Mathematics - 2016-11-01 (page 2 of 3)

user4612744

11:03

@BalarkaSen could you extend upon your idea?

Ali Caglayan

11:29

@user4612744 what is cot?

user4612744

cos x/sin x

Ali Caglayan

So what is cot^2?

user4612744

Simply (cos x/sin x)^2 ?

Ali Caglayan

yup so what is 1 + (cos x / sin x)^2?

user4612744

hmm

Ali Caglayan

11:35

Lets simplify this to a problem you should know

user4612744

I'm sorry, my trig is terrible.. a little brain-freeze right now

Ali Caglayan

What is 1 + a / b

How would you write that as a single fraction?

user4612744

a + b/b ?

Ali Caglayan

yup

So what would a and b be in this context

user4612744

cos^2 x and sin^2 x

Ali Caglayan

11:37

Yes

So What would 1 + (cos x / sin x)^2 be

user4612744

(cos^2 x + sin^2 x)/(sin^2 x)

Ali Caglayan

You know something about sin^2 x + cos^2 x

what does it equal?

user4612744

1!

Ali Caglayan

So then 1/ that would be

user4612744

ah so we have 1/sin^2 x

Ali Caglayan

11:40

yes but we still haven't finished

remember that is the bottom part of the fraction

user4612744

indeed

Ali Caglayan

so what do you get

user4612744

1 multiplied by sin^2 x

Ali Caglayan

Is that equal to the right hand side of the equation you are trying to prove?

user4612744

so basically sin^2 x/ 1 on the left side

indeed :D

Ali Caglayan

11:41

Ta daa

That's all there is to it

user4612744

you are a brilliant tutor, thanks for helping out

Ali Caglayan

No problem, any time

robjohn

@Sophie have a look at this answer.

Sophie

12:13

damn this thing is amazing

robjohn

12:31

@Sophie You can also show that $\sum\limits_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}=\pi\csc(\pi z)$

Akiva Weinberger

@Sophie If you look at the graph of cot(x) and the graphs of 1/(x-nπ) for a bunch of n it makes a bit of sense

Ramanujan

0

Q: Area of triangle formed from equation of of its sides.

The area of triangle formed by the lines $$a_1x + b_1y + c_1z=0$$ $$a_2x + b_2y + c_2z=0$$ $$a_3x + b_3y + c_3z=0$$ Is $$\frac {\Delta^2}{2\lambda_1 \lambda_2 \lambda_3}$$ Where $$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & a_2 & c_2 \\ a_3 & b_3 & c_3...

@DHMO help (^_-)

Ramanujan

12:55

@robjohn hi

I need help in straight line

Sophie

do you know the shoelace formula?

Shoelace formula

The shoelace formula or shoelace algorithm (also known as Gauss's area formula and the surveyor's formula) is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. The user cross-multiplies corresponding coordinates to find the area encompassing the polygon, and subtracts it from the surrounding polygon to find the area of the polygon within. It is called the shoelace formula because of the constant cross-multiplying for the coordinates making up the polygon, like tying shoelaces. It is also sometimes called...

Ramanujan

Ok,t hen?

DHMO

13:23

@Kaumudi You don't need two equations to get the focus of the parabola. You can use the fact that the focus lies on the same line as the shorted distance between the origin and the directrix. You have found the point $\left({1,1}\right)$ on the directrix from which you calculated that $a=\sqrt2$. Therefore, the focus would be at $\left({-1,-1}\right)$. However, it is good to learn from mistakes, so please include the two equations you found.

user228700

@DHMO: I don't even remember the question :| I'll get back to you.

user228700

Thanks :-)

DHMO

yesterday, by Kaumudi

@DHMO: I have a quick homework-tsy question about parabolas.

yesterday, by Kaumudi

I've been asked to find the equation of a parabola whose vertex is at the origin and the equation of its direction is $x+y=2$. I figured it out, but the answer isn't correct :/ I found $a=\sqrt{2}$. Then, using the fact that the focus of this parabola is at a distance of $\sqrt{2}$ units from the vertex and $2sqrt\2$ units from the directrix, I found two equations to find the coordinates of the focus but I'm not obtaining real roots :/

yesterday, by Kaumudi

If I find the coordinates of the focus, since I already know the equation of the directrix, I can find out the equation of the parabola...

yesterday, by Kaumudi

@DHMO: Also, I've done all this only because from the equation of the directrix, it is clear that this parabola is not a standard parabola and will be tilted w.r.t the coordinate axes.

yesterday, by Kaumudi

I've also been asked to find the length of the latus rectum of this parabola, and this is given by $4a$, so I get $4\sqrt2$, which matches the answer given in my textbook.

yesterday, by Kaumudi

So Ik the value of $a$ is correct...

yesterday, by Kaumudi

Thoughts?

user228700

Omg. Dude!

user228700

I will get back to u. I am caught up with something at the moment.

DHMO

13:32

alright

Semiclassical

Hi chat

DHMO

@Semiclassical Hi God

Semiclassical

Pfft

Oh hey parabolas

That particular question seems to be in the realm of tedious but straightforward if you have a strategy

Use directrix + vertex to find the axis of symmetry, use that to get the focus, and then write down the definition of the parabola

Sophie

is the notation $[x]$= the integer part of x common? I think $\lfloor x \rfloor$ is better

DHMO

@Sophie both are often used

Semiclassical

14:31

@Sophie Depending on convention, for $x<0$ one doesn't necessarily have $\lfloor x\rfloor =[x]$

For instance, is the integer part of -2.5 equal to -2 or to -3? The former is "the part of -2.5 before the decimal point", the latter is "the largest integer that doesn't exceed -2.5"

Mike Miller

the latter is imo the correct convention; the fractional part should never be negative

Semiclassical

So you'd want to confirm how the integer part is defined for negative numbers.

DHMO

@MikeMiller convention cannot be correct

Sophie

this is probably the mathematical equivalent of vim vs emacs

Semiclassical

Well, the former has $[x]$ and $x-[x]$ as odd functions of $x$

Mike Miller

14:39

@DHMO Sure it can. While two choices can give mathematically equivalent theories as a result, one might be consistent with older conventions or mean you have to use minus signs less.

Semiclassical

just because a convention isn't "illegal" doesn't mean it should be used

Ramanujan

@DHMO hi

DHMO

hi

Ramanujan

I have a little problem

Semiclassical

I tend to think the integer part should be defined so that $[-x]=-[x]$, but only so that it fulfills a different function than the floor funvtion

Otherwise it's redundant.

Ramanujan

14:44

0

Q: Area of triangle formed from equation of of its sides.

The area of triangle formed by the lines $$a_1x + b_1y + c_1z=0$$ $$a_2x + b_2y + c_2z=0$$ $$a_3x + b_3y + c_3z=0$$ Is $$\frac {\Delta^2}{2\lambda_1 \lambda_2 \lambda_3}$$ Where $$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & a_2 & c_2 \\ a_3 & b_3 & c_3...

triangle area

I dunno why I should use this formula,this formula like so strange to me

DHMO

I know nothing about this

Semiclassical

A formula being true doesn't force it to be useful

It might be useful in some contexts, but there's nothing mandating it

Ramanujan

@Semiclassical it's useful

robjohn

@Sophie you can easily compute $\zeta(2n)$ for $n\in\mathbb{Z}$, but it is an open question whether $\zeta(3)$ (Apéry's constant) is transcendental.

Semiclassical

Then why are you asking why you should use it?

DHMO

14:47

@robjohn this is wizardry

Ramanujan

@Semiclassical When you get 3 equation of straight lines,to get area,you first need to solve equations in couple and get each points and then apply it on equation of area of triangle

Sophie

@robjohn no, Apery proved it is irrational

that's why it's named after him

DHMO

> It is still not known whether Apéry's constant is transcendental.

Sophie

Apéry's theorem

In mathematics, Apéry's theorem is a result in number theory that states the Apéry's constant ζ(3) is irrational. That is, the number ζ ( 3 ) = ∑ n = 1 ∞ 1 n 3 = 1 1 3 ...

but that is what I meant anyway

Semiclassical

I'm not sure whether any other odd zeta values have been proven to be irrational

DHMO

14:49

> Indeed, recent work by Wadim Zudilin and Tanguy Rivoal has shown that infinitely many of the numbers ζ(2n+1) must be irrational, and even that at least one of the numbers ζ(5), ζ(7), ζ(9), and ζ(11) must be irrational.

Sophie

at least one of zeta(5), zeta(7), zeta(9) and zeta(11) is irrational

Semiclassical

It seems unlikely that any of them would be rational, but that's not a proof

robjohn

@DHMO The same can be done for Dirichlet's Beta Function at odd integers, $\beta(2n+1)$.

teadawg1337

15:15

I'm currently writing up a question to post on MSE. If I've approached the problem in question in two ways, should I include both of them in the question, even if it's detrimental to the length of the post?

Semiclassical

@teadawg1337 Can't really hurt. It's up to you how much detail to show, of course---you might give one of them in depth, and the other in a more cursory way

teadawg1337

@Semiclassical My draft currently includes the rigorous method, and I'm explaining it in detail. I'm thinking about adding in the elegant method afterwards, but using less detail

I'm just worried about not garnering enough attention if it's too long

DHMO

15:36

How am I supposed to state the negation of this? $\forall \epsilon > 0: \exists n \in \Bbb N: \forall i, j > n: \left|{x_i - y_j}\right| < \epsilon$

$\neg \left( \forall \epsilon > 0: \exists n \in \Bbb N: \forall i, j > n: \left|{x_i - y_j}\right| < \epsilon \right)$
$\implies \exists \epsilon > 0: \neg \left(\exists n \in \Bbb N: \forall i, j > n: \left|{x_i - y_j}\right| < \epsilon \right)$
$\implies \exists \epsilon > 0: \forall n \in \Bbb N: \neg \left(\forall i, j > n: \left|{x_i - y_j}\right| < \epsilon \right)$
$\implies \exists \epsilon > 0: \forall n \in \Bbb N: \exists i, j > n: \neg \left(\left|{x_i - y_j}\right| < \epsilon \right)$

Sophie

that whole thing is one statement?

DHMO

yes

I mean

12 mins ago, by DHMO

How am I supposed to state the negation of this? $\forall \epsilon > 0: \exists n \in \Bbb N: \forall i, j > n: \left|{x_i - y_j}\right| < \epsilon$

This whole thing

Semiclassical

@teadawg1337 You could approach it like this, then: Put the statement of the question in the text, and then include your partial progress as an answer to that same question

teadawg1337

15:56

That's a pretty good idea!

Wouldn't that void me from being able to accept a different answer, though? It's been a while, I don't remember exactly how it works

DHMO

Is anyone familiar with Cauchy Sequence as definition of Real Number?

I have a doubt about it

I have a sequence $x_n = \displaystyle\sum_{i=1}^{n} \frac1i$.

By the definition of Cauchy sequence, $\langle x_n \rangle$ is a Cauchy sequence

Mike Miller

That is not correct.

DHMO

Of course I know that is not correct, but my question is why.

> A sequence $a_1$, $a_2$, $...$ such that the metric $d(a_m,a_n)$ satisfies
$\displaystyle \lim_{min(m,n)\to\infty} d(a_m,a_n)=0$.

mathworld.wolfram.com/CauchySequence.html

@MikeMiller Why does my sequence not match this definition?

Akiva Weinberger

16:11

@DHMO You're only considering $d(a_m,a_{m+1})$

DHMO

@AkivaWeinberger sure, let's consider $d(a_m,a_{m+n})$...

Akiva Weinberger

You'd get a different response for $\lim_m\lim_nd(a_m,a_{m+n})$ and $\lim_n\lim_md(a_m,a_{m+n})$.

By which I mean $\lim_{n\to\infty}$ etc

DHMO

multivariable limits...

Akiva Weinberger

The former is $\infty$, the latter is $0$

See, the deal is that the distances between $m$ and $n$ can get arbitrarily large.

DHMO

so what is the formal way to state it?

Akiva Weinberger

16:16

The limit is larger than $A$ for any $A$ (and hence infinite), since for any index $N$ you give me, I can find $m$ and $n$ both greater than $N$ such that $d(a_m,a_n)>A$

Mike Miller

the key point is that you can't fix $n$

which is what you wanted to do

Mees de Vries

You could make multivariable limits formal. Honestly, I feel that just makes it more difficult than it needs to be. Really, the good way to state the definition of Cauchy is the negation of what Akiva just said.

Akiva Weinberger

…Which I can prove by considering $m=N+1$ and $n=$the first number such that $a_n>a_m+A$, which exists because the series diverges to infinity

(re my last comment)

DHMO

@MeesdeVries So $\forall \epsilon > 0 : \exists c \in \Bbb N : \forall m , n > c : d(a_m,a_n) < \epsilon$ ?

Akiva Weinberger

I believe so (but double-check me)

Mike Miller

16:19

correct

DHMO

Let's say $\langle x_n \rangle$ is a Cauchy sequence in $\Bbb Q$ and its limit is $x$ where $x \in \Bbb R$.

Mees de Vries

Yes. With the caveat that I would use $N$ instead of $c$, but of course that does not matter.

DHMO

How could I prove that $\exists N \in \Bbb N : \forall n > N : |x_n| > 0$ if $x\ne0$? (Sorry for the edit)

I cannot simply substitute $\epsilon = x$ because it is in $\Bbb Q$, right?

Mike Miller

by proving that in a cauchy sequence with limit $x$, $\forall \varepsilon > 0, \exists N : \forall n > N, d(x_n, x) < \varepsilon$

Nathan Drieling

Hello everyone. I have been here for a while now but how can I make the fancy math format/notation when I post a question?

DHMO

16:24

@MikeMiller I thought $d$ is also only in $\Bbb Q$

@NathanDrieling read room description

Mees de Vries

Take $\epsilon = |x/3|$.

DHMO

@MeesdeVries still not rational

Nathan Drieling

oh ;p sorry

Mees de Vries

Uh, fine. Then take some rational in $(|x/4|,|x/3|)$.

Mike Miller

The distance function is defined on the reals.

DHMO

16:26

@MeesdeVries the existence of rational between reals is dependent on this

Mike Miller

Since you're working with an element of the completion of the reals, I see no reason not to use the induced distance function.

DHMO

@MikeMiller but I thought everything about the Cauchy sequence is only defined in $\Bbb Q$ for now (except its limit)

Mees de Vries

Be precise about what your conditions are then. Have we constructed the reals yet? If not, what is $x$?

DHMO

the limit of the sequence

Mees de Vries

So we do have the reals?

DHMO

16:26

@MikeMiller how would you define?

@MeesdeVries but you do not have the fact that a rational exists between two reals

Mike Miller

I'm with Mees on this. You should either tell us everything we need to know to answer your question, or not expect an answer.

Mees de Vries

Do we have the archimedean property of the reals?

DHMO

I'm proving the existence of inverse

@MeesdeVries yes but you don't have inverse

Mees de Vries

No inverses of reals, but inverses of (non-zero) rationals, right?

DHMO

yes

Mees de Vries

16:28

So, then, pick some $\epsilon$ in $(0,|x|/4)$.

By the Archimedean property.

DHMO

wait

Mees de Vries

Then you get an $N$ out of the definition of Cauchy, and then comes some triangle inequality magic.

DHMO

@MeesdeVries how to use Archimedean property here?

Mees de Vries

That... is precisely the Archimedean property? If $a > 0$ then there is an $n$ such that $\frac1n \in (0,a)$.

So take $\epsilon = \frac1n$.

DHMO

wait

the form I know is that there is $n$ such that $n > \dfrac1a$

Mees de Vries

16:31

OK. Well, do we have ordered field axioms yet for $\mathbb R$? Because you can derive my version pretty easily then.

DHMO

but we don't have the existence of $\dfrac1a$ for that matter

Mike Miller

You do for rationals.

DHMO

but $a$ is not rational

Mees de Vries

Then you need to formulate what it means to have the archimedean property for you, if you haven't assigned meaning to $\frac1a$ yet.

Maybe that there always exists an $n$ such that $an > 1$?

DHMO

@MeesdeVries the thing is

we are proving that $\Bbb R$ is a field

so whence ordered field

Mees de Vries

16:34

Uh, OK. Ordered ring then.

I think that does not matter too much, probably.

We do have addition, subtraction, multiplication, right?

Also, side note: this is why I recommend defining $\mathbb R$ by Dedekind completion, and not via Cauchy sequences.

DHMO

@MeesdeVries I believe so

Mees de Vries

OK. Then what is the Archemedean property that we have?

DHMO

For any $a$ there exists $n>a$

Mees de Vries

That... doesn't sound like the Archimedean property to me. So we have to prove an "actually useful" version first?

DHMO

my mind is now all over the place

Mike Miller

16:39

what book are you learning this from, @DHMO

DHMO

@MikeMiller the internet

Mike Miller

so I'm going to suggest that if you want to learn analysis, find an analysis book

DHMO

@MeesdeVries why isn't it obvious that there exists $N$ such that for all $n > N$ we have $|x_n|>0$

@MikeMiller sure

Mike Miller

it is obvious, but you still need to write down a proof

DHMO

yes

and how on earth am I supposed to prove that

Mees de Vries

16:40

It depends on what you mean by "obvious".

DHMO

I mean "intuitive"

Mees de Vries

The difficulty with writing proofs of very basic things is often being very clear about the things you are allowed to assume/have already proved, and go from there.

OK, but you say I am not allowed to use (my version) of the Archimedean property. Is that not "obvious" for you?

DHMO

alright, it wasn't a point.

Mees de Vries

Well, once you are allowed to pick $\epsilon$ some number in $(0,|x|/3)$, the proof becomes pretty easy. One or two applications of the triangle inequality.

DHMO

yes

provided that you can pick

Mees de Vries

16:43

And you do not find this obvious?

DHMO

I do

but I cannot

How do you induce $d$ to $\Bbb R$ anyway

Mees de Vries

What do you mean "I cannot"?

DHMO

never mind

Mees de Vries

If you have defined $d$ on $\mathbb Q$ and $\mathbb R$ and limits properly, then you can define $d(x,y)$ for reals $x,y$ by taking representing Cauchy sequences $\langle x_n\rangle, \langle y_n\rangle$ and setting $d(x,y) = \lim_{n \to \infty} d(x_n,y_n)$.

But, if you're doing this to construct the reals, of course you cannot do that yet.

DHMO

exactly

Mees de Vries

16:48

You can work around that, which is where we've been going. But, yeah: that is a lot of work.

Unsurprisingly, constructing the reals from first principles, a project which took historical mathematicians decades,is not trivial ;-)

DHMO

indeed

Mike Miller

@BalarkaSen What are you doing?

DemCodeLines

$5^x$ = -1 mod 27. How would you find the discrete log x?

Mees de Vries

17:04

Trial and error? :p

The answer is 9, if you are curious

DemCodeLines

17:15

What's a Ring?

In Discrete Math?

Mees de Vries

Probably the same thing as in algebra. A set of things you can add and multiply, and which obey some sensible rules about addition and multiplication.

DemCodeLines

Is Z18 a ring?

Krijn

Z18 is a vague term

If you mean $\mathbb Z/18 \mathbb Z$, then yes, it is

Any $\mathbb Z/n \mathbb Z$ is

DemCodeLines

How do you know it's a ring?

Krijn

Because we can define addition just as we do in $\mathbb Z$ and the same with multiplication

Then we check the axioms for a ring

Mees de Vries

17:28

By defining operations $+,*$ on it, and giving elements $0,1$ in it, and showing that the ring axioms are satisfied for these.

Krijn

It's a bit tedious, but clearly doable

DemCodeLines

I just started learning it and am terribly confused.

Krijn

Do you know what a group is

DemCodeLines

Yeah

Krijn

So you can see $\mathbb Z /18 \mathbb Z$ as an Abelian group under addition

Now a ring, you can see as a group with an extra operation: multiplication

It is probably clear to you how we should multiply two elements from $\mathbb Z/ 18\mathbb Z$

It is not immediatly clear why this is a ring, then, as it needs to satisfy some extra axioms

For some reason, the Dutch wikipedia article for Ring is linked to the English wikipedia article for Rng. Confusing

saturatedexpo

17:33

if i prove that injectivity and surjectivity of f,g implie inj./surj. of $g\circf$ is the same for bijectivity automaticly proven?

$g\circ f$

Krijn

@DemCodeLines Wikipedia explains it a lot better then me, really

@saturatedexpo Not automatically, but easily, yes.

privetDruzia

Hello I have an extremely stupid question regarding complex numbers. I always thought $i^2 = -1$. So how is the following possible:

$\frac{i}{i-0.5} = \frac{1}{\sqrt{1^2 + 0.5^2}}$

Am I missing something?

Balarka Sen

@MikeMiller Schoolwork.

Krijn

@privetDruzia The rule $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ holds for $a,b \in \mathbb{R}_{\geq 0}$, not in general

DemCodeLines

I wish there a place that showed step-by-step for these problems.

privetDruzia

17:37

@Krijn I understood what you wrote but don't see how that relates to my question.

Krijn

@privetDruzia Oh, sorry, I misunderstood your question, I thought you were doing something different

I don't see how you get that equation though?

privetDruzia

@Krijn neither do I...

Krijn

It's not true, is what I mean

privetDruzia

Indeed but that's what is written here in front of my eyes.

Haven't smoked any weed as far as I know.

Krijn

I can't help you with that, then

Akiva Weinberger

17:41

@privetDruzia Yeah, that equation's not true

privetDruzia

@AkivaWeinberger that's what I think as well, but for some strange reason our teacher wrote that. It had to do with vector magnitudes.

Akiva Weinberger

$i/(i-0.5)=0.8-0.4i$

saturatedexpo

let me rephrase my question: if i, speratly, prove that injectivity of f,g implied injectivity of $g\circ f$. And then prove the same for surjectivity. Why dont i have proven the same for bijectivity already?

Akiva Weinberger

@privetDruzia Are you sure there aren't supposed to be absolute value bars around the left hand sign

Like,$$\left\lvert\frac i{i-0.5}\right\rvert$$

privetDruzia

analyzing it...

cannot decypher it...

Akiva Weinberger

17:44

Do you know what absolute value means for complex numbers?

privetDruzia

the magnitude of $\frac{i}{i-1/2} = 1,12$ ?

Does that make any sense?

Not to me.

@AkivaWeinberger No sorry, I am really not used to work with complex numbers in that way.

Akiva Weinberger

Do you know the complex plane?

privetDruzia

yes

ofc

Akiva Weinberger

The absolute value of a complex number (usually called its "magnitude") is defined to be the distance between it and $0$ on the complex plane.

Notice that that is the same as the absolute value you're used to when we're taking the absolute value of a real number.

privetDruzia

ok makes sense, in other words the magnitude.

Akiva Weinberger

17:49

By the Pythagorean Theorem, we can show that $|a+bi|=\sqrt{a^2+b^2}$.

privetDruzia

ok

Akiva Weinberger

Also, the rule $\lvert a\rvert\lvert b\rvert=|ab|$ still works, though it's not obvious why.

privetDruzia

but in my case why are we only applying pythagoras theorem to the denominator?

to find the magnitude.

Akiva Weinberger

We also have $|a/b|=|a/b|$. Thus, $|1/z|=|1|/|z|=1/|z|$.

privetDruzia

$\frac{i}{i-1/2}$ .... $\sqrt{1^2 + (1/2)^2} = 1.12$

Akiva Weinberger

17:51

It's the reciprocal of that

It ends up being $1/\sqrt{1.25}=2\sqrt5/5=.89443$

privetDruzia

@AkivaWeinberger Could you please explain what happens at "..." ?

nothing more.

I know the length of the i vector = 1.

Akiva Weinberger

$ $\begin{align}\left|\frac i{i-0.5}\right|&=\frac{|i|}{|i-0.5|} \\&=\frac1{\sqrt{1^2+(-0.5)^2}}\\ &=\frac1{\sqrt{1.25}}\end{align}

One moment

privetDruzia

np.

Akiva Weinberger

There. Is it rendering now?

privetDruzia

yes.

seems incorrect. The answer is 1,12

not 0.89

Akiva Weinberger

17:56

You sure?

privetDruzia

yes

very sure.

Akiva Weinberger

That's the reciprocal…

Maybe it's supposed to be $|(i-0.5)/i|$

privetDruzia

I don't want any side roads. Just the straight and immediate way to get 1,12.

Akiva Weinberger

@privetDruzia The right-side of the equation you wrote here is 0.89

privetDruzia

No it is not because it s written twice the same. I wrote two times the same thing on different days.

Akiva Weinberger

17:58

23 mins ago, by privetDruzia

Hello I have an extremely stupid question regarding complex numbers. I always thought $i^2 = -1$. So how is the following possible:

$\frac{i}{i-0.5} = \frac{1}{\sqrt{1^2 + 0.5^2}}$

Hm. Maybe you didn't do the last step of the division. I dunno, ask a classmate

privetDruzia

I have another example based on the same principle. I think that might clarify it.

$H = \frac{1}{2}(1-i)$
$|H| = \sqrt{2}$

@AkivaWeinberger does that make sense to you?

Akiva Weinberger

Um. That's the reciprocal of the right answer again… or you forgot the $\frac12$

saturatedexpo

Hi can you look on my proofsketch? Already proven: f,g are injective$\to$ $g\circ f$ is injective. f,g are surjective$\to$ $g\circ f$ is surjective. To show: f,g are bijective$\to$ $g\circ f$ is bijective. Proof: f,g are bijective$\to$ (f,g are surjective) and (f,g are injective). Therefore $g\circ f$ is also injective AND surjective. Bijectivity follows from definition.

Akiva Weinberger

It's also twice the right answer

privetDruzia

@AkivaWeinberger Could you please show me how you calculated it for the last one?

I tried $\frac{1}{2}\sqrt{1^2 - 1^2}$ but that obviously results in zero...

Akiva Weinberger

18:04

$|H|=|\frac12(1-i)|=|\frac12||1-i|=\frac12\sqrt{1^2+1^2}=\frac12\sqrt2$

Remember $(-1)^2=1$

privetDruzia

oh ok I seem to mess that up every time

Akiva Weinberger

$1-i=(1)+(-1)i$. You add the squares of those.

Also $\frac12\sqrt2$ can also be written as $\frac1{\sqrt2}$

but people don't like roots in the denominator so the first is better

privetDruzia

shouldnt the square root 2 be on top in stead of in the denominator?

Akiva Weinberger

Yeah

They're still equal though, $\frac12\sqrt2$ and the other thing

Balarka Sen

Maybe I'll write down a proof of the fundamental theorem of algebra in my school project.

saturatedexpo

18:10

@BalarkaSen do you explain it in Laymans terms to the class?

privetDruzia

After further "analysis" I'd rather write it as: $'|\frac{1}{2}| |1-i| = \frac{1}{2}\sqrt{1^2 -(i)^2} $

$ = \frac{1}{2} \sqrt{1^2 -(-1)}$
$= \frac{1}{2} \sqrt{2}$

Do you concur?

Balarka Sen

@saturatedexpo The statement of the theorem is quite understandable.

saturatedexpo

the statement yes, the proof probably not @BalarkaSen

Mike Miller

Someone tell me a bedtime story.

Balarka Sen

@MikeM: Strongly recommend Hedgehog In The Fog for a short, nonmathematical bedtime story.

Akiva Weinberger

18:14

@BalarkaSen Maybe write down six

Anubhav

@MikeMiller I've a doubt

Akiva Weinberger

@privetDruzia To see if you understand — compute $|3+4i|$ and $|3-4i|$

They should both be equal to $5$.

privetDruzia

not the first one in my case...

$\sqrt{3^2 + 4^2(i)^2}$ = $\sqrt{9-16}$

Anubhav

for a fixed genus consider $S_g$ bundle over $S^1$. For given any element $f\in Mod(S_g)$ we can have a maping torus $M_f$ corresponds to this. Now It is easy to show that if %f,g$ are conjugate in $Mod(S_g)$ then they $M_f$ and $M_g$ are actually homeomorphic.

but is the converse true.

Akiva Weinberger

@privetDruzia You're not supposed to have $i$ in there.

Anubhav

18:19

Moreover can we classify all these $S_g$ bundle over $S^1$ by some speical class related to Mod(S_g)?

Akiva Weinberger

The first is computed as $\sqrt{3^2+4^2}$ and the second as $\sqrt{3^2+(-4)^2}$. @privetDruzia

It will help to draw them on the complex plane.

Remember that the magnitude is the distance to $0$.

saturatedexpo

@AkivaWeinberger why 6 in particular? (number of proofs)

Anubhav

@MikeMiller there was a latex typo in my last msg... As you can see if $f,g\in Mod(S_g)$ are conjugate then $M_f$ and $M_g$ are homeomorphic. So is the converse true?

privetDruzia

@AkivaWeinberger Are you supposed to just neglect the 'i' ?

Akiva Weinberger

I found a book in a library once that was six proofs of the FTA and the necessary background to each proof @saturatedexpo

Two from algebra, analysis, and topology each (if I recall correctly)

@privetDruzia Yes. Drawing it will make it more clear

Anubhav

18:23

If not, can we give some more stronger restriction on $f,g$ s.t $M_f$ and $M_g$ are homeomorphic?

Akiva Weinberger

Gtg

@privetDruzia You might as well also neglect the $+$/$-$ sign, it changes nothing

Mahmoud

Hello.

Tonepoet

Hi, I am a lowly intruder who knows nearly nothing about mathematics. I hail from English Language and Usage with some questions regarding regarding the word vinculum I think you folk might know better than my peers: First would you please tell me if any of you would recognize the word vinculum offhand? Also, if you can recognize it, would you also give me a personal opinion as to whether it can be used as a synonym for fraction bar, or if it is restricted to binding the order of operations?

Mees de Vries

I recognise the word vinculum, but only know what it means after you've mentioned it. (But, I am an ESL speaker.)

Actually, on opening Wikipedia to be sure, I did not know what it meant. I thought you meant that it was a word for the symbol $\div$.

Akiva Weinberger

18:40

Use "fraction bar" instead @Tonepoet

Wait, no, that's not what vinculum means @Tonepoet

Use overbar.

Or just bar.

Mahmoud

Is $y=f(x) \iff (x,y)\in f$ ?

Mees de Vries

For the common encoding of functions in ZFC, yes.

Mahmoud

Does it mean that $f$ is the set of all possible pair constructed from the domain and range of the function ? @MeesdeVries

Mees de Vries

If you have $f : X \to Y$, then $f$ typically does not consist of all pairs $(x,y)$ where $x \in X, y \in Y$ (that would be $X \times Y$), it consists of all pairs $(x,y)$ for which $f(x) = y$.

Mahmoud

So can we deduce that $f=\{(x,y) / f(x)=y\}$ ?

And please correct my symbol for ''Such that''

Tonepoet

18:48

Hmm, the reason I'm asking is because I am getting some conflicting data from different sources. I'm not sure if I know any better now than I did before but I think your opinions will be helpful for my purpose regardless. Thank you @MeesdeVries and @AkivaWeinberger .

Mees de Vries

@Tonepoet, for what it's worth, at least two mathematicians do not know the word, so stay away from it in text that is supposed to be easy to understand, if you can.

Mike Miller

@Anubhav I think so, but it's nontrivial.

Mees de Vries

@Mahmoud, sure, if you want. But really, in formal ZFC, $f(x) = y$ is just a shorthand for $(x,y) \in f$, so this is pretty trivial. Also, you can make the vertical bar in your set builder notation with \mid.

Mike Miller

Or maybe I don't think so.

Mahmoud

$\mid$ ^_^ Thank you @MeesdeVries But if $f$ is considered to be a set, why do we write $f(x)$ this way ?

Mees de Vries

18:53

For convenience. ZFC came after most of mathematics. It makes more sense to think of ZFC as a language in which we encode (already existing) mathematics, than to think we started from ZFC and built mathematics from there.

Mahmoud

Thanks @MeesdeVries again $:)$

PVAL-inactive

I think I'd prefer "overline" in text and "bar" in real life.

Alessandro

Everything is a set in ZFC but it's usually not very productive to think about mathematical objects in the way they are formalized there (also because there could be other formalization where, just to make an example, "function" is taken as primitive) @Mahmoud

Mahmoud

Thank you @Alessandro $:)$

Mike Miller

@Anubhav I think a discussion of the Thurston norm will show that you can have many different ways of fibering over the circle, even with fixed fiber. See eg here for an intro.

0celo7

19:31

Why should it be "obvious" that a linear map between finite-dimensional normed spaces is bounded before one has discussed the operator norm or equivalence of norms?

Mees de Vries

What does it mean to be bounded without discussing the operator norm?

That is to say, I would say $F: V \to W$ is bounded if there is a constant $K \in \mathbb R$ such that for all $v \in V$ with $\| v \| \leq 1$ we have $\|F(v)\| \leq K$, but that is already basically the definition of the operator norm.

0celo7

19:51

@MeesdeVries $||L(x)||\le c||x||$.

I mean, without knowing that the operator norm is the largest singular value

PearlSek

Did you know that scientists found the constant of the best duration to cook pasta ?

It's called the Euler-Macaroni constant

bye

Transcript for

$Mathematics$ Mathematics