Alfred Chern
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 Apr6 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! Apr6 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! Apr6 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? Apr6 revised Generalize Gauss-Bonnet Formula to non-simple closed curves added 192 characters in body Apr6 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. Mar31 asked Generalize Gauss-Bonnet Formula to non-simple closed curves Dec28 revised The functional $\int_0^{2\pi}\frac{\sqrt{1-\varphi^2-(\varphi')^2}}{1-\varphi^2}d\theta$ added 377 characters in body Dec28 asked The functional $\int_0^{2\pi}\frac{\sqrt{1-\varphi^2-(\varphi')^2}}{1-\varphi^2}d\theta$ Dec9 awarded Caucus Nov12 answered Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}$ Nov10 answered A function is convex and concave, show that it has the form $f(x)=ax+b$ Nov10 comment Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ @Parhs: See my edit. Nov10 revised Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ added 218 characters in body Nov10 answered Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ Nov10 answered Solving Trigonometric Limits Nov10 answered Existence of a sequence with prescribed limit and satisfying a certain inequality Nov10 revised If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$ added 111 characters in body Nov10 comment If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$ @Rono: Note that $f(x)=\sum_{i=1}^k[f(\frac{i}{k}x)-f(\frac{i-1}{k}x)]+f(0)$. Nov10 answered How find this $\lim_{n\to\infty}\frac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}$ Nov10 answered If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$