m2 一問

MONEYER

P(x)=(x+1)(x+2)(x+3)(x+4).........(x+n)
P'(1)=?
我想問答案係唔係 0 ?

小皇子．明

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

Naozumi

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)]

36661124

補充一點: take log 要諗下舊野係咪>0

Naozumi

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 ]