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MONEYER一星新手會員
2011-1-13 00:33#1
P(x)=(x+1)(x+2)(x+3)(x+4).........(x+n)
P'(1)=?
我想問答案係唔係 0 ?
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小皇子．明三星高級會員
2011-1-13 13:02#2
原帖由 MONEYER 於 11-1-13 12:33 AM 發表 
P(x)=(x+1)(x+2)(x+3)(x+4).........(x+n)
P'(1)=?
我想問答案係唔係 0 ?
唔會,呢個有排計,
你試下做P'(x)出黎,之後再代x=1,
呢個數有n係度:victory:
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NaozumiVIP會員
2011-1-13 15:05#3
就咁計 P'(x)會好痛苦,可以借用logarithmic differentiation (對數微分法)
P(x)=(x+1)(x+2)(x+3)(x+4).........(x+n)
ln P(x) = ln (x + 1) + ln (x + 2) + .... + ln (x + n)
Differentiate 兩邊
P'(x)/P(x) = 1/(x + 1) + 1/(x + 2) + .... + 1/(x + n)
P'(1)/P(1) = 1/2 + 1/3 + ... + 1/(n + 1)
P'(1) = [3 x 4 x 5 x ... x (n + 1)][1/2 + 1/3 + ... + 1/(n + 1)]
= (n+ 1)!/2! [1/2 + 1/3 + ... + 1/(n + 1)]
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36661124二轉會員
2011-1-13 20:15#4
補充一點: take log 要諗下舊野係咪>0
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NaozumiVIP會員
2011-1-13 20:54#5
好認真咁講,條式應係 d/dx (ln |x|) = 1/x
其實反而擔心 P(x) = 0
不過而家d M2教科書,教logarithmic differentiation時都冇提呢點, 唔知乜野原因
[ 本帖最後由 Naozumi 於 11-1-13 08:56 PM 編輯 ]