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I am trying to expand the modified Bessel function of the second kind $K_{\nu}(z)$ for small values of the argument $z$ keeping $\nu$ fixed.

Mathematica 12.2.0 says that it is

  In[1]:=  Normal[Series[(BesselK[\[Nu], \[Alpha]*z]), {z, 0, 1}, 
       Assumptions -> {\[Alpha] > 0, Abs[\[Nu]] > 0, 
         Abs[z] > 0}]] // Expand 
    
  Out[1]:=  2^(-1 - \[Nu]) z^\[Nu] \[Alpha]^\[Nu] Gamma[-\[Nu]] + 
     2^(-1 + \[Nu]) z^-\[Nu] \[Alpha]^-\[Nu] Gamma[\[Nu]]

So, Mathematica essentially tells me that there are two terms, viz.,

$$ K_{\nu} \sim \dfrac{1}{2}\Gamma\bigl(\nu\bigl)\Bigl(\dfrac{1}{2}z\Bigl)^{-\nu}+\dfrac{1}{2}\Gamma\bigl(-\nu\bigl)\Bigl(\dfrac{1}{2}z\Bigl)^{\nu}, ~ z \to 0.$$

However, this seemingly does not agree with result quoted in Abramowitz and Stegun (and in Wikipedia). According to Abramowitz and Stegun, it is

$$K_{\nu} \sim \dfrac{1}{2}\Gamma\bigl(\nu\bigl)\Bigl(\dfrac{1}{2}z\Bigl)^{-\nu}, ~z \to 0, ~ \mathrm{if} ~ \nu>0. $$

What is the reason behind this discrepancy? Does this have something to with the fact $K_{-\nu}(z)=K_{\nu}(z)$?

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    $\begingroup$ If I am not mistaken, the modulus of 2^(-1 - \[Nu]) z^\[Nu] \[Alpha]^\[Nu] Gamma[-\[Nu]] approaches zero as z tends to zero, whereas the modulus of \[Nu]) z^-\[Nu] \[Alpha]^-\[Nu] Gamma[\[Nu]] tends to infinity as z tends to zero. The result of MMA is in accordance with Abramowitz and Stegun. Don't worry, be happy. $\endgroup$ – user64494 Jan 28 at 19:42
  • $\begingroup$ It is because you only looked at pt. 9.6.9 of Abramowitz and Stegun where only the main term is given. If you look at pt. 9.6.11 you find a more complete expression. $\endgroup$ – Alexei Boulbitch Jan 28 at 19:45
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    $\begingroup$ I don't see a discrepancy. One is restricted to \[Nu]>0, the other is not. $\endgroup$ – Daniel Lichtblau Jan 28 at 20:19
  • $\begingroup$ @DanielLichtblay: If $\nu>0$, then $$ K_{\nu} \sim \dfrac{1}{2}\Gamma\bigl(\nu\bigl)\Bigl(\dfrac{1}{2}z\Bigl)^{-\nu}+\dfrac{1}{2}\Gamma\bigl(-\nu\bigl)\Bigl(\dfrac{1}{2}z\Bigl)^{\nu}, ~ z \to 0,$$ implies $$K_{\nu} \sim \dfrac{1}{2}\Gamma\bigl(\nu\bigl)\Bigl(\dfrac{1}{2}z\Bigl)^{-\nu}, ~z \to 0,$$ as is noticed in my comment. $\endgroup$ – user64494 Jan 28 at 20:50

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