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12:21 AM
@anon: I don't know that stuff, but I think you're going to have to do more solid reading than just Wiki pages. You might start with MathOverflow and Robert Bryant (my good friend who's an expert on all things exceptional and holonomy). For example, there's this as a starting point.
Bryant's Park City lectures on Lie groups might have stuff for you, for example.
3
 
MGA
Is it true that if v_1,...,v_m are linearly independent and w is not equal to any of the v's, then v_1 - w, ... , v_m - w are also linearly independent?
 
@AlecTeal: Yes, I've done it by hand several times. The shape generated by a rotating cube appears as an exercise in three of my books.
 
@Ted the MO thread seems to be about exceptional groups, not exceptional isomorphisms, but the answer there looks fun.
 
Right, @anon, it's not exactly what you want, but Bryant is a fabulous source for all such questions.
 
@MGA in 3D, pick three noncollinear points from an affine plane not containing 0, then translate them until they're in a 2D subspace
 
MGA
12:29 AM
@anon Awesome, thanks!
 
hey actually @anon
@anon
 
one @anon will suffice
 
For someone who wishes to remain anon, even one anon is one too many.
 
Our prof said something weird today. That the integers are not a subset of rationals because in rationals you can represent the same number with different ways(equivalence classes)
 
Hi @UserX !! Sounds like a lie to me.
 
12:32 AM
I actually found that there is at least a sequence of successor of length 2
for example
suppose X is finite
consider P(X)
 
It made a little sense, but then he said that N isn't a subset of Z either which sounded a lot more weird
 
Hello Professor emeritus @TedShifrin :-)
 
consider all of the topologies of X that is less than than P(X) and it is bigger than any other topologies
 
Maybe I'm misunderstanding but this goes exactly opposite of what I've been studying about sets and their constructions
 
Well, perhaps he's being 6000% pedantic, @UserX. There are isomorphic copies that are subsets. I don't like it.
 
12:33 AM
one could equally well say natural numbers are not a subset of integers because you can represent the same integer different ways (we use equivalence classes to construct negative integers too)
 
ha ha @skull
 
take $T_0$ in this topology to be one of them
 
That's exactly what he said next @anon
 
and $T_1$ to be P(X)
 
So, @UserX, even with equivalence classes, the equivalence class of $(n,1)$ is a well-defined rational number, and it happens to be an integer.
 
12:34 AM
If X is finite
 
Hey @TedShifrin
 
what is the topologies that are less than P(X) in the lattice of topologies ?
@anon?
 
the fact that two different ordered pairs of integers can represent the same rational number is not really a reason for saying ordered pairs of integers are technically not the same kind of thing as integers.
 
maybe also @TedShifrin you would like to discuss such thing
 
@KarimMansour all topologies are finer than P(X) (other than P(X) itself)
 
12:35 AM
Not I, @Karim.
 
you mean coarser
 
@anon is it not a good enough reason if you're nitpicking enough?
 
@KarimMansour no, I mean finer
 
How can you be finer than the discrete topology?
 
finer means has more open sets, Karim
 
12:37 AM
well, in my book $\tau_1$ is finer than $\tau_2$ iff $\tau_2 \subset \tau_1$
 
I'm totally confuzled.
 
@TedShifrin wait what? If a set is A is isomorphic to a set B and set A is a subset of a set X, doesn't that imply that set B is also a subset of set X?
 
@UserX the fact that equivalence classes have different representatives is irrelevant.
 
That's backwards, @Karim. But I don't understand @anon, either.
 
@UserX do you think Q is a subset of Z?
 
12:38 AM
I mean the other way
 
No, @UserX, but in my mind I identify $B$ with $A$ and I don't make a huge deal out of it.
 
@anon Is it a trick question? :P I've heard a lot of counter intuitive stuff today
 
oops, I did mean coarser originally, then mentally switched to thinking about finer somewhere along the way
 
reprimands @MikeM for his appearance
 
sorry @Karim
 
12:39 AM
Hi @MikeMiller
 
@UserX it's a question designed to make you wonder if you really do believe what you said
 
@anon Generically I know precisely about $n \leq 4$. Robert Bryant knows everything about interesting/exceptional Lie groups so you might hunt down MO answers he's posted, or otherwise I'm sure this has been covered on MSE. PS, since I said Robert Bryant, Spin(7) is apparently interesting for reasons I don't understand. (It's one of the rather few possible holonomy groups.)
@TedShifrin: I was requested, approximately.
 
LOL
I already wrote paragraphs about Robert :)
 
Same suggestion? Cool. You would know better than I, I've just seen his MO answers and know he's originally famous for G_2.
 
its okay. I want to know something because I want to use it in my proof is there always topologies in the lattice of topologies that is less than $P(X)$ but bigger than any other topologies ?
bigger in the sense of finer
 
12:41 AM
I also linked to his Park City lectures on Lie groups, etc.
No, @Karim, the discrete topology is the maximal element.
 
Haven't seen those, might be worth putting in the bookmark graveyard.
 
but bigger than any other topologies other than P(X) @TedShifrin
that is
 
I had a huge folder of a ton of Robert's preprints and reprints, which, of course, I had to throw out in the retirement festivities.
yes, @Karim.
 
As an ignoramus I don't really know what Spin(5) or Spin(6) are good for.
 
hahhahaha ted I know how to do it
hhahaha so cool
 
12:43 AM
@Anon what set is Q isomorphic to that is a subset of N?
 
presumably they show up in physics somehow, @MikeM.
 
@anon you will like this too
 
@UserX any infinite subset of N
 
How's the air refresher working? @TedShifrin
 
@TedShifrin did you see my question that I posted before
????
 
12:44 AM
I still smell smoke, @skill, but it's helping, I think. I got a cold — dammit — and so I feel crummy.
 
@TedShifrin: Well, Spin(3) and Spin(4) do, but since when has something physical been 5 or 6 dimensional? :)
 
That was meant to be Z but still yea
 
State space, @MikeM.
 
OK, I stand by 5.
 
LOL ... some sort of contact reduction.
 
12:45 AM
Yeah, was about to say. Boo.
 
LOL ... struts for the moment
 
I'm not supposed to be out of the salt mines yet. See ya.
 
Bye
 
@TedShifrin your system is going to have to build up a whole new defence against California viruses :-)
 
@anon But Z is cyclic. It's not isomorphic to Q
 
12:47 AM
well, I had to take my car to the shop and the one day I walked 2 miles home, it was raining .... I think that got it started, @skull.
We don't mean isomorphic as groups, @UserX. Only as sets.
 
10 mins ago, by UserX
@TedShifrin wait what? If a set is A is isomorphic to a set B and set A is a subset of a set X, doesn't that imply that set B is also a subset of set X?
you said sets, not groups
 
@anon my claim is the following that there is always atleast a sequence of length 2 of unique successors and @TedShifrin here is how I construct it
suppose X is any set
it could be countable or uncountable
 
or finite
 
Ohhhhh
 
12:48 AM
although I suppose countable could include finite
 
yeha
Now let ${x_\alpha}_{\alpha \in I}$ be singletons elements that enumerate X. consider $T_0 = \{\phi,X\}$, fix a specific $\lamda \in I$ , let $T_1$ be the topology generated by the subbasis $T_0 U {x_\lamda}$ and we keep doing this process repeatadly until we hit the P(X)
now
The topology generated before the powerset
 
What does that mean?
 
The power set most likely does not have an immediate successor, let alone a unique one (even in the finite case).
 
I will give you an example
take X = {a,b}
$\tau_{1} = \{X,\phi,\{a\}\}$
add b to this $\tau_{1}$
and consider the topology generated by that set
it will be P(X)
 
12:54 AM
So you're assuming some sort of listing (enumeration) of all of $X$, which is not necessarily possible.
 
yeah
I am adding singletons to topologies and move on
 
you will need some sort of transfinite induction, and I stand by my comments.
 
yeah that is why I used
I to be some index set
 
I is just another name for X.
 
Hi @TedShifrin
 
12:58 AM
heya mr eyeglasses; how be you?
 
@TedShifrin Kind of terrible because I'm doing so poorly in my classes
How about you
 
well, mr eyeglasses, I'm sorry for that, but maybe it's time you realized that grad school in math is not for everyone.
I have a cold, so I feel like ****.
 
Awww
 
if I can help on a question or two, let me know, mr eyeglasses
 
1:01 AM
Why are there so few proofs from geometry on standardized tests @TedShifrin
 
how do you expect them to test proofs on multiple choice, @skill?
besides, they've all but 100% disappeared from the US curriculum
 
A semantics question; We can follow these paths in undergrad; Analysis-Pure math-Geometry-applied math which is vague by itself but anyway where does number theory fall under?
 
that makes no sense to me, @UserX
 
Me neither. All of these are somewhat connected
 
hey @TedShifrin or @anon I want to prove that for finite case that there is a sequence of length two for successor as I am seeing a list of topologies I am noticing something quite picular
 
1:06 AM
number theory falls mostly under algebra at the undergrad level ... so pure math, I guess, although cryptography/computational number theory is applied.
 
that topologies even the ones that aren't comptaible
always have the same size
is that true ?
 
What does number theory have to do with algebraic structures other than working in some of them?
 
I doubt that seriously, @Karim.
You use ring theory, group theory, etc., in number theory, @UserX.
 
that is that can you a topology of cardanility n aside from the P(X) that is unique for the topologies on a set X
?
that is
there doesn't exist a topology that have same cardinal
 
Rewrite that sentence, @Karim.
 
1:08 AM
ok I will
 
I'm definitely not that advanced in NT
 
some schools teach a totally elementary number theory course, but it's better if you know some algebra, @UserX.
 
I thought NT was about finding relationships between primes
 
It's about all sorts of things.
 
Suppose you have a set X and now suppose $\tau_1$ is a topology that is neither the discrete or the indiscrete topology, then does there always exist a topology $\tau_2$ such that $|\tau_1| = |\tau_2|$ ?
@TedShifrin ?
 
1:09 AM
Are you assuming finiteness, @Karim?
 
Well, then take the action of the permutation group on the set of topologies :P
 
coz I have a idea of using group action of $S_n$
yeah exactly
 
good idea
 
I was gonna prove that there is only a sequence of length two on finite topology that by using action of $S_n$ on the topology
yeah for finite case there should be only sequence of successor of length 2
 
1:12 AM
ok, dinnertime for me ...
g'night all
 
Later
 
g`night @TedShifrin
I will be thinking of this problem
I need to take a shower though
I am stinky
 
2:09 AM
Howdy.
Ya know, math is pretty awesome. You just gotta look at it from the right angle.
 
@Pies omg
ಠ_ಠ
 
:D
Well then again, it can be a triggy subject...but no one can deny its absolute value in life
 
ಠ_ಠ
 
Anyway...
"Hello everyone, my interests are mathematics and"
Erm...ya finish that sentence or...?
 
Yes
 
2:23 AM
So, ya in high school?
 
No, college
I should be in high school though
 
Freshman?
 
Senior
 
.-.
So then how old are ya?
 
I'm too dumb to be a college student
 
2:24 AM
Well what're ya majoring in?
 
26
Math
I'm failing pretty miserably though
 
That sounds fun though...
You get to learn about the Universe
 
It's fun when you can understand the material
 
At least you get to learn. It's my junior year in high school and my calc teacher just left a week ago
Now we don't know if we'll learn it
 
I'm not even learning anything
 
2:25 AM
?
 
I'm trying to learn but I'm not smart enough to learn anything :(
 
Ah come on. Ya just gotta have confidence
Maybe a tutor?
 
I'm too poor to afford a tutor
 
Ya got a job?
 
No
 
2:28 AM
Try to fit one into your shedule
 
Last time I tried to work while studying I failed all my classes
 
Hmm...ya tried working in a study group?
 
No but I'm trying to make friends
I haven't spoken to any classmates yet and I promised myself I would before I graduate
 
Work with someone in your class. You'll help each other out
That's what I did and my grades went from Cs to As
 
Well I don't care about my grades, I just want to know the material properly
I got all As in my classes but the classes here are bad so I didn't really learn anything
 
2:31 AM
Better grades will help ya out a bit later. But groups will definitely help with the latter
Anyway, it's almost 2:00am. Gotta go, 'night
 
Night
I wish I was born in Europe instead
 
Why?
 
Grass is greener
 
On the other side of the ...
 
world
 
3:05 AM
and it's a big world out there.
 
3:15 AM
@anon in few hrs I will present a proof atleast for the finite case that its only possible to have a list of sequence of two successors only
I like my proof alot so far
 
 
1 hour later…
Huy
4:44 AM
@morphic: You can compensate by asking many questions here!
 
5:28 AM
@Huy
how r u doing
 
Huy
hungry, haven't had breakfast yet
u?
 
I am so happy btw I proved that for finite set X we can only have a sequence of of length 2 for the successors
and their is only one way to obtain it
 
Huy
5:50 AM
cool
what are you studying atm, @Karim. still munkres?
 
Huy
I borrowed it when I was studying topology because I didn't want to pay at that time. Now I want to have the book myself but now it's 20 bucks more expensive.
As soon as I'm working 100% I'll add so many great textbooks to the few I have already.
 
6:09 AM
why point set topology
I can't wait until I finish point set topology to study algebraic topology
 
Huy
to improve my foundations
 
yeah your right
 
Huy
I forget stuff I don't use on a regular basis very quickly
and I get sad when I realize that so I want to revise
 
yeah
I wanted to ask a question
 
Huy
what is that?
 
6:12 AM
if we take maybe the standard topology
and union that with a singeleton
do we get a topology that is finer than standard topology?
 
Huy
so is the singleton open or?
sorry I just woke up maybe I'm being slow
 
that is consider B1 to the the basis of the standard topology
take $B_1 U {2}$
will it be finer than $B_1$?
 
Huy
intuitively I'd say no
you mean strictly, right?
 
 
2 hours later…
8:08 AM
@AlecTeal that reality is really annoying, do you think im satisfied of the fact that a 15 min answer in reverse engineering brought me more points than a 1 week long answer ????? folks just like visual
 
$\sum_{n=1}^{\infty}(n^3)$ diverges right?
 
@Rememberme Yes, of course
 
of course
to infinity
 
@Tobias So is it true that $\sum_{n=1}^\infty(n^3) \neq (\sum_{n=1}^\infty (n))^2$
 
@Rememberme Neither side is defined
 
8:17 AM
So it is not true.
 
they both diverge to infinity
but not at the same speed
 
So they cannot be said to be equal?
 
woops, actually at the same speed
both are polynomials of degree 4 in $n$
well, with proper interpretations to make things actually make sense
 
In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice: n3 = n × n × n. It is also the number multiplied by its square: n3 = n × n2. This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the one-third power. Both cube and cube root are odd functions: (−n)3 = −(n3). The cube of a number or any other mathematical...
 
@Agawa001 ??
 
8:19 AM
yes same speed i was late :p
 
I know it is true when taken sum to a finite value is it true when sum is taken to infinity ?
 
@Rememberme that question does not make sense
 
Well I mean to say that $\sum_{n=1}^N{n^3}=(\sum_{n=1}^N{n})^2$ but is this true when the sum goes to infinity@Tobias
 
@Rememberme The sum does not "go to infinity"
 
I have to go for lunch that is why I am using such weird terms . Is this true
$\sum_{n=1}^\infty(n^3) \neq (\sum_{n=1}^\infty (n))^2$
 
8:23 AM
Unless you mean as formal power series (I am fairly sure you are not familiar with those). And for formal power series I think it might be true, but I would have to do the calculation
@Rememberme they are either trivially equal, or you need these to be formal power series to be able to compare them
 
8:35 AM
^ i doubt really if this pile of lines have already been posted
thats same as saying, $lim_{x->\infty} (x+1)^2 \neq lim_{x->\infty} (x^2+2x+1)$
 
8:57 AM
@Agawa001 Well, I think the question is better asked about formal power series, but I also think that is beyond the level of @Rememberme at the moment
 
@TobiasKildetoft i think not, he had the craving of posting something here and go. he could elaborate in less sophisticate examples, asking if $(\infty)^2 == \infty$
 
@Agawa001 Right, in the stated form, the question is either meaningless or trivial. But as a question about formal power series I think it is a fine question (which I haven't checked yet)
 
@TobiasKildetoft iv checked it before, that sum of powers found by bernoulli numbers.
$\sum(n^3)$ is just a case
 
@Agawa001 There are no bernoulli numbers in that power series (just $1$s)
 
9:17 AM
i was reading this starred post by skillpatrol, lol, tbh, the lady has right to distrust the vaccination
 
@Agawa001 What? Please don't start spouting anti-vax stuff here
 
More importantly @Agawa001, we should take pause before making anything mandatory in such an extreme regard. Strongly recommended? Sure. Ostracizing people in the form of restricting access to public property? Sure. But actually giving people a very real choice between getting vaccinated or getting executed is ridiculous.
 
hi @KajHansen, @TobiasKildetoft
 
@BalarkaSen Hi
 
Pretty much every problem that has ever arisen in policymaking has found its inception when short-term good, excitement, and zealousy have been allowed to overshadow, obfuscate, and make harder to predict long-term disaster.
 
9:30 AM
@KajHansen i gave the lady the right to abstain, nothing more to tell here, its her own blood, and her own children. i know people who do so, like my neighbor doctor he never let her daughter take that school-vaccination. im not intimidating people, but any such medical procedure souldnt be obligatory, thats all, people have right to choose independently of their level of knowledge or "not being allowed science" or "allowed citizen" or anything like this, back to my maths now.
 
That's what I'm saying @Agawa001
Refusing someone participation in as intrinsically human of an activity as democracy, science, or mathematics is a very extreme and cruel move.
 
@Agawa001 That is a terrible argument. People refusing vaccines are not only hurting their own children but also potentially immuno-compromised children whose parents would have liked nothing better than for their children to be vaccinated.
(also, since when is it ok to do something that hurts your own children, just because they are yours?)
 
Anyone with the proper command of the language they speak can be arbitrarily powerfully persuasive if only they worked hard enough. We should strive for the betterment of society via good rhetorical skills and peaceful persuasion rather than the force of law, in the sense that @Agawa speaks (forcing procedures upon people)
 
Just a warning @TobiasKildetoft:
in The 2nd Monitor, Oct 2 at 13:14, by rolfl
He is a known troll, a pretty good one, and if you engage with him you get all the crap and frustration you deserve
 
Another thing that would go a LONG ways is for the government to stop doing shit that makes people distrust it. The CIA and related organizations have a long history of experimenting on people, and this is definitely not just relegated to the conspiracy theories.
 
9:37 AM
@JoshuaA which of them?
 
@TobiasKildetoft That message is in regard to Agawa
 
Cry wolf too many times, and people stop trusting you. In the same way, it's now harder for the government to convince people vaccines are safe, and that is not something these skeptics should be blamed for. That's an entirely rational response.
 
@JoshuaA I see
@KajHansen There is nothing rational about ignoring all scientific studies
 
I know this @TobiasKildetoft. However, when a largely distrustful and dishonest organization is the one doing a lot of the salesmanship for something that is as good as vaccines, therein lies the problem.
 
@KajHansen which organization is this?
 
9:41 AM
The government @Tobias. The US government has a long history of ghastly, unethical human experimentation that has been done as recently as the 80's, and probably since then as we could find out when more recent documents become declassified in the future.
 
@KajHansen But the recommendation is coming from multiple sides, most of them unrelated to the government (and please, let's not make "the government" into a single uniform entity here).
but anyway, this is getting way off topic
 
Consider, for example @TobiasKildetoft, that the US Public Health Service sponsored infecting minorities with syphilis at Tuskegee university as recently as 1970 under the guise of vaccinations.
Anyways, I have a serious amount of cryptography to get done over the next 4 hours.
 
We have that t, n, b consist the orthonormal basis of R^3. We know that the vector a is perpendicular to n, and the angle between a and t is theta. How can we write a using the orthonormal basis? Which are the coefficients?
 
As a final note, my goal here was not necessarily to defend anti-vaxxers, but rather I think it is immeasurably valuable to understand where misguided opinions are coming from. Understanding why your opponent feels a certain way is the first step towards actually effecting change within people.
 
@KajHansen Can you tell me what a Berkovich space is?
 
9:48 AM
@MaryStar So $a$ is in the plane spanned by $t$ and $b$ and you know the angle to $t$. This does not tell you $a$ completely
it only gives you the direction of $a$, not the length
 
I have been planning to have a discussion with someone about that.
 
@MaryStar, I don't have time to explain in detail, but take dot products to yield a system of two equations that your vector coordinates of $a$ must satisfy, recalling the relationship between the dot product of two vectors and their lengths / angle between.
@BalarkaSen, absolutely, but not when I have crypto due so soon. I'm trying to not be distracted as I am with chat right now :P
 
ah, sure. let me know when you get time.
I have exams ahead of me too, I am just procrastinating.
 
@BalarkaSen, one consideration of many I've thought about recently: math.stackexchange.com/questions/1470083/…
 
The reason I am interested in Berkovich spaces is two fold : (1) I have heard that one can do ergodic theory with them in the utmost arithmetic sense (2) I have heard they are some kind of p-adic analytic manifolds or a generalization of them, and I have been wondering about whether one can have actions of Gal(\bar Q_p/Q_p) over those
 
9:52 AM
oh please im off for a couple of days
 
@KajHansen I'm having a look.
Interesting question, @KajHansen. Maybe there's some intrinsic property of the conics that make them appear frequently.
 
 
2 hours later…
11:38 AM
@Chris'ssistheartist are you here? I summon you
 
Huy
12:26 PM
@Hippalectryon: did you ever start looking at the book on functional analysis I sent you yet ?
 
1:01 PM
@UserX Yes, now.
 
1:54 PM
J.M.? Is that you?? You're back!!! :-D @J.M.isback.
Welcome.
 
I'm way better looking than Gabe Kaplan. :P
 
@Chris'ssistheartist any faster ways for $\int_0^1 \frac{x^2}{1+x^4}dx$ other than the classic computations?
@Chris'ssistheartist any faster ways for $\int_0^1 \frac{x^2}{1+x^4}dx$ other than the classic computations?
 
2:07 PM
@UserX Is it from $0$ to $1$? If it's from $0$ to $\infty$ you can apply Cauchy-Schlomilch transformation.
That is
$$\int_0^{\infty} \frac{x^2}{1+x^4}dx=\int_0^{\infty} \frac{1}{x^2+2}dx=\frac{\pi}{2\sqrt{2}}$$
@UserX Did you think of using geometric series?
Using geometric series all reduces to calculating $$\frac{1}{4}\sum _{k=0}^{\infty } \frac{(-1)^k}{k+3/4}$$
Not sure if you recognize the series but it's a classical one.
$$\sum_{n=0}^{\infty} (-1)^{n} \frac{1}{n+1+x}=\log(2)-\psi(x)+\psi(x/2)+1/x$$
On this page you also have the needed special values
In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: It is the first of the polygamma functions. == Relation to harmonic numbers == The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as == Integral representations == If the real part of x is positive then the digamma function has the following integral representation...
that can be derived with the formulae presented on the page (if you wanna do that).
Gauss's digamma theorem
That's all.
Now let me address to myself the following question: can I get the answer in a more brilliant way? Maybe, but I need some more time.
Of course, one does not need to use special functions since you have a nice primitive there. For a complete elementary way I might suggest you using a system of equations in integrals.
That is use $\displaystyle I=\int_0^{\infty} \frac{1}{1+x^4}dx$ and $\displaystyle J=\int_0^{\infty} \frac{x^2}{1+x^4}dx$ and add them and subtract them. All flows beautifully with a bit of work.
I mean the indefinite ones too
$\displaystyle I=\int \frac{1}{1+x^4}dx$ and $\displaystyle J=\int \frac{x^2}{1+x^4}dx$ (that is you use the classical substitutions in Cauchy-Schlomilch transformation)
 
2:34 PM
Is there a name for the shape that's a sphere contained within a larger sphere? Like an annulus rotated about an axis through its center?
 
@Addem "spherical shell"?
 
For instance, we can write that $$I+J =\int \frac{1 + 1/x^2}{(x - 1/x)^2 + 2} \ dx$$ whence all is clear. And we proceed similarly for $\displaystyle I-J$.
OK, let me see another way, a 3rd way. Well, beta function can be used too.
 
@Chris'ssistheartistc it's 0 to 1. 0 to inf would hint me to complex analysis
How do you come up with all that lol
Any brilliant ways for 0 to 1?
And no I've never heard that transformation before
 
@UserX You can use the way with system of equations or beta function.
 
2:57 PM
(beta function doesn't seem to work directly)
But rather what is known as incomplete beta function according to the definition here
@UserX ^^^
$$\frac{1}{4}\int_0^1\frac{x^{1/4}}{1+x} \ dx$$ where you use formula $(1.1)$ - but this is more or less what I was suggesting at the beginning.
 
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