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12:17 AM
@r9m I have fantastic news here ...
 
r9m
@Chris'ssis :D
 
@r9m Do you guess what's about?
 
r9m
@Chris'ssis umm .. you solved some uber cool problem ? :D
:15968438 Mother of god !! how do you see these stuff ?? !! .. honestly I didn't try that one ... looked too difficult to me from the get go :| ..
 
@r9m Tomorrow you have that proof (complete).
I need some sleep now.
@r9m It's just a matter of practice (for someone with no background in mathematics). :-)
 
r9m
@Chris'ssis maybe .. but you are too talented !!!! .. I guess you see these stuff with your eyes closed !!
 
12:28 AM
@r9m Not really, but I like to bend things in my mind (you know, like graphs of the functions) and imagine many things.
 
r9m
@Chris'ssis LOL :P
 
@r9m Yeah ... that's it! :-)
I'm out for some sleep. Have fun! ;)
 
r9m
@Chris'ssis Gdn8 :)
 
 
2 hours later…
2:05 AM
@Chris'ssis I hope you slept well. I was very busy today offline. I will be busy again tomorrow. I won't be back much until Monday morning.
 
r9m
2:30 AM
@robjohn very nice answers here and here :D
 
@r9m thanks. The big-O answer came to me just as I was falling asleep.
 
r9m
@robjohn is it easier to show it with contradiction ? .. I mean is there a direct way of deriving the result ?
 
@r9m I don't know. If so, I would like to see it. Some problems are much easier to handle with contradiction. Are you against contradiction?
 
r9m
@robjohn nothing as such ... the construction is mind blowing !! I don't think I would have learnt such a construction if you proved the result directly (instead of the contradiction proof) :-)
@robjohn functions that satisfy that property seems interesting ... does it have deeper properties ?
 
@r9m which property?
 
r9m
2:45 AM
@robjohn for any absoulutely convergent series $\sum_{n=1}^{\infty}a_{n}$,the series $\sum_{n=1}^{\infty}f(a_{n})$ converges
 
@r9m I believe that $O(x)$ is iff in that case.
 
r9m
@robjohn okay ..
 
@r9m absolute convergence is necessary.
 
r9m
@robjohn what happens if it converges but not absolutely ?
 
@r9m I think the situation if you replace "absolutely convergent" with "convergent" is much more complicated.
 
r9m
2:52 AM
@robjohn interesting .. !
 
@r9m for example $f(x)=|x|$ is not good
 
hey guys. I was wondering could you take a look at this really quickly.
1
A: alternate proof to alternating series test

TomSome intuition why your proof isn't quite correct: When you used the triangle inequality $$|(-1)^{n+2}a_{n+1} + \cdots + (-1)^{n+k'+1}a_{n+k'}| \leq |(-1)^{n+2}a_{n+1}| + \cdots + |(-1)^{n+k'+1}a_{n+k'}|\\ = a_{n+1} + \cdots + a_{n+k'} $$ you removed the alternating nature of the series, so if y...

 
r9m
@robjohn yes .. thats the first thing that comes to mind ofcourse :)
 
I'm just wondering if what I have is correct. the user used the definition of cauchy series. I need to use cauchy convergence
 
3:06 AM
hello?
 
@user60887 what you have is correct, but a bit more complicated than it needs to be.
 
ok thanks
i have problems with over complicating things unfortunately.
 
3:42 AM
@MikeMiller It seems you made a great question on MO after all!!!
 
4:00 AM
Ah shaddup.
 
@MikeMiller I'm serious.
I saw Zev and Mariano share it on Facebook.
 
It's the answers that are great.
 
Well, you must have made a good question to trigger good answers.
 
 
1 hour later…
r9m
5:10 AM
@Sawarnik In case you still care :P I added an alternative proof (based on trig ineq) here .. I also proved a few related Jack Garfunkel's inequalities .. tell me if you are interested in the proofs :D :)
@Chris'ssis @robjohn I have a nice limit problem :D $$\lim\limits_{n\to\infty} \frac {n}{\ln^2 n}.\left[\left(\sqrt[n]{H_n}\right)^{H_n}-1\right]$$, where $H_n = 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}, n \ge 1$ :)
 
5:34 AM
@r9m, i get 0 on that
 
5:54 AM
Heya.
 
hola @Studentmath
 
How are you @David?
 
good good.. did family stuffs in the morning, went to a tattoo convention with my girlfriend in the afternoon, and doing math now in the evening
 
Oh, it's evening now..
Sounds like an un-wasted day
 
yup, so what are you working on tonight?
 
6:06 AM
Continuing with Java, hopefully will be done by the 10th and move back to math
No idea how I iwll make it all in time :P
 
r9m
@DavidKirby thats right :D .. how did you approach it ? :)
honestly .. it saddened me when I found the answer is 0 .. expected something like a more flashy constant :P lol
 
6:26 AM
@r9m, well, it was easy to see that the log went to infinity, and the harmonic series goes to 1, and the '-1' made it zero. so 0/infinity = 0. l'hopital's rule i think?
 
r9m
@DavidKirby irk .. I don't get what you say :| (sorry)
 
i split it up like this : $$\lim_{n\to \infty } \, \frac{n \left(\lim_{n\to \infty } \, \left(\left(H_n\right){}^{1/n}\right){}^{H_n}-1\right)}{\log ^2(n)}$$
 
r9m
@DavidKirby okay .. that is $0/0$ form :)
$\lim\limits_{n\to\infty}\ \left(\sqrt[n]{H_n}\right)^{H_n} = 1$
 
6:53 AM
i love the harmonic series.. here's a fun identity i noticed the other day : $$e^z=-1+\frac{2}{1-\frac{\pi z}{-i z H_{\frac{i z}{2 \pi }}+i z H_{-\frac{i z}{2 \pi }}+2 \pi }}$$
 
Finally.
I finished the problem I have been trying for weeks.
 
congrats
 
Thanks, @DavidKirby.
 
care to share or are you in the midst of writing a paper (et al) on the problem?
 
@DavidKirby I am not writing a paper, no. But I'd have to do that soon.
Well, here's what I did, in short :
A quintic is not solvable by radicals, which comes from the basics of Galois theory. What I proved is that the Poincare autforms satisfy a general quintic, which comes from the fact that a certain algebraic curve $X(5)$ "looks like" a Riemann sphere with exactly 12 points removed from the vertices of some circumscribed icosahedra.
So, strictly speaking, the riemann surface of $x^5 - x - Z$ "looks like" the curve $X(5)$, which is quite interesting.
Alongside, I derived a complete different proof of the exceptional iso $A_5 \cong PSL_2(\Bbb F_5)$ and solved the inverse galois problem for $PSL$ groups.
 
7:10 AM
Greetings
 
@Studentmath Morning. :)
 
@r9m I compute that limit without pen and paper
$$\lim\limits_{n\to\infty} \frac {n}{\ln^2 n}.\left[\left(\sqrt[n]{H_n}\right)^{H_n}-1\right]$$
 
r9m
@Chris'ssis how ? :D
 
@Balarka Sen, that's beautiful.. intuitively i can understand but i wouldn't know where to begin to prove it! nice
 
Heya @Ilan
 
7:13 AM
@r9m The limit is $1$.
 
r9m
@Chris'ssis irk .. I get $0$ !! .. maybe I'm wrong .. lemme double check
 
@r9m Let me check it again ...
 
@DavidKirby Thanks. I suspect that no abstract proof of solvability of quintics over function fields $\Bbb C(j)$ has been stated anywhere (though this fact was known for years), but even if it is not the case, my understanding increases 10fold and it was an extreme pleasure to discover this wonderful connection between algebra and geometry.
 
@r9m I did somehting wrong ... wait
@r9m You're right, the limit is $0$.
@r9m let me write the proof
@r9m $$\lim\limits_{n\to\infty} \frac {n}{\ln^2 n}.\left[\left(\sqrt[n]{H_n}\right)^{H_n}-1\right]=\lim\limits_{n\to\infty} \frac {H_n \log H_n}{\log(n) \log(n)}\cdot \lim_{n\to\infty} \frac{\left[\left(\sqrt[n]{H_n}\right)^{H_n}-1\right]}{\displaystyle \frac{H_n}{n}\log(H_n)}=0$$
 
r9m
@Chris'ssis meesa jaw drops ! .. Awesome !!
 
7:24 AM
:D
 
@Balarka Sen, ... playing around with this surface and it's wild. could you explain just a smidgen more about the inverse?
 
Which inverse, @DavidKirby?
 
@Balarka Sen, specifically i was wondering about the inverse of $x^5-x-Z$, but then i realized that 'automorphic' was the word i was looking for
@Balarka Sen, but yeah, no need to explain further. i need to find the right questions before getting the right answers, so reading up on the whole shebang at the moment. ;)
 
@DavidKirby Yeah, well, that's the whole point, proving that the inverse is algebraically dependent of poincare automorphic forms. What is more, one can transform the quintic into a sextic. This is a consequence of $A_5 \cong PSL_2(\Bbb F_5)$ (the former acts of 5 points and the later on six $\{0, 1, 2, 3, 4\} \cup \infty$.
This sextic is the desired surface we look for. Indeed, it's the modular equation for $X(5)$ and is parameterized by ellmod functions.
 
r9m
@Chris'ssis Lemme bounce a integral inequality back :P .. $$ \displaystyle \frac{1}{2} + (1-\frac{2}{\pi})\int_0^{\pi/4} \cos(\sin x)\,dx \le \int_0^{\pi/4} \dfrac{\sin(\sin x)}{\sin x}\,dx \le \dfrac{\pi}{6} + \frac{1}{3}\int_0^{\pi/4} \cos(\sin x)\,dx$$ :D .. looks nice :-)
 
7:39 AM
@DavidKirby Sure. If you are interested, you can look at some texts like R. B. King or so.
 
@r9m :-)
 
r9m
@Chris'ssis have you seen this one ? :)
 
@r9m did you use this one to solve my one?
 
r9m
@Chris'ssis nope :-)
 
@r9m OK
 
7:47 AM
@Balarka Sen, ok, so that makes total sense now.. quite curious to read the paper, please let all us chat dwellers know when it's released! :)
 
@DavidKirby haha, i don't think it has enough importance to be published but will see.
first i have to show my calculations to a few people like my professor and professor V. Kumar Murty, who proposed the problem and asked me to write it up.
 
r9m
@N3buchadnezzar @Chris'ssis How's this $\displaystyle \int^{\pi}_0 x\left(\sin^2(\sin x)+\cos^2 (\cos x)\right) \,dx$ ? :D
 
@Balarka Sen, I often wonder about the results that seem unimportant or not worth publishing -- a web app where those could be published and reviewed/refined/appropriated/etc could be pretty neat.
One person's statistically insignificant regression coefficient is another person's treasured additive white Gaussian noise.
 
@DavidKirby do you know of arxiv?
it's probably the kind of place you want.
 
@Balarka Sen, ah i suppose preprints have gotten quite informal these days
 
7:59 AM
@r9m This is an old story ... $\pi^2/2$
 
r9m
@Chris'ssis :-)
 
@r9m let $x\mapsto \pi-x$ and use symmetry
 
r9m
@Chris'ssis right-o-right :) (y)
 
But @r9m might be interested.
 
@r9m let me recheck that inequality ...
 
r9m
8:06 AM
@BalarkaSen looks like an older result .. atleast seen it b4 2014 :| .. but I'm not sure ^^'
 
@r9m For real $n$?
 
r9m
@BalarkaSen ah I see .. it has [3] Mitrinovich Elementary Ineq reference :) .. nice result !!!
 
@r9m lol, your inequlities are stronger than mine ...
 
r9m
@Chris'ssis are you talking about the integral inequalities ?
 
@r9m aha
 
r9m
8:13 AM
<- grins :D ;) @Chris'ssis
 
@r9m What did you use there?
 
r9m
@Chris'ssis its a one liner .. If I tell the trick the entire tree falls down in one swoop :-)
 
sigh I hate this course..
 
@r9m I'm preparing to make an announcement ... I got a stronger inequality ...
 
r9m
8:28 AM
@Chris'ssis I'm all eyes .. O-O
 
@r9m $$\frac{1}{24}+\frac{11 \pi}{48}\le \int_0^{\pi/4} \dfrac{\sin(\sin x)}{\sin x}\,dx\le \frac{19}{480}+ \frac{883 \pi}{3840}$$
 
r9m
@Chris'ssis Mother of God :D
 
@r9m However, this form doesn't contain $\int_0^{\pi/4} \cos(\sin x)\,dx$
 
@r9m LOL Where do you take these pictures from? :-)
 
r9m
8:37 AM
@Chris'ssis try 'mother of god troll meme' on google :P
rage comics
 
@r9m maybe you think I don't know what you used above ... but you'd be wrong thinking like that :-)
 
How do I calculate historical yearly market worth knowing only its yearly growth rate and value for year 2015?
"According to MarketLine, the global home improvement market is expected to grow at 2.5% yearly rate between 2010-2015 and is predicted to be worth almost $678 billion by 2015"
 
@r9m Everitt's inequalities (I just found my old paper ...) :D
 
I have 2.5% and 2015 value but what I need to know is value from 2010 until 2014
 
r9m
@Chris'ssis OMG ... !! that has a name ????!!!! .. whats the Everitt's inequality ?
 
8:45 AM
@Boris_yo for each year?
That would be regular algebra, generally, you can define 2015 as your 0 point for X axis, and then you get the function $f(x)=1.025x+678$ considering every point on y axis is in billions
 
@r9m This one? :-) $$\frac{4}{5}-\frac{x^2}{15}+ \frac{1}{5}\cos(x)\le \frac{\sin(x)}{x}\le \frac{2}{3}+\frac{1}{3} \cos(x)$$
 
If you just want the year values, recall it grows 2.5% yearly, so you can have the 'magic square' to do the rest of the work for you. But it's the same as the function.
 
0
Q: How to calculate historical data knowing only yearly percentage increase and data for year 2015?

Boris_yoHow do I calculate historical yearly market worth knowing only its yearly growth rate and its martket worth for year 2015? Example: According to MarketLine, the global home improvement market is expected to grow at 2.5% yearly rate between 2010-2015 and is predicted to be worth almost $6...

 
@r9m and now -> youtube.com/watch?v=LaTGrV58wec :D (everybody dance now)
 
r9m
@Chris'ssis the upper bound is that one .. ;-) .. the lower bound is $\frac{\sin(x)}{x} \ge \frac{2}{\pi} + (1-\frac{2}{\pi})\cos x$
 
8:51 AM
@Studentmath Is this for Excel?
 
r9m
@Chris'ssis hahahahahaha !!!! U the devil man :P
 
@r9m lol, at my last job I was spoiling like that "little evil&genius" :D Well. just a spoiling ... :D
 
@Studentmath I am making presentation not for someone specific but I want it to be on SlideShare website. I only have yearly 2.5% and 2015 economical data. If I calculate to discover historical economical data, will it be legit?
 
@Boris disregard what I wrote, it's wrong
 
@Studentmath What you wrote is wrong?
 
r9m
8:53 AM
@Chris'ssis haha .. I don't know what you speak of :-)
 
Yeah, a bit tired. You should go with compund interest to compute it.
 
@r9m that's because you said above "devil man" ... of course, I'm not. Just a shy girl here ;)
 
Okay what if I had 2 data points - year 2010 and 2015?
 
r9m
@Chris'ssis well the shy girl is the ideal avatar for the devil man :P lol
 
hahahaha :D
 
r9m
9:01 AM
@Chris'ssis you said old paper .. did you write a paper on that inequality ?
 
@r9m No, it's just an inequality I wrote it down in some notebook (well, I said "paper") a while ago.
 
r9m
@Chris'ssis I see .. eagle eyes !! :)
 
r9m
@Chris'ssis how do yo propose we get to the lower bound $\frac{\sin(x)}{x} \ge \frac{2}{\pi} + (1-\frac{2}{\pi})\cos x$ ? :)
 
hmmm. I'd like to explain that geometrically
 
r9m
9:12 AM
@Chris'ssis Geometry ?!! .. I'm gonna love it ! :D
 
@r9m I think I also saw it here some time ago arxiv.org/pdf/1105.0859.pdf. I didn't search for it now, but I think it is there or something very similar.
 
r9m
@Chris'ssis okay .. lemme check :) Thanks :)
 
@r9m There you have a bunch of awesome inequalities. By the way, that integral inequality I showed you some time ago only uses the elementary stuff in high school. No need for powerful inequalities.
 
r9m
@Chris'ssis okay .. :-) .. ya just calculus .. :-)
gotta go for lunch .. bbl
 
In analogy to perfect numbers, let's say I wanted a set of numbers whose average equals the sum of their common divisors. Can such a thing even exist?
 
9:28 AM
How do you calculate this using Windows calculator?
 
@r9m Enjoy your lunch! ;)
 
9:46 AM
@Chris'ssis Another not-so-important paper on arxiv.
 
@BalarkaSen Well, for my art is very important.
 
@r9m I get $0$ as well.
I see that others have gotten the same thing, now that I read further
 
r9m
10:02 AM
@robjohn how did you approach it ? :D
 
10:28 AM
@r9m did you know the Problems in mathematical analysis by Kaczor ?
 
r9m
@G.T.R ya ,, there are 3 volumes right ?
 
@r9m yes, I wish I found about them earlier
 
r9m
@G.T.R its full of awesome problems .. I came to know about them from a comment in a question in the main .. a few months ago :)
 
@Chris'ssis I meant in general mathematics.
Your "art", integral and series computation, is fun but not that much important in mathematics, right? =)
By "important" I don't mean useless, but rather an "important tool". Integrals and series closed forms are used, but not frequently as they aren't usually a good tool in general branches of mathematics.
 
@r9m I estimated $H_n\sim \log(n)$, then $$\log\left(H_n^{H_n/n}\right)\sim\frac{\log(n)}{n}\log(\log(n))$$ Then for small $x$, $e^x-1\sim x$
 
10:41 AM
@BalarkaSen No, you're not right. Could you imagine life without integrals? Would the progress be possible without integrals?
 
@Chris'ssis I said not much important (of a tool).
I didn't say completely useless. Life without integrals would be a pain.
 
r9m
@robjohn ah .. Nice :) !!
 
@BalarkaSen Glad to read that! :-)
 
@r9m so the whole thing goes as $\frac{\log(\log(n))}{\log(n)}\to0$
 
@Chris'ssis Well, that's the truth.
 
10:42 AM
@BalarkaSen I know.
 
In any case, I am getting a little involved with integrals lately, to be frank. Have you read about this?
It's a recent on-development branch in transcendental number theory.
 
Heyall
 
@N3buchadnezzar Ahoy
 
Are there any examples where
$$ \int_a^b f(x)g(x) \mathrm{d}x =\left( \int_a^b f(x) \mathrm{d}x \right) \left( \int_a^b g(x) \mathrm{d}x \right) $$
?
=)
 
@N3buchadnezzar Have you tried Fubini?
 
10:47 AM
Lemme give it a try
 
I'd actually look for global function before looking at local ones. $$\int f(x)g(x) \mathrm{d}x = \left ( \int f(x) \mathrm{d}x \right ) \left ( \int g(x) \mathrm{d} x \right )$$
 
It's respectively easy to solve differential equations than evaluating definite integrals.
 
@BalarkaSen Seems tough though
 
@N3buchadnezzar Definitely.
Convert it into a standard differential equation first. What do you get?
Eh, just do this $$F'(x)G'(x) = F'(x) G(x) + F(x) G'(x)$$
$F'(x) = f(x)$ and $G'(x) = g(x)$
now solve it.
 
10:55 AM
The righthandside looks like the product rule
 
It is the product rule, silly. smack on the head
 
:p
So the idea would be to find an f, then F, and use the above relation to find G and eventually g
 
My idea is to solve the differentio-functional equation above than doing that.
 
Can you solve it explicitly though?
 
At least solve it implicitly first. Then think about being explicit.
 
11:00 AM
$$ F'(x)/F(x) = \frac{G'(x)}{G'(x)-G(x)} = \frac{1}{1-G(x)/G'(x)} $$
 
That's what I have.
But then I can't. Thank you!
 
The lefthandside is just the derivative of log F(x) right?
 
@N3buchadnezzar Yes.
But the RHS is weirdo.
 
Yeah
$$F(x) = \exp \int \frac{\mathrm{d}x}{1-G(x)/(G'(x))}$$
Something like this?
 
That'd do the trick.
You have a relation between $F$ and $G$. (It'd have been nice to have a simpler expression though)
Well, I guess you can expand that in a series.
But, we are drifting away from the original problem/
Let $G(x) = x$
 
11:06 AM
The righthandside integral seems impossibru to solve for even for the simplest class of functions though
hence not very usefull ?
 
$$F(x) = \exp \int \frac{dx}{1-x}$$
And what's that?
 
And not correct even for the simplest functions?
 
@N3buchadnezzar isn't a relation between $F(x)$ and $G(x)$ sufficient?
just have an arbitrary $G(x)$ and plug in $F(x)$ get $f(x)$ and $g(x)$ and convert to definite integrals.
$G(x) = x, F(x) = \frac1{1-x}$
Eh, we can just check the diff equation
 
Okay, but I found the following pair ^^
\begin{align*}
f(x) & = x^2 + ax + 1 \\
g(x) & = x^2 + \frac{1}{1+a}\left( \frac{16}{15} - a \right) x + 1
\end{align*}
 
Is my $F$ and $G$ above OK?
 
11:12 AM
I can check
 
Do they satisfy $(FG)' = F'G'$?
Of course. We forgot the constant of integration.
In this case, $(FG)' = -F'G'$
 
They do not look the same to me
 
@N3buchadnezzar ?
$FG = x/(1-x)$
 
@BalarkaSen -x-ln(-1+x) and -(1/2)*ln(1-x)*x^2
 
@N3buchadnezzar What's that?
Why the hell are you integrating?
 
11:14 AM
fg and int f \cdot ing g respectively
Silly me, forgot I said anything
 
Hey @N3b, @Balarka
 
Hey @Studentmath
 
@Studentmath Yo
 
How goes it?
 
@Studentmath We integrate do.
Differentialing solve equation.
 
11:17 AM
Sounds like fun
 
@Studentmath Fun at not all.
@N3buchadnezzar We where were, so?
Should I cut it out?
 
Seems like only the constant is missing
 
@BalarkaSen Divide by $F(x)G(x)$ and things become simpler
 
Addicting is it, no?
Well, more fun than trying to open canned food without a canopener, Balarka... always remember that
 
@robjohn You do that.
I don't think it gets simple at all.
OH!
$$\frac{F'}F \frac{G'}{G} = \frac{F'}{F} + \frac{G'}{G}$$
 
11:24 AM
@BalarkaSen yes
 
$A(x) B(x) = A(x) + B(x)$
That's the functional equation.
You're such a genius, @robjohn.
 
so you have $(u'-1)(v'-1)=1$
 
@robjohn Uh?
 
$u=\log(F)$ and $v=\log(G)$
 
Oh.
And that's much simpler.
 
11:27 AM
@BalarkaSen given an $F$, you can integrate to get $G$
 
@robjohn Yes. But is it any simpler than our formula?
 
@robjohn Is that not what we did above?
 
25 mins ago, by N3buchadnezzar
$$F(x) = \exp \int \frac{\mathrm{d}x}{1-G(x)/(G'(x))}$$
 
25 mins ago, by N3buchadnezzar
$$F(x) = \exp \int \frac{\mathrm{d}x}{1-G(x)/(G'(x))}$$
 
@N3buchadnezzar let me look. I saw the original equation, but didn't see everything
 
11:28 AM
What the...?
 
@N3buchadnezzar I guess that is equivalent. Never mind :-)
 
@robjohn Equivalent but useless for any set of interesting functions beyond the trivial ones:p
 
@N3buchadnezzar Not useless.
 
Do a quadratic polynomial, I dare you
 
11:30 AM
Sure.
$G(x) = x^2 + x + 1$
$G'(x) = 2x + 1$
Crap. Do me a favour and compute the integral for me, @N3buchadnezzar
 
-3*ln(x-1)+ln(x)
 
I get $$\log(F(x))=x+\int\frac{\mathrm{d}x}{G'(x)/G(x)-1}$$
 
@robjohn No.
 
@BalarkaSen it is the same
 
I don't think so.
@robjohn huh?
@N3buchadnezzar You sure?
 
11:32 AM
@BalarkaSen bring the $x$ inside the integral as $1$ and add
 
@robjohn Oh, ok.
$F(x) = \exp(-3\log(x-1) + \log(x))$, right, @N3buchadnezzar?
$= \frac{1}{(x-1)^3} \cdot x$
So $FG = x(x^2+x+1)/(x-1)^3$
 
Missing the constant agian
 
Righto. but that can be thought of later.
do me a favour and compute the derivatives for me.
i don't have papers in front of me.
@N3buchadnezzar Jinx like confundus?
 
Here is an interesting pair: $F(x)=e^{x+e^x}$ and $G(x)=e^{x-e^{-x}}$
 
@robjohn OK. Did you get that through that formula?
 
11:37 AM
@BalarkaSen Nope... I used $\left(\frac{\mathrm{d}}{\mathrm{d}x}\log(F(x))-1\right)\left(\frac{\mathrm{d}}{‌​\mathrm{d}x}\log(G(x))-1\right)=1$
 
@robjohn Oh, ok. But it can be derived through that one too, probably.
The formula is useful, @N3
What if... I chose $G(x) = \exp(x)$?
 
I thought you were looking for pairs, $F$ and $G$. The formula can give you a $G$ given an $F$, but not all $F$'s yield an elementary $G$.
 
No corresponding $F$ for that one.
@robjohn Yes, but some doesn't yield a $G$ at all.
Like the one I showed above.
 
@robjohn They do not look the same to me =(
 
@BalarkaSen yes. The equation I was using yields many pairs.
 
11:40 AM
@N3buchadnezzar They are.
I checked.
 
@robjohn Right. That's superior.
 
@BalarkaSen if you're looking for pairs of functions, I think so.
 
Our works provide some realistic appllications, @N3buchadnezzar. Ad :

Tired correcting the usual high-school mistake $(uv)' = u'v'$? Need a hint? TRY OUT OUR NEW LIST OF NIFTY FUNCTIONS $(F, G)$!
Contact blah blah and blah blah for getting one. A pound each pair.
Gotta go.
 
r9m
12:21 PM
I love it when the OP has already done more than half the job :D .. feels nice (saves effort) ;)
 
such as?
 
r9m
@skullpatrol hi :) .. the OP had already proposed a conjecture .. all I had to do is connect the dots :)
 
yes, that is a pleasant task indeed :-)
 
r9m
difficult questions are those which I have to build from scratch and explain everything :P
 
agreed
that's the job of the textbook, in my opinion
 
r9m
12:54 PM
right .. :)
 
1:05 PM
@Chris'ssis it's about to begin
 
@G.T.R Yeahhh :-)
 
@robjohn is there a way to get the list of questions posted by a deleted user ?
 
r9m
@robjohn [this](http://math.stackexchange.com/a/160352/129017) is a very 'powerful' answer !! (tells a lot) **awesome** :D ..

@Chris'ssis nice question this one ! .. is it one of your creations ? :)
 
@r9m Yes. Well, some might have thought of it before me. I also saw it in other parts.
 
r9m
@Chris'ssis Cooool !!!
 
1:16 PM
@r9m There is a big problem ... no elementary answer ...
 
r9m
@Chris'ssis ah ... okay :-)
 
@r9m I mean there is no elementary answer to that question I posted (from the link you showed me).
@r9m Maybe you find one ... :-)
 
r9m
@Chris'ssis I'm on it :D .. but the answers already provided are so rich .. I feel poor :|
@Chris'ssis what was your motivation when you created the problem ? :)
 
torture? :D
just kidding
 
r9m
@skullpatrol lol :P .. thats always microtext/ written in invisible ink :P
mee kidding too :P
 
1:31 PM
no pain, no gain
2
I guess that's why they call it a discipline...
 
r9m
indeed
@Chris'ssis we could go for finding $O(R(n))$, where $ \displaystyle R(n) = 1 + \frac{e^n.n!}{2n^n} - \frac{n!}{n^n}\left( 1+ \frac{n}{1!}+ \cdots +\frac{n!}{n!} \right)$ :-) .. but thats like tracing back robjohn :P
 
:D
@r9m It did out of curiosity while working on a different problem.
 
r9m
@Chris'ssis curiosity !! .. you sound like Hannibal Lecter :P
 
@r9m Ah, wait a second then ...
This song is magnificent though. :-)
 
r9m
1:47 PM
@Chris'ssis I'm Hans Zimmers fan ... die hard fan !!!!!!
 
@r9m I'm a Hans Zimmers fan too. :-)
 
Me three!!!
 
@skullpatrol Welcome to the club ... :-)
 
@r9m thanks.
@G.T.R not that I know of...
 

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