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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 $p+q\ne 0$.
Oct
18
comment The product of a uniformly continuous function and a bounded continuous function is uniformly continuous
You may start with $g\equiv 1$.
Oct
18
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
@EricAuld: I don't know, but if you are familiar with Jensen's inequality, then you can apply it to different convex functions to produce kinds of inequalities like the one you mentioned as you wish.
Oct
18
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
@EricAuld: You are welcome!
Oct
18
answered If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational
Oct
18
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:
@Sanchez: You are welcome and thank you for your help in advance!
Oct
17
revised Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
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Oct
17
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
The flaw is you didn't obtain any control of the integral of $\frac{1}{|f(\sigma)|}$ over $S^{n-1}$.
Oct
17
answered Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
Oct
17
comment Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$
By Fubini's theorem, the integral related to the GM will diverge when $\alpha_i\le n$ for some $i$.
Oct
17
reviewed Approve complete metric on a Riemann Surface
Oct
17
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:
Hi, Sanchez. Since the question is closely related to $T_n$ and since $T_n$ satisfies a recurrence relation $T_n(x)=2xT_{n-1}(x)-T_{n-2}(x)$, I wonder if you have considered the following statement. If $f$ is a polynomial with $\deg f=n\ge 2$, then there exist polynomials $g$ and $h$, such that $f(x)=2xg(x)-h(x)$, $\deg g\le n-1$, $\deg h\le n-2$ and $\|g\|,\|h\|\le \|f\|$. Here $\|f\|$ denotes the maximum modulus of $f$ on $[-1,1]$. I have no clue whether statement is true or not. Sorry for bothering.
Oct
16
revised Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$
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Oct
16
answered Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$
Oct
15
comment An abstract definition of the cotangent space to a smooth manifold.
What about GTM 94, Foundations of Differentiable Manifolds and Lie Groups, by Frank W. Warner?