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6:01 PM
<--- enjoys being too quiet
 
@Ted I've got a question.
Ah, @Robert, I've got a question for you too :P
@TedShifrin Sanity-check : It is true that if X is a CW complex, with homotopy equivalent subcomplexes A, B, then X/A and X/B are also homotopy equivalent, right?
I guess it is.
 
Would someone kindly look at my attempted answer to this question: math.stackexchange.com/questions/1089188/…
 
Hello!! Which are the elements of $\mathbb{F}_{2^2}$ ??

There are $4$ elements, right??

For each element $a$ in $\mathbb{F}_{2^2}$ it stands that $a^2=a \Rightarrow a^2 \in \mathbb{F}_{2^2}$, right??
 
6:16 PM
someone downvoted my answer. can i get an explanation?
 
Recommended reading for PDE is Hormander's first PDE book, @Ted. He and I meet again...
 
Hello @BalarkaSen!! Do you maybe have an idea??
 
People.
How are you doing today?
 
6:26 PM
crappy, @Lord_Farin
 
i ask for an evalutation of my answer in chat and then just get a downvote with no explanation. frustrating!
 
@BalarkaSen Why's that?
 
@EricGregor I didn't downvote, and I don't think there's much reason to believe anyone here did. It could be coincidental.
 
@RobertCardona I have a question about a problem on Atiyah-MacDonald.
 
@MikeMiller, how long did it take you to work through it? Did you do most/all of the exercises?
 
6:29 PM
Shall I go to your room and ask it there?
 
@BalarkaSen what's the question ?
Yeah! If it's still open!
 
If what is open?
You mean the room?
 
Yeah the room
 
I worked on it on and off for a while, probably about a full year in total. I think with concentrated study it shouldn't take more than 3 months. I did most of the exercises. Part of the reason the book is so good is the exercises; I wouldn't suggest bothering with it unless you do them,.
 
I think it's not freezed yet ;)
 
6:31 PM
Yeah, I gave myself 6 weeks to do it. I'm halfway through but have been slacking on the exercises :/ some take longer than expected and I end up having to develop lots of lemmas for them.
 
@MikeMiller it is most likely someone here because the question had no activity or comments for a day until a minute after i posted here
 
@RobertCardona the exercises are very hard
 
Yeah. The exercises in the first chapter were much easier, so I planned my schedule based on that.
 
That's good though, @Robert. (Not the slacking part, but for everything else.) It's a difficult book, super worth the effort.
 
The exercises in the first chapter are hard. I have just finished it. ;)
 
6:34 PM
I had seen most of the chapter one results already, so it wasn't to bad for me
 
ah. well, i have just started it and zariski topology is so weird.
 
K, I'll keep at it mike and I'll double my effort on the exercises!
 
@RobertCardona i have asked it.
 
Yeah! They are at first
 
Anyone has a function $f:[0,1]\to[0,1]$ so that $\forall0\le x\le1,f^{-1}(x)$ is a tuple (2 elements) ?
@BalarkaSen That doesn't work
 
6:39 PM
Yeah.
 
@BalarkaSen I'm pretty sure that the functions isn't continuous
Whoever starred that, could you star the message above (my question) rather than random posts ?
 
My answer was downvoted here: math.stackexchange.com/questions/1089188/… . Can someone explain it?
 
There is no continuous function that does what you want, @Hippa. There are lots if you're just demanding a function. e.g., pick your favorite bijection $[0,1/2) \to [0,1]$; and define your map to be that for $x < 1/2$, and $2x-1$ for $x \geq 1/2$.
 
$f(x) = 2x$ for $x \in [0, 1/2]$ and $f(x) = 2x - 1$ for $x \in [1/2, 1]$.
Darn you @Mike
:P
 
@EricGregor Post your attempt in your question, not as an answer : "Here is what I have tried : bladibla"
@EricGregor Many here consider that the answers section is just for 'real' answers
 
6:44 PM
ok, thanks @Hippalectryon
 
@BalarkaSen doesn't work
@BalarkaSen $f(.5)=?$
 
It should have been [0, 1/2) and [1/2, 1]
 
@BalarkaSen $f^{-1}(0)$=?
@MikeMiller Could you give me an explicit formula ? ;_;
I don't know any such bijection
 
Darn you @Hippa
:P
 
@BalarkaSen :P
@BalarkaSen seemingly "obvious" solutions probably won't work. I've searched quite a bit before asking
 
6:48 PM
well, i haven't.
 
7:05 PM
@Hippa Nah, i bet you can find one though. Why do you want an explicit formula?
 
@MikeMiller Why not :D Because I can't find one
 
So?
 
So i'd like to find one ;D
It's bugging me
 
Ok, if that bugs you, will it bug you that you can't personally write down a bijection between $[0,1]$ and $\Bbb R^n \setminus \Bbb Q^n$?
 
7:21 PM
@Hippalectryon I try to explain to myself why people are so silent these days ... (it seems harder to answer than the questions from my research)
 
@Chris'ssis Hi. How are you?
 
@Chris'ssis What is your theory ?
Ted has eaten all the children ?
 
@Hippalectryon lol :-))))))
@JasperLoy Hello! I'm working on my book! How about you? :-)
 
smacks @Hippa
 
:O he's here
 
7:23 PM
@Chris'ssis I am trying to make sense of the past 33 years of my life so that I can fix it.
 
hi @Chris'ssis, @Jasper
 
@TedShifrin Hello. It's interesting that Theodore shortens to Ted.
 
So does Edward :P
Why is it interesting?
 
@TedShifrin You wouldn't have (still the same question) found a function f:[0,1]→[0,1] so that ∀0≤x≤1,f−1(x) is a tuple (2 elements) by any chance ?
 
@TedShifrin Well, I find things interesting all the time.
 
7:25 PM
Nope, @Hippa. Hard to write down things like well-orderings :P
 
Find a bijection between $[0,1)$ and $[0,1]$, @Hippa.
 
@MikeMiller But I can't find one :/
I've already searched
 
@JasperLoy Is there something you would have liked to do and you didn't do (yet)?
 
Maybe there is none, @Mike :P
 
7:27 PM
@JasperLoy What has happened after 33 years of your life so that it should be fixed?
 
@mikeonly the 34th
 
@mikeonly Shit, shit and more shit.
@Chris'ssis Yes, many things. But, as you know, I have to get well first before doing those things.
 
@Hippalectryon Is it always 34th that crashes everything that was before?
 
@JasperLoy Sometimes we never get rid of some things, but at least we might try to learn how to live with them.
 
No idea ;_; I'm not too old yet
 
7:31 PM
@Chris'ssis Yes, but sometimes, we just cannot live with some things.
 
@JasperLoy Did you try your best?
 
That seems likely @Ted
 
@Chris'ssis Well, that is hard to answer. I would like to say yes, but maybe I am wrong. Anyway, I will keep trying to achieve what I hope.
 
MSE then q_q
0
Q: Bijection between $[0,1]$ and $[0,1)$

HippalectryonCould one give me a bijection between $[0,1]$ and $[0,1)$ ? I am trying to find a function $f:[0,2]\to[0,2]$ such that $\forall0\le x\le2,f^{-1}(x)$ is a tuple (2 elements), and such a bijection would give me an answer easily.

 
@JasperLoy You infuse me with hope, mango.
 
7:34 PM
@mikeonly I don't think so. I have not said anything hopeful here. Most things I say here are so miserable.
 
@JasperLoy You think that problem of yours blocks you from making progress in all the areas of your life? For instance, does it prevent you from going jogging? Maybe not. I might try to improve in those areas where I'm able to improve with the resources I have.
 
@Chris'ssis How much of your book is done now?
 
@Chris'ssis where will you publish your book? (in which edition house, e.g. springer) or will it just be createspace.com like the book elementary applied topology, see math.upenn.edu/~ghrist/notes.html , and Sir Ghrist gives details to how much you can win this way
 
@user153330 Where?
 
Hi @JasperLoy
 
7:37 PM
Hello @user130018. I am sad. Maybe I will email you to talk soon.
 
Okay @JasperLoy
 
@JasperLoy This is somewhat secret for some reasons. I prefer not to give these details.:-)
 
I heard Springer gives bad deals to authors
 
@JasperLoy Springer
 
I thought I did terrible in my classes but somehow I got all 'A's
 
7:40 PM
@Chris'ssis As long as it's buyable by the end of 2015, and not 5k pages long... :D
 
is there no value of $\arctan\left(\tan \frac{18\pi}{5}\right)$ such that the answer is in the bounds $(-\frac{\pi}{2}, \frac{\pi}{2})$?.
 
@MathyPerson $\arctan\left(\tan \frac{18\pi}{5}\right)$ is one value...
 
yes, but $\frac{18\pi}{5}$ isn't in the bounds to do arctan(tanx) = x
 
@user153330 Springer (my first thought)
 
@Hippalectryon
 
7:43 PM
@MathyPerson Hence $\arctan(\tan(18\pi/5))\neq18\pi/5$
 
@Hippalectryon yes, but how does that help us?
 
@MathyPerson I'm just saying that your question does not make sense
arctan(tan18Ï€/5) is always in $-\pi/2,\pi/2$
 
@Hippalectryon i meant to say that the range of function arctan must be in $-\pi/2, \pi/2$
 
No
Well, what do you mean by 'range'
 
@user153330 Interesting that way.
 
7:46 PM
Arctan is defined over $\mathbb{R}$, to $]-\pi/2,\pi/2[$. And so ?
 
to $(-\pi/2,\pi/2)$ or, for you Frenchies, $]-\pi/2,\pi/2[$ :D
As if by magic ...
 
And for Ogres ? >.>
:P
 
@Hippalectryon the $\mathbb{R}$ is the domain
 
@MathyPerson I got that. But I still don't get your question.
 
@MathyPerson: $\tan (x-\pi) = \tan x$ ... so how does this help you?
 
7:50 PM
@TedShifrin $x=\frac{23\pi}{5}
 
you're going the wrong way, @Mathyperson :P But apply that principle.
 
$\tan(\frac{18\pi}{5} - \pi) = \tan \frac{18\pi}{5}$? @TedShifrin
 
it is very interesting
 
@user153330 I've been already reading it. Thanks! :-)
 
Right, @Mathyperson: Is $18\pi/5-\pi$ in your desired interval yet?
 
7:53 PM
@TedShifrin no, it's gives me $\frac{13\
oops
@TedShifrin $\frac{13\pi}{5}$
 
ok, @MathyPerson, so what do you do now?
 
@TedShifrin use the formula again?
 
Let me know when you're done :)
 
@TedShifrin $\frac{3\pi}{5}$ and $-\frac{2\pi}{5}$, but these are greater than the bounds, if i'm not mistaken
 
Both?
 
7:57 PM
@TedShifrin if you subtract $\pi$ from the first one it results in the second one
oh wait, $-\frac{2\pi}{5}$ is in the bounds
 
RIGHT!
There's your arctan.
 
@TedShifrin ah, i see now. i actually had that answer before, but i miscalculated and thought it was outside of the bounds. thank you for the assistance. :)
 
No problem :)
 
@TedShifrin if $f$ is continuous and $\forall 0\le x,y\le1,f((x+y)/2)\le(f(x)+f(y))/2$, how do I show that $f$ is convex ? (Hints only please)
 
Gee, I don't know. Do you know a bunch of equivalent criteria for convexity?
 
8:02 PM
Oh and, the question before (same exercise) asks to show that $\{k/(2^n)|n\in\mathbb{N},k\in\{0,\dots,2^n\}\}$ is dense in $[0,1]$ but I don't see how to use that
@TedShifrin Increasing derivative
 
So think about (i.e., prove) what should be true about the chord joining $(x,f(x))$ and $(y,f(y))$ if $f$ is convex.
I guess this is more fun than my probability final, anyhow :D
 
It's above the curve, I know that.
@TedShifrin >:c I'll do it, I told you. I just have more things to do than planned.
@TedShifrin I hope to have finished 30 exercises for tomorrow
Out of 90 q_q
 
Oh, so use your exercise and continuity to show that the chord is above the curve because of that condition.
Do all the students at your school work this hard, @Hippa?
 
@Chris'ssis but you should most probably find a good graphic design professional who could make for you an excellent cover as well as an excellent overrall formatting (with Tex using tikz-pgf, asymptote, etc...)
 
@user153330 I don't like that much the way the book looks like inside. I'd like something in the style of Ovidiu Furdui's book.
@user153330 That's definitely true.
 
8:08 PM
@TedShifrin Not in maths, at least (but remember, I'm in one of the 3 best prep schools, not all are like that. Besides, we don't have to do all the 90 exercises, our teacher just gives us some sheets of exercises per chapter, and we have 4h/week where we correct in them in class (students go to the board) with usually half the people in the class. What is more, I'm by far more interested in maths than the average student in my class (I'm in PC Physics-Chemistry, not MP Maths-Physics))
 
@Chris'ssis there's also this great cover for this book by our Ittay Weiss (overall springer makes some excellent covers)
user image
10
 
@user153330 lllllloooooollllllllllllllllllllllllllllllll :-))))))))))))))))))
@user153330 I'm on the floor :-))))))))
 
@Chris'ssis worth the price : P
 
@TedShifrin But doesn't the hypothesis work on one point at a time (i.e. I only get information about the middle, not about the whole chord)
 
@user153330 You missed to add "Hippalectryon" somewhere in the title of the book ... (or should it have a chapter called like that?) :-)
 
8:13 PM
:-D
 
good grief, @Hippa, all the math you do isn't even for the math-intensive students? Comme tu es fou!! Why aren't you in the math group?
Using the preceding exercise, you extrapolate to all $k/2^n$ points on the chord, @Hippa.
 
@Chris'ssis Undergraduate Texts in Mathematics for Hippies
 
smacks @user153330 :D
 
@user153330 I might try to publish my first 2 volumes at Springer, and then I'll see ...
 
Hi all
 
8:14 PM
hi @JohnDoe
 
@JohnDoe heya
 
@user153330 :D
 
@TedShifrin Well we still have 13h maths/week (out of 40 hours of class)
@TedShifrin I'm in PC because I couldn't resolve myself to abandon chemistry :/
The MP's chemistry is not even chemistry anymore
 
Does anyone here every feel like they need a good long break from maths?
 
ah, I adore physical chemistry ... not so much organic stuff which I've not studied.
What level are you at, @JohnDoe?
 
8:16 PM
@TedShifrin Quantum chemistry FTW :D
@TedShifrin He's on the 5th floor
flees
 
I can still move, @Hippa: I'm not yet dead.
 
Good question, masters @TedShifrin
 
The Shifrin Dojo. Meet Master Shifrin. He will demonstrate his uber spanking move of the death.
 
I tell undergrads not to go to grad school in math unless they feel like they cannot live without it, @JohnDoe ... It's just not a good thing to do because you can't think of anything else to do :P
But grad students often need a day or two off :D
Je ne te donnerai plus de hints, @Hippa!
 
Some people need a day of two off.. per day :P
@TedShifrin Ah non :3
cute cat face
Ah damn that's not a cat, that's a narwhal
 
8:19 PM
What if you would really hate doing anything else @TedShifrin
 
you'll end up like Moby Dick, @Hippa
 
@TedShifrin I'm afraid I've never read that book.
 
If that's true, @JohnDoe, then you take a few days' break, and go back to work :)
 
:) Yeah, would be nice to feel like I can't live without maths, sure one could get a lot done with ease with that state of mind.
 
@JohnDoe if i can't do maths then i do phys, if i can't do phys then i do maths, but in general maths takes up my whole time
 
8:22 PM
@TedShifrin Do you mind looking at this to see if I am going in the right direction? math.stackexchange.com/questions/1092039/…
 
@user153330 You one of those who can't live without...taking a break from maths to do physics...
 
@MathyPerson: I certainly don't like talking about $\tan X$, as $X$ is a point on a ray. You should have angles $\alpha$, $\beta$ from the $x$-axis to points $X$, $Y$ respectively
 
Has anyone encountered the convention in any book or notes of function spaces $C^{0}(\bar{\Omega})$ and $C(\bar{\Omega})$ being defined as different spaces?
 
@TedShifrin Oh, that was a typo on my part. It should have said "Angle 1" instead of X and "Angle 2" instead of Y, but alpha and beta can be used instead
 
nope, @JohnDoe: Unless there's some convention to the contrary $C$ denotes $C^0$.
 
8:26 PM
@TedShifrin I still don't get what you hint me at. Let $0\le z\le1,z=\lim k_n/2^{n+1}$. The only thing I can think of to use the property is $f(k_n/2^n+k_n/2^n)\le\dots$, but I can't really use the RHS (it involves $f(k_n/2^n)$)
 
Think about what the midpoint rule tells you for different intervals, @Hippa.
Don't do limits yet. Try to see what's going on with the various chords.
So I don't understand what the problem is, @Mathyperson. $X$ and $Y$ aren't uniquely determined points, so I can't solve the problem.
 
Yeah I don't know this book I am uses both. In your experience, is the norm $\| u \| = \text{max}|u(x)|$?
 
That's the usual $C^0$ norm on a compact set, @JohnDoe, sure. Of course, if you put in derivatives, norms change :P
 
Yeah okay kewl.
 
@TedShifrin edited my post..maybe that would help?
 
8:31 PM
No, as I commented there, the line segment $\overline{XY}$ will vary (as will its slope) as you vary along the terminal rays.
Do you have more information about the particular points $X$ and $Y$?
Or is the question really asking you for $\tan(\beta-\alpha)$ and you've mucked it up? :D
 
@user153330 You don't know me, I don't know you, so, I wanna tell you a thing: I don't give a f**k on any cent I can get from those books, I mean my only aim is to publish a very good book, at least, (and, yes, I need money, but I simply cannot think of making money by publishing math books). I mean my aims are far more noble, it's something related to my past, it's about promises, about things I cannot share entirely here but they are very important to me.
 
@Chris'ssis Then send me the book for free when it's done :P (kidding of course)
 
@Hippalectryon Sure.
 
Let's give maths a break from all the poking and prodding that goes on here.
 
@Chris'ssis I don't even know how much it costs to send a book from France to Romania, but that must cost quite a handful, really.
That's why I said I was kidding :P
 
8:37 PM
I always thought your profile pic was Sonic @Hippalectryon
 
@JohnDoe OH NO
 
poking and prodding, @JohnDoe?
 
Of course it's not Sonic
it's SANIC
 
Are you referring to the hatred some of us have for one another?
 
haha
Yeah it's sonic
 
8:38 PM
Nono $sonic\neq sanic$
 
It's sonic before he shaves his arms.
 
@Hippalectryon You'll have a book from me for sure if everything goes well (I'm afraid the book won't be at your level that is much higher than mine I suppose ...) :-)
 
@TedShifrin Well, word for word, it is: Point X is the terminal point of $\alpha$ and point B is the terminal point of $\theta$. Point X is in the first quadrant and point Y is in the second quadrant, while $\tan \alpha = 1$ and $\tan \beta = -7$. Find the slope of $\overline{XY}$.
 
@MathyPerson: What is the terminal point of an angle?
 
8:39 PM
@Chris'ssis How many ties must I tell you I am but a student :/
 
:) Yeah exactly sonic on meth.
 
@Mathyperson: To me, an angle is given by two rays. You can choose points on the initial and terminal rays, but there is no particular one.
 
@TedShifrin the terminal point is the point that rotates from the original one on the unit circle
 
Ohhhhhh, @Mathyperson, so $OX$ and $OY$ both are $1$?
OK, now the problem makes sense.
Do you know trig? What are we supposed to be using?
 
8:41 PM
Why are people even starring sanic pictures in a math chat -__-
 
@TedShifrin the more exact definition is: "the point that results when the point (1,0) is rotated counterclockwise by "theta" around the origin"
 
By the way, all your arctan stuff was not giving you the right angle. Arctan gives you angles in quadrants I, IV, not I, II.
 
yep, trig
 
Got it now, @Mathy, but we don't all know that.
 
@TedShifrin lol, yes
 
8:43 PM
So, knowing the tangent values, you can certainly find (in terms of square roots, if necessary) the coordinates of both $X$ and $Y$. What's wrong with that?
Alternatively, you know the base angles of $\triangle OXY$, so you can find the angle that $\overleftrightarrow{XY}$ makes with the $x$-axis.
 
@TedShifrin would using cos x = x coordinate help? would they both have the same y coordinate?
 
I highly doubt they have the same $y$-coordinate.
Just use the tangent value and think about similar triangles.
 
@TedShifrin for alpha, would alpha = pi/4?
 
Sure, although we don't need that. We just need a right triangle with legs $1$ and $1$ ... scaled appropriately. What about $\beta$?
 
@Chris'ssis so publish it in springer then, you'll have a better marketing overall and your audience will be able to localize your book better than if you self-published it
 
8:48 PM
@user153330 To get a best seller one probably needs to use the language of Paul J. Nahin in his book! I mean the cold mathematical language will never beat his language, that is clear, at least for the large audience.
 
@TedShifrin Both legs are 1 for beta too?
 
NOOO, @Mathy.
 
@TedShifrin wait, i see my mistake. silly me
@TedShifrin -1 for x coord?
because of the unit circle?
 
not because of the unit circle. Draw me a right triangle where you'll see $\tan\beta = -7$.
Start with $+7$ if that's easier.
 
@TedShifrin not sure how to add drawings, but i drew it on a piece of paper
 
8:55 PM
OK. So when you have both points $X$ and $Y$ on the unit circle, tell me. I actually like other ways of doing the problem better, but this is probably best for you.
 
@TedShifrin sorry for the delay, my computer froze. i drew the unit circle
 
It's most convenient to work with triangles that are too "big" and then use similar triangles to get them back onto the unit circle.
 
@TedShifrin x coord is -1 and y coord root 7?
 
no root
the root comes into the hypotenuse
or was $\tan\beta = \sqrt7$?
 
@TedShifrin nevermind, it
'*it's -7, not - root 7
y coord is 7 and x coord -1?
 
9:04 PM
ok
 
@TedShifrin so x has coords (1,1) and Y has coords (-1,7)
 
No, silly. First you must put them onto the unit circle :)
 
@TedShifrin how would it change it? wouldn't it just stay the same? or does the unit circle limit it in some way?
 
That's why I talked about similar triangles. To have a point on the unit circle, the hypotenuse of the right triangle must be what?
 
@TedShifrin the length of how much the point is shifted from (1,0) counterclockwise
 
9:09 PM
no no ... If you draw the side $OX$, how long is it?
 
noo, I capped
 
don't be greedy, @Jorge :D
 
@TedShifrin root 2
 
but $X$ has to be on the unit circle, so that's no good!
 
@TedShifrin oh, i see. so pi/4 is incorrect for alpha?
 
9:12 PM
LOL, no, it's fine.
What's the point on the unit circle at angle $\pi/4$?
 
@TedShifrin terminal point, you mean?
 
yes
 
@TedShifrin (1,1)
 
Hmm, so let's see: $1^2+1^2 = $? On unit circle, really?
 
@TedShifrin no, so it has to be different
@TedShifrin but how can it be pi/4 and not (1,1) for the coords?
 
9:15 PM
Because you know that $\cos(\pi/4)$ and $\sin(\pi/4)$ aren't $1$ !!
 
@TedShifrin oh, they are root 2 /2
 
uh huh
Now do you understand why?
 
@TedShifrin I believe so
 
Now work out point $Y$.
BTW, $\sqrt2/2$ = $1/\sqrt2$. That's where it really comes from.
 
@TedShifrin wait, just to confirm: X is (root 2/2, root 2/2) for the coordinates?
 
9:20 PM
yes, which is $\dfrac1{\sqrt2}(1,1)$.
 
@TedShifrin (1 / 5*root 2, - 7 / 5*root 2)?
 
Your minus sign is in the wrong coordinate, but otherwise good.
 
@TedShifrin (- 1 / 5*root 2, 7 / 5*root 2), then
 
uh huh, quadrant II, right?
 
yes @TedShifrin
 
9:24 PM
Hello!! Let $f(x) \in \mathbb{Z}_p[x]$. Show that there is a finite field $\mathbb{F}_{p^n}$ at which $f(x)$ splits.

And if $f(x)$ is also separable, show that $f(x) \mid x^{p^n}-x$.

Could you give me some hints how I could do that??
 
@MaryStar: You need to start working with your professor on your homework. This material is highly dependent on exactly how it is presented in the book or in class, and we cannot know that.
 
@TedShifrin I don't suppose we could use $\frac{y_2 - y_1}{x_2 - x_1}$ on that?
* I suppose
 
Well, slope is slope is slope, @Mathy.
 
@TedShifrin 1/2?
 
That cannot be right, @Mathy.
 
9:28 PM
@TedShifrin -1/3
 
Righto.
If you think about the picture, you know the slope must be negative, and intuition should tell you not as steep as slope $-1$.
 
@TedShifrin yep, i see now. i made a miscalculation to get 1/2 instead of -1/3
thanks again!
 
You're welcome, @Mathy.
 
What books on classical physics can mathematicians suggest?
No resources? Well, I am hopeless in front of tomorrow's exam then.
 
That's harsh, a class that gives you less than 24 hours before being tested on new material.
 
9:42 PM
@mikeonly what type of classical physics?
 
All you need to know is $F=ma$ and you can't push with a rope.
 
@user153330 Isn't it a conventional name for all kind of mechanics?
 
@mikeonly there is newtonian classical mechanics, analytic mechanics (lagrangians, hamiltonians, paths..)
@Chris'ssis forgot to mention it, if you want to publish for springer then you may want to send them (the editors of a series) 2 chapters and see their feedback
 
@Arkamis Thanks for a joke. Well, I was prepared for this exam. At least I have been thinking so until today.
 
@user153330 The content of 2 chapters put in the format of a book or the only thing that matters is the content (no matter the format - of course, using latex)?
 
9:52 PM
@user153330 Non inertial reference frames are what I want to be mathematically explained. Explanations are shrank in most of books.
 
@mikeonly: My personal favorite for a serious first-year mechanics text is Kleppner and Kolenkow.
Ah, non-inertial reference frames is doing covariant derivatives in differential geometry :P
 
@Ted Thanks. I'll have a look on it. Have never heard about it before.
 
@Chris'ssis the content i guess, but it should be formatted in latex
 
It's the MIT text for the serious course that uses multivariable calculus, @mikeonly. I took it from the authors as it was being written. Fabulous course.
Superb exercises, too.
 
if you want to be sure contact one of the editors of the series, or contact Elizabeth Loew
(elizabeth.loew@springer.com or Dr. Eugene Ha (eugene.ha@springer.com
@TedShifrin +1 that's a nice text, do you have the second edition? for me it's just the first ed. the typesetting is a bit outdated
 
9:57 PM
@user153330 It would be nice if I'd prepare the content in latex, and then they'll create the book. I'm not good at creating books in latex, no experience here!
 
I don't know if there was a second edition. Complaining about typesetting is something I ignore.
 
@Ted So it looks like it's much deeper sea of mechanics and math beneath this topic. I hate those squeezed semesters and not fully embraced themes.
 
@Chris'ssis springer already offers the latex template you just have to download it
 
@Chris'ssis: It took me three books to get really good at it, but my first book was produced (indeed, I had it printed) from my LaTeX. I don't like it anymore :P
I understand, @mikeonly. Here we say, "a mile wide and an inch deep." :P
 
@TedShifrin Thanks for encouragements! :-)))
 
9:59 PM
@mikeonly: I would not read that book by arnold, he is a player hater
 
@user153330 Yeah, I know.
 

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