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Why does Multinomial[-1/2, -1/2, 1] give Indeterminate rather than 1/Pi?

GROUPS:

Is this a bug? Or there might be something I don't understand.

In[1]:= Multinomial[-1/2, -1/2, 1]
"During evaluation of In[1]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered."
Out[1]= Indeterminate

None of the arguments are located at poles of the factorial function (negative integers), so I expected this answer:

In[2]:= (-1/2 - 1/2 + 1)! / ((-1/2)! (-1/2)! (1)!)
Out[2]= 1/Pi

Checking with Simplify using Assumptions:

In[3]:= Assuming[a == -1/2 && b == -1/2 && c == 1,
            Simplify[ Multinomial[a, b, c] == 1/Pi ]
        ]
Out[3]= True

Thanks! (I'm running 11.2)

In[4]:= $Version
Out[4]= "11.2.0 for Mac OS X x86 (64-bit) (September 11, 2017)"
POSTED BY: Brad Chalfan
Answer
2 months ago

It seems it might be using the identity

Multinomial[a, b, c] == Binomial[a + b, b] Binomial[a + b + c, c]

The identity is only generically true, but Indeterminate for a = b = -1/2 and c = 1.

A possible workaround:

Multinomial[a, -1/2, 1] /. a -> -1/2
POSTED BY: Michael Rogers
Answer
2 months ago

A possible workaround:

Binomial[a + b, b]*Binomial[a + b + c, c] // FunctionExpand // FullSimplify

(* Gamma[1 + a + b + c]/(Gamma[1 + a] Gamma[1 + b] Gamma[1 + c])*)

as a function:

MultinomialUSER[a_, b_, c_] := Gamma[1 + a + b + c]/(Gamma[1 + a] Gamma[1 + b] Gamma[1 + c])
MultinomialUSER[-1/2, -1/2, 1]

(* 1/\[Pi] *)
POSTED BY: Mariusz Iwaniuk
Answer
2 months ago

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