Alfred Chern
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 Nov 7 asked Prove that $P(x)$ is a polynomial. Sep 27 awarded Yearling Apr 6 comment Generalize Gauss-Bonnet Formula to non-simple closed curves @RyanBudney: Yes, I think so. But I haven't find this version in Milman-Parker. Thanks very much! Apr 6 comment Generalize Gauss-Bonnet Formula to non-simple closed curves @AlexDegtyarev: Thanks! I assume the curve is non-simple closed, means an immersed closed curve, it can have self-intersections and corners. For $w(p)$, I means the winding number of the curve with respect to the point $p$, and you can define this from a topological viewpoint. I also think its "freshmen differential geometry", but could you give me any reference? Thanks very much! Apr 6 comment Generalize Gauss-Bonnet Formula to non-simple closed curves Thanks very much! Could you give me any reference where I can find this version of Gauss-Bonnet Formula? Apr 6 revised Generalize Gauss-Bonnet Formula to non-simple closed curves added 192 characters in body Apr 6 comment Generalize Gauss-Bonnet Formula to non-simple closed curves @LiviuNicolaescu: Yes, for non-simple closed curve, I means an immersed closed curve, it can have intersections and corners. Mar 31 asked Generalize Gauss-Bonnet Formula to non-simple closed curves Dec 9 awarded Caucus Nov 12 answered Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}$ Nov 10 answered A function is convex and concave, show that it has the form $f(x)=ax+b$ Nov 10 comment Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ @Parhs: See my edit. Nov 10 revised Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ added 218 characters in body Nov 10 answered Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ Nov 10 answered Solving Trigonometric Limits Nov 10 answered Existence of a sequence with prescribed limit and satisfying a certain inequality Nov 10 revised If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$ added 111 characters in body Nov 10 comment If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$ @Rono: Note that $f(x)=\sum_{i=1}^k[f(\frac{i}{k}x)-f(\frac{i-1}{k}x)]+f(0)$. Nov 10 answered How find this $\lim_{n\to\infty}\frac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}$ Nov 10 answered If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$