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How can I find the expansion of $$(1-a^k-a^{-k})^m$$ where $m$ is a positive integer?

I know that I can split the sum and use the binomial theorem but this seems really ugly. Does anybody recognize instantly what will this be, or, even better, can tell me how to find it with Mathematica (if possible)?

Thanks.

Ѕᴀᴀᴅ
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Zugzwang14
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3 Answers3

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You may be interested in the multinomial expansion $$(x_1 + x_2 + \cdots x_m)^n = \sum_{\overset{i_1,i_2,\ldots,i_m \geq 0}{i_1 + i_2 + \cdots + i_m = n}} \dfrac{n!}{i_1! i_2! i_3! \cdots i_m!} x_1^{i_1}x_2^{i_2} \cdots x_m^{i_m}$$ For trinomial expansion, this simplifies to $$(x_1 + x_2 + x_3)^n = \sum_{\overset{i_1,i_2 \geq 0}{i_1 + i_2 \leq n}} \dfrac{n!}{i_1! i_2! (n-i_1-i_2)!} x_1^{i_1}x_2^{i_2} x_3^{n-i_1-i_2}$$

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Following up on my comment, you can use $a^k = e^{k \ln a}$ to simplify the expression to

$$(1 - a^k - a^{-k})^m = (1-2 \cosh [k \ln a])^m$$

From here, you can use the binomial theorem. Is this what you're intended to do? No idea, but it's mathematically valid.

Muphrid
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  • Interesting, thanks! Actually, I am trying to conjecture a relation between the sum of the $2m$th powers of the diagonals of a regular $n$-gon and $n$. My approach is through the $n$th roots of unity: the diagonals are given by $$|1-\omega^k|$$ and hence the sum is given by $$\sum_1^{n-1} |1-\omega^k|^{2m}=\sum_1^{n-1} (2-\omega^k-\omega^{-k})^m$$. That is why I want the expansion so I can see exactly how, if I split the sum, the inner sums will start to cancel out to $-1$ or $n$ because of the basic properties of the roots of unity. – Zugzwang14 Jan 28 '13 at 00:30
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If this helps at all use it,

$$(a-(a^k+a^{-k}))^m=\sum_{k=0}^m\sum_{r=0}^k(-1)^k{m \choose k}{k \choose r}a^{-k+2r}$$

ReverseFlowControl
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