UTF8gbsn
先来描述一下问题.对于系数为属于
C
\Bbb{C}
C ,而非常数的多项式
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P(z)=\sum_{j=0}^{n}\left(a_{j} z^{j}\right)=a_{0}+a_{1} z+a_{2} z^{2}+\cdots+a_{n-1} z^{n-1}+a_{n} z^{n}
P ( z ) = j = 0 ∑ n ( a j z j ) = a 0 + a 1 z + a 2 z 2 + ⋯ + a n − 1 z n − 1 + a n z n
那么
p
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p(z)
p ( z ) 含有n个根,且属于
C
\Bbb{C}
C .也就是说有n个
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z_i
z i 使得
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p(z_i)=0
p ( z i ) = 0
note
先来明确一点,constant polynomial的定义.当
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1
z^k,k\geqslant 1
z k , k ⩾ 1 的系数都为0的时候,这个多项式就为constant polynomial.而constant polynomial实际上就是只剩常数的多项式,比如
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p(z)=a_{0}+0\cdot z+0\cdot z^{2}+\cdots+0\cdot z^{n-1}+0\cdot z^{n}
p ( z ) = a 0 + 0 ⋅ z + 0 ⋅ z 2 + ⋯ + 0 ⋅ z n − 1 + 0 ⋅ z n
如果,那么
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p(z)=a_{0}+0\cdot z+0\cdot z^{2}+\cdots+0\cdot z^{n-1}+0\cdot z^{n}
p ( z ) = a 0 + 0 ⋅ z + 0 ⋅ z 2 + ⋯ + 0 ⋅ z n − 1 + 0 ⋅ z n 有哪些根呢?假如
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p(z)=a_{0}+0\cdot z+0\cdot z^{2}+\cdots+0\cdot z^{n-1}+0\cdot z^{n}\neq 0
p ( z ) = a 0 + 0 ⋅ z + 0 ⋅ z 2 + ⋯ + 0 ⋅ z n − 1 + 0 ⋅ z n = 0 , 那么
z
z
z 取多少的值可以满足
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p(z)=a_{0}+0\cdot z+0\cdot z^{2}+\cdots+0\cdot z^{n-1}+0\cdot z^{n}=0
p ( z ) = a 0 + 0 ⋅ z + 0 ⋅ z 2 + ⋯ + 0 ⋅ z n − 1 + 0 ⋅ z n = 0 呢?很明显是0.没有任何的
z
z
z 可以满足这一点.但是如果
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p(z)=a_{0}+0\cdot z+0\cdot z^{2}+\cdots+0\cdot z^{n-1}+0\cdot z^{n}=0
p ( z ) = a 0 + 0 ⋅ z + 0 ⋅ z 2 + ⋯ + 0 ⋅ z n − 1 + 0 ⋅ z n = 0 ,那么什么样的
z
z
z 可以满足这一点呢?很明显所有的
z
z
z 都可以满足这一点儿.
归纳法
归纳法中有一点是前提条件.也就是对于非常数的多项式.一定会有一个根
∈
C
\in \Bbb{C}
∈ C .这个命题是叫做 fundamental theorem of algebra.另外一个需要用到的东西是factor theorem.有了这个命题我们就很容易证明原命题了.
当k=1时,
p
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0
p(z)=a_1z+a_0,a_1\neq 0
p ( z ) = a 1 z + a 0 , a 1 = 0 含有一个根
z
=
−
a
0
a
1
z=-\frac{a_0}{a_1}
z = − a 1 a 0
假设k=n-1时,
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p(z)=a_{k}z^k+a_{k-1}z^{k-1}+\cdots + a_0
p ( z ) = a k z k + a k − 1 z k − 1 + ⋯ + a 0 有n-1个根.
有对于k=n,按照 fundamental theorem of algebra,我们可以断言
p
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z
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p(z)
p ( z ) 含有一个根
z
λ
z_{\lambda}
z λ ,我们可以根据factor theorem来将
p
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p(z)
p ( z ) 改写为
p
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z
λ
)
g
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p(z)=(z-z_{\lambda})g(z)
p ( z ) = ( z − z λ ) g ( z ) 其中
g
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z
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g(z)
g ( z ) 是一个
n
−
1
n-1
n − 1 次的多项式.我们知道
g
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z
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g(z)
g ( z ) 含有n-1个根.那么
p
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z
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p(z)
p ( z ) 就一定含有n个根.
可见原式得证.