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

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5:00 PM

@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

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

Morning, by the way.

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 $i \neq j$

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

@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.

(Marvellous results all over the place)

:D

@r9m ^^^

5:09 PM

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

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

@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

@r9m hahahahaha :-))))))

@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.

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

5:11 PM

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!

@Chris'ssis like this one :P

@r9m :D:D:D

Hmm

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.

@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 $\log (...)^2$

5:14 PM

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?

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

@r9m It looks really nice.

Okay, thank you!

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

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

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

5:16 PM

@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

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

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

I stumbled upon it :)

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

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

5:19 PM

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

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

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

I can suggest more interesting linear algebra proofs than that.

@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).

@TedShifrin Hmm, ideas?

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

5:21 PM

@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 :)

@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

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

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

That was actually an interesting question, I thought.

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

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."

5:29 PM

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 $\ker(T)\oplus\text{im}(T) = \Bbb R^n$.

@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.

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

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

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

@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.

5:38 PM

@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 ...

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 $U$.

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

@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 .. :)

@r9m I have the key to it!!!

@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?

5:48 PM

@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)

@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 $\mathscr{V}$?

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

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

oh! indeed!

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

I'm out now.

@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"

5:51 PM

yess!!

yess!!

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

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

@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

Yes, assume everything finite-dimensional, @Clarinetist.

@MikeMiller sure, I have too

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

5:54 PM

@anon ok

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

Hi guys

@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 $U$.

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

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?

5:57 PM

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.

Now how to approach this...

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: $$t^\theta\leq \theta t + (1-\theta)$$

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

@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.

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 :)

6:01 PM

@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

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

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

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

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

@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

6:04 PM

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 ...

@TedShifrin nope, there's nothing like that there

Did you take single-variable calculus?

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

Correct.

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...

6:06 PM

First, what function are you trying to minimize?

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*

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

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 ...

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 $m \geq 0$.

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

6:09 PM

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

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

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

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

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

OH DUH

@TedShifrin Stupid question, sorry :P

@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

6:17 PM

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

@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.

@TedShifrin Ah okay

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

@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

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

since U and V are subsets of U+V

6:27 PM

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]

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

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

take a peek at the definition

also, have some geometric intuition for inner products too

Oh wait

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

Hmm

use my logic

6:31 PM

hi guys

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.

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

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

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

yes

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

6:33 PM

well it will put sin at top @anon

@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 $(U+V)^{\perp} \subset U^{\perp} \cap V^{\perp}$.

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

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

what do you want to reach @Owatch

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

6:37 PM

@anon Ah, k. Thank you!

Oh, my bad.

yeh @Owatch

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

just out with it

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

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 $u \in U$, $v \in V$, $\langle x, u\rangle = \langle x, v\rangle = 0$.

6:40 PM

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

yeah

It's where it came from.

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

I was about to suggest that too

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

6:40 PM

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.

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)

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.

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

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

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

6:44 PM

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

:(

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

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

maybe try by parts

I was doing exactly that..

I'll try again

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

6:47 PM

It is.

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

1/2tan(2x)

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

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

yeah

it should be in 1 step

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.

6:52 PM

$\sec^2(t)$ 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)

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

yeah

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

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

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

Thanks for help .

7:14 PM

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 ?

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

$\int{dx}$ ?

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

are you trying to communicate?

Is that the integral of dx?

no

7:19 PM

It's x

Maybe?

its x yes

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

mmhmm

@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?

@TedShifrin !

7:20 PM

¿ @Karim ?

@TedShifrin heh, probably not :)

:D

hi

Heya

I didn't study dual space in my LA class

or annihilators

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

7:22 PM

I see

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

so what is a dual

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

oh I see

interesting

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

7:24 PM

yeah

interesting concept

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.

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

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

I guess thats where representation theory comes into play ?

Eh.

Nevermind

I can check if with mr.wolf

7:29 PM

@TedShifrin ?

or @MikeMiller

Sure, among other things.

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

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

yeah

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

i can't really parse the question

7:31 PM

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

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

Maybe category theory is exclusively about objects and maps :P

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

oh I see seems very interesting to study that

shrug :D

7:32 PM

Three 'Aye's for category theory :)

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

not usually but sometimes

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

well

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

7:35 PM

Depends what you have enabled.

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

that is what I do.

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

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

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.

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

7:37 PM

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

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

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

It could well be the App.

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

Anything else, Mike?

7:37 PM

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

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

user image

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

I thought everyone took one with them on their walks.

thanks, @Owatch :)

7:38 PM

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

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

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

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.).

oh I see makes sense @TedShifrin

What happens if your phone breaks?

Then you lose your wireless.

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

7:42 PM

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

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

no, no, not at all ...

It's fine then.

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.

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

7:47 PM

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

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

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

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

I agree! It's awful :P

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

7:50 PM

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

ugh

Lol

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

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

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

7:53 PM

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)

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

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

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

are you of Asian heritage, Clarinetist?

How do you think, @Maximilian?

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

7:55 PM

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

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.

@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

3^1.77=7

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

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

Calculator

and the instructions say use calculator

7:58 PM

What did you punch into the calculator?

"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

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

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