Higher order derivatives of Composition of Dirac delta distributions












2












$begingroup$


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.










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 16:24









      Qmechanic

      5,17711858




      5,17711858










      asked Jan 3 at 7:08









      MathwuMathwu

      113




      113






















          1 Answer
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          $begingroup$

          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.






          share|cite|improve this answer











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            1 Answer
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            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$



                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 18:36

























                answered Jan 5 at 16:23









                QmechanicQmechanic

                5,17711858




                5,17711858






























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