2014 年第 2 期的 “The American Mathematical Monthly” 文章较多, 一共有 17 篇, 其中至少 5 篇是对旧定理–诸如 Stirling’s Formula, 余弦定理, Clairaut’s Theorem(Symmetry of second derivatives 二阶导数的对称性)这样的经典结论–的新证明.

Symmetry of second derivatives If $$f_{xy}$$ and $$f_{yx}$$ are continuous at any given point, then they are equal at the point.

Lemma  Let $$f_{xy}$$ and $$f_{yx}$$ be continuous on rectangle $$R=[a,b]\times[c,d]$$. Then

$\iint_Rf_{xy}\,\mathrm dA=\iint_Rf_{yx}\,\mathrm dA=f(b,d)-f(b,c)-f(a,d)+f(a,c).$

$\iint_Rf_{xy}\,\mathrm dA=\int_a^b\left(\int_c^d f_{xy}(x,y)\,\mathrm dy\right)\,\mathrm dx.$

Suppose they are not identically equal. Then at some point $$(a, b)$$, they differ; say

$f_{xy}(a, b)-f_{yx}(a, b)=l\gt0.$

Note that, since $$f_{xy}$$ and $$f_{yx}$$ are continuous, there is some small $$\triangle x\times\triangle y$$ rectangle, centered at $$(a, b)$$, on which

$f_{xy}(x, y)-f_{yx}(x, y)\geqslant\frac l2,$

Hence,

$\iint_R\left(f_{xy}-f_{yx}\right)\,\mathrm dA\geqslant\iint_R \frac l2\,\mathrm dA=\frac l2\triangle x\triangle y\gt0.$

this contradicts the lemma.                 $$\Box$$

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