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location Beijing, China
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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 $|f(x)|\leq a|x|+b$
Nov
9
awarded  Nice Answer
Nov
9
revised Simplfiying $x(5xy+2x-1)=y(5xy+2y-1)$
added 31 characters in body
Nov
9
answered Simplfiying $x(5xy+2x-1)=y(5xy+2y-1)$
Nov
9
answered Proving inequality to be true
Nov
9
revised Prove that a positive polynomial function can be written as the squares of two polynomial functions
added 1 character in body
Nov
9
comment Case of convex function
@markich: $\mathbb{D}^n(1)=\{x||x|\leq1\}$, the unit disc.
Nov
9
answered $\forall v \in \mathbb R^n , v^tAv \ge 0 $ then $Au=\theta$ iff $A^tu=\theta$?
Nov
9
comment $\forall v \in \mathbb R^n , v^tAv \ge 0 $ then $Au=\theta$ iff $A^tu=\theta$?
Should it be $Au=\theta$ iff $A^tu=-\theta$?