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12:11 AM
Assume I have a group G with finite many elements and each with order m = 1,2,3,4. A possible order for G is 4!, right?
 
How do I get the residue of this?
@TedShifrin I can’t figure it out via Laurent expansion, but I’m sure that probably I’d the best way
 
 
1 hour later…
1:35 AM
Am trying to show $n^3$ congruent modulo 9 is either 1,-1 or 0
any proof to this?
I took $n $ congruent to 0 to 8 mod 9
and then cubed it to find the congruences mod 9
but seems not much rigorous?
 
well, it's a proof by cases
you've got 9 cases, and you've tested them all
whether that's the most elegant proof is another question, but it's perfectly sound
 
tbh proof by doing every possible case is like.. the most sound
 
it's proof by exhaustion, both in that you're exhausting all options and in that it's typically rather tiresome
 
2:00 AM
cool
sounds good
 
@baymax one thing that may simplify your proof, though. do you notice anything about which numbers 0-8 end up with cubes 1,-1,0?
 
which numbers end up with 1,-1,0?
all numbers from 0 to 8 will end up in either 1,-1,0 after applying modulo 9right?
 
2:22 AM
You don't have to try all of them, for what it's worth, because (-x)^3 = -x^3
So really just try 0-4, and 0 and 1 are easy
So just 2^3 = 8 = -1, 3^2 = 0 so 3^3 = 0, and then 4^3 = (2^3)(2^3) = (-1)(-1) = 1
 
3:11 AM
cool!
 
3:25 AM
@JakeRose When you have a simple zero in the denominator, you should just know (derive once and for all) the basic formula: $\text{Res}_{z=a}\dfrac{f(z)}{g(z)} = \dfrac{f(a)}{g'(a)}$. Can you derive that it one or two lines?
 
yo @Ted
 
3:49 AM
Henlo
 
What websites are trustworthy about reading uni reviews before applying ?
 
what does that even mean
 
just looking for recommendations where to read about universities before applying
Bsc math courses
 
i wouldnt really trust much besides actually talking to people w direct experience w whatever school
 
4:35 AM
I'd probably run with a marketing and or public relations degree that seems to not doom people to a life living on the poverty line anyway ah, how is it that in two different versions of maple, one disagrees and one agrees with my evaluation of these series really makes me wonder math.stackexchange.com/questions/3168362/…
 
don't they have a physics SE?
 
They do
but some do have mathematical questions
like on the existence of solutions for a given equation
there exist even tags for that on MSE
 
Well I'm bound to miss something, I can only afford to be functional for 50 percent of every fortnight. The Australian government decided I need to pay $400 buying medication off the streets that used to cost me $5, yay for less than human rights while living in the developed world
 
@Adam the market-liberal countries like Switzerland, US maybe also AU, always scared me with respect to social and health security. Whats the point to participate to a society if it can't provide you with relative standard means of health service?
When Arnold Schwarzenegger entered the US one the first things he did was to care for a proper health insurance (that was of course a private one, but maybe was relatively well affordable by him back then).
wise step
 
4:59 AM
Well that's the catch, I know several people that are in the exact same circumstance as me, in the same locality, treated drastically differently to me. I have no explanation as to why I have been placed on a lifetime ban, where as many others are not even subject to the scrutiny or restrictions I have been.
 
I read that in AU you pay 2% from your income to the public health care system. In my native country its around 16%. Alone that explains a lot. I am relatively young and quite healthy but I pay that money happily since I know it will safe my life or at least make everything easier in case any larger problems will occur. A good health system is very important for a society.
 
so much so that a lucrative black market exists for those with access to prescriptions, to sell them to people like me who have been restricted from such healthcare, despite having had access for the 15 years prior to that
 
We see all kind of super rich people from distant places like Arabia, US and so on travelling to our country for getting medical services. The overall standard is quite high precisely since everyone contributes much.
So the society as a whole profits from that.
 
We have a good healthcare system overall one of the best in the world actually, but in my specific situation, it's quite a suspicious turn of events
 
@Adam sad to hear. Ever tried to consult any administratives on your issue?
 
5:12 AM
yes I have made several appeals over the past 3 years naturally
 
Zee
what medication is it?
pain killers?
 
no
Adderall
 
Zee
you dont need that poison
 
lol
ok
 
Zee
amphetmines have been shown to be nuerotxic in monkey brain at theraputic doses
 
5:15 AM
do you think you are funny?
 
As a chemist I am astonished that SUCH a simple compound can be that expansive.
 
Zee
not funny , its reality
 
no it's not
 
Zee
its the patents that make them expensive
and amphetamines are dirt cheap to produce
 
My students can cook that in 100g quantities for costs at about a 50 $ I'd say.
 
5:17 AM
you have no idea what you are talking about. It's the legal system that makes them expensive
 
@Zee sure
@Adam really?
 
yes I was only paying $5 for 200 5 mg tablets up until 2016.
but on the black market $600 per gram has been the average cost for the past 3 years
 
Zee
Just drink coffee my man
 
@Zee this is called side effect. Erdös was running also on amphetamines i.I.r.c. and it didn't harm him too much as not to conjecture a hell.
 
I'm sure you are well and truly educated in medicine yes I shall take your advice random person
 
5:21 AM
@Adam which is still hillariously expansive
 
That's correct dexamphetamine is not neurotoxic where as methamphetamine is, Zee is just trying to bait me I believe
 
Zee
Erdos was developing symptoms of OCD and was not able to do mathematics without them, which he was able to before
No buddy, Meth is very toxic but amphetmines have been shown to be toxic as well in monkeys
ill find the paper
 
The substance itself is essentially as cheap as sugar or salt, it'll mostly the purification which costs a bit of worktime.
@Zee but the problem for a pharmacological layman like us (?) is to judge what that really means,
and most of all to balance the pros and cons
thats why there are physicians
 
they restrict them with legal red tape because almost half of the police for is employed to prosecute dealers and users, if they decriminalized it there is a profit margin lost for federal government and the private contractors they tender
 
Zee
physicians dont really know pharmacology too well
 
5:25 AM
and to become one you need to learn and experience a lot
 
Zee
most pharmacologists work in research
 
@Zee your confusing what I tried to sort out.
 
Zee
Explain please
 
The pharmacologist can read the paper on the toxicology and interpret that.
the physcian in turn is able to weigh the pros and cons when it comes to the question if it will be good or bad for the patient
 
and they are never used to treat OCD lol
 
Zee
5:29 AM
Erdos developed AFTER using
 
Eg. the pharmacolgist would be able to compare the side effects of amphetamin with those of say ethanol.
and your medical doctor would be able to decide on a proper basis if specifically you should take it for your case.
 
Zee
And @Rudi_Birnbaum the studies are done on animals, I am not aware of any studies that examined the effects of amphetmines on the human brain
 
But there is no way from the tox. paper to proper decision for a layman
@Zee I was just explaining on how I think you have confused what I sorted out.
 
Zee
here is a paper discussing how the oxidation of dopaime in the brain is actually toxic to some extent
which amphetmines release (unlike methylphenidate)
 
" After 1971 he also took amphetamines, despite the concern of his friends, one of whom (Ron Graham) bet him $500 that he could not stop taking the drug for a month.[25] Erdős won the bet, but complained that during his abstinence, mathematics had been set back by a month: "Before, when I looked at a piece of blank paper my mind was filled with ideas. Now all I see is a blank piece of paper."[citation needed] After he won the bet, he promptly resumed his amphetamine use. "
 
5:33 AM
@Zee So what is your conclusion from that?
 
Zee
that amphetmines are nuerotoxic
 
for a monkey
 
Zee
lol ok
 
@Zee as is dopamine, but does it mean sports is bad for your brain?
 
Zee
monkey is the gold slandered for research in this field
 
5:35 AM
can I give you a gold star for online trolling?
 
If the conclusion is that dopamine damages the brain it is even more difficult to really judge properly what that all means - for us laymen
 
Zee
yes, anything that produces dopamine will be toxic
but the amout compared to amphetmines is tiny
 
So here you go with Churchill: "no sports"
@Zee How you know? Take for instance a professional runner ..
And someone using very low doses of amphetamines
 
Zee
you need to show me that running has a big effect on dopamine release
 
And now how to judge for a layman what that study really means
random second google hit: healthyliving.azcentral.com/…
 
Zee
5:41 AM
BIG
 
"big" is the key word though. How could
we judge what big means without dwelling deeply into toxicology and pharmacology
 
Well there are people that will say anything to keep red tape up on certain substances, the monopoly exists only when there are laws that are readily enforced, for example I know how to make it and would, but I face up to 20 years in prison for doing so
 
Zee
here is the abstract of a paper, stating it is toxic to primates and not known clearly for human beings
 
So... thankyou for clarifying how weak your original argument actually is Zee
 
again " have the potential to damage central monoaminergic neurons. " for the layman that is close to meaningless
since practically anything has that potential
especially lots of endogenic substances
 
5:45 AM
So... just meaningless full stop
 
Zee
If you are ok with using chemicals that was shown to be toxic to primates and not known to be safe for humans , then go ahead
 
@Zee ever heared that slogan that its the dosis makes the toxin?
 
Zee
yes
 
Well no I don't want to be arrested for the manufacture of an illicit substance, facing a jail term far longer than I would for first degree murder...
 
@Zee so what about alcohol?
 
Zee
5:47 AM
thats terrible but we have the statistics to back it up, over 3000 years of it
 
You see
 
Zee
amphetamines, maybe 50 years
 
in that way you arrive at completely wrong concusions
how sad would that world be if there would be no beer?
so you can't simply say its terrible,
that would be what one calls a proper ideology
which is of course a main field of mental activity in young people ;-)
 
Zee
I smoke and drink, but i know they are bad, i dont lie to myself
 
and assume every human brain can be considered identical to another, but hey it's not like ethanol has been subject to irrational prohibition laws in the same way as I have stated historically speaking... oh no wait yes it has
 
5:52 AM
@Zee ok, that makes sense
(no irony)
 
if we assume a philosophy of purely optimizing our longevity, as grazing mammals from a strictly biological sense, yes, all of these things are "bad"
 
@Adam not sure about the grazing, though
there exists stuff like orthorexia ..
 
Zee
Adam, just consider the idea, you dont have to quit but just keep in mind, that YOU COULD
 
I was forced to quit. I did for 12 months. I went back to sleeping 20 hours per day as I was before the age of 16 and I was originally prescribed them, but yes I DID
I cant afford street prices with what I earn at the moment, so every second week, I quit
 
Zee
I sleep long too, thats why I put a big cup of coffee next to bed so I can just drink it, But, I hope things get better :)
 
6:00 AM
but yes I have been forced to interact with criminals on a regular basis, I've had knives put to my throat, a gun put to my head at one point, all for something I had collected from the pharmacy for the previous 15 years. What is your health rating on that circumstance?
on several occasions I have just been robbed, or the bottle filled with paracetamol.
leaving me with around $20 to spend on food for the next 14 days
 
Zee
Just buy it on the dark web you fool...
 
There is no guarantee it will come. Dream market is in fact a total sham, it never arrives
have you ever bought off the dark web?
I tried several times, and nothing arrived
you're a simpleton
 
Zee
yes and it worked
but that was a couple years ago
make sure to check the ratings and keep the amounts small
 
6:37 AM
Helo! Can somebody point out what does $x\in Q^c$ mean?
 
6:49 AM
It depends on the context
 
7:13 AM
@Astyx I read your message after a while, so didn't respond on time
 
So $Q$ is the set of rationals
$Q^c$ is complementary set of the set of rationals, that is irrationals
 
And the last $Q$ in the proof should be $Q^c$ here
 
@Zee point is, I shouldn't have to, and besides, I've tried several times and only been ripped off in a decentralized currency rather than the local currency.
The ratings mean nothing
there is 0 accountability
and law enforcement agencies ensure that stays the way it is as I am sure you well know
 
 
1 hour later…
8:28 AM
Hello !
I have seen the use of generating coefficients in solving permutations and combinations problems.
Can someone help me in understanding its application properly ?
25
Q: 6-letter permutations in MISSISSIPPI

George ZhaiHow many 6-letter permutations can be formed using only the letters of the word, MISSISSIPPI? I understand the trivial case where there are no repeating letters in the word (for arranging smaller words) but I stumble when this isn't satisfied. I'm wondering if there's a simpler way to computing a...

Like ^
 
 
1 hour later…
9:46 AM
Morning-ish everyone
 
10:01 AM
Gallian asks to find a mistake in the statement following picture:
 
$xyz \neq 0$
 
10:36 AM
Heile @Rudi :)
 
10:58 AM
@Silent heh, $(-1)^n+1^n=0^n$ for all odd $n$
 
11:28 AM
[Random]
Let $r\in \Bbb{R}$ and let $P \in \Bbb{P}(\Bbb{Z})$
where $P(x) = \sum_{k=0}^n a_kx^k$ for some $n \in \Bbb{N}$
if $r \in \Bbb{A}$, then:
$P(r) = \sum_{k=0}^n a_kr^k = 0$
else if $r \in \Bbb{T}$, then:
$P(r) = \sum_{k=0}^n a_kr^k \neq 0$ for all $P$
Let $[r]_m$ be the mth order approximation of $r$ under some protocol $\text{Proc}$. For example $\text{Proc}$ can be defined as follows:
$$\text{Proc} : [r]_m = \min_{(p,q)} |r - \frac{p}{q^m}|$$
Thus if $r$ is algebraic, then there is some $M$ such that $ P([r]_M) < \epsilon$ for some $\epsilon$ expression (to be determined), otherwise, it is transcendental
Currently figuring how to relate this to Diophantine approximations and Liuoville's theorem
The basic idea is that only approximations of transcendentals cannot be zeroed out quick enough by any fixed polynomial, whereas algebraics can reach zero after a polynomial mapping and its approximation should approach zero quick enough to do so
That is, if $r$ is transcendental, then the sequence $P_m([r]_m)$ will only have some terms zero if $P_m\neq P_s$ for some $m,s \in \Bbb{N}$
likewise if $r$ is algebraic, there exists a unique polynomial $P$ such that $\lim_{m\to \infty }P([r]_m) = 0$
Hmm...
$$P([r]_{m+1}) - P([r]_m)$$
$$\frac{\partial}{\partial y} P(x) = \frac {\partial a_k(y)}{\partial y}$$
typo
$$\frac{\partial}{\partial y} P(x) = \frac {\partial a_k(y)}{\partial y} x^k$$
Hmm... given a set of vectors:
$\{1,x,x^2,x^3,\cdots\}$
where $x \in \Bbb{R}$. Then $x$ is transcendental if:
$a_0+a_1x+a_2x^2+\cdots = 0$ implies $a_0=a_1=\cdots = 0$
Thus algebraic independence is similar to the good ol' linear independence of vector spaces
except, not really, because the vectors can add to each other...
Hmm...
Let $r > 1$, Then $\frac{1}{r} < 1$. $[\frac{1}{r}]_m$ should start to truncate higher order terms in any given polynomial $P$
 
12:36 PM
The above writing also showed why it is in general hard to determine whether a number is transcendental:
In computer science, the subset sum problem is an important decision problem in complexity theory and cryptography. There are several equivalent formulations of the problem. One of them is: given a set (or multiset) of integers, is there a non-empty subset whose sum is zero? For example, given the set { − 7 , − 3 , − 2 , 5 , 8 } {\displaystyle \{-7,-3,-2,5,8\}} , the answer is yes because the subset { − 3 ...
It is really a special case of the subset sum problem where the polynomial coefficients we want to solve are the integer subsers we need
Or more accurately, a subset of the knacksack problem
The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items. The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science...
where the weightings $w_i$ has the form $r^i$ for the given number $r$ and the target sum is 0
Thus, the crux of the criteria of transcendental number can be rephrased as follows:
Given a sequence $(x^n)_{n\in \Bbb{N}}$ can you find a nonzero integer sequence $(a_n)_{n \in \Bbb{N}}$ such that they sum to zero
if yes, then $x$ is algebraic, if no, then $x$ is transcendental
The most general form of these problems is the function series problem:
$$\sum_{k=0}^{\infty} a_k f_k(x)=0, \text{solve for $a_k\in \Bbb{Z}$ given $f_k, x$}$$
might study this later, it might had something to do with integrals
Anyway, it basically means that any direct methods to deal with transcendence determination will not work since the knapsack problems are NP hard and some are NP complete
thus that explains why all the proof relies on approximation schemes
 
1:14 PM
In conclusion: Transcendental number theory in terms of functional analysis is a subset of the following problem:
> Find the orthgonal complement with integer components of any given vector $v$ in the hilbert space $\Bbb{P}(C)$ with the usual inner product
if $v=(x^n)_{n \in \Bbb{N}}$ which has finite support, this is precisely the transcendental number problem
Thus $x$ is transcendental if $v^{\perp}$ is trivial, otherwise it is algebraic
 
 
2 hours later…
3:34 PM
If there some users with good knowledge of this area are around, they might be able to give some advice on tag related discussion on meta and in chat: Should we divide (dimension-theory) into 2 tags, one for topology, and one for algebra?, About (dimension-theory)
 
4:26 PM
@ÍgjøgnumMeg Hi
 
Hey @Jacksoja
 
I have a small question and you helped me before
it is a linear algebra question
 
I can try :P
 
if F^S denoted the maps from the set S to the field F
I dont understand how the vector space F^n
can be viewed as a map from the set {1,2,...n} --> F
this is in sheldon axler book linear algebra done right
@ÍgjøgnumMeg still here? :)
 
elements in F^n are n tuples. Thus each entry in the tuple is indiced by a natural number, and hence is really a function f : {1,2,3,...,n} --> F
 
4:38 PM
@Secret but if we take an n tuple, is the outout an element in F?
or is it in F^n ?
{1,2,3,...,n} --> ( x_1, ...,x_n ) in F^n
 
ah sorry, I mean for each entry in the tuple, each natural number get mapped into some F, thus f takes n arguments and give n outputs
so like you written there
 
okay so component wise it is a map from {1,2,3,...,n} --> F ?
 
yeah, something like for each i in {1,2,3,...,n}, f(i) in F
thus you have (f(1),f(2),f(3),...,f(n)) in F^n
 
okay thanks
 
4:53 PM
is the sum of zeta(n) from n=2 to N approximately N?
 
This article gives some specific values of the Riemann zeta function, including values at integer arguments, and some series involving them. == The Riemann zeta function at 0 and 1 == At zero, one has ζ ( 0 ) = B 1 − = − B 1 + = − ...
Even zeta integers do tend to 1 as it goes to infinity, but odd values are not written, probably also do similar behaviour
 
 
1 hour later…
6:03 PM
Let $E$ be a regular expression over an alphabet $\Sigma$. Let $[x^n y^m] f(x,y)$ be the number of strings in $\Sigma^*$ of length $n$ with $m$ prefixes matching $E$. Is there a systematic way to obtain $f$ from an arbitrary $E$?
 
2
Q: Surface measure as a finite Borel measure in $\Bbb R^n$

Impending UncertaintyGiven the result in http://www.math.jyu.fi/research/pspdf/236.pdf (theorem 1.2) I want to prove that for $ f \in L^2 (S^{n-1}) $ we have $$ \| \hat{f} \|_{L^q( \mathbb{R^n} )} \lesssim_{n,q} \| f \|_{ L^2 ( S^{n-1})} $$ for $ q > 2 \frac{n+1}{n-1} $ I know that $|\hat{\sigma}(x)| \lesssim_n...

 
@Jacksoja No. The ith component is telling you where the integer $i$ in $\{1,2,\dots,n\}$ gets mapped.
 
any ideas on the above post ?
 
 
2 hours later…
7:43 PM
I have two sequences of RVs that are independent. If they converge in distribution, are the limit Rvs independent? Intuitively it seems obvious that they are
 
7:53 PM
To be clear: You've got $\{X_n\},\{Y_n\}$ such that $\lim_{n\to\infty} F_{X_n}(t)=\lim_{n\to\infty} F_{Y_n}(t)$ for any $t$, and you know $F_{X_n,Y_n}(x,y)=F_{X_n}(x) F_{Y_n}(y)$ for all $n$. Must it be the case that $$\lim_{n\to \infty} F_{X_n,Y_n}(x,y) = (\lim_{n\to\infty} F_{X_n}(x))(\lim_{n\to\infty} F_{X_n}(x)) \quad ?$$
that notation is a bit yuck
$F^{(n)}_{X}\equiv F_{X_n}$ might be easier
 
no theyre not necessarily converging to the same distribution
 
Oh, they converge separately
 
yes but otherwise your setup is what I'm asking
 
So the first two limits need to exist, but not necessarily equal
 
yup, in distribution
I'm gonna assume yes
 
7:56 PM
also, last limit should have been over $F_{Y_n}(y)$
 
right
 
I want to say it's true as well: it amounts to whether $\displaystyle \lim_{n\to\infty} \left(F_{X_n}(x) F_{Y_n}(y)\right) = \left(\lim_{n\to\infty} F_{X_n}(x)\right) \left(\lim_{n\to\infty} F_{Y_n}(x)\right)$
Which seems like it's just the product rule for limits
 
thanks Semi
 
The only weird thing I could imagine is that the joint distributions fails to converge in distribution despite the individual RV's doing so
 
gonna assume that's not gonna happen
 
8:00 PM
that may be forbidden tho
my intuition when it comes to these kinds of limits is limited (no pun intended)
 
ugh I'm not seeing how to do this
 
Is there a larger problem that this is part of?
 
yes and I dont think this is the way to go, it's this
well, this is the part of the larger problem I'm stuck on
N(t) is an RV with pdf u(t) = λexp(-λt)
define ΔN = N(t+Δt) - N(t)
 
I'm confused. In what sense is N a function of t?
 
in the sense that it's a process
a drift or whatever it's called
 
8:04 PM
Poisson something something?
 
it defines a family of distributions parameterized by t
so its image on the time line is a path
 
Okay. Are you given how that family is parametrized by t?
It's not clear from what you've said what the sample space of N(t) would be, for instance.
 
damn it, I hate the way these questions are worded
and there may be typos with this prof
 
do you have the original question?
 
let me reread it from the beginning, yes
$T_1, T_2, \ldots, T_M \sim U([0..R])$ independently
I was mixing things up before, let's start from scratch
Define $T_{[1]} = \min T_k, T_{[2]} = \min T_k \text{ such that } T_k > T_{[1]}$, and so on
define $N_t = \#\left\{ { T_k < t} \right\}$
 
8:10 PM
So you start with M random samples of a uniform RV, and then you order that
\# maybe?
 
right
first part, which I did:
Show that in the limit as $R \to \infty$ we have
 
also, shoould that be $T_k<t$ or $T_{[k]}<t$
 
the professor makes typos so I think it should be the latter
I showed that as $R \to \infty$ we have $\Pr(N_t = n) = (\lambda t)^n e^{\lambda t} \frac 1 {n!}$
 
huzzah for poisson
 
How do you find the maximum value of $\sin\frac{\theta}{2}(1+\cos\theta)$ when $0<\theta<\pi$
?
 
8:13 PM
except that's not what he wrote, he wrote $t^n$ instead of $(\lambda t)^n$ and when I did the proof I didn't get that so it had to be a typo
 
Well, I'd actually expect that $\lambda$ to be in there
 
it's smooth MrAP so if it has a max then the derivative vanishes, so that's your starting point
oh
 
otherwise you'd have $\text{Pr}(N_t=n)\propto t^n/n!$
 
right you're right
 
and $\lambda$ really would have no reason to be in there at all
Also, I'm guessing it should be $e^{-\lambda t}$?
Should have $\sum_{n=0}^\infty \text{Pr}(N_t=n)=1$
 
8:15 PM
yes that's what I wrote down but I typed it here incorrectly, I got
 
I want to do it without using the concept of maxima and minima, or in other words, calculus,@GFauxPas.
 
wait no
ugh I need to take a break, these typos are frustrating me too much
 
mmkay
 
talk to you later
 
But basically I'm seeing that you've got a poisson RV $N_t$ with rate parameter $t$
 
8:16 PM
I have tried to do it using calculus. The equation you have to solve after taking the first derivative and equating it to 0 is not simple to solve.
This is BTW a question carrying 2 marks.
 
@MrAP try using some trigonometry. Your first factor is in terms of the half-angle and your second is in terms of the whole angle
 
I have tried.
 
But you can write the latter in terms of half-angles if you use some trig
 
Yes I have done that.
 
okay. What did you get?
 
8:20 PM
$2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}$
 
Right. It's convenient to note that $\cos^2(\theta/2)=1-\sin^2(\theta/2)$
 
Yes. I was going the type that.
 
So that amounts to $2\sin\frac{\theta}{2}-2\sin^3\frac{\theta}{2}$
 
Yes.
 
Now, doing calculus at this point is not so bad.
it's particularly not-bad if you think in terms of the chain rule
 
8:29 PM
I am getting $\frac{4\sqrt{3}}{9}$
@Semiclassical
 
same
if you're writing this up as an exercise, you should make sure to check the endpoints. just because your function has a critical point there, doesn't mean it achieves its maximum there
 
but beyond that you're good
 
how many hours a day do you guys study math?
 
I had got two values for $\theta$ after solving the equation obtained after equating the first derivative to 0. Then took the second derivative, substituted both values separately and found that value of $\theta$ for which the second derivative is less than 0. After that I substituted that value in the original expression to get the maximum value.
 
8:38 PM
yeah, that should do it
it's a critical point and a local maximum, and since it's the only such local maximum (and because the function is continuous on this range) you're okay to conclude that it's the maximum on that interval
 
I study math for $x^3+3x^2+9x+59$ hours everyday
 
Yeah.
 
8:53 PM
I'd like to elaborate. I used to study math whenever, but then I got serious about calculating precisely how many hours I should study in order to maximise my efficiency. Over the course of several years I painstakenly pored over thousands of math documents but something wasn't quite right...however after countless revisions, foreclosures, bankrupcties, and cleaning the street with a tooth brush to scrape by, I found THE polynomial that maximized my efficiency. The end.
 
seems legit
 
$x\in R$ I hope?
 
it's got complex roots, so maybe not
 
Nah, $x\in\mathbb{Z}/17\mathbb{Z}[y]$
 
most schedules are complex, anyways, since they typically contain both real and imaginary parts
 
8:59 PM
Woh! Which dimension do you live in?
@Semiclassical, Thats also possible.
 
(an old joke, but still a good one)
 
Haha
 
Hi all, i have a quick question. In Ahlofor's Complex Analysis, I have come across the phrase about a collection of functions being "uniformly bounded on compact sets". Does this mean the same bound should work for all compact sets? Or, rather, the bound depends on the compact set?
Btw, this was in the statement of Montel's Theorem
 
9:51 PM
Is it true for any field that the sum of the roots of a polynomial equals the coefficient of the second highest degree term?
 
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