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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)

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You can use double dollar signs to get displayed equations; they're a lot easier to read. – joriki Dec 12 '11 at 10:47
What recursion relation did you start from, and what was the series you said you got? – J. M. Dec 12 '11 at 10:50
As a tiny note: Bessel functions of one-third order are expressible in terms of Airy functions. – J. M. Dec 12 '11 at 10:53
The recursion relation is quite long, and difficult to explain. I do not have a closed form for the series (which is what I'm looking to get), but I can easily calculate using the recursion relation. This is a problem that arises from the WKB method of solving the 1d Schrodinger equation. Here is the paper that I'm working with - . However it will contain a lot of extra information. The above problem is from a special case of equation 18, the differential equation being given in equation 31. – VolatileStorm Dec 12 '11 at 11:07

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

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



For instance, this yields

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


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

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My apologies, I'd just noticed a typo in my solution! However, at best approximation this does appear to simplify your final answer even more! – VolatileStorm Dec 12 '11 at 11:18
@VolatileStorm: In fact it simplifies both results and removes all the ugly factors of $2$. I've edited accordingly. – joriki Dec 12 '11 at 11:22

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:



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These are the correct coefficiants! I am interested to know why Maple can do this, but my Mathematica cannot. However, as stated, I am looking for a closed form for these (if it exists). I'm trying solve a recursion relation, so calculability isn't a problem. – VolatileStorm Dec 12 '11 at 18:14
Knowing absolutely no maple, but trying your method I am pleased to see that Maple can do it spectacularly fast. Which makes me think that maple is doing something clever and analytic that I don't know of. – VolatileStorm Dec 12 '11 at 18:20

$$ f(z) = 1-{\frac {1}{4}}z-{\frac {5}{32}}{z}^{2}-{\frac {15}{64}}{z}^{3}-{ \frac {1105}{2048}}{z}^{4}-{\frac {1695}{1024}}{z}^{5}-{\frac {414125} {65536}}{z}^{6}-{\frac {59025}{2048}}{z}^{7}-\dots $$
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 ...

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Following the lead in OEIS, I note that Series[1 + z + AiryAiPrime[z^2]/AiryAi[z^2], {z, Infinity, 25}] in Mathematica reproduces the coefficients (but with wrong sign and multiplying the wrong powers). This shouldn't be a surprise, as Airy functions and Bessel functions of one-third order are related after all. – J. M. Dec 13 '11 at 0:39

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