Re: [單變]微積分
※ 引述《GBRS ()》之銘言：
: Let f(x) be a differentialbe function on |R satisfying
: ╭x^2
: f(x^2)=1+│ f(y)( 1-tan y )dy for all xε|R
: ╯0
: Then f(π)=?
f'(x^2)*2x=f(x^2)(1-tanx^2)*2x
f'(x^2)=f(x^2)(1-tanx^2)
設x^2=k
f'(k)=f(k)(1-tank)
f'(k)-(1-tank)f(k)=0
設g(k)=[e^-(k+1/2lncos^2k)]f(k)
g'(k)=-[e^(k+1/2lncos^2k)]*(1-tank)f(k)+[e^(k+1/2lncos^2k)]f'(k)
g'(k)=e^(k+1/2lncos^2k)(f'(k)-(1-tank)f(k))=0
即g(k)為常數=[e^-(k+1/2lncos^2k)]f(k)=u
又由題設知f(0)=1=k*e^0
k=1
f(k)=e^(k+1/2lncos^2k)
f(π)=e^(π+1/2lncos^2π)
=e^π
ans:e^π
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 220.130.182.146
※ 編輯: kcuricky 來自: 220.130.182.146 (09/08 23:40)
討論串 (同標題文章)