Mathematics - 2017-05-08 (page 4 of 4)

Semiclassical

20:01

But I've rambled enough for now :P

Eric

@Daminark why are you set to die exactly

I really hope Benson teaches AG again, since I'm already endeared to him and won't be nervous going to his office lol

Arrow

20:16

Hello

Let $f:A\to B$ be a monoid homomorphism. Suppose the kernel acts transitively on every fiber. Does this imply $A$ is a group?

Sha Vuklia

@Semiclassical ohh I'm so sorry, I clicked away the chat and continued reading!

(I think I need a short break, so I'll read it in 5-10 minutes)

Semiclassical

np

Sylent Nyte

Guys, I need some help

I've been given the derivative of a function $f'(x) = 4x+ 3$ and the second derivative $f''(x) = 4$ and that $f(1) = 6$, how could I use all of this information to find out $f(2)$

WITHOUT using integrals

Semiclassical

You pretty much can't.

Sylent Nyte

But even an approximation?

Semiclassical

20:24

At least, you can't do so without doing an antiderivative in disguise.

Just to make clear what I mean: Can you guess what $f(x)$ should be if $f'(x)=4x+3$?

Sylent Nyte

Educate me pls

@Semiclassical $2x^2 + 3x + 1$

but thats using integral stuff

Semiclassical

Well, 2x^2+3x+c. But once you plug in f(1)=6, yeah.

Sylent Nyte

But, apparently you can get an approximation for the values of f(2) and f(3) without needing the actual function

Semiclassical

@SylentNyte Beyond that, hrm.

Sylent Nyte

only the derivatives

Semiclassical

20:32

Well, okay. What you can do is argue that $f(x) \approx f(1)+f'(1)(x-1)+f''(1)(x-1)^2/2.$

That's the Taylor series of order 2 around $t=1$.

In your case, that'd give $f(x)\approx (1)+(7)(x-1)+(4/2)(x-1)^2=1+7(x-1)+2(x-1)^2$.

And therefore $f(2)\approx 1+7(1)+2(1)^2=10$ and $f(3)\approx 1+7(2)+2(2)^2=19.$

Sylent Nyte

"NO computers, NO Taylor's theorem, NO integration theory" - teacher

Semiclassical

...

Sylent Nyte

:|

he only put in that no computers because of me and my scripts :(

Semiclassical

"Well, yes, getting from one side of the room to the other is hard if you tie my feet together."

Zophikel

Does anyone know what the contour would look like for $\frac{1}{2i} \int_{-\infty}^{\infty}\frac{e^{iz} -1}{x}dz$

Semiclassical

20:35

That's what this question feels like.

Sylent Nyte

You couldd...

@Zophikel try wolframalpha

Semiclassical

I mean, you need to have some way of connecting $f(x)$ with $f'(x)$.

Sylent Nyte

well

if you take the average of f'(1) + f'(2)

and add that onto f(1), you get f(2)

Semiclassical

hmm.

Sylent Nyte

because thats the approximation of the rate of change in that portion of the graph

Semiclassical

20:37

True. I think that's basically a trapezoid approximation in disguise.

Sylent Nyte

but apparently the second derivative is needed to make the approximation "better", however here its useless

(in the case of this equation and all, it gives the exact values....?)

Zophikel

@Synlent Nyte I found it would be an indented semicircle

Found the definition on another site

Semiclassical

Well, if you take that as writ, you get $f(2)\approx f(1)+(f'(1)+f'(2))/2\cdot 1$.

Sylent Nyte

@Zophikel sick, the equation looks horrible as I have not come across it as of yet but Im glad youve got it.

Semiclassical

Which I guess you could also argue as $(f(2)-f(1))/(2-1)\approx (f'(1)+f'(2))/2$.

The $f''(x)=4$ bit is rather superfluous, since it follows from $f'(x)=4x+3$ directly.

Zophikel

20:41

@SylentNyte i'm going to have to sketch out a proof apporch to this one and fill in any algebraic manuiplations later :)

Semiclassical

In the present case, that gives $f(2)\approx 6+(7+10)/2=14.5$

By comparison, the exact value is $f(2)=2(2)^2+3(2)+1=15.$

So, not bad as an approximation.

To make a better approximation, you'd want to use more points between $x=1$ and $x=2$.

For instance, in what's known as Simpson's rule you'd have $f(2)\approx f(1)+\frac{1}{6}(f'(1)+4f'(1.5)+f'(2))=6+(7+36+10)/6 = 14.8\overline{3}$. That lowers the absolute error by a factor of 3.

Sylent Nyte

@Semiclassical regarding the $f''(x) = 4$ being superfluous

it was literally a case of pointing about the obvious

Semiclassical

Ah.

Sylent Nyte

because I also thought that the second derivative in this case is pointless as we can see it from the first derivative

Semiclassical

Sure.

Sylent Nyte

20:48

but its used to see how the derivative before it changes (apparently idk why thats significant tho)

Semiclassical

The point to make, I guess, is that if you only use $f'(1)=7$ and $f'(2)=10$ then you're not really taking advantage of the fact that you know $f'(x)$ for every $x$ between 1 and 2.

There's a bunch of functions that'd have those values but would behave differently between 1 and 2. Those would have different values of $f(2)$.

Sylent Nyte

Yea of course

Daminark

Hey everyone!

Sylent Nyte

Hello!

Semiclassical

Therefore, if you want to take advantage of the fact that you know $f'(x)$ for all $x$, you'll either need to 1) evaluate $f'(x)$ at more points than just 1,2 or 2) make use of the second derivative.

Daminark

20:50

@Eric Mariannaplex, Wombo combo, and Emerton algebra = gg

Mike Miller

you make so many terrible puns

Semiclassical

I think, though, that said use of the second derivative is basically Taylor's theorem in disguise :P

Mike Miller

how do you have friends

Semiclassical

That's what sorta annoys me about this question. The only way I know to justify these approximations is by appeal to integration.

To pretend otherwise just feels silly.

Sylent Nyte

Is it appropriate to ask how the taylors theorem is derived here or to look it up online

Semiclassical

20:51

That could be a long discussion, yeah.

Lozansky

@Semiclassical Could I annoy you with a problem regarding tensors?

Semiclassical

You can try :P

Lozansky

It's quite simple I guess

Sylent Nyte

@Semiclassical Thanks for your help!, ill be back later because im gonna get something related as work tomorrow, which i will need help on understanding, thank you tho

Lozansky

I just wanna make sure I understand it

Semiclassical

20:52

mmkay.

Daminark

@Mike Friends who are able to remain friends even if it involves dealing with my puns must be good, that's for sure

Legend has it that a subset are even amused though I'm skeptical

Fargle

I've become obsessed with polyrhythms and odd time signatures in music. Does that mean I'm destined to become an algebraist? thinking emoji

Lozansky

A tensor in the cartesian coordinate system $K$ has the components $(T_{ik}) = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1& 1 \\ 0 & 0 & 1 \end{bmatrix}$. Does it exist a cartesian coordinate system $K'$ such that $(T'_{ik}) = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2& 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}$?

Mike Miller

gross

Lozansky

@Semiclassical

Jason Bourne

20:56

Hello @MikeMiller.

Daminark

(Wombo combo wasn't even my idea)

Lolol, @Fargle come be an algebraist with me!

Mike Miller

Hi Jasper

Jason Bourne

@Fargle It depends on how you eat your corn.

Fargle

@Daminark Only if I get to start a band called Torsion Group.

Lozansky

What an odd time signature?

Fargle

20:58

@JasonBourne Haha, I remember reading that.

@Lozansky In most music, the "beat" is felt in 2, 3, or 4.

Lozansky

Uh huh

Fargle

Lately I've been listening to music where measures are divided into 5, 7, 9, 10, 11, 13...

Jason Bourne

@Fargle It seems they don't care about geometers in the corn experiment, but I guess geometry just reduces to algebra and analysis. =)

Daminark

@Fargle I'll allow it :P

Lozansky

@Fargle Like clair de lune

Daminark

20:59

Though if you're talking about measures, maybe you should be an analyist

Fargle

@Lozansky Exactly!

Lozansky

I think some chopin nocturnes are in odd signatures as well

Jason Bourne

@Fargle The most important thing about the score is that it is only a tool to write the music. Music is music and need not be confined to a specific time signature, and the score is just an imperfect representation of how a piece should be performed.

Lozansky

I know fantaisie impromptu is 3 against 4s

Since I've played it :P

Fargle

@JasonBourne Agreed. I guess it should be understood that I mean in our modern system of written conventions, and in our twelve-tone equal temprament system of notes.

Jason Bourne

21:01

On my good days, I can reach low low C and middle G, so that's two and a half octaves on my vocals.

Fargle

I can hit low low Db and middle E most of the time.

Jason Bourne

That means we are both bass.

Fargle

I wish the genres I'd like to write in were better suited to bass/baritone vocals. IT seems like every prog band has a tenor/mezzo vocalist.

Jason Bourne

Bass range is E to E, baritone A to A and tenor C to C.

Fargle

Mm.

I never knew the ranges exactly.

Jason Bourne

21:03

Just a rough guideline. Every voice is different.

Fargle

Of course, I could take a page out of the book of people like Sithu Aye, Plini, and the current iteration of Intervals, and just go instrumental.

But I feel like instrumental music needs something extraordinary (in the most literal definition of the term) to really stand out.

Jason Bourne

But you should be able to sing at least 2 octaves perfectly well on bad days, otherwise you can't sing.

Lozansky

@Semiclassical Any ideas on my question?

Fargle

I was formally trained in vocals and piano for some time, but I'm out of practice.

Jason Bourne

I think it's OK for pop singers to use lip syncing on bad days, voice can go bad so easily.

Fargle

21:19

Lately I've been playing with 7 over 4 and 9 over 4 polymeter, because holy cow, it sounds groovy.

Santosh Linkha

How to show that $\{p+q\alpha + r \alpha^2+s\alpha^3: p,q,r,s \in\Bbb Q, \alpha^4 = 1 \}$ is a field?

Fargle

21:41

@SantoshLinkha I think you should just be able to exhibit inverses for $1, \alpha, \alpha^2, \alpha^3$, but don't quote me on that.

Actually, what I just said feels wrong.

arctic tern

21:55

need all nonzero linear combinations of them having inverses

(and of course closure under multiplication)

(presumably closure under addition and scalar multiplication is obvious)

Santosh Linkha

22:08

@Fargle I need to prove $\Bbb Q[i, \sqrt 2]$ is a field. I thought I would transform it into $\alpha$

Fargle

@SantoshLinkha Wait, if you're working with that field, isn't $\alpha^4 = -1$? Or am I just crazy because I woke up recently?

Santosh Linkha

hmm ... it's 1

arctic tern

which element of $\Bbb Q[i,\sqrt{2}]$ is $\alpha$ supposed to be

Santosh Linkha

I put $\alpha = \sqrt 2 + i $

arctic tern

$(\sqrt{2}+i)^4\ne1$ because the two sides have different norm

Fargle

22:11

Would recommend $\alpha = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$

Santosh Linkha

hold on I'll put the whole question

Fargle

Whether you pick $\alpha, -\alpha, \overline{\alpha}, or -\overline{\alpha}$, though, in all cases $\alpha^4 = -1$

Alessandro Codenotti

@SantoshLinkha Prove that $\Bbb Q[i]$ is a field, call that $\Bbb K$, then prove that $\Bbb K[\sqrt{2}]$ is a field. Even better prove that if $\Bbb F$ is a field and $\alpha$ is algebraic over $\Bbb F$ then $\Bbb F[\alpha]$ is a field

Santosh Linkha

what strategies do you suggest? I alraedy solved the lattter part of the problem ... just need to show that it it field.

arctic tern

To show F[a] is algebraic over F when a is a root of a polynomial, you need to show closure under multiplicative inverses. consider linear map "multiplication by a nonzero element of F[a]"

standard proof

user193319

22:33

What does it mean for two parametrizations $C_1(t)$ and $C_2(t)$ to represent the same curve? My guess is that the images $C_1(\mathbb{R})$ and $C_2(\mathbb{R})$ are equal. Does this sound right?

arctic tern

sounds right

user193319

@arctictern Thanks for the response!

Alessandro Codenotti

22:46

You probably want the parametrizations to have the same orientation and not just the same image

Zophikel

$\frac{|a+b|}{|c|} \leq \frac{a+b}{min|c|}$

                       ^ Is this inequality vaild ?

arctic tern

min|c| ?

user193319

@AlessandroCodenotti Would you know of a better formulation? Also, I am trying to prove that $L_1(t)$ and $L_2(s)$ are parametrizations of a line, then $L_1$ and $L_2$ are the same line if and only if they intersect at one point and their direction/slope vectors are parallel. But this is turning out a bit harder than I thought.

Simply Beautiful Art

xD Jesus, people must really find the "Frivolous theorem of arithmetic" interesting

user379685

23:30

Hello, could some help me with the lim n->inf $(arccos(1/n^2)^n$ ?

Simply Beautiful Art

@user379685 Hint: $\arccos(x)\sim x-\frac13x^3$

arctic tern

that's sin(x) ~ x-(1/3)x^3

Zophikel

23:50

Anyone know a good Computer Albegra System one can get started with

well y naught

sage? pari-gp?

(the latter is a component of the former)

List of computer algebra systems

The following tables provide a comparison of computer algebra systems (CAS). A CAS is a package comprising a set of algorithms for performing symbolic manipulations on algebraic objects, a language to implement them, and an environment in which to use the language. A CAS may include a user interface and graphics capability; and to be effective may require a large library of algorithms, efficient data structures and a fast kernel. == General == These computer algebra systems are sometimes combined with "front end" programs that provide a better user interface, such as the general-purpose GNU TeXmacs...

Simply Beautiful Art

@arctictern Oh, oops

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$Mathematics$ Mathematics