OK, so you are using Kummer's transform to recover E4

@BalarkaSen I'd say I recognized E4 via that transform

Hmm. I don't digest the fact that something expressible in terms of Eisenstein series is the inverse of j.

@BalarkaSen You know the olden trick via the AGM? That is the same with Jacobi's $\Theta_{00}^2$

Seems... unexpected.

@ccorn which trick?

@robjohn :D !!

@BalarkaSen Well, using $\Theta_{00}^2(\tau) = {}_2F_1(1/2,1/2;1;\lambda)$ and computing the latter via the AGM. Then replacing $\tau\mapsto -\tau^{-1}$, knowing how that will transform $\Theta_{00}^2$, and dividing.

@Chris'ssis Since you seem to be gone for the day... I'll say 27 kg

Now I am off for a bit BBL

@BalarkaSen That gives an inversion formulae for the $\lambda = k^2$. So, traditionally, one used the algebraic relation between $j$ and $\lambda$, then the AGM quotients for $1-\lambda$ and $\lambda$ to recover $\tau$. Yet, some overly self-confident people seem to claim that no one would have known how to invert $j$ before 2011. I don't need to tell how much I'm disgusted.

what is $U_{77}$?

the unitary group of $77\times 77 $ matrices?

presumably $(\Bbb Z/77\Bbb Z)^\times$, no?

the cyclic group of order 77?

oh, the multiplicative group mod 77?

Greetings

@robjohn If we add up those kilos, we have each of the pets $2$ times, and then all we need is to divided all by $2$. :-)

@Chris'ssis there are many ways to go...

@robjohn I think this is the best way.

@Chris'ssis I noted that the dog weighed 10 lb more than the cat, then it was easy to see that the cat weighed 7 and the dog 17. The rabbit then weighs 3. I guess for a solution book, your way might be slickest.

@robjohn Indeed, there are many ways.

@ccorn Well, people seemingly forget that there have been developments on modular forms even after Ramanujan.

Ramanujan was only one person with millions of (impressive, I won't deny) identities sprouting from his head. The formal connection and the analogies have been developed through years of effort of algebraic number theorists.

@ccorn That seems interesting.

@Bananarama I have seen that notation mostly in number theoretic context. One calls the group of invertible elements in some ring the "group of units". Thus the U.

U(Z/nZ) then just refers to the multiplicative part of Z/nZ.

@BalarkaSen Inverting $j$ would have seemed trivial to Gauss and Jacobi. They were used to inverting $\lambda$ by means of AGM quotients, and from that to inverting $j$ one just has to invert a biquadratic and a cubic, where explicit formulae have also been publicised at least since Cardano. I'm irritated that there are claims to novelty of such formulae in the 21st century.

Yet, I am impressed by the 2F1 quotients that involve $j$ directly; these absorb the previously necessary algebraic transformations into the 2F1 expressions themselves. But that's not from the ArXiv preprint I am ranting against. The formula there is the classic approach via $k^2$.

$$\int_0^1 x^x \sin^2\left(\frac{\pi}{2}x\right) \ dx \le\frac{5}{12}$$

:D

$$\sum^\infty_{n=1}\frac{1}{n^22^n}$$

@VarunIyer This is for kids

$$\frac{1}{2}\left(\zeta(2)-\log^2(2)\right)$$

How would you solve then

@VarunIyer That series it's simply $ \operatorname{Li_2}(1/2)$

Err, Barlaka isn't here..

The solution made the substition of $x^n = \frac{1}{2^n}$ and then used a power series, and tried to integrate

Why though

@VarunIyer mathworld.wolfram.com/Dilogarithm.html

that's what they attempt to do

sumo.stanford.edu/pdfs/smt2014/calculus-solutions.pdf

number 10

@VarunIyer I see. It's a very easy question though (just a matter of little practice).

@Chris'ssis From your experience, I can see that. I'm only a high school student, please understand. Can you show me how to continue from the series you previously mentioned?

@VarunIyer Use the identity $(5)$ here mathworld.wolfram.com/Dilogarithm.html where you set $x=1/2$ and you're done.

Maybe anyone here could help me: need to show that if $q$ is odd, $a\in GF(q)$ is a square iff $a^{(q-1)/2}=1$. I think I got it the 'easy way', not even sure of that though, and clueless about the other way around.

@Chris'ssis thanks for your help.

@VarunIyer Welcome. How about the problem $9$? Can we finish it in a different way?

Oh I solved number 9, lots of work

but its a simple derivative and substiutions

@Chris'ssis the only one I could not solve was number 10

But thanks to you I understand a solution a bit more, so thanks again

@Chris'ssis is there a different way to solve it?

@VishwaIyer Yeah, but I just observed now that solution (I thought of) also appears there, it's the solution $2$.

@Chris'ssis I just have one more question. How can you see that the series equals

@VarunIyer I used the series representation of the well-known dilogarithm, it's about the definition mathworld.wolfram.com/Dilogarithm.html . Check $(1)$.

@Chris'ssis oh I understand now thanks

The problem $5$ is very funny.

@Chris'ssis when I did it by simply rationalizing the two fractions and adding them you didn't even need to use limits

@Chris'ssis its a misnomer, you didn't need to use an "improper" integral

@Chris'ssis okay $(9)$ is interesting .. how about $\displaystyle \int_0^{\infty} W\left(\dfrac{1}{x^2}\right)\,dx$ 'WOPORPAP'? :)

@r9m $\sqrt{2\pi}$?

@Chris'ssis :D !!

@r9m This is well-known though ...

@r9m Did you finish this one? $$\int_0^1 x^x \sin^2\left(\frac{\pi}{2}x\right) \ dx \le\frac{5}{12}$$

:D

@Chris'ssis that one is behaving like an integral inequality gangster :| .. no progress yet :|

@r9m hehe, yeah (it's very hard to do a bit of progress)

timecube.com

@Chris'ssis @robjohn Any progess with the double power of 2 sum?

@sarah It is related to a series by Ramanujan that doesn't have a closed form.

oh dear

@sarah I got an answer yesterday, but it was in terms of another sum that doesn't have a nice closed form, though it does converge quickly.

@robjohn I spent a while trying to get a good numerical ISC on it.

$$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{2^k+2^n} =\frac12+2\sum_{n=1}^\infty\frac1{2^n+1}$$

@robjohn Did the alternating version lead any better results?

Odd, I don't find the use for $q$ being odd..

@Alizter Your sum isn't looking good :(

@sarah do you mean $\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\frac{(-1)^{k-1}}{k+2^n}$?

I haven't worked on it

@robjohn No same sum just alternating.

Alizter posted that one first

@sarah what is the exponent of $-1$?

$\displaystyle \sum^\infty_{i=0}\sum^\infty_{j=0}\frac{(-1)^{i+j}}{2^i+2^j}$

@robjohn

That should be $1+4\sum\limits_{n=1}^\infty\frac{(-1)^k}{2^k+1}$

which is approximately $0.17720297785390431660$

@robjohn Does that also not have a closed form?

@sarah not that I know of.

oo dear

@Alizter is that a problem?

@robjohn I hoped it would have a nice closed form. That's all :)

Thank you for your help @robjohn @Chris'ssis @sarah

Upvote my question to counter the trolls math.stackexchange.com/questions/881504/…

Lets crush them

@BalarkaSen I see, thank you

how i can prove this

prove $\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$

Anyone know a simple way to show $\tan x>x$ on $(0,\pi/2)$?

I have the bound $\sin x>x+\frac{x^3}3$, $\cos x<1+\frac{x^2}3$ for $x\le1$ which yields the inequality for $x\le1$, but outside that I don't see any nice approximations

@Studentmath Hmm?

@MarioCarneiro tan(x)'s derivative at 0 is the same as the derivative of x, and increases from there

@r9m see the new A-bomb ... $$\sum_{n=1}^{\infty} \sum_{i=1}^{n} \sum_{j=1}^{n}\frac{1}{n^3(i+j)}$$

There's a geometric proof, with $2\tan(x)$ being the length of a pair of line segments that go from $(\cos 2x,\sin 2x)$ to $(1,0)$ outside the circle, and a general theorem about paths outside a convex set being longer than the shortest path on the boundary... @MarioCarneiro

@Chris'ssis ^^

@r9m :D

The segment $(\cos 2x,\sin 2x)$ to $(1,\tan x)$ is length $\tan x$, and likewise for $(1,\tan x)$ to $(1,0)$. @MarioCarneiro

Visually, it is clear this is greater than $2x$, but you need a difficult theorem about convex sets to prove it.

I'm going for an algebraic proof, hopefully. I'm posting a question on it, will link

@Balarka had some Finite fields questions

I wouldn't call any argument that uses infinite series "algebraic." But that's a quibble with terms. :) @MarioCarneiro

Still have :P

@ThomasAndrews Fair enough. The main issue is that I don't have derivatives and integrals defined, so it's tough to use those sorts of arguments directly. They usually have "algebraic" alternatives though - for example, I have a proof that $\sin x$, $\cos x$, $\tan x$ are monotonic on their respective intervals using trig identities instead of calculus.

What is the mathematical term for thoroughly writing out each step you take when doing a regular calculation or rewriting equations? Where you don't skip steps, assuming that the receiving party (the reader) understands what happens next.

Perhaps you can prove that $\frac{\tan x -\tan 0}{x-0} = \sec^2 z$ for some $z\in(0,x)$ without derivatives and intermediate value theorem?

Actually I have the IVT, just not the MVT

I'd call it "Painful?" :) @sammyg

Yeah, meant MVT @MarioCarneiro :)

@ThomasAndrews You are entitled to you your own opinion. :) But some might call it "elegant". ;)

@sammyg I would call it a "rigorous" proof

Alternatively, if I want to emphasize that theorems are not being used, I call it a direct-from-axioms proof

Depends really on what you are talking about. I was assuming you were talking about "formal" proofs, where every step is verifiable automatically. "Rigorous" is a good term. Neither is elegant nor inelegant, but formal proofs are often very painful, even for basic theorems. @sammyg

As an (ironic) aside, my proof of $\tan x>x$ is actually a formal proof

us2.metamath.org:88/mpegif/sin01bnd.html

No, it is not. It might be one that can be formalized, but your proof is not formal. @MarioCarneiro

well, it's not yet cause it's not done

but when I see how to solve the problem, I really will code it up into a real formal proof

the linked proof that $\sin x>x-x^3/3$ is a formal proof

I know there is a special word for it my math teacher used, but I just can't remember what it is. I haven't heard it for over eight years, so... :) Hehe, I might ask my old teacher. Yeah, it could be "rigorous"... there's even a Wikipedia entry on this... "Mathematical rigour".

Apparently, mathematical "rigour" in classrooms is the stuff of hot debates. Didn't know that...

What is {} sign for in math.stackexchange.com/questions/601314/… ?

Sure, most people learn calculus non-rigorously first, and only learn the rigorous parts with care. I learned about unique factorization into primes long before I saw a proof. Even Euclid, most people's introduction to proof, is actually greatly lacking in "rigor," with lots of terms assumed to have intuitive meanings. @sammyg

Use \{ and \} @MrWho

@ThomasAndrews What does it mean?

@ThomasAndrews what does for instance, ${t}$ , mean?

Oh, misread :) It usually means "fractional part" - that is $\{x\}=x-\lfloor x\rfloor$ @MrWho

@MarioCarneiro I wish you luck with a formal proof. I was looking for one myself, years ago

@ThomasAndrews THanks

I actually enjoyed watching my math teacher do his "rigour" thing! :P That's the main reason I liked him the best of all the math teachers. Hi was killing that whiteboard! :) Changing pens by the minute! You had to really tune in and focus, and follow what he's doing. He didn't talk too much. Not excessively like some teachers do who can go on and on for hours without giving a single actual example on the board. Now that I think about it... it was mostly like a trance session... hypnotizing! :P

@sammyg to adapt the geometric proof to a formal proof, you'd probably need to define arc length. is there already some "formal" geometric framework?

Once he was pounding that whiteboard so hard, it fell off the wall! :P

Haha, just kidding...

@MarioCarneiro Check math.stackexchange.com/questions/98998/… . I think this is like what I found.

Rigor does not require multiple colors. Multiple colors are for reader clarity. Not that it is a bad thing, just not part of rigor. There's a famous story of a teacher teaching a very rigorous version of calculus without any pictures. In the middle of a proof, he gets stuck on the next step, draws a picture on the board with his coat hiding it from the students, then quickly erasing it so that his students never saw it. :)

When I heard that story, it was about Claude Chevalley.

Blargh. Every element $a\in GF(q)$ has an element $b\in GF(q)$ so that $b^k=a$ iff $\gcd(k,q-1)=1$. I managed to get to $kx=1, ky=q-1$ where $0<x,y\le q-1$. But that doesn't help me much.

@cxseven Is that similar to my answer? using areas instead of lengths.

In fact I know that it is true that $kx=y$ for every $0<y\le q-1$, where $x$ is between these ranges too but could be anything. Feel stuck.

@robjohn apparently

@robjohn Yes, I've seen that proof of yours before, it's very slick. Unfortunately it's hard to define these geometrical things formally without getting into a circular (no pun intended) argument :(

@MarioCarneiro yes, it is more of a justification to satisfy the doubt, rather than a formal proof.

you have to deal with angles sweeping out proportional areas in a circle.

thinking about approximating a sector of a circle using little squares and using that to define trig functions gives me hives

@MarioCarneiro small triangles is better than small squares.

If you approximate the circle with an n-gon, archimedes style, I think you end up with exactly the same algebra equalities in the end

I just found out that Harlan Brothers calculation of Exp(1) = 2.71828182845... works also for complex numbers, but not the way one would expect.

From the Pascal triangle.

@ThomasAndrews Haha, funny story that is! What gave him away? :) No, colors are not part of rigor, of course not. It would be silly to think otherwise. It's not exactly what I meant. He was using up the pens that fast! Same color! ;) See why rigorous explanations do matter! :) The "painful" truth of rigorousness. ;)

Ah, too much time for me at work whiteboards, with multicolor meant to add clarity, but often adding noise instead.

Whiteboards are pernicious. Blackboards are delicious.

Last office had specially painted walls that could be used as white boards, but they required special pens, and you had to shake them. Seemed pointless :)

@Daniel mind hinting me in the right direction with yet another question?

@Studentmath Not if I can. If I cannot, I mind.

Every element $a\in GF(q)$ has an element $b\in GF(q)$ so that $b^k=a$ iff $\gcd(k,q-1)=1$. I got to $kx=1, ky=q-1$ when $0<x,y\le q-1$, but not sure how to go from there. I know actually that I can have $kx=y$ for every $0<y\le q-1$ for some $x$ in the same range, but still.. no idea how to proceed

@Studentmath For problems like that, with powers in finite fields, since $0^k = 0$ for $k > 0$, you can ignore $0$ and concentrate on the nonzero elements. For those, unless an addition appears somewhere, it is mostly clearer if you forget that you have a field, and concentrate on the cyclic group of nonzero elements under multiplication. The order of the group is $q-1$. So, in a cyclic group of order $n$, when is the map $a \mapsto a^k$ surjective?

@Daniel Thanks! Will look up what a cyclic group is.. new to this field, no pun intended

@Studentmath A cyclic group is one that is generated by a single element, so that all elements have the form $g^k$ for some integer $k$.

I thought you knew, since you used that in your proof in the question earlier.

@Daniel Oh! So that's what that primitive element does, forms a cyclic group.

@Daniel ah, a bit of research shows me iff $k$ is coprime with $q-1$. I will look up the proof of it, perhaps it's in some additional textbook or something I can manage myself. Thanks a lot!

@Studentmath A map from a finite set to itself is surjective if and only if it is injective. Since cyclic groups are abelian, $a\mapsto a^k$ is a group homomorphism. When is a group homomorphism injective?

@ThomasAndrews Having to shake the walls in order to write on them sounds mighty inconvenient.

@900sit-upsaday Just to let you know, I liked your previous names much better. No Bull.

@Daniel iff the kernel is the identity, that means that we can't have $a^k=1$ for any $a\neq 1$, which means that $k$ can't be a multipicity of $q-1$ or else some element $g\in C_{q-1}$, the primitive element (which is different than 1) would hold the propert of $g^k=1$ as well! (?)

@Studentmath The kernel being trivial means that no element has an order dividing $k$. What are the orders of elements of a cyclic group of order $n$?

Hooray going back to Turkey in the morning.

@Studentmath OK

I see Daniel already answered it.

@ThomasAndrews Haha. Very nice anecdote.

@Alizter Had luck with your differential fields?

Is it possible to look for specific integral like $\int_{0}^{1}\frac{\tan^{-1}(x)}{x}dx$ in the MSE search bar?

@DanielFischer ^

@MrWho You're asking the wrong person. My search-fu is almost nonexistent.

@DanielFischer Do you know the right person for the question to be asked?

Sorry, no.

what is search-fu?

@Bananarama Like master-fu I think :D

@Bananarama en.wiktionary.org/wiki/-fu

fu-shore

@r9m: I have extended the Reverse Cauchy-Schwarz to a reverse Hölder...

the constant is $$\frac1{p^{\frac1p}q^{\frac1q}} \left[ \left(\frac{M_f}{m_f}\right)^{\frac1q}\left(\frac{M_g}{m_g}\right)^{\frac1p} + \left(\frac{m_f}{M_f}\right)^{\frac1p}\left(\frac{m_g}{M_g}\right)^{\frac1q} \right]$$

note that when $p=q=2$, we get the old formula

@robjohn Nice !! :D

@r9m I might post it as a separate answer. It would be kind of confusing to combine the two.

@robjohn That will be great !! :-)

peace :)

@robjohn there is already a reverse/converse to Holder's Inequality mentioned in this comment. The particular form we used to prove the reverse CS is mentioned as Diaz-Metcalf Ineq in this link

@robjohn You may give it another name .. like robjohn inequality =)

@MAZux peace :)