f(x)在x处二阶可导,求lim{[f(x+h)-2f(x)+f(x-h)]/h^2},h趋向于0
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f(x)在x处二阶可导,求lim{[f(x+h)-2f(x)+f(x-h)]/h^2},h趋向于0
f(x)在x处二阶可导,求lim{[f(x+h)-2f(x)+f(x-h)]/h^2},h趋向于0
f(x)在x处二阶可导,求lim{[f(x+h)-2f(x)+f(x-h)]/h^2},h趋向于0
利用级数展开到二阶即可
h趋向于0时:
lim{[f(x+h)-2f(x)+f(x-h)]/h^2}
=lim{[f '(x+h)-f '(x-h)]/(2h)} ................运用洛必达法则
=lim{[f '(x+h)-f '(x)+f '(x)-f '(x-h)]/(2h)}
(注意:因为没有f(x)在x处二阶导数连续这个条件,
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h趋向于0时:
lim{[f(x+h)-2f(x)+f(x-h)]/h^2}
=lim{[f '(x+h)-f '(x-h)]/(2h)} ................运用洛必达法则
=lim{[f '(x+h)-f '(x)+f '(x)-f '(x-h)]/(2h)}
(注意:因为没有f(x)在x处二阶导数连续这个条件,
若再运用洛必达法则, 则之后的limf ''(x+h)及limf ''(x-h)没法求出)
=(1/2)lim{[f '(x+h)-f '(x)]/h+[f '(x-h)-f '(x)]/(-h)}................f(x)在x处二阶可导,运用导数定义
=(1/2)[f ''(x)+f ''(x)]
=f ''(x)
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