证明：因为f(x)具有连续的二阶导数,由拉格朗日定理
f(x+h)-f(x)=hf'(x+t1h)①
f(x)-f(x-h)=hf'(x-t2h)②
(0①-②得f(x+h)+f(x-h)+2f(x)=[f'(x+t1h)-f'(x-t2h)]h
对y=f'(x)在(x-t2h,x+t1h)上使用拉格朗日定理
[f'(x+t1h)-f'(x-t2h)]h=f''(k)(t1+t2)h^2,
(x-t2h所以[f(x+h)+f(x-h)-2f(x)]/h^2 =f''(k)(t1+t2)
因为h趋于0,所以k趋于x
故[f(x+h)+f(x-h)-2f(x)]/h^2 =f''(x)(t1+t2)
只能到这了.