Confluent hypergeometric function: Difference between revisions

Revision as of 11:21, 11 October 2016

Revision as of 07:19, 10 December 2016

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:<math>M(a,b,z)\sim\Gamma(b)\left(\frac{e^zz^{a-b}}{\Gamma(a)}+\frac{(-z)^{-a}}{\Gamma(b-a)}\right)</math>

The powers of {{mvar|z}} are taken using <math>-\tfrac 3 2\pi<\arg z\le\tfrac 1 2\pi</math>.<ref>This is derived from Abramowitz and Stegun (see reference below), [http://people.math.sfu.ca/~cbm/aands/page_508.htm page 508]. They give a full asymptotic series. They switch the sign of the exponent in exp(''iπa'') in the right half-plane but this is unimportant because the term is negligible there or else ''a'' is an integer and the sign doesn't matter.</ref> The first term is onlynot needed when {{math|Γ(''b'' − ''a'')}} is infinitefinite (that is, when {{math|''b'' − ''a''}} is not a non-positive integer) or whenand the real part of {{mvar|z}} isgoes to non-negative infinity, whereas the second term is onlynot needed when {{math|Γ(''a'')}} is infinitefinite (that is, when ''a'' is a not a non-positive integer) or whenand the real part of {{mvar|z}} isgoes to non-positive infinity.

There is always some solution to Kummer's equation asymptotic to <math>e^zz^{a-b}</math> as {{math|''z'' → −∞}}. Usually this will be a combination of both {{math|''M''(''a'', ''b'', ''z'')}} and {{math|''U''(''a'', ''b'', ''z'')}} but can also be expressed as <math>e^z(-1)^{a-b}U(b-a,b,-z)</math>.