Mathematics - 2017-03-18 (page 3 of 4)

Learning user

19:00

actually now two equations

for first down cycle

x*(1+a/100) = x-441

for second down cycle

x*(1-a^2/10000) = 1944.81

Semiclassical

The change of 441 was after price had been raised/lowered once, wasnt it?

Learning user

then what are the two equations

Semiclassical

Right now you've only included the effect of the increase in price for computing x-441.

Learning user

change of 441 after the price is lowered

Semiclassical

Right. You need to include both the increase and the decrease.

Learning user

19:04

yes

correct

Semiclassical

So the first equation should have a factor of 1-a/100 as well.

Learning user

x*(1-a^2/10000) = x-441

Semiclassical

Right.

Learning user

then another equation

Semiclassical

And the second equation is after two such cycles.

So you multiply by both factors twice, not once.

Learning user

19:06

so much big process

which equals

1944.81

correct

Semiclassical

What's the whole equation?

Learning user

wait

x*(1-a^2/10000)*(1-a^2/10000) = 1944.81

correct

Semiclassical

right.

Aka x*(1-a^2/10000)^2=1944.81

Learning user

Now i have to substitute and find the value of x

i am not able to get it

right

Semiclassical

Yeah. In this case, though, you'll want to find a first.

No, that's not right.

Learning user

19:10

why

Semiclassical

That would work if the second equation had x twice not once.

Learning user

i am not able to get it

Semiclassical

lets do one thing to make the algebra less horrid.

Learning user

no, you are right

Semiclassical

Define t=1-a^2/10000.

Learning user

19:12

sorry

thank you so Much

@Semiclassical

for your patience

Semiclassical

Then the first equation is xt=x-441 and the second is xt^2=1944.81

Np.

Learning user

if possible another question

Semiclassical

a short one, perhaps

Learning user

0

Q: Mixture and Alligation Problems::

A 10 litre container holds a mixture of water and sugar,the volume of sugar being 15% of total volume.A few litres of the mixture is released and an equal amount of water is added.Then the same amount of the mixture as before is released and replaced with water for a second time.As a r...

algebra-precalculus

try with this one please

Semiclassical

That's not short, so I'll have to pass

Learning user

19:17

ok

my way of doing right

@Semiclassical please upvote my two questions , if you like it

Don't disturb

Prove that $$3\zeta(2)\zeta(5)+\frac{3}{4}\zeta(3)\zeta(4)-6 \zeta(7)>0$$

Anybody already done with it?

Learning user

@Semiclassical, I like your patience and attitude, you deserve it, thank you so much I have learnt lot from you

ALannister

0

Q: Product of Principal Ideals is Principal Ideal of the Product $(x)(y) = (xy)$ when $R$ is commutative, but not necessarily unital

When $R$ is a ring (not necessarily commutative, and not necessarily with unity), I have a result that tells me that for $x \in R$, the ideal generated by $x$, $(x) $, is $= I_{x} = \langle RxR \rangle + Rx + xR + \mathbb{Z}x$. If I define $(y)$ in a similar way, namely that $(y) = I_{y} = \lan...

abstract-algebra ring-theory ideals

@arctictern we talked about this a little last night, but now I'm using a different definition of principal ideals.

This one works for rngs

I actually had to look up "alligation" in order to find out it has nothing whatsoever to do with alligators, @Learninguser

SoumyoB

19:37

is anyone here who has done a PhD or is doing one?

in math?

Learning user

@ALannister you know to find alligation problem

I have liitle strucked

Meow Mix

@SoumyoB I want to do one if that counts... /s

Hi @Dami

Semiclassical

Helping with algebra problems: easy

Grading these lab reports: ughhhhhhhh

SoumyoB

@Semiclassical so you're in PhD then?

Semiclassical

Physics, but yes

SoumyoB

19:41

since grading must mean TA-ship

how much of a load is it, academically?

as in, how many hours go into academics-be it studies or grading papers?

Semiclassical

the first year or two it's tough.

SoumyoB

shit

Liad

i need to prove that if $f : R \ ^ n \to R \ ^ n $ is $K \gt 0$ lipshitz function then $f$ maps zero measure sets to zero measure sets.
someone can help with that?

Semiclassical

After that you typically are doing research rather than taking classes

SoumyoB

I'm moving to the US, into a university that's offering me a fully funded PhD with stipend and from what I've heard it's hell on earth

instead of being happy about it I'm just horrified about it

Semiclassical

19:44

The thing you probably want to think in terms of is what you intend to do with a PhD

SoumyoB

I just want to discover something new, that's all

Semiclassical

If your ambition is academic research, there's really no way around it.

SoumyoB

it's.... fun

not really, I eventually want to get into research and development in industry

Semiclassical

Hmm.

ALannister

@SoumyoB I'm doing it right now. Semi's right the first year or two is tough. It's been especially tough for me b/c I also have a disability.

Semiclassical

19:47

Can't advise you re: industry. I just dunno

ALannister

But, make connections with your professors and the important administrators. They can help you out of a bind when you need it.

SoumyoB

the thing is, I thoroughly believe I'm highly undeserving of the offer letter I've received

Semiclassical

I think it helps to always have in mind what you really intend to do with a PhD.

SoumyoB

as in, if I convert my university's GPA into the typical US GPA it's just 2.72-ish

ALannister

@SourmyoB everybody feels that way. It's called "Imposter Syndrome". And you know, your university is probably way tougher than the ones in the states.

As long as you're not from Greece.

SoumyoB

19:49

I just have a good summer project work and made good impressions on my professors-they think I can do good research but they well know academics-taking classes and performing in exams-is not my strong suit

ALannister

Impostor syndrome

Impostor syndrome (also known as impostor phenomenon or fraud syndrome) is a concept describing high-achieving individuals who are marked by an inability to internalize their accomplishments and a persistent fear of being exposed as a "fraud". The term was coined in 1978 by clinical psychologists Pauline R. Clance and Suzanne A. Imes. Despite external evidence of their competence, those exhibiting the syndrome remain convinced that they are frauds and do not deserve the success they have achieved. Proof of success is dismissed as luck, timing, or as a result of deceiving others into thinking they...

Well, potential as a researcher is the most important thing they look for.

SoumyoB

@ALannister but my GPA is quite a good indicator of my performance in academics

ALannister

Not necessarily. Like I said, your school could be a lot tougher academically than most schools in the US. Some schools give out A's like candy.

SoumyoB

2.72

ALannister

Which is a B?

B+?

SoumyoB

19:51

even if you accounted for the toughness in my school, I don't think it can come any close to a respectable US GPA

ALannister

If you tell me you go to that school in Switzerland, ETF or whatever it's called, you're probably a frigging genius.

SoumyoB

well we're marked out of 8, and my GPA is 5.3, which is a C

my university is Indian Institute of Science

ALannister

What country are you from?

I had a friend from the US who went to University of Madras. She said it kicked her ass.

If you're worried, why don't you find out what courses you are going to be expected to take your first year, get your hands on some books, and spend the summer preparing?

SoumyoB

8 is S, <=7 is A, <=6 is B, and so on up to <=4, which is a D, and then F

you're right about that, and I plan on doing that

ALannister

I think that is a very good strategy.

SoumyoB

19:54

but I still don't think it'll be good enough and I'll have to improve my time management capabilities by leaps and bounds

ALannister

<-- time management not my strong suit either. For me, it's part of my disability. I have a neurological disorder that affects that.

SoumyoB

I see

ALannister

But, as of now I've got a 4.0, which is like an 8 for you guys.

SoumyoB

congratulations dude

ALannister

I've needed to work really hard and hustle a lot ;)

SoumyoB

19:56

it's not exactly time management

ALannister

The hustling is. I'm constantly needing to ask for more time to do things.

SoumyoB

you spelled it for me-what I can't do is hard work on end for more than 3 consecutive days

my fuel runs out real fast

ALannister

So, you need to take a break in between.

And make sure you eat right.

SoumyoB

do they give breaks?

ALannister

Gosh, I sound like such a mom.

No, I mean, just take an hour out of your studying every couple of days to relax.

SoumyoB

19:57

an hour of rest for every couple of days?

4

!!

you're kidding me right?

ALannister

No, you get more than that.

What the heck do you think graduate programs are like, prison?

SoumyoB

I imagine so

ALannister

I just mean make time for yourself every now and then so you don't burn out.

Well, they're not.

SoumyoB

I mean they're paying me a salary which is four times that of my dad, obviously they'll expect just as much of an output from me

ALannister

Not really. You'd be surprised.

SoumyoB

19:59

I really hope so

ALannister

I'll tell you what. In addition to going tuition free, my university pays me about $25,000 a year.

That's barely minimum wage in the US.

I imagine they're probably giving you about the same. I've never known someone who's become a millionaire as a TA.

SoumyoB

well I'm an immigrant there so minimum wage laws don't apply to me

of course

ALannister

Yeah they do!

You have rights as a non-resident alien.

SoumyoB

really? NOW that's surprising

ALannister

But you're not allowed to work other than what your visa is for.

You can look up all this stuff online.

SoumyoB

20:01

I see

ALannister

And some schools have graduate student unions.

SoumyoB

thanks for all the optimism and information buddy

I gotta go now, I'll catch you later

ALannister

uscis.gov/working-united-states/students-and-exchange-visitors/…

Sure. Look at that link though. it's helpful!

Hey @arctictern

arctic tern

hey

Liad

i need to prove that if f:R n→R nf:R n→R n is K>0K>0 lipshitz function then ff maps zero measure sets to zero measure sets.
someone can help with that?

ALannister

20:06

@arctictern I'm looking at your answer right now.

I'm going to try multiplying $(x)$ as you wrote it out with $(y)$ and see if that helps.

Meow Mix

If linear map $A: \Bbb R^n \to \Bbb R^n$ has rank $k$, the dimension of the subspace of vectors satisfying $A\mathbf{x} = 0$ will be $n-k$, right?

arctic tern

@ALannister you mean $(x)=Rx+\Bbb Zx$?

@MeowMix yes

ALannister

Is that what it reduces to if it's commutative?

What is $\langle RxR \rangle$?

arctic tern

$\langle RxR\rangle$ is the additive subgroup generated by the subset $RxR=\{rxs: r,s\in R\}$

ALannister

I know that, I mean how do I write it out so I can multiply it with stuff?

arctic tern

20:10

Which is comprised of sums of elements of the form $rxs$

ALannister

oh, is that what the middle sum you wrote in the answer is?

And then what is $r_{0}x + xs_{0}$ there?

arctic tern

About the "is that what it reduces to if it's commutative" - (a) You should be able to determine that yourself in a matter of moments. Not trying to be judgmental, telling you not to fear thinking about it out of fear of wasting mental energy. (b) Read my answer.

ALannister

@arctictern just did. Found the answer to what I asked.

Going to digest it a bit and work things out. Thanks, man.

arctic tern

@ALannister what do you mean what is $r_0x+xs_0$? it is what it is,

ALannister

I just meant which terms in what I originally wrote did it come from?

arctic tern

20:12

$Rx+xR$

ALannister

Thank you. Thought so.

arctic tern

If $(x)$ is an ideal of a non-unital not-necessarily-commutative ring $R$ that contains $x$, then it must contain things that look like $rx$, must contain things that look like $xs$, and must contain things that look like $rxs$, and must contain things that look like $nx$ when $n\in\Bbb Z$, and must contain any sum of those things. then you can just check the sums of those things is closed under addition and ambient (left or right) multiplication.

ALannister

Thanks!

You're a good person. And you get me. Thank you.

arctic tern

You're welcome.

Daminark

I just realized that scalar multiplication of vectors is a group action

There's more to it ofc but damn this just came to mind now and I'm far more surprised by this than I ought be

arctic tern

20:26

nonzero scalars

and the orbit space is projective space

Daminark

Is projective space the set of one-dimensional subspaces of a space?

arctic tern

yes (although you can make it a manifold, not just a set)

you can also imbue it with incidence relations (projective geometry) and layer it to include k-subspaces for each k

it also gets an induced action from GL, which factors through PGL

so a point in projective space is a line in affince space, a line in projective space is a plane in affine space, etc.

weird things happen with $k^4$, you get the klein quadric in $\Bbb P(\Lambda^2k^4)\cong\Bbb P(k^6)$

Daminark

Wow, that's cool

What's the $\Lambda$ here referring to?

arctic tern

exterior power

so like $\Lambda^2V$ is sums of things that look like $v\wedge w$, where $\wedge$ is has the distributive property and is antisymmetric, so $(u+v)\wedge w=u\wedge w+v\wedge w$ and $u\wedge v=-v\wedge u$

Daminark

Oh is it that construction which differential forms come out of?

arctic tern

20:39

over $\Bbb R$, it's like a linearization of the set of all planes. indeed, the collection of all $u\wedge v$ (with $u,v$ orthonormal) inside of $\Lambda^2\Bbb R^n$ is diffeomorphic to the manifold of all oriented 2D subspaces

@Daminark yes

Daminark

Huh, we had only ever done wedge things with forms in the plane

Our professor wanted to focus on stuff like homotopy of curves rather than the exterior algebra

arctic tern

not sure how related those two things are

Liad

if the boxes $\{P_i \}$ covers a measure zero set ,$A$ , and $\sum m(P_i) < \epsilon $ and $f$ is $k>0$ lipschitz, how can i cover$f(A)$ with boxes that their vol will be less than $\epsilon$ ?

Daminark

I mean, he said that to do things in $\mathbb{R}^n$, we'd need a full exterior algebra formulation. By keeping things to the plane, he could keep to what was necessary there, and then focus on talking about stuff like homotopy and winding number

alan2here

Given an infinite list of integers L, from a given position P, add to a finite number of subsequent values by given amounts, e.g the next 5 values {1, 0, 0, 5, -2}, and move the position P alone by some amount, e.g -3.

As in, start with the list L, and transform L by applying the additions, then moving P.

such transformations are not assoceative or commutative, and is not a group

sorry, I missed a point, the transform repeats the process until P = 0, and can transform the positions before and at P as well as after.

programs formed of nested sequences of thease transformations of L are in effect the entirety of an important esoteric programming language goofely named BrainFuck, and in my opinion it's amazing that this process is turing complete, meaning that it can perform any calculation

but, these transformations are not assoceative, commutative or distrubative

Joe Stavitsky

21:15

How can I show values of the sides of a right triangle with hyp = 1 without appealing to trigonometry? It seems like you have to learn trig to prove trig.

arctic tern

what does it mean to show a value?

what is it exactly that you're trying to prove?

Joe Stavitsky

the known values of sin and cos for the unit circle

Meow Mix

What, given an angle?

Joe Stavitsky

as in, sin (pi/6), cos (pi/6), sin (pi/4), cos (pi/4)... etc

yes

arctic tern

you can do those with geometry

Meow Mix

21:21

You can use simple geometry to get some of those values

Here's pi/6

We reflect the right triangle to get an equilateral triangle

And use simple geometry to find the side lengths of the triangle i.e. sine, cosine

Joe Stavitsky

I mean, pi/4 and pi/2 are obvious, I agree

and pi and 2pi

Meow Mix

Refer to pi/6

Joe Stavitsky

@MeowMix, so you are saying just graphically?

Balarka Sen

how do you do pi/5

Joe Stavitsky

@MeowMix o it is equilateral n/m that is rather cool actually

Random Variable

21:26

@DanielFischer Do you know if unbounded entire functions exist such that $\lim_{R \to \infty} f(Re^{i\theta}) = 0$ for every $\theta$ between $0$ and $2 \pi$ inclusive?

Balarka Sen

@RandomVariable Funny question.

Forever Mozart

complex numbers still freak me out. I avoid working with them

alan2here

complex numbers are the states found from the sucessor function and 0.

successor(0) we tend to call 1.

successor(1) we tend to call 2.

etc...

the sucessor gives us new values from existing values, always the case, always new values from previously unseen values.

Forever Mozart

those are the natural numbers

alan2here

yes

your with me so far, with sucessing, meaning new values, that are always the same given the same value, and naming them?

Random Variable

21:38

@BalarkaSen Why is it a funny question?

Balarka Sen

@RandomV I mean, first guess is no but I don't know a proof.

I'd be very surprised if there exists such a thing

Daniel Fischer

@RandomVariable I Don't know, but I think such beasts exist.

Forever Mozart

complex number is pair of real numbers

alan2here

I was going to go to on all the way to complex numbers the explaine why thats the the furtherest you can go.

Simple

in topology, what does $\doteq$ stand for

Forever Mozart

21:39

$\langle a,b\rangle$

i forget the property they have that the reals do not have

a field under multiplication?

no

Balarka Sen

@ForeverMozart fundamental reason why C has more structure than R^2 is because you're choosing a natural automorphism J : R^2 --> R^2 such that J^2 = -I

C is also a field, yeah.

Alessandro Codenotti

@ForeverMozart it's an algebraically closed field

Forever Mozart

usually when I read a paper with complex numbers, they are only being used to describe a rather intuitive topological object

cause I only read topology papers:)

alan2here

@ForeverMozart sort of, and yes, but your with me so far when it comes to sucessing?

Meow Mix

hi

Forever Mozart

21:45

@alan2here yes

Alessandro Codenotti

Hi Zach

Is $\pi_1(\Bbb R^2\setminus\Bbb Q^2)$ free? It sure looks like it @Balarka

Eric

@RandomVariable such entire functions do exist, I recall this being mentioned in lindelof's book calcul des residus

Can't find a reference atm as I'm on my phone in the airport but I believe that's where I saw it anyway.

Balarka Sen

@Alessandro That's a good question. I think it should be, but I don't know a proof.

I am worried about stuff like what happens in Hawaiian earring, whose fundamental group is not free

Random Variable

@Eric Thanks.

Balarka Sen

I think this chap has an earring subgroup

Alessandro Codenotti

21:51

@BalarkaSen It doesn't have a free group with countably many generators as fundamental group? This earring gets weirder every day

Balarka Sen

No.

wildtopology.wordpress.com/2013/11/23/the-hawaiian-earring

The point is you can compose infinitely-long loops and cancel from infinity

Meow Mix

what are open sets in the hawaiian earring

sets that are open in $\Bbb R^2$

alan2here

@ForeverMozart functions, or more generally, relations, such as the sucessor have inverses

Balarka Sen

open sets of the earring are exactly open sets of R^2 intersection earring

Ali Caglayan

@MeowMix no subsets of the hawaiian earing are open in R^2 with usual topology

Alessandro Codenotti

21:53

I think there are earrings in $\Bbb R^2-\Bbb Q^2$, you only need to find a point $(a,b)$ and a sequence of points $(x_i,y_i)$ with $\{x_i,y_i\}$ linearly independent over $\Bbb Q$, such that $(x_i,y_i)\to(a,b)$ right? Because then the circle of center $(x_i,y_i)$ through $(a,b)$ won't have rational points for any $i$

Random Variable

@Eric Is there a version of the textbook translated into English?

Balarka Sen

@Alessandro Ya

Meow Mix

@AliCaglayan what?

Ali Caglayan

but yeah as Balarka says intersect them with open sets of R^2 and you're good

Balarka Sen

You just pick an accumulating sequence of rational lattice points and take it's complement in R^2

(that deformation retracts to the earring)

Meow Mix

21:55

i'm confused, ill just stick with my ted-cercises

Eric

Uhh not sure if there is one @RandomVariable I just read the original language

Alessandro Codenotti

Hmm, interesting, I'm reading that blog post you linked now, thanks for the reference

alan2here

@ForeverMozart the inverse of the successor, which is in effect +1 is -1. success(0) = 1, success(x) = 0, we tend to call 'x' -1

Forever Mozart

The hawaiian earring $H$ does not contain an open subset of $\mathbb R ^2$. it is a subspace of $\mathbb R ^2$, which means you intersect open sets in $\mathbb R ^2$ with $H$ to get the open sets in $H$.

you intersect a disc with $H$ and you get some intervals

Balarka Sen

or some hawaiian earring if you intersect stuff near the bad point :P

alan2here

21:57

@ForeverMozart now, if you repeat addition X times you get multiplication

Meow Mix

But, the hawaiian earring contains a bunch of points in $\Bbb R^2$?

Balarka Sen

It does.

Meow Mix

OH, wait.

Nevermind...

I realized some open path in $\Bbb R^2$ isn't open...

alan2here

@ForeverMozart and it's inverse division yeild some new values, which we must name and gery naturally become the ratonal numbers

Forever Mozart

the topology on $H$ must be a collection of subsets of $H$

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$Mathematics$ Mathematics