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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)

  • 0
    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 - http://arxiv.org/PS_cache/nlin/pdf/0003/0003069v1.pdf . 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 31.2011-12-12

3 Answers 3

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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} $

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    @VolatileStorm: In fact it simplifies both results and removes all the ugly factors of $2$. I've edited accordingly.2011-12-12
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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);

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

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    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.2011-12-12
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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 ...

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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.2011-12-13