推 immortal8305:謝謝 03/07 12:58
※ 引述《immortal8305 (immortal)》之銘言:
: ※ [本文轉錄自 trans_math 看板 #1DSwZ0RD ]
: 作者: immortal8305 (immortal) 看板: trans_math
: 標題: [微分] 勘根定理
: 時間: Sun Mar 6 23:31:09 2011
: Let p(x) be a polynomial of odd degree with real coefficient.Then equation
: p(x)=0 has at least one real root.prove this.
n n-1
Set p(x) = a x + a x + ... + a x + a has odd degree.
n n-1 1 0
WLOG, assume that a > 0. Let M = max {|a |, |a |, |a |, |a |, 1.}.
n n n-1 1 0
1
Since a > 0, there exists a positive integer N such that a > ── .
n n N
n n-1 n-2
So p(kM) ≧ a (kM) – M(kM) – M(kM) – ... – M(kM) – M =
n
n n
n n n k -1 n n n k -1 1 n
a k M – M ──── ≧ a k M – M ──── > (a – ──) k
n k-1 k N k N
for k-1 ≧ N. So p( (N+1)M ) > 0. Similarly, p( -(N+1)M ) < 0.
By intermediate value theorem, p(c) = 0 for some c in ( -(N+1)M, (N+1)M ).
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