Higher order derivatives of Composition of Dirac delta distributions
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
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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
add a comment |
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
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 2 days ago
Qmechanic
4,87711854
4,87711854
asked Jan 3 at 7:08
MathwuMathwu
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113
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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.
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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.
add a comment |
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.
add a comment |
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.
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 12 hours ago
answered 2 days ago
QmechanicQmechanic
4,87711854
4,87711854
add a comment |
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