Complicated differential operator and application of the Zassenhaus formula

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I have a differential operator

$$D_x = frac1x left[ (x^2 - a^2) fracddx left( frac1x fracddx right) + 2 fracd dx right], $$

and I would like to compute

$$ e^i lambda D_x ~j_0(x) = sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x), $$

where $~j_0(x) = sin(x)/x~$ is the spherical Bessel function.
Is there some nice way I could do this?

I tried using the Zassenhaus formula, using the splitting

$$ D_x = A_x + B_x, ~~rm where~~
A_x=frac(x^2 - a^2)xfracddx left( frac1x fracddx right), ~~B_x = frac2x fracd dx,$$
but it does not seem like a promising path, given that higher commutators $[A_x,B_x],~[A_x,[A_x,B_x]],ldots$ get more and more complicated.

I also tried just brute force derivation that would give me a
series representation as
$$sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x)
= sum_n=0^infty frac(ilambda)^nn! left[ f_n(x) sin(x) + g_n(x) cos(x) right],$$
but I couldn't find close formulas for $f_n(x)$ and $g_n(x)$ functions.
What else could I do?

derivatives noncommutative-algebra

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asked Jul 24 at 14:49

z.v.

177111

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up vote
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I have a differential operator

$$D_x = frac1x left[ (x^2 - a^2) fracddx left( frac1x fracddx right) + 2 fracd dx right], $$

and I would like to compute

$$ e^i lambda D_x ~j_0(x) = sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x), $$

where $~j_0(x) = sin(x)/x~$ is the spherical Bessel function.
Is there some nice way I could do this?

I tried using the Zassenhaus formula, using the splitting

$$ D_x = A_x + B_x, ~~rm where~~
A_x=frac(x^2 - a^2)xfracddx left( frac1x fracddx right), ~~B_x = frac2x fracd dx,$$
but it does not seem like a promising path, given that higher commutators $[A_x,B_x],~[A_x,[A_x,B_x]],ldots$ get more and more complicated.

I also tried just brute force derivation that would give me a
series representation as
$$sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x)
= sum_n=0^infty frac(ilambda)^nn! left[ f_n(x) sin(x) + g_n(x) cos(x) right],$$
but I couldn't find close formulas for $f_n(x)$ and $g_n(x)$ functions.
What else could I do?

derivatives noncommutative-algebra

share|cite|improve this question

asked Jul 24 at 14:49

z.v.

177111

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I have a differential operator

$$D_x = frac1x left[ (x^2 - a^2) fracddx left( frac1x fracddx right) + 2 fracd dx right], $$

and I would like to compute

$$ e^i lambda D_x ~j_0(x) = sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x), $$

where $~j_0(x) = sin(x)/x~$ is the spherical Bessel function.
Is there some nice way I could do this?

I tried using the Zassenhaus formula, using the splitting

$$ D_x = A_x + B_x, ~~rm where~~
A_x=frac(x^2 - a^2)xfracddx left( frac1x fracddx right), ~~B_x = frac2x fracd dx,$$
but it does not seem like a promising path, given that higher commutators $[A_x,B_x],~[A_x,[A_x,B_x]],ldots$ get more and more complicated.

I also tried just brute force derivation that would give me a
series representation as
$$sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x)
= sum_n=0^infty frac(ilambda)^nn! left[ f_n(x) sin(x) + g_n(x) cos(x) right],$$
but I couldn't find close formulas for $f_n(x)$ and $g_n(x)$ functions.
What else could I do?

derivatives noncommutative-algebra

share|cite|improve this question

asked Jul 24 at 14:49

z.v.

177111

I have a differential operator

$$D_x = frac1x left[ (x^2 - a^2) fracddx left( frac1x fracddx right) + 2 fracd dx right], $$

and I would like to compute

$$ e^i lambda D_x ~j_0(x) = sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x), $$

where $~j_0(x) = sin(x)/x~$ is the spherical Bessel function.
Is there some nice way I could do this?

I tried using the Zassenhaus formula, using the splitting

$$ D_x = A_x + B_x, ~~rm where~~
A_x=frac(x^2 - a^2)xfracddx left( frac1x fracddx right), ~~B_x = frac2x fracd dx,$$
but it does not seem like a promising path, given that higher commutators $[A_x,B_x],~[A_x,[A_x,B_x]],ldots$ get more and more complicated.

I also tried just brute force derivation that would give me a
series representation as
$$sum_n=0^infty frac(ilambda)^nn! [D_x]^n ~j_0(x)
= sum_n=0^infty frac(ilambda)^nn! left[ f_n(x) sin(x) + g_n(x) cos(x) right],$$
but I couldn't find close formulas for $f_n(x)$ and $g_n(x)$ functions.
What else could I do?

derivatives noncommutative-algebra

share|cite|improve this question

asked Jul 24 at 14:49

z.v.

177111

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 14:49

z.v.

177111

asked Jul 24 at 14:49

z.v.

177111

asked Jul 24 at 14:49

z.v.

177111

177111

add a comment | 

add a comment | 

2 Answers
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The answer can be represented as an integral which appears at the bottom of this answer.
The first step is to transform the operator $D_x$ to one $D_w$ via the transformations $x=sqrtu$ and then $w=u-a^2.$ We find that
$$D_x=(x^2-a^2)frac1xfracddxfrac1xfracddx + frac2x fracddx to 4D_w,, D_w=w,fracd^2dw^2+fracddw .$$
The $D_w$ operator is related to the Laguerre polynomials which allows us to make significant progress. The OP's problem then becomes
$$exp(ilambda D_x), j_0(x) to exp(c D_w), j_0(sqrtw+a^2) ,,,, (c=4ilambda). $$
Expand the sinc function, using the abbreviation $y=a^2$,
$$ j_0(sqrtw+y) = sum_k=0^infty (-1)^kfrac(w+y)^k(2k+1)!=
sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m,w^m .$$
We now need to see how $exp(c,D_w)$ behaves on monomials $w^m.$ Expand the exponential and note how powers of $D_w$ behave:

$$ big(D_wbig)^0 w^m = w^m $$
$$ big(D_wbig)^1 w^m = big(w,fracd^2dw^2+fracddwbig)w^m=m^2, w^m-1 $$
$$ big(D_wbig)^2 w^m = big(w,fracd^2dw^2+fracddwbig), m^2w^m-1=(m,(m-1))^2, w^m-2 $$
$$ big(D_wbig)^j w^m = Big( j!binommj Big)^2 , w^m-j $$
Thus it is easy to see that
$$ exp(c,D_w) w^j = sum_j=0^m c^j j! binommj^2 w^m-j =
m! sum_j=0^m c^m-j binommj fracw^jj! $$
where in the last step a series rearrangement has been performed. Use the definition of the Laguerre polynomials and we have
$$ exp(c,D_w) w^j = m!c^m , L_m(-w/c) .$$ Thus we have a representation of a solution,
$$ exp(c,D_w) j_0(sqrtw+y) = sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m c^m , m!,L_m(z) , , , (z=-w/c).$$
This representation can be transformed to a integral. Use the integral that can be found on Wolfram's website
$$ m!L_m(z) = e^z int_0^infty e^-t t^m J_0(2sqrtt,z) dt. $$
Pull the integral through the finite sum, then the infinite sum (justifiable by analytic continuation in a region that the integral at the end makes sense).
$$exp(c,D_w) j_0(sqrtw+y) =e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m (ct)^m$$
$$=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! big(y + ctbig)^k
=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) fracsin(sqrty+ct)sqrty+ct $$
In terms of the original variables this becomes
$$exp(ilambda D_x), j_0(x) = e^big((x^2-a^2)/(-4ilambda)big)
int_0^infty dt,e^-t J_0Big(2sqrtt,frac(x^2-a^2)-4ilambdaBig) fracsin(sqrta^2+4ilambda t)sqrta^2+4ilambda t.$$
Not knowing the domains of the parameters, I can't say if the integral is finite; it is probable that it is considering the nice functions in the integrand.

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answered Jul 27 at 19:15

skbmoore

99026

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up vote
1
down vote

Use the integral form of the $j_0(x)$ that is given by
$$
j_0(x)=int_-infty^inftyPi(k)e^-2ikxdk
$$
where
$$
Pi(t) = begincases
0 & mboxif |t| > frac12 \
frac12 & mboxif |t| = frac12 \
1 & mboxif |t| < frac12. \
endcases
$$
Then, your problem reduces to the evaluation of the derivative of an exponential.

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answered Jul 24 at 15:36

Jon

4,20011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
  – z.v.
  Jul 26 at 6:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Yes, it is a possible avenue.
  – Jon
  Jul 26 at 6:40
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The answer can be represented as an integral which appears at the bottom of this answer.
The first step is to transform the operator $D_x$ to one $D_w$ via the transformations $x=sqrtu$ and then $w=u-a^2.$ We find that
$$D_x=(x^2-a^2)frac1xfracddxfrac1xfracddx + frac2x fracddx to 4D_w,, D_w=w,fracd^2dw^2+fracddw .$$
The $D_w$ operator is related to the Laguerre polynomials which allows us to make significant progress. The OP's problem then becomes
$$exp(ilambda D_x), j_0(x) to exp(c D_w), j_0(sqrtw+a^2) ,,,, (c=4ilambda). $$
Expand the sinc function, using the abbreviation $y=a^2$,
$$ j_0(sqrtw+y) = sum_k=0^infty (-1)^kfrac(w+y)^k(2k+1)!=
sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m,w^m .$$
We now need to see how $exp(c,D_w)$ behaves on monomials $w^m.$ Expand the exponential and note how powers of $D_w$ behave:

$$ big(D_wbig)^0 w^m = w^m $$
$$ big(D_wbig)^1 w^m = big(w,fracd^2dw^2+fracddwbig)w^m=m^2, w^m-1 $$
$$ big(D_wbig)^2 w^m = big(w,fracd^2dw^2+fracddwbig), m^2w^m-1=(m,(m-1))^2, w^m-2 $$
$$ big(D_wbig)^j w^m = Big( j!binommj Big)^2 , w^m-j $$
Thus it is easy to see that
$$ exp(c,D_w) w^j = sum_j=0^m c^j j! binommj^2 w^m-j =
m! sum_j=0^m c^m-j binommj fracw^jj! $$
where in the last step a series rearrangement has been performed. Use the definition of the Laguerre polynomials and we have
$$ exp(c,D_w) w^j = m!c^m , L_m(-w/c) .$$ Thus we have a representation of a solution,
$$ exp(c,D_w) j_0(sqrtw+y) = sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m c^m , m!,L_m(z) , , , (z=-w/c).$$
This representation can be transformed to a integral. Use the integral that can be found on Wolfram's website
$$ m!L_m(z) = e^z int_0^infty e^-t t^m J_0(2sqrtt,z) dt. $$
Pull the integral through the finite sum, then the infinite sum (justifiable by analytic continuation in a region that the integral at the end makes sense).
$$exp(c,D_w) j_0(sqrtw+y) =e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m (ct)^m$$
$$=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! big(y + ctbig)^k
=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) fracsin(sqrty+ct)sqrty+ct $$
In terms of the original variables this becomes
$$exp(ilambda D_x), j_0(x) = e^big((x^2-a^2)/(-4ilambda)big)
int_0^infty dt,e^-t J_0Big(2sqrtt,frac(x^2-a^2)-4ilambdaBig) fracsin(sqrta^2+4ilambda t)sqrta^2+4ilambda t.$$
Not knowing the domains of the parameters, I can't say if the integral is finite; it is probable that it is considering the nice functions in the integrand.

share|cite|improve this answer

answered Jul 27 at 19:15

skbmoore

99026

add a comment | 

up vote
1
down vote



accepted

The answer can be represented as an integral which appears at the bottom of this answer.
The first step is to transform the operator $D_x$ to one $D_w$ via the transformations $x=sqrtu$ and then $w=u-a^2.$ We find that
$$D_x=(x^2-a^2)frac1xfracddxfrac1xfracddx + frac2x fracddx to 4D_w,, D_w=w,fracd^2dw^2+fracddw .$$
The $D_w$ operator is related to the Laguerre polynomials which allows us to make significant progress. The OP's problem then becomes
$$exp(ilambda D_x), j_0(x) to exp(c D_w), j_0(sqrtw+a^2) ,,,, (c=4ilambda). $$
Expand the sinc function, using the abbreviation $y=a^2$,
$$ j_0(sqrtw+y) = sum_k=0^infty (-1)^kfrac(w+y)^k(2k+1)!=
sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m,w^m .$$
We now need to see how $exp(c,D_w)$ behaves on monomials $w^m.$ Expand the exponential and note how powers of $D_w$ behave:

$$ big(D_wbig)^0 w^m = w^m $$
$$ big(D_wbig)^1 w^m = big(w,fracd^2dw^2+fracddwbig)w^m=m^2, w^m-1 $$
$$ big(D_wbig)^2 w^m = big(w,fracd^2dw^2+fracddwbig), m^2w^m-1=(m,(m-1))^2, w^m-2 $$
$$ big(D_wbig)^j w^m = Big( j!binommj Big)^2 , w^m-j $$
Thus it is easy to see that
$$ exp(c,D_w) w^j = sum_j=0^m c^j j! binommj^2 w^m-j =
m! sum_j=0^m c^m-j binommj fracw^jj! $$
where in the last step a series rearrangement has been performed. Use the definition of the Laguerre polynomials and we have
$$ exp(c,D_w) w^j = m!c^m , L_m(-w/c) .$$ Thus we have a representation of a solution,
$$ exp(c,D_w) j_0(sqrtw+y) = sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m c^m , m!,L_m(z) , , , (z=-w/c).$$
This representation can be transformed to a integral. Use the integral that can be found on Wolfram's website
$$ m!L_m(z) = e^z int_0^infty e^-t t^m J_0(2sqrtt,z) dt. $$
Pull the integral through the finite sum, then the infinite sum (justifiable by analytic continuation in a region that the integral at the end makes sense).
$$exp(c,D_w) j_0(sqrtw+y) =e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m (ct)^m$$
$$=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! big(y + ctbig)^k
=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) fracsin(sqrty+ct)sqrty+ct $$
In terms of the original variables this becomes
$$exp(ilambda D_x), j_0(x) = e^big((x^2-a^2)/(-4ilambda)big)
int_0^infty dt,e^-t J_0Big(2sqrtt,frac(x^2-a^2)-4ilambdaBig) fracsin(sqrta^2+4ilambda t)sqrta^2+4ilambda t.$$
Not knowing the domains of the parameters, I can't say if the integral is finite; it is probable that it is considering the nice functions in the integrand.

share|cite|improve this answer

answered Jul 27 at 19:15

skbmoore

99026

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The answer can be represented as an integral which appears at the bottom of this answer.
The first step is to transform the operator $D_x$ to one $D_w$ via the transformations $x=sqrtu$ and then $w=u-a^2.$ We find that
$$D_x=(x^2-a^2)frac1xfracddxfrac1xfracddx + frac2x fracddx to 4D_w,, D_w=w,fracd^2dw^2+fracddw .$$
The $D_w$ operator is related to the Laguerre polynomials which allows us to make significant progress. The OP's problem then becomes
$$exp(ilambda D_x), j_0(x) to exp(c D_w), j_0(sqrtw+a^2) ,,,, (c=4ilambda). $$
Expand the sinc function, using the abbreviation $y=a^2$,
$$ j_0(sqrtw+y) = sum_k=0^infty (-1)^kfrac(w+y)^k(2k+1)!=
sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m,w^m .$$
We now need to see how $exp(c,D_w)$ behaves on monomials $w^m.$ Expand the exponential and note how powers of $D_w$ behave:

$$ big(D_wbig)^0 w^m = w^m $$
$$ big(D_wbig)^1 w^m = big(w,fracd^2dw^2+fracddwbig)w^m=m^2, w^m-1 $$
$$ big(D_wbig)^2 w^m = big(w,fracd^2dw^2+fracddwbig), m^2w^m-1=(m,(m-1))^2, w^m-2 $$
$$ big(D_wbig)^j w^m = Big( j!binommj Big)^2 , w^m-j $$
Thus it is easy to see that
$$ exp(c,D_w) w^j = sum_j=0^m c^j j! binommj^2 w^m-j =
m! sum_j=0^m c^m-j binommj fracw^jj! $$
where in the last step a series rearrangement has been performed. Use the definition of the Laguerre polynomials and we have
$$ exp(c,D_w) w^j = m!c^m , L_m(-w/c) .$$ Thus we have a representation of a solution,
$$ exp(c,D_w) j_0(sqrtw+y) = sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m c^m , m!,L_m(z) , , , (z=-w/c).$$
This representation can be transformed to a integral. Use the integral that can be found on Wolfram's website
$$ m!L_m(z) = e^z int_0^infty e^-t t^m J_0(2sqrtt,z) dt. $$
Pull the integral through the finite sum, then the infinite sum (justifiable by analytic continuation in a region that the integral at the end makes sense).
$$exp(c,D_w) j_0(sqrtw+y) =e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m (ct)^m$$
$$=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! big(y + ctbig)^k
=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) fracsin(sqrty+ct)sqrty+ct $$
In terms of the original variables this becomes
$$exp(ilambda D_x), j_0(x) = e^big((x^2-a^2)/(-4ilambda)big)
int_0^infty dt,e^-t J_0Big(2sqrtt,frac(x^2-a^2)-4ilambdaBig) fracsin(sqrta^2+4ilambda t)sqrta^2+4ilambda t.$$
Not knowing the domains of the parameters, I can't say if the integral is finite; it is probable that it is considering the nice functions in the integrand.

share|cite|improve this answer

answered Jul 27 at 19:15

skbmoore

99026

The answer can be represented as an integral which appears at the bottom of this answer.
The first step is to transform the operator $D_x$ to one $D_w$ via the transformations $x=sqrtu$ and then $w=u-a^2.$ We find that
$$D_x=(x^2-a^2)frac1xfracddxfrac1xfracddx + frac2x fracddx to 4D_w,, D_w=w,fracd^2dw^2+fracddw .$$
The $D_w$ operator is related to the Laguerre polynomials which allows us to make significant progress. The OP's problem then becomes
$$exp(ilambda D_x), j_0(x) to exp(c D_w), j_0(sqrtw+a^2) ,,,, (c=4ilambda). $$
Expand the sinc function, using the abbreviation $y=a^2$,
$$ j_0(sqrtw+y) = sum_k=0^infty (-1)^kfrac(w+y)^k(2k+1)!=
sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m,w^m .$$
We now need to see how $exp(c,D_w)$ behaves on monomials $w^m.$ Expand the exponential and note how powers of $D_w$ behave:

$$ big(D_wbig)^0 w^m = w^m $$
$$ big(D_wbig)^1 w^m = big(w,fracd^2dw^2+fracddwbig)w^m=m^2, w^m-1 $$
$$ big(D_wbig)^2 w^m = big(w,fracd^2dw^2+fracddwbig), m^2w^m-1=(m,(m-1))^2, w^m-2 $$
$$ big(D_wbig)^j w^m = Big( j!binommj Big)^2 , w^m-j $$
Thus it is easy to see that
$$ exp(c,D_w) w^j = sum_j=0^m c^j j! binommj^2 w^m-j =
m! sum_j=0^m c^m-j binommj fracw^jj! $$
where in the last step a series rearrangement has been performed. Use the definition of the Laguerre polynomials and we have
$$ exp(c,D_w) w^j = m!c^m , L_m(-w/c) .$$ Thus we have a representation of a solution,
$$ exp(c,D_w) j_0(sqrtw+y) = sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m c^m , m!,L_m(z) , , , (z=-w/c).$$
This representation can be transformed to a integral. Use the integral that can be found on Wolfram's website
$$ m!L_m(z) = e^z int_0^infty e^-t t^m J_0(2sqrtt,z) dt. $$
Pull the integral through the finite sum, then the infinite sum (justifiable by analytic continuation in a region that the integral at the end makes sense).
$$exp(c,D_w) j_0(sqrtw+y) =e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! sum_m=0^k binomkm y^k-m (ct)^m$$
$$=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) sum_k=0^infty frac(-1)^k(2k+1)! big(y + ctbig)^k
=e^z int_0^infty dt,e^-t J_0(2sqrtt,z) fracsin(sqrty+ct)sqrty+ct $$
In terms of the original variables this becomes
$$exp(ilambda D_x), j_0(x) = e^big((x^2-a^2)/(-4ilambda)big)
int_0^infty dt,e^-t J_0Big(2sqrtt,frac(x^2-a^2)-4ilambdaBig) fracsin(sqrta^2+4ilambda t)sqrta^2+4ilambda t.$$
Not knowing the domains of the parameters, I can't say if the integral is finite; it is probable that it is considering the nice functions in the integrand.

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answered Jul 27 at 19:15

skbmoore

99026

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share|cite|improve this answer

answered Jul 27 at 19:15

skbmoore

99026

answered Jul 27 at 19:15

skbmoore

99026

answered Jul 27 at 19:15

skbmoore

99026

99026

add a comment | 

add a comment | 

up vote
1
down vote

Use the integral form of the $j_0(x)$ that is given by
$$
j_0(x)=int_-infty^inftyPi(k)e^-2ikxdk
$$
where
$$
Pi(t) = begincases
0 & mboxif |t| > frac12 \
frac12 & mboxif |t| = frac12 \
1 & mboxif |t| < frac12. \
endcases
$$
Then, your problem reduces to the evaluation of the derivative of an exponential.

share|cite|improve this answer

answered Jul 24 at 15:36

Jon

4,20011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
  – z.v.
  Jul 26 at 6:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Yes, it is a possible avenue.
  – Jon
  Jul 26 at 6:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

Use the integral form of the $j_0(x)$ that is given by
$$
j_0(x)=int_-infty^inftyPi(k)e^-2ikxdk
$$
where
$$
Pi(t) = begincases
0 & mboxif |t| > frac12 \
frac12 & mboxif |t| = frac12 \
1 & mboxif |t| < frac12. \
endcases
$$
Then, your problem reduces to the evaluation of the derivative of an exponential.

share|cite|improve this answer

answered Jul 24 at 15:36

Jon

4,20011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
  – z.v.
  Jul 26 at 6:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Yes, it is a possible avenue.
  – Jon
  Jul 26 at 6:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Use the integral form of the $j_0(x)$ that is given by
$$
j_0(x)=int_-infty^inftyPi(k)e^-2ikxdk
$$
where
$$
Pi(t) = begincases
0 & mboxif |t| > frac12 \
frac12 & mboxif |t| = frac12 \
1 & mboxif |t| < frac12. \
endcases
$$
Then, your problem reduces to the evaluation of the derivative of an exponential.

share|cite|improve this answer

answered Jul 24 at 15:36

Jon

4,20011022

Use the integral form of the $j_0(x)$ that is given by
$$
j_0(x)=int_-infty^inftyPi(k)e^-2ikxdk
$$
where
$$
Pi(t) = begincases
0 & mboxif |t| > frac12 \
frac12 & mboxif |t| = frac12 \
1 & mboxif |t| < frac12. \
endcases
$$
Then, your problem reduces to the evaluation of the derivative of an exponential.

share|cite|improve this answer

answered Jul 24 at 15:36

Jon

4,20011022

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 15:36

Jon

4,20011022

answered Jul 24 at 15:36

Jon

4,20011022

answered Jul 24 at 15:36

Jon

4,20011022

4,20011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
  – z.v.
  Jul 26 at 6:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Yes, it is a possible avenue.
  – Jon
  Jul 26 at 6:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
  – z.v.
  Jul 26 at 6:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. Yes, it is a possible avenue.
  – Jon
  Jul 26 at 6:40
  

  
  
  
  

Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
– z.v.
Jul 26 at 6:27

Thanks Jon, this approach indeed simplifies things relative to what I had, although computing $[D_x]^n e^i2kx$ is still nontrivial, given that $exp$ it is not the eigenfunction of $D_x$. I guess I should try to find the eigenfunctions of $D_x$ and than try to represent $j_0$ in this basis.
– z.v.
Jul 26 at 6:27

You are welcome. Yes, it is a possible avenue.
– Jon
Jul 26 at 6:40

You are welcome. Yes, it is a possible avenue.
– Jon
Jul 26 at 6:40

add a comment | 

 

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