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[函数] 一道复数题目

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dahool Posted 2020-4-28 09:44 |Read mode
已知多项式$P(x)=(1+x+x^2+\cdots++x^{13})^2-x^{13}$有26个复根,$z_k=r_k(\cos 2\pi\alpha_k+i\sin 2\pi\alpha_k)$,其中$k=1,2,\cdots,26,0<\alpha_1 \leqslant \alpha_2 \leqslant \cdots \leqslant\alpha_{26} <1,r_k>0$,求$\alpha_{16}$

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kuing Posted 2020-4-28 12:56
为方便撸代码记 `t=x^{13}`,则
\[P(x)=\left( \frac{1-t}{1-x}+t \right)^2-t=\frac{(1-t)(1-tx^2)}{(1-x)^2}=\frac{(1-x^{13})(1-x^{15})}{(1-x)^2},\]下略。

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 Author| dahool Posted 2020-4-28 14:59
回复 2# kuing

厉害,谢谢!

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