delta^{prime}(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{1}{4x^2}frac{d^2}{dx^2}-frac{1}{4x^3}frac{d}{dx}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8x^2}sum_{pm}delta^{primeprime}(x!mp!1)
-frac{1}{8x^3}sum_{pm} delta^{prime}(x!mp!1),cr
&~~~vdotsend{align}tag{C}$$
which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.
--
$^1$ In this answer, we make repeated use of the following distribution identities:
$$delta(f(x))~=~sum_{i}^{f(x_i)=0}frac{1}{|f'(x_i)|}delta(x!-!x_i), tag{D}$$
$$ {f(x)-f(y)}delta(x!-!y)~=~0, tag{E}$$
and derivatives thereof:
$$ left(frac{d}{dx}right)^k[{f(x)-f(y)}delta(x!-!y)]~=~0. tag{F}$$
$^2$ We have for convenience shifted OP's definition $k-1to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.
$endgroup$
add a comment |
$begingroup$
TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.
We calculate$^2$
$$ u_k(x)~:=~(-1)^kdelta^{(k)}(1!-!x^2)
~=~delta^{(k)}(x^2!-!1)~=~left.delta^{(k)}(y) right|_{y=x^2-1}$$
$$~=~left.left(frac{d}{dy}right)^kdelta(y)right|_{y=x^2-1}
~=~left(frac{1}{2x}frac{d}{dx}right)^kdelta(x^2!-!1)
~=~left(frac{1}{2x}frac{d}{dx}right)^kfrac{1}{2}sum_{pm}delta(x!mp!1). tag{A}$$
By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d}{dx}frac{1}{2x}+frac{1}{2x^2}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4}sum_{pm}pm delta^{prime}(x!mp!1)+frac{1}{4}sum_{pm}delta(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d^2}{dx^2}frac{1}{4x^2}+frac{d}{dx}frac{3}{4x^3}+frac{3}{4x^4}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8}sum_{pm}delta^{primeprime}(x!mp!1)+frac{3}{8}sum_{pm}pm delta^{prime}(x!mp!1)+frac{3}{8}sum_{pm}delta(x!mp!1),cr
&~~~vdotsend{align}tag{B}$$
By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4x}sum_{pm}delta^{prime}(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{1}{4x^2}frac{d^2}{dx^2}-frac{1}{4x^3}frac{d}{dx}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8x^2}sum_{pm}delta^{primeprime}(x!mp!1)
-frac{1}{8x^3}sum_{pm} delta^{prime}(x!mp!1),cr
&~~~vdotsend{align}tag{C}$$
which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.
--
$^1$ In this answer, we make repeated use of the following distribution identities:
$$delta(f(x))~=~sum_{i}^{f(x_i)=0}frac{1}{|f'(x_i)|}delta(x!-!x_i), tag{D}$$
$$ {f(x)-f(y)}delta(x!-!y)~=~0, tag{E}$$
and derivatives thereof:
$$ left(frac{d}{dx}right)^k[{f(x)-f(y)}delta(x!-!y)]~=~0. tag{F}$$
$^2$ We have for convenience shifted OP's definition $k-1to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.
$endgroup$
add a comment |
$begingroup$
TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.
We calculate$^2$
$$ u_k(x)~:=~(-1)^kdelta^{(k)}(1!-!x^2)
~=~delta^{(k)}(x^2!-!1)~=~left.delta^{(k)}(y) right|_{y=x^2-1}$$
$$~=~left.left(frac{d}{dy}right)^kdelta(y)right|_{y=x^2-1}
~=~left(frac{1}{2x}frac{d}{dx}right)^kdelta(x^2!-!1)
~=~left(frac{1}{2x}frac{d}{dx}right)^kfrac{1}{2}sum_{pm}delta(x!mp!1). tag{A}$$
By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d}{dx}frac{1}{2x}+frac{1}{2x^2}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4}sum_{pm}pm delta^{prime}(x!mp!1)+frac{1}{4}sum_{pm}delta(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d^2}{dx^2}frac{1}{4x^2}+frac{d}{dx}frac{3}{4x^3}+frac{3}{4x^4}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8}sum_{pm}delta^{primeprime}(x!mp!1)+frac{3}{8}sum_{pm}pm delta^{prime}(x!mp!1)+frac{3}{8}sum_{pm}delta(x!mp!1),cr
&~~~vdotsend{align}tag{B}$$
By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4x}sum_{pm}delta^{prime}(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{1}{4x^2}frac{d^2}{dx^2}-frac{1}{4x^3}frac{d}{dx}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8x^2}sum_{pm}delta^{primeprime}(x!mp!1)
-frac{1}{8x^3}sum_{pm} delta^{prime}(x!mp!1),cr
&~~~vdotsend{align}tag{C}$$
which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.
--
$^1$ In this answer, we make repeated use of the following distribution identities:
$$delta(f(x))~=~sum_{i}^{f(x_i)=0}frac{1}{|f'(x_i)|}delta(x!-!x_i), tag{D}$$
$$ {f(x)-f(y)}delta(x!-!y)~=~0, tag{E}$$
and derivatives thereof:
$$ left(frac{d}{dx}right)^k[{f(x)-f(y)}delta(x!-!y)]~=~0. tag{F}$$
$^2$ We have for convenience shifted OP's definition $k-1to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.
$endgroup$
TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.
We calculate$^2$
$$ u_k(x)~:=~(-1)^kdelta^{(k)}(1!-!x^2)
~=~delta^{(k)}(x^2!-!1)~=~left.delta^{(k)}(y) right|_{y=x^2-1}$$
$$~=~left.left(frac{d}{dy}right)^kdelta(y)right|_{y=x^2-1}
~=~left(frac{1}{2x}frac{d}{dx}right)^kdelta(x^2!-!1)
~=~left(frac{1}{2x}frac{d}{dx}right)^kfrac{1}{2}sum_{pm}delta(x!mp!1). tag{A}$$
By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d}{dx}frac{1}{2x}+frac{1}{2x^2}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4}sum_{pm}pm delta^{prime}(x!mp!1)+frac{1}{4}sum_{pm}delta(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{d^2}{dx^2}frac{1}{4x^2}+frac{d}{dx}frac{3}{4x^3}+frac{3}{4x^4}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8}sum_{pm}delta^{primeprime}(x!mp!1)+frac{3}{8}sum_{pm}pm delta^{prime}(x!mp!1)+frac{3}{8}sum_{pm}delta(x!mp!1),cr
&~~~vdotsend{align}tag{B}$$
By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $kinmathbb{N}_0$:
$$begin{align}u_{k=0}(x)&~=~frac{1}{2}sum_{pm}delta(x!mp!1),cr
u_{k=1}(x)&~=~frac{1}{2x}frac{d}{dx}frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{4x}sum_{pm}delta^{prime}(x!mp!1),cr
u_{k=2}(x)&~=~left(frac{1}{2x}frac{d}{dx}right)^2frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~left(frac{1}{4x^2}frac{d^2}{dx^2}-frac{1}{4x^3}frac{d}{dx}right)frac{1}{2}sum_{pm}delta(x!mp!1)cr
&~=~frac{1}{8x^2}sum_{pm}delta^{primeprime}(x!mp!1)
-frac{1}{8x^3}sum_{pm} delta^{prime}(x!mp!1),cr
&~~~vdotsend{align}tag{C}$$
which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.
--
$^1$ In this answer, we make repeated use of the following distribution identities:
$$delta(f(x))~=~sum_{i}^{f(x_i)=0}frac{1}{|f'(x_i)|}delta(x!-!x_i), tag{D}$$
$$ {f(x)-f(y)}delta(x!-!y)~=~0, tag{E}$$
and derivatives thereof:
$$ left(frac{d}{dx}right)^k[{f(x)-f(y)}delta(x!-!y)]~=~0. tag{F}$$
$^2$ We have for convenience shifted OP's definition $k-1to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.
edited Jan 7 at 18:36
answered Jan 5 at 16:23
QmechanicQmechanic
5,17711858
5,17711858
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%2f3060318%2fhigher-order-derivatives-of-composition-of-dirac-delta-distributions%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