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11:54 AM
hello
 
 
6 hours later…
6:17 PM
Time for some more mathfitti: Let's build numbers using just +, *, and 1's. (And parentheses if you must.) So 7 = 1 + (1+1)*(1+1+1) and 18=(1+1)*(1+1+1)*(1+1+1) . For n>1 let c(n) count the smallest number of ones necessary to form n. It is easy to show 3lg n <= c(n) < 5 lg n, where lg is log base 3. Is c(n) <= 4 lg n for all n>1? So far the extreme case seems to be n=1439.
 
mh
1439 is a prime
how did you check those cases ?
and why exactly do you calculate in log base 3 ?
 
7:21 PM
The answer is nicely expressed in terms of lg. The calculations are done via program, not using base 3 arithmetic. See OEIS for links to a nice version in C.
 
ah it was a puzzle not a question ?
 
For checking the inequality, it suffices to check it only for those n such that c(n)> c(m) for all m < n.
If it is a puzzle, it is one to whch I want to know the answer and currently do not.
 
8:15 PM
Anyone around? @WillJagy? I have some goodies for you guys, but I need some input on it
Basically, I have a script that adds mathjax buttons and keyboard shortcuts
(Ctrl-M for $...$, Ctrl-D for $$...$$, applies on selected text)
I was wondering if there are any other feature you would like for that
 
oh that's cool
haha
 
(For example, Chem.SE has $\ce{...}$)
 
i don't have any input tho
 
Any common mathjax snippets that you find yourself using a lot?
 
\mathbb{..}
 
8:17 PM
Like maybe \mathbf?
@JonBeardsley for number sets? That could be a button
 
or, well, for integers, reals, complexes, etc. we frequently write $\mathbb{R}$
or whatever
and i dunno, a lot of people have this shortened to \R, \Z, \C, \P, whatever
 
So something like Alt-S and you get \mathbb{...} with the cursor at the ... (or selected text gets enclosed in mathbb)
 
in their personal LaTeX stuff
 
@JonBeardsley Ah, that too
 
but yeah i dunno, i don't want to be the only one making suggestions, haha
 
8:19 PM
You can use \newcommand in MathJax, so that helps :)
 
oh really? what about chatjax?
i mean, is it permanent?
 
@JonBeardsley Same thing
Chatjax is just a bookmarklet that loads mathjax
 
i see
 
@JonBeardsley If newcommand is on the page, it gets loaded
 
haha interesting, sooo i could do something like
\newcommand{\R}{\mathbb{R}}
is that syntax right
 
8:20 PM
Yep
 
No, like this
$\newcommand{\R}{\mathbb{R}}$
$\R$
^^ in all its glory
 
$\renewcommand{\R}{\mathbb{ISMELLBAD}}$
$\R$
 
ah crap it keeps disappearing
 
It works!
 
8:22 PM
Actually, an interesting userscript would be one that lets you manage your own \newcommands, and inserts them into posts whenever you try to use them
 
yeah that'd be really cool, or like, yeah a button you press that just quickly loads all your newcommands
\newcommand{\R}{\mathbb{R}}
what am i doing wrong here
 
You need dollar signs.
 
ohhhhhhh
sure
 
Also, since it's already defined, you need to do \renewcommand.
 
hahaha
 
8:23 PM
@EvanJenkins moneymoneymoneymoney
 
$\renewcommand{\R}{\mathbb{R}}$
 
@EvanJenkins NO! Nobody touches my beloved \R
 
$\R$
BOOM
what does \R usually do?
 
nothing
It's just that I had defined it before you
 
oh
but wait a second... so... we're all running the same newcommands?
 
8:24 PM
$\R$
 
Oops I broke it.
$\mathbb$
 
dangitz
dudebro
 
$\newcommand{\a}{a}$
$\a$
$\newcommand{\a}{b}$
 
$\R$
 
8:25 PM
$\a$
 
cool
 
$\to$
 
$\R$
 
No need for \renewcommand
 
8:26 PM
I broke chat.
Sorry.
 
$\newcommand{\G}{\mathfrak{GODZILLA}}$
$\G$
noice!
 
Stop breaking mathbb, people! Poor Cantor is turning uncountably infinite times in his grave!
 
what the hell am i doing, i'm supposed to be making "slides" for my calculus babies
uhoh, look out, here comes $\G$
that's super convenient
$\mathfrak{F}$, $\mathfrak{G}$
in demazure and gabriel's groupes algebrique, Tome I, their F's and G's look identical
 
@JonBeardsley One of my friends was learning LaTeX. He was very fond of using custom commands. One day, I created a python program that overwrites all newcommands and makes them into various nonsense things. I then aliased latex to it. Poor guy was completely confused :P
 
hahah
 
8:30 PM
$\mathbb{R}$
 
that's great. i wish i knew how to do things like that
WHAT HAVE YOU DONE EVAN
$\diamondsuit$
 
Can somebody delete my post where I broke mathbb?
 
lol
$\mathbb{R}$
 
I flagged it for moderator attention.
Are there any moderators around?
It's probably not good that some malefactor can come and ruin various LaTeX commands.
 
i was thinking about that
i wish it only changed it for YOU, not for everyone in here
$\R$
$\renewcommand{\R}{\mathrm{I'm~a~dummy}}$
 
8:34 PM
$\mathbb{R}$
 
@DominicMichaelis didn't use dollar signs
 
Hello all: have popped in to address flag!
 
$\mathbb{R}$
 
Any good?
 
8:35 PM
Evan broke \mathbb
 
Praise the Lord!
An angel from on high!
 
@EvanJenkins Working?
 
$\mathbb{R}$
 
@evan give me 2 mins
 
$\mathbb{R}$
:(
 
8:35 PM
@JosephWright let me try it 2 mins
 
I think somebody re-broke it.
 
somebody killed \mathbb
 
@DominicMichaelis OK
@DominicMichaelis I can't actually tell :-)
 
oh boy
 
$\mathbb{R}$
 
8:37 PM
nope
 
Should I just kill all of the \renewcomands?
 
@JonBeardsley OK
Will work backward
 
Evan, you owe this guy a nice dinner
 
$\mathbb{R}$
Hrm
 
8:39 PM
Have killed some: any joy?
 
$\mathbb{R}$
 
@JonBeardsley What about this one?
 
is anyone left?
 
that was just to demonstrate a point
 
OK
It works
I reloaded chat.
 
8:40 PM
@EvanJenkins Cool
 
All right, I won't do that again.
 
@EvanJenkins :-)
 
Thank you @JosephWright.
 
should we ban renewcommand from chat ?
 
Probably.
 
8:46 PM
$\R$
oh
that one is still set
apparently
$\renewcommand{\R}{\mathbb{R}}$
$\R$
 
$\R$
$\let\mathbbb = \mathbb$
$\mathbbb{Q}$
 
$\mathcal{R}$
 
$\renewcommand{\mathbb}{\mathfrak{ISMELLBAD}}$
$\R$
 
$\mathscr{F}$
$\mathscr{R}$
$\mathscr{I}$
$\mathscr{C}$
$\mathscr{K}$
 
$\let\mathbb = \mathbbb$
$\R$
Sehr gut.
I'm not going to rest until I find all of the ways to abuse ChatJax.
 
8:54 PM
did you ever redefine \fi in latex ?
 
That sounds like a bad idea.
Hm
$\mathbb{R}$
Maybe I shouldn't try this.
 
try it at home not in public ;)
 
It worked in real LaTeX.
By "worked" I mean I was successfully able to break and then unbreak things.
Maybe this would go over better at the tex.stackexchange chatroom.
 
this is not your mother's LaTeX
 
They've probably already figured out all the ways to break things.
 
9:02 PM
which sounds weird
 
2 days ago, by Anna Lear
"No math was hurt during the production of this chatroom."
well that didn't last long
 
@JosephVanName Thanks again for answering my Question :)
 
I think I found a bug on this domain's robots.txt. It says "Allow: /transcript/" when clearly everybody present would prefer "Disallow: /transcript/"
 
lol
clearly?
what would happen if that were disallowed>?
 
9:10 PM
SE chat transcripts have always been indexed
 
If it were disallowed, chat transcripts wouldn't show up in searches.
 
oh isee
 
I just don't want "ISMELLBAD" to be the first thing people see when they Google me.
It will hurt my job prospects.
 
but
i mean. you don't want to be a liar do you?
 
There's a big difference between lying and simply omitting truths.
 
9:13 PM
lol. man, i think that'd be some information i'd really like if i was going to hire someone. their odor level
i mean, it would at least sort of have to overshadowed by math powers
 
This is the primary reason for in-person interviews.
 
haha! dear sir or madam, due to your outrageous stank, we will not be able to hire you this year
 
@anon Lol. All @EvanJenkins's fault
 
Oh, sure. Blame me.
 
9:26 PM
in case you didn't know.... $\mathbf{D}f(\mathbf{x})\mathbf{v}&=\mathrm{grad}f(\mathbf{x})\cdot\mathbf{v}=\n‌​abla f(\mathbf{x})\cdot\mathbf{v}\\&=\left[\frac{df}{dx}(\mathbf{x})\right]v_1+\left[\‌​frac{df}{dy}(\mathbf{x})\right]v_2+\left[\frac{df}{dz}(\mathbf{x})\right]v_3$
well that just didn't work
 
Take out the ampersands.
 
$\mathbf{D}f(\mathbf{x})\mathbf{v}=\mathrm{grad}f(\mathbf{x})\cdot\mathbf{v}=\na‌​bla f(\mathbf{x})\cdot\mathbf{v}=\left[\frac{df}{dx}(\mathbf{x})\right]v_1+\left[\fr‌​ac{df}{dy}(\mathbf{x})\right]v_2+\left[\frac{df}{dz}(\mathbf{x})\right]v_3$
lol
 
Close.
 
\mathbf{D}f(\mathbf{x})\mathbf{v}\mathrm{grad}f(\mathbf{x})\cdot\mathbf{v}=
\nabla f(\mathbf{x})\cdot\mathbf{v}=\left[\frac{df}{dx}(\mathbf{x})\right]v_1+\left[\frac{df}{dy}(\mathbf{x})\right]v_2+\left[\frac{df}{dz}(\mathbf{x})\right]v_3
whoops
lol, nvm. :(
 
@JonBeardsley That's what I call a... puts on glasses
Math Overflow!
 
9:29 PM
Now, now
No need to pun-ish him any further.
 
 
1 hour later…
10:40 PM
hello
 

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