Confluent hypergeometric function: Difference between revisions

Revision as of 23:47, 26 August 2020

Revision as of 05:52, 27 August 2020

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then the differential equation gives

:<math>z^{2-b}\frac{d^2v}{dz^2}+2(1-b)z^{1-b}\frac{dv}{dz}-b(1-b)z^{-b}v + (b-z)\left[z^{1-b}\frac{dv}{dz}+(1-b)z^{-b}v\right] - az^{1-b}v = 0</math>

which, upon dividing out {{math|''z''<sup>1−''b''</subsup>}} and simplifying, becomes

<!--:<math>z\frac{d^2v}{dz^2}+2(1-b)\frac{dv}{dz}-b(1-b)z^{-1}v + (b-z)\left[\frac{dv}{dz}+(1-b)z^{-1}v\right] - av = 0</math>-->

:<math>z\frac{d^2v}{dz^2}+(2-b-z)\frac{dv}{dz} - (a+1-b)v = 0.</math>

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Note that the solution {{math|''z''^1−''b''''M''(''a'' + 1 − ''b'', 2 − ''b'', ''z'')}} to Kummer's equation is the same as the solution {{math|''U''(''a'', ''b'', ''z'')}}, see [[#Kummer's transformation]].

For most combinations of real or complex {{mvar|a}} and {{mvar|b}}, the functions {{math|''M''(''a'', ''b'', ''z'')}} and {{math|''U''(''a'', ''b'', ''z'')}} are independent, and if {{mvar|b}} is a non-positive integer, so {{math|''M''(''a'', ''b'', ''z'')}} doesn't exist, then we may be able to use {{math|''z''<sup>1−''b''</subsup>''M''(''a''+1−''b'', 2−''b'', ''z'')}} as a second solution. But if {{mvar|a}} is a non-positive integer and {{mvar|b}} is not a non-positive integer, then {{math|''U''(''z'')}} is a multiple of {{math|''M''(''z'')}}. In that case as well, {{math|''z''<sup>1−''b''</subsup>''M''(''a''+1−''b'', 2−''b'', ''z'')}} can be used as a second solution if it exists and is different. But when {{mvar|b}} is an integer greater than 1, this solution doesn't exist, and if {{math|1=''b'' = 1}} then it exists but is a multiple of {{math|''U''(''a'', ''b'', ''z'')}} and of {{math|''M''(''a'', ''b'', ''z'')}} In those cases a second solution exists of the following form and is valid for any real or complex {{mvar|a}} and any positive integer {{mvar|b}} except when {{mvar|a}} is a positive integer less than {{mvar|b}}:

:<math>M(a,b,z)\ln z+z^{1-b}\sum_{k=0}^\infty C_kz^k</math>

When ''a'' = 0 we can alternatively use: