Hello @did :)

Hello @Fermésomme

Are you from France @Fermésomme ?

@evinda No, I got the handle from a Measure theory book.

A ok.. Where are you from? @Fermésomme

@evinda That is a secret

A ok.. Can I guess? :D @Fermésomme

@evinda Yes

Sweden? @Fermésomme

@evinda No

@evinda How is your algorithms course going on, BTW?

@evinda Yes, I do have such a course.

And what do you do at this moment at the course? We are doing graphs.. @Fermésomme

@evinda We are also doing graph algorithms.

You too? Nice... @Fermésomme

And do you like them? @Fermésomme

@evinda Hmm... Are we classmates? ;p

Hmm... I don't think so... @Fermésomme

@evinda We just started graph algorithms. So far things are interesting and I like them.

I'm so proud of myself. In a calculus test, I was asked to prove that $3x+\cos(x)=0$ has only one real solution, and I was able to generalize the problem and find a value of $m$ such that $mx=-\cos(x)$ has exactly two solutions and $y=mx$ is tangent to $y=-\cos(x)$ at one of these points

Also, good morning everybody

@teadawg1337 It seems pretty easy.

@Chris'ssis Everything is easy for you.

@Chris'ssis It was easy, but the main point is that I approached the problem in a unique way :)

@JasperLoy :D

Also, I'm not sure if the value can be computed in a closed form, I was only able to find an approximation to four decimal places

@Fermésomme How may students are there in class at this subject? :p

@teadawg1337 if $f(x)=3x+\cos(x)$, then since $f'(x)>0$ and $f(-\infty)=-\infty$, $f(\infty)=\infty$ the equation has only one real solution. Q.E.D.

@Chris'ssis I know, but I generalized the problem and found a tangent line to $y=-\cos(x)$ that passes through the origin and intersects $-\cos(x)$ at one other point besides the point of tangency

Which implies that any value $a$ of the function $y=ax+\cos(x)$ greater than the slope of the tangent line described above forces said function to have exactly one real root

@Chris'ssis I have a tendency to make things far more complicated than they have to be...

@teadawg1337 It's easy to see how one should approach the problem. $y$ passes through $(0,1)$.

@Chris'ssis That's the method I used on the test, I revisited the problem afterwards and used the method I described above.

@teadawg1337 OK ;)

@Chris'ssis Graphically: here

@teadawg1337 Yeap.

@teadawg1337 I really like to see you work on math. I don't see such people every day, studying all kind of approaches.

Quora is boring when it comes to getting math questions answered ... :| there's no way I can promote my questions (no bounties) :(

@r9m Hey, how are you doing these days? :-)

:21001115 upside-down you mean? :-)

yes .. precisely :)

@r9m hehe, it happens once in a while! Not all days are great days. ;)

@r9m hahaha, OK! :D

@teadawg1337 I was asking myself if I'm able to prove that $$\frac{\pi}{4}\le \frac{1}{1^2+1}+\frac{1}{2^2+1}+\frac{1}{3^2+1}+\cdots $$ without computing the sum.

@Chris'ssis so using only the first three terms isn't allowed? $S_3=0.8\gt \frac{\pi}{4}$

@teadawg1337 lol, OK :-)

I guess I'm asking what you mean by "sum." Infinite sum and/or partial sums?

@teadawg1337 Your answer is very good.

@Chris'ssis That would give a lower bound on $\coth \pi$ then ?

@r9m It's just a stupid (since there are no stupid questions) question that came to mind.

@Chris'ssis No such thing as a stupid question, and I've asked far worse

@Chris'ssis we could compare term by term with Liebnitz series for $\pi/4$ as well :} .. ugh that might not work ,...

@Chris'ssis How about $$\frac{\pi^2}{9}\geq\frac{1}{1^2+1}+\frac{1}{2^2+1}+\frac{1}{3^2+1}+\dots$$?

On second thought, that's not readily apparent without computing the sum...

@teadawg1337 It's a stronger one.

@teadawg1337 @r9m is the master of the sharp inequalities. :-)

@teadawg1337 you mean $\frac{\pi \coth \pi - 1}{2} < \frac{\pi^2}{9}$ ? :o

@Chris'ssis no one is the master of sharp inequalities :P

@r9m Indeed

Disclaimer: I somewhat cheated. I used a calculator. Turns out the difference between the two is less than $0.02$

@teadawg1337 What kind of calculator do you have

How about doing that without using a calculator ... I just exploit an elementary way ...

Well, I came up with the idea without using a calculator. As you know, I've been working quite a lot recently with dilogarithms and the like, so expressions similar to $\frac{\pi^2}{9}$ occur quite often.

Long story short, I've become quite familiar with the approximate value of $\pi^2$

Done.

@ᴇʏᴇs I have a TI-36X Pro, but I use WolframAlpha to verify more complicated computations.

@teadawg1337 I'm done with an elementary approach.

$$\frac{\pi^2}{9}\geq\frac{1}{1^2+1}+\frac{1}{2^2+1}+\frac{1}{3^2+1}+\dots$$

$$\sum _{k=8}^{\infty } \frac{1}{k^2}+\sum _{n=1}^8 \frac{1}{n^2+1}$$

:D

?? I have no idea how you concluded that .. :|

@r9m It's perfect.

@r9m It's $\approx 1.09283$.

calculator ?

Ah, I see. The second line is also less than $\frac{\pi^2}{9}$, and $\frac{1}{n^2}\gt \frac{1}{n^2+1}$. Therefore the inequality holds by the comparison test

How can we say the second line is less that $\pi^2/9$ ? :o

$$\frac{\pi^2}{9}\ge\sum _{k=8}^{\infty } \frac{1}{k^2}+\sum _{n=1}^8 \frac{1}{n^2+1}$$

@r9m ^^^

okay but how do we get that ? I mean how do we decide we have to consider the first $8$ terms ?

@r9m One can even check by hand the evolution of the first terms like above, split into 2 sums.

@r9m Checking. Well, it's not a perfect way.

O_O okay ... I thought there was something slick going on :|

It works, though. I don't understand how the result can be proven without using a calculator at some point

I mean, $$\sum_{n=1}^8 \frac{1}{n^2+1}$$ is a nasty fraction...

okay ... I needed to use $\pi > 3+ \frac{1}{10}$ and no calculators :P (I am beginning to think I have some sort of mild OCD :P)

@Chris'ssis did you see T. Andeerscu's problem in the latest math reflections journal .. ? the one with $\displaystyle \prod\limits_{n \ge 0} \left(1 - \frac{2^{2^n}}{2^{2^{n+1}}+1}\right)$ ? :)

@r9m No, but it looks like a boring telescoping product.

@Chris'ssis partly :)

I have very little experience with infinite products...

I think its the cutest problem in this issue :) (the other one being his inequality which is cute too)

@Chris'ssis I've heard back from my academic advisor. One of the professors interested in working with me is the mathematics department head at MTSU :)

what is MTSU ?

Hi. math.stackexchange.com/questions/1225276/… gets downvotes, but no critical comments, so I don´t know if the question is offtopic, not precise enough, or too trivial, or what. Help?!? Otherwise I´ll delete it (and possibly leave the site for good).

@r9m It's not hard as I can see ...

Middle Tennessee State University

@teadawg1337 Good for you!

@Chris'ssis hard ? not at all .. its 'kawaiii' :D

@teadawg1337 :) okay !!! Nice :)

@r9m What is kawaiii?

@TheBlastOne It could be people who dislike the field of statistics

@Chris'ssis ah .. means 'cute' (Jap)

@r9m Ah, I see. :-)

@Eyes That´d include me, but I can´t downvote my own question ;)

should i re-post at the statistics SE site?

@TheBlastOne I always get lots of downvotes on my questions without any feedback or anything, so don't take it personally

It already helps if you ack that the question is wellformed enough to exist. I can live with downs as long as I have a chance for an answer (or constructive comment).

BBL

$$\displaystyle \prod\limits_{n \ge 0} \left(1 - \frac{2^{2^n}}{2^{2^{n+1}}+1}\right)=\displaystyle \prod\limits_{n \ge 0} \left(1 - \frac{1}{2^{2^{n}}+2^{-2^{n}}}\right)$$

The giant Pi symbol is for multiplying?

Yes @ᴇʏᴇs

How come there is $n \geq 0$ at the bottom of the symbol instead of $n=0$ and then $\infty$ above it

Then $$\prod\limits_{n \ge 0} \left(1 - \frac{1}{2^{2^{n}}+2^{-2^{n}}}\right)=\prod\limits_{n \ge 0} \left(\frac{2^{2^{n}}+2^{-2^{n}}-1}{2^{2^{n}}+2^{-2^{n}}}\right)$$

Then we can think of the simple fact that $(x-1)(x+1)=x^2-1$ or $\displaystyle x-1=\frac{x^2-1}{x+1}$.

Now, set, say , $x=2^{2^{n}}+2^{-2^{n}}$, and Mathematica showed me that (well, no need for Mathematica)

(excepting the latex part)

$$x^2-1=2^{2^{n+1}}+2^{-2^{n+1}}+1$$

and $$x+1=2^{2^{n}}+2^{-2^{n}}+1$$

On the other hand

if we denote $y=2^{2^{n}}+2^{-2^{n}}$, then ... (wait a bit)

@Chris'ssis please .. no spoilers please :P it's a journal problem :-) others might be trying it :D

Ready to continue now since we do nothing but what we did before, slightly modified.

@r9m What do you mean? I can show you such products I attended in the past with the date when I posted them. :-)

@Chris'ssis I did too :-) (I answered a similar question once .. close enough)

@r9m Well, I already put the main problem. I'm sure people already sent tons of solutions.

@Chris'ssis yes ofcourse !! :-)

So, we have that $$y=2^{2^{n}}+2^{-2^{n}}=\frac{2^{2^{n+1}}-2^{-2^{n+1}}}{2^{2^{n}}-2^{-2^{n}}}$$

From here the problem is almost done.

I think the final result is $3/7$ (unless I made some calculation mistake)

$$\prod\limits_{n \ge 0} \left(\frac{2^{2^{n}}+2^{-2^{n}}-1}{2^{2^{n}}+2^{-2^{n}}}\right)=\lim_{N \to \infty}\prod\limits_{n \ge 0}^N\frac{2^{2^{n+1}}+2^{-2^{n+1}}+1}{2^{2^{n}}+2^{-2^{n}}+1}\cdot \prod\limits_{n \ge 0}^N \frac{2^{2^{n+1}}-2^{-2^{n+1}}}{2^{2^{n}}-2^{-2^{n}}}$$

Something is not good. WAIT!

$$\prod\limits_{n \ge 0} \left(\frac{2^{2^{n}}+2^{-2^{n}}-1}{2^{2^{n}}+2^{-2^{n}}}\right)=\lim_{N \to \infty}\prod\limits_{n \ge 0}^N\frac{2^{2^{n+1}}+2^{-2^{n+1}}+1}{2^{2^{n}}+2^{-2^{n}}+1}\cdot \prod\limits_{n \ge 0}^N \frac{2^{2^{n}}-2^{-2^{n}}}{2^{2^{n+1}}-2^{-2^{n+1}}}$$

use $x_n = 2^{2^n}+2^{-2^n}$, for ease of writing :-)

@r9m I use f(n) and g(n)

okay .. whatever is easier to typeset for you :)

@r9m I got $3/7$. Sorry, a wrong sign.

@Chris'ssis :D I got that too ! :D

@r9m I learned one simple thing about these questions, for all of them: the most powerful identity to kill them all is $(x-1)(x+1)=x^2-1$. :-)

@Chris'ssis :D yes !!!!

@r9m :-)

@r9m hmmm, let me show you something but it's a bit to search for.

@r9m see this one that is done in the same spirit. Anyway, I have more like that solved and some are done by me and @robjohn here on channel.

@Chris'ssis :D Nice !! :D

@r9m Thanks. :-)

@Chris'ssis spoiler my solution :) hee

@r9m Is there a problem with latex? I cannot read.

@Chris'ssis is robjohn's render mathjax not working ?

@r9m NO, that's suprising!

@Chris'ssis which browser are you using ?

@r9m firefox, but it worked before when you gave me links.

@r9m I think you can make the LaTeX on your website render automatically without the visitor having to do anything on their part

@Chris'ssis ah! firefox has that problem ! that script does not see 'https' well :P and firefox opens every page in https by default :|

lemme check something

@r9m I put the whole thing in a texworks page.

@Chris'ssis oh! that always works :P lol

@r9m Nice! More or less, we used the same tools. You explained things better than me.

@Chris'ssis :) :)

@r9m Again, that identity I mentioned above is the key for all such questions. :-)

@Chris'ssis yes ! that's the master killer :D

@r9m lol, indeed!

@r9m I wonder if one can also add there some trigonometric functions ...

I'll think of such a version to be out of order.

@Chris'ssis trig functions ? hmm .. that would be interesting indeed :)

@r9m If I ever meet Titu I'll ask him when he plans to propose a problem like the one below

:-)))))))))

@Chris'ssis Nice !! is it very difficult ? :o

@r9m It's a matter of research. It depends, you need to know some auxiliary things.

okay :D

@r9m I'd like to see from Romanian people problems like the ones proposed by Ramanujan. I'd like to see again and again the need of the use of new ingenious tools.

@Chris'ssis :O .. I haven't seen another Ramanujan fan like you !! :O (I mean it as a complement :D .. )

@r9m Yeah. In a few days ... it's 26th April ...

My final exams start from 26th April ... God help me :P lol

hi @r9m

@r9m :-)

@BalarkaSen hello fellow :)

22 December 1887 – 26 April 1920

Anyway.

somehow I'm glad his life was short-lived :P

@r9m what have you been studying lately, then?

@BalarkaSen only exam er jonno gantacchi :P micro-economics and Electrodynamics :P

heh.

@r9m Because he also let us some things to discover? :-)

@Chris'ssis lul .. yes :P

@r9m Indeed! :-)))))

@r9m yikes, not good. i thought you were going to take mathematics exams.

@BalarkaSen ah! math I have to write algebraic NT, Sieve Theory etc

how much algebraic number theory did you learn, then?

i barely know the basics about class numbers, etc. studied kummer's proof of FLT for regular primes a long time ago.

hmm .. not much :| only basic stuff :|

forgot most of it...

I'm glad ..

:P

i have to study them at some point of time. i didn't take the classes on algebraic number theory this winter because i wanted to study homology thoroughly.

I took alg NT because I didn't have an other choice .. :P the other choice was non-linear analysis :P (now I have no interest in that :P)

i thought CMI provides a fair amount of subjects for one to choose.

yes ,, this is an even semester .. and most are algebra courses that I hate :P

oh. then you should be struggling with algebraic NT :( algebra is wonderful.

sigh

@r9m you're taking stuff like topology, etc, then?

I already have (5th sem)

i am not sure if schools provide specialized course on topology for undergrads.

e.g., diff. topo.

We have differential topology

How long does it take for one to become productive in mathematics? Almost whatever I discover myself, after searching through google, specifically wikipedia has been discovered in 18-20th centuries :|

These mathematicians of 20th century were really ***holes

or it could be the other way round as well :P

bbl

@r9m You pointing at me?

@ᴇʏᴇs I have you.

@JasperLoy I think you made a typo

@JasperLoy The 'v' should be a 't'

@ᴇʏᴇs I think it helps me to think that you said I have 100 per cent chance.

@r9m MathJax is not rendering on that page.

Hi @teadawg1337

@robjohn It is not your fault.

howdy, @Jasper, mr eyeglasses, @teadawg

Hi @TedShifrin

@JasperLoy why do you say that?

@robjohn firefox https problem ? :|

hi @robjohn and @r9m

@r9m I used to be able to replace the https with http, and it would work. Now, it forces the https.

@TedShifrin hey there... How long until you are in SD?

@robjohn ya ,, since the latest update firefox is acting like that :( I use google chrome .. It works once I disable security on the page ..

@robjohn It is my favourite movie line and also the line that keeps me alive whenever I feel guilt. Meant as a joke to you, perhaps.

@JasperLoy hi....Superman

@robjohn You have clearly not been following the transcript much these days, lol.

Hi @TedShifrin professor

@Ted!!

Where is skullpatrol?

I hope he is still alive.

Poets of the Fall ~ The Poet and the Muse // Lyri…: youtu.be/zZiYFFDGoh0 Listen to this song @JasperLoy you will like it I think

@BalarkaSen can I assume the function the function injective

Hi @teadawg1337

i dunno what you mean.

oops ... was off talking to colleagues.

planning to apartment-hunt end of May, move in July, @robjohn. Fingers crossed

hi @Sayan, @Balarka

Hey @Ted, I heard back from my advisor; two professors from MTSU and one from Tennessee Tech are interested in meeting with me. One of the professors from MTSU is the mathematics department head :D

moving must be such a pain, @Ted.

Very cool, @teadawg. Very cool, indeed :)

Are those reasonable drives for you?

@Balarka: I've been in the same house for about 30 years, so I've forgotten :)

You're moving to California, right?

@BalarkaSen I move every day.

Yup, where there's no water.

MTSU is only about half an hour away, whereas TTU is an hour and a half away (without traffic)

Any plans what you'd do after retirement?

Oh, that's still not bad, @teadawg. Very exciting. If you ever get interested in geometry, I'll be glad to talk to you :P

@TedShifrin I will be glad to talk to you if you are interested in cooking for me. =)

Not worrying about plans at this point, @Balarka. I'll wait and see if I miss teaching. Right now, I'm fed up.

Gee, how generous of you, @Jasper :D

indeed, a class full of D's is not encouraging for the professor.

@Ted Who says I can't do both geometry and analysis? :P

The division into different branches of math is artificial.

@teadawg: That's called PDEs :)

@BalarkaSen the theorem you gave which I have have to prove can I assume that the function is injective

? repeat the question. i forgot.

Somewhat, @Jasper, but there are people who work in very specialized corners of subjects. There are also some of us who work in areas which overlap 3 or 4 fields of math. It just depends.

I am using my phone I can't write LaTeX

@TedShifrin Yes, for me, I work on an area called 'mental problems' now, lol.

That's a very specialized corner, @Jasper.

@SayanChattopadhyay don't use latex.

@Ted realization : number theory is the branch where almost every branch of math interacts.

f is a function from a set A to itself o have to prove that this function is bijective and set A is finite

that was not the question :P

and not all functions from A to itself are bijective either.

you've either misread the question or forgot what it is

I am glad @BalarkaSen is advising @SayanChattopadhyay.

I am not.

I'll burn out very quickly if I don't experiment with other fields of mathematics. I hope to move on to differential geometry at some point, since it relies heavily on higher dimensions

(I'm starting to lose interest in working with only two dimensions)

That, I suppose, requires a lot of mult. calc, @teadawg1337.

Which in turn requires linear algebra.

@Balarka Indeed, which shouldn't be a problem for me.

btw, @Ted : no one ever told me rank nullity theorem is just splitting lemma. such stupid. :P

@Ted Should I take a multivariable calculus course before linear algebra, after, or concurrently?

@BalarkaSen is it given it is injective......I can't remember

the problem was to count the number of bijective functions from {1, 2, ..., n} to itself.

@TedShifrin I think you might like to know how I see the other "participants" to this amazing game of computing integrals, series and limits. My profound desire is to see them all extremely good, like Ramanujan or even higher, and all learning from each other, never having in mind the idea of comparing to anyone of them. In my case I only compare with myself, and try to beat myself. I would be very bored if I were the best one in all I do. It's very hard for you to ever understand me.

you had a good idea, but now you've forgotten everything...

Oh no I found this one .....the other one......the other question

This is n!

Oh, hello @Sayan : I forgot to respond earlier

@SayanChattopadhyay Prove it.

@TedShifrin I'm not upset on you for the last message, but I only say you don't understand me. In art, this is what I do, all admire the masterpieces, not fight against each other for showing which one is the greatest.

There was no other question.

Right, you're done.

Assuming this is an integer-valued function, yes

Or it maps from the set of integers to all reals

Oh I mixed up your question with someone else's sorry for that :p@BalarkaSen

@teadawg1337 Not integer valued. It's a map from {1, 2, ..., n} to itself.

Yup

@Balarka I see, I didn't know the original context of the problem

Yes that's why

@Chris'ssis Nobody understands you because you are a genius.

@SayanChattopadhyay Right, ok. Now let $\text{Map}(n)$ be the collection of all bijective maps from $\{1, 2, ..., n\}$ to itself.

You've just proved that $|\text{Map}(n)| = n!$

@JasperLoy Just a usual kid from the countryside. ;) (You said it's "a usual")

@JasperLoy I understand her to a certain extent, but I think most of her prowess comes from experience in addition to natural aptitude.

@Sayan Let $f, g \in \text{Map}(n)$. What can you say about $f \circ g$?

@Teadawg: Most multivariable courses don't use any linear algebra. (Unless you follow my lectures, where it's all integrated.) Typically, multivariable is more plug-and-chug, and linear algebra starts to have a few proofs in it.

BBL (I need to run a bit)

lol

a fair answer.

Nicolas Bourbaki

lol

@TedShifrin That's what I figured. I'm signing up for summer and fall classes this week, so I wanted to hear your input before potentially screwing myself over :P

@teadawg1337 I'm going to study multivariable calculus the next month too.

It's fun stuff, @Balarka

I guess so.

@Huy Rank-nullity is just splitting lemma :P

Hey @Huy

Hi @BalarkaSen

Hey there, @evinda

@evinda Aha!

hi

@BalarkaSen: Yeah, pretty much

@TedShifrin Do you cover the spectral theorem in your course

How are you? @BalarkaSen @Huy

@Huy i've skipped the proof.

i can prove splitting lemma by diagram chasing so don't see the point of reading the tedious proof of a much easier case.

@JasperLoy ^^

:D

@evinda ok.

Hi, How many elements are in $\mathbb{Z[i]/<3+i>$? I think infinite, what do you guys think? just confirming

Z[i]/<3+i>

sorry, deleted by accident. @zed111 as a start, when do we have $n \in <3+i> $ when $n \in \mathbb{Z}$?

Hi @Hippalectryon

Anyone interested ?

@ᴇʏᴇs :D Hellu

@JC574 I didn't get what you mentioned. For n \in Z, we can have infinite number of elements of the form n + <3+i>, so I thought infinite?

Yes, @zed111. But you have to make sure that two integers are nonequal mod 3 + i

@zed111 yeah. but make sure you reason it carefully. For example $10+<3+i> = 0 +<3+i>$

@JC574 really? how did you get that?

@zed111 10 - (3 + i)(3 - i).

<3 + i> is an ideal of Z[i], notice that.

Thanks @BalarkaSen @JC574