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Macdonald function evaluation

  1. Apr 26, 2008 #1
    It is common knowledge that 1st order Macdonald function [tex]K_{1} (z)[/tex] has following asymptotic behavior if [tex]Re (z) > 0[/tex]

    [tex]z K_{1} (z) \rightarrow 1 \ \mathrm{if} \ z \rightarrow 0[/tex],
    [tex]\sqrt{z} e^{z} K_{1} (z) \rightarrow \sqrt{\frac{\pi}{2}} \ \mathrm{if} \ z \rightarrow \infty[/tex]

    I am interesting in a question, how to find following constraint from only asymptotic relations which are indicated above

    [tex]|K_{1} (z)| \leq \mathrm{const} \cdot (|z|^{ -1/2 } + |z|^{ -1 }) e^{ - Re (z) }[/tex]

    ( I cannot find an explanation in the 'H.Bateman, Higher Transcendental Functions'. Thanks)
     
    Last edited: Apr 26, 2008
  2. jcsd
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