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seen Oct 8 at 8:44

Oct
22
answered A Problem on Improper Integrals
Oct
22
comment For a closed plane curve, showing some inequalities.
@JeongNam-ho: Yes, $\langle,\rangle$ denotes inner product. I thought it was a conventional notation, so I didn't mention it. Sorry about the confusion. For the remark, I think it's not very hard to prove, but it need quite a few words to write down all the details.
Oct
22
comment For a closed plane curve, showing some inequalities.
@JeongNam-ho: It's just the linearity of integral. To see this more clearly, you may write write $v=(v_1,v_2)$, $\gamma'(t)=(x'(t),y'(t))$, and evaluate both sides of the equality.
Oct
22
revised For a closed plane curve, showing some inequalities.
added 1495 characters in body
Oct
22
answered For a closed plane curve, showing some inequalities.
Oct
21
comment Computing a limit almost surely using the strong law of large numbers
@Shanks: You are welcome! :)
Oct
21
revised Computing a limit almost surely using the strong law of large numbers
added 17 characters in body
Oct
21
answered Computing a limit almost surely using the strong law of large numbers
Oct
21
revised convergence of total variation measure
added 512 characters in body
Oct
21
answered convergence of total variation measure
Oct
20
comment Could someone explain chirality from a group theory point of view?
You are welcome and thank you for your understanding.
Oct
20
revised Order of Double Coset
deleted 175 characters in body
Oct
20
answered Order of Double Coset
Oct
20
comment Is $x_1^{\alpha_1} + \dotsb + x_n^{\alpha_n}\geq x_1^{h/n}\dotsb x_n^{h/n}$ an example of power means?
Compare $\lambda_1y_1+\cdots+\lambda_ny_n\ge \left(y_1^{\lambda_1}\cdots y_n^{\lambda_n}\right)^r$ with the inequality in my last comment. Then we can find the following correspondence: $\frac{h}{n\alpha_i}\leftrightarrow \lambda_i$ and $x_i^{\alpha_i}\leftrightarrow y_i$.
Oct
20
comment Is $x_1^{\alpha_1} + \dotsb + x_n^{\alpha_n}\geq x_1^{h/n}\dotsb x_n^{h/n}$ an example of power means?
It might be noticed that the inequality in your second paragraph can be replaced by a more precise one, $\frac{h}{n}(\frac{x_1^{\alpha_1}}{\alpha_1} + \dotsb + \frac{x_n^{\alpha_n}}{\alpha_n})\geq x_1^{h/n}\dotsb x_n^{h/n}$, which is equivalent to the inequality between weighted means.
Oct
19
comment Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
Thank you. If I cannot figure it out in a couple of days, I will consider to post an question later.
Oct
19
comment Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
You are welcome. I should thank you for solving this problem, because it puzzled me for a few days. By the way, are you interested in the question I asked in a comment to Sanchez's answer?
Oct
19
comment Number of sigma algebras for set with 4 elements
Doesn't the answer in the linked question solve your problem?
Oct
19
comment Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
(+1) Nice answer! It seems that you only used the fact that $|P|\le 1$ at the extremal points of the Chebyshev polynomial, which is much weaker than $|P|\le 1$ on $[-1,1]$; maybe you could highlight it. By the way, there are some typos, such as the coefficient of Lagrange interpolation $B$ in $(10)$, and the missing summation in $(14')$.
Oct
18
comment If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational
As reminded by Vedran Šego and wilsonw in their comments, this answer is valid under the assumption that $p+q\ne 0$.