Re: [微積] f(x+y) = f(x)f(y) implies f(x) = a^x

[venson0502 (dd)]

※ 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.









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※ 編輯: venson0502      來自: 111.248.48.26        (02/14 23:14)