Mathematics - 2015-05-17 (page 3 of 5)

r9m

17:00

@Chris'ssis This appears in conjunction with $\sum\limits_{n=1}^{\infty} \frac{H_n^{-}}{n^2}$ and can be easily dealt with using symmetry (following one of @robjohns answers) :-)

@Chris'ssis interesting :D I'll try computing it later :D

Mike Miller

@TedShifrin: There are plenty of contractible compact 4-manifolds.

Morning, by the way.

Clarinetist

Okay, so I never learned Gram-Schmidt when I took Linear Algebra 4-5 years ago. This book says

Let $\mathscr{N}$ be a space with basis $\{x_1, \dots, x_r\}$. There exists an orthonormal basis for $\mathscr{N}$, say $\{y_1, \dots, y_r\}$ with $y_s$ in the space spanned by $x_1, \dots, x_s$, $s = 1, \dots, r$.

I am to complete this proof.

It gives me the first step, saying

Define the $y_i$s inductively: $y_1 = \dfrac{x_1}{\|x_1\|}$, $w_s = x_s - \sum\limits_{i=1}^{s-1}\left(x^{\prime}_s y_i\right)y_i$, $y_s = \dfrac{w_s}{\|w_s\|}$ (I assume this is meant for $s > 1$). So how do I complete…

$\prime$ denotes the transpose

So now you have to somehow show that $y_i \cdot y_j = 0$ for $i \neq j$

Chris's sis

@r9m :-) By the way, I'm surrounded by very nice stuff. :D

Mike Miller

@TedShifrin: I answered the question. If he just wants simply connected, every 3-fold bounds a simply connected thing. This is not the interesting question; the interesting question asks what bounds a contractible 3-fold.

Chris's sis

(Marvellous results all over the place)

:D

@r9m ^^^

Clarinetist

17:09

And... I have a feeling this is not going to be pretty (the proof)

Ted Shifrin

I figured you'd polish it off quickly, @MikeM :)

r9m

@Chris'ssis when someone says I'm surrounded by very nice stuff I think of a scene from Uncle Scrooge cartoon .. lemme find it :P lol

Chris's sis

@r9m hahahahaha :-))))))

Ted Shifrin

@Clarinetist: There's really nothing going on. Just project off using an orthonormal basis for the $k$-dimensional subspace and then move up one more dimension.

Chris's sis

@r9m It's bad not all mathematicians are as funny as you are. ;)

Mike Miller

17:11

I have absolutely no idea what bounds a contractible 4-fold, and I don't think anybody does, @TedShifrin. Because if you have a 3-manifold bounding a contractible 4-fold, you now have an embedding of your 3-fold into a homotopy sphere. Unfortunately pretty much the only known obstructions to embedding into $S^4$ are obstructions to bounding a homology ball!

r9m

@Chris'ssis like this one :P

Chris's sis

@r9m :D:D:D

Owatch

Hmm

Mike Miller

I suspect if one could give a full answer to what 3-folds bound a contractible 4-fold, it would give a line of attack for SPC4.

r9m

@Chris'ssis have you seen this thread ? :) It's an AMM problem from recent issues (possibly the latest issue or the one before that) .. it's giving me a real hard time :O .. there's a slight mistake in the Q though .. it should be $\log^2 (...)$ not $\log (...)^2$

Owatch

17:14

What notation should I use when representing increasing/decreasing values?

$(-inf, 1)U(5,+inf)$

Is that how it is normally done, if you are increasing between -inf and 1, and 5 to +inf?

Clarinetist

@Owatch The code you want is (-\infty, 1) \cup (5, +\infty). Yes, it is usually done this way IME

Chris's sis

@r9m It looks really nice.

Owatch

Okay, thank you!

$(-\infty, 1) \cup (5, +\infty)$

Clarinetist

I have not missed being an actuarial company's $\LaTeX$ expert.

Chris's sis

@r9m I solved such problems, it shouldn't be hard.

r9m

17:16

@Chris'ssis yea .. it looks so stunning that I was planning to use it to get rid of the gals if I got bored at any point during a date :P

@Chris'ssis oho!!! :D

Chris's sis

@r9m The left side is clear, so no concern. (as galactus noted)

Mike Miller

Thanks for linking that, @Ted, never would have seen it otherwise.

Ted Shifrin

I stumbled upon it :)

r9m

@Chris'ssis yes .. the right side is the trouble maker ,,, I have difficult times with generating functions :|

Clarinetist

@TedShifrin Looks like the proof uses induction. Dang, I'm rusty. I haven't used induction or even thought of induction in two years.

Ted Shifrin

17:19

Yeah, @Clarinetist, but just understand it, rather than worrying about formalities.

The algorithm one does is recursive. That's all induction is.

Clarinetist

I'm doing proofs mainly because I think my mathematical chops are awful.

Ted Shifrin

I can suggest more interesting linear algebra proofs than that.

Chris's sis

@r9m I'd recommend yoou to use some of the integral representations of the harmonic number, the ones usually coming from digamma function. Then you're immediately done (I think).

Clarinetist

@TedShifrin Hmm, ideas?

Ted Shifrin

I have a book full of 'em, @Clarinetist. :) How far have you gotten on topics?

r9m

17:21

@Chris'ssis I'm stuck at a point where if I knew the generating function for $\displaystyle a_n = \int_0^1 (1-x)^{n-1}(1+x)^{n}\,dx$ I'll be done immediately :)

Clarinetist

@TedShifrin Just reviewing what a basis is, about to hit orthogonal complements, null spaces, eigenvalues, and gonna learn about positive/nonnegative definite matrices, and generalized inverses. I'm really surprised statistics people don't use linear maps, but I'm definitely up for a linear map proof

Mike Miller

@Ted: He has now asked a completely different question in the comments. :P

Ted Shifrin

LOL @MikeM .... Well, that's how learning happens around here :)

That was actually an interesting question, I thought.

pjs36

Please tell me this Lickorish theorem is about torus knots, @MikeMiller... I just want it to be true.

Clarinetist

I read the worst piece of actuarial propaganda yesterday:

"Even if you take a non-actuarial role mid-career or never actually end up working as an actuary, you will still be better off for pursuing actuarial science. Actuaries build an incredible skill set throughout the process of taking exams. There is no question that the exams build your analytical skills and work ethic."

Ted Shifrin

17:29

Well, here are a few for you, @Clarinetist. 1. Prove that $\dim (U+V) = \dim U + \dim V - \dim (U\cap V)$ ($U,V$ subspaces of a given vector space). Hint: Start with a basis for $U\cap V$. 2. Prove that $(U+V)^\perp = U^\perp \cap V^\perp$. 3. Suppose $T\colon\Bbb R^n\to\Bbb R^n$ is a linear map with $\text{rank}(T) = \text{rank}(T^2)$. Prove that $\ker(T^2)=\ker(T)$ and that $\ker(T)\cap\text{im}(T) = \{0\}$. Deduce that $\ker(T)\oplus\text{im}(T) = \Bbb R^n$.

Mike Miller

@pjs36 No, if you want to realize a 3-manifold as surgery on knots, those knots will usually be very complicated, and indeed you can't realize every 3-manifold as surgery on knots. Sometimes you need to use links.

Clarinetist

Oh boy. @TedShifrin Thanks. Got some definitions to relearn but I need to do this!

Ted Shifrin

I can give you lots more, @Clarinetist. These aren't totally trivial.

r9m

Time to watch F&F7 :D .. (waited for 12 hrs to download this :P acursed internet speed :[ ..)

Clarinetist

@TedShifrin Suppose $\mathscr{V}$ is a vector space, $U \leq \mathscr{V}$ and $V \leq \mathscr{V}$. It can be easily shown that $U \cap V \leq \mathscr{V}$ (don't feel like it lol). Thus $U \cap V$ is a vector space and has a basis.

Chris's sis

17:38

@r9m Well, there is a mistake in the right side. The sum starts from 0? I have some research that it might help to finalize that gloriously, but I need to go out soon.

@r9m can you tell me who posed that question? The author ...

Ted Shifrin

Sure, @Clarinetist. And you also should prove that a given set of linearly independent vectors in $U$ can be extended to form a basis for $U$.

Chris's sis

The key there might be the use of one of the logarithmic integrals I created in the past ... (well, almost sure)

r9m

@Chris'ssis see Problem 11832 - 04 - D. Knuth (USA) Here .. I don't have access to latest issues .. I follow Roberto Tauraso's page for the problems :)

at least that way I get to know the ones he solved .. :)

Chris's sis

@r9m I have the key to it!!!

r9m

@Chris'ssis alrighty!! :D

damn AMMs should atleast make the problem section free :( who the hell's gonna pay $12 for each f***ing page to read it from jstor?

Chris's sis

17:48

@r9m I'm going to use the fact that $(H_{2n} -H_{n})/n$ can be written as an elementary logarithmic integral. :-) (when I'm back home)

Clarinetist

@TedShifrin So I've never understood the "can be extended" phrase (sadly). Does this mean that if you have a linearly independent set of vectors $W \subset \mathscr{V}$, there is some $W^{\prime} \subset \mathscr{V}$ (also linearly independent) such that $W \cup W^{\prime}$ is a basis for $\mathscr{V}$?

Ted Shifrin

Yes ... @Clarinetist ... or, more simply, you can take your list in $W$ and add some more vectors so as to obtain a basis.

r9m

@Chris'ssis oo! it can be? cool!!

oh! indeed!

Chris's sis

@r9m with log(1+x) ::D::D:D:D:D:

I'm out now.

anon

@Clarinetist your comment suggests you've seen the phrase before. full disclaimer: when I hear the phrase "can be extended," I think of a different usage of the phrase. namely, choosing one vector in Y for each basis vector in X wrt a chosen basis determines a unique linear map X->Y which sends the basis vectors of X to the chosen vectors in Y. we say that our original function on the basis has been "extended linearly"

r9m

17:51

yess!!

@Chris'ssis bye!! take care :-)

Clarinetist

@TedShifrin Am I allowed to assume $U$ is finite-dimensional?
@anon Thanks, that does make sense

Mike Miller

@anon: I've heard this phrase before to mean you can add more vectors to a basis for a subspace to make a basis for the whole space

Ted Shifrin

Yes, assume everything finite-dimensional, @Clarinetist.

anon

@MikeMiller sure, I have too

Ted Shifrin

That's what Clarinetist and I were doing, Mike, of course.

Mike Miller

17:54

@anon ok

Ted Shifrin

Extending functions (whether linear or not) is another matter. But perhaps a more important matter. Even Balarka was trying to do that :P

EvilGoat

Hi guys

Clarinetist

@TedShifrin Suppose $U$ has dimension $n$. Then every basis for $U$ has $n$ vectors. Consider $W \subset U$ of $m < n$ vectors. What I need to show is that there exists $W^{\prime}$ with $n-m$ vectors such that $W \cup W^{\prime}$ is a basis for $U$.

Ted Shifrin

I don't know if we need an evil goat. Enough gets our goat in here ! :D

EvilGoat

I need some help with an assignment because I cannot understand how to aproach this: "Express a positive number A as a product of three positive numbers so that their sum is minimum." This is the most vague thing I have encountered in my life, any ideas?

Ted Shifrin

17:57

Assuming the vectors in $W$ are linearly independent, of course, @Clarinetist.

@EvilGoat: Write a math equation that says $A$ is the product of three numbers.

Clarinetist

Now how to approach this...

JMoravitz

If someone is familiar with this problem, I would appreciate a hint as to how to approach it. I'm studying $l^p$ spaces and am trying to prove Holder's Inequality. The notes I'm using to study presents it as a 3 step exercise and ashamedly the first step is giving me some trouble. It asks to prove that if $0<\theta<1$ and $t\geq 0$ then the following inequality holds: $$t^\theta\leq \theta t + (1-\theta)$$

EvilGoat

@TedShifrin $A$=xyz . . . so I take the smallest possible x,y and z I guess? I thought of this, it seems dumb

anon

@JMoravitz the RHS is 1+(t-1)theta, so probably the newton-binomial expansion. probably want different restrictions on t for that to work though.

Ted Shifrin

No, @EvilGoat, the smallest possible would be $0$, and that wouldn't work. So, given $xyz=A$, what are you trying to do?

heya @JMoravitz :)

EvilGoat

18:01

@TedShifrin well i take the smallest possible x,y,z that are positive as in 0.'a trillion zeroes here"...001, I don't get it honestly

Ted Shifrin

No, @EvilGoat: Surely you've done max/min problems before?

JMoravitz

Hi @ted, I'll look into that @anon, so you suggest applying the log to pull the exponent down and try working with the rhs into some sort of expansion (mercator / binom, etc). I'll fiddle around with those and see what happens

anon

yeah, using t=1+(t-1)

r9m

@Chris'ssis Holy Crap!! I was over complicating things! One word from you and I'm done! :D Awesome!!!!!

EvilGoat

@TedShifrin well the theory and problems up to this point in the book (at least in this chapter - equations with two variables) are mostly about finding the infimums and supremums of the equations and stuff like that, so I figure it has to do something with that, but I can't figure out a proper approach

Ted Shifrin

18:04

But you did dozens of max/min problems where you took an equation like $xyz=A$ and substituted to get rid of one of the variables ...

EvilGoat

@TedShifrin nope, there's nothing like that there

Ted Shifrin

Did you take single-variable calculus?

EvilGoat

it's only with two variables and most stuff is done with derivatives

Ted Shifrin

Correct.

EvilGoat

So, I should derivate for z to get it out of the way for example? Still what I am left with is about the same problem with two variables...

Ted Shifrin

18:06

First, what function are you trying to minimize?

Clarinetist

Hmm, @TedShifrin, it looks like Insel proves this statement using some "replacement theorem," which I have never heard of. Looks like an induction argument. Nevertheless I'm going to try to dissect it.

The idea is, we have this linearly independent set $W$ with $m < n$ vectors. Perform induction on $m$.

If $m = 0$, boring: we have $W = \varnothing$ and take $W^{\prime} = U$.
If $m = 1$... wait, it looks like it uses... is it strong induction? Dang. *starts reviewing*

EvilGoat

I am not given any function, but what I assume from the problem I should minimize: $f(x,y,z,)=x+y+z$ where all x,y,z are positives

Ted Shifrin

Yes, the replacement theorem is one of the elegant ways of proving that dimension makes sense.

Correct, @EvilGoat. Except now it'll turn into a function of just $x$ and $y$ because ...

Clarinetist

Wait no, this is just regular induction

Never mind

with $m = 0$ as the base case

Now assume that the statement is true for some $m \geq 0$.

Owatch

Just raided Staples, took the last of my preferred notepads.

Clarinetist

18:09

Consider a linearly independent set with $m + 1$ vectors.

Remove one of those vectors, you have a linearly independent set yet again.

EvilGoat

@TedShifrin so when I derivate, I get $df/dz=x+y+1$ and then I should find a local minimum or something like that?

Clarinetist

Call this set of $m$ vectors $V$. By the induction hypothesis, there exists a set $V^{\prime}$ with $n - m$ vectors such that they form a basis for $U$. [Insel goes on to say that $m \leq n$, why?] @TedShifrin

Ted Shifrin

No, @EvilGoat, you have $F(x,y) = x+y+A/xy$.

Clarinetist

OH DUH

@TedShifrin Stupid question, sorry :P

EvilGoat

@TedShifrin uhmmm can you elaborate on how that happens? This actually helps me though, I can see how to do this afterwards probably

@TedShifrin nvm see what you did

took me a minute

columbus8myhw

18:17

Can you guys find a function $f:\Bbb R\to\Bbb R$ that's continuous and infinitely differentiable such that $f(-x)+f(x)=f(x^2)$?

Ted Shifrin

@Clarinetist: Things seem muddled to me. Seems like you want to start with your linearly independent set $V$ and a basis $B$ for the subspace, and decide which things in $B$ to add to $V$ or which things to remove from $B$ when replace them with things in $V$. ... To be honest, I was more interested in your using this result.

Clarinetist

@TedShifrin Ah okay

Ted Shifrin

OK, guys, I'm going for a walk. Back later.

Clarinetist

@TedShifrin So now back to the original problem

2. Prove $(U + V)^{\perp} = U^{\perp} \cap V^{\perp}$

This is screaming a containment proof

By definition, $(U + V)^{\perp} = \{x \in \mathscr{V} \mid x \cdot y = 0, y \in U + V\}$... I think starts looking

Okay, that looks right

and $U^{\perp} = \{x \in \mathscr{V} \mid x \cdot y = 0, y \in U\}$, $V^{\perp} = \{x \in \mathscr{V} \mid x \cdot y = 0, y \in V\}$

Let $x \in (U + V)^{\perp}$.

Then for all $y \in U + V$, $x \cdot y = 0$.

Now here's where I might mess up...

Since $y \in U + V$, there exist vectors $u \in U$ and $v \in V$ such that $y = u+v$

So we have $x \cdot (u+v) = 0$.

Now is the inner product distributive? Hmm checks

anon

if x annihilates U+V, then in particular it annihilates U and it annihilates V

since U and V are subsets of U+V

Clarinetist

18:27

We have $x \cdot (u + v) = x \cdot u + x \cdot v = 0$ and furthermore, $x \cdot u \geq 0$ and $x \cdot v \geq 0$, so that $x \cdot u = x\cdot v = 0$. Hence $(U+V)^{\perp} \subset U^{\perp} \cap V^{\perp}$, right?

[dang, I have not looked at the definition of an inner product in years]

anon

what makes you think $\langle x,u\rangle$ or $\langle x,v\rangle$ are nonnegative? (I refuse to use $\cdot$ for inner product)

Clarinetist

@anon Yeah, I gotta lookup the code for those. Isn't it by definition?

anon

take a peek at the definition

also, have some geometric intuition for inner products too

Clarinetist

Oh wait

$\langle x, x\rangle \geq 0$

Hmm

anon

use my logic

Karim Mansour

18:31

hi guys

Owatch

How should I rewrite $\frac{1}{1-2sin^{2}t}$ ?

So that sin is on top?

I know I can write sin as csc, but am not sure about the rest.

Karim Mansour

multiply by sin^2 top and button so you get try that

anon

don't see how s^2/(s^2-2s^4) is any help @Karim

Clarinetist

@anon Ah, I see, you're using the fact that $u + v \in U + V$ is arbitrary and $U, V \subset U + V$, right?

anon

yes

you don't need to mention the first thing, just that $U,V$ are subsets

Karim Mansour

18:33

well it will put sin at top @anon

Clarinetist

@anon So basically, since $U, V \subset U + V$, it follows that $\langle x, u+v \rangle = 0$ implies that for any $u_1 \in U$ and $v_1 \in V$, $\langle x, u_1\rangle = \langle x, v_1 \rangle = 0$. Hence $(U+V)^{\perp} \subset U^{\perp} \cap V^{\perp}$.

Owatch

@KarimMansour Well, that gives me $\frac{1-2sin^{2}t}{(1-2sin^{2}t)^{2}}$

anon

@Clarinetist there is really no need to mention u+v or put subscripts on u or v

Karim Mansour

what do you want to reach @Owatch

anon

@Owatch Karim said to multiply by sin^2, not by 1-2sin^2

Clarinetist

18:37

@anon Ah, k. Thank you!

Owatch

Oh, my bad.

Karim Mansour

yeh @Owatch

anon

but it still remains to be seen what it is you truly seek @Owatch

just out with it

Owatch

I can't tell you, it's top secret.

Clarinetist

Quite clever... but of course, I haven't really done a proof in a year :P

Now we must show $U^{\perp} \cap V^{\perp} \subset (U+V)^{\perp}$.

Let $x \in U^{\perp} \cap V^{\perp}$. Then for all $u \in U$, $v \in V$, $\langle x, u\rangle = \langle x, v\rangle = 0$.

pjs36

18:40

@Owatch Recall: $1 - 2\sin^2(x) = \cos(2x)$, or is that where it came from?

Karim Mansour

yeah

Owatch

It's where it came from.

Clarinetist

Doesn't this follow immediately from the inner product definition? @anon $\langle x, u \rangle + \langle x, v \rangle = \langle x, u + v\rangle = 0$?

Karim Mansour

I was about to suggest that too

anon

@Clarinetist yes, x annihilates any u+v in U+V because it annihilates the u and the v

Karim Mansour

18:40

but without knowing what he wants there is noway to tell which way is better to suggest @pjs36

anyway I will go afk to solve my physics assignment.

Clarinetist

Hence $U^{\perp} \cap V^{\perp} \subset (U+V)^{\perp}$ so that we can now say $U^{\perp} \cap V^{\perp} = (U+V)^{\perp}$. Success!! (with part 1 of 4 if I recall, lol)

Owatch

There's nothing secret about it, I'm just trying to solve a problem myself, and don't want to have it all answered for me. However, I just might give it all since I'm frustrated and not really getting too far.

Clarinetist

First containment proof that I've done in maybe 2-3 years. I am definitely not going back to actuarial science

pjs36

Well, you could try factoring out the $\sin^2t$, so that $1 - 2\sin^2t = \sin^2t(\csc^2t - 2)$... then I guess you'd have $\csc^2t$ in the top

Owatch

$\int{tsec^{2}2t *dt}$ . . .

Clarinetist

18:44

Anyway, as sad as this is, I need to take a break. Thanks @anon @TedShifrin . I shall hopefully remember to try that problem again later

Owatch

:(

Can I replace $sec^{2}2t$ with $1+ (tan2t)^{2}$?

pjs36

You should be able to evaluate $\int \sec^2(2t)\ dt$, so I'd just try integrating by parts.

Karim Mansour

maybe try by parts

Owatch

I was doing exactly that..

I'll try again

Karim Mansour

well just set the table for your derivative as t and the integral one should be $sec^2(2t)$

Owatch

18:47

It is.

Karim Mansour

so not integrate $sec^2(2x)$ that should give you

1/2tan(2x)

Owatch

I don't know how to solve it.

I just don't

I've tried like three identities, I've got a whole bunch of fractions that go nowehre

I can't find anything to substitute

abel

@Owatch, integrate by parts once. you don't need any subs.

Karim Mansour

yeah

it should be in 1 step

Owatch

What do you mean IBP? It's an IBP question, I can't integrate $dv$ to get v.. .

I can't do it unless I have v.

pjs36

18:52

$\sec^2(t)$ is one of those "basic" functions whose antiderivative you just have to remember: $\frac{d}{dx} \tan(x) = \sec^2(x)$. The $2x$ instead of $x$ just means that you need a factor of $1/2$ (from $u$-substitution, if you like)

abel

$\int t \sec^2 2t = \frac 12 \int t d\, tan 2t = \frac 12 t \tan 2t - \frac 12 \int \tan 2t dt $

Karim Mansour

yeah

@Owatch pages.pacificcoast.net/~cazelais/187/tabular.pdf

Owatch

$\frac{ttan2t}{2}-\frac{1}{4}(ln|sec2t|)+c$

I had forgotten about $\int{sec^{2}x}$ . . .

Thanks for help .

Silenttiffy

19:14

What would be a good approach for proving: GCD(2^a * (2^b - 1), 2^a + 2^b - 1) = 1 for all integers a,b > 0 ?

anon

first off, there's cancellation: you can get rid of any factors of one side that are coprime to the other. in particular the right argument is odd, and so you can get rid of the 2^a on the left

see if you can go from there

Owatch

$\int{dx}$ ?

$\frac{d^{2}x}{2}$?

anon

are you trying to communicate?

Owatch

Is that the integral of dx?

Karim Mansour

no

Owatch

19:19

It's x

Maybe?

Karim Mansour

its x yes

Silenttiffy

GCD(2^b - 1, 2^a + 2^b - 1) = GCD(2^b - 1, 2^a) = 1. Anon, thank you very much!

anon

mmhmm

Ted Shifrin

@anon: Although dual space and annihilator are fundamental concepts, typically in most beginning linear algebra classes we do $V^\perp$ with the inner product :) Was Clarinetist keeping up with your annihilator stuff?

Karim Mansour

@TedShifrin !

Ted Shifrin

19:20

¿ @Karim ?

anon

@TedShifrin heh, probably not :)

Karim Mansour

:D

hi

Ted Shifrin

Heya

Karim Mansour

I didn't study dual space in my LA class

or annihilators

Ted Shifrin

No, I don't teach it in linear algebra, but I do do it in my multivariable math class because of differential forms.

Karim Mansour

19:22

I see

Chris's sis

@r9m Just back. :-) Since I wanna become like Ramanujan, I need to behave like him in terms of solving problems. :-)

Karim Mansour

so what is a dual

Ted Shifrin

the dual of a vector space is the vector space of all linear maps from it to the field of scalars.

Karim Mansour

oh I see

interesting

Ted Shifrin

So, if you have a dot product, every linear map is given by dot product with some vector, so it's no biggie :)

Karim Mansour

19:24

yeah

interesting concept

Ted Shifrin

The general premise is that one way to study spaces (linear algebra spaces or topology spaces) is to study maps in and out of them.

Karim Mansour

yeah to see how the spaces gets deformed and how the action on our object behaves yeah seems logical

Mike Miller

there are many ways in which spaces are entirely determined by the maps out of them

Karim Mansour

I guess thats where representation theory comes into play ?

Owatch

Eh.

Nevermind

I can check if with mr.wolf

Karim Mansour

19:29

@TedShifrin ?

or @MikeMiller

Ted Shifrin

Sure, among other things.

Mike Miller

i don't really think i'd say that's representation theory. representation theory is about actions; here we're just talking about maps

pjs36

Indeed, @KarimMansour, with groups that happens, and not necessarily just with representations; normal subgroups are just kernals of homomorphisms, from one viewpoint

Karim Mansour

yeah

is there a field that just looks at maps that studies math objects regardless of whatever it is ?

Mike Miller

i can't really parse the question

Ted Shifrin

19:31

If I understand the question, all fields incorporate these ideas.

Karim Mansour

I mean that studies maps into and out of spaces of any math object but in more generalized settings.

Ted Shifrin

Maybe category theory is exclusively about objects and maps :P

Mike Miller

actually if I understand it "category theory" really is the right answer

Karim Mansour

oh I see seems very interesting to study that

Ted Shifrin

shrug :D

pjs36

19:32

Three 'Aye's for category theory :)

Ted Shifrin

hey, @MikeM, do you listen to music on your iPhone?

Mike Miller

not usually but sometimes

Ted Shifrin

I just did on my walk and noticed it ate up a large amount of battery :(

Karim Mansour

well

teadawg1337

iPhones aren't exactly remarkable in terms of battery life, @TedS :(

Owatch

19:35

Depends what you have enabled.

Karim Mansour

you need to make your brightness not that high of your phone it will eat your battery.

that is what I do.

Ted Shifrin

Right, I know about brightness. And I have lots of stuff turned off.

Owatch

I don't have bluetooth, or any wireless on when out with my phone.

Ted Shifrin

I wasn't complaining except the 1/2 hour of music zapped about 20% of the battery.

Oh, maybe I should have turned wifi off.

Owatch

I'll see how much I lose when I go on my walk later.

Ted Shifrin

19:37

But I leave wifi and data on when I drive and don't notice any particular battery drainage.

Mike Miller

@TedShifrin: I tend to just bring my charger with me everywhere.

Ted Shifrin

LOL, a charger doesn't help in the woods, @MikeM :D

Owatch

It could well be the App.

Mike Miller

You just need to bring a generator too, @Ted.

Ted Shifrin

Anything else, Mike?

pjs36

19:37

@TedShifrin You need to stop listening to all that heavy metal; smooth jazz is way better for battery life.

Ted Shifrin

How 'bout a gourmet kitchen, while we're at it?

Owatch

Ted Shifrin

Well, I listen almost exclusively to classical, @pjs36, but this time it was Wicked, since I just bought that.

Owatch

I thought everyone took one with them on their walks.

Ted Shifrin

thanks, @Owatch :)

Mike Miller

19:38

That'd probably be nice, @Ted, but I think it won't help with the music.

Owatch

You'll get enough power from that, though you might not hear the music over the roar.

Karim Mansour

probably an ipod would be a better choice

to get

I don't have a phone since I don't have people to call aside I just have ipod for music whenever I go for a run

Ted Shifrin

I have my iPod, but I'd rather carry fewer things, and I have way more control with the phone than with the pod.

I didn't used to have a modern phone, @Karim, but when I move I plan to use it for everything (no landline, GPS, etc.).

Karim Mansour

oh I see makes sense @TedShifrin

Owatch

What happens if your phone breaks?

Then you lose your wireless.

Then you can't come tell us your phone broke.

Ted Shifrin

19:42

I intend to still have TV and wifi when I've moved, @Owatch.

Owatch

I though you planned to substitute it in place of your router.

Ted Shifrin

no, no, not at all ...

Owatch

It's fine then.

Ted Shifrin

I intend to run it off my home WiFi, in fact, once I'm settled ... but will use it in the car for GPS, music, etc.

OK, I didn't mean to derail the whole room. Back to math.

Owatch

I'm going to do one of those harder looking integrals.

Clarinetist

19:47

Ah, the magic number 165. Cooking chicken breast without burning it is difficult.

Ted Shifrin

I can give cooking lessons, @Clarinetist.

Of course, my first advice is that if you're going to pick a part of a chicken to cook, you should go with thighs :P

Clarinetist

@TedShifrin Neat! What do you recommend cooking thighs with? Parents used to always boil them in water

Ted Shifrin

yikes ... don't boil any meat in water ... unless you're making stock.

Clarinetist

I agree! It's awful :P

Ted Shifrin

all sorts of things ... braise with garlic, onions, mushrooms, wine ... do Asian style ... broil with a mustard, herb, breadcrumb topping ...

Clarinetist

19:50

My parents aren't the best at cooking. Usually they just turn the heat all the way up

Ted Shifrin

ugh

Clarinetist

Lol

Ted Shifrin

maybe I'll run a cooking tutorial service for math geeks :)

Clarinetist

The Mathematics of Cooking Hmm. :) I think that would be an interesting Math SE blog post

Ted Shifrin

to go along with the ribbing I've been getting, it'll have to be Sophisticated Cooking: A Geometric Approach :D

Clarinetist

19:53

Yeah, so since I left college... I've been trying to improve my chicken cooking

Generally I cook chicken in one of three ways:

1) Slice chicken breast, use in Asian stir-fry (easy)

2) Get some organic-ish chicken thighs, use for soup

3) Directly fry chicken breast onto pan

(w/o cutting)

Ted Shifrin

chicken breasts should be sautéed to brown the skin, but then braised in chicken stock or wine, for not very long, or they'll get dried out and yucky.

don't forget garlic and herbs and stuff :P

Maximilian

How would I start $3^{x-2}=7$?

Karim Mansour

I should also increase my cooking abilities I don't want to be those later those grad students who always eat outside the house and have bunch of noodles :D

Ted Shifrin

are you of Asian heritage, Clarinetist?

How do you think, @Maximilian?

Maximilian

I don't think i can change 7 to get 3^x so i can remove the bases

Clarinetist

19:55

Herbs ALWAYS. I have ALWAYS used herbs since I left for college. Yep, I'm Asian @TedShifrin

Ted Shifrin

Sure you can, @Maximilian, using what we discussed yesterday. $7 = 3^?$?

Tarragon is good with chicken breast, Clarinetist. So is a bit of thyme or lemon thyme.

Clarinetist

@TedShifrin Complete amateur here... what I usually do is use olive oil and lime to cook chicken breast at medium to medium-high heat (rather than extremely high heat like my parents did). Came to the the realization that chicken is best cooked at medium rather than extremely high heat about... 7 months ago? Hah

Yep, I use Tarragon, sea salt, pepper, maybe some paprika, and some basil

Maximilian

3^1.77=7

Ted Shifrin

Try braising after your first sauté. It'll be much more juicy. And do not overcook.

No calculators, @Maximilian. How did you find that?

Maximilian

Calculator

and the instructions say use calculator

Ted Shifrin

19:58

What did you punch into the calculator?

Maximilian

"Solve each equation. use calculator and round to tenths if necessary".. and I did 3^x and did random numbers until i got close enough

Ted Shifrin

What did we teach you about $\log_3(8)$ yesterday?

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$Mathematics$ Mathematics