@meer2kat Who?

@Sawarnik ...the horse. duh

clearly the doctor who fans haven't gathered here today

Hello

I am given a pattern, and I found a corresponding function $f(n)$, where the $n\mbox{th}$ element of that pattern is equal to $f(n)$. How would I prove that the function does indeed correspond to the pattern? Or am I overthinking it?

Goodnight !!!!

That was random...

Intutively and logically I would say that if you show the nth element of the pattern is equal to $f(n)$ then you are done..

But then again I might be wrong.

@Studentmath hmm, perhaps.

Initially, I was thinking of using the principle of mathematical induction, but all the examples online involve equations. Mine isn't really an equation, if I think about it. Or is it...

I am unsure what problem you speak of, but if you have a pattern with n elements, and you can prove via careful induction that for every n the nth elemnt of that pattern is equal to $f(n)$, you proved the correspondence. I think.

Given that the variables of a symmetric polynomial sum to some constant K, can we say that the polynomial will have an extremum when all the variables are equal?

For the record, the pattern looks like this: $1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 16, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 1\ldots$, and the corresponding function is $f(n) = 2(n - 2^{\lfloor\log _2 \left(n\right)\rfloor}) + 1$

I wonder if there exist a nice expression $\varphi(u)$, such that $$\frac{\displaystyle \int_0^{u} \int_0^{\tan(x)} \frac{\arctan(t)}{t} \ dt}{\displaystyle \int_0^{u} \int_0^{\sin(x)} \frac{\arcsin(t)}{t} \ dt}=\varphi(u), \space u \in (0, \pi/2]$$

Note that $$\varphi(\pi/2)=2$$

I have one question: what does this integral means: $\int d^4x$ does it mean to take integral 4 times like this: $\iiiint dxdxdxdx = \frac{x^4}{24}$ ($c=0$)?

The pattern is, for all $n$ being powers of two (i.e. $n = 2^k$, where $k \in \mathbb{Z}$), $f(n) = 1$. When $n$ isn't a power of two, then $f(n)$ is an odd number between two powers of two.

@GigiButbaia It typically means an integral over a 4-dimensional domain. The domain is unspecified here. You seem to assume that the domain is $(0,x)^4$ and that the integrand is $1$.

(I'd probably better say region instead of domain here)

@Chris'ssis I literally said out loud: "Jesus Christ!" upon reading this

Although it does seem like there could be a tricky little answer in there!

@GigiButbaia It is not an iterated integral. If you want an integral of $f(t)$ over $(0,x)$ iterated $n+1$ times, you could write that coolly as $\frac{1}{n!}\int_0^x (x-t)^n f(t) dt$ (use partial integration to see that)

Again, I would say there needs not to be any proof once you show via induction that the function corresponds. If you wanna be accurate and thorough, at least. Can skip that too..

As it seems

Given the field $Z_5[x]/<x^3-x^2-1>$ and an element $\alpha = x + <x^3-x^2-1>$, how do I express \alpha^4 with the basis {1, x, x^2}?

gotta go afk for a while BBL

@AndrewThompson Let $I=\langle x^3-x^2-1\rangle$. Then $x^3-x^2-1\equiv0\pmod{I}$ and $\alpha\equiv x\pmod{I}$. Thus $\alpha^4\equiv x^4\equiv (x^3-x^2-1)x + x^3+x \equiv x^3+x \equiv (x^3-x^2-1)+x^2+x+1 \equiv x^2+x+1\pmod{I}$.

@ccorn I'm having some trouble seeing the step $x^4\equiv (x^3-x^2-1)x + x^3+x \equiv x^3+x$, otherwise it was very clear.

Sorry, I understand $(x^3-x^2-1)x + x^3+x \equiv x^3+x$

I do not understand $ $x^4\equiv (x^3-x^2-1)x + x^3+x$

@AndrewThompson Just expand. I could have written $=$ instead of $\equiv$ there

Hm, I don't follow. How does one "expand" in this scenario? Sorry if I'm being slow, I've never seen the material presented in this way before.

It's one step of a polynomial division. Write $x^3$ as $(x^3-x^2-1)+x^2+1$ and then mod out the parenthesized part.

I'll give the task a try on paper.

You have $x^4$, so multiply by $x$, and do the same trick again.

BTW, the $(\ldots)$ are just that. Parentheses. Not ideals like $\langle\ldots\rangle$.

Sure, but our polynomial is a maximal ideal since it is a field. If I understand it correctly, we are treating our polynomial as zero in the quotient ring, making $x + \langle p(x) \rangle$ simply a way to express a coset. By the logic you are writing out here, that I agree upon, is that our field is of order 125 since we can always factor out something from polynomials of degree higher than two, giving an order of $5^3$. I understand this in theory, its the computation that confuses me,

likely because I lack experience with congruence arithmetic, although I do know the basics by know.

@AndrewThompson So you know that $x^3\equiv x^2+1\pmod{I}$. Thus you can recursively replace polynomials of degree 3 or higher with those of lower degree.

Essentially that's polynomial division by $x^3-x^2-1$. You keep the remainder.

Oh, yes I got it now!

That last sentence saved me, I can't believe how simple it was. Thanks a lot!

Ah :-)

@Chris'ssis Wait a minute! Isn't $\tan(\frac{\pi}{2})$ infinite?

@ccorn, sorry for being buggy, when dividing $x^4$ by $p(x)$ I get $x^2+x-1$, did I mess up or did you just mistype something?

@MickLH that fact doesn't affect my expression (in a bad sense). There we have a double integral ...

@AndrewThompson Pari/GP agrees with me :-)

@MickLH To understand me, let me show you something.

@MickLH did you take it?

Haha, yes, 1*-1 = -1. Thanks again :)

@Chris'ssis I've opened it

thanks everyone

@MickLH You see now that $\pi/2$ doesn't "badly" affect my expression? :-)

@MickLH Oh, let me give you another version ....

@Chris'ssis Yep :)

@MickLH that one is a bit harder ...

@GigiButbaia Enjoy!

done

@Chris'ssis WOW !! that series expansion of $\log(\tan x)$ is aWesome !!

@GigiButbaia Wait... as I review my answer to your question, there is some stuff in it that I wanted to rule out explicitly... if the region were $[0,x]^4$ and the integrand $1$, the integral would amount to $x^4$ of course. It's not an iterated integral, but an integral over $[0,x]$ of $1$ iterated $4$ times would give $x^4/24$.

Yeah buddy!

:15216434 There was a quantifiable improvement in my math skills that happened while reading that, thank you! :P

ok thanks @ccorn

@r9m thank you! Please keep it for you. :-)

I'll be back a bit later.

Someone stars all my utterances, including the erroneous ones. Ouch.

But the iterated-integral formula is cool indeed. One can derive the Taylor formula from it, with the remainder term. And if you replace (x-t)^n with a Bernoulli polynomial, you end up with the Euler-Maclaurin summation formula, again including the remainder term.

@Chris'ssis Too bad I didn't see the previous link you posted .. I just closed my eyes for half an hour and I feel like I missed a lecture ..

@ccorn, what about 1/(\alpha + 1)? I don't need a full solution, just a finger in the right direction would be great.

This girl is wearing a blue shirt that says "pink" in large letters

@AndrewThompson Since $x+1$ is coprime to $x^3-x^2+1$, they can be combined to form $1$. I'll take $3$ instead: $(x^2-2x+2)(x+1) - (x^3-x^2-1) = 3$

In other words, $(x^2-2x+2)(x+1)\equiv3\pmod{I}$.

Alright, I think I had a similar idea, I set $1/(\alpha +1) = \psi$, so $1 = \psi(\alpha + 1)$

I'll give it a go first.

@AndrewThompson Indeed. A systematic method would be the extended euclidean algorithm applied to polynomials.

Sentences like that make me doubt my choice of doing abstract alg as a first-year :) Haha, jk, thanks for everything, I've gained a lot of intuition today.

@AndrewThompson I can recommend the online text of Robert B. Ash. He seems to be devoted to teaching, and in his foreword he says "I never write 'it is easy to see'"

@r9m I wonder if we can employ it here to prove that $$\int_0^{1/4} \log^2(\cot(\pi x)) \ dx= \frac{\pi^2}{16}$$

Hm, I'll give it a shot. Currwntly working through Fraleigh.

brb

@Chris'ssis Okay .. I will try to think about it :)

I have a series of exams for the next 5 days .. (summer vacation starts after that) .. planning to start MSEing furiously after that :D

Hello, is it ok if I ask for a paper?

It's this one, ams.org/books/memo/0451

I don't have acces in my library.

@Chris'ssis I think that should be doable using the fourier expansion of $\log(\sin(x))$ and $\log(\cos(x))$

@robjohn Yeah. I just created this one $$\int_0^{\pi/6} \log(\cot(x)) \ dx=\frac{\pi \log(3)}{12}+\operatorname{Ti}_2\left(\tan\left(\frac{\pi}{6}\right)\right)$$

I looks so awesome!

Let me add it to my special collection.

@robjohn did you manage to take a look at the problem for that I gave 500 points bounty?

@Chris'ssis I did... do you think it has a simpler answer?

My feeling is that the nice answer comes from the elliptic integrals-related identities.

@Chris'ssis Oh, so a nice answer in terms of elliptic integrals.

:-)

@Chris'ssis What is $\mathrm{Ti}_2$?

@robjohn mathworld.wolfram.com/InverseTangentIntegral.html. This notation is pretty known. Was it a joke, right? :-)

@robjohn Invented by Wolfram so that chris's sis can name that the "Titan formula"

@Chris'ssis other than a temporary molecule of two Titanium atoms

@robjohn :-)))))

@Chris'ssis no, I've never seen $\mathrm{Ti}_2$ before

@Chris'ssis Okay, so it is the alternating form of the $\mathrm{Li}_2$ series

@robjohn just a simple notation for the inverse tangent integral.

di-Titanium

@EnjoysMath I don't think it would hold together very long.

@EnjoysMath not like di-Lithium :-)

*nice

@Chris'ssis Indeed, by a complicated bit of application of complex analysis, one can derive a generalization.

Through elliptic integrals, yes.

It's somewhere in one of the posts of a member of Integral&Series.

@BalarkaSen that integral you mean or something similar?

A generalization.

@BalarkaSen hmmm, that sounds interesting.

$$\int_{0}^{\pi \alpha} \log(\cot(x)) dx$$

$\alpha \in [0, 1)$

I don't quite remember the closed form, but I have definitely seen this one.

@BalarkaSen I'm afraid one needs to make use of the multiple gamma function for giving the generalization of this.

Barnes $G^n$ was used, as I remeber it, @Chris'ssis

@BalarkaSen Yeah. Anyway, I know to derive that. No need to look for that.

I am not looking =)

I have some pressing matters here on my desk for now.

@BalarkaSen For instance, I wouldn't use that formula for $$\int_0^{\pi/4} \log(\cot(x)) \ dx$$

There should be too much work to do ... (and I wouldn't like to memorize that formula, not even to write it down)

@Chris'ssis It's $3/8 \cdot \zeta(2)$

Isn't it?

@BalarkaSen It's Catalan's constant.

OK, I see.

I think I forgot the closed form for it.

Nevertheless, Catalan is expressible in terms of tan integrals.

@BalarkaSen $$\operatorname{Ti}_2(1)$$

Exactly.

An interesting formula I saw once expressed Catalan in terms of Clausen.

That was done by one of the members of I&S.

$$C=4\pi \log\left( \frac{ G(\tfrac{3}{8}) G(\tfrac{7}{8}) }{ G(\tfrac{1}{8}) G(\tfrac{5}{8}) } \right) +4 \pi \log \left( \frac{ \Gamma(\tfrac{3}{8}) }{ \Gamma(\tfrac{1}{8}) } \right) +\frac{\pi}{2} \log \left( \frac{1+\sqrt{2} }{2 \, (2-\sqrt{2})} \right)$$

@BalarkaSen Nice.

@Chris'ssis was your past username something like Chris'ssisandcompany ?

@r9m lol, no. I never had such an username. :-)

Q: What is the difference between Ignorance and Apathy?

@BalarkaSen what do you think are common aspects?

Your decision.

@ccorn I don't know and I don't care

=D

@BalarkaSen ;)

@Chris'ssis I saw it a few days ago ... Aha .. its "Chris'ssisterandpals" !!

@r9m Yeah.

@Chris'ssis so are you like multiple users of the same account ?

@r9m Covering spaces.

@r9m Every one of the hypothesized users probably feels as a single user

@r9m No.

Are the pals those of Chris or of his sister?

@Chris'ssis who are the pals then ? :)

@r9m Some friends of mine. The meaning was that some questions I posted in the past came from them.

@Chris'ssis Thats really nice :) .. (y)

I expected stars on

sigh

@r9m I worked very much, no matter the user that is typing right now. ;)

@BalarkaSen Don't sigh

@ccorn Would it be a good pun here to star this message of yours?

Or perhaps it's not a good idea, as it reminds me of Gruber's PSA

@BalarkaSen anything is random

@ccorn What's your favourite xkcd strip?

I like xkcd.com/356

@BalarkaSen No preference. They are all good.

@ccorn Any chance you know of what-if.xkcd.com?

@BalarkaSen yes, but I recall having read only two entries

This is the best

Hi @Mike.

@BalarkaSen Try that one

@ccorn Wonderful. That goes to my list.

@PedroTamaroff Have you seen my finite group theory problem?

Just noticed something, wonder why it is so. I mean, the question may sound stupid, but.. in probablity, when an event is dependent of another, the probablity of the intersection is larger than the multipication of the probabilities of the two events. Is there a meaning to this certain difference between the two results?

@r9m is it possible that $$\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}} \ dx_1 \ dx_2 \cdots \ dx_n $$ has a nice closed form?

@Chris'ssis umm .. $\dfrac{1}{\sqrt3}$ maybe :D

@r9m wow, I didn't even think of that. How did you get that? :-)

@Chris'ssis one thing is $\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \left(\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\right) \ dx_1 \ dx_2 \cdots \ dx_n = \frac13$, so my guess is Weierstrass approx/appropriate taylor series of $\sqrt{x}$ in $[0,1]$ should give the answer

@r9m Indeed, it makes sense! Good job! :-)

@Chris'ssis similarly for power-$p$-means and taking limit $p \to \infty$ gives $\max\{x_i\}$ case that I said about earlier today :D

@r9m Impressive! Glad to learn from you.

@Chris'ssis r9m BLUSHES <o^_^o> .. and tries to dive off the window .. ;)

@r9m ;)

I have some proofs here I need to finish ... but I'd like to create some more before attending them.

@Balarka Trivial problem there.

Got two more random downvotes.

Wow.

@r9m the first one is $$\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}} \cdot \frac{n}{\displaystyle \frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_n}} \ dx_1 \ dx_2 \cdots \ dx_n$$

Then, the second one is $$\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}} \cdot \sqrt[n]{x_1 x_2\cdots x_n} \ dx_1 \ dx_2 \cdots \ dx_n$$

Hello

@Chris'ssis you showed that the Harmonic mean integral is $0$, should follow from sqeeze theorem

@r9m based upon what you taught me today I conjecture at first sight that the second limit evaluates to $1/(e \sqrt{3})$

@r9m Yeah.

Let's write another one

$$\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}} \cdot \frac{x_1+x_2+\cdots+x_n}{n}\cdot \sqrt[n]{x_1 x_2\cdots x_n} \ dx_1 \ dx_2 \cdots \ dx_n$$

hmmm, I conjecture this one tends to $1/(e 2 \sqrt{3})$.

This is the math I like. :-)

@r9m Ah, wait! There is some more ...

Ugh

@robjohn, remove the bottom "answer" please

@r9m $$\lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \max\{ x_1, x_2, \cdots, x_n\} \cdot \sqrt{\frac{ x_1^2 + x_2^2 + \cdots + x_n^2}{n}} \cdot \frac{x_1 + x_2 + \cdots + x_n}{n} \cdot \sqrt[n]{x_1 x_2\cdots x_n} \ dx_1 \ dx_2 \cdots \ dx_n$$

I just look at it and know the answer (or might I be wrong?).

@Chris'ssis is this your subtle plan of making me jump from the roof ? :P

Now, by using Weierstrass approximation, one can create hundreds and thousands of mind-blowing questions. :-)))

@r9m No, we need you here! I like the way you think!:-))))))

I have so much fun working on these. :-)

@Chris'ssis $\displaystyle \lim_{n \to \infty}\int_0^1 \int_0^1 \cdots \int_0^1 \sqrt[n]{\dfrac{x_{1}(x_{1}+x_{2})(x_{1}+x_{2}+x_{3})\cdots(x_{1}+x_{2}+\cdots+x‌​_{n})}{n!}}\,dx_{1}\cdots\,dx_{n}$ :D ;)

@r9m let me give it a try without pen and paper ...

@r9m $\displaystyle \frac{1}{2}$ (by Cauchy d'Alembert)

This should be the answer!

hmmm, I think some additional tricks are needed ... (to make that rigorous on paper)

@Chris'ssis what is Cauchy d'Alembert ? link please :)

@r9m Can you tell me if my answer is correct?

@r9m a criterion for limits.

@Chris'ssis I dont know .. I saw that expression in one of chinamath's inequality questions this

Hi all, is every map from Klein bottle to Torus induces trivial map on cohomology?

@r9m It is $1/2$ I think.

@Chris'ssis okay .. using equality of root and ratio test criteria ?

@Chris'ssis awesome .. I see it now :)

Gah, this question.. I have three events, A,B,C. Given that two of them occured (A,B or A,C or B,C), what's the chance that the third occured as well. So basically it's the union of three conditional situations. But I can't just compute for each conditional situation and add them up, that'll go over 1 (they are not disjoint)

And I have no idea how one is supposed to get to the supposed intersection of these events.. can't have two conditional situations in one probability..

Someone that wanted to make my day brighter sent me this picture. You have to see it! :-)

OK, let me add some more questions to my collection.

brb

@r9m btw, aren't you interested in taking my bounty here? math.stackexchange.com/questions/737311/…

Hah, got it! And nice pic.

@Chris'ssis that integral is beyond the reach of my knowledge .. :'(

@Mike @seaturtles

@KarlKronenfeld

hmm, just and and or apparently en.wikipedia.org/wiki/Logical_connective

@PedroTamaroff yes?

@seaturtles Let me recall my question.

Ah!

It is known $1+X+\cdots+X^{p-1}$ is irreducible for a prime $p$ over $\Bbb Q$.

Yet, say $1+X+X^2$ is reducible over $\Bbb Z/3\Bbb Z$ by $1=-2$ and $=(X-1)^2$.

In fact, $1+X+\cdots+X^{p-1}=(X-1)^{p-1}$ in $\Bbb Z/p\Bbb Z$.

mmhmm

you want to know how cyclotomics factor in finite fields?

What happens when we look at other primes?

I know I got a paper from arxiv a year or two about that

i.e. not the same $p$, say $q$.

@seaturtles Ah, OK.

Yes.

brb

Hi

k back

@PedroTamaroff your question is more specific though

@seaturtles Yes, I guess it is easier...?

yes

you can determine relatively easily if $\Phi_p$ is reducible or not over $\Bbb F_\ell$

OK.

Maybe I should think about it for a while?

consider roots of unity

@r9m you mean you tried it?

@Chris'ssis ya .. but the answer provided involves Hypergeometric series and stuff that I don't understand

@seaturtles OK.

@r9m Hypergeometric series? I proposed this question some weeks ago ...

(I mean I proposed it to myself)

@Chris'ssis wow .. you got yourself a Balrog to slay :O

@r9m hehe :-) This reminds me of one of the Ramanujan's series-integral question that can be nicely computed by observing that the sum in the right nicely telescopes, but there it's a matter of powers.

@r9m you get a nice series over there if you sum over $n$ ...

@Chris'ssis I'm not sure what the expressions stand for ..

@r9m I mean summing from $0$ to $\infty$ over $n$ in the expression in the left, the one containing the hypergeometric function.

@r9m you get $\displaystyle \frac{\pi^2}{4}$

Hi @Chris

@user127001 Hi! How are you doing? :-)

@seaturtles

?

Why does $X^p-X-1$ irred over $\Bbb Z/p\Bbb Z$ implies it is irred over $\Bbb Q$?

Let me try something.

And I will get back at you.

@Chris my exam got pushed back but now I have 2 math exams in one day :/

@user127001 don't worry. I had many exams in school, some of them were really horror (like 6, 7 exams a day).

@seaturtles Nevermind.

@chris I'm going to fail and then I will have to quit math :(

@user127001. Study like you've never studied before.

From 9am - 3am.

Every day.

That's what you have to do to succeed.

@user127001 Well, there are 2 important things: 1). to think like a winner, like one that never fails and be highly confident, you can pass any exam successfully, you know, thinking like a gladiator 2). then you need to work on math, to struggle yourself with the problems and never give up, you want to understand things better and better. Learning is a process that never stops, it depends on you how much time assign to it. :-)

@seaturtles I have an idea to show $X^p-X-1$ cannot be reducible over $\Bbb F_p$.

Might help.

Suppose $X^p-X-1=h(X)g(X)$ with $\deg h,\deg g >0$.

Then for every $t\in \Bbb F_p$; $$-(g(t)-1)(h(t)-1)=h(t)+g(t)$$

@seaturtles Are you there?

I see this won't get me anywhere, though.