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Eric Kvaalen (talk | contribs) z^{1-b}U(a+1-b,2-b,z) = U(a,b,z). And z^{1-b}M(a+1-b,2-b,z) is a closed-form solution to Kummer's equation when a+1-b is a non-positive integer. |
Eric Kvaalen (talk | contribs) Rewrote part on how to get two independent solutions for any combination of a and b (more accurate). Added U(c-n,c,z) to list of special cases. |
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and many of the properties of the confluent hypergeometric function are limiting cases of properties of the hypergeometric function.
Since Kummer's equation is second order there must be another, independent, solution. The [[indicial equation]] of the method of Frobenius tells us that the lowest power of a power series solution to the Kummer equation is either 0 or
:<math>w(z)=z^{1-b}v(z)</math>
then the differential equation gives
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Although this expression is undefined for integer {{mvar|b}}, it has the advantage that it can be extended to any integer {{mvar|b}} by continuity. Unlike Kummer's function which is an [[entire function]] of ''z'', ''U''(''z'') usually has a [[singularity (mathematics)|singularity]] at zero. For example, if ''b''=0 and ''a''≠0 then <math>\Gamma(a+1)U(a,b,c)-1</math> is asymptotic to <math>az\ln z</math> as ''z'' goes to zero. But see [[#Special cases]] for some examples where it is an entire function (polynomial).
Note that the solution <math>z^{1-b}U(a+1-b,2-b,z)</math> to Kummer's equation is the same as the solution <math>U(a,b,z).</math> (See [[#Kummer's transformation]] below.)
For most combinations of real (or complex) ''a'' and ''b'',
:<math>
When ''a'' = 0 we can alternatively use:
When <math>b=1</math> this is the [[exponential integral]] E<sub>1</sub>(''-z'').
A similar problem occurs when ''a''−''b'' is a negative integer and ''b'' is an integer less than 1. In this case <math>M(a,b,z)</math> doesn't exist, and <math>U(a,b,z)</math> is a multiple of <math>z^{1-b}M(a+1-b,2-b,z).</math> A second solution is then of the form:
:<math>
▲:<math>w(z)=z^{1-b}\int_{-\infty}^z(-u)^{b-2}e^u\mathrm{d}u.</math>
===Other equations===
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::<math>U(a,a,z)=e^z\int_z^\infty u^{-a}e^{-u}du</math> (a polynomial if ''a'' is a non-positive integer)
::<math>\frac{U(1,b,z)}{\Gamma(b-1)}+\frac{M(1,b,z)}{\Gamma(b)}=z^{1-b}e^z</math>
::<math>U(c-n,c,z)=\frac{\Gamma(c-1)}{\Gamma(c-n)}z^{1-c}M(1-n,2-c,z)</math> when ''n'' is a positive integer. A closed form with powers of ''z''.
::<math>U(a,a+1,z)= z^{-a}</math>
::<math>U(-n,-2n,z)</math> for non-negative integer ''n'' is a Bessel polynomial (see lower down).
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