# Limit of Hypergeometric Functions

Mathematica (12) seems to have a lot of trouble with taking limits involving hypergeometric functions. A simple example I am interested in is the following

Limit[HypergeometricU[1/2 - I c \[Epsilon], 1, I b \[Epsilon] ] /
HypergeometricU[3/2 - I c \[Epsilon], 2,
I b \[Epsilon] ] , \[Epsilon] -> 0, Assumptions -> {c > 0, b > 0}]


For me this runs until eventually it crashes. Series[] just spits back the same function. Evaluating at zero gives Indeterminate.

Any help is much appreciated!

• The free parameters are $c$ and $b$. They are both strictly positive, as above. If it helps, I'm willing to restrict to $b>c$. Jan 4 at 14:51
• No, no. Sorry that was my bad. You can make some progress numerically, but analytically I have not figured out anything. Not sure how interesting that would be for your purposes.
– bmf
Jan 4 at 14:54
• Yeah, unfortunately I would need analytics. Jan 4 at 14:55

Here it is in 13.3.1 on Windows 10:

MaxLimit[HypergeometricU[a, b, eps], eps -> 0,
Assumptions -> a > 0 && b > 0, Direction -> "FromBelow"]


DirectedInfinity[-(1/(Sign[-1 + a] Sign[Gamma[-1 + a]]))]

MinLimit[HypergeometricU[a, b, eps], eps -> 0,
Assumptions -> a > 0 && b > 0, Direction -> "FromBelow"]


DirectedInfinity[1/(Sign[1 - a] Sign[Gamma[-1 + a]])]

MinLimit[HypergeometricU[a, b, eps], eps -> 0,
Assumptions -> a > 0 && b > 0, Direction -> "FromAbove"]


ConditionalExpression[Infinity, (-1 + a)*Gamma[-1 + a] > 0]

MaxLimit[HypergeometricU[a, b, eps], eps -> 0,
Assumptions -> a > 0 && b > 0, Direction -> "FromAbove"]


ConditionalExpression[Infinity, (-1 + a)*Gamma[-1 + a] > 0]

Taking into account the result of

NMinimize[{(-1 + a) Gamma[-1 + a], a >= 0}, a]


{0.885603, {a -> 1.46163}}

, we may draw the conclusion that the right limit is infinity and the left limit is minus infinity.

• ComplexPlot3D[HypergeometricU[3, 2, z], {z, 0.01}, PlotLegends -> Automatic, PlotRange -> All, WorkingPrecision -> 30] confirms it. Jan 4 at 16:38
• Hypergeometric1F1 is the solution of the hypergeometric equation with two parameters, regular in (0,1), regular at x=0, the hypergeometric series. The U function is the second independent solution, naturally having a singularity at z=0: en.wikipedia.org/wiki/Confluent_hypergeometric_function Jan 4 at 18:32
• @RolandF: Let us quote Wiki: " U(z) usually has a singularity at zero. For example, if b = 0 and a ≠ 0 then Γ(a+1)U(a, b, z) − 1 is asymptotic to az ln z as z goes to zero. But see #Special cases for some examples where it is an entire function (polynomial)" to be exact. Did you pay your attention to "usually"? Jan 4 at 20:20
• Yes, as always in this field, the eigenvalue problems with hypergeometric series make a discontinuous switch in function space in case a square integable solution exists. The choice of the function in these cases is delicate and notation depends on detailed study of Whitacker/Watson Treatise. Playing around with free parameters in hypergeometric series and its second linear independent solution of the hypergeometric equations is fruitless, the more so poking around in the fog with Mathematica. Jan 4 at 22:01
• @RolandF: Sorry, all those empty words are off-topic. Jan 5 at 5:10