Re: 一個函數問題
※ 引述《klsh520 (殺手無義，婊子無情)之銘言：
: 若f(x)、g(x)皆為R→R的函數；且滿足
: f(x+y)=f(x)g(y)+g(x)f(y)
: g(x+y)=g(x)g(y)+f(x)f(y)
^應該是減吧 g(x-y)=g(x)g(y)+f(x)f(y)
: g(x)^2-f(x)^2=1
^應該是加吧 g(x)^2+f(x)^2=1
: 試證明：g(0)=1、f(-x)=-f(x)、g(-x)=g(x)
: 小弟有看出來這是三角函數的公式
: 可是不知道要怎麼證明
: 煩請各為大大幫幫忙
g(0) = g(0-0) = g(0)g(0) + f(0)f(0) (By 第二個條件式)
= [g(0)]^2 + [f(0)]^2
= [g(0)]^2 + {1 - [g(0)]^2} (By 第三個條件式)
= 1
f(0) = f(0+0) = f(0)g(0) + g(0)f(0)
= 2f(0)g(0)
= 2f(0) (By 上一個結論g(0)=1)
=> f(0) = 2f(0)
0 = f(0) (By 等量公理，兩邊同減掉f(0))
即 f(0) = 0
g(-x) = g(0-x) = g(0)g(x) + f(0)f(x) (By 第二個條件式)
= 1．g(x) + 0．f(x) (By 上面兩個結論，g(0)=1，f(0)=0)
= g(x)
f(-x) = 忽然想不到~~~~想到再補上~~或是其他強者補充~~
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※ 編輯: ahongyeh 來自: 218.164.75.248 (08/12 03:26)
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