If $X$ is a metric space and $A$ is a subset of $X$, is the function $\rho(p)$, where $p\in X$ so well-known that $\rho $ need not be defined? My only guess is that $\rho(p)=1$ when $p\in A$ and $\rho(p)=0$ otherwise.
@TheSubstitute I would use $\chi_A$ for the characteristic aka indicator function. I would also explicitly call it that, and not just use the symbol without defining it. I don't think I've ever seen the letter $\rho$ used for the characteristic function.
@Alyosha the splitting field of a family of polynomials is obtained by adjoining all of their roots. adjoining roots yields an algebraic extension (no matter how many you adjoined).
@skullpatrol The only controversial thing I noticed was that she was discussing what seemed to be religious beliefs with some others. That can be messy on the internet with a lot of varied beliefs.
@VibhavPant oh! there is a pattern ! .. my advice is to go through the olympiad books they recommended ! .. 80% of the questions are framed keeping olympiad problems in mind !
@BalarkaSen nope .. its testing how fast you can write stuff down in exam hall .. there should be 14 subgrps to the naked eyes, proving that its actually 13 is slightly trickey business (especially when the prof hasn't proved the required lemmas in class .. so that one has to write it down in exam hall .. which IMHO is 'sad' )
$$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}\sum_{j=1}^{i} \frac{(-1)^{j+1}}{j}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{j+1}}{j}\right)^2$$ and also this form $$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}\sum_{j=1}^{i} \frac{(-1)^{j+1}}{j}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{j+1}}{j}\right)^3$$
If it could be proven that for certain sequences of complex numbers $A_n$ and $a_n$: $$\prod _{n=2}^{\infty} \frac{1}{1-\frac{1}{A_n}} = \lim_{n\to \infty } \, \frac{A_n}{a_n} $$
@robjohn $\sqrt{1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}} = l_n$, then $\sqrt{n}(l - l_n)^{1/n}$ has a nice closed form :D .. where $l = \lim\limits_{n \to \infty} l_n = l$ (problem by @Chris'ssis :-) )
@r9m Have you ever seen this one? $$\lim_{ n\to \infty} \sqrt{n}\left(l-\sqrt{ 1+ \sqrt{2 + \cdots +\sqrt{n}}}\right)^{1/n}$$ where $l$ is the limit of the nested radical.
@r9m Looking at the derivative of $\sqrt{1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}}$ I get $$\lim_{n\to\infty}\sqrt{n}\left(\frac1{2^n\sqrt{n!}}\right)^{1/n}=\frac{\sqrt{e}}{2}$$
@r9m The next term to be added to the $n$ will be between $\sqrt{n+1}$ and $\sqrt{n+1}+1$ which when multiplied by the derivative and the $n^\text{th}$ root is taken is insignificant.
If I had obtained the job I applied for, my time for doing this kind of activities would have been dramatically reduced. At any rate, sooner or later, my time won't permit doing this, and a blog really requires a lot of time.
(that's why I said it's an idea to use somewhere in the future)
I've come across the term "pairwise distinct" in many research papers. But, I don't understand how it differs from just saying that the elements of a set are unique instead of saying that they are pairwise distinct.
Can someone please explain the difference, if any, to me?
@Alizter By the way, I also created $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{n+1}}{n}\right)^2= \frac{3}{2}\zeta(2)\log(2) + \frac{1}{3}\log^3(2) - \frac{1}{2}\zeta(3)$$
@robjohn I'm inclined to believe we can make use of $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{n+1}}{n}\right)$$ to elementarily compute $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} H_n}{n}$$
I've come across the term "pairwise distinct" in many research papers. But, I don't understand how it differs from just saying that the elements of a set are unique instead of saying that they are pairwise distinct.
Can someone please explain the difference, if any, to me?
