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seen Apr 9 at 18:25

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 suggested edit on 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?
Oct
11
answered Functions satisfying $(b-a)f'(\tfrac{a+b}{2}) = f(b)- f(a)$
Oct
8
comment If $f(x)$ is positive and decreasing, can $xf(x)$ have more than one maxima?
An alternative example: $f(x)=1-x$ when $x\in[0,\frac{1}{3}]\cup [\frac{2}{3},1]$; $f(x)=\frac{2}{9x}$ when $x\in[\frac{1}{3},\frac{2}{3}]$.
Oct
8
comment Question about convolution
What if $f\equiv 1$ and $g$ is not integrable?
Oct
7
comment How prove this $f(a)\le f(b)$
@nanchangjian: Once again, you are welcome.
Oct
7
comment How prove this $f(a)\le f(b)$
@nanchangjian: You are welcome. By definition, $x_y$ is at least a lower bound of the set $E=\{x\in[a,b]: f_\epsilon(x)<y\}$, which means that if $x<x_y$, $f(x)\notin E$, i.e. $f(x)\ge y$. Then by continuity, $f(x_y)\ge y$(actually, since $x_y$ is the greatest lower bound, $f(x_y)= y$).
Oct
7
answered How prove this $f(a)\le f(b)$
Oct
7
comment How to prove that there exists $g(x)$ such $\int_{0}^{1}g(x)dx\ge\frac{1}{2}\int_{0}^{1}f(x)dx$
@BettyMock: You are welcome.
Oct
6
revised Harmonic Function bounded by a linear function
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Oct
5
revised How to prove that there exists $g(x)$ such $\int_{0}^{1}g(x)dx\ge\frac{1}{2}\int_{0}^{1}f(x)dx$
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