※ 引述《bisconect (隨便你叫)》之銘言：
: if
: 1) f(0) = g(0)
: 2) f'(0) = g'(0)
: 3) for any t, f''(t) = h(f(t)) and g''(t) = h(g(t))
: then
: for any t, f(t) = g(t)
f"-g"=0; ∫(f"-g")dt=∫0dt=0=f'-g'+C1 ==>f'(0)=g'(0) ==>C1=0 ==>f'-g'=0
同理可證f-g=0 ==>f=g
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.112.242.149