$f(x)$六个零点求六连乘积

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若 $f(x)=(x-5) (x+2) (2 x-3) (3 x-1) (4 x+1) (5 x-4)-\lambda(x-2) (x-1) x (x+1) (2 x-1)$ 有六个实根，分别为 $x_1,\,x_2,\dots,x_6$ 且 $x_1>\dots>x_6$，则 $\big(2x_1-x_2-1\big)\big(2x_2-x_3-1\big)\dots\big(2x_6-x_1-1\big)$ 的值为___.

hbghlyj

设$\varphi(x)=\frac{2 x_1-1}{x_1+1}$，则$x_1,\varphi(x_1),\varphi\circ\varphi(x_1),\dots$都是$f$的根.
$f(2)=-3240<0$，因为$x_1$是$f$的最大根，所以$x_1>2$，所以$$x_1>\varphi(x_1)>\varphi\circ\varphi(x_1)>\dots>\varphi\circ\varphi\circ\varphi\circ\varphi\circ\varphi\circ\varphi(x_1)$$
因为$x_1>\dots>x_6$，所以
$$x_2=\frac{2 x_1-1}{x_1+1},x_3=\frac{2 x_2-1}{x_2+1},\dots,x_6=\frac{2 x_5-1}{x_5+1},x_1=\frac{2 x_6-1}{x_6+1}$$
为了求$\prod_{i=1}^6(2x_i-x_{i+1}-1)$的值，
$$2x_i-x_{i+1}-1=2 x_i-\frac{2 x_i-1}{x_i+1}-1=x_ix_{i+1}$$
$f(x)$的首项系数$2\cdot3\cdot4\cdot5=120$
$$\prod_{i=1}^6x_i=\frac{f(0)}{120}=1$$
从而
$$\prod_{i=1}^6(2x_i-x_{i+1}-1)=\prod_{i=1}^6x_ix_{i+1}=1^2=1$$

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取$λ=\sqrt2$验证一下：

1. λ=Sqrt[2];
2. roots=ReverseSort[x/.Solve[(x-5) (x+2) (2 x-3) (3 x-1) (4 x+1) (5 x-4)-λ(x-2) (x-1) x (x+1) (2 x-1)==0,x]];
3. Product[2roots[[i]]-roots[[Mod[i+1,6,1]]]-1,{i,1,6}]//RootReduce

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hbghlyj

hbghlyj 发表于 2024-6-5 08:44
若 $f(x)=(x-5) (x+2) (2 x-3) (3 x-1) (4 x+1) (5 x-4)-\lambda(x-2) (x-1) x (x+1) (2 x-1)$ 有六个实根

$\lambda$取任何实数时，$f(x)$总是有6个不同的实根。而且
$$x_1>2>x_2>1>x_3>\frac12>x_4>0>x_5>-1>x_6$$