- 证明多项式 f ( x ) = 1 + x + x 2 2 ! + ⋯ + x n n ! f(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} f(x)=1+x+2!x2+⋯+n!xn 无重根。
f ( x ) − f ′ ( x ) = x n n ! f(x) - f'(x) = \frac{x^n}{n!} f(x)−f′(x)=n!xn
( f ( x ) , f ′ ( x ) ) = ( f ( x ) , f ( x ) − f ′ ( x ) ) = ( f ( x ) , x n n ! ) = x k \left( f(x), f'(x) \right) = \left( f(x), f(x) - f'(x) \right) = \left( f(x), \frac{x^n}{n!} \right) = x^k (f(x),f′(x))=(f(x),f(x)−f′(x))=(f(x),n!xn)=xk
其中 k k k 是小于等于 n n n 的非负整数，由于 0 显然不是 f ( x ) f(x) f(x) 的根，所以只能是 k = 0 k=0 k=0，即 ( f ( x ) , f ′ ( x ) ) = 1 \left( f(x), f'(x) \right) =1 (f(x),f′(x))=1
故 f ( x ) f(x) f(x) 无重根
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求行列式
D n = ∣ x + a 1 a 2 a 3 ⋯ a n a 1 x + a 2 a 3 ⋯ a n a 1 a 2 x + a 3 ⋯ a n ⋮ ⋮ ⋮ ⋱ ⋮ a 1 a 2 a 3 ⋯ x + a n ∣ D_n = \begin{vmatrix} x + a_1 & a_2 & a_3 & \cdots & a_n \\ a_1 & x + a_2 & a_3 & \cdots & a_n \\ a_1 & a_2 & x + a_3 & \cdots & a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & a_3 & \cdots & x + a_n \\ \end{vmatrix} Dn= x+a1a1a1⋮a1a2x+a2a2⋮a2a3a3x+a3⋮a3⋯⋯⋯⋱⋯ananan⋮x+an
D n = ∣ x + a 1 a 2 a 3 ⋯ a n a 1 x + a 2 a 3 ⋯ a n a 1 a 2 x + a 3 ⋯ a n ⋮ ⋮ ⋮ ⋱ ⋮ a 1 a 2 a 3 ⋯ x + a n ∣ = ∣ x E n + [ 1 1 1 ⋮ 1 ] [ a 1 a 2 a 3 ⋯ a n ] ∣ = x n − 1 ∣ x + [ a 1 a 2 a 3 ⋯ a n ] [ 1 1 1 ⋮ 1 ] ∣ = x n − 1 ( x + ∑ k = 1 n a k ) . \begin{aligned} D_n &= \left| \begin{array}{ccccc} x + a_1 & a_2 & a_3 & \cdots & a_n \\ a_1 & x + a_2 & a_3 & \cdots & a_n \\ a_1 & a_2 & x + a_3 & \cdots & a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & a_3 & \cdots & x + a_n \end{array} \right| \\ &= \left| xE_n + \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] \left[ \begin{array}{ccccc} a_1 & a_2 & a_3 & \cdots & a_n \end{array} \right] \right| \\ &= x^{n-1} \left| x + \left[ \begin{array}{ccccc} a_1 & a_2 & a_3 & \cdots & a_n \end{array} \right] \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] \right| \\ &= x^{n-1} \left( x + \sum\limits_{k = 1}^n a_k \right). \end{aligned} Dn= x+a1a1a1⋮a1a2x+a2a2⋮a2a3a3x+a3⋮a3⋯⋯⋯⋱⋯ananan⋮x+an = xEn+ 111⋮1 [a1a2a3⋯a