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Maybe this is a rather stupid question. The multinomial theorem tells us that $$ (x_1+x_2+\ldots+x_n)^k=\sum_{k_1+k_2+\ldots k_n=k}\binom{k}{k_1,k_2,\cdots, k_n}x_1^{k_1}\cdot x_2^{k_2}\cdot\ldots\cdot x_n^{k_n} $$ for non-negative integers $k_1,\ldots,k_n$, $k=k_1+\ldots+k_n$ and real numbers $x_1,\ldots,x_n$.

Now, if I have, for instance,

$$(1+x+x^{-1})^k,$$

this polynomial is not defined for $x=0$. Does this mean that for all $x\in\mathbb{R}\setminus\{0\}$, we have $$ (1+x+x^{-1})^k=\sum_{k_1+k_2+k_3=k}\binom{k}{k_1,k_2,k_3}x^{k_2}x^{-k_3}? $$

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  • $\begingroup$ What are your doubts? $\endgroup$
    – I was suspended for talking
    Dec 7, 2017 at 19:01
  • $\begingroup$ @Sisyphus The multinomial theorem says nothing about the domain of the polynomial on the left hand side. So I am not sure if we have a polynomial (as my example) which is not defined on the whole reals, the multinomial theorem holds on this domain. $\endgroup$
    – John_Doe
    Dec 7, 2017 at 19:03
  • $\begingroup$ It should hold at all $x$ other than $x=0,$ but that value is already excluded by the left side anyway. $\endgroup$
    – coffeemath
    Dec 7, 2017 at 19:15

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Yes, you are correct. Notice that the RHS is not defined in $x=0$ either, with (for instance) $k_1= 0,k_2 = 0,k_3 = k$ the corresponding term is $x^{-k}$, and this term is not defined in the origin.

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