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13:00
@BalarkaSen Em. Here is a very basic problem i was thinking about. Any idea how to show $|\sum_{i} \binom{n}{3i+1}-\binom{n}{3i}|\leq 1$??
Can someone tell about some resources to learn "smoothing" an inequality?
@BalarkaSen BTW, are you planning to join a undergrad or grad school somewhere?
@S.C. That's an interesting problem. Not sure.
@BalarkaSen No issues :) It's nice to see that you are learning sophisticated math at such an young age.
Thanks. I try to learn stuff.
13:07
@BalarkaSen Great. Are planning to pursue math?
Yeah, something like that.
@BalarkaSen Like where? Have you thought of any places?
CMI or ISI perhaps.
I am generally clueless about these things.
@BalarkaSen Yeah those 2 are the only good places in the country for undergrad math :)
So I heard.
13:10
@BalarkaSen Are you into olympiad math?
No, not really. I am generally bad at that stuff, and it also doesn't inspire me too much. I like to think about interesting problems occasionally though, but I am not into it officially.
@BalarkaSen But then these ISI/CMI questions are of those calibre. So to get into these places you need to somehow crack those exams :)
Yeah that's what I am worried about.
@BalarkaSen BTW, do you discuss such sophisticated math with your schoolmates? Pheww. I don't even udnerstand your questions even aften being a grad student myself :P
No, I don't; that'd be sort of jerkish to do. I think there's a lot of interesting things to talk about in high school math, so that's what I do.
I value solving problems more than knowing theory in general (I am not good at the former, but trying to get better).
13:16
@BalarkaSen I am neither good at those 2
@S.C. you can calculate $\sum_i \binom{n}{3i}$ by adding $1+(1+\omega)^n+(1+\omega^2)^n$
@SoumyoB Yeah where w is the cube root of unity. But isn't there any other way?
not that I know of
For $\mathbf{v} = \hat{e_{\phi}} = (-sin \phi, cos \phi, 0)$, what is meant by $v_{r}, v_{\theta}$ and $v_{\phi}$?
@SoumyoB and how do you calculate $\binom{n}{3i+1}$?
13:18
something along the lines of $1\cdot \omega^i+(1+\omega)^n\cdot\omega^j+(1+\omega^2)^n\cdot\omega^k $
It seems like that sum is $\sum \chi_i \binom{n}{i}$ where $\chi_i$ is 1 for i = 1 mod 3, -1 for 0 mod 3 and 0 for 2 mod 3
you're gonna have to find $i,j,k$ suitably
I can't arrange that into a binomial sum
the $\omega$ idea is interesting though
just tricks they taught me in high school for IIT JEE preparation
credit goes to them
13:19
don't wanna sound like a plagiarist
I can't prove this for all $n$ but there are values of $n$ for which \sum\binom{n}{3i+1} and $\sum\binom{n}{3i}$ are same. say like n=7
They follow from Pascal's identity. Not able to solve such problems makes me feel terribly unworthy of doing math :(
Sob sob :(
My solution for that crisis is to accept that I am bad at math and consider it as a layman's exploration. I think that's effective.
@BalarkaSen Thats a good talisman to have :)
@BalarkaSen BTW, i understand via your questions you have a good understanding of Galois theory? Would you mind teaching me? Helping me with the basics?
I learnt Galois theory at one point and liked it. I think I still remember a few things.
(I am more into topology.)
@BalarkaSen No issues. I am quite bad at Topo as well :) So it's ok can you help me with the basics then? Or is it too immature to ask?
13:27
Nah, it's never bad to ask for help. I think we can help each other in one way or another.
Feel free to ask questions :)
@BalarkaSen Thanks. How do i contact you then? Would you mind sharing your email?
Balarka = Rising sun it means in Samskrit :)
@S.C. I am rather more active in this chat than on my email. We can talk here! I think it's a great place to discuss things.
@BalarkaSen Okay! I will get back if i have any questions :)
Definitely!
Also yeah, that's what it means in Sanskrit.
@BalarkaSen In fact it should be balArka. the big "A"should be present in your name since we are combining (doing Sandhi) 2 words :)
13:31
Yeah but that would just sound strange
:P
Hehee :
)
13:45
So imagine I had a new number type. The idea is simple: It is an array (possibly with infinitely many elements) that you can do addition on it. For example, {1,2}+3 and {1,2}+{3,4} would be a valid thing. Note, however, that order is important, and that {1,2}+{3,4} is not equal to {3,4}+{1,2}. I call these numbers "ordinals".

I then have a function. It takes an ordinal and gives an output through some algorithm. I will be using this function:

1) f(0,n) = n
2) f(ß+1,n) = f(ß,n+1)
3) f(ß,n) = f(ß[n],n)
@Secret ^ That may interest you.
Ok, if I understood correctly,
1) is the final evaluation
2) drain the left argument to add onto the right
3) For any limit ordinal reached, pick the term in its fundemental sequence correspond to what is in the RHS

Therefore, 2) grows in increments of 1, and 3) will only grew quickly if the RHS is already big

Also the function is computable, since the ordinals decreases for each run through any of the 3 rules
Very brief look at it, it sounds like a + version of ackerman
Now what about f(ω,ω)?, then we got stuck
In fact, for any two limit ordinals $\alpha > \beta$, $f(\alpha,\beta)$ will get stuck
@Secret The function should be with one ordinal and one finite natural integer.
ok then, there will be no problems
@Secret That is the Hardy Hierarchy, and the fact that it starts out growing much slower than the fast growing hierarchy makes it easier to explain.
Ah I see
13:59
f(ω^ß,n) = fgh(ß,n) for ß < ε0
is the notation ß[n] related to the cofinal? @SimplyBeautifulArt
@LeakyNun Sure, its usually mentioned in the context of fundamental sequences.
14:44
hi guys
I know the following identity cosh(x) + sinh(x) = e^x
I was wondering if it is possible to use somehow the inverse hyperbolic functions to obtain log_e(x)
If $f: G \rightarrow H$ is a homorphism, $H$ is abelian and $N$ is a subgroup of $G$ containing $\ker f$, then $N$ is normal in $G$. Proof: Let $g \in G$ and $n \in N$. Then $f(gng^{-1}) = f(g) f(n)f(g^{-1}) = f(n)$ implies $f(n^{-1}gng^{-1}) = e$. Hence, $n^{-1}gng^{-1}$ is in $\ker f \subseteq N$. Therefore, by closure $n \cdot n^{-1}gng^{-1} = gng^{-1}$ is in $N$, proving that $gNg^{-1} = N$.
Does this sound right?
@user8469759 It's usually done in the other way around.
namely?
14:45
namely you define inverse hyperbolic functions in terms of ln(x)
ok, but the question is
cosh_-1(x) = log(x + \sqrt(x^2 - 1))
sinh^-1(x) = log(x - sqrt(x^2 - 1))
guys, would anyone know how to write this more neatly (the last paragraph) :
is there any combination of those two functions, such that I can get log(x) as result?
Let $(V,\mathbb K,\langle{.,.}\rangle)$ be an inner product space with basis $\gamma=\langle c_1,\dots,c_n\rangle$. Consider the nested subspaces $U_1\subset\dots\subset U_n$ induced by $\gamma$, where $U_k=\operatorname{span}\{c_1,\dots,c_k\}$. Then there exists an orthonormal basis $\beta$ that induces the same nested subspaces.

The proof starts as follows:

Define: $U_0=\{0\}$. Voor each $k\in\{1,\dots,n\}$, choose $b_k\in U_k$, such that $b_k\perp U_{k-1}$ and $\Vert b_k\Vert=1$.

This not equivalent to saying: chose $b_k\in U_k$, such $b_k=\lambda c_k$ (for some $\lambda\in\mathbb K$)
(sorry I misswrote the sinh^-1)
14:48
maybe I could write: $b_k'=\lambda c_k+b$ $(\lambda\neq 0)$, where $b$ is a linear combination of vectors in $U_{k-1}$, and $a\in U_k\setminus U_{k-1}$.
shit, I can't delete it.
I could have written: $c_k=c_k'+b$, where $c_k'\in U_k\setminus U_{k-1}$ and $b\in U_k$. And therefore $b_k=c_k-b=\mu_1c_1+\dots+\mu_kc_k$.
yea alright
@user193319 No time to read your proof, but perhaps math.stackexchange.com/questions/542264/… can help?
15:09
Evening.
15:23
Can somebody help me with math.stackexchange.com/questions/2253737/… . 10 views in almost a day $\ddot \frown$.
15:39
Hello!! We have 48 Cube. To find different Cuboids of different form, do we make some cubes next to each other?
How many cuboids can you make with 48 cubes?
15:55
I wonder how I can write these 3 mathematical structure more rigorously...?
(Elaboration: The mathematical structure I am concerning is an abstraction of the 3 boundaries you saw here)
Hmm... Let's see... So I have a region partitioned into two. I also have some particle or fluid that is in the region such that it is free to move about (or if not moving like a field, has some nonzero extent)
1. The leftmost case corresponding to some kind of one way membrane. I wonder how can this be modelled?
2. Is easy, I just impose a constant vector field that point to the right, and then define my boundary that partition the two regions
3. That's some kind of asymptotic boundary that can be crossed when special conditions are reached, not sure about this yet...
How are you doing cute user @ShaVuklia?
@Secret, are you done with some of my problems?
16:12
Nah, I got stuck with decompositing some of the zeta functions into products of zetas, I need to think about them more
It seems I cannot always do that
Trying to use the do nothing technnique and using the shffle identity of zetas (let me check the actual name of the identity again...) and I still end up with extra products of zeta terms thus not really simplifying the inequality
Yeah so basically...
$$3\zeta(2)\zeta(5)+\frac{3}{4}\zeta(3)\zeta(4)-6 \zeta(7)>0$$
Initially, I thought the zeta 2 and 5 and zetas 3 and 4 can be made into zeta 3 5 and zeta 3 5 via the reflection formula, so I can group those terms together
but it appears it just spawn more zeta term that cannot be grouped nicely, thus not a viable method
@Secret I didn't do much math lately.
Same, besides regular maths seminars in my uni (where we went through applying ergodic theory on number theory just today) and simpleart's growing function and ordinal stuff, I am mostly wrestling with my chemistry PhD bash coding in the computing cluster to get things to calculate
@Secret which zeta? Riemann?
16:20
yup
@Semiclassical hey mate
Though honestly, I am not sure if riemann zeta can handle ordinal arguments
@Secret In general my problems are meant to be solved in very simple ways. That simplicity of the problem put ppl in trouble.
The tricky thing with 'simple' problems is that it can be hard to find the right mindset.
4
16:22
Maybe.
The issue is that what seemed simple to most of us, i.e. symmetry, is either hard to came by without enough experience, or we thought there's some kind of symmetry in the problem, yet it isn't
Unless...
Plus, if the answer is "it's simple if you know the right integral representation" then I don't really count that as simple.
You dealt with a type of simplicity thought process which most of us not aware of, hence hard to came by except with a mindset similar to yours. (to be continued)
@MaryStar Look at triplets of factors of 48.
I have actually checked you question and answer history. From what I observed, your solutions make use of a lot of properties of polylogs and zetas
16:24
@TimTheEnchanter Ahh so every possible factorization of 48 into 3 numners are the possible dimensions, right?
But I suspect you have other methods in mind, that can transform the problem into something tractable, and that, I am not sure what it is
@Secret I have lots of problems with special functions, that's true.
@Secret Yes, I find no pleasure if I have a problem where others could think of my precise solution.
Most of the time I have some special approach (nowhere published).
which two smallest numbers?
@TimTheEnchanter Thank you!! :-)
16:28
@Waiting did you see the integral $2\pi=\int_0^{2\pi}e^{\cos(x)}\cos(x-\sin(x))\ dx$?
@MaryStar No problem
@SimplyBeautifulArt aha
I'd probably default to using the Jacobi-Anger expansion for that.
@Semiclassical there is an easier approach
Didn't say there wasn't :P
16:31
:-) you just need the right mindset
One reason I found the symmetries of infinite series nonintuitive is that they are hard to visualise (well, I am a rather visual person...). My current strategy is just to keep doing them and hopefully some patterns started to click
Main reason I'd use Jacobi-Anger, tbh, is because I'd want to know if it generalizes
e.g. with $e^{a \cos x}$ instead.
@Semiclassical it can generalize, again, using simple approaches
Paul Nahin has an integral like that (in his book) which he calculated by means of contour integration and that can be finalized using series.
Let me find it ...
16:33
I'd be surprised if generalizing it appropriately didn't convert it into some Bessel function.
Nope, it should end up real simple
$$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x}\textrm{d}x$$
Very easy.
(actually)
@Waiting is this a question or a challenge?
(or neither?)
Use the same idea needed in the previous question.
16:35
@LeakyNun It's related to the previous talk. If I challenged anyone here, no one would answer the question.
So I'm supposed to solve it?
It would be of no use to challenge anyone.
Nah, @SimplyBeautifulArt you showed it to me before.
16:36
@LeakyNun It's a question just for fun.
Hmm, it's a very bumpy function
I wonder if it has cylindrical symmetry in the complex plane?
(that will provide a way to solve it via contour integration, I guess...?)
Btw, one of the past results involving integrals I am most proud of deriving is this:
in SecretLabs (SE Branch), Mar 19 at 11:32, by Secret
General form of any exponential integrands with bounded derivatives $i.e. a^{(k)}(x)< \infty, \forall k \in \Bbb{N}$

$$\int a(x)e^{-nx}dx=-e^{-nx}\sum_{k=0}^{\infty}\frac{a^{(k)}(x)}{n^{k+1}}$$
However...
while the sin and cos case works, the series will become divergent. How that divergence get cancelled in this formula is still to be investigated
This formula will work for any $a(x)$ that is a polynomial, and thus explains 2 of the exponential integrals in the table of integrals
How could we compute the roots of 2x^3 - 4x^2 + 2=0 ?
@Waiting should be $\frac\pi2(e-1)$ or something
@MaryStar same way all cubic polynomials
@SimplyBeautifulArt $$\frac{\partial}{\partial p}(e^{p \cos(x)} \cos(p \sin(x)))= e^{p\cos(x)}\cos(x+p\sin(x))$$
@SimplyBeautifulArt Yeap
@SimplyBeautifulArt To we have to find the divisors of 2 and apply the euclidean algorithm?
@MaryStar $x=1$
@Waiting how'd you solved it?
@SimplyBeautifulArt Use $$\sum_{n=1}^{\infty} \frac{p^n \cos(nx)}{n!}$$ which is equal to the expression we differentiate above.
Hi everyone please help. Is continued fraction expansion (Simple) of any irrational unique? Can someone direct me to a proof?
16:51
2
Q: On the uniqueness of a simple continued fraction

LiebeExpanding an irrational number in form of a simple continued fraction each of the terms $a_i$ in $[a_0;a_1, a_2, a_3, \dots ]$ are 'uniquely' determined since the lowest integer less than or equal to some positive number is unique. But does the uniqueness hold in the reverse direction? That is, i...

[Random, possible ovrgeneralisation] Investigate the following:
$$\frac{dF(x)}{dx}=\sum_{i=1}^{\infty}a_if_i(x)$$
@Secret Thanks a lot.
Goal: If $F(x)$ is given, determine $a_i$ and $f_i$, if $a_i, f_i$ are given, determine $F(x)$
(NB This is not really a question, because it is uncountably big)
@Semiclassical Can please you help me with math.stackexchange.com/questions/2253737/… ?
I can point out some of what's going wrong, yes. (Not going to write an answer out right now due to time.)
16:55
Some approach(?) Consider the partial sum: $$\sum_{i=1}^{n}a_if_i(x)$$
and find patterns
First, E=4pi*sigma is only valid at the surface of a conductor. (I presume you've got k=1 here so that this is just sigma/epsilon0 in disguise.)
But the electric field of the spherical shell is defined throughout all of space. You have to consider this entire field in order to compute the total electrostatic energy correctly.
@Semiclassical Yes I have $k = 1$.
Do I need to integrate ?
from $0$ to $\infty$ like $\int^\infty_0 E \cdot E dv$ ?
well, dv is the differential volume element
16:58
Thank you, I will try that.
Yes I know it is.
i.e. it's dx dy dz in cartesian. so those aren't the right limits of integration
If you know it is then it doesn't make sense to say "from 0 to infinity"
I think one interesting thing we can ask is that, what is the partial inverse of arbitrary summation

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