若f(x)有二阶导数,证明f''(x)=lim(h→0)f(x+h)-2f(x)+f(x-h)/h^2.
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若f(x)有二阶导数,证明f''(x)=lim(h→0)f(x+h)-2f(x)+f(x-h)/h^2.
若f(x)有二阶导数,证明f''(x)=lim(h→0)f(x+h)-2f(x)+f(x-h)/h^2.
若f(x)有二阶导数,证明f''(x)=lim(h→0)f(x+h)-2f(x)+f(x-h)/h^2.
给你提供三种方法,都读研的人了,本来不想做的,不给加分没良心.
key1:洛必达法则
lim(h→0)f(x0+h)+f(x-h)-2f(x) / h^2
=lim(h→0)f '(x+h)-f '(x-h) / 2h
=lim(h→0)f ''(x+h)+f ''(x-h) / 2
=f ''(x)+f ''(x) / 2=f ''(x)
为什么2f(x)可以消去,为什么减号能变成加号呢?
注意哦:::::::上下都对h求导.不能上面对x,下面对h,这叫啥嘛、
key2:拉格朗日定理:
f(x+h)-f(x)=hf'(x+th),0<t<1 (1)
f(x)-f(x-h)=hf'(x+sh),0<s<1 (2)
.自己想
或者:
泰勒中值定理:
f(x+h)=f(x)+f'(x)h+(f''(x)h^2/2)+o(h^2) (1)
f(x-h )=f(x)-f'(x)h+(f''(x)h^2/2)+o(h^2) (2)
(1)+(2),
f(x+h)-2f(x)+f(x-h)/h^2=[f''(x)h^2+o(h^2)]/h^2→f''(x)(h→0)
key3:定义.要很清楚导数的定义.
将f(x)在零点泰勒展开,f(x+h)=f(x)+f'(x)h+(f''(x)h^2/2)+o(h^2)
f(x-h)=f(x)-f'(x)h+(f''(x)h^2/2)+o(h^2)
故有f(x+h)-2f(x)+f(x-h)/h^2=[f''(x)h^2+o(h^2)]/h^2→f''(x)(h→0)
看不懂