This is from mathcad, anybody want to hazard a guess as to what is going on? goo.gl/XMDmeO

@JoeStavitsky it's doing its job. what about it would you like to understand?

What job was it assigned to do

@anon, the results don't match?

oh, the values are off

How do I find tangential acceleration $a_t$

those values are 120/n^2 instead of 1+2/n. was that summation your input?

@SethKitchen you're going to need more context than that

@anon, nooo. You mean the sigma one, right? You saw the simplified one below it, yes?

The simplified one below it I think is correct, I just need to change the decimal place setting

yea that fixed it

@JoeStavitsky oh, in that one summation you have 10 up top while you're letting n vary, so there's that

the problem with computers is they do exactly what you tell them to

wow, this is like stupid week for me. Now I am actually scared of the final :(

@rubito your parentheses are all out of whack. figure out how to type it correctly. also you can use \binom{n}{m} for binomial coefficients.

I have position function $r(t)$ and I calculated the velocity $r'(t)$ and acceleration $r''(t)$, speed $|r'(t)|$ unit tangent vector $v/|v|$ Normal Vector $T'/|T'|$, but it asks for $a_t$

Good Afternoon

@Owatch Happy Mother's Day!

@SethKitchen Happy Mother's Day

Mom upside down is Wow!

Wow

Why does the iterative inversion method work so well?

I am asked to solve for differential equation: $\frac{dy}{dx} = \frac{e^{y}sin^{2}x}{ysecx}$. I managed to change it to the form $\frac{y}{e^{y}}dy = \frac{sin^{2}x}{secx}dx$, then tried to differentiate both sides.

What is uniformily continuous?

@Owatch You mean "integrate both sides"?

Yes

@anon &P(x>=3)=1−[(\binom{6}{0}((0.2)^0)*(1−0.2)^6)+(\binom{6}{1}((0.2)^1)(1−0.2)^5)+(‌​\binom{6}{2}((0.2)^2)((1−0.2)^2))]=.0989$ fingers crossed

Note that $\dfrac{\sin^2x}{\sec x}=\sin^2x\cos x$.

I know that.

It is the prior I have trouble with.

$\int{\frac{y}{e^{y}}}dy$

So, you want to find $\int ye^{-y}\operatorname{d}\!y$ and $\int\sin^2x\cos x\operatorname{d}\!x$, right?

ye^-y

yeah

For the first one, try integration by parts.

I did.

u = y

$u=y,\operatorname{d}\!v=e^{-y}\operatorname{d}\!y$...

du = 1

dv = $ 1/e^{y}$

$\operatorname{d}\!u=\operatorname{d}\!y,v=-e^{-y}$

$\int ye^{-y}\operatorname{d}\!y=uv-\int v\operatorname{d}\!u$

du = dy?

You said du=1, but it's really du=1dy. You still need a d-something term.

Oh, yes.

I got a different v

I had $y + c$

Well, if $\operatorname{d}\!v=e^{-y}\operatorname{d}\!y$, then you just have to integrate wrt $y$ there. (You can ignore the constant, it cancels out later anyway)

So $v=-e^{-y}$.

@robjohn Why did you choose to work with $\lfloor nx\rfloor$ in this answer math.stackexchange.com/questions/54448/combinatorial-proof/… ?

And $\int ye^{-y}\operatorname{d}\!y=y(-e^{-y})-\int(-e^{-y})\operatorname{d}\!y$.

$=-ye^{-y}+\int e^{-y}\operatorname{d}\!y$

$=-ye^{-y}-e^{-y}$

Standby.

I am rewriting some stuff

I got all of this.

What are the amount of possible matchups in League of Legends?

I just wanted to repost it so that it's all in one place.

However, I originally hadn't rewritten e on the numerator

5v5 matches with ~150 different characters

So I got something with ln which was confusing

I end up with, for the first integral: $-ye^{-y}-e^{-y}+c1$

That's what I got, too. (Damn, forgot the $+c_1$!)

$-ye^{-y}-e^{-y} + c1 = \frac{sin^{3}x}{3} + c2$

Seems to make sense.

(Wait, what was the original problem?)

Yeah

Is there a problem?

Did I rearrange wrong?

No, looks fine.

Ok good!

Is that OK as a final answer, or do you need it in $y={}$ form?

I think I need it in y =

So I just finished that part

$ y = \frac{-1}{e^{-y}}*(\frac{sin^{3}x}{3} + C) - 1$

I would think that, if the teacher wants it in $y={}$ form, they'd want the right-hand-side to not have any $y$s...

It looks crude. And so I'm worried I did not do it right.

Oh right. .

Well then I don't know how to solve it.

WA says that there's no elementary form for $y$. So I think that $-ye^{-y}-e^{-y}=\dfrac{\sin^3x}3+C$ is probably fine.

(I just collected the constants onto one side, doesn't really change anything)

Okay, well then I guess that is good enough.

Oh my.

$\frac{dz}{dt} + e^{t+z} = 0$

hi @Ted, @mixedmath

rehi @Balarka ...

no, that would be "rerehi"

almost your bedtime

nah, my bedtime is around 1:00 AM.

$\dfrac{\operatorname d\!z}{\operatorname d\!t}=-e^{t+z}$, I guess

@Owatch

Can I split $e^{t+z}$ into two parts?

I think so

@Gato so that the binomial theorem could be used, no?

$e^{t+z}=e^te^z$

$e^{t}*e^{z}$

Ye

So, $\dfrac{\operatorname d\!z}{\operatorname d\!t}=-e^te^z$?

And that looks pretty separable.

inordinately separable

$e^{-z}\operatorname d\!z=-e^t\operatorname d\!t$, yeah?

$\frac{1}{e^{z}}dz = -e^{t}dt$

I should write like you do.

Makes integration easier

(I like writing $e^{-z}$ because fractions are annoying.)

$\displaystyle\int e^{-z}\operatorname d\!z=-\int e^t\operatorname d\!t$

(just slapped an $\int$ sign in front of everything)

First part becomes -, second remains unchanged

Yeah.

$-e^{-z}=-e^t+C$

+C

+C !!

V?

I delete problem

Typo

I don't know anymore

I meant $C$

I can't do this.

@Owatch: This is way easier than stuff you were doing a month ago.

I was joking.

ha!

right?

ok, good

$e^{-z}=e^t+C$, I guess

(I could write $-C$, but minus a constant is a constant)

Verynice.

And than $-z=\ln(e^t+C)$, or $z=-\ln(e^t+C)$, and we're done!

I forgot to do that.

(Check: $\dfrac{\operatorname d\!z}{\operatorname d\!t}=-e^t\dfrac1{e^t+C}=-e^te^z=-e^{t+z}$)

Morning, @Ted. I've got a question. Want to give it a shot?

goodnight, @Mike. I dunno. How smart do I have to be?

Hello guys, is anyone here familiar with hypothesis testing I would like to ask a question too !

Sorry, I learned some probability last fall but I don't know statistics.

At least smarter than me, @Ted, but I think you're fine

Setup: $Q$ is odd-dimensional. Consider $Q \times (-\varepsilon,\varepsilon)$ with some symplectic form $\omega$. Assume I have two vector fields $X, X'$, both transverse to $Q \times 0$, that are Liouville: that is, $\mathcal L_X \omega = \mathcal L_{X'} \omega = \omega$.

I want to show they point in the same direction. My current strategy is to assume not: then (because they're transverse), if $X'$ points in the opposite direction, $X-X'$ can't possibly be zero anywhere on $Q \times 0$. So I want to show it is.

@TedShifrin We covered it in about an hour because we're behind.

What do you mean by "point in the same direction"?

But I think I get it now.

@TedShifrin: $Q$ separates $Q \times (-\varepsilon,\varepsilon)$ into two pieces.

Anyone know any nonstandard analysis?

oh, same half-space

Oh, fair point.

So assuming the above, I know $\mathcal L_{X-X'} \omega = 0$; that is, $(di+id)\omega = 0$; that is, $di\omega = 0$; that is, $\omega(X-X', \cdot)$ is a closed form. This is pretty much all I know how to say.

So what do we learn from using the normal form for $\omega$? Have you tried doing $\dim Q= 1$?

Is the fear of similar things-- like twins, triplets, and peas in a pod-- called homeophobia?

Not sure what you mean by normal form here. Write it down in a neighborhood of $Q$ using one of our symplectic neighborhood theorems?

Oh, I see, we'd have to decide what $dt$ looks like in terms of $dp$ and $dq$, but, persisting ...

yeah, that's what I meant

Hello @TedShifrin and @MikeMiller

hi mr eyeglasses ... you get your fixed point settled?

Yes

Cool.

Can I ask a question, guys? I fear it may be stuped to be asked in front page :S ?

@SpawnKilleR Don't ask to ask please

Just ask.

When they ask me to find a solution for a differential equation that satisfies an initial condition, are they asking me to integrate, and test if Y(x) = (result-of-conditoon) in my integrated form of the original equation given?

Ok So i have this weird fraction

No, @Owatch, they're asking you to use the initial condition to determine $C$.

Okay, so integrate, and solve for C?

@TedShifrin You should be a mod. You're always in here

No, I'm really not here that often. And no thanks.

(1/4)*3^x*log(3) + (3/4)*log(1/3)*(1/3)^x is the numerator and (1/4)3^x + (

(1/4)*3^x*log(3) + (3/4)*log(1/3)*(1/3)^x is the numerator and (1/4)3^x + (3/4)*(1/3)^x the denominator

and I set it equal to y

Can I solve it for x in close form ? :(

You should put it in ChatJax syntax if you can.

It helps readability.

Its like reqular latex?

Yes, it's very possible. You're basically dealing with $(3^x)*(constant_1) + (3^(-x))*(constant_two) = y.

oh wait

didnt see denominator

i have a denominator too

clear denominators and collect terms and you get a(y)e^x + b(y)e^-x

[3^x*c_1 + 3^(-x)c_2]/[(3^x)*(c_3) + (3^(-x))*(c_4)] = y

y this thing

is not a sentence

@SpawnKilleR don't worry, im drunk too

Still, @Don?

Can you solve this for x actually???

@SpawnKilleR yes. are you following along with what we're telling you?

@TedShifrin: The main issue is that the best version I know says "Suppose you've got two forms in $M$ that are equal on the tangent space of $Q$, and nondegenerate on $Q$. (here I mean nondegenerate on $T_qM$ for all $q \in Q$). Then a neighborhood of $Q$ is symplectomorphic to a neighborhood of $Q$ (with the second symplectic form).

No I don't think i understand

The very nicest one - it's just a nbhd of the zero section in the cotangent bundle! - is only for Lagrangian submanifolds. Certainly that works in your suggested 1-dimensional case.

@SpawnKilleR let's start with (A u^x + B u^-x) / (C u^x + D u^-x) = y. what's the first thing I told you to do?

Why are you messing with two symplectic forms, @Mike? It's two vector fields and one symplectic form?

@TedShifrin: You want me to get a normal form. If there is such a normal form, the above thing provides a symplectomorphism to it.

Clear denominators but what do you mean by that, I am not native speaker so can't fully understand :(

I just wanted you to use Darboux normal coordinates and investigate.

Do you mean in a neighborhood of a point, rather than a neighborhood of $Q$?

Yes ...

@SpawnKilleR that means multiply the equation by whatever you need to to get rid of denominators. multiply both sides by C u^x + D u^-x and what do you get?

What you're asking is local anyhow

Yeah the denominator goes away

@TedShifrin: I don't see why Darboux coords have to preserve $Q$ in any nice way, e.g. why should I locally be able to set $Q$ as the $x_1=0$ hyperplane or something? It seems like I'd have to if I wanted to do this.

Right, @Mike, I realized that was an issue ... hence my $dt$ comment way up there ^^^

Symplectomorphic is my favorite new $5 word, by the way.

ah

@pjs36: How about contactomorphic?

For what value of $c$ do we have $\displaystyle\cos(x^\circ)\ge c(90^2-x^2)$?

But, still, I wanted to get intuition, and I don't even see it with $\Bbb R\times (-\epsilon,\epsilon)$ yet.

That's a close second, for sure

So I just read a very strange post on an actuarial website... someone's about to finish their M.S. statistics degree and is asking how they can get an actuarial internship to prepare them for a statistician role...

@pjs36 clearly hasn't learned enough morphing yet.

@anon So then I write it two terms of U^x and U^-x

?

@SpawnKilleR so we have A u^x + B u^-x = Cy u^x + Dy u^-x. what next?

I'm in the -hedra game: Associahedra, permutohedra, cyclohedra, and a few more in the works :P

I knew you were a monster, @pjs36.

But you're right, @TedShifrin, I do need to step my -morphic game up!

write it as (A-Cy) u^x + (B-Dy) u^-x, or equivalently (A-Cy) u^x = (Dy-B) u^-x.

@anon i think i got it

k

Would you rather: 1) Never eat watermelons ever again, or 2) Unlimited free watermelons for life, but for each watermelon you eat you also have to eat the green part of it

And I also read a blog post about how someone calls using Excel "programming."

ah, mr eyeglasses is turning into a philosopher.

@anon you get a logarithmic function right?

LOL @Clarinetist

@TedShifrin: I'm sorry. $Q$ is closed. I think I screwed up and didn't say that at the start. In any case, here's how to do the 2-dim case. A 1-dim submanifold of a 2-dim symplectic manifold is automatically Lagrangian, so one of the symplectic neighborhoods says it's isomorphic to a neighborhood of the zero section in $T^*S^1$, which is isomorphic to $S^1 \times (-\varepsilon,\varepsilon)$ with form $-dt \wedge d\theta$.

For the law of cosine, if the number on the right becomes negative (resulting in a^2+b^2+2abcosC), do I make it still subtract sometimes? or do i always follow the formula?

@TedShifrin My professor was discussing philosophy with a student who says his major is philosophy but he's taking graduate level math courses and they got into some really math-dense philosophy, I think

@SpawnKilleR you get x = (1/2) log_u (Dy-B)/(A-Cy)

It is very confusing because in earlier problems, if cosC is negative then i would subtract the number from 180 to get the positive value from cos and then i would get the correct answer that the book says

@TedShifrin In less than 24 hours, resignation letter will be in. I can't express in words how excited I am to leave

Yeah, @Mike, I was starting with $dx\wedge dt$ ... but the Lie derivative condition is complicated enough that it isn't obvious to me that the $\partial/\partial t$ components have to have the same sign.

@anon y thats what i got too!!! Thnx !!!

but the problem i just did I had to add them together and then i got the answer this time.

@TedShifrin: I'll fiddle with this case for a bit and tell you how it goes.

Congrats, @clarinetist. You're following in @AlexW's footsteps.

Anyone know?

Oh really @TedShifrin? What did AlexW do?

To 'program' in Excel carries about as much authenticity as programming your coffee machine does.

Quit his job to go to grad school in math, @Clarinetist.

Well, that's a little harsh. It's definitely a bit better than that.

Programming in LaTeX lol

And you can program in Excel, actually.

@OWatch: not if all you do is click and drag in Excel :P

@Owatch VBA is a disgusting programming language :P

If you start using VB for your functions.

@Owatch do you know

Dim [everything in the world you need] As Blah

@Maximilian I tried reading what you wrote, but I don't understand the question.

What number on the right?

@ᴇʏᴇs Speaking of which, I was once hired to teach $\LaTeX$ to actuaries

The managers, before I had even drafted a lesson, were concerned that I would teach their underlings too much information.

You're poisoning their minds with your nonsense.

So what I did was give them real-world :O :O $\LaTeX$ problems

The law of cosine ($c^2=a^2+b^2-2abCosC$) is giving me strange answers. During some homework problems, If cos C resulted in a negative number, I would subtract 180-C to get a positive number and then continue to get the correct answer

that I used for that position

and they liked it apparently, but retained none of it. I left after 2 months

No, @Maximilian: Negative cosine is negative cosine.

but then sometimes I would continue with the negative number to get the correct answer. SO I don't know if my book is just tripping up or if there is a reason to do it

That whole position was nonsense. Complete nonsense.

@Clarinetist: I sure hope you develop some patience once you've quit :)

@TedShifrin I hope the "real world" isn't like what I've seen in actuarial.

The real world is full of incompetence, too.

I don't know how extensive Latex is, beyond what I use right now. But I learnt it in about 5 mintues.

It's very much more extensive, @OWatch.

I've typeset 4 books in it.

Oh, then I stand corrected.

They had the nerve, after asking me to pay back study hours (which is basically asking for your paycheck back) with free labor (they wouldn't even accept a check!!), to ask me to come back to do part-time work with them.

@tedshifrin thanks, i figured out why now

I didn't do much, but I'm glad you helped yourself, @Maximilian :)

In any case, I've integrated the problem in order to solve for that C, and I got something like: $y + ln(y) = -xcosx+sinx+c$

it was for vectors that when using it sometimes I would have to subtract it by 180 to get the other pulling side

and so i confused it for when just using law of cosine

oh, with vectors you have to be careful about where their tails are.

However, should I just equate the left side of the equation to zero? It is $y + ln(y)$

@Owatch This place... they used some sort of hybrid of HTML and LaTeX. It was bizarre.

But putting 0 in for y doesn't sit well with ln()

What do you mean Ted?

And of course, there was 0 documentation on this language.

@TedShifrin: The Lie derivative condition says that $di_X \omega = \omega$. Let $X = a(t,\theta) \frac{\partial}{\partial t} + b(t,\theta) \frac{\partial}{\partial \theta}$. Then because $\omega = -dt \wedge d\theta$, $i_X \omega = bdt - ad\theta$. Exterior derivative gives $$\frac{\partial b}{\partial \theta} d\theta \wedge dt - \frac{\partial a}{\partial t} dt \wedge d\theta;$$ aka, $\left(\frac{\partial a}{\partial t}+\frac{\partial b}{\partial \theta}\right) \omega$.

So the desired condition is that that mess is 1.

Yup, @Mike, that's correct.

This seems problematic?

And I don't see why that dictates that the two $a$'s have the same sign.

We only need them to have the same sign on $S^1$... maybe that helps

I don't see why. Set $a=1$ and $a'=-1$. Clearly we can arrange for $b$ and $b'$.

The initial condition does not seem to work. If $y(0) = 1$, then shouldn't $y + ln(y) = -xcosx+sinx+C$ become $y + ln(y) = -(0)cos(0)+sin(0)+C$, which becomes $y+ln(y) = C$ . .

I'm very confused, @Ted.

That's what you get for talking to me, @Mike.

There must be some issue in my basic assumptions. Maybe this is only valid for higher dimensions. :P

Oh, another $\dim>4$ thing? I doubt it.

No, dim > 2. This is really a result about contact hypersurfaces, and there's only one contact manifold of dimension 1, but it's not very interesting. I dunno.

We both came to the same conclusion so the calculations aren't wrong.

I'll ask Jacob tomorrow. He probably did this exercise at some point.

OK, let me know the answer.

Oh and tell him to let me know when he's where :P

Is there a way to re-write $y+lny = ...$ in terms of y only?

Nope, @Owatch.

Have you asked him yourself first, @Ted?

If Y(o) = 1, then is $y+ln(y)$ to become 1, if I am to solve for C?

@Mike, the same sort of thing will happen with $\Bbb R^3\times (-\epsilon,\epsilon)$, I'm pretty sure.

No, he rarely answers me, @Mike, but I will.

@TedShifrin: $Q$ is supposed to be closed. We'd be working with $S^3$.

Which I suppose doesn't help, because that's still parallelizable.

Locally there's no difference. Are you telling me this has to be a global phenomenon.

Hell if I know.

It sounds pretty local to me.

An explicit assumption in the exercise is that $Q$ is compact.

Hmm....

But this is probably just because the equivalence of two definitions of "hypersurface of contact type" depends on being compact, maybe.

I think most contact topology happens on compact things.

I really don't know this stuff :(

Maybe you and Jacob can give me a lesson ...

Me neither!

I had no idea how recent symplectic & contact topology were as fields. I guess it was essentially invented by Arnold and developed by Gromov and some others (Eliashberg should probably be on that list?)

yes, Eliashberg for sure. And probably McDuff

yeah

It seems to me like contact topology might have actually been invented in the 80s. Astonishingly recent for something that feels very approachable.

I guess $y + ln(y) = C$, since everything else becomes zero.

I still don't know what y is, except that x produces 1 if it is zero. .

So I guess C = 1

Google says: Lie introduced the notion of contact transformation in 1872. Contact geometry has been around for a while. Topological methods started showing up in the 1970s; global topological results essentially didn't exist until the mid-80s

contact geometry came from some Hamiltonian mechanics in physics, too

right... working on $T^*M \times \Bbb R$, the "extended phase space" that includes a time variable

I have no idea what you're talking about, but what does that odd looking X mean?

it's ok, @Owatch, we're a few years ahead of you :)

funny: 'According to A. S. Besicovitch, as quoted in Littlewood's A Mathematician's Miscellany [60, p. 59], "a mathematician's reputation rests on the number of bad proofs he has given," for pioneer work is clumsy.'

Thanks for the explanation.

it means product of sets

Thanks.

I'm asked to find the orthogonal trajectory of: $y^{2} = kx^{3}$. So, I followed what the book does in it's example, and differentiated it to find this 'single differential equation satisfied by all members of the family'. I get $2yy' = 3kx^{2}$, then $y' = \frac{3kx^{2}}{2y}$

But I'm not sure if that's correct.

Another questions guys.. :) I have an inequality ([e^(ax) -1 ] / [ax] ) > y. Can this be solved for x ?

I don't think there is a closed form solution is there ?

There is such a drama on main, no (complete) answer to my question ...

@BalarkaSen maybe you wanna give it a try.

Is that the integral of an integral?

It's a double integral.

I guess the hard part is solving the nested integral?

IDK, I'm not trying to solve it.

@Owatch The hard part is to make it easy to compute. :-)

I just imagine once you got that, that it would be a number, so easy to integrate a second time.

oh.

I can't even do orthogonal trajectory problem easily.

Quick puzzle: For what value of $c$ doe we have$$\cos(x\text{ degrees})\ge c(90^2-x^2)$$

Guys is there anyone here familiar with hypothesis testing that I could ask a question ?

today is my birthday thus we should double the amount of math that we do !

hi @TedShifrin

Happy Birthday!

thanks spawn :D