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12:12 AM
How do I prove recurrence relations by induction? Do I assume true for$k-1$ and $k$?
 
that is often how induction works
 
It’s often called complete induction.
 
12:30 AM
Complete induction is when you assume true for two successive values instead of the one?
 
You get to assume all previous cases.
 
1:00 AM
@TedShifrin Good to know. But this question doesn't seem to be working out no matter what I try :(
 
1:20 AM
@LearningCHelpMeV2 it’s tricky to see how to proceed. Want a hint?
 
@TedShifrin Please
 
Start with $a_n -3a_{n-1}+a_{n-2}$. Put in the formula for $a_n$ and do algebra.
You will need the formula again (for $a_{n-1}$) and the inductive hypothesis.
 
1:35 AM
Does anyone know how to convert an integral to an infinite sum?
 
You can divide the interval of integration
 
1:49 AM
The redesigned Questions page was briefly displayed on Mathematics Stack Exchange, and it appeared to have the same MathJax bug as the Activity page.
 
What is the derivative of $\frac{4}{3}\pi r^2$ With respect to t, is
It $4\pi r^2$?
 
2:10 AM
How in the world did you do that?
There’s no $t$ anywhere, so the answer is $0$.
 
He divided by zero obviously and make a $t$ out of nothing.
 
And multiplied by $3$, mysteriously.
 
Turned out, $\pi r := e^{3t/2}$.
 
Yeah it's a random side effect of multiplying by the null operator. It requires knowing the position of a particle over time, so the probability that it is multiplied by $3$ varies in proportion to the probability density function describing the location of that particle in space, and you don't know if it was multiplied by $3$ or not until you evaluate the expression.
 
3:12 AM
@robjohn MathJax is still broken on the Activity page, and now it's also broken on the Top Questions page.
 
@RandomVariable I'd noticed
 
3:37 AM
@TedShifrin My math teacher did that, i'm so confused...
 
since ${4 \over 3} \pi r^2$ has no $t$, the derivative wrt $t$ is zero. as already mentioned
 
ok, bt how did my teacher get the above result? something they did wrong?
I know you can't differentiate with respect to t since there is no t.
that's why i'm so confused
 
well, the derivative of a constant is zero.
you can differentiate a constant.
 
i'm talking about $4\pi r^2$...
 
3:54 AM
Do you know chain rule and implicit differentiation? Regardless, your teacher is wrong or you copied it wrong.
What if $r=r(t)$?
 
No, the question is just to differentiate $\frac{4}{3}\pi r^2$ with respect to t.
 
Space after \pi.
I will bet seriously this is the question. Differentiate $\frac43\pi r^3$. The answer is $4\pi r^2 \frac{dr}{dt}$.
Otherwise, fire the teacher or fire you.
 
But the question printed on the paper was $\frac43\pi r^2$ not $\frac43\pi r^3$.
so is it a typo error then?
 
Typos everywhere.
Answer’s wrong, too.
 
But then what would the correct answer be?
$4\pi r^2$
 
4:05 AM
ask your instructor, how would we know?
 
Ok.
 
I already told you. If you ignore what I type, I won’t talk to you.
 
Me? @TedShifrin
 
Yes, you.
Is reading too hard?
 
No, I don't even know what I ignored?
 
4:08 AM
@copper I’m joining you with losing patience.
@mohan You can’t go back and read what I’ve said? Leave us alone.
 
They don't call it Schrödinger's equation for nothing.
That's all for tonight folks! (Good night!)
 
@TedShifrin :-)
 
@TedShifrin I did read what you said, I don't get what you mean by ignoring you.
@TedShifrin Cab you at least tell me what I have ignored?
Oh, I didn't know you were talking about the answer for $\frac43\pi r^3$. Sorry.
 
4:44 AM
@TedShifrin Did I do some weird circular argument or is this acceptable?
 
Hi chat! suppose $a,b,c,d,e,f$ are all positive
 
missing a few $=0$s above.
that is a nice positive way of looking at the first 6 letters of the alphabet
 
Then I think that the inequality is always true
 
the suspense is killing me
 
$$xy + bx^3 + ax^2 + f x + c y+ u+ e x u < ax^4 + dx^2y + y^2$$
 
4:55 AM
@copper.hat despite that, is the proof sound?
 
for $x,y \in R$? what if it is bounded in some regions?
If it is true, to prove it I think i need to make some squares!
 
@LearningCHelpMeV2 the idea is fine, the exposition could be improved. for example, instead of saying the "numerator equals zero" (what numerator) i would have said something like "multiplying both sides by $a_{k-1}$..."
@BAYMAX is there some origin for this or just a formula that popped into your head?
 
@copper.hat Ok, I see. Thanks for the advice.
 
just a suggestion
@BAYMAX if you take $y=1$, $u=0$ and $x=2$ you can choose $a,b,c,d,e,f$ to be very small positive numbers (essentially zero) and you have essentially $2 < 1$.
but why would you think it true in the first place???
 
5:20 AM
Actually it's the rate of change of a Lyapunkv function
Lyapunov function
The rate of change to be negative so that trajectories of a system is bounded
 
you need a different Lyapunov function then...
 
They have shown a similar kind of proof there
 
clearly there must be some other constraints on the parameters
 
Like equation 8 how to prove
Yeah I will check.. Thank you!
Was thinking if some graphical analysis could help
 
sry, no idea what equation 8 is
 
5:26 AM
= eight
 
In the paper I shares above
 
paywall
that's my supply of patience for the day
 
@copper.hat some run out of fear or a sense of danger, you have run out of patience.
 
:-) i'm tired, i did my little 5km run around the cemetery
and the market gyrations added a little excitement to the day
 
5:45 AM
@copper.hat just in case you were interested in equation $8$, here it is:
$$
\begin{align}
&\left(\sqrt{a}x^2-\frac{bx}{2\sqrt{a}}\right)^2+(xy)^2+\frac{9y^2+z^2+I^2+c^2}4\gt\left(x+\frac y2\right)^2+\left(x+\frac I2\right)^2\\
&+\left(y+\frac c2\right)^2+\left(x+\frac z2\right)^2+\left(xy-\frac{dx}2\right)^2+\frac{16ka+4b^2-4ad^2-48a}{16a}x^2
\end{align}
$$
more of an inequality, really.
 
@robjohn thanks!
 
@copper.hat My reaction was "WTF?"
 
i was thinking this is what happens when you drink & derive
to be fair, many of my neurons did exhibit extreme multistability today
this is what maple & the like are for
i'm looking for a nice soothing problem to solve to put me to sleep
 
@copper.hat until wave function collapse
 
:-) could happen soon. i did have a glass of cabernet at a bar a short while ago
 
5:53 AM
@copper.hat None of the residents joined you I hope.
 
we sat outside in the balmy 50f
 
Are Riemann surface and Riemannian manifold different objects?
 
6:12 AM
Absolutely.
A Riemann surface is a 1-dimensionsl complex manifold.
 
I see. Latter is basically real manifold.
 
Of any dimension, yes, with extra structure.
 
 
1 hour later…
7:31 AM
@copper.hat that's so of dynamical systems!
 
 
2 hours later…
9:32 AM
@AMDG Here's the error plot for the 10th degree Chebyshev polynomial for $2^x$, i.e., the curve shows $cheb(x)-2^x$, where cheb(x) is the poly. The minimax poly is very similar. As you can see the maximum absolute error is ~2e-16, so around 1 bit for standard 53 bit floats. The blue curve is for the poly computed using 160 bit real coefficients, the green curve is the approximation using standard floats.
 
 
5 hours later…
2:37 PM
@PM2Ring 54-bit*
That's quite nice. Only order 10.
@PM2Ring I also just realized that since my function for quotients is actually valid for positive rational inputs as well (which is deduced from the definition of ratios by definition), it can be used to approximate $2^x$ directly which is in the integer portion since it is of the form $Q = 2^x + \sum$. Therefore, $Q - \sum = 2^x$.
So for the infinite sum (we'll call it $I$), we have $I\equiv Q - 2^x = \frac{2^{x +\lfloor\log_2(y)\rfloor + 1}}{y} - 2^x$.
 
Hello, if $g(x)$ at some point $x_o$ not equal to $g(x_0)=0$ and g is continious, why is it that 1/g strict monoton??
 
2:53 PM
Now, being that we restrict our domain to $x,y\in \Bbb{N}$, we apply argument reduction to obtain an approximation of the quotient and then set equal to the RHS and solve for $2^x$, then substitute back into the above equation and solve once again for $2^x$ to get an identity of $2^x$ which is suitable to be approximated efficiently.
 
Yes of course there exist a neigbourhood around $x_o$ such that all the points is unequal to $g(x)= 0$ but why is it strict?
 
@AMDG No, 53 bit. ;) en.wikipedia.org/wiki/Double-precision_floating-point_format binary64 Significand precision: 53 bits (52 explicitly stored)
 
@PM2Ring But... $\log_2(1024) = 10$. $10 + 1 = 11$. $64 - 11 = 53$. significand = 54 bits (53 stored) because of the implicit one always at the front.
I guess I must have forgotten the size of a 64-bit double. Oof.
Sorry, my bad.
Still, I don't really care for 53 bit precision.
 
No worries.
 
I only care about machine precision.
For the record, I'm not a fan of IEEE754 anyways.
Not very performant in software.
Anyways...
 
3:00 PM
Yes, you've mentioned that before. But most machines use binary64 floats. And a vast amount of software uses them.
 
Dunncar
I use whatever is efficient for the platform.
Float is slow on x86
So technically 80-bit floats are what x86 supports, but that never seems to be seen unless of course those floats are in intermediate registers with undocumented register names that I can access.
Theirs uses 64 whole bits of significand and the rest is sign and exponent.
 
Sage is very flexible. It's very easy to specify whatever precision you want. I often use 80 bits for general work. To compute Chebyshev coefficients, it's nice to use at least double the precision of what you want for the coefficients. And for minimax, the Remez algorithm often gets flaky if you don't use a lot of precision. It's not unusual for me to use 240 or 320 bits, or even more.
 
I can imagine
 
My old Amiga had 80 bit Motorola float hardware. It was disappointing to downgrade to a PC.
 
At least there's still OpenCL and GPUs.
At the low cost of 150c or more
 
3:16 PM
FWIW, I posted an integer sqrt algorithm in Python that uses no division, only bitshifts. stackoverflow.com/a/70841843/4014959 Of course, it's not as fast as implementations in C, but I mostly wrote it to illustrate the algorithm. Think of it as runnable pseudocode. ;) It'd be easy to adapt it to C, if you have code for basic arbitrary precision integer arithmetic.
 
Yes, that's the one thing I like about Python and which I intend to solve with Project Aquinas.
Python is great if you just want to get something up and running, i.e. prototyping, due to the interpreted nature and the fact that it can execute statements and code instantly in a shell.
Of course the shell, performance, and readability are then the bottlenecks with Python for such a usecase.
If I had a C REPL for all the ideas I had...
 
Python can be very readable, when you're used to it. And you don't indulge in writing cryptic one-liners.
 
That's the issue though. A language shouldn't allow for writing "cryptic one-liners", and of course, every language is readable to its users who use it day in and day out.
It's just a design flaw to have too many degrees of freedom without necessity.
That's also why I don't like shader nodes. They're ugly and cringe compared to the ordered and structured nature of the linear, 2D text plane of programming languages. I would like them more if the degree of freedom were restricted to fixed directed graphs on a low-precision grid.
 
I don't think a general-purpose language can avoid the ability to write cryptic code. The users of the language need to adopt and promote good practices.
 
@PM2Ring I agree on the latter, and it is impossible to remove all possibility of cryptic code, but at the same time, the language itself should facilitate that and not make it easy to make a mess.
 
3:25 PM
But this is the maths chat, not a programming chat. We should only talk about programming stuff incidentally, preferably in relation to something mathematical.
 
Yes, such as the set of programming languages.
 
:)
 
Anyways, I'm going to finish working on the $2^x$ I mentioned above and I'll get back to you hopefully in a bit. It isn't much to work on after all.
 
Good luck!
 
This isn't it, but cell 37 is essentially what happens when we simplify the identity of $\frac{2^x}{y}$ by removing the common factor of $2^x$. desmos.com/calculator/sy5jp6biq1
 
3:32 PM
hey everyone, I'm hoping there's a commutative algebraist listening again today
I have no sense of how divisorial ideals work, and I'm trying to see why the ring of algebraic integers is not a Mori domain. How does one see this?
 
@rschwieb hey it's you again!
@rschwieb integral divisorial ideals are just ideals, I think
 
@LeakyNun Hey, long time.
how are things
 
and $\overline{\Bbb Z}$ does not satisfy ACC because you can keep taking square roots of $2$
 
I was aware it wasn't NOetherian, but i was not having high hopes that all ideals would be divisorial :)
 
Just a reminder because it now disappeared:
https://chat.stackexchange.com/transcript/36?m=60251048#60251048
 
3:47 PM
And "integral ideal" is the same as fractional ideal?
uhh, yeah no I'm still sorting out these definitions
So divisorial ideals are special fractional ideals
It sounds like I'm supposed to conclude that a principal divisorial ideal for the algebraic integers is already contained in the algebraic integers (i.e. is 'integral')?
 
@rschwieb integral just means it is contained in the ring of integers, I think
 
@LeakyNun Yeah, I think I got that through my skull now
 
@rschwieb you can prove that every principal ideal is divisorial
so my argument still works
 
4:03 PM
@LeakyNun Frankly I don't see that you presented an entire argument. But from the last comment I do see one: principal ideals are divisorial, and the algebraic integers does not satisify the ACC on principal ideals, so it is not Mori.
You aren't eliding "fractional" from any of those, are you? Obviously principal fractional ideals are divisorial but I'm still looking at the claim that principal (integral) ideals are divisorial...
 
@rschwieb but all integral ideals are fractional ideals
 
ok yes
if you've got the field of fractions K for a domain D, the fractional ideals are the D submodules of K
including the integral ideals
principal integral ideals are therefore principal fractional
ok, that makes sense then, and is simpler than i was worreid about! thank you
@LeakyNun Part of the reason I'm asking is that someone was looking for "really badly behaved integral domains" as in, "one without any of the other nicer conditions"
From the data I store, that woud involve avoiding Mori, N-1, atomic, maybe also Noetherian
@LeakyNun Wait... this also means that all more domains satisfy the ACC on principal ideals... I wonder why i haven't seen that before
waitaminit. The wiki says " ascending chain condition on integral divisorial ideals."?
I thought it was just "ascending chain condition on divisorial ideals."
hrm, there DO exist some papers only claiming "divisorial" without "integral." cambridge.org/core/services/aop-cambridge-core/content/view/…
 
4:31 PM
@PM2Ring Hey look! I found something! desmos.com/calculator/ksltyxb90u
Number of values computable in the domain doubles as x doubles.
 
@rschwieb you can multiply the whole chain by a number to make the chain integral
wait nvm maybe you cant
 
@LeakyNun Either way I'm good now: thanks for explaining@
 
$(1) \subset (1/2) \subset (1/4) \subset (1/8) \subset \cdots$ would be a problem
 
it's also revealed a few missing things I can add to the site...
 
so is $\Bbb Z$ a mori domain?
 
4:41 PM
Yes, any PID is
 
but what about my chain
oh, integral
 
or any noetherian domain even
This is an alcove of ring theory I haven't ever really explored :)
 
@Leaky So you're permanently in the algebra vortex? :P
 
@TedShifrin I guess what I'm studying now counts as algebraic geometry
 
I guess it depends who's counting :D
 
5:01 PM
@TedShifrin oh, i meant arithmetic geometry
 
Ah, makes more sense.
 
5:14 PM
To find oblique asymptotes of this curve $f(x) = \frac{x(x-2)(x-a)}{x^2-1}$ do you just consider when $x$ gets very large $f(x) \approx \frac{x^3}{x^2} \approx x$?
 
That is a first approximation. But it more likely will be $y=x+c$ that is the asymptote for an appropriate $c$.
What should you do instead?
 
@TedShifrin What about some integral like $\int [f(x) - (x+c)] dx$ and take a limit as the area between the curves approach 0
 
Would that work with $f(x)=1/x$?
 
@TedShifrin hmm, doesn't appear so
 
5:30 PM
Nope. Besides, you could only do that integration by first simplifying all the algebra.
So what does the algebra say you should do?
If I give you $\dfrac{x^3-2x+1}{x^2-1}$, what do you do?
 
@TedShifrin Ohhh. You'd just express it in the form $Q(x) + \frac{R(x)}{x^2-1}$ Second term would approach $0$ as $x$ approaches infinity and the oblique asymptote would be $Q(x)$
 
Bingo.
 
Thanks :)
 
You do well with my non-answers :)
 
5:52 PM
Is it just me or is $\frac{\ln(2)}{x}$ an asymptotically good approximation to $2^{\frac{1}{x}}$?
For $x>0$
It would make sense, logically, now that I think about it. If you already knew the $x$th root of $2$, you could just divide by the root raised to an integer power such that the quotient is $2^{\frac{1}{x}}$.
 
i presume you mean $1+{ \ln 2 \over x}$?
 
Oh yes, right, thank you.
So as x goes to positive infinity, we get a better and better approximation of $2^{\frac{1}{x}}$ for $1 + \frac{\ln(2)}{x}$ according to the plot.
 
it is just $e^x \approx 1+x$.
 
I had a student ask yesterday "What does '$\approx$' mean?"
Because it was used in the book.
 
It obviously means the waves are rough at the beach today.
 
5:59 PM
It was annoying, because the author of the text explained exactly what it meant in the sentence directly before it was used.
 
@copper.hat Not sure I follow. Those are the first two terms of the taylor expansion for $e^x$ about $x=0$.
Oh wait now I see it
Substitution.
Funny how that works... 1+x is a nice approximation for x near 0 but 1 + ln(2)/x is a nice approximation for roots as x goes to infinity.
Must be something about the asymptotic behavior of 1/x itself
Something like $\lim_{x\to\infty^{+}} e^{\frac{1}{x}} = \lim_{x\to\infty^{+}} \frac{1}{x}$.
Aight, I need a break. Back to the elevator.
 
@Xander That is not unusual behavior. We find plenty of examples in this chat of people who "forget" to read things we have just told them seconds before.
 
I don't find this a surprise given that public schools aren't schools.
 
@AMDG This is totally muddling. The point is that as $x\to\infty$, $1/x\to 0$ and so we can use Taylor polynomials at $0$.
 
Oh, I see. The way you put it is much better.
Yeah and that limit for $e^{\frac{1}{x}}$ is $1$ I now notice.
 
6:22 PM
You could do the same thing with $e^{x-1}$ as $x\to 1$, or with $\sin(1/x)$ as $x\to\infty$, etc.
 
noice
 
6:38 PM
@robjohn Well, they've now messed up the display on the main question page, too.
Gotta love progress.
(I posted about this — my first "question" — in meta.)
 
7:00 PM
@TedShifrin Yeah, Random Variable mentioned that yesterday. Wonderful.
 
I only noticed it this morning.
 
There are so many bugs in the new, improved interface that I don't know where to put my efforts in complaining.
You'd think there would be a QA cycle where they actually looked at these things and had a roomful of monkeys typing away to find the bugs.
 
I don't know if meta was the right place for me to post. Probably not.
@robjohn They are wanting to close my "question" because it's a duplicate. Asinine.
 
@TedShifrin It was, meta or math.meta?
 
math meta
 
7:06 PM
@TedShifrin it probably is a duplicate. It might be better on meta.SE
because this is probably the same for all sites that use MathJax
 
Well, the cases pictured in the one I'm a duplicate of are far less disruptive than my two.
Do we know that it's gone wrong globally?
I wouldn't want to be presumptuous.
 
yeah, and the problem has gotten worse, at best
 
LOL, then I'll get censured for cross-posting.
 
@TedShifrin As far as I know, it's not just math.SE
 
I'm sure it'll be a duplicate over there, too.
 
7:08 PM
@TedShifrin delete the one on math.meta if you post on meta
@TedShifrin best to search there
 
LOL, the guy complained about my snark.
 
I have to look...
 
@robjohn A MathJax developer responded here.
 
@RandomVariable That seems to be about our old subject, not the new one.
@robjohn Is there an easy way for me to migrate the question, or do I have to do it all manually?
 
@RandomVariable Yes, I have communicated with Davide Cervone many times about MathJax issues.
@TedShifrin Let me look to see if this has already been reported there.
But I think a mod could migrate if it's not already there.
 
7:17 PM
The guy I was "commenting" with says there's already a post there.
 
@TedShifrin The post there is not about the main page, but I commented on that post that it is showing up on our main questions page and linked to your question.
so leave your question on math.meta for now.
Davide Cervone will look at it. He is responsive (unlike the new responsive interface).
 
I have recollections of interacting with Davide when I was deeply ensconced in WeBWork (and wrote close to 1000 questions for my multivariable course).
We're also mathematical relatives — he's my nephew, I guess. :)
 
7:43 PM
@robjohn @leslie Here is a more interesting question for you. Offhand, I don't see in my head that that norm is comparable to the usual Sobolev norm, but maybe I'm missing something.
 
@TedShifrin the question was deleted while I was reading it.
 
hate when that happens
 
I think that bounding the first derivative and knowing the value at a point, should give an equivalent norm, but the function has to be continuous to use $f(0)$.
 
Crazy.
 
I'll have to check this, however.
 
7:47 PM
Well, the $L^2$ integral of the first derivative doesn't bound it pointwise.
Only $L^1_{\text{loc}}$ assumed.
I thought it was not an uninteresting question.
I guess we could still have a different, inequivalent norm and call it a Sobolev space?
(rather than "the")
Hi @copper
 
Hi @TedShifrin Just finished a 3hr long work meeting. i need a nap
 
You decrepit old men need your naps.
You're almost as bad as leslie :D
 
8:02 PM
i am less tired after cycling for 3 hrs
 
Interesting. Physical as opposed to psychic stress.
 
i am an addict. endorphins. beats psychotherapy & other drugs.
makes life bearable
 
Whereas conversation in this chatroom makes it unbearable :D
 
i have a chronic need for interaction
 
I never did ask you what the two word message from your son was which changed your mood entirely.
 
8:07 PM
"miss you"
 
Aw. That was nice.
 
my poor kids, such a needy dad.
 
in this propositional formula ((p∧q)→¬(r∧s)) the connective with the largest scope is what?
what it;s meant wiith the largest scope?
 
don't know, but would guess the implies since it is at the top of the expression tree
 
What is the scope of a connective? So many words.
What is a connective?
 
8:17 PM
well it can\t be a negation
or conditional
 
Please how to prove that if f is continuous on (0,1) then it is bounded?
 
so between a dis-junction and a conjunction
I am not sure
 
I see four connectives: $\land$, $\to$, $\lnot$, and $\land$.
@Vrouvrou For goodness sake. Stop and use your brains before you ask questions.
 
more scope
or more importance
 
@imbAF I assume that "scope" means how many terms of the sentence it interacts with, or some such.
 
8:19 PM
I have to choose between
a negation, a conditional, a disjunction or a conjunction
And I don't get the scope part
like what it wants me to do?
 
Well, your course should have defined that word for you.
 
which obviously
 
Is there a negation in your sentence?
 
they did not
¬
this one
 
No, that is not obvious to me. Students ask questions all the time when they haven't read their material.
 
8:21 PM
not the case
 
OK, there is a negation. What is the negation acting on?
 
(r^s)
I did the table
the boolean table of it
 
OK, so that's the scope of $\lnot$.
The boolean table is irrelevant for "scope."
 
I write the definition of continuity:
 
@Vrouvrou what about the mapping $x\mapsto \frac 1x$?
 
8:23 PM
Think of scope as the formulas in the sentence that are connected or impacted by the connective. The scope of $\land$ in $r\land s$ is $r$ and $s$.
 
And if I let $x\in(0,1)$ then from the definition of continuity
 
@Koro This is why I rudely said to use brains before asking.
 
@Koro it is not continuous on 0
 
So what connects or impacts the most terms?
 
yes I know
the and
between each case
 
8:24 PM
Hi professor Ted, sorry if I interfered. I thought you were helping imbAF.
 
The $\land$s only impact the letters (statements) immediately adjoining, right?
 
yes
 
The $\lnot$ only impacts what immediately follows it.
So we haven't talked about $\to$. Tell me about that.
 
Ah, I saw your earlier message, Ted. :)
 
$\to$ concerns p,q with r,s
 
8:26 PM
@Koro The strange thing is that in France they usually use $]a,b[$ for what we write as $(a,b)$.
 
implication
 
@imbAF Right. It connects everything in the whole sentence.
 
yes
 
So, that's the answer, right? It has largest scope because it interacts with the most.
 
a so
this is what it's meant with scope in this case
biggest scope**
 
8:28 PM
We read in school that $(a,b)$ is also written as $]a,b[$ but the latter was almost never used.
 
I thought it was basically limited to France :P
 
the one symbol that includes the majority @TedShifrin
 
not includes, but interacts with or impacts. shrug
 
well forgive my caveman
 
If I write $(p\to q)\to (r\to s)$, which of the three connectives has largest scope?
 
8:29 PM
way of articulating
 
Thank you 1/x is continuous on(0,1) but not boundef
 
the 2nd arrow
 
Right.
 
ok i got it thank
 
You're welcome.
 
8:38 PM
Please how to prove that this limit do not exist using sequence $x^x/[x]^{[x]}$ ?
At infinity
Where [x] is the floor function
 
is this homework central?
 
No
I just want to remember some basic idea
The question of boundeness is in a cours and this is a remark
 
@imbAF That's a fancy word :) I wasn't criticizing; I was trying to clarify it for you.
 
Is such a function $f:S\to \mathbb R$ that satisfies $f(x+y)+\sqrt{6+f(y)}=f(x)f(y)$ for all $x,y\in S$ even well defined?
 
@TedShifrin Regarding this bug report, we can mark meta questions with the tag in order to put it into the debugging queue at a network level. Thus bug reports on meta contribute to the bug tracking system. Marking a bug as a duplicate (even if there isn't an answer) is a way of indicating that the bug has been appropriately reported, and that someone will get to it.
 
8:44 PM
Interchanging roles of x and y, and then solving the two equations, we get: $f(x)=f(y)$ for all x and y.
That is, f is a constant function.
But I doubt if the constant will satisfy the given equation.
Oh, I think it will.
 
If I take $x=n$ then the limit is 1 and if I take $x=n+1/2$ the limit is different is it enough to conclude?
 
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