Mathematics - 2015-03-25 (page 4 of 5)

7:00 PM

Something else... when we have that $f(0)=0$ and $f'(0)=0$, can we deduce something from that?? For example, can say something about the sign of $f'(x)$ ??

Hello @evinda @KajHansen !! Do you maybe have an idea??

Owatch

C = -4, B = -8, A = 9

teadawg1337

Your value for A is wrong

Owatch

Well.

A+B = 1

B = -8

A - 8 = 1

A = 1 + 8

That is how I got it. .

teadawg1337

No, the equation to be working with is $x^2+1=A(x-2)^2+B(x-3)(x-2)+C(x-3)$ for $\frac{A}{x-3}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$

Owatch

I was working with that.

teadawg1337

7:07 PM

Then how did you get to $A+B=1$??

Owatch

$A(x^{2}-4x+4)+B(x^{2}-5x+6)+Cx-3C$

= $Ax^{2}+Bx^{2} - 4Ax - 5Bx + Cx + 4A + 6B -3C$

teadawg1337

Leave the equation in its factored form, it's easier to work with that way

Owatch

Oh.

$x^{2}+1 = x^{2}(A+B)-x(4A+5B-C)+4A+6B-3C$

Well, that is what I was doing at the time.

teadawg1337

Not like THAT

Leave the equation as I stated, you're making it more complicated than it has to be

Owatch

You did ask how I got those answers, so I was trying to show you. I can just stop though.

teadawg1337

7:11 PM

Continue, I suppose

Owatch

In any case, A+B was matched to the coeffecient of $X^2$, which is 1.

-x has no coeffecient. So I made $4A + 5B -C = 0$

teadawg1337

No.... That's not how to use partial fraction decomposition at all...

Owatch

And then $4A+6B-3C$ = 1

teadawg1337

Maybe it's valid, I dunno.

I've never seen it done that way before.

Owatch

I would do it your way.

But it's all done on the notebook I have not gotten back, so I could not recall exactly what it was.

So I did it the way I was taught. . .

Which is annoying.

In any case, those three equations are to be solved for A, B, and C.

Which is what I was trying to do at the time.

But let me go back to the point where you do it differently and start there.

Okay I am back.

@teadawg1337 To this point

Vrouvrou

7:17 PM

someone has an idea please:math.stackexchange.com/questions/1206342/…

Owatch

Now I think you told me to find whatever would make the other two equations = to zero

And then solve for the remaining one

For A: x = 2 will make it zero. For B: x = 2, and x = 3, will make it zero. And for C, x = 3 will make it zero.

I cannot solve for B though? Because nothing will make both A and C zero.

teadawg1337

In some cases, you'll have to use the other values obtained and plug in a random number for $x$ and solve for the last coefficient

Owatch

What do you mean? I have B(x-3)(x-2) as the remaining piece to solve.

So I should plug in anything for x?

I will choose zero then.

B(-3)(-2) = 6B

Now how could I use A and C to solve for B?

Oh wait

Perhaps I can choose one of the ones that makes A or C zero

Then plug in A or C depending on which remains?

teadawg1337

Use the equation $x^2+1=A(x-2)^2+B(x-2)(x-3)+C(x-3)$. Since $A$ and $B$ are now known, plugging a value in for $x$ will turn this into an equation with only one variable to solve for

And no, using either $x=2$ or $x=3$ will make $B(x-2)(x-3)=0$

Owatch

You're right.

teadawg1337

7:28 PM

My first message should read $A$ and $C$ instead of $A$ and $B$

Owatch

I got -6.5 for B

teadawg1337

sighs That means either or both of your values for $A$ and $C$ are wrong

Owatch

I got A = 10

teadawg1337

And $C=?$

Owatch

C = -5 . .

Must be C then

teadawg1337

7:29 PM

No, both of those are correct.

Owatch

$x^{2}+1 = 10 + B(x-3)(x-2) - 5$

teadawg1337

No...

$A\ne A(x-2)^2$....

Likewise with $C$...

Owatch

Oh my bad.

$x^{2}+1 = 10(x-2)^{2} + B(x-3)(x-2) - 5(x-3)$

teadawg1337

... Do you really need my approval before continuing?

Owatch

I'm doing it

B = -9

teadawg1337

7:36 PM

Correct

Owatch

Standby

Mary Star

Something else... when we have that $f(x) \leq 0$, $f(0)=0$ and $f'(0)=0$, can we deduce something from that?? For example, can say something about the sign of $f'(x)$ ?? @DanielFischer @DavidWheeler

teadawg1337

@MaryStar It can be deduced that $f(x)$ has a global maximum at $x=0$

Well, no, the function could have other zeros as well...

Owatch

$10ln(x-3)-9ln(x-2)-\frac{5}{2-x}$

teadawg1337

Incorrect @Owatch, but close, try again

Mary Star

7:40 PM

That means that $f(0)$ is the maximum value of $f$. Since $f \leq 0$ can we say that $f=0$?? @teadawg1337

teadawg1337

@MaryStar No. For example, $f(x)=-x^2$ satisfies those conditions.

And it has a global maximum at $(0,\,0)$

Mary Star

Oh sorry...it is $f(x) \geq 0$ not $f(x) \leq 0$. @teadawg1337

Owatch

I can't copy paste it

But was it + $\frac{5}{2-x}$

Not -

teadawg1337

Then it has a global minimum at $(0,\,0)$ @MaryStar

Owatch

That's the only thing I think I caught wrong.

teadawg1337

7:44 PM

@Owatch You're still missing one tiny detal

Mary Star

What do get from the fact that $0$ is the minimum value of $f$ ?? @teadawg1337

teadawg1337

$f(x)$ must be an even function @MaryStar

Owatch

Goddamnit Tee dog

That doesn't Count

+c

teadawg1337

It doesn't count without the constant of integration @Owatch

Owatch

Is that a question?

teadawg1337

7:46 PM

??

It was a statement, implying that you will get points deducted for not including $+C$

Balarka Sen

@KajHansen You know anything about Rado's theorem?

Owatch

It does count.

+C, never forget. o7

Mary Star

So, can we not conclude somehow that $f(x)=0$ or something about the derivative $f'(x)$, for example the sign?? @teadawg1337

teadawg1337

@MaryStar Using the information given, no.

Owatch

I have put together a sheet to bring to the test.

teadawg1337

7:48 PM

Well... Let me think on it for a second

Owatch

He has permitted us to bring trigonometric identities, and so forth.

Balarka Sen

@Owatch What does calcII consist of?

Owatch

It consists of pain.

teadawg1337

Definitely cannot conclude that $f(x)=0$, I haven't thought about the derivative...

Balarka Sen

@Owatch Calculi hurts, definitely, but what's the syllabus?

Basic differentiation/integration/series etc?

Kaj Hansen

7:50 PM

A little @BalarkaSen

Mary Star

Ok... @teadawg1337

Owatch

7.1: Integration by parts. 7.2: Integration of trig. 7.3: Integration using Trig substitution. 7.4: Integration of rational functions by partial functions

We are now doing approximate integration.

Kaj Hansen

@MaryStar, not for functions $\mathbb{R} \rightarrow \mathbb{R}$. E.g. $f(x) = x^2$ and $f(x)=x^3$ both satisfy those properties.

teadawg1337

@KajHansen But $f(x)=x^3$ does not have a global minimum at $(0,\,0)$

Kaj Hansen

oh, I didn't know that needed to be a stipulation

Well x^2 and x^4 then

Owatch

7:52 PM

Time to practise trig substitution.

Balarka Sen

@KajHansen The number theorist guy I met a few days ago is working on something called monochromatic number theory, and said a lot of stuff about arithmetic combinatorics that went above my head. He mentioned this one Rado's theorem, though. What's this stuff about?

Kaj Hansen

The idea is that you color (partition) subsets of certain sets and see if they have a given property.

So let's start with Schur's theorem, a specific case of Rado's theorem.

Mary Star

The function that I am looking for is $E(t)$ of the answer of math.stackexchange.com/questions/1205836/… @teadawg1337 @KajHansen

teadawg1337

Disregard my statement that $f(x)$ must be even, I didn't think it through properly @MaryStar

Kaj Hansen

@BalarkaSen, you can see my discussion of Schur's theorem here: youtube.com/watch?v=VO4EFKctSFE

Balarka Sen

7:56 PM

let me have a look

teadawg1337

I have yet to study differential equations @MaryStar

Owatch

Solved one!

Another now.

Mary Star

Ok... @teadawg1337

Kaj Hansen

If I'm not mistaken, Rado's theorem generalizes Schur's theorem to give sufficient conditions for when that holds for polynomials in $\mathbb{Z}[x_1, x_2, ..., x_n]$. Schur's theorem is just one of these w/ the polynomial being x+y-z @BalarkaSen

Balarka Sen

this looks like a very interesting claim. i am gonna watch this video. rubs hands

Kaj Hansen

8:02 PM

I don't know how much Ramsey theory you've seen @BalarkaSen, but you may need to watch earlier videos. In particular, you need that $R(x_1, x_2, ..., x_n)$ is finite for $x_i \in \mathbb{N}$.

Balarka Sen

ah, i see you talk about a proof in the video. that's ok, i am not planning to read a proof since i am not familiar with Ramsey theory, but the theorem is surprising.

wonder how possibly can this interact with number theory...

Owatch

Another solved!

Balarka Sen

yeah, so i just looked at the paper and what they prove is that given an r-coloring of the integer lattice points in $\Bbb R^2$, there is always a triangle of area $T_r$ with monochromatic vertices @Kaj

Kaj Hansen

Sounds about right. Link?

Balarka Sen

actually, it seems it's a known result but they've given a different proof.

siba-ese.unisalento.it/index.php/notemat/article/viewFile/12864/…

ah, in here they prove something related about trapezoids.

ah, i recall : the number theoretic thing comes from van der waerden's theorem. it states that given an r-coloring of {1, 2, ..., n} there always exists a monochromatic k-tuple the elements of which are in arithmetic progression.

fun stuff.

Kaj Hansen

8:20 PM

Green-Tao theorem

Such wut

Owatch

And another done.

Balarka Sen

@KajHansen indeed. i never knew that it comes from such a combinatorial background.

Owatch

I will go back and do more partial function problems.

Owatch

8:32 PM

$\int{\frac{1}{x^{3}-1}}dx$

How should I approach this problem?

teadawg1337

$x^3-1=(x-1)(x^2+x+1)$

Owatch

I tried factoring it on my calculator

Is just gave me: x1 = 1

Mary Star

Is someone familiar with computability, and especially with automata??

Owatch

Which I guess is (x-1) = 0

Could you explain how you factored that?

I guess it makes sense if you distribute it.. but

teadawg1337

It's a simple property of polynomials of higher order

$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\dots+x^{n-k}+\dots+x^2+x+1)$

Vrouvrou

8:37 PM

can someone help me on functional analysis ?

teadawg1337

Factoring on your calculator will just give you the zeroes of the function @Owatch...

Owatch

Yes it did not give me much.

Because there was only one zero.

The rest has no roots I guess.

teadawg1337

$x^2+x+1$ has two complex roots @Owatch.

Owatch

No real roots. ..

teadawg1337

Anyways.... Solve $1=A(x^2+x+1)+B(x-1)$

Owatch

8:39 PM

Isn't it Bx+c(x-1)?

teadawg1337

@Owatch Yes, my mistake.

Owatch

I was starting to distribute the Bx+C a bit, but I suppose I shouldn't do that right?

teadawg1337

I told you before, leave the equation in its factored form...

Owatch

I had changed it to $A(x^{2}+x+1) + B(x^{2}-x) + C(x-1)$ But I will not go that way.

David Wheeler

There is a famous "bogus proof" that goes: Suppose $a^2 + a + 1 = 0$. Then $(a - 1)(a^2 + a + 1) = a^3 - 1 = (a - 1)0 = 0$, so that $a^3 = 1$. Thus $a = 1$, and thus $a^2 + a + 1 = 3 = 0$.

ᴇʏᴇs

8:43 PM

Hi @DavidWheeler

Long time no see

Vrouvrou

@DavidWheeler hello, can you help me in functional analysis please ?

Owatch

I cannot make A zero.

teadawg1337

That is true, so plug a value in for $x$ once you have found $B$ and $C$.

David Wheeler

What should be done, is to solve $(Ax + B)(x - 1) + C(x^2 + x + 1) = 1$. This can be re-written as: $(A + C)x^2 + (B - A + C)x + (C - B) = 0x^2 + 0x + 1$.

Owatch

I got: A = 1/3, B = -1/3, C = -2/3

teadawg1337

8:51 PM

@DavidWheeler That works as well, I suppose it comes down to personal preference

@Owatch correct

Now integrate.

David Wheeler

I prefer to write the numerator of a quadratic factor as linear, and the numerator of the linear factor as constant.

Owatch

What a terrible integral.

teadawg1337

@Owatch I agree, it's not pretty by any means.

Owatch

$\int{\frac{1}{x-1}}dx + \int{\frac{\frac{-1}{3}x+\frac{-2}{3}}{x^{2}+x+1}}dx$

what did you say about my integral???

teadawg1337

I said it was incomprehensible, that was before you fixed the former into a fraction

Owatch

8:56 PM

Oh.

Jaken Herman

@Owatch in all fairness he redacted his statement haha

But I still think he wants to fight.

teadawg1337

Move the factor of $\frac{1}{3}$ in front of both integrals, and solve

@Jaken Nah

hi guys

Hello @Karim

Hey @KarimMansour

8:58 PM

I made it $\int{\frac{\frac{-1}{3}(x+2)}{x^{2}+x+1}}dx$

Shouldn't I move a three out front

David Wheeler

So $\int\dfrac{1}{x^3 - 1}\ dx = \dfrac{1}{3}\left(\int\dfrac{1}{x - 1}\ dx - \int\dfrac{x+2}{x^2 + x + 1}\ dx\right)$

Owatch

Thank you dahvid. But tee dog has already assisted me with this part.

teadawg1337

He's helping me explain, it's alright

Recall that $A=\frac{1}{3}$ @Owatch

David Wheeler

@Owatch I was explicitly illustrating a 1/3 comes "out in front", not a 3

Owatch

Ohh

David Wheeler

9:02 PM

A mistake like that can mean your answer is off by a factor of 9, not so good.

Owatch

So how did you move out the 1/3 from the second integral?

Isn't $\frac{\frac{1}{3}}{1}$

1/3 . . . . .. .

teadawg1337

=3? HELL no

David Wheeler

what is ANYTHING divided by 1?

Owatch

Oh my,

David Wheeler

$a/1 = a$ (even if $a$ is a fraction)

Owatch

9:04 PM

Edited

That was such a silly mistake, by bad.

David Wheeler

You can easily check: $\dfrac{1}{3}\left(\dfrac{1}{x-1} - \dfrac{x + 2}{x^2 + x + 1}\right) = \dfrac{1}{3}\left(\dfrac{x^2 + x + 1}{x^3 - 1} - \dfrac{(x + 2)(x-1)}{x^3 - 1}\right)$

$ = \dfrac{1}{3}\left(\dfrac{x^2 + x + 1 - x^2 - x + 2}{x^3 - 1}\right) = \dfrac{1}{3}\dfrac{3}{x^3 - 1} = \dfrac{1}{x^3 - 1}$

Owatch

Down to $\int{\frac{1}{x^{2}+x+1}}dx$

David Wheeler

I always "back-substitute" to test for errors.

Where did the x + 2 part go?!?

Owatch

I split it.

That is the second piece, the 2 was moved out in front

$\int{\frac{x+2}{x^{2}+x+1}}dx$ = $\int{\frac{x}{x^{2}+x+1}} + \int{\frac{2}{x^{2}+x+1}}dx$

David Wheeler

If you split it, you should have $-\dfrac{1}{3}\left(\int\dfrac{x}{x^2 + x + 1}\ dx + \int \dfrac{2}{x^2 + x + 1}\ dx\right)$

But I don't really recommend splitting them up

teadawg1337

9:13 PM

I wouldn't recommend splitting them either

Owatch

Oh

I have to go.

Goodbye.. :<

Thanks for all the help today

teadawg1337

Ok... Be sure to finish that integral

Jaken Herman

So long and thanks for all the fish

teadawg1337

I'm off to play some Torchlight II. I need to blow off some steam by killing hordes of monsters

Vrouvrou

9:26 PM

@DavidWheeler hello

meer2kat

9:38 PM

A wild meer appeared! What will you do?

Mike Miller

about the same as I was before

David Wheeler

David uses: charm!

meer2kat

@DavidWheeler It was super effective!

Vrouvrou

if someone have an idea :math.stackexchange.com/questions/1206617/weakly-convergence

thank you

Huy

@MikeMiller: Basic geometry question, if I enlarge a straight pyramid by some factor $\lambda$, does the volume correspondingly increase by $\lambda^3$?

Eulerian Adventurer

9:51 PM

Hello

Mike Miller

Yes, @Huy. This is linear algebra.

Eulerian Adventurer

Can we post problems here?

Huy

@MikeMiller: I thought so. Can you give me a pointer? I've been thinking about it for a bit but don't quite see it.

Mike Miller

Given a matrix $A$ and an "object" $S \subset \Bbb R^n$ (a closed $n$-manifold should be good enough), then $\text{vol}(AS) = \det(A)\text{vol}(S)$. The idea is to approximate $S$ by cubes and then apply $A$ to this approximation; by taking a limit as the cubes get smaller we obtain the Riemann integral.

Huy

@MikeMiller: I need a more basic argument without the determinant / integral.

Mike Miller

9:54 PM

Bah humbug.

Huy

(I know what you just said though)

Mike Miller

The answer is yes, I don't know how one of the ancient Greeks would prove it.

Ark

Apparently no one can solve this problem. Would anyone like to give it a shot? math.stackexchange.com/questions/1206376/…

Huy

I have an exercise for high schoolers where a certain pyramid and its volume is given and I need to find the pyramid with a fraction of the other pyramid's volume. And I don't quite know how to do it with the tools of a high schooler.

(the height would suffice)

Mike Miller

I don't understand. You're scaling the pyramid by $\lambda$. Is it not obvious that the height also scales by $\lambda$?

Huy

9:57 PM

@MikeMiller: It is somewhat obvious to me but I doubt it will be to them which is why I'll need an argument they understand.

Mike Miller

I mean, you're going to have to trade off "too wordy" for "not rigorous enough". You could rephrase my previous sentence as "Compose the embedding $\Delta^3 \hookrightarrow \Bbb R^3$ with the projection $\Bbb R^3 \to \Bbb R$. That's the height map. Fitting in $\lambda I: \Bbb R^3 \to \Bbb R^3$ in the middle, we obtain the map $\lambda h$, and thus the height scales by $\lambda$."

Huy

lol

Now you're just making fun of me.

Mike Miller

No. I just think that if you have a scale to balance here. There are two ends: the above, which is a proof, and the previous thing, which is "well, it's obvious".

I dunno what the best balance is. You'd know better than me.

Ark

The function $f:\mathbb{Q}\rightarrow \{0,1\}$ is defined as $f(a)=0$ if there do not exist integers $x$, $y$ and $z$ for which $a$ is expressible in the form $$\frac{xy+yz+zx}{x^2+y^2+z^2},$$ and $f(a)=1$ otherwise. Show that $$f(a)=f(0.5-a).$$

Huy

I have a proof from my years in high school where I used the theorem on intersecting lines but it's not so nice.

I thought there would be an obvious proof using some ratios.

Ark

10:04 PM

@Huy What is a straight pyramid. I only know high school level math so any solution I can come up with is what you're looking for.

David Wheeler

If $V = \dfrac{lwh}{3}$ and all of those factors scale by $\lambda$ (that is, we have simiiar pyramids) then $V_1 = \dfrac{\lambda l\lambda w\lambda h}{3} = \lambda^3 V$

Huy

@Ark: The axis of the pyramid is perpendicular to its base area.

Mike Miller

He seems unconvinced that those factors scale by $\lambda$, @David.

Ark

@Huy Thanks, I'll give it a go

Huy

@MikeMiller: I'm not unconvinced but I doubt any of my students would be convinced.

Danu

10:05 PM

Hi guys, I'm looking for a name of a professor that I'm trying to remember

Mike Miller

I know, @Huy. Bad phrasing on my part.

Danu

It's a Japanese guy who, recently (at least post-2000) developed an entirely new (branch of?) Teichmuller theory

He says that it can be applied to solve a very famous problem---forget the name :(

But he refuses to really travel to spread his ideas

Someone must know who I'm talking about

Ark

@Huy Wait by pyramid do you mean square-based?

Huy

@MikeMiller: Something else entirely: Do you know which contributions of Nash lead to his Abel prize? Would be fun to tell my students in their Game theory course tomorrow.

@Ark: No, it's actually triangle based, but should work for any base.

Mike Miller

You're thinking of Shinichi Mochizuki. It's related to part of Grothendieck's anabelian geometry program. I think it's wildly unfair to say that he refuses to really travel to spread his ideas; he prefers to stay in his home country, yes, but he is working with mathematicians there to help them better understand it.

@Huy: Nothing related to game theory.

Danu

10:07 PM

Nash's paper that established the concept of "Nash equilibrium" is VERY funnily short :D

(less than a page)

@MikeMiller Yes, awesome, thanks :)

Huy

@MikeMiller: I know, some connection between PDEs and analytic geometry or so?

Mike Miller

I'll find the excerpt the AMS sent me.

Oops; my game theory claim was wrong.

The Norwegian Academy of Sciences and Letters will award the 2015 Abel Prize to John F. Nash Jr. (above left), Princeton University, and Louis Nirenberg (above right), Courant Institute, New York University, "for striking and seminal contributions to the theory of nonlinear partial differential equations and its applications to geometric analysis."

Although Nash and Nirenberg did not formally collaborate on any papers, they influenced each other greatly. The Abel committee notes that: "Their breakthroughs have developed into versatile and robust techniques that have become essential tools …

The biography on the Abel Prize website should help with specifics.

Ark

@Huy I'm definitely missing something. For convenience let x denote lambda. If you let the volume of the initial pyramid be V=bh/3, where b=base area and h=length of altitude from vertex opposite said base to the same base, then scaling everything by x means the new base area is x^2b and new height is xh. So new volume is V_1=(x^2b)(xh)/3=x^3(bh/3)=x^3V.

Huy

@MikeMiller: Sorry for my ignorance but what exactly is geometric analysis about?

Mike Miller

Hell if I know. The catchphrase I've heard is in two directions: using geometric ideas to study PDE (someone told me there was recent progress on a certain hyperbolic equation arising from physics that came from studying foliations?), and using PDE to study geometry (gauge theory, i.e. Yang-Mills and Seiberg-Witten theory, goes here).

Huy

10:15 PM

Wikipedia claims Geometric analysis is a mathematical discipline at the interface of differential geometry and differential equations.

I see.

Doesn't sound like anything I'd be good at.

:-(

Mike Miller

Boo to that. Don't put yourself down.

Huy

Nah, I can safely say Differential Geometry is not my strength.

Mike Miller

At present, the starboard has something with 0! stars, 1! stars, 2! stars, and 3! stars, and some random dude with 1 star at the bottom.

Huy

And PDEs, I don't even want to talk about.

Mike Miller

I at least need to try to be good at it. I'm morphing into a low-dimensional topologist, and gauge theory is a key tool in the theory of 4-manifolds.

Huy

10:19 PM

Good luck.

Danke schoen.

Bitteschön.

Thanks for the stars

@MikeMiller This always "sounds" hilarious to me as a Dutchman, since "schoen" is an actual Dutch word: It means "shoe" :) Of course, I'm aware of the German umlaut alternative spelling thing

Mike Miller

Pronounced the same?

Danu

10:28 PM

No, totally different

You can try typing "shoe" into google translate and using the audio option

Mike Miller

I think it's actually incorrect to write schoen instead of putting an umlaut above the o. I just don't have that on my keyboard.

Danu

The pronunciation is pretty okay :)

@MikeMiller No, it's pretty widespread in Germany as well, don't worry about it

Also, it's easy to make it possible to switch between e.g. US and GER keyboards (I do it all the time, now that I live in Germany) using some settings on your computer

Mike Miller

I use a chromebook, which is very lightweight; there are probaby ways to do it if I download extensions.

infinitesimal simplicio

Why did you choose to go to school in Germany @Danu?

ᴇʏᴇs

I would go to school in Germany if I could

Danu

10:33 PM

@infinitesimalsimplicio My school is significantly better than what I could've gotten in Holland, I think.

Also, I liked the prospect of an internationally oriented degree in mathematical physics.

infinitesimal simplicio

I see.

Gato

someone for complex analysis ? :)

Huy

10:53 PM

@MikeMiller: A Chromebook? Really? I'm looking at different possibilities to replace my current laptop and I haven't heard good things about the Chromebook. Can you recommend it?

Mike Miller

Yes, it's the best purchase I've made in a while. The only downsides are that it can't do high-power things (playing games, eg), and that to do most anything you need an internet connection, and that some native windows apps don't have chromebook counterparts. (The only ones of this last thing that has affected me is photoshop and gp-pari.)

Danu

Using apps on a computer is just the worst prospect for me.

Mike Miller

You use apps on your computer every day...

Huy

@MikeMiller: Can you install MikTeX or alike?

Mike Miller

Not as far as I know. I use online equivalents; one that turns google docs into a latex editor and compiler, and overleaf for compiling complicated documents.

This has not affected me.

Huy

10:58 PM

@MikeMiller: So no offline compiling.

Mike Miller

No. If you're regularly in places without an internet connection it might be trouble.

Huy

Dropbox integration?

Mike Miller

This is, by far, the biggest problem with a Chromebook. I have no problem with it.

Look at the google apps store. I bet there's a dropbox app.

Huy

How are they integrated then? Can you save the stuff locally or just view it in the cloud?

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$Mathematics$ Mathematics