Reputation
12,494
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
15 42
Newest
 Enlightened
Impact
~65k people reached

Jun
14
awarded  Enlightened
Jun
14
awarded  Nice Answer
May
7
answered $\log|x|\in\text{BMO}(\mathbb R^n)$
May
2
comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$?
I have noticed that Theorem 3 in sos440's answer to the original post about this topic has solved the problem in my last comment. Thank you!
May
2
accepted Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$?
May
2
comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$?
Dear David Speyer, thank you for your excellent answer! I still have one question. The displayed equality in your answer also seems to hold for some $f$ that fails to be holomorphic somewhere, say $f(z)=\frac{1}{1+|z|^\alpha}$ for real $\alpha>2$. Could such $f$ be covered as a simple corollary of your result?
May
1
awarded  Enlightened
Apr
30
awarded  Nice Answer
Apr
29
comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$?
@DavidSpeyer: Thank you for your comments. I agree with your observation. I think in fact the statement for general positive real $a_k$ can be deduced from the statement for integers. First, by change of variables for $x$, the statement holds for rational $a_k$; second, use rationals to approximate reals.
Apr
29
comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$?
@SamratMukhopadhyay: Thank you, but I have no idea how to apply argument principle for this problem.
Apr
29
comment On the inequality $ \int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$
@DavidSpeyer: I have posted the problem. Please refer to the linked question to this post.
Apr
29
asked Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$?
Apr
29
comment On the inequality $ \int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$
@DavidSpeyer: You are welcome. Yes, I also got stuck in showing the independence of $x_k$. It seems to work for many $f$'s; for example $f(t)=\frac{t^{2m}}{1+t^{2n}}$, $1\le m\le n$ are integers. Would you mind if I post a new question for this problem?
Apr
29
comment On the inequality $ \int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$
@DavidSpeyer: The statement in your last comment is equivalent to $\int f(\sum_{k=1}^n\frac{1}{x-x_k})dx=n\cdot\int f(\frac{1}{x})dx$ for $f(t)=\frac{t^2}{1+t^2}$. It seems that the same equality holds for more general $f$, but I am completely clueless about the reason.
Apr
28
answered On the inequality $ \int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$
Jan
28
comment Order of Double Coset
@Sushil: The natural 1-1 and onto map is $axb\leftrightarrow axbx^{-1}$ for $a\in A$ and $b\in B$.
Jan
6
comment Annoying Polynomial Inequality
@SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial.
Dec
27
comment How to show that $f'(x)<2f(x)$
@Hans: Sorry, I don't know any other similar question, and after some failed attempts, I have to admit that I cannot explain my motivation of the "by (1)" step better than the post itself suggests.
Dec
25
comment Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear?
@VictorWang: You are welcome.
Dec
25
comment Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear?
@VictorWang: Sorry, I just read your comment because it is very long time since the last time I visited this website. Please note that since $\varphi(\Omega)\subset \Omega$, $\varphi^k$ is a well-defined holomorphic function on $\Omega$ for every $k$. Then from the choice of $r$ we know the radius of convergence of the Taylor expansion of $\varphi^k$ around $z_0$ is no less than $r$.