Mathematics - 2017-03-29 (page 3 of 3)

Akiva Weinberger

20:00

Bring one of them halfway around one of the loops that the arc is made of.

Owatch

Can someone point out to me why: $- \frac{e^{-\frac{\pi}{2}} \cdot e^{14 \pi * i * t}}{2} == \frac{e^{-\frac{\pi}{2}} \cdot e^{14 \pi * i * t}}{2}$

Akiva Weinberger

I'd probably need to draw this to explain it better, but you end up with two loops on the same vertical plane that are each looped around different sections of the arc

Balarka Sen

(In fact $\epsilon$-nbhd of that is easily seen to be a manifold because it's itself a manifold with two isolated non-tame point; so obv boundary is also a manifold.)

Akiva Weinberger

You then combine them into one loop

and then bring that big loop around one of the ends of the arc.

Balarka Sen

I see it.

Akiva Weinberger

20:02

Arright

Balarka Sen

Why does that say if each of those loops are individually boundaries?

I don't get it

You're just giving a boundary for $a + (-a)$

Alessandro Codenotti

I talked with my TA, who talked with a couple of professors, about $\Bbb R^2\setminus \Bbb Q^2$ with or without the origin and they all agree that they should be homeomorphic, but we didn't come up with a proof

I think I'll ask about it on MSE

Akiva Weinberger

@BalarkaSen I'm starting with $0=a+(-a)$ and ending up with a nontrivial-looking loop.

Balarka Sen

dude why do you like to think about weird stuff

Alessandro Codenotti

Because it's interesting

Akiva Weinberger

20:06

So, you can do that backwards, and show that the nontrivial-looking loop does bound.

Balarka Sen

@Alessandro "You're such a wonderful person / But you've got problems"

Alessandro Codenotti

Let me guess, is that Bowie?

Balarka Sen

Ya

Mike Miller

took me a sec to get the tune

BUT YOU'VE GOT PROBLEMS

what an album

Balarka Sen

Low is pretty awesome

like it the most out of the Berlin trio

Mike Miller

20:13

agreed, though i have a soft spot for lodger

Balarka Sen

Lodger is my 2nd

got african night flight stuck on my head

Mike Miller

can't possibly write out the lyrics even if i can hear them right now

Balarka Sen

haha

Sha Vuklia

Consider $\begin{align}f(x,y)=\frac{x^3-xy^2}{x^2+y^2}\end{align}$ and $f(0,0)=0$. My book says that $D_1f(\vec 0)=1$, but I would think it it 0? Didn’t they just differentiate $f$ with respect to $x$ only? And then plug in $x=0,y=0$. That can’t give 1?

Adeek

what does it mean for cardanility to be locally constant ?

Is it is constant on each path component ?

Sha Vuklia

20:27

oh never mind

I forgot I could easily use the definition of the partial derivative

could=should

skull petrol

should = can

Sha Vuklia

no, right? I can't do anything else

because the rules for $f$ are different at 0

so I can't just do $\begin{aligned}\frac{\partial f}{\partial x} \end{aligned}$, in the sense that I differentiate $\begin{aligned}\frac{x^3-xy^2}{x^2+y^2}\end{aligned}$ with respect to $x$, and leave $y$ constant

Akiva Weinberger

Right

LeGrandDODOM

@Akiva are you familiar with complex logarithm ?

Akiva Weinberger

Yeah, what about it?

@ShaVuklia Also, notice that $f(x,0)=x$

LeGrandDODOM

20:41

@AkivaWeinberger is it true that $Log(z^n)=nLog(z)$ when $z$ is complex and $n$ an integer ?

Akiva Weinberger

$\rm Log$ is the principal branch, right?

I think $z=-1$ gives a counterexample.

${\rm Log}((-1)^n)={\rm Log}(-1)=i\pi$, but $n{\rm Log}(-1)=n\cdot i\pi$.

LeGrandDODOM

ok thanks

A---B

Is there a way to find roots of $f(x) = x^3 - x^2 + 1$ ?

I can prove that only roots are between $[-1,0]$

and there is only one real root.

Akiva Weinberger

@A---B Do you know of Cardano's formula? Aka the cubic formula

It's quite complicated.

A---B

@AkivaWeinberger I searched on wikipedia before asking here and there were some formulae but I don't know much.

I need to find a good approximation not exact solution.

Balarka Sen

20:54

@A---B You can prove that the only roots lie in that interval by seeing there are no other intervals where it changes sign, I suppose

Akiva Weinberger

Oh, approximations are easier

A---B

@BalarkaSen Yes. I used IVT.

Akiva Weinberger

Wolfram Alpha knows the exact form (click on "exact form" where it says "real solution")

@A---B

A---B

@AkivaWeinberger Yes I know wolfram alpha but I want a way that I can do by hand.

Balarka Sen

Nevermind, your question is to find the roots approximately.

Akiva Weinberger

20:56

@A---B So, like, you want the root to a few decimal places?

A---B

This was in chemistry book and I am not allowed to use external softwares.

Akiva Weinberger

Newton's method might be applicable

A---B

@AkivaWeinberger Three will be fine.

Akiva Weinberger

@A---B Are you sure you have the right polynomial, by the way?

A---B

@AkivaWeinberger Yes I am sure because this was in "Additional Problems" section where we get such type of problems.

And this was a problem to find $K_c$ for a ion in solution. So it was easy to find the polynomial.

Akiva Weinberger

20:59

Look up Newton's method (if you know derivatives), then. You'd need a calculator

LeGrandDODOM

@AkivaWeinberger please have a look at my question math.stackexchange.com/questions/2209302/…

skull petrol

Newton's method

== English == === Etymology === Named after Isaac Newton. === Proper noun === Newton's method (mathematics) A method for finding successively better approximations to the roots (or zeroes) of a real-valued function. === Synonyms === Newton-Raphson method...

Akiva Weinberger

I don't know if I'll have time (I need to leave in like a minute), but maybe later @LeGrandDODOM

A---B

@AkivaWeinberger That sounds fair since I have Newton's method in this semester. Do you know any other method that does not require calculators ? Anyway I am fine with Newton's method. Thank you very much.

Akiva Weinberger

@A---B I don't think you can do this without a calculator. (Or perhaps a log table, I suppose)

Balarka Sen

21:02

There are a bunch of other increasingly complicated methods for approximating roots of a polynomial.

Akiva Weinberger

@LeGrandDODOM Looks interesting. I'd have to think

A---B

@AkivaWeinberger I guess I have to use Log table anyway.

Felix.C

Hey guys some questions about complex analysis:
1.) Wikipedia says: "In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point." $1/{1-z}$ is not holomorphic right? Still we can find an Taylorexpansion for $\vert z \vert < 1$ with the geometric series. This seems strange to me -.-
2.) Also I don't quite understand the difference between the Laurent and the Taylor series. Or what the Laurent series is used for. Is it true that we use the Laurent series to approximate a func…

A---B

@skullpetrol How did you do that ?

skull petrol

"define:"

It works in all rooms.

A---B

21:07

me

Pronoun: me (first-person singular pronoun, referring to the speaker)

As the direct object of a verb.
Can you hear me?
(obsolete) Myself; as a reflexive direct object of a verb.
1819, John Keats, La Belle Dame sans Merci,
And I awoke, and found me here.

(52 more not shown…)

Adjective: me m (feminine mee)

insufficient, scanty, not full

Noun: me

mother
me
male
husband
me

(3 more not shown…)

Conjunction: me

and
me
and

Interjection: me

baa (representing the bleating sound sheep make)

Wow, nice.

skull petrol

No need for the quotation marks.

Balarka Sen

"me
baa (representing the bleating sound sheep make)"

Sounds like me indeed

skull petrol

Interjection

interjection

Noun: interjection (plural interjections)

(grammar) An exclamation or filled pause; a word or phrase with no particular grammatical relation to a sentence, often an expression of emotion.
Some evidence confirming our suspicions that topicalised and dislocated constituents occupy different sentence positions comes from Greenberg (1984). He notes that in colloquial speech the interjection man can occur after dislocated constituents, but not after topicalised constituents: cf.
(21) (a) Bill, man, I really hate him (dislocated NP)(21) (b) ✽Bill, man, I really hate (topicalised NP)
An interruption; something interjected
interjection f (plural interjections)

(3 more not shown…)

Balarka Sen

sheepish interjections are the worst

skull petrol

Indeed.

A---B

21:12

I like sheeps.

skull petrol

sheeps

Noun: sheeps

(nonstandard or obsolete) plural of sheep
sheeps
(Unquachog) a sheep

define: unquachog

Balarka Sen

oops lol

A---B

Seems like we don't need English SE now.

skull petrol

:-/

Unquachog: the Eastern Algonquian language of the Unquachog.

A---B

define : Algonquian

Owatch

21:24

Is anyone familiar with signal processing here?

A---B

There is a SE for it, no ?

Owatch

I guess, but it doesn't seem very active.

A---B

It is a promoted site, so I think you will get your question answered.

skull petrol

Algonquian

Adjective: Algonquian (not comparable)

Relating to a group of North American languages.

Noun: Algonquian (plural Algonquians)

An Algonquin.

Owatch

Alright, I suppose I'll post it there as a question then.

skull petrol

21:35

You put a space between define and : @A---B

Meow Mix

Hi

skull petrol

21:51

Lo

Meow Mix

Med?

Will Njundong

Find the points on the parabola $y=x^2$ that are closest to the point (0,44)

Forever Mozart

Lo dHi - Hi dLo

Will Njundong

how?

Forever Mozart

what is the distance btwn 2 points in the plane $(x,y)$ and $(0,44)$?

$\sqrt{x^2 +(y-44)^2}$ right?

$\sqrt{ x^2-(x^2-44)^2}$

Will Njundong

21:55

yup

Forever Mozart

now minimize that

skull petrol

First draw a sketch.

Forever Mozart

$\frac{1}{2\sqrt{x^2+(x^2-44)^2}} \cdot (2x+2(x^2-44)(2x))$

$x+2x^3-88x=0$

$x(2x^2-87)=0$

3 solutions

Evinda

Hello

I have a question

Will Njundong

ok so how do i determine which values takes the place $x2, x1, y2, y1$?

Forever Mozart

22:09

somehow $(0,0)$ cannot work

so I think it's the other two points

$x=\pm\sqrt{87/2}$

Will Njundong

i made an error posting the original question: it shoudl have been (0, 4) not (0, 44)

but thats not really an issue

Forever Mozart

ok then $x+2x^2-8x=0$

$x(2x^2-8)=0$

$x=0,\pm 2$

Will Njundong

so basically how this works is, i use distance formula then the derivative of its result

Forever Mozart

but again I think 0 cannot work

you will have to decide why

Evinda

We have that $d_p \equiv d \pmod{p-1}$ and $d_q \equiv d \pmod{q-1}$.

Also $m_p \equiv c^{d_p} \pmod{p}$ and $m_q \equiv c^{d_q} \pmod{q}$.

I found that $d=(d_p+d_q) \pmod{\phi(n)}$.

We have that $m=c^d \pmod{n}$, so $m= c^{d_p} c^{d_q} \pmod{n}$.

But how do we relate the latter to $m_1$ and $m_2$ ?

Forever Mozart

22:12

hint: I think $(0,0)$ has MAX distance from $(0,4)$

also my calculation above is wrong

Will Njundong

i think zero should work because $f(x) = x^2$ when f(0) = 0 is as close as can get to (0, 4)

Forever Mozart

but you get the idea

check the sign of the derivative and you will see that there is not a minimum there

$x=0$ will be a local max

Evinda

Hey @BalarkaSen
Do you maybe have an idea?

n=pq, where p,q primes

Will Njundong

hmm i see

CausingUnderflowsEverywhere

What does multiplying a polynomial's coefficients by the attached exponents of the term have to do with getting the derivative?

Simply Beautiful Art

22:22

@MeowMix Say, any progress on the longest sequence you could write problem I gave you?

CausingUnderflowsEverywhere

Hey Neev why are you imitating my profile picture? Trying to impersonate me?

Simply Beautiful Art

lol

How do you get those images made? I'd like to generate a list of SE starter profile pictures and choose one that I like

Forever Mozart

alex jones says the government is poisoning me

Neev Parikh

Ahahahaha @CausingUnderflowsEverywhere mate, I didn't realize I had a profile picture xD

CausingUnderflowsEverywhere

thats what they all say Neev

Simply Beautiful Art

22:32

*dun dun dun*

Neev Parikh

any IB math HL people?

Semiclassical

Hi chat

Lazy question of the day: math.stackexchange.com/q/2209396/137524

Akiva Weinberger

@CausingUnderflowsEverywhere You mean, why does $(x^n)'=nx^{n-1}$?

It essentially comes from the first two terms of the binomial formula

(Specifically, $(x+h)^n=x^n+h\cdot nx^{n-1}+\dotsb$)

Simply Beautiful Art

@AkivaWeinberger What does \dotsb stand for?

Akiva Weinberger

More or less the same as \cdots, I think

Dots binary (as in binary operator, like $+$), I think.

Not sure.

And that's why he had to die

Simply Beautiful Art

22:42

#leavenotrace

xD

Daminark

@Neev Was

Neev Parikh

@Daminark How was it?

@Daminark what was your option?

Daminark

Calc, and meh

Neev Parikh

Are you in uni now?

Simply Beautiful Art

23:05

@NeevParikh No, I am not

Meow Mix

23:21

Hi

Daminark

23:35

@Neev Yeah I am

I liked IB at the time but I've realized the math level, in terms of content, is iffy

Akiva Weinberger

What's IB?

Meow Mix

Hi @Akiva, @Dami

CausingUnderflowsEverywhere

Hi meowmix

I've

Meow Mix

I've?

Akiva Weinberger

You've what?

CausingUnderflowsEverywhere

23:46

In what language does adding an a to end of the word make it plural

Akiva Weinberger

Don't know

MickLH

@CausingUnderflowsEverywhere the set of languages that are correct responses to this question

Meow Mix

@Akiva I always wanted to make my own constructed language

Akiva Weinberger

/r/conlangs

Meow Mix

I know.

I'm just saying

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$Mathematics$ Mathematics