f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0
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![f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0](/uploads/image/z/14770102-22-2.jpg?t=f%28x%29%E5%9C%A8X0%E5%A4%84%E4%BA%8C%E9%98%B6%E5%8F%AF%E5%AF%BC%2C%E8%AF%81lim%28h-%3E0%29%5B+f%28x-h0%29%2Bf%28x0%2Bh%29-2f%28x0%29%5D%2Fh%5E2%3Df%60%60%28x0%29+%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E8%83%BD%E8%BF%99%E4%B9%88%E5%81%9A%3F%E5%8E%9F%E5%BC%8F%3Dlim%28h-%3E0%29%5Bf%28x0%2Bh%29-f%28x0%29%5D%2Fh%5E2+%E2%80%94+lim%28h-%3E0%29%5Bf%28x0%29-f%28x0-h%29%5D%2Fh%5E2%3Df%27%28x%29%2Fh-f%27%28x%29%2Fh%3D0)
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?
原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0
极限是不能随便的写成2个极限和或差的,比如最简单的 (tanx - sinx) / x^3 = 1/2 这个极限如果写成2个极限差就得到0-0=0的错误答案,这样直接分成2个之差会直接导致高阶无穷小的丢失而造成结果的错误.
这一个可以用f(x)在x0处的泰勒展开式
f(x0+h) = f(x0) + f'(x0) h + f''(x0) h^2/2 + ...
f(x0-h) = f(x0) - f(x0) h + f''(x0) h^2 / 2 + .
所以f(x0+h) + f(x0-h) - 2f(x0) = f''(x0) h^2 + O(h^3)
所以[ f(x-h0)+f(x0+h)-2f(x0)]/h^2 = ( f''(x0) h^2 + O(h^3) ) / h^2 = f'' (x0)
题目应是:f(x)在X0处二阶可导,证lim(h->0)[ f(x0-h)+f(x0+h)-2f(x0)]/h^2=f ''(x0)
所给解答问题是:当h->0时,lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2这两个极限都不一定存在;
此外,当h->0时,极限的结果中已无h,怎会还有f'(x)/h-...
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题目应是:f(x)在X0处二阶可导,证lim(h->0)[ f(x0-h)+f(x0+h)-2f(x0)]/h^2=f ''(x0)
所给解答问题是:当h->0时,lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2这两个极限都不一定存在;
此外,当h->0时,极限的结果中已无h,怎会还有f'(x)/h-f'(x)/h。
证明:f(x)在X0处二阶可导,则f(x)在x0处及某邻域内连续,在x0处及某邻域内一阶可导,
显然lim(h->0)[ f(x0-h)+f(x0+h)-2f(x0)]=0,lim(h->0)h^2=0.
由洛必达法则,lim(h->0)[ f(x0-h)+f(x0+h)-2f(x0)]/h^2=lim(h->0)[ -f ‘(x0-h)+f '(x0+h)]/(2h)
因为f(x)在X0处二阶可导,由导数定义,
所以lim(h->0)[ -f ‘(x0-h)+f '(x0+h)]/(2h)
= lim(h->0)[ -f ‘(x0-h)+f '(x0+h)-f(x0)+f(x0)]/(2h)
=(1/2) lim(h->0)[ -f ‘(x0-h)+f '(x0+h)-f(x0)+f(x0)]/h
=(1/2) lim(h->0)[ -f ‘(x0-h))+f(x0)+f '(x0+h)-f(x0)]/h
=(1/2) lim(h->0){ [ f ‘(x0-h))-f(x0)]/(-h)+[f '(x0+h)-f(x0)]/h }
=(1/2) {f ''(x0)+f ''(x0)}
= f ''(x0)
综上所述,
于是lim(h->0)[ f(x0-h)+f(x0+h)-2f(x0)]/h^2=f ''(x0)
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