作者jacky7987 (忆)
看板Math
标题Re: [微积] f(x+y) = f(x)f(y) implies f(x) = a^x
时间Sat Feb 12 13:39:53 2011
※ 引述《GSXSP (Gloria)》之铭言：
: 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)
: 但是还是看不出来 ----- 是constant
: f(x)
: 请问这题要怎麽做呢?
提供另外一个方法试试看好了
2
f(0)=f(0+0)=f(0)
=>f(0)=0 or 1
Case 1:
If f(0)=0 them f(x)≡0 for all x in R.....done
Case 2:
if f(0)=1
for n \in N n
f(n)=f(1+1+...+1)=f(1)
￣￣￣￣n times
1=f(0)=f(1+(-1))=f(1)*f(-1)
-1
=> f(-1)=f(1)
Hence for n be negtive integer
-n -1 -n n
f(n)=f(
(-1)*(-n))=f(-1)*f(-n)=f(-1)*f(1) =f(1) * f(1) =f(1)
1 1 1 1 m
Moreover, f(1)=f(--- + --- + ...+---) = f(---)
m m m m
￣￣￣￣￣￣￣￣m times
1
(---)
1 m
hence f(---)=f(1)
m
1 n
(---) ---
n 1 n m
finally, f(---)=f(---) =f(1)
m m
到此我们做完全部的有里数的部份
therefore, for r be a irrational number there exists a seq {x_n} ,where x_n
are all rational number such that x_n->r
and f is diff => f is conti
Hence
x_n r
f(r)=lim f(x)= lim f(x_n)= lim f(1) =f(1)
x->r n->inf n->inf
let f(1)=a, we are done
＝＝＝
这作法是模仿线性的XDD
f(x+y)=f(x)+f(y)
--
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◆ From: 114.37.124.220
※ 编辑: jacky7987 来自: 114.37.124.220 (02/12 13:40)
1F：推 GSXSP :推 02/12 16:04