Re: [微積] f(x+y) = f(x)f(y) implies f(x) = a^x
※ 引述《GSXSP (Gloria)》之銘言：
: f(x) differentiable on (-∞,∞)
: with f(x+y) = f(x)f(y)
: prove that f(x) = a^x for some a
: (Hint: ln f(x) must have constant derivitive)
: 我做 f'(x) f'(y)
: grad ln f(x+y) = (------ , ------ )
: f(x) f(y)
: f'(x)
: 但是還是看不出來 ----- 是constant
: f(x)
: 請問這題要怎麼做呢?
f(0)=f(0)f(0) -> f(0)=0 OR 1
For f(0)=0 : for all real r, f(r+0)=f(r)f(0)=0=f(r), so a=0
For f(0)=1 :
f(x)(f(y)-1) 1
(lnf(x))'=lim ____________ _____
y->0 y f(x)
f(y)-f(0)
=lim ____________
y->0 y-0
=f'(0)
By Fundemental Theorem of Calculus, lnf(x)=C+f'(0)x
f(x)=(e^C)*(e^(f'(0)x))
But, f(0)=e^C=1,
that is f(x)=e^f'(0)x, here we can let e^f'(0)=a.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.248.48.26
※ 編輯: venson0502 來自: 111.248.48.26 (02/14 23:14)
討論串 (同標題文章)
本文引述了以下文章的的內容：
完整討論串 (本文為第 4 之 4 篇)：