- #1

Pere Callahan

- 586

- 1

Hi all,

I am trying to find an expression for the values of the derivates of the Bessel-[tex]J_1[/tex] functions at two.

The function is defined by

[tex]J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\left(\frac{x}{2}\right)^{2k+1}}[/tex]

this I can differentiate term by term, finding for the n^th derivative at two:[tex]J_1^{(n)}(2)=\sum_{k=\left\lfloor\frac{n}{2}\right\rfloor}^\infty{\frac{(-1)^k(2k+1)!}{(k+1)!k!(2k+1-n)!2^n}}[/tex]

Does anybody know a nice expression for this? It can probably be written as the sum of two Besselfunctions evaluated at two, but I have no idea how to find the coefficients...

Thanks

-PereEdit:

I found that for n even the result can be written as a linear combination of [tex]J_1(2)\text{ and } J_2(2)[/tex]. When multiplied by 2^n the coefficients are all integral ... (Experimental results).

I tried the sequence of these integers in that famous integers sequence database, but nothing ...

I am trying to find an expression for the values of the derivates of the Bessel-[tex]J_1[/tex] functions at two.

The function is defined by

[tex]J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\left(\frac{x}{2}\right)^{2k+1}}[/tex]

this I can differentiate term by term, finding for the n^th derivative at two:[tex]J_1^{(n)}(2)=\sum_{k=\left\lfloor\frac{n}{2}\right\rfloor}^\infty{\frac{(-1)^k(2k+1)!}{(k+1)!k!(2k+1-n)!2^n}}[/tex]

Does anybody know a nice expression for this? It can probably be written as the sum of two Besselfunctions evaluated at two, but I have no idea how to find the coefficients...

Thanks

-PereEdit:

I found that for n even the result can be written as a linear combination of [tex]J_1(2)\text{ and } J_2(2)[/tex]. When multiplied by 2^n the coefficients are all integral ... (Experimental results).

I tried the sequence of these integers in that famous integers sequence database, but nothing ...

Last edited: