Taylor Series of Ratio of Bessel Functions

Question

In attempting to solve a recursion relation I have used a generating function method. This resulted in a differential equation to which I have the solution, and now I need to calculate the Taylor series around $z=0$. The solution involves Bessel functions of $1/z$, and hence I am not sure how to calculate the complete Taylor Series (or if it is even possible). The function is

$$f(z) = \frac{z}{2}+\frac{I_{-\frac23}(\frac2{3z}) + I_{\frac43}(\frac2{3z}) }{2 I_{\frac13}(\frac2{3z})}$$

where $I_{\alpha}(x)$ is the modified Bessel function of the first kind. I have attempted to calculate the Taylor coefficients numerically and it appears that they do exist (and are roughly what they recursion relation gives).

Is it possible to calculate the complete Taylor series of this function, and if so how would you go about it?

Edit: My apologies, I've made a typo in the function! There should be a 2 in the bessel function denominator. (Which there now is)

Answer 1

This isn't a complete solution, but it might help: You can rewrite this using the recurrence relations

$$I_\nu(z)=\frac z{2\nu}\left(I_{\nu-1}(z)-I_{\nu+1}(z)\right)$$

and

$$I'_\nu(z)=\frac12\left(I_{\nu-1}(z)+I_{\nu+1}(z)\right)\;.$$

For instance, this yields

$$ \begin{eqnarray} f(z) &=& \frac z2+\frac{I'_{\frac13}\left(\frac2{3z}\right)}{I_{\frac13}\left(\frac2{3z}\right)} \\ &=& \frac z2+\left(\log I_{\frac13}\right)'\left(\frac2{3z}\right) \end{eqnarray}$$

and

$$ \begin{eqnarray} f(z) &=& \frac z2+\frac{2I_{-\frac23}\left(\frac2{3z}\right)-zI_{\frac13}\left(\frac2{3z}\right)}{2I_{\frac13}\left(\frac2{3z}\right)} \\ &=& \frac{I_{-\frac23}\left(\frac2{3z}\right)}{I_{\frac13}\left(\frac2{3z}\right)}\;. \end{eqnarray} $$

Answer 2

Maple 15 gives

f:= z/2 + (BesselI(-2/3,2/(3*z)) + BesselI(4/3,2/(3*z)))/(2*BesselI(1/3,2/(3*z))); series(f,z,10);

$$(1-{\frac {1}{4}}z-{\frac {5}{32}}{z}^{2}-{\frac {15}{64}}{z}^{3}-{ \frac {1105}{2048}}{z}^{4}-{\frac {1695}{1024}}{z}^{5}-{\frac {414125} {65536}}{z}^{6}-{\frac {59025}{2048}}{z}^{7}-{\frac {1282031525}{ 8388608}}{z}^{8}-{\frac {242183775}{262144}}{z}^{9}+O \left( {z}^{10} \right) ) $$

I don't know if there's a closed form for the series. But the continued fraction looks interesting:

numtheory:-cfrac(f);

$$1-z/(4-5z/(2-7z/(4-11z/(2-13z/(4-17z/(2-19z/(4-23z/(2-25z/(4-29z/(2+\ldots))))))))))$$

Answer 3

If
$$ f(z) = 1-{\frac {1}{4}}z-{\frac {5}{32}}{z}^{2}-{\frac {15}{64}}{z}^{3}-{ \frac {1105}{2048}}{z}^{4}-{\frac {1695}{1024}}{z}^{5}-{\frac {414125} {65536}}{z}^{6}-{\frac {59025}{2048}}{z}^{7}-\dots $$
as Robert says, then $$ \frac{1}{2} - \frac{1}{2}\,f\left(8y\right) =y + 5 y^{2} + 60 y^{3} + 1105 y^{4} + 27120 y^{5} + 828250 y^{6} + 30220800 y^{7} + \dots $$
and these coefficients seem to match A062980 ...