# Discrepancy in the series expansion of BesselK[ν, z]

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

Out:=  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 $$K_{-\nu}(z)=K_{\nu}(z)$$?

• 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. – user64494 Jan 28 at 19:42
• 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. – Alexei Boulbitch Jan 28 at 19:45
• I don't see a discrepancy. One is restricted to \[Nu]>0, the other is not. – Daniel Lichtblau Jan 28 at 20:19
• @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. – user64494 Jan 28 at 20:50