Mathematics - 2019-04-18 (page 1 of 2)

topologicalmagician

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?

Jake Rose

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

BAYMAX

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?

Semiclassical

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

Érico Melo Silva

tbh proof by doing every possible case is like.. the most sound

Semiclassical

it's proof by exhaustion, both in that you're exhausting all options and in that it's typically rather tiresome

BAYMAX

2:00 AM

cool

sounds good

Semiclassical

@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?

BAYMAX

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?

Daminark

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

BAYMAX

3:11 AM

cool!

Ted Shifrin

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?

Érico Melo Silva

yo @Ted

Daminark

3:49 AM

Henlo

flowian

What websites are trustworthy about reading uni reviews before applying ?

Érico Melo Silva

what does that even mean

flowian

just looking for recommendations where to read about universities before applying

Bsc math courses

Érico Melo Silva

i wouldnt really trust much besides actually talking to people w direct experience w whatever school

Adam

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/…

Rudi_Birnbaum

math.stackexchange.com/questions/3191983/… any contributions welcome.

Adam

don't they have a physics SE?

Rudi_Birnbaum

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

Adam

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

Rudi_Birnbaum

@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

Adam

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.

Rudi_Birnbaum

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.

Adam

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

Rudi_Birnbaum

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.

Adam

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

Rudi_Birnbaum

@Adam sad to hear. Ever tried to consult any administratives on your issue?

Adam

5:12 AM

yes I have made several appeals over the past 3 years naturally

Zee

what medication is it?

pain killers?

Adam

no

Adderall

Zee

you dont need that poison

Adam

lol

ok

Zee

amphetmines have been shown to be nuerotxic in monkey brain at theraputic doses

Adam

5:15 AM

do you think you are funny?

Rudi_Birnbaum

As a chemist I am astonished that SUCH a simple compound can be that expansive.

Zee

not funny , its reality

Adam

no it's not

Zee

its the patents that make them expensive

and amphetamines are dirt cheap to produce

Rudi_Birnbaum

My students can cook that in 100g quantities for costs at about a 50 $ I'd say.

Adam

5:17 AM

you have no idea what you are talking about. It's the legal system that makes them expensive

Rudi_Birnbaum

@Zee sure

@Adam really?

Adam

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

Rudi_Birnbaum

@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.

Adam

I'm sure you are well and truly educated in medicine yes I shall take your advice random person

Rudi_Birnbaum

5:21 AM

@Adam which is still hillariously expansive

Adam

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

Rudi_Birnbaum

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

Adam

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

Rudi_Birnbaum

5:25 AM

and to become one you need to learn and experience a lot

Zee

most pharmacologists work in research

Rudi_Birnbaum

@Zee your confusing what I tried to sort out.

Zee

Explain please

Rudi_Birnbaum

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

Adam

and they are never used to treat OCD lol

Zee

5:29 AM

Erdos developed AFTER using

Rudi_Birnbaum

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

Rudi_Birnbaum

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)

ousar.lib.okayama-u.ac.jp/files/public/3/30980/…

Adam

" 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. "

Rudi_Birnbaum

5:33 AM

@Zee So what is your conclusion from that?

Zee

that amphetmines are nuerotoxic

Adam

for a monkey

Zee

lol ok

Rudi_Birnbaum

@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

Adam

5:35 AM

can I give you a gold star for online trolling?

Rudi_Birnbaum

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

Rudi_Birnbaum

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

Rudi_Birnbaum

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

Rudi_Birnbaum

"big" is the key word though. How could

we judge what big means without dwelling deeply into toxicology and pharmacology

Adam

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

researchgate.net/publication/…

Adam

So... thankyou for clarifying how weak your original argument actually is Zee

Rudi_Birnbaum

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

Adam

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

Rudi_Birnbaum

@Zee ever heared that slogan that its the dosis makes the toxin?

Zee

yes

Adam

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...

Rudi_Birnbaum

@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

Rudi_Birnbaum

You see

Zee

amphetamines, maybe 50 years

Rudi_Birnbaum

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

Adam

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

Rudi_Birnbaum

5:52 AM

@Zee ok, that makes sense

(no irony)

Adam

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"

Rudi_Birnbaum

@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

Adam

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 :)

Adam

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...

Adam

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

Anti-American Anti-Zionist

6:37 AM

Helo! Can somebody point out what does $x\in Q^c$ mean?

Astyx

6:49 AM

It depends on the context

Anti-American Anti-Zionist

7:13 AM

@Astyx I read your message after a while, so didn't respond on time

Astyx

So $Q$ is the set of rationals

$Q^c$ is complementary set of the set of rationals, that is irrationals

Anti-American Anti-Zionist

Thank you!

Astyx

And the last $Q$ in the proof should be $Q^c$ here

Adam

@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

Jasmine

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

How 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 ^

ÍgjøgnumMeg

9:46 AM

Morning-ish everyone

Silent

10:01 AM

Gallian asks to find a mistake in the statement following picture:

ÍgjøgnumMeg

$xyz \neq 0$

ÍgjøgnumMeg

10:36 AM

Heile @Rudi :)

TheSimpliFire

10:58 AM

@Silent heh, $(-1)^n+1^n=0^n$ for all odd $n$

Secret

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 $P$

Secret

12:36 PM

The above writing also showed why it is in general hard to determine whether a number is transcendental:

Subset sum problem

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

Knapsack 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

Secret

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

Martin Sleziak

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)

Jacksoja

4:26 PM

@ÍgjøgnumMeg Hi

ÍgjøgnumMeg

Hey @Jacksoja

Jacksoja

I have a small question and you helped me before

it is a linear algebra question

ÍgjøgnumMeg

I can try :P

Jacksoja

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? :)

Secret

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

Jacksoja

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

Secret

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

Jacksoja

okay so component wise it is a map from {1,2,3,...,n} --> F ?

Secret

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

Jacksoja

okay thanks

Ultradark

4:53 PM

is the sum of zeta(n) from n=2 to N approximately N?

Secret

Particular values of the Riemann zeta function

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

user76284

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 $E$?

Impending Uncertainty

2

Q: Surface measure as a finite Borel measure in $\Bbb R^n$

Given 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...

functional-analysis fourier-analysis lp-spaces harmonic-analysis

Ted Shifrin

@Jacksoja No. The ith component is telling you where the integer $i$ in $\{1,2,\dots,n\}$ gets mapped.

Impending Uncertainty

any ideas on the above post ?

GFauxPas

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

Semiclassical

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

GFauxPas

no theyre not necessarily converging to the same distribution

Semiclassical

Oh, they converge separately

GFauxPas

yes but otherwise your setup is what I'm asking

Semiclassical

So the first two limits need to exist, but not necessarily equal

GFauxPas

yup, in distribution

I'm gonna assume yes

Semiclassical

7:56 PM

also, last limit should have been over $F_{Y_n}(y)$

GFauxPas

right

Semiclassical

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

GFauxPas

thanks Semi

Semiclassical

The only weird thing I could imagine is that the joint distributions fails to converge in distribution despite the individual RV's doing so

GFauxPas

gonna assume that's not gonna happen

Semiclassical

8:00 PM

that may be forbidden tho

my intuition when it comes to these kinds of limits is limited (no pun intended)

GFauxPas

ugh I'm not seeing how to do this

Semiclassical

Is there a larger problem that this is part of?

GFauxPas

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)

Semiclassical

I'm confused. In what sense is N a function of t?

GFauxPas

in the sense that it's a process

a drift or whatever it's called

Semiclassical

8:04 PM

Poisson something something?

GFauxPas

it defines a family of distributions parameterized by t

so its image on the time line is a path

Semiclassical

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.

GFauxPas

damn it, I hate the way these questions are worded

and there may be typos with this prof

Semiclassical

do you have the original question?

GFauxPas

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 $N_t = \#\left\{ { T_k < t} \right\}$

Semiclassical

8:10 PM

So you start with M random samples of a uniform RV, and then you order that

\# maybe?

GFauxPas

right

first part, which I did:

Show that in the limit as $R \to \infty$ we have

Semiclassical

also, shoould that be $T_k<t$ or $T_{[k]}<t$

GFauxPas

the professor makes typos so I think it should be the latter

I showed that as $R \to \infty$ we have $\Pr(N_t = n) = (\lambda t)^n e^{\lambda t} \frac 1 {n!}$

Semiclassical

huzzah for poisson

MrAP

How do you find the maximum value of $\sin\frac{\theta}{2}(1+\cos\theta)$ when $0<\theta<\pi$

?

GFauxPas

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

Semiclassical

Well, I'd actually expect that $\lambda$ to be in there

GFauxPas

it's smooth MrAP so if it has a max then the derivative vanishes, so that's your starting point

oh

Semiclassical

otherwise you'd have $\text{Pr}(N_t=n)\propto t^n/n!$

GFauxPas

right you're right

Semiclassical

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 $\sum_{n=0}^\infty \text{Pr}(N_t=n)=1$

GFauxPas

8:15 PM

yes that's what I wrote down but I typed it here incorrectly, I got

MrAP

I want to do it without using the concept of maxima and minima, or in other words, calculus,@GFauxPas.

GFauxPas

wait no

ugh I need to take a break, these typos are frustrating me too much

Semiclassical

mmkay

GFauxPas

talk to you later

Semiclassical

But basically I'm seeing that you've got a poisson RV $N_t$ with rate parameter $t$

MrAP

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.

Semiclassical

@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

MrAP

I have tried.

Semiclassical

But you can write the latter in terms of half-angles if you use some trig

MrAP

Yes I have done that.

Semiclassical

okay. What did you get?

MrAP

8:20 PM

$2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}$

Semiclassical

Right. It's convenient to note that $\cos^2(\theta/2)=1-\sin^2(\theta/2)$

MrAP

Yes. I was going the type that.

Semiclassical

So that amounts to $2\sin\frac{\theta}{2}-2\sin^3\frac{\theta}{2}$

MrAP

Yes.

Semiclassical

Now, doing calculus at this point is not so bad.

it's particularly not-bad if you think in terms of the chain rule

MrAP

8:29 PM

I am getting $\frac{4\sqrt{3}}{9}$

@Semiclassical

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

PutnamLegend

hai

Semiclassical

but beyond that you're good

PutnamLegend

how many hours a day do you guys study math?

MrAP

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.

Semiclassical

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

Ultradark

I study math for $x^3+3x^2+9x+59$ hours everyday

MrAP

Yeah.

Ultradark

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.

Semiclassical

seems legit

MrAP

$x\in R$ I hope?

Semiclassical

it's got complex roots, so maybe not

Rithaniel

Nah, $x\in\mathbb{Z}/17\mathbb{Z}[y]$

Semiclassical

most schedules are complex, anyways, since they typically contain both real and imaginary parts

MrAP

8:59 PM

Woh! Which dimension do you live in?

@Semiclassical, Thats also possible.

Semiclassical

(an old joke, but still a good one)

MrAP

Haha

Nicholas Roberts

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

user193319

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?

Transcript for

$Mathematics$ Mathematics