※ 引述《PttFund (批踢踢基金)》之銘言：
: Suppose that h > 0 and f' exists and is finite for every x
: in (a-h,a+h). Also suppose that f is continuous on [a-h,a+h].
: If f"(a) exists, show that the limit
: f(a+h) - 2f(a) + f(a-h)
: lim ------------------------- = f"(a)
: h→0 h^2
: Note: 請順便給一個反例 f(x), 使得左邊的極限式存在, 但 f" 不存在.
這個用 L'Hopital's rule 做就可以了:
f(a+h) - 2f(a) + f(a-h) f'(a+h) - f'(a-h)
lim ------------------------- = lim -------------------
h→0 h^2 h→0 2h
f'(a+h) - f'(a) f'(a-h) - f'(a)
= 1/2 { lim ----------------- + lim ----------------- }
h→0 h h→0 -h
= 1/2 { f"(a) + f"(a) } = f"(a).
反例也很簡單, 可以想一想 :p
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