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While reading this paper this paper, I was stumble upon a strange relationship between the modified Bessel function of the second kind and the Psi function that is:

$c{K_1}\left( c \right) \approx 1 + \frac{{{c^2}}}{2}\left( {\ln \left( {\frac{c}{2}} \right) + {{\text{C}}_0}} \right) + \frac{{{c^4}}}{{16}}{{\text{C}}_1}$

Where ${C_0}$= −φ(1) /2 − φ(2) /2 and C1 = −φ(2)/2 − φ(3)/2

The paper said that φ(•) denotes the Psi function and point a citation to

. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals Series and Products, New York, NY, USA:Academic, 2007.

Which yield the following definition and integral representation

enter image description here

Please help me understand where this relationship come from ?

Also, is it possible to extent this approximation for a modified Bessel function of the second kind of order $M$ like this ${K_M}\left( {2\sqrt x } \right)$ ?

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    $\begingroup$ For reference purposes, $\psi(x)$ is known as the digamma function $\endgroup$ Jan 19, 2021 at 2:27

1 Answer 1

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The DLMF (eq. 10.3.1) provides the following sum formula for $K_n(x)$ with nonnegative integer $n$:

\begin{align} K_n(x) &=\frac12 \left(\frac{x}{2}\right)^{-n}\sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(-\frac{x^2}{4}\right)^k+(-1)^{n+1}\ln\left(\frac{x}{2}\right)I_n(x)\\ &\qquad\quad\quad +(-1)^n \frac12 \left(\frac{x}{2}\right)^n \sum_{k=0}^\infty\left[\psi(k+1)+\psi(n+k+1)\right]\frac{(x^2/4)^k}{k!(n+k)!} \end{align}

For the case of $n=1$, this reduces to $$ K_1(x) =\frac1x+\ln\left(\frac{x}{2}\right)I_1(x) - \frac{x}{4} \sum_{k=0}^\infty\left[\psi(k+1)+\psi(n+k+1)\right]\frac{(x^2/4)^k}{k!(k+1)!} $$

Since $I_1(x)\approx x/2$ for small $x$, we find

\begin{align} x K_1(x) &\approx 1+\ln\left(\frac{x}{2}\right)\cdot \frac{x^2}{2}-\frac{1}{4}(\psi(1)+\psi(2))x^2-(\psi(2)+\psi(3))\frac{x^4}{32}\\ &=1+\frac{x^2}{2}\left(\ln\left(\frac{x}{2}\right)-\frac12 \psi(1)- \frac12 \psi(2)\right) -\left(\frac12 \psi(2)+\frac12 \psi(3)\right)\frac{x^4}{16} \end{align} which is the stated result.

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  • $\begingroup$ When x is small why does the infinite sum that contain the Psi function becomes finite ? $\endgroup$ Jan 19, 2021 at 4:54
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    $\begingroup$ Because it's an expansion in powers of $x$, so I've only kept powers up to $x^4$. $\endgroup$ Jan 19, 2021 at 5:58
  • $\begingroup$ Does this series converge faster than just using taylor expansion? $\endgroup$ Jan 19, 2021 at 7:57
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    $\begingroup$ Probably not, but that's because the above isn't that far from a Taylor series. Indeed, if you rearrange the main sum formula as $K_n(x)+(-1)^n \ln(x/2)I_n(x)$, then what remains on the right-hand side is just the Taylor series. $\endgroup$ Jan 19, 2021 at 14:28

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