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Given $x+\dfrac1x=n$, I derived several expressions in terms of $n$ to solve for $x^k+\dfrac1{x^k}$ and put them in a chart as shown below. My questions is how are the coefficients of these polynomials related? I have observed that the first coefficient is $1$, the second coefficient is $-k$, and the signs always alternate.

Chart of polynomials

Martin Sleziak
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asdf
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  • In case the chart is unclear, if $x+\dfrac1x=3$ (third column where $n=3$) then $x^3+\dfrac1{x^3}=18$ which I put into the second row where $k=3$. – asdf Mar 10 '16 at 18:12

1 Answers1

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If $x + 1/x = n$, writing $x = \exp(it)$ we have $\cos(t) = n/2$. Then $x^k + 1/x^k = \exp(ikt) + \exp(-ikt) = 2 \cos(kt) = 2 T_k(n/2)$ where $T_k$ is the $k$'th Chebyshev polynomial of the first kind. Thus for $k=11$, $$T_{11}(x) = 1024\,{x}^{11}-2816\,{x}^{9}+2816\,{x}^{7}-1232\,{x}^{5}+220\,{x}^{3}- 11\,x$$ so you get $$ 2 T_{11}(n/2) = {n}^{11}-11\,{n}^{9}+44\,{n}^{7}-77\,{n}^{5}+55\,{n}^{3}-11\,n$$

Robert Israel
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  • Excellent, thank you. I have determined from the definition $T_{n}(x) = \dfrac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}2$ that if $x+\dfrac1x=n$, $x^k+\dfrac1{x^k}=(a+\sqrt{a^2-1})^k+(a-\sqrt{a^2-1})^k$ where $a=\dfrac n2$. – asdf Mar 10 '16 at 19:07
  • Simplified and in terms of $n$ this becomes $\dfrac{(n+\sqrt{n^2-4})^k+(n-\sqrt{n^2-4})^k}{2^k}$. – asdf Mar 10 '16 at 19:17