00:00 - 16:0016:00 - 00:00

12:00 AM
@PeterTamaroff Not young enough!
user19161
@PeterTamaroff I think that would be a secret.
@amWhy You should give me a hard integral I should solve txo know it
@PeterTamaroff hahahahaha! I'll work at it!
It should be the solution to some arbitrary kind of problem.
user19161
@amWhy I have thought of what secrets to share with you and how to go about it, you will see in a few days time. =)
12:03 AM
@WillHunting 8)
secrets are for girls
user19161
Exactly.
where is the garbage collector?
user19161
Gone.
why?
user19161
12:10 AM
@amwhy It is very interesting that both John Lee and Jeffrey Lee wrote books on differential geometry. They have the same initials as me.
user19161
@aDangerousIdea Hmm, she might return, I don't know.
user19161
@WillHunting Rihanna
user19161
Oh I thought that was Victoria. =)
12:17 AM
@WillHunting Victoria who?
user19161
@aDangerousIdea Victoria's Secrets. Get it? =)
user19161
HAHAHAHAHAHA
user19161
@robjohn Also aim for Legendary.
@WillHunting Interesting there is one for 50 times and 150 times, but not 100
user19161
12:23 AM
@robjohn Interesting you used interesting like I always do.
@WillHunting interesting...
user19161
I think I will go to bed now, when I wake up it will be 1 Dec.
user19161
Christmas is coming!
@amWhy Hey. This looks crazy, but I think it works, @robjohn
Howdy, I have a minor question. Let $\alpha \in [0,1]$. $\forall n, \vert \alpha - \frac{p}{q} \vert <\frac{1}{q^n}$ has $\alpha$ many solutions. How is this the set of all Liouville numbers in [0,1]?
12:29 AM
@LinearMan $\alpha$ many? What if $\alpha = 1/\sqrt 3$?
@robjohn $$s_n=\sum_{k=1}^n \lambda_{n,k} \chi_{\lambda(n,k)}$$
Where $\lambda_{n,k}=\inf \{f(x):x\in \lambda(n,k)\}$
and $\lambda(n,k)=\{x\in[a,b]:t_{k-1}<f(x)<t_k\}$
and $t_k=m_f+k \Delta_n$
And $\Delta_n=(M_f-m_f)/n$, with $m_f=\inf f$ ,$M_f=\sup f$ over $[a,b]$
Lebsgue, stahp!
I think $\Lambda_{n,k}=\sup \{\cdots\}$ would have worked too.
Or any inbetween values
Now, I'm thinking that $f(x)-s(x)$ can be at most $$\max_{1\leq k \leq n}\{\lambda_{n,k}-\lambda_{n,k-1}\}$$
and I can make that as small as I please, right?
@robjohn
12:55 AM
rooooooooooooooooooooooooob
1:05 AM
$${s_n} = \sum\limits_{k = 1}^n {{\lambda _{n,k}}} {{\bf{1}}_{\lambda \left( {n,k} \right)}}$$
@robjohn Am I left or right?
@BrianMScott
1:33 AM
@amWhy
@PeterTamaroff hey, there! I just "saw" you (and your post)!
@amWhy Hahaha yes. It is a funny example.
@amWhy COuld you look up here what I wrote?
@PeterTamaroff here?
@amWhy Yes, more or less.
Yes, I meant the whole "chunk"...I'll take a look, but first, will upvote your answer!
1:36 AM
@amWhy Oh, thanks for the former! (I don't really mind about the later =P)
@PeterTamaroff What you wrote looks good!
@amWhy GAWD, it does get messy though, right?
@PeterTamaroff Yes, it gets messy. I'm not very quick with reading unrendered formatting. I can write it fairly quickly, but I wish mathjax were enabled in chat!
@amWhy We have it!!!
50

This bookmark processes the current page with MathJax. It has been modified from the bookmark on this page to handle $$...$$ and to handle \$...\$ properly, to include AMS additions, and to update automatically. The COPY TO CLIPBOARD link on pastebin.com should copy the bookmark to your clipbo...

@PeterTamaroff Ahhhh! Thanks for pointing that out!
1:52 AM
My work today! =)
2:19 AM
@PeterTamaroff nice!
@leo Ei! Que bueno que estas por aca
@PeterTamaroff what is $1_{\lambda(n,k)}$?
@robjohn indicator function?
@PeterTamaroff ya no vengo muy seguido :-P
@robjohn There was an overloa d of greek letters
2:21 AM
@PeterTamaroff yes, of what?
@robjohn $\lambda(n,k)=\{x \in [a,b]:t_{k-1}<f(x)\leq t_k\}$
@PeterTamaroff then $\lambda_{n,k}$ is different from $\lambda(n,k)$
@robjohn yep
that is confusing
$\lambda_{n,k}$ is the infimum of $f$ over $\lambda(n,k)$
2:23 AM
I'm off to pick up dinner... bbl
@robjohn I made a mess unnecessarily
I'll try and unwind all this tomorrow
I get the idea, now I have tco write it down
3:12 AM
2
anyone know wolfram?
conocer? no.
2
don't know that there were Gauss Facts
3:28 AM
@leo I am trying to add a condition in my inequality that says the variables involved must be integers
like if i write a<=X, I want to make it clear that a is integral
@KaliMa can you provide a link so we can see what are you requesting to WolframAlpha
or just paste here the code
Obviously, an ad hoc way is to add the condition $a=[a]$.
But I do not use wolfram alpha, so I would not know whether even that is feasible.
-X<=a+b<=X, -X<=-ab<=X, X>0, a,b,X integers
@KaliMa Try Reduce[-X<=a+b<=X&&-X<=-ab<=X&&X>0,{a,b,X},Integers]
trying now
hm i am not sure
one of the resulting solutions is 0<a<1
which shouldn't be considered since a is an integer
3:37 AM
@KaliMa are you trying in WolframAlpha or Mathematica?
i have both
should i be doing this in mathematica?
and if so what should i be typing in?
@KaliMa In Mathematica it should work
i just tried the above line and it gave me some weird results
@KaliMa Like what?
C[1] | C[2] | C[3] | C[4] | C[5] | C[6] | C[7] | C[8] | C[9] | C[10] |
C[11] | C[12] | C[13] | C[14] | C[15] | C[16] |
C[17]) \[Element] Integers && C[1] >= 0 && C[2] >= 0 &&
C[3] >= 0 && C[4] >= 0 && C[5] >= 0 && C[6] >= 0 && C[7] >= 0 &&
C[8] >= 0 && C[9] >= 0 && C[10] >= 0 && C[11] >= 0 && C[12] >= 0 &&
C[13] >= 0 && C[14] >= 0 && C[15] >= 0 && C[16] >= 0 &&
C[17] >= 0 && ((X ==
1 + C[1] + C[2] + C[3] + C[4] + C[5] + C[6] + C[7] + ...
etc
3:42 AM
@KaliMa I see. I don't know how to interpret that
yeah it's completely different output from wolfram
4:07 AM
@KaliMa looks like at least a missing paren
I just posted a subsection
just the sort of output it provides
4:23 AM
@KaliMa can you post the full output
(C[1] | C[2] | C[3] | C[4] | C[5] | C[6] | C[7] | C[8] | C[9] |
C[10] | C[11] | C[12] | C[13] | C[14] | C[15] | C[16] |
C[17]) \[Element] Integers && C[1] >= 0 && C[2] >= 0 &&
C[3] >= 0 && C[4] >= 0 && C[5] >= 0 && C[6] >= 0 && C[7] >= 0 &&
C[8] >= 0 && C[9] >= 0 && C[10] >= 0 && C[11] >= 0 && C[12] >= 0 &&
C[13] >= 0 && C[14] >= 0 && C[15] >= 0 && C[16] >= 0 &&
C[17] >= 0 && ((X ==
1 + C[1] + C[2] + C[3] + C[4] + C[5] + C[6] + C[7] + C[8] +
C[9] + C[10] + C[11] + C[12] + C[13] + C[14] + C[15] &&
i've added more conditions Reduce[-X <= a + b <= X && -X <= -ab <= X && X > 0 && (a + b)^2 - 4 ab = (b - a)^2 && b >= a, {a, b, X}, Integers]
still errors out though
"(-a + b) && b >= a is not a quantified system of equations and inequalities"
@KaliMa you must write equations with ==
ah ok
specifically (a + b)^2 - 4 ab == (b - a)^2
yeah
it's still not working properly
agh
just lags, no output
4:31 AM
@KaliMa paste what you input
user19161
Hey @jay. I just woke up, did not sleep much.
@WillHunting At what time did you sleep? :-D
the output you posted should be interpreted as $c_1,\ldots,c_{17}\in\Bbb Z_{\geq 0}$, $X=1+\sum_{1=1}^{15} c_i,\quad a=1+c_1+c_2+c_3-c_{10}-c_{11}-c_{12}+c_{16}-c_{17}$ and so on
@leo I tried input Reduce[-X <= a + b <= X && -X <= -ab <= X && X > 0 && (a + b)^2 - 4 ab == (b - a)^2 && b >= a, {a, b, X}, Integers]
i am basically trying to say, given a $a$, what are the bounds of $b$
@KaliMa So you want the $X$ in terms of the other variables
?
or the $b$ in terms of the others?
4:43 AM
for example i want it to show that when a<0, then -a<=b<=-X/a
if that makes sense
@KaliMa Try Reduce[Abs[a+b]<=X&&Abs[ab]<=X&&X>0&&(a+b)^2-4ab==(b-a)^2&&b>=a,{b},Integers]
hm for some reason still not showing it
@KaliMa what it gives?
@KaliMa yes in WolframAlpha it don't works
if you have Mathematica try on it
4:54 AM
doesnt show it here either
@KaliMa what Mathematica gives?
frustrating
(a | ab | X | b) \[Element]
Integers && ((a <= -2 && ((X == -(a^2/(-1 + a)) &&
ab == -a^2 + a^3/(-1 + a) &&
b == (-a^2 + a^3/(-1 + a))/a) || (-(a^2/(-1 + a)) <
X <= -(a^2/(1 + a)) && -X <= ab <= -a^2 - a X &&
b == ab/a) || (-(a^2/(1 + a)) < X <= a^2 && -X <= ab <= X &&
b == ab/a) || (X > a^2 && -X <= ab <= a^2 &&
b == ab/a))) || (a == -1 && ((X ==
1 && ((ab == -1 && b == 1) || (ab == 0 && b == 0))) || (X ==
2 && ((ab == -2 && b == 2) || (ab == -1 &&
b == 1) || (ab == 0 && b == 0) || (ab == 1 &&
@KaliMa don't feel frustrated yet. I'm sure what you want can be computed with Mathematica. You might ask how achieve that in Mathematica.SE
pretty sure they'll help
maybe i am understanding my own problem wrong
that -X <= a+b <= X and -X<=-ab<=X and X>0 and a<=b
and i asked you the bounds of b of when a<0
what would you say?
@KaliMa well when $a\lt 0$: $$\begin{gather*}\frac{X}{a}\leq b\leq -\frac{X}{a}\\ |b|\leq-\frac{X}{a}\end{gather*}$$
5:10 AM
right, but now what about when a>0?
that is the tough one
@KaliMa if $a\gt 0$ you can divide by $a$ each member in your second restriction and preserve the inequalities so you'll end up with $$|-b|\leq\frac{X}{a}$$
which is of course $$|b|\leq\frac{X}{a}$$
not a<=b<=min(X-a,X/a)?
@KaliMa to prove that it's enough to show that $b\leq X-a$
which follows from you first constraint
so $|a+b|\leq X,\quad -X\leq -ab\leq X,\quad X\gt 0,\quad a\leq b,\quad$ and $a\gt 0$ implies $a\leq b\leq \min\{X-a,X/a\}$
5:28 AM
what about the bounds of a when b<0?
when $b\lt 0$ you can divide by $-b$ preserving inequalities. The bounds of $a$ can be deduced in an analogous way as we deduced those of $b$
user19161
@JayeshBadwaik Hmm, never mind. Anyway I will try to sleep from 2 am to 10 am from now...
but i mean
@leo I get -min(-X/b,X+b) <= a <= b, not sure if true
user19161
@jayesh Sent you an email.
5:48 AM
Hello, does anyone know the name of this formula? latex.codecogs.com/gif.latex?f_p=f_0+p\Delta%20f_0%20+%20\dfrac{p%28p%20-%201%29\Delta^2f_0}{2!}%20+%20\dfrac{p%28p-1%29%28p-2%29\Delta^3f_0}{3!}%20+%20\dfrac{p%28p-1%29%28p-2%29%28p-3%29\Delta^4f_0}{4!}+\dots It is part from the topic of numerical differentiation, but I don't know its name
f_p=f_0+p\Delta f_0 + \dfrac{p(p - 1)\Delta^2f_0}{2!} + \dfrac{p(p-1)(p-2)\Delta^3f_0}{3!} + \dfrac{p(p-1)(p-2)(p-3)\Delta^4f_0}{4!}+\dots
My eyes!!!!
I'm sorry, I'm new here, I don't know how to put formulas, does this works? $f_p=f_0+p\Delta f_0 + \dfrac{p(p - 1)\Delta^2f_0}{2!} + \dfrac{p(p-1)(p-2)\Delta^3f_0}{3!} + \dfrac{p(p-1)(p-2)(p-3)\Delta^4f_0}{4!}+\dots$
Yes!!
@hinafu Look at the starboard on right on how to use LaTeX in chat
I can't see the formula with $x^2$
6:12 AM
Ok, I found it, it is called Newton forward formula
user19161
6:22 AM
@JayeshBadwaik Yes, not exactly what you think, but never mind. Just felt like telling you. Anyway, you love to say hmm don't you?
@WillHunting Yes. Not exactly? Okay, sorry then. Will you please elaborate in the mail?
user19161
@JayeshBadwaik Hmm, never mind, I will say more when I feel like. Anyway, what does your hmm in the email mean? Does it mean something like "no comments"?
Hi, guys.
Is there anyone knowing Cantor's function?
user19161
@FrankScience Long time no see.
@WillHunting It means "okay".
6:31 AM
@WillHunting Nice to meet you.
user19161
@JayeshBadwaik I see. I realise I tell you many things which I don't explain, it's more to just talk to someone rather than ask for advice, so I don't always explain everything. So you don't need to worry about that. =)
user19161
@FrankScience We have met before...
@WillHunting Okay. =) :D
@WillHunting Yeah, I see. You've changed your nickname but your image is not changed.
6:33 AM
Do you mean to ask if it is differentiable on one side??
right differentiable / left differentiable.
Not at all points
It has a derivative = 0 almost everywhere.
Yesterday I proved a proposition.
Wait, do you mean either a left derivative or a right derivative at all points?
6:38 AM
Suppose $f$ is continuous and $f_+^\prime(x)<c$ on interval $[a,b)$, then $f(x)-f(a)\le c(x-a)$ for all $x\in[a,b)$.
The condition could be reduced to $f_+^\prime(x)\le c$. We could add $\epsilon>0$ to $c$ then apply the proposition, and let $\epsilon\to0$.
@FrankScience Ahh, yes it is semi-differentiable.
user19161
I am purposely not installing chatjax to prevent myself from discussing complicated math in chat. That would be reserved for the main site. =)
user19161
@JayeshBadwaik Is $C_+^\prime(x)$ increasing where $C$ is Cantor's function?
6:42 AM
@FrankScience No, $C_+^'(x)=0$
@FrankScience Here
@WillHunting One is on 9th Dec.
Then next ones are in Feb, March and May accordingly (also they are progressively more difficult, so the one in May expects students to be completely familiar with Real and Complex Analysis, Undergrad Algebra, Linear Algebra and Topology.
@JayeshBadwaik If $f_+^\prime(x)$ and $f_-^\prime(x)$ are increasing, can we conclude that $f$ is convex?
@FrankScience No, the first counter-example that comes to my mind is the bloch periodic function.
just a sec.. I will explain.
Basically consider a function $f(x) = x^2$ in $[-1,1]$
$f(x) = x^2 - 4$ in $[-2,-1] \bigcup [1,2]$ and so on.
@JayeshBadwaik But the point of $x=\pm1$ is conflict.
6:52 AM
Ahh sorry, make the intervals open-closed.
@JayeshBadwaik Then it's discontinuous, not semi-differentiable.
@FrankScience Hmm, I thought you were only assuming existence of right and left and derivatives in the second problem.
@JayeshBadwaik Yes, we're only assuming the existence of right and left derivatives, but if $f_-^\prime(x_0)$ and $f_+^\prime(x_0)$ exists, $f$ should be continuous at $x_0$.
$f_+^\prime(x_0)$ exists, so $f(x)=f(x_0)+o(x-x_0)$ for $x\to x_0+0$.
Hmm, consider this function then
$f(x) = \lvert x \rvert$ in $[-1,1]$
$f(x) =\lvert x/100000 \rvert$ othewise.
No, wait, sorry, I am messing up.
$f(x) = x^2$ in $[0,1]$
and $f(x) = x^2/1000 + 999/1000$ otherwise. Does this satisfy our conditions in [0,\infty]?
$f_+^\prime(x)=2x$ for $0\le x<1$ but $f_+^\prime(x)=x/500$ for $x\ge1$, so $f_+^\prime$ is not increasing.
7:03 AM
ohh, like that, hmm, then I think there are no counter examples. The function should be convex then.
@JayeshBadwaik Somebody told me that there's some counterexample, just a modification of Cantor's function, but I think I've proved that.
@JayeshBadwaik So if there's some counterexample, my proof is wrong somewhere.
@FrankScience Ohh. I must think on that then.
@JayeshBadwaik I'll show my sketch here.
@FrankScience Okay. I am going out for lunch in a few minutes. May be I will see it later.
@JayeshBadwaik First, if $f$ is continuous and $f_+^\prime(x)<c$ on interval $[a,b)$, let $S=\left\{\;x_0\;\big|\;\forall x(a\le x\le x_0):\,f(x)-f(a)\le c(x-a)\;\right\}$, we can prove that $\sup S=b$.
@JayeshBadwaik Then $S=[a,b)$ so $f(x)-f(a)\le c(x-a)$ for all $x\in[a,b)$. We can reduce $f_+^\prime(x)<c$ to $\le c$ because we can replace $c$ with $c+\epsilon$ and let $\epsilon\to0^+$.
@JayeshBadwaik Next we can prove such stronger proposition: suppose $f$ is continuous and $f_+^\prime(x)$ is increasing on some open interval $I$, then $f$ is convex.
@JayeshBadwaik Ah, there's something addition to the preceding lemma. If $f$ is continuous and $f_+^\prime(x)\ge c$, then $f(x)-f(a)\ge c(x-a)$ on interval $[a,b)$. The proof is similar.
@JayeshBadwaik Come back to the stronger proposition. Suppose $x<y<z$ are arbitrarily on the interval $I$, then $f_+^\prime(t)\le f_+^\prime(y)$ when $t\le y$, and $f_+^\prime(t)\ge f_+^\prime(y)$ when $t\ge y$, so the lemma works, and we have $(f(x)-f(y))/(x-y)\le f_+^\prime(y)\le (f(y)-f(z))/(y-z)$, proved.

1 hour later…
8:30 AM
Help!!! I don't know why I can't get into openstudy.com
Good day!
8:36 AM
@Nimza Hi!
@FrankScience Just came, will look at it now.
@JayeshBadwaik I have a short question, If positive measure $\mu$ is finite, then it's Fourier transform is well defined, right?
@Nimza No measure theory my friend! I learnt my fourier without taking any measures.
8:39 AM
@JayeshBadwaik Any ideas on how to get around the problem with the network?
@JayeshBadwaik I think that's it is an easy question, look: $\int\limits_{0}^{\infty} \mu(dx) < \infty$. Then $\left| \int\limits_{0}^{\infty} e^{i \xi x} \mu(dx) \right| \leqslant \int\limits_{0}^{\infty} \mu(dx) < \infty$
@JayeshBadwaik I've just to post it in article, so I'm afraid a little of everything
Ahh, I am not sure, so probably a wrong person to ask. :-(
8:44 AM
putnam exam in T minus 6 hrs 15 min
@anon Enjoy.
@JayeshBadwaik Thanks I'll give it a try.
@JayeshBadwaik I opened another account at openstudy and got in right away.
Putnam?
9:18 AM
morning all
hi
how are you?
@OldJohn Do you use TOR?
42 mins ago, by Jayesh Badwaik
Hi - I'm fine, thanks
I have used TOR - but not at the moment, as I have not got round to installing it after re-installing Linux
Would you recommend it?
I've never heard of it...
I found it very useful for getting some sites blocking features depending on where they come from - but I have found it slow, so I only switched it on when I really needed it
9:33 AM
hmm... I don't like to be slowed down
Why might you need TOR?
1 hour ago, by a Dangerous Idea
Help!!! I don't know why I can't get into openstudy.com
@aDangerousIdea Does openstudy block people from certain countries?
@OldJohn No, I think they installed some updates that will not load for me...
In that case, I am not sure that TOR will help - but it is probably worth a try
9:37 AM
I've been using the site for a couple of weeks.
it is very easy to switch in on and off - I used to surf normally for 99% of the time, and just switch it on for a few minutes when I really needed it
OK thanks I'll give it a try.
But if the problem is due to some updates, TOR might not be the solution
But I went to openstudy and opened a new account and got in fine.
TOR can also be useful to protect yourself from nasty governments :)
9:40 AM
Maybe they are blocking my original account?
not sure - but it might be a possibility
since the new one works fine
@aDangerousIdea That probabkly indicates that TOR would not be the real solution
thanks
@aDangerousIdea have you tried emailing them explaining what the problem is from your end?
9:51 AM
@OldJohn no
@aDangerousIdea that might be worth a try - unless you might have done something on the site to get yourself blocked :)
@OldJohn I'll give it a try.
10:38 AM
@OldJohn I generally use TOR to see if there is a routing problem between me and the website or is it something else. :-)
But I agree about it being futile with the new email address working fine.
@JayeshBadwaik Now I'm having the exact same problem with the new account :(
I did send them an email/ suggestion.
All I get is the message "loading more..." and a spinning wheel...
10:57 AM
Yes, seems to be a problem with atleast some part of the site. It is not loading the site after selecting the area for me too.
@JayeshBadwaik OK thanks for checking, all I can do is wait for them to reply.
The site has some nice ideas for group study.
hmm.
Mathematics Subtopics
Linear Algebra
Statistics
Precalulus
Geometry
Probability
Algebra
Discrete Math
Pre-Algebra
Calculus1
Collaborative Statistics
Trigonometry
Meta-math
Differential Equations

2 hours later…
1:01 PM
@robjohn I found another interpretation of an imaginary Erdos number:

"My Erdös Number is 2i where i is the imaginary number: âˆš-1.

This is because I post-doc'ed with Stanislaw Ulam 1967-68, who published with Paul Erdös (Erdös&Ulam, 1968), but I didnâ€™t see eye **to eye** with Ulam at all, and thus published nothing with him...."

1 hour later…
2:15 PM
hi all, where is the difference between a line integral and a curve integral?
user19161
@PeterSheldrick We usually call these line and surface integrals, not curve integrals in calculus.
okay
user19161
@anon Good luck. I am with you in spirit!
user19161
I always cannot access lepointdufle.net. I think there is some connection problem between that site and here. Tried different computers at different places already.
user19161
It's a French language site with great free lessons.
2:59 PM
@anon T minus ?
@aDangerousIdea "T minus 1 hour" = 1 hour to the deadline, in common parlance
@aDangerousIdea No - net yet ...
@OldJohn Classic.
looking now - it is possible I saw it years ago
@aDangerousIdea I must have missed it when it was aired on UK TV - it seems very good
3:08 PM
@OldJohn Yes it is.
ErdÅ‘s was great, wasn't he :)
I agree.
Good evening
@Nimza Hi there
3:10 PM
hi @robjohn! Is it true that if $f(x)$ is convex on $(0,\infty)$ then $f''(x)$ is a nonnegative ditribution in general?
hi @OldJohn!
@robjohn that is, $\int\limits_{0}^{\infty} f(x) \varphi''(x) dx \geqslant 0$ for any nonnegative smooth compactly supported $\varphi$ and for any convex $f$
hello
can anyone help me with a bounding question
@KaliMa If you ask - someone is likely to try to help
I have -X<=a+b<=X and -X<=-ab<=X with X>0, a<=b
i am trying to find all the ways to find the bounds of $b$ given $a$
since the bounds change based on where $a$ is
3:23 PM
@Nimza I haven't worked out a rigorous proof, but it looks true.
@KaliMa I might be tempted to look at that graphically -
@robjohn I think it may be done via approximation of $f$ by $C^2$ functions
@KaliMa Maybe look at the region in the ab-plane where both inequalities hold
i couldnt get it to graph
3:25 PM
was trying "plot b, X<=a+b<=X ,-X<=-ab<=X ,X>0, a<=b " in wolfram
@KaliMa just draw a sketch graph on paper - does not have to be accurate
i tried
and i think i kept making manual errors
i already have a rough understanding of the answer, i just want to get there myself
@KaliMa from my very quick sketch, I believe the max of $a$ will be where the curves $a+b=X$ and $ab=X$ intersect - so just solve them simultaneously
@robjohn True, I don't see eye to eye with too many people.
They intersect at b=-a/(a+1) and a = -b/(b+1)
hyperbola
3:30 PM
Look the Pedros are here :-D
@KaliMa no - solve the two equations I gave simultaneously for the unknowns $a$ and $b$ - you have just eliminated $X$
you really want to eliminate $b$
This is rumba, baby.
you mean b=X-a, -a(X-a)=X, a = (1/2)(X-sqrt(X+4)sqrt(X)) and a=(1/2)(X+sqrt(X+4)sqrt(X))?
(fixed)
@KaliMa yep - that is the sort of thing
@aDangerousIdea yes, I saw that earlier :)
another factor of i on my Erdos number
3:37 PM
are there any online graphers that will handle these inequalities
i can't get wolfram to do it
@KaliMa I think it is much more useful to be able to quickly produce a sketch graph of these things on paper - it is very easy - and very quick
i am not good at graphing complex inequalities in tandem with each other, constraints and all
for $a+b=X$, you just draw a line between $(X,0)$ and $(0,X)$ in the ab-plane
similar for $a+b=-X$
and for $ab = X$, you just have a hyperbola
oh here we go, i think wolfram just needs a number for X
is this accurate? tinyurl.com/cet6y6p
i just gave arbitrary value for X
@KaliMa hmm - doesn't look like my sketch ...
3:42 PM
hmm
let me try again, i'll post a picture of what i tried
@KaliMa OK - I have just noticed that you also restricted $a$ - it makes more sense now
does -X<=a+b<=X make a sort of diamond?
a square with vertices fixed on each axis?
(scraaaaaaaaaaatch that)
@KaliMa It is actually the area between 2 parallel lines - it goes on for ever in the north-west and south-east directions
light blue is what i assume but am uncertain about
oh ok
take the line from $(0,X)$ and $(X,0)$ and continue it forever in both directions ... then do the same for the parallel line $a+b = -X$, and the region you want is between the 2 lines
3:57 PM
yellow being the final area
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