Mathematics - 2017-09-12 (page 2 of 3)

YOUSEFY

08:06

@MaryStar we have 4 e. Thus, instead of writing only "10 choose 3" you should write also "4 choose 3" plus "10 choose 3".

Mary Star

Could you explain to me the part that we have to subtract? @YOUSEFY

Secret

> the union of a countable set of countable sets is itself countable. So ℵ 1 {\displaystyle \aleph _{1}} \aleph _{1} cannot be written as the sum of a countable set of countable cardinal numbers, and is regular.

Hmm interesting

That naively sounds like if AC fails, there exists countable cardinalities $S$ such that S+S+... =uncountable

Tobias Kildetoft

@Secret Right. Or rather, in the absense of AC it is consistent with ZF that such sets exist

Secret

or in other words, in cases where AC fails, cardinal sum can allow reaching from a lower level of infinity to the next higher one using only the lower levels of infinities

which is impossible if AC is true since $\aleph_1$ will be regular

Tobias Kildetoft

Though for the countable one here, countable choice suffices, and that is a version considered a lot weaker than full choice

Secret

08:22

I think I will find ZF+countable choice (CC) pretty interesting once I get my head around what cofinality is

> Without the axiom of choice, there would be cardinal numbers which were not well-orderable. Moreover, the cardinal sum of an arbitrary collection could not be defined. Therefore only the aleph numbers can meaningfully be called regular or singular cardinals.

hmm, need to be careful...

> Sets with cardinality ℶ 2 {\displaystyle \beth _{2}} \beth _{2} include:

> The set of all functions from Rm to Rn

Hmm, so that means the set of all mxn matrices has cardinality $\beth_2$

(to be proved or find counterexample: The underlying set of the general linear group over the reals $GL_n(\Bbb{R})$ has cardinality $\beth_2$)

Tobias Kildetoft

08:40

@Secret No, that group has same cardinality as the reals

Alessandro Codenotti

@Secret why?

Secret

An mxn matrix is a linear map from $\Bbb{R}^m$ to $\Bbb{R^n}$?

Balarka Sen

no, R^n --> R^m

Secret

o wait, forgot the nonlinear maps...

Alessandro Codenotti

indeed

Secret

08:46

Ok, so we know that $\mathcal{C}(\Bbb{R})$ has cardinality $\beth_1$ because it must be continuous at rational points and there are only countably many rationals. Now to figure out why the set of all linear maps between two finite dimensional vector spaces also have cardinality continuum. I wonder:

The set of all invertible functions $f: \Bbb{R}\to \Bbb{R}$ have cardinality $\beth_1$ as well :?

let me think...

Alessandro Codenotti

An $n\times m$ matrix is determined by $nm$ real numbers, so there is a nice surjection $\mathcal{P}^{\text{fin}}(\Bbb R)\to\text{Hom}(\Bbb R^n,\Bbb R^m)$

Secret

Ah right so all power sets of reals of finite support will form a bijection with the mxn matrices

which there are continuumly many of them

Alessandro Codenotti

I think $[\Bbb R]^{<\omega}$ is the standard notation for the set of finite subsets?

Secret

In that case, I am not very sure what $\mathcal{P}^{\text{fin}}$ means

Alessandro Codenotti

the same thing, I've seen both before

Secret

08:54

ok

As for invertible, not necessary continuous functions in $\Bbb{R}^{\Bbb{R}}$, given each point in the domain, its image is unique, so these functions necessary form a bijection from the reals to some subsets of the reals (if the functions themselves are also surjective, then they form a bijection from the reals to itself), and the number of bijections in $\Bbb{R}^{\Bbb{R}}$ is... (computing...)

should be equinumerous to the number of permutations of the reals which is $2^{\mathfrak{c}}=\beth_2$

YOUSEFY

@MaryStar It should be something like: $\frac{10!}{2!\cdot 2!\cdot 1!\cdot 1!\cdot 4!}-{ \binom{4}{3} * \binom{10}{3}$. Sorry it should be multiplicative rule instead of addition rule.

Secret

Or in other words, the symmetric group over the reals has cardinality $\beth_2$

Hmm...

$S_{\Bbb{R}} = \{\sigma|\forall r,s \in \Bbb{R},\sigma (r) = \sigma (s)\}$?

Tobias Kildetoft

There, Marie Curie proposal submitted. Now to let it lie for a day and see if I realize something that should have been added (I can still update it until the deadline on Thursday).

Secret

09:12

$S_{\Bbb{R}} = \{\sigma | \forall r \in \Bbb{R}, s \in A \subseteq \Bbb{R}, \sigma (r)=\sigma (s) \implies r=s\}$

seemed to be it

Proof:
So... for each $\sigma(r)$ there are continuumly many possible images, and there are continuumly many possible $r$. Therefore in total there are $\mathfrak{c} \cdot \mathfrak{c} \cdots = \mathfrak{c}^{\mathfrak{c}} = 2^{\mathfrak{c}}=\beth_2$ many $\sigma$

Of course a more foolproof proof will be simply showing that $S_{\mathbb{R}}$ bijects with the powerset of reals, but I have no idea how to start the proof without somehow counting the number of possible $\sigma$ s

and unlike a countable set, I don't have sequences to enumerate the elements and use some kind of induction proof

Secret

09:29

How would you guys prove that $|S_{\Bbb{R}}|=\beth_2$?

Alex K Chen

09:59

Are you hair, @LeakyNun ?

Leaky Nun

10:30

@Secret you need the function to be surjective

ACuriousMind

@AlexKChen I think they're feet.

Leaky Nun

10:47

@Secret The cardinality of the bijective functions from $X$ to $X$ is $|X^X|$. For $X=\Bbb N$, see this and this. For arbitrary $X$, see this.

Tobias Kildetoft

@LeakyNun Well, for infinite $X$

Leaky Nun

11:14

right

user193319

Why isn't the indefinite integral of $\ln (t+1)$ equal to $(t+1) \ln(t+1) - (t+1) +c$? I mean, isn't $\int \ln (x) dx = x \ln x - x + c$?

Secret

11:40

The only reason I can think of is the +1 get absorbed into the c

actually no

When you integrate by parts, the second term is very different than compared to the case of $\ln x$

because the second term is $-\int \frac{t}{t+1}dt$

and that changes things

Leaky Nun

@user193319 who said it isn't?

Secret

$$\int ln x dx = x \ln x - \int \frac{x}{x} dx = x \ln x - x + C$$

$$\int (x+1) dx = x \ln (x+1) -\int \frac{x}{x+1}dx=x\ln (x+1) -\int 1-\frac{1}{x+1}dx =(x+1)\ln (x+1) - x + C$$

Leaky Nun

just let x=t+1

and reduce the problem to a solved case

Secret

Yes, inverse chain rule works for this problem

Leaky Nun

huh, inverse chain rule

Secret

11:55

32

Q: Is there a chain rule for integration?

I know the chain rule for derivatives. The way as I apply it, is to get rid of specific 'bits' of a complex equation in stages, i.e I will derive the $5$th root first in the equation $(2x+3)^5$ and continue with the rest. I wonder if there is something similar with integration. I tried to integr...

Special case of integration by substitution

i.e. change of variable so that the integral becomes $\int f'(x)g(f(x))dx = G(f(x)) + C$

Leaky Nun

right

@Secret have we tried finding out the cardinality of continuous real functions?

Secret

We do roughly 3 months ago. Alessandro and steamy root was also there. The proof hinge on the continuity need to hold at rationals and there are onnly countably many rationals

Leaky Nun

right

ponder at this function $\Bbb Q \to \Bbb Q$:

$$f(x) = \begin{cases} 1 & x^2<2 \\ 0 &x^2>2 \end{cases}$$

Convince yourself that it is continuous

Secret

under the subspace topology?

Leaky Nun

right

aka the usual topology on $\Bbb Q$

metric, order, etc.

SteamyRoot

12:04

@Secret If I was, I can't guarantee I paid any attention to it :P

Leaky Nun

then proceed to ask more bizarre questions

Secret

It's ok, I have uncountably too many rambles for people to keep track of

SteamyRoot

Though the argument is rather simple, indeed using the rationals

since they form a countable dense subset of the reals

Secret

there's just one problem with this question...
I have yet to read the chapters on the notion of continuity in topology

SteamyRoot

Doesn't matter

Just use the epsilon-delta

abenthy

12:15

I'm trying to get a feeling for the distinction between locally symmetric and globally symmetric Riemannian manifolds. So I'm wondering, is there an example of a locally symmetric manifold which is isometrically embedded in $\mathbb{R}^3$ and is not globally symmetric?

sashas

Kelly criterion

In probability theory and intertemporal portfolio choice, the Kelly criterion, Kelly strategy, Kelly formula, or Kelly bet, is a formula used to determine the optimal size of a series of bets. In most gambling scenarios, and some investing scenarios under some simplifying assumptions, the Kelly strategy will do better than any essentially different strategy in the long run (that is, over a span of time in which the observed fraction of bets that are successful equals the probability that any given bet will be successful). It was described by J. L. Kelly, Jr, a researcher at Bell Labs, in 1956....

0

Q: Question related to kelly fractions

I have been studying about Kelly fractions. Background I have wealth $W$. There is a game which I win with probability $p$ and lose with $1-p$. If I win the game then I get back whatever I invested plus $o$ times whatever I invested. If I lose I lose what I invested and get nothing. What is t...

probability probability-theory expectation

any help is appreciated

thank you

abenthy

I decided to post it as a question now: math.stackexchange.com/questions/2426530/…

Secret

12:30

$\lim_{x\to a^+} x^2 = \{ 0, a^2 < 2 ; 1, a^2 > 2 =\lim_{x\to a^-}x^2$

in fact, the only point where the upper and lower limits don't coincide is $x=\sqrt{2} \not\in \Bbb{Q}$ thus it is continuous in $\Bbb{Q}$

user307388

How can I cover the torus $\mathbb R^2 / \mathbb Z^2$ with charts?

Leaky Nun

@Secret you can view it that way...

new puzzle: partition the rationals into two dense (sub)sets

Secret

any dedekind cut at an irrational will partition the rationals into two sets

which are dense subsets of the rationals

Leaky Nun

but they are not dense

Secret

How is $\Bbb{Q} \cap \{x|x < \sqrt{2}\}$ and $\Bbb{Q} \cap \{x|x > \sqrt{2}\}$ not dense?

Leaky Nun

12:45

2 is not a limit point of the former

-2 is not a limit point of the latter

SteamyRoot

I think he means the union?

Leaky Nun

he doesn't

SteamyRoot

Though, to be correct, you have to specify dense in what set

A set on its own is not dense or not. It's dense or not in something else

Secret

hmm...

Leaky Nun

in the rationals :)

Secret

12:50

How about:

the subset of rationals with odd numerators and even denominators, and the subset of rationals with even numerators and odd denominators?

SteamyRoot

rip 1/3

Secret

......

is the set of prime fractions dense?

i.e. both numerators and denominators being prime

Leaky Nun

what is prime fraction?

i would say yes

Secret

Then partition Q into rationals made of prime numerators and denominators, and its complement (either or both numeroator and denominator being composite)

SteamyRoot

You'll want to throw a minus sign in there somewhere, though.

user307388

12:54

Thanks for your answers.

Secret

ah yes, including negative numbers for each set

SteamyRoot

It's not too hard to prove the fractions p/q are dense in [0,1]

Secret

but I seriously think there's a more "even" partition of the rationals, but I am not terribly good with the structure of rationals

SteamyRoot

I don't think your idea of odd and even things was bad.

It just wasn't a partition.

Secret

There should be 4 distinct (insert word) of rationals, namely even/even, even/odd, odd/even, odd/odd

but the problem are things like these:

4/2=6/3=8/4

these screwed the partition

SteamyRoot

12:59

even/even always simplifies to something else.

Simply Beautiful Art

Not meaning to interrupt, but is anyone in here familiar with the Euler-Maclaurin formula?

SteamyRoot

Not me.

Secret

Simpleart: A little bit, what do you want to know about it?

Simply Beautiful Art

If my error term is correct

6

A: How to accelerate the convergence of $1 + \frac{1}{2^2} + \frac{1}{3^2} + \ldots$?

By the Euler-Maclaurin formula, $$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^{a-1}\frac1{n^2}+\underbrace{\int_a^\infty\frac1{x^2}~\mathrm dx}_{=1/a}+\frac1{2\times a^2}+\sum_{k=1}^p\frac{B_{2k}}{2ka^{2k}}+R_p$$ $$|R_p|\le\frac{2 (2p)!}{(6a)^{2p}}$$ Choosing large enough $a$ will result in very f...

I tried a = 5 and p = 7, which should give me something like 1.7 times 10^-11 error

but its off by 2/1000

Simply Beautiful Art

13:13

:-/ maybe my calculator's being weird

Secret

I cannot seemed to derive the last two terms using the formula...

Simply Beautiful Art

You mean the part with the Bernoulli numbers?

Secret

yup

Simply Beautiful Art

It should come down to finding the nth derivative of 1/x^2

SHHOOOOT

Welp, that's my mistake

I took the nth derivative of 1/x

Secret

2

Q: Show that the ring of all rational numbers, which when written in simplest form has an odd denominator, is a principal ideal domain.

Show that the ring of all rational numbers $m/n$ with $n$ an odd integer is a principal ideal domain. We haven't really discussed principal ideal domains. I've heard that this is easy, but I just lack the basic knowledge of what a principal ideal domain is.

abstract-algebra commutative-algebra principal-ideal-domains rational-numbers

Semiclassical

13:17

@SimplyBeautifulArt Here's a fun facet of the Euler-Maclaurin formula.

Secret

Steamy and Leaky: PID might provide the relevant partition

Test:
$\Bbb{Q} = \{\frac{p}{q} \in \Bbb{Q}, \gcd(p,q)=1\} \cup \{\frac{p}{q} \in \Bbb{Q}, \gcd(p,q)\neq 1\}$

Semiclassical

One way to interpret Euler-Maclaurin is that it tells you not only that the sum $\sum_{n=a}^b f(x)$ serves as a trapezoidal approximation of the integral $\int_a^b f(x)\,dx$, but a series of smaller and smaller corrections to this approximation.

and that these corrections are in terms of how $f(x)$ and its derivatives behave at one endpoint versus another.

Secret

Unrelated side note: The sum $\sum_{n=1}^{k} \frac{H_n^p}{n^q}$ can be computed in closed form entirely in terms of generalised harmonic number except it is quite tedious when done directly. More details in the Mathworks room

Semiclassical

@Secret I'd personally be happy to know the leading-order behavior of that sum when $k$ is large.

(That's not in the sense of 'i want to know it for my own stuff' but "I'd be satisfied with knowing that much.")

user307388

The rotation group $SO(3)$ is a 3-manifold?

Simply Beautiful Art

13:25

@Semiclassical well ofc

Secret

Let me double check my workings first... (so far I have $\sum H_n$ and $\sum H_n^2$ computed, but I have not checked whether they equal to the known values of $\sum H_n^2$). It does seemed computing the infintie sum by starting with the partial sum and take limits seemed to be the safest bet but I need to be sure

Semiclassical

okay. Now consider a special case of that: Suppose $f(x)$ is a function of $x$ such that $f$ and its first $k$ derivatives at one endpoint matches the other endpoint.

then you can show that the error term signified by those corrections is of order $O(1/n^k)$. (maybe k+1, maybe k-1. i don't actually remember the precise detail)

user307388

It seems that I have been ignored by almost all users here.

Secret

NB: To be announced later: In light of the recent dangerous signal of Ted is starting to lose tolerance to filter my rambles, and hence a potential risk of him falling off the timeline, and that after 4 years of observation, my rambles seemed to be produced in a rate steady enough to keep any new rooms alive. Mathworks room is set up to handle that so that Ted will not fall off the timeline

and interested users (we knew it is nonempty) can still read my rambles

Semiclassical

Now suppose that $f$ is a periodic and analytic function of $x$, such that $f$ and all of its derivatives match at the endpoints. Then all Euler-Maclaurin error terms vanish. @SimplyBeautifulArt

Does it follow that the sum is exactly the same as the integral?

Secret

13:30

@MathAminPhysics .

Charts on SO(3)

In mathematics, the special orthogonal group in three dimensions, otherwise known as the rotation group SO(3), is a naturally occurring example of a manifold. The various charts on SO(3) set up rival coordinate systems: in this case there cannot be said to be a preferred set of parameters describing a rotation. There are three degrees of freedom, so that the dimension of SO(3) is three. In numerous applications one or other coordinate system is used, and the question arises how to convert from a given system to another. == The space of rotations == In geometry the rotation group is the group of...

Yes it is a 3-manifold

some details on how to understand its structure here

I am unable to say anything else as I have almost no background on diff geom

Simply Beautiful Art

@Semiclassical No idea

On what interval?

user307388

@Secret Thanks. It seems that you have not ignored me yet.

Alessandro Codenotti

@MathAminPhysics it's homeomorphic to $\Bbb P^3(\Bbb R)$

Semiclassical

some finite interval. If one wants to make this as simple as possible, one would probably take the interval to be $[0,2\pi]$ and the function to be something like $f(2\pi k/n)$ where $f(x)$ is $2\pi$-periodic.

(the usual statement of Euler-Maclaurin is geared towards the sum form more than the integral.)

Euler-Maclaurin in that case would take the following form:

user307388

@AlessandroCodenotti Thanks. Why?

Alessandro Codenotti

13:36

There is an intuitive explanation on wiki, in the "topology" section

Semiclassical

$$\sum_{k=1}^n f(2\pi k/n)=\int_{0}^n f(2\pi x/n)\,dx +\frac{f(2\pi)-f(0)}{2}+\sum_{k=1}^{\lfloor p/2\rfloor} \frac{B{2k}}{(2k)!}(f^{(2k-1)}(2\pi)-f^{(2k-1)}(0))+R$$

dangit

Simply Beautiful Art

lol

Semiclassical

phew

which we can reorganize to

$$\frac{1}{n}\sum_{k=1}^n f(2\pi k/n)-\frac{1}{2\pi}\int_{0}^{2\pi} f(x)\,dx=\frac{f(2\pi)-f(0)}{2}+\sum_{k=1}^{\lfloor p/2\rfloor} \frac{B{2k}}{(2k)!}(f^{(2k-1)}(2\pi)-f^{(2k-1)}(0))+R$$

Simply Beautiful Art

Uh

Shouldn't there be a 1/n on the RHS?

Semiclassical

(>_<)

yes.

Simply Beautiful Art

13:43

:-P

Semiclassical

too late to change it now, alas.

I am disappoint.

Simply Beautiful Art

$$\frac{1}{n}\sum_{k=1}^n f(2\pi k/n)-\frac{1}{2\pi}\int_{0}^{2\pi} f(x)\,dx=\frac1n\left[\frac{f(2\pi)-f(0)}{2}+\sum_{k=1}^{\lfloor p/2\rfloor} \frac{B_{2k}}{(2k)!}(f^{(2k-1)}(2\pi)-f^{(2k-1)}(0))+R\right]$$

Semiclassical

thx

Simply Beautiful Art

:P

Semiclassical

anyhoo. if $f(2\pi)\neq f(0)$, then the leading term at large $n$ on the RHS is the 1/n term.

so if $f$ fails to be periodic, then the error goes as 1/n.

if f is periodic but f' is not, then the error will be 1/n^2. and so forth.

Simply Beautiful Art

13:46

nifty

Semiclassical

now suppose that $f(x)$ is $2\pi$-periodic and analytic for all $x$. Then $f$ and all its derivatives will match at $x=2\pi$ versus $x=0$. Hence every term in the above sum would seem to vanish!

Simply Beautiful Art

Mhm

Semiclassical

Well, except for that $R$. And that's actually the escape clause here.

Simply Beautiful Art

:P

Semiclassical

The above reasoning is enough to conclude that R should decrease with $n$ faster than any $n^{-k}$, though.

Since if it didn't, then it should show up in the Euler-Maclaurin terms.

Simply Beautiful Art

13:48

Mhm... and that itself would be good enough for me

Lol, I'm trying to calculate $\pi$ by hand rn

Semiclassical

And this kind of thing can actually show up in the real world. Suppose you were doing a complete elliptic integral.

so, for instance, $\int_0^{\pi/2}\sqrt{1-m \sin^2 x}\,dx$. up to a factor of four that's just the integral from 0 to 2pi, and then the punchline is that the integrand is periodic and analytic.

Hence the error from an $n$-point trapezoidal approximation goes down faster than any $1/n^k$. In fact, the error goes like $a^{-n}$ for some $a>1$.

Simply Beautiful Art

Oh, that reminds me

Semiclassical

(the precise $a$ depends on which m you're doing.)

Simply Beautiful Art

Since the integrand is even, you can make it 1/2 times the integral from $-\pi/2$ to $+\pi/2$

Semiclassical

sure.

Simply Beautiful Art

13:52

And I heard applying trapezoidal rule to such an integral converges fairly fast

Semiclassical

Right. That's pretty much this.

Simply Beautiful Art

Oh xD

Semiclassical

It'll decrease geometrically with $n$ rather than algebraically.

The best way to analyze the error in that case is actually to think of $f(x)$ in terms of its Fourier series

Simply Beautiful Art

That's cool. thanks @Semiclassical

Semiclassical

the integral over [0,2\pi] then filters out all Fourier coefficients besides the zeroth one

whereas the filters out all Fourier coefficients except $a_0,a_n,a_{2n},$ etc

hence you end up only caring about $a_n$ to get the error.

and that'll fall pretty fast.

i came across this in a review article in one point and thought it was cool.

(and actually used it to motivate my own research, lol)

Simply Beautiful Art

13:57

Mmm, I should learn Fourier analysis soon

Semiclassical

yeah, it's good stuff

what I always find neat is the following: If you've got a $2\pi$-periodic function with Fourier series $\sum_{k=-\infty}^\infty c_k e^{i k \theta}$

then the change of variables $z=e^{i\theta}$ turns that into $\sum_{k=-\infty}^\infty c_k z^k$, which is just a Laurent series. So Fourier analysis on a finite interval is just a different presentation of Laurent series on the unit circle.

Simply Beautiful Art

How many places of $\pi$ should I attempt by hand?

Semiclassical

dunno. I've never really caught the $\pi$ bug.

Simply Beautiful Art

Lol, I'm just bored

Akiva Weinberger

3.141592…903

Gabriel Romon

14:02

@SimplyBeautifulArt math.stackexchange.com/questions/2426581/…

Simply Beautiful Art

@GabrielRomon Uh... Riemann sums?

Oh, no

:-/ no idea atm

Gabriel Romon

it checks out numerically

I don't think it's related to equidistribution

Akiva Weinberger

Oh hm yeah so you essentially want to prove that the integers mod pi are evenly distributed or something I guess

'cause you're saying that the average value of $|\sin k|$, $k\in\Bbb Z$ equals the average value of $|\sin k|$, $k\in[0,1]$

Simply Beautiful Art

Oh wait

I think that's a dupe

But brain is fuzzy on where I saw it

Akiva Weinberger

A duplicate question?

Simply Beautiful Art

14:07

Probably

Approach0 has no hits though

Semiclassical

Seems like one could use Euler-Maclaurin, appropriately enouhh

Simply Beautiful Art

lol

BAYMAX

Lets do a probability question!

There are two bags $A$ and $B$

Faust

Morning everyone

Gabriel Romon

@AkivaWeinberger you think $|\sin(1)|,\ldots, |\sin(n)|$ are evenly distributed in the intervals $[\left|\sin(\frac{i}{n})\right|[, \left|\sin(\frac{i+1}{n})\right|[$ ?

BAYMAX

14:13

Bag $A$ contains $n$ white and $2$ black balls and $B$ contains $2$ white and $n$ black balls , now one of te two bags is chosen at random and $2$ balls are drawn from it without replacement .If both the balls drawn are white and Probability that $A$ was used to draw is $\frac{6}{7}$ , then what should be the value of $n$ ?

Faust

i want bag B

BAYMAX

You have it!

Akiva Weinberger

@GabrielRomon No, I think $1\!\mod\pi$, $2\!\mod\pi$, $\dots$, $n\!\mod\pi$ end up evenly distributed in $[0,\pi]$

Dodsy

Helllooooo

Faust

Morning dodgy

Dodsy*

i guess Dodsy could be dodgey?

Gabriel Romon

14:24

@Faust you don't like discreet math ?

Faust

i like graph theory?

all the other discreet math i have done is counting something im not intrested in counting

thats boring

why cant people in discrete math count somthing useful or intresting?

Anonymous

Does anyone know a book which describes the conditions for differentiability of multivariable functions?

Gabriel Romon

Discreet math would put me to sleep as well

Faust

or at least tell us of tings u can count that are intresting

Gabriel Romon

But I don't mind discrete math

Faust

14:26

but grpah theory is intresting

Dodsy

@Faust lmfao

Faust

^^

Dodsy

my calc teacher today told us that he wants us to love graphing functions by friday

Faust

O.o

Dodsy

>_>

Faust

14:27

there easy to graph...

BAYMAX

Well then you must learn to like Desmos :)

Dodsy

no desmos allowed.

Gabriel Romon

This was a pun on discreet vs discrete but nevermind

Faust

whats a Desmos?

Dodsy

like graphing software

Faust

14:29

ah i dont use that

Gabriel Romon

there are lots of surprising techniques in combinatorics

Dodsy

you just do it by hand?

BAYMAX

oh

Faust

yeah i can draw pretty much any function by hand

multiple defined funtions of t is hard sometimes

like f(t) +g(t) +h(t) can be hard to draw

Gabriel Romon

Hello @MaryStar, how did you come up with math.stackexchange.com/questions/357030/… ?

Faust

14:30

it helps to be able to futz with cords to move around the idea whqat ur drawing

Dodsy

yeah he wants us to be able to see functions without even graphing them, i think.

Faust

but you will never be able to do diff geo if you cant draw basically all the elementry funtions and combonations

imo anyay

Dodsy

I believe the question was $\{x \in \mathbb{R} : \frac{x^2 + 3x}{x^2-1} \geqslant 1\}$

Faust

@Dodsy math.stackexchange.com/questions/339280/…

Dodsy

jesus haha

Faust

14:34

its missing a bunch fo the bictures i drew of the bifurcations in 3 dimensions :\

was a random project i did

well it as no roots in that range yes?

so u know its positive

as x gets big its acts like a straight line going up

with a slope of 3

Semiclassical

@Dodsy where I'd start on that is to note that x^2 shows up in both num and denom

Faust

@Semiclassical why do you use english so much better than i?

Semiclassical

So if I subtract one from both sides then the x^2 in the numerator goes away

shrug

Are you a native English speaker?

Dodsy

@Semiclassical he first noted that $x \neq \pm 1$

Faust

yeah...

look at my picture

Gabriel Romon

14:38

@Faust have you taken the GRE yet ?

Semiclassical

Hmm. Dunno what to tell you then @Faust

Faust

where u think i from china?

Semiclassical

@Dodsy sure, those aren't in the domain of the LHS

Faust

@GabrielRomon no i didnt even know what that was till a couple days ago im in canada

Semiclassical

Could have been somewhere in Europe

Gabriel Romon

14:39

@Faust there's a ton of fancy words to know for the verbal section

Faust

hmm i guess that make sense i have blood form all over europe

Semiclassical

I mean, Faust makes me think of Germany more than Canada :)

ACuriousMind

@Faust You could well be any Western European, so it would be a bit presumptous to suppose you are a native English speaker purely from your picture

Faust

yeah most Germans speak better english than me anyway

ACuriousMind

Also, yeah, Faust would indicate German affinity :P

Faust

14:41

@GabrielRomon even for a math student?

Dodsy

@Semiclassical and x cannot be equal to zero or when you divide x on both sides of the inequality you could be dividing by zero.

so there's two cases

Gabriel Romon

@Semiclassical how common would you say are the words "sycophant" and "mawkish" ?

Dodsy

one were x is greater than zero

and one when x is less than zero.

Semiclassical

@dodsy anyways, if you subtract 1 from both sides you get (1+3x)/(x^2-1) >= 0

Dodsy

oh that's interesting :P

Gabriel Romon

14:42

@Faust yeah, top programs will take that into consideration, though they might focus more on the quant part

Faust

i Graduated high school with 92% physics 88% biology 99% Chem 94% math and 52% in English

Semiclassical

Yep. For one, you can immediately read off that you have equality only when 1+3x=0

Faust

when i arrived at university i scored so poorly on the english placement test

it said i was supposed to take a ESL course

Dodsy

@Semiclassical sorry, back up just a tad, when did $x^2 + 3x$ become $1+3x$

Faust

but becuase my highschool was here in canada i couldnt register for it

Semiclassical

14:44

(x^2+3x)/(x^2-1)-1 = ?

Faust

oh oh can i guess?

Semiclassical

Lol

Faust

:)

Dodsy

that would be $1 + 3x$ :P

Faust

?

Semiclassical

14:45

Yep, in the numerator

Dodsy

over 1

Faust

=

Semiclassical

?

Dodsy

?

Now I'm confused

Semiclassical

Denominator in what I wrote above is not 1

Dodsy

14:47

$\frac{x^2+3x}{x^2-1)-1}=\frac{x^2+3x}{x^2-2}=\frac{1+3x}{-2}$ ?

is that it

Semiclassical

That is not at all what I wrote.

(A/B)-1 = ?

Faust

no dodsy

Semiclassical

I'm subtracting 1 from the entire fraction

Faust

whats $ \frac{1}{3}-1 \neq \frac{1}{2}$

Dodsy

I see.

okay so continue.

what's your next step

we have

(1+3x)/(x^2-1) >= 0

really it's your formatting that confused me there, semi.

Semiclassical

14:51

Right. The only way for that fraction to be zero is when x=-1/3

Dodsy

right, yes.

Faust

so i got it backwards

Semiclassical

So all we need worry about is > i.e. When is that fraction positive

And for that to be true, both numerator and denominator must have same sign

So, when is x^2-1 positive and when is it negative?

Faust

they should add a draw picture button into chat

Semiclassical

Here's where the cases your instructor gave are handy

Dodsy

14:54

Are we talking about when parts of the graph are in the first quadrant?

?

anyways the answer as a disjoint union is $(1,\infty)\cup(-1,\frac{-1}{3}]$

wow

$\mathbb{SEMICLASSICAL}$

Leaky Nun

15:13

Not only is that terrible in general, but you just KNOW Billy's going to open the root present first, and then everyone will have to wait while the heap is rebuilt.

Simply Beautiful Art

Say, does anyone happen to know if $\frac{d^k}{dx^k}\frac1{(x^N-1)^n}$ looks nice?

Leaky Nun

@SimplyBeautifulArt wolframalpha knows

Simply Beautiful Art

Does it?

Oh, it does

Wow

Leaky Nun

how does it look like?

Simply Beautiful Art

Horrible

Leaky Nun

15:21

rip

Simply Beautiful Art

:P

Trying to do a series thingy

Leaky Nun

@SimplyBeautifulArt do you mind disclosing the full question

Simply Beautiful Art

$$x=\sum_{n=1}^Nr^n$$

Trying to write $r$ as a series of $x$

Balarka Sen

15:44

@Blue No but I can tell you if you want.

Leaky Nun

16

Q: What is the proper way to ask a "find the domain" question?

A function is not really a function unless it's defined everywhere on its domain. So consider these three questions: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the square root function $f(x) = \sqrt{x}$. What is its domain? Let $f: [0, \infty) \rightarrow \mathbb{R}$ be the square root funct...

undergraduate-education functions

"Let $f:X \to \Bbb R$ be the square root function $f(x) = \sqrt{x}$, where $X \subseteq \Bbb R$. Find the union of all $X$ that makes the function well-defined." is obviously the proper way.

Anonymous

@BalarkaSen Could you come over the the other room?

Balarka Sen

Sure.

Transcript for

$Mathematics$ Mathematics