Mathematics - 2015-03-15 (page 4 of 5)

Ted Shifrin

15:00

Do you know about the universal vector bundle on the Grassmannian, for example?

Tim Davids

@TedShifrin This might be a silly question but why does this imply a tangent with infinte slope. The left and right limits don't agree so the limit doesn't exist.

Saal Hardali

"manifolds and differential geometry"

Tim Davids

@evinda Hi how's it going?

Saal Hardali

by lee

evinda

Fine, thanks... and you? @TimDavids

Saal Hardali

15:00

no i dont no that... :(

Ted Shifrin

But you have a vertical line, regardless of whether you think of the slope as $\plus\infty$ or $-\infty$, @TimD.

Committing to a challenge

I am so tired I can't remember why $A_{ij}=$ that LOL

Ted Shifrin

Who wrote that book, @Saal?

Saal Hardali

jeffery lee

Ted Shifrin

In the real case you're done, @Committing. You have to remember what hermitian inner products satisfy for the complex case.

Ah, I don't actually know that book, @Saal. I can't be a mind-reader. So you don't know things like $Hom(E,F)$ for $E$ and $F$ vector bundles. What is your definition of the Grassmannian?

Saal Hardali

15:02

so far it covered tangent maps and bundles and vector fields and flows.

Tim Davids

@TedShifrin So it was a silly question? Yeah I see cos it's the same thing when you consider it as a measure of slope.

Ted Shifrin

well, bundles in general or just tangent bundles, @Saal?

Not a silly question, @TimD.

Saal Hardali

tangent bundles

the first chapter was about covering manifolds and regular submanifolds

Ted Shifrin

There must have been some discussion in the book or a previous exercise leading you to this, @Saal.

Saal Hardali

and fundamental groups

Tim Davids

15:03

@evinda Alright, maths is hard, which makes life hard...but alright overall, forging ahead.

Saal Hardali

the grassman manifold is defined

analouges to the projective space

Ted Shifrin

Intuitively, you can understand how to change a line through the origin by moving the direction vector for the line orthogonal to the line — think about the tangent space to the unit sphere. You can do analogous things for $k$-dimensional subspaces of $\Bbb R^n$.

evinda

@TimDavids Are you an undergraduate student?

Tim Davids

@evinda Yeah. Yourself?

Committing to a challenge

So $(A)_{ij}=\langle T(b_j),b_i\rangle=\langle b_j,T(b_i)\rangle$ implies A is symmetric?

I can't see it in this funky notation tbh

Ted Shifrin

15:06

@Committing: If you have a real inner product, how can you rewrite the second term?

Committing to a challenge

commute them?

Ted Shifrin

right.

Committing to a challenge

Oh lol

Oh LOL

Ted Shifrin

And if it's hermitian?

Committing to a challenge

Complex conjugate?

Was it

Ted Shifrin

15:08

Does that tell you the matrix $A$ is hermitian, then?

David Wheeler

sesquilinear is such a strange term

Ted Shifrin

sesqui = 1 1/2

Saal Hardali

@TedShifrin I think the problem is i'm not sure what's considered a "concrete description". I'm pretty sure those tangent bundles aren't trivial so i'm looking for charts. But that's a rather abstract description since i'm basing all of it on the definition of the underlying space.

Ted Shifrin

or maybe it's sesque

David Wheeler

Latin vowels do weird things in combos

Ted Shifrin

15:09

I honestly don't know what your professor expects, @Saal. I teach this stuff in my courses, but I have talked about the universal sub- and quotient-bundles on projective space and the Grassmannian.

evinda

@TimDavids Me too..

David Wheeler

So do consonants, for that matter

Tim Davids

@TedShifrin Is it challenging keeping track of all the questions directed at you? I am using this result of the existence of the tangent with infinte slope to state that the surface $f(x,y)$ cannot be differentiable at $(0,0)$ since the partial derivatives $f_{x}(0,0)$ and $f_{y}(0,0)$ lie in a plane different to the tangent line of the curve of intersection of $f(x,y)$ and $y=x$ at $(0,0)$.

evinda

@TimDavids In which semester are you?

Ted Shifrin

My advice, @Saal, is to talk with your professor.

Saal Hardali

15:10

@TedShifrin thanks.

Ted Shifrin

@TimD: So do you have the rigorous definition of differentiable (in terms of limits and a linear map or matrix)? Or just the tangent plane?

Tim Davids

@evinda First semester and already tired, trying to work in the day so that I can drink at night.

Ted Shifrin

Being a serious math student doesn't go along with being a drunkard, @TimD. :D

evinda

@TimDavids Do you go out every night?

Committing to a challenge

Complex conjugate just does:

$$\langle o_j,T(o_i)\rangle = \overline{\langle T(o_i),o_j\rangle }$$?

Commutes and conjugate bar thing?

Ted Shifrin

15:12

Yes, @Committing

Tim Davids

@TedShifrin I have a definition in Stewart which uses limits but I am using the differentiability at a point implies existence of tangent plane at a point, and the tangent plane contains all tangent lines at the point.

Committing to a challenge

Which looks like $(A)_{ji}$ with bar thingy over it so $\overline{A_{ji}}$

Ted Shifrin

OK, oh, it's Stewart. OK. Not a more rigorous treatment. Then you're fine.

nods @Committing

Committing to a challenge

Even when I am tired I can do things(actually I can only do things tired from past experience + right now)

Tim Davids

@evinda No I don't, maybe once or twice a week.

evinda

15:13

Aha! @TimDavids

Ted Shifrin

finish this problem and go to sleep, @Committing

Committing to a challenge

But I have another afterwards

And this is worth 10% of my grade for some reason

Ted Shifrin

Nope. I quit.

Committing to a challenge

Buttt Teeeddd

I think I am part way to solving it

Tim Davids

@TedShifrin I thought there were a lot of drunk over achieving mathematicians?

Committing to a challenge

15:14

I think it was the one you called a shit problem(my words not yours)

Ted Shifrin

Not really, @TimD.

Committing to a challenge

Or was that someone else

Mike Miller

I like a good drink, @Ted. Do I count?

Ted Shifrin

I didn't use those words. I said that it was very confusingly worded. Go to bed.

I have a huge cabinet full of booze I have to give away before I move, @Mike, so I do my share of drinking. But ...

Mike Miller

If you've good whiskey, maybe you should give it away after you move... ;)

evinda

15:16

@TimDavids What subjects do you have this semester?

Ted Shifrin

I'm not a whiskey person. There is a bottle of scotch and maybe one other.

I have about 10 kinds of gin, but I'm going to finish them off :P

David Wheeler

@Ted Commiting's prof is teaching LinAlg "two ways at once"-abstractly (algebraically) using linear transformations, vector spaces, and concretely (computationally) using matrices, and $n$-tuples.

Ted Shifrin

I do that in my serious multivariable course, too, @DavidW. Not in introductory linear algebra for the masses.

Mike Miller

You enjoy gin, @Ted? I've never been able to like it except as mixing fodder.

Committing to a challenge

Even though I am doing crap right now tired, I don't have time to do it after resting, so I will do it until done

Ted Shifrin

15:17

I love good gins, @Mike.

Tim Davids

@evinda maths, astronomy, chemistry, applied maths and physics....what are you studying?

Ted Shifrin

Well, @Committing, I'm out of here, so you're on your own.

Committing to a challenge

NOOOOOOOOO

David Wheeler

@TedShifrin My understanding is it's actually an advanced algebra class, not just linear algebra per se.

Committing to a challenge

PLZ TED MA GR8 FR&

Ted Shifrin

15:17

My understanding, too, @DavidW.

ᴇʏᴇs

Later @Ted

evinda

@TimDavids I am studying applied mathematics.. you too?

Ted Shifrin

Hi/bye mr eyeglasses

teadawg1337

Cya @Ted

Ted Shifrin

@TimD, where are you?

Committing to a challenge

15:19

@TedShifrin It is an advanced algebra class lol

It has 10 students

Tim Davids

@TedShifrin Sub-Saharan Africa.

ᴇʏᴇs

So does it like use Artin or something

Committing to a challenge

We had 200 in linear algebra for the masses

Ted Shifrin

Ah, very cool, @TimD.

Committing to a challenge

200 math majors and 400+ engineers

Ted Shifrin

15:19

You can tell me more about the course another day after you've slept, @Committing.

Committing to a challenge

:'(

Now I am proving the other direction :))

This is week 1 assignment if that makes my class not look trivial lol

Tim Davids

@TedShifrin It's okay, a bit too hot for my liking. If you have a surface $f(x,y)$ and a plane $y =x$ and a curve of intersection $C$, if there is a point on $C$ which is a singularity of the curve (curve is not differentiable at the point), would it be correct to say that $f(x,y)$ is not differentiable at the point?

Committing to a challenge

:\ @Ted @DavidW can you look at meta.math.stackexchange.com/a/4726/142198 and give me some clue

evinda

Hallo @Theorem

David Wheeler

15:36

@Committingtoachallenge Need a bit more on the $\leftarrow$...

Committing to a challenge

LOL

I sure do

\Longleftarrow

David Wheeler

namely, that $\overline{\langle T(o_i),o_j\rangle} = \langle o_j,T(o_i)\rangle$

Committing to a challenge

Is that what I am trying to show?

Is that what it means here for T to be self-adjoint in this notation?

I am so confused

David Wheeler

Then you have that $\langle o_j,T(o_i)\rangle = \langle T(o_j),o_i\rangle$, which means self-adjoint.

Well, self-adjoint on basis vectors, but we can extend by linearity.

I really think you're over-thinking this-what an inner product IS, in an orthogonal basis, is:

$\overline{(x^T)}y$

Committing to a challenge

orthogonality on all

Man I really need water but I don't want to wake people up with the loud taps brb

David Wheeler

15:46

How do you do a "dot-product"? you take the conjugate-transpose on the left vector and take the MATRIX product of the two.

How do you find the $ij$-th coordinate of a matrix in an orthogonal basis? you take the $j$-th column, and take the inner product with the $i$-th basis vector (which picks out the $i$-th coordinate in the linear combination: $T(o_j) = a_{1j}o_1 + \cdots + a_{nj}o_n$).

Technically, it should be an orthonormal basis, so that $\langle o_i,o_i\rangle = 1$.

Committing to a challenge

and otherwise $0$

Well that's the standard orthonormal basis

Oh wait nvm they are meant to be unit vectors

Okay yep

So $[T(a)]_\mathcal O = A[a]_\mathcal O$

David Wheeler

Other bases can be orthonormal besides the standard basis, for example $\{(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}),(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{‌2})\}$ in $\Bbb R^2$

Committing to a challenge

True true

Is that it then?

2

A: Sandbox for drafts of long, complex posts

$2.\quad$ Translate the notion of symmetric matrices into the language of linear operators. A linear operator $T$ on a $\operatorname{Dim} n$ inner product space $V$ over the field $\Bbb R$ or $\Bbb C$ with an orthonormal basis $\mathcal O$ is called self-adjoint $\iff$ The representing matrix $...

jarlemag

Hi everyone: I've posted a question about a simple stats problem dealing with determining minimum sample size. If anyone can check my thinking it would be much appreciated: math.stackexchange.com/questions/1190848/…

David Wheeler

Now you haven't used the Hermitian property at ALL.

ᴇʏᴇs

15:57

I like hermits

David Wheeler

If $A$ is hermetian, then $A = \overline{A^T}$, you need to use that somewhere.

$A = [T]_{\mathcal{O}}$, by the way, it's $T$ we are representing, matrices are just arrays, it's associating a basis that makes the arrays represent $T$

Committing to a challenge

My proof works for the real case atleast right?

Nah I think I just don't get it

Not this tired

I don't wanna give up though

David Wheeler

Your argument is circular-you used self-adjointness to prove self-adjointness.

Committing to a challenge

Sorry, I didn't use the conjugate transpose right

Look now and I don't know how to fix the red yet

David Wheeler

That's still circular.

Committing to a challenge

16:08

I used the complex conjugate though

I just need to fix the red part which doesn't follow except if real

If it were real the complex conjugate does nothing and it isn't circular

(except transposes)

David Wheeler

$\langle T(o_j),o_i\rangle = A_{ij} = \overline{A_{ji}} = \overline{\langle o_j,T(o_i)\rangle} = \langle T(o_i),o_j\rangle$

Committing to a challenge

Does that work without transposing the inner product on the right?

Or it doesnt matter for complex inner products

Do complex inner products commute aswell?

I thought we only had conjugate symmetry

David Wheeler

In the complex inner-product $\langle x,y\rangle = \overline{\langle y,x\rangle}$

Committing to a challenge

So don't we need $\langle T(o_j),o_i\rangle = \langle o_j, T(o_i)\rangle$?

Which is equivalent in your proof if we were using a real inner product at the end

David Wheeler

Gah!, all this latex is driving me nuts

Committing to a challenge

16:18

Me too lol

David Wheeler

Let me start over:

$\langle T(o_j),o_i\rangle = A_{ij} = \overline{A_{ji}} = \overline{\langle T(o_i),o_j\rangle} = \langle o_j,T(o_i)\rangle$

There.

Committing to a challenge

omg

They always make me feel dumb haha

I blame being overworked :P

David Wheeler

Now write $x = x_1o_1 +\cdots+x_no_n$, and we have:

$\langle T(x),o_i\rangle = \langle T(\sum_k x_ko_k),o_j\rangle = \sum_k x_k \langle T(o_k),o_j\rangle$

$= \sum_k x_k\langle o_k,T(o_j)\rangle$

N3buchadnezzar

It always fun when students you are TA'ing for is asking questions a few hours before the deadline math.stackexchange.com/questions/1190927/…

David Wheeler

Finally, writing $y = \sum_m y_mo_m$ (I'm running out of letters)

Committing to a challenge

16:26

Can we just write these sums as a vector in the base $\mathcal O$

$\langle T(x),o_i\rangle = \langle T([x]_\mathcal O), o_j\rangle $

David Wheeler

We have $\langle T(x),y\rangle = \langle T(x),\sum_m y_mo_m\rangle = \sum_m \overline{y_m}\langle T(x),o_m\rangle$

giving us $\langle T(x),y\rangle = \sum_m \overline{y_m}\langle T(x),o_m\rangle = \sum_m \sum_k x_k\overline{y_m} \langle o_k,T(o_m)\rangle$

$= \langle x, T(y)\rangle$

Committing to a challenge

Wow

That's a little harder to follow so give me a min

David Wheeler

That horrendous bit of "subscript tracking" is what I mean by "extend by linearity"

Committing to a challenge

That giant block can't be simplified using basis notation at all? Isn't that y_m,o_m sum sort of just a vector $[y]_\mathcal O$?

David Wheeler

We're "linear in the first coordinate" and a vector is just a linear combination of basis vectors (basis vector coordinates)

We proved the "self-adjoint property" for basis vectors ONLY. We need it "for any two vectors"

Sure, but what is $[y]_{\mathcal{O}}$ in terms of the $o_j$?

Committing to a challenge

16:37

Whatever you want it to be in the span

David Wheeler

Ok, let's do this just for a two-dimensional space (it will be easier to follow).

So we have two basis vectors $o_1,o_2$.

Committing to a challenge

Oh it is 1

Oh nvm I think I am dead now

David Wheeler

So our relation $\langle T(o_i),o_j\rangle = \langle o_i, T(o_j)\rangle$ is four relations in all.

A typical vector $x = x_1o_1 + x_2o_2$.

A typical vector $y = y_1o_1 + y_2o_2$.

Now $\langle T(x),y\rangle = \langle T(x_1o_1+x_2o_2),y_1o_1+y_2o_2\rangle = x_1\langle T(o_1),y_1o_1+y_2o_2\rangle + x_2\langle T(o_2),y_1o_1+y_2o_2\rangle $

$=x_1\overline{y_1}\langle T(o_1),o_1\rangle + x_1\overline{y_2}\langle T(o_1),o_2\rangle + x_2\overline{y_1}\langle T(o_2),o_1\rangle + x_2\overline{y_2}\langle T(o_2),o_2\rangle$

Now we apply our proven rule to the four "simple" inner products, and stuff the coordinates back in.

Committing to a challenge

Okay, I see

David Wheeler

It's tedious, and far EASIER to say, and the result follows by linearity.

To continue (just for completeness' sake)

$= x_1\overline{y_1}\langle o_1,T(o_1)\rangle + x_1\overline{y_2}\langle o_1,T(o_2)\rangle + x_2\overline{y_1}\langle o_2,T(o_1)\rangle + x_2\overline{y_2}\langle o_2,T(o_2)\rangle$

(there we've finally used what we PROVED)

Committing to a challenge

16:49

xD

David Wheeler

$= x_1\langle o_1,T(y_1o_1 + y_2o_2)\rangle + x_2\langle o_2,T(y_1o_1 + y_2o_2)\rangle$

$= \langle x_1o_1 + x_2o_2, T(y_1o_1 + y_2o_2)\rangle = \langle x,T(y)\rangle$

When we extend it all out, we get "a lot of terms" ($n^2$ in fact)

LeGrandDODOM

50 close votes waiting in queue, wtf ?

David Wheeler

But nothing "special" is going on, because $T$ is LINEAR.

Committing to a challenge

Fair enough

David Wheeler

With linear maps, you typically just prove something for the basis, and that's enough.

The reason you want to do the Hermitian case, is that if everything is real, then everything is self-conjugate, and you just get the transpose (symmetric case).

So proving the Hermitian case proves the symmetric case with no extra work.

Committing to a challenge

16:59

True true

David Wheeler

Physicists get lazy, and don't even write the summation sign, they just write $x^io_i$

Committing to a challenge

I don't blame them

Huy

I do.

3

evinda

Could I ask someone something about the calculation of errors of Euler method?

Silent

17:18

How to show that triangular inequality holds for metric $$d(\mathbf x,\mathbf y)=\left[\sum_{i=1}^\infty(x_i-y_i)^2\right]^{1/2}$$ satisfies triangular inequality?@DanielFischer

evinda

That is my question:

0

Q: How to calculate the errors of single and double precision

We consider the initial value problem $$\left\{\begin{matrix} y'=y &, 0 \leq t \leq 1 \\ y(0)=1 & \end{matrix}\right.$$ We apply the Euler method with $h=\frac{1}{N}$ and huge number of steps $N$ in order to calculate the approximation $y^N$ of the value of the solution $y$ at $t^N, \ y(t^N)=...

Committing to a challenge

3:30am and I get up at 5:30am

evinda

@robjohn @DanielFischer Do you maybe have an idea?
http://math.stackexchange.com/questions/1191072/how-to-calculate-the-errors-of-single-and-double-precision

ᴇʏᴇs

I think you might need Cauchy-Schwartz @Silent

N3buchadnezzar

Heh. It is always fun having an accepted answer with zero up votes.. =P

Silent

17:26

Does that work for $\mathbb R^\omega$ @ᴇʏᴇs?

Committing to a challenge

17:44

yawnnn

How do I get surjectivity on this?

3

A: Sandbox for drafts of long, complex posts

$6.\quad$ Let $k$ and $N$ be positive integers and suppose $k\leq N$. Let $W(k)$ denote the vector space spanned by vectors $V_S$, where $S$ ranges over subsets of $\{1,\dots,N\}$ of size $k$. $6.i.\quad$ Compute the dimension of $W(k)$ Assuming linear independence, we want to find the number o...

$\forall v_{A^c}, \exists v_A | \phi(v_A)=v_{A^c}$

4am, one sub question to go

robjohn

18:01

@evinda I added a comment

evinda

@robjohn Could you explain it further to me? :/

JMoravitz

18:22

@N3buchadnezzar gogo unsung hero

robjohn

@evinda I am assuming that you are numerically solving a differential equation. You have not provided the algorithm used to arrive at your data, but look at how many terms you are adding up and how much error is incurred for each single-precision operation.

@evinda as $N$ increases, the error gets larger. The error in the numerical approximation will decrease with $N$, but the roundoff error will increase

N3buchadnezzar

@JMoravitz :p

evinda

@robjohn I haven't understood for which reason the error increases for $N>10^5$ when the calculations are done with single-precision, but gets half when the calculations are done with double-precision. Also I haven't understood how we can calculate the difference $|y^N-y(t^N)|$ for a specific $N$.

robjohn

@evinda what algorithm are you using to compute $y$?

evinda

18:37

@robjohn It is given that the exact solution of the initial value problem is $y(t)=e^t$ and we find the approximations by the formula $y^{n+1}=y^n+hf(t^n,y^n)$, from which we get in this case that $y^n=(1+h)^n$.
Also, we know that $t^n=t^0+nh$, where $h=\frac{b-a}{n}$.

Sawarnik

@robjohn In the American system, is it necessary for the president to be a member of one of the houses? :/

robjohn

@evinda Yes, so do you see that with single precision arithmetic, as you increase the number of steps, you will increase the amount of roundoff error that get introduced?

@Sawarnik houses?

Sawarnik

@robjohn Umm, like the Senate and one more?

robjohn

@Sawarnik Oh, no. Anyone meeting certain criteria, like age, being born American, etc, can be president. Usually they are a politician of some sort, but that is not necessary.

@evinda sorry, I have to leave for a bit.

Sawarnik

@robjohn Oh thanks :)

Daniel Fischer

18:49

@Silent Yes, Cauchy-Schwarz, as @ᴇʏᴇs suggested, is the usual thing.

evinda

@robjohn I don't really see it.. Why does it happen?

columbus8myhw

19:05

Given that $y''+y=0$, $y(0)=0$, and $y'(0)=1$, is there an epsilon-delta proof that $-1\le y\le1$?

(The solution being $y=\sin(x)$, of course.)

(I suppose that proving $y^2+(y')^2=1$ would immediately give us the result, I guess...)

I need to go, but I'll think about the problem...

Saal Hardali

Hey guys! Any feedback you can give would be very helpful.
http://math.stackexchange.com/questions/1191205/global-description-of-a-vector-field-to-a-non-trivial-tangent-bundle

Chris's sis

19:29

Furious - Greatest Motivational Video 2015 ft. Les Brown

►

@ABeautifulMind ^^^

@ABeautifulMind We don't give up ... We don't surrender ... We don't quit ... (until we get all we wanna get)

That's the attitude!

ᴇʏᴇs

Hello @Mike

Tim Davids

19:45

@DanielFischer How's it going? Could I ask just one question. Would it be true to state that if a surface, say $z= f(x,y)$ has a curve of intersection which has a point with no tangent line, then the surface is not differentiable at that point. For example the surface $f(x,y) = \sqrt{|xy|}$ and $y = x$ has a curve of intersection $f(t,t) = |t|$ which is not differentiable at $(0,0)$, hence has not tangent line there. Could I then conclude that this surface is not differentiable at $(0,0)$?

Daniel Fischer

@TimDavids Yes. If the surface is differentiable, then all of its [non-empty] intersections with planes are differentiable curves, excluding intersections that consist of isolated points [which aren't curves at all]. So conversely, if you have an intersection curve that isn't differentiable, the surface can't be differentiable.

Tim Davids

@DanielFischer Thanks a lot. Is this a difficult result to prove?

Sawarnik

@hippa le hello

Hippalectryon

@Sawarnik le hi

Sawarnik

le what le are le you le doing le these le days? @hippa

Hippalectryon

19:53

@Sawarnik le stuff le le le

Sawarnik

:D

ᴇʏᴇs

Why @Saw

Sawarnik

tomorrow is my history exam .. and there is a 30 page chapter on french revolution that i am too scared to study :'(

ᴇʏᴇs

Awww

Daniel Fischer

@TimDavids Not really difficult. Are you already familiar with submanifolds?

Sawarnik

19:57

this is nice: danlarremore.com/faculty

Tim Davids

@DanielFischer No unfortunately not.

ᴇʏᴇs

Hi @TheEmperorofIceCream

Daniel Fischer

@TimDavids And I forgot to exclude intersections that are two-dimensional above, if a surface has a flat part, that is of course not a differentiable curve.

@TimDavids Well, you are, I think, at the moment only interested in surfaces of the form $z = f(x,y)$, and intersections with planes of the form $\alpha \cdot x + \beta \cdot y = \gamma$, which is much simpler.

The Emperor of Ice Cream

@ᴇʏᴇs Hello, How are you doing?

Ted Shifrin

@TimDavids I just saw your question. The answer is yes, and it follows from the chain rule.

evinda

20:04

Hello @TheEmperorofIceCream

Tim Davids

@TedShifrin That seems plausible but how?

The Emperor of Ice Cream

@evinda Hello!

Huy

@TedShifrin: You're using an iPad, sometimes, right?

evinda

@TheEmperorofIceCream Did you write the next tests?

Hello @Huy

Huy

hey, @evinda

evinda

20:05

How are you? @Huy

The Emperor of Ice Cream

@evinda I am currently in fifth and I need to get fourth, I have written the first two test and I have to write two others. They are next saturday and then the saturday two weeks after that.

evinda

@TheEmperorofIceCream How do you prepare for it?

The Emperor of Ice Cream

I am working through the book Putnam and Beyond.

Tim Davids

@TedShifrin Did I ask you the same question I asked Daniel Fischer, does a surface being differentiable imply the curves of intersection are differentiable?

Why can't I add pictures to chat, do I not have enough reputation for that?

Huy

@evinda a bit tired, played a lot of football today, so I'm off to bed soon. you?

Ramanewbie

20:12

@Hippalectryon Great you took your old pseudo back !

Chris's sis

Prove that $$\log(2)\sum_{n=2}^{\infty} \frac{1}{n \log(n)\log(2n)}+\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n\log(n)}=\frac{1}{\log(2)}$$ by Ramanujan.

evinda

@Huy Today I have done some exercises in PDEs and now I am studying Numerical Solution of Differential equations.

Huy

I see. I never managed to enjoy numerical maths.

A Beautiful Mind

@Chris'ssis Just watched your video, thanks. =)

ᴇʏᴇs

@ABeautifulMind Hi

A Beautiful Mind

20:16

@ᴇʏᴇs Hey. It's 4 am here and I just woke up after sleeping for a few hours.

Chris's sis

@ABeautifulMind ;)

A Beautiful Mind

My cousin is now taking meds for depression. He doesn't want to go back to the army.

The things I don't want to get starred get starred, and the things I want to get starred don't get starred.

How did I get 6 stars for a stupid message?

I figure that this room likes sophisticated puns while I like silly things.

You guys are too serious man!

Ramanewbie

@Hippalectryon irc

@Chris'ssis can you ask hippa if he put me on ignore please ?

Hippalectryon

20:32

@Chris'ssis no need ^

Chris's sis

@Hippalectryon: @Ramanewbie: @Chris'ssis can you ask hippa if he put me on ignore please ?

OK

Ramanewbie

@Hippalectryon Oh you here ! ^^

Khallil Benyattou

20:55

Long time no see everyone!

(I see you've changed your name to another great film name, @ABeautifulMind!)

I have a quick question on integration by substitution. If $t \mapsto t + 2$ written in the calculation of an integral in terms of $t$, is that equivalent to using the variable change $t = u - 2a$?

Chris's sis

@robjohn did you ever try the nice question by Ramanujan?

$$\log(2)\sum_{n=2}^{\infty} \frac{1}{n \log(n)\log(2n)}+\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n\log(n)}=\frac{1}{\log(2)}$$

robjohn

@Chris'ssis Haven't seen it, but I think I have an idea of how to evaluate the alternating sum.

Hippalectryon

Anti-Pi Rant, 3/14/15

►

a day too late though

Chris's sis

@robjohn OK. I doubt though you can get some closed form.

ᴇʏᴇs

What is this @Hippalectryon

Hippalectryon

21:08

@ᴇʏᴇs A video

**le troll**

ᴇʏᴇs

Oh you're supposed to click on it

Daniel Fischer

Rickrolling 101, @Hippa?

Hippalectryon

@DanielFischer :D

robjohn

@Chris'ssis Now that I think of it, the method I was thinking of works for $$\sum_{n=1}^\infty(-1)^n\frac{\log(n)}n$$ but won't work as nicely for this one

Chris's sis

@robjohn That question is of the type: more sums, each one very ugly, add them and get something nice. :-)

robjohn

21:12

@Chris'ssis However, if I am lucky, the method might work for the sum you ask about

Chris's sis

@robjohn That would be amazing if you're referring to getting the closed form for each series.

robjohn

@Chris'ssis That happens a lot. It happened in something I did last week. I think you saw it.

@Chris'ssis no

Huy

Does anyone TeX on iPad?

Chris's sis

@robjohn Not sure. Where?

I miss Ramanujan ... His life was so damn short ... He could have published tons of integrals, series and limits if he lived longer.

A Beautiful Mind

@Chris'ssis I believe you are Ramanujan.

Chris's sis

21:15

@ABeautifulMind No ...

ᴇʏᴇs

I think Galois lived too short

Chris's sis

Also Abel lived far too short ... The great Abel.

user43418

Does anybody know how to integrate using branch cuts ?

0

Q: Integration: Branch cuts

Can someone show me how to calculate this integral using branch cuts ? $$\int_0^{\infty}\Big(\frac{x}{1-x}\Big)^{\frac{1}{3}}\frac{1}{1+x^2}dx$$

integration branch-cuts

A Beautiful Mind

Just cut a branch, lol.

I think there are some forms of Christianity that teach rebirth instead of everlasting heaven or hell.

user43418

@ABeautifulMind Can you do it for me ?

I will give you .1 BTC :D

A Beautiful Mind

21:20

I have forgotten all my math, sorry. Those who can help will.

Committing to a challenge

Assignment is almost done

A Beautiful Mind

@Committingtoachallenge Did you get sleep?

Alec Teal

Hey

Committing to a challenge

An hour

Ted Shifrin

21:39

@TimD: I linked my response to your original question :)

@Committing: You're fired.

Committing to a challenge

@TedShifrin What why

Ted Shifrin

@Huy: I've given up chatting on the iPad, but I do use the MSE app on it.

Committing to a challenge

@TedShifrin Unfire me

Ted Shifrin

You need sleep to do mathematics, @Committing.

Committing to a challenge

@TedShifrin I am doing the extension of that stupid question btw.

I need no sleep last night to get grades in mathematics

3

A: Sandbox for drafts of long, complex posts

$6.\quad$ Let $k$ and $N$ be positive integers and suppose $k\leq N$. Let $W(k)$ denote the vector space spanned by vectors $V_S$, where $S$ ranges over subsets of $\{1,\dots,N\}$ of size $k$. $6.i.\quad$ Compute the dimension of $W(k)$ Assuming linear independence, we want to find the number o...

I have done the stupid part here

Ted Shifrin

21:41

Oh, no, @Huy, no TeX on the iPad.

Committing to a challenge

I am fairly sure it is correct, although my notation is a little sloppy

Ted Shifrin

@Huy: The iPad is, after all, not a computer.

Oh, @Hippa is back ... for the moment.

Ramanewbie

@ted hi

Ted Shifrin

Salut, @Ramanewb.

Ramanewbie

@ted thanks god, hippa changed back his pseudo

Ted Shifrin

21:43

Well, he is equally obnoxious with all pseudos

Committing to a challenge

From Iwriteonpeoplewhowriteonbananas?

Halp @Ted :)

$6.iii.\quad $ Define a map $e: W(k)\to W(k+1)$ and another map $f:W(k+1) \to W(k)$ by

$$e(v_S)= \sum_{S \subset {S'}} v_{S'},\quad\quad f(v_{S'})=\sum_{S\subset S'} v_S$$

Let $h=(2k-N)I_{W(k)}$. Show that we have the following equalities:

$$ef-fe=h,he-eh=2e,hf-fh=-2f$$

Ramanewbie

@ted lol... That's mean -___-

Ted Shifrin

I don't want to think about it, @Committing.

Committing to a challenge

It defines a lie algebra me thinks

Ted Shifrin

Sometimes I'm mean, @Ramanewb.

Ramanewbie

21:45

@ted I had plenty oaacsions to see that ! ^^

Ted Shifrin

And you will see more.

Ramanewbie

@ted OK

Tim Davids

@TedShifrin The same one I asked Daniel Fischer?

Ted Shifrin

Yes, @TimD, as far as I know.

You will see the chain rule soon. If $f(x,y,z)$ is differentiable and $g(t)=(x(t),y(t),z(t))$ is differentiable, then $f\circ g(t) = f(x(t),y(t),z(t))$ is differentiable.

Tim Davids

@TedShifrin Oh okay, so how would you use the chain rule to show that?

Huy

21:46

@TedShifrin: Point being you don't want to use it as a computer or you don't think you can?

Ted Shifrin

It has limitations, @Huy. And I don't want to try.

Huy

Ok.

Committing to a challenge

Has anyone got experience with lie algebras?

Ted Shifrin

So @TimD, it's the contrapositive of that.

Tim Davids

@TedShifrin $f \circ g$ is the curve of intersection?

nullgeppetto

21:58

Hi everyone, I need your help! Let S = {s_1, s_2, ..., s_n} a set of characters. I would like to find the number of all possible strings I could construct using every character once. The order of each character is significant. Also, strings may be of length at least 1. Could you help me please? Thanks a lot in advance!

Committing to a challenge

@KAj do you like lie algebras :)?

Kaj Hansen

@evinda, a bit. Sorry I left last night; was in dire need of sleep

@Committingtoachallenge, maybe? I have to try them out first

nullgeppetto

For instance, with N=2, I have 4 possible strings, and if I am not mistaken, with N=3 I have 15 possible strings...

Mike Miller

morning, Ted

Committing to a challenge

@KajHansen E.g. you don't like them xD

Kaj Hansen

21:59

Oh hey, I can finally ping @Hippalectryon again

Transcript for

$Mathematics$ Mathematics