Mathematics - 2018-01-21 (page 3 of 6)

Balarka Sen

07:00

There is no homeomorphism of $S^2 \times S^2$ taking the meridian to the diagonal

Whereas there is one, of course, for $S^1 \times S^1$

Akiva Weinberger

@BalarkaSen Kinda feels like how the index of any vector field on the sphere is 2

Balarka Sen

That's precisely it

Akiva Weinberger

I see

I'm still kinda mystified by how the self-intersection number is defined. But I guess it's cohomological nonsense

Balarka Sen

You can define it entirely differential topologically, I mean

Akiva Weinberger

Yeah, but, $\sf Top$ category

Balarka Sen

07:03

fuck TOP lol

yeah you need cohomological nonsense for that

Akiva Weinberger

I imagine for PL it's easy as well

Either that or the concept doesn't exist

Nah, it should be doable. Not that I know how.

Balarka Sen

Hmm. It should be doable.

Daminark

@BalarkaSen did I hear E-M spaces?

:P

Akiva Weinberger

It's interesting how the smooth category is like, "it's made of infinitely many infinitesimal linear pieces" and PL is "it's made of linear pieces"

They're similar in that way

Locally linear versus piecewise linear.

Balarka Sen

@AkivaWeinberger On that note, it's an illuminating exercise to see that orientability of the manifold in the smooth sense and in this sense agree

Akiva Weinberger

07:08

It's strange how much of topology is built on the machinery of homology, despite being so non-obvious to define

(I mean, I wouldn't be able to come up with it, absent of being told about it)

Balarka Sen

It's a cool trick, tracing your space with simplices to detect topology

Akiva Weinberger

I've been able to convince myself that the idea behind singular homology is, "compact spaces are nice, Euclidean spaces are nice, let's map Euclidean compact spaces into our space so that we can take the preimages of stuff and then they'll be nice too"

Balarka Sen

It's really all combinatorics

Akiva Weinberger

> A global orientation is a section/trivialization of the orientation double cover.

What does he mean by section or trivialization

Connected component?

Balarka Sen

Yep (also that answer is by me)

Akiva Weinberger

07:14

scrolls down

Wow, you managed to be surprisingly coherent

Balarka Sen

lmao

Akiva Weinberger

I don't understand your isomorphism there

Like, $H_2$ of the Klein bottle is zero, right?

Balarka Sen

Indeed true

Akiva Weinberger

But $\Gamma$ would be… $\Bbb Z$?

Balarka Sen

No! $\Gamma$ is generated by sections of the orientation double cover

In this case, the orientation double cover is the torus

Which is connected

So it has no sections

Akiva Weinberger

07:17

Ah, OK

So by "section" you mean like a homeomorphic copy of the thing

Balarka Sen

Yeah. Maybe I should be explicit about what a section is. If $p : E \to B$ is a covering map, a section $s : B \to E$ is a map such that $p \circ s = \text{id}_B$

Akiva Weinberger

I see

But if $M$ is something sane like a torus, wouldn't $\Gamma$ be $\Bbb Z^2$? 'Cause there are two sections

And $H_2(M)$ would just be $\Bbb Z$

Balarka Sen

$\Bbb Z$, because it's generated by $1$ and $-1$.

I shoved some details under the rug in defining $\Gamma$. Maybe I should expand on that

Akiva Weinberger

@BalarkaSen Where does this fall apart for disconnected $M$?

Balarka Sen

But the point is the two sections correspond to the two generators $1$ and $-1$ of $\Bbb Z$

Akiva Weinberger

07:23

I see

Balarka Sen

@AkivaWeinberger Hmm

Well you could have like an oriented copy and a nonorientable copy

Akiva Weinberger

But like you prove $H_n(M)=\Bbb Z$ if it's orientable

Or I guess the problem is, if it has two oriented components then it has four sections?

Balarka Sen

That's right

$1_a, -1_a, 1_b, -1_b$

Akiva Weinberger

So it breaks down there

Balarka Sen

So it's a $\Bbb Z^2$

Akiva Weinberger

07:24

Yeah

Hey, I just realized—orientation can't be defined in this way on the boundary of a manifold with boundary, can it

only on the manifold points

Balarka Sen

Yeah but boundary of manifolds can be nonorientable!

As in a nonorientable manifold of one lower dimension

Akiva Weinberger

Ah right that's weird

Can an orientable manifold-with-boundary have a nonorientable boundary?

I'm guessing no.

Right, yeah, shouldn't be possible

Balarka Sen

As in the interior is orientable? I doubt.

Akiva Weinberger

'Cause you can take an epsilon neighborhood of the boundary, which should be $B\times[0,1]$ which is nonorientable

Balarka Sen

Yeah good point

Akiva Weinberger

07:28

And you can take such a neighborhood, 'cause, uh, local compactness and blah blah and magic

Balarka Sen

I was thinking about doing a homological argument using $(M, \partial M)$

$0 = H_n(\partial M) \to H_n(M) \to H_n(M, \partial M) \to H_{n-1}(\partial M) = 0 \to \cdots$

So $H_n(M) \cong H_n(M, \partial M)$

And hm

$H_n(M, \partial M)$ has to be $\Bbb Z$ then

Akiva Weinberger

As one would expect

Balarka Sen

That does not seem to contradict much

Akiva Weinberger

But $H_n(M)$ should be zero, no? 'Cause it's got a boundary

Balarka Sen

Oh fuck good point

I was thinking about closed manifolds

$H_n(M, \partial M) \cong 0$

Akiva Weinberger

07:31

I don't know how to prove it but I think the homology of a manifold with boundary should be zero

Balarka Sen

It's true

Akiva Weinberger

(I mean, assuming it's connected and does in fact have at least one boundary)

Balarka Sen

Take a collar neighborhood and glue it on the other side to make it a noncompact manifold homotopy equivalent to it

Noncompact manifold has top homology 0

That's the whole compactly supported theory

Akiva Weinberger

('cause due to a weird quirk in terminology, manifolds-with-boundary do not need boundaries)

@BalarkaSen Ah, cool

Arright, nice answer

Balarka Sen

Ah, ok. You can take the double $M \cup_{\partial M} M$ which is an orientable manifold with a nonorientable codimension 1 submanifold, and that's impossible

But I guess that's the same argument as yours

Akiva Weinberger

07:35

I suppose, in higher dimensions, the analogue for the Möbius strip would be a solid cylinder with the ends identified the wrong way

Balarka Sen

Because you go through the taking tubular neighborhood thing

Akiva Weinberger

$S^{n-1}\times[0,1]/(\langle a,0\rangle\sim\langle\bar a,1\rangle)$

Balarka Sen

What is $\bar{a}$? Antipodal map or reflection?

The latter I suppose

Akiva Weinberger

Reflection. I wanted to write $-a$ but then realized that's not gonna work in odd dimension

Balarka Sen

Mhm

The antipodal map of $S^{n-1}$ has degree $(-1)^n$

Akiva Weinberger

07:37

Yup

But the same argument you had should work in higher dimension, then

Without hyper-Möbius strips, you can extend a local orientation to a global one with paths.

If two paths conflict, join 'em and you get a hyper-Möbius.

Balarka Sen

Correct. Now I am thinking that normal bundle of submanifolds of orientable manifolds should be oriented bundles

(Oriented bundle means it has an orientation fiberwise)

Akiva Weinberger

What's the normal bundle of the center circle of a Möbius band?

Or is it defined intrinsically?

Balarka Sen

The moebius band ;)

Hm, how does the normal bundle of RP^2 in RP^3 look like

Akiva Weinberger

Mhm. So the normal bundle of the orientable submanifold was nonorientable, but only because it was in a nonorientable space @BalarkaSen

Balarka Sen

@BalarkaSen This was bollocks

RP^2 in RP^3 is the counterexample

Akiva Weinberger

07:42

RP3 is orientable?

Balarka Sen

Yes.

Akiva Weinberger

Hm, I guess RP1 is

Right, I see.

@BalarkaSen Hm, that's nonorientable but one-sided

Balarka Sen

@Akiva I am no longer sure about your argument then

Akiva Weinberger

whereas in $M\cup_{\partial M}M$ it's nonorientable and two-sided

Balarka Sen

You have to prove that wouldn't you

Akiva Weinberger

07:44

Nah, one side's in one copy of $M$ and the other side's in the other copy

My original argument is cleaner

Balarka Sen

No I mean, why would $\partial M$ have a $\partial M \times [0, 1]$ neighborhood on one side?

It's not clear to me if $\partial M$ is nonorientable

Why can't it be a fucked line bundle?

That the collar neighborhood is trivial is believable but proof-worthy imo

Akiva Weinberger

Fine

My instinct is to give it an open cover, each of which looks like a Euclidean half-space

but then neighboring half-spaces could fail to be compatible or some shit I dunno

It would have to look really bad.

Balarka Sen

Yeah it may twist. It shouldn't

Hatcher proposition 3.42 page 253

Reading it now. I never bothered looking before...

Akiva Weinberger

Ah, that says that orientable manifolds have orientable boundaries?

Balarka Sen

It says any compact manifold with boundary has a nbhd of the boundary which looks like $\partial M \times [0, 1)$

Akiva Weinberger

07:49

Ah, OK

Balarka Sen

Weird proof!

Akiva Weinberger

That step in the second paragraph (with the functions) is weird, it took me a bit to justify it

but I guess you take a neighborhood that looks like a half-space, and then in a half-space you definitely have such a function that's 1 near your favorite point and 0 far away enough

Balarka Sen

So my understanding is that they are constructing an exhaustion $U_1 \subset U_2 \subset \cdots$ of $\partial M$ such that each $U_k$ has a collar extension $U_k \times [0, 1)$

You can do that for the center circle of the moebius strip too...

Keep exhausting closer and closer to the twist

But never hitting the twist

So what fails?

Well it's a finite exhaustion...

Oh I see

It's the one-sidedness.

Oh duh

@AkivaWeinberger Choose boundary charts $V_1, \cdots, V_n$ near $\partial M$ like you said. Choose a smaller cover by $U_i$'s such that each $\partial U_i$ has a collar extension $\partial U_i \times [0, \epsilon) \subset V_i$, say.

Akiva Weinberger

So Hatcher is kinda pushing out the manifold piece by piece

Balarka Sen

These products are glued by homeomorphisms $f_i : \partial (U_i \cap U_{i+1}) \times [0, \epsilon) \to \partial (U_i \cap U_{i+1}) \times [0, \epsilon)$

But $f_i$ is always orientation-preserving on the $[0, \epsilon)$ component.

If you glue a bunch of product spaces $X_i \times I$ with transition functions that are orientation preserving in the $I$-component, your life's set

You have to have a trivial bundle in the end, a product space $X \times I$

Make that $[0, \epsilon]$, whatever

Akiva Weinberger

08:00

@BalarkaSen What do you mean?

Balarka Sen

@Akiva Like, a neighborhood of the boundary is an $I$-bundle over $\partial M$, yeah?

The transition functions of that $I$-bundle are all orientation-preserving

Akiva Weinberger

For the $f_i$ bit I meant

Balarka Sen

$f_i$ are the transition functions.

Akiva Weinberger

I guess

I think I need to go to bed now though

I'll think about Hatcher's proof, it definitely is a weird one

Balarka Sen

It's just reiterating what I said in a specific set up. Interval bundle with transition functions orientation preserving is the trivial product bundle

Akiva Weinberger

08:03

(Very technical, like an analysis proof, as well)

Good night

Balarka Sen

If you "transit" from one local product $U \times I$ to another $V \times I$, the map $\varphi : (U \cap V) \times I \to (U \cap V) \times I$ doesn't do anything weird

Night, @Akiva

Akiva Weinberger

08:20

OK, I know I said I was going to bed, but I had to get up to write this down because it's weird

So say you have a plane in 3-space, and a shape on that plane with area $A$

and say $A_{xy}$ is the area of its projection onto the $xy$-plane, and similarly for the other two planes

Do you have $A^2=A_{xy}^2+A_{yz}^2+A_{xz}^2$??

Arright, going back to bed now

skullpatrol

08:36

cya

Secret

09:14

in The h Bar, 2 mins ago, by Captain Bohemian

@Slereah are you talking to me? I don't know much about number theory. I originally considered number theory has no application in physics until the recent, when I read from quanta magazine number theory has application in scatter amplitude.

skullpatrol

in The h Bar, 3 mins ago, by Slereah

I just mean no numbers at all

;-)

jublikon

09:36

I have to prove by induction, that

$\sum_{k=1}^{2n+1}{\frac{(-1)^{k+1}}{k}} \ge \frac{1}{2}+\frac{1}{2n+1}$

I have there:

$\begin{align}
\sum_{k=1}^{2(n+1)+1}{\frac{(-1)^{k+1}}{k}} &= \sum_{k=1}^{2n+1}{\frac{(-1)^{k+1}}{k}} + \frac{(-1)^{2n+2}}{2n+}+\frac{(-1)^{2n+3}}{2n+1} \\
&= \frac{1}{2} + \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{2n+3}
\end{align}$

But the solution sheet says:

$\begin{align}
\sum_{k=1}^{2(n+1)+1}{\frac{(-1)^{k+1}}{k}} &= \frac{1}{2} + \frac{1}{2n+1} - \frac{1}{2n+2} + \frac{1}{2n+3}\\

jublikon

09:49

had some typo errors:

here the question again

I have to prove by induction, that

$\sum_{k=1}^{2n+1}{\frac{(-1)^{k+1}}{k}} \ge \frac{1}{2}+\frac{1}{2n+1}$

I have there:

$\begin{align}
\sum_{k=1}^{2(n+1)+1}{\frac{(-1)^{k+1}}{k}} &= \sum_{k=1}^{2n+1}{\frac{(-1)^{k+1}}{k}} + \frac{(-1)^{2n+2}}{2n+2}+\frac{(-1)^{2n+3}}{2n+3} \\
&\ge \frac{1}{2} + \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{2n+3}
\end{align}$

But the solution sheet says:

$\begin{align}
\sum_{k=1}^{2(n+1)+1}{\frac{(-1)^{k+1}}{k}} &= \frac{1}{2} + \frac{1}{2n+1} - \frac{1}{2n+2} + \frac{1}{2n+3}\\

never mind

i have found my problem

Tuki

10:58

@akiva you there ?

hello

if i have $f(x,y)=x^2y^3$ and i would like to evaluate this in point $(3.1,0.9)$ using differential as approximates

total derivative ?

$D(x^2y^3)=3x^2y^2dy+2xy^3dx$ ?

or jacobs matrix ?

usukidoll

11:21

hi

skullpatrol

heya

usukidoll

got a complex analysis question
http://prntscr.com/i3fpxl
here's the graph on desmos
http://prntscr.com/i3fpib
is what I got so far and the boundary points are when x = 2 and y = 4
Re(z) = 2 and Im(z) = 4

The interior points should be in that purple box but not sure what else to put for this

mercio

11:55

what else to put for what ?

Mohan Sharma

Can i ask question about general topology?

usukidoll

like it's not closed... there's no exterior points

if it's open, there's interior points.. . those points are probably in that purple box

Mohan Sharma

Yes

mercio

maybe write down the definition of "is open" ?

usukidoll

A set D is called open if all of its points are interior points

that's in my complex analysis book..yeah I'm being exposed to the same definitions as last semester for topology @_@!!!!!

Alessandro Codenotti

12:00

@MohanSharma just ask, don't ask to ask

usukidoll

@MohanSharma general topology is hard...get ready to read a bunch of vocabulary in one paragraph

skullpatrol

Vocabulary is always the starting point :P

usukidoll

@___@ errr

mercio

then what's the definition of interior point ?

usukidoll

prntscr.com/i3g2x6

Mohan Sharma

12:06

usukidoll

there's a disc with a point centered... x_X

mercio

well you have everything you need there

Mohan Sharma

This definition looks like saying every interior point isadherent point

usukidoll

I still think the disc is in that purple box in the graph on desmos since it's both x < 2 and y < 4

mercio

I don't see any disc

usukidoll

12:08

maybe all the points in the purple box are interior points

mercio

that's what you want to prove

skullpatrol

\o @MatsGranvik

mercio

because that's what it means for that purple box to be open

usukidoll

so I have to find a disc in that purple box

mercio

no

you have to show that every point is an interior point

start with an easy point maybe

can you show $(-1,-2)$ is an interior point of the purple box ?

(and now this is about finding a disc)

(but you have to do that for all the points not just one)

usukidoll

12:11

If I plot (-1,-2) then it's an interior point for the purple box... oh so all the points involved are in the purple box . Like
(-2,-2) could be another interior point of that purple box

mercio

how do you know it's an interior point

usukidoll

errrrrrrrr by graphing x_X we have interior points if the set is open

mercio

no

you want to prove that the purple box is open

Mats Granvik

@skullpatrol Hi

usukidoll

if the purple box is open then there exists a disc with a point centered in the disc?

mercio

12:13

yeah but you don't know that the purple box is open

usukidoll

then I have to prove that the purple box is open first
then there exists interior points
if the purple box is closed, there is only exterior points

mercio

no

and no

usukidoll

.-.

mercio

write down the definition of interior point

usukidoll

hands are too sore from typing so much earlier

Leaky Nun

12:27

looks like someone is getting lost in definitions in topology

Abcd

Hello

In how many ways can 10 persons take seats in a row of 24 seats such that no two persons take consecutive seats?

Complement method makes it really lengthy.

Leaky Nun

stars and bars?

Abcd

but our objects aren't identical..

Leaky Nun

$\times 10!$

Abcd

stars and bars is for identical things..

mercio

12:38

imaginea person as taking 2 seats at once

then it almost becomes a simple problem

Abcd

@LeakyNun Someone on Math.SE told me to not use stars and bars for non identical things.

@mercio why

Leaky Nun

I don't care what Someone on Math.SE said

Abcd

okay so you mean stars an bars is applicable for non identical objects too?

Leaky Nun

I mean that when doing maths you need to adapt, instead of relying on what Someone on Math.SE said

OSAS tells his FATHER and MOTHER names - Ovuvuevuevue enyetuenwuevue ugbemugbem OSAS

►

@BalarkaSen

Balarka Sen

12:55

Eh, not funny.

2

Abcd

@LeakyNun is stars and bars valid when objects are distinct

Balarka Sen

Imagine someone bullying you for your and your family's name. Wouldn't be very funny, would it? (I am not even sure if this is genuine or troll in which this guy takes part of, I just don't like the way it's depicted) The original Osas video was funny because it's such a long name. This one's not

4

BeginningMath

can someone please help me checking if a claim is true or not?

${v1,...,vn}$ base of V and T:V→V follows:$T_v1=0$, $T_{vi}=v_{i−1}$ (2≤i≤n), so there exists $T^k=0$ in $1≤k<n$

so from what i udnerstand this question is about a nilpotent transformation

also knowing that $T_{v2}=v_1$ and that $T_{v1}=0$

how can i check if it's a true claim or not? does it have anything to do with the degree?

Narcissus jewel

13:13

Why not write down the matrix?

BeginningMath

if $k \leq n$ is the minimal polynomial it seems to be nilpotent

Narcissus jewel

You can see immediately what the matrix is

BeginningMath

i have a question

can it be inferred that the trace is always 0 here?

Narcissus jewel

Well, write the matrix down, does it have any term on the diagonal?

BeginningMath

i don't know how to check for $T_3$

$t_3$ and on... i am, not ure if they're all 0

Narcissus jewel

13:15

Are you rendering your latex?

I can interpret most of it, except for that T_3

BeginningMath

nope, just typing it as if latex for your convenience. whenever i try to run latex renderrer it crashes or doesn't show it at all (in the chat)

$T_v3$, sorry

Narcissus jewel

Well you forgot to subscript things, and then you subscripted other things for no reason :P

It's actually probably strictly worse as you are texing it :P

BeginningMath

lol

Narcissus jewel

You should really try to write the matrix down

BeginningMath

i don't know how to write it for $v_3$ and afterwards

Narcissus jewel

13:19

Why not?

What does Tv_3 = v_2 mean?

It takes a vector (0,0,1,0,0,0,....,0) to (0,1,0,0,...,0)?

BeginningMath

i didn't understand that before

Narcissus jewel

Write T with arbitrary letters (t_{ij}) and then work out what they are?

That's okay, this stuff is tough at first

BeginningMath

i-1j-1

but how can i know if the other elements are 0 or not? if the trace is not zero, then it's not a nilpotent matrix and the laim is false, from whati understand

Narcissus jewel

Yashas

Is there an intuitive explanation of the difference between right-continuous and left-continuous?

Narcissus jewel

13:26

So letting b=1,a=c=d=e=0

You can compute this as you know the result @BeginningMath, and you can obtain information about the t_{ij} that appear in the matrix

A very large amount of information indeed ;)

BeginningMath

13:37

@Narcissusjewel, there ws a power outtage in my building

so basically because the supediagonal moves one to the left, now the trace is full of 0's if i understand correctly

Narcissus jewel

There will be a power outage in my building after a few more drinks

BeginningMath

which makes it a nilpotent matrix and the claim is true

lol

Narcissus jewel

So you got that the matrix is just 1s on the super diagonal, and that T^2 has 1s on the higher diagonal, on and on, until it dies

The trace of the matrix is just a number, so its not really full of zeros, but I know what you mean, it's the sum of a bunch of zeros

BeginningMath

yes, which make the matrix nilpotent, and because of that the claim seems to be true

am i right?

Narcissus jewel

Well, now you have the matrix, so you can verify it yourself right?

For some k, T^k will have a single nonzero entry in it, and then applying it again, you get?

Oh, you meant the traces of each T^k perhaps for each k, yes, you've shown that is zero with this sure

but of course, the trace of a matrix an be zero, without the matrix being nilpotent, since perhaps its square doesn't have trace zero

So you may as well do the calculation

BeginningMath

13:45

ywes, for each t^k

Narcissus jewel

@BalarkaSen You might like the album 'Fri' from the band 'Make a change...' (except perhaps the first 5 minutes or so of the first song)

Anderson Felipe Viveiros

Can someone help me with this one?

https://math.stackexchange.com/questions/2612101/conditions-on-a-lipschitz-function-fu-subset-bbb-rm-to-bbb-rn-which-guara/2614683#2614683

Tuki

13:56

any idea how you would define tangent plane at point $f(2,1)$ when $f(x,y)=x^3+3y^4$ ?

gradient can give the direction the plane needs to face but it doesn't define the angle it needs to be

Akiva Weinberger

Well, what is the gradient there?

Tuki

or gradient length $=$ angle ?

Silent

I can't understand here what b does not imply c means

I will appreciate a counterexample

Tuki

The gradient is $\nabla f(2,1)=\begin{bmatrix} 12 \\ 12 \end{bmatrix}$

Akiva Weinberger

Oh, wait, that's wrong @Silent

Tuki

14:01

and then i can define vector that is perpendicular to the gradient which would be any vector that satisfies a=-b when $\begin{bmatrix} a \\ b \end{bmatrix}$

Akiva Weinberger

Hm. I'm not sure, actually @Tuki

Tuki

so to simplify i have two vectors that make plane $$ \hat{a}=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \hat{b}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$$

now i would need the angle that this plane is in

probably can notice that these don't have $z$ component

@AkivaWeinberger Well how would you try to define tangent plane at this point ?

Silent

@AkivaWeinberger, do we interpret $d$ as gcd(a,b) or just a number? because if d is gcd then c holds anyway, right?

Akiva Weinberger

@Tuki I think you get the plane $z=12x+12y+C$ for some $C$

(plugging in $x=2$, $y=1$, and $z=(2)^3+3(1)^4$ For find that constant)

Notice that it has the same gradient that $f$ does

Tuki

@AkivaWeinberger hmm okey how did you come up with this ?

also length of the gradient is same as slope ?

$$||\nabla f(2,1)||= \sqrt{12^2+12^2} =12\sqrt{2}$$

Akiva Weinberger

14:13

Should be, yeah

Narcissus jewel

@Tuki f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)

Tuki

so $$ \hat{a}=\begin{bmatrix} 1 \\ 1 \\ 12\sqrt{2} \end{bmatrix}, \hat{b}=\begin{bmatrix} 1 \\ -1 \\ 12\sqrt{2} \end{bmatrix} $$ plane between these should also be considered as the tangent plane ?

or mayby not

@Narcissusjewel what you mean with $f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)$ ?

Akiva Weinberger

$f_x$ is $\frac{\partial f}{\partial x}$

which is $12$ in this case

Abcd

2 hours ago, by Abcd

In how many ways can 10 persons take seats in a row of 24 seats such that no two persons take consecutive seats?

Tuki

combinatorics ?

${24}\choose{10}$ ways ?

Abcd

14:26

My friend got the right answer using:
Choose 10 seats
Put them in left 15 gaps.
Then account for arrangements using $10!$

I don't get my friend's method at all :(

Because seats are fixed.

How can we "choose and put"?

@Tuki Certianly not.

Tuki

not exactly good with this ^^

Abcd

The answer is $\dbinom{15}{10}\times 10!$

@Tuki what do you mean?

Tuki

@abcd that i am not good with combinatorics ?

Abcd

Ok

if anyone can help me understand my friend's method...

philmcole

@Abcd math.stackexchange.com/questions/1170021/…

I believe your friend got the wrong answer since Tuki said the right answer

Abcd

14:31

@philmcole But the answer given in the book is my friend's answer

philmcole

Ah sry didn't read "consecutive"

You can look here: math.stackexchange.com/questions/1643085/…

Tuki

14:50

hmm the constant for my problem should be $C=(2)^3+3(1)^4=11$

so tangent plane is $z=12x+12y+11$ ?

now $z(2,1)=f(2,1)$ ?

but they don't match

Abcd

@philmcole the chairs are numbered and fixed.

Tuki

if i plot this it would seem like it is ok ?

Abcd

how can we set them out like the answerer did?

Never mind. I got it and feel great :-) !

Tuki

i think i have wrong constant in this ?

@AkivaWeinberger the constant in this should be $C=11$ ?

Narcissus jewel

@Tuki $z= f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)$

Tuki

15:05

@Narcissusjewel how did you come up with this ?

Narcissus jewel

@Tuki Hold $y$ constant, and see how the function varies with $x$, you obtain some slope. You want a linear approximation of the surface at a point, so you can take a plane given by the $x$ and $y$ slopes, but first you must translate this plane to the correct place on the surface

One can think of it like this

You have (x,y,z). Where z is a function of x and y

So if your surface is f(x,y)=x^3+3y^4

You can write (x,y,x^3+3y^4)

Tuki

so $f_x(2,1)$ would be slope respect to x axis, $f_y(2,1)$ is slope respect to y axis. (i mean slope between x or y axis and z)

and $(x-2)$ and $(y-1)$ are for what ?

@Narcissusjewel another thing is i dont understand that why this wouldn't work with gradient ?

Gradient gives us direction and the angle for the plane

Huy

hey, I'm looking at an exercise about solving a differential equation using the Laplace transform. I get the (correct) result, that
$$\mathcal{L}[y] = \frac{1}{(1+s^2)^2} e^{-\pi s} + \frac{1}{(1+s^2)^2}.$$
now, the standard solution suggests to "know" the value of
$$\mathcal{L}^{-1} \left[\frac{1}{(1+s^2)^2}\right]$$
which I don't. Instead, I rewrote my result as
$$\mathcal{L}[\sin(t)] \cdot \mathcal{L}[\sin(t-\pi)] + \mathcal{L}[\sin(t)]^2,$$
which is
$$\mathcal{L}[\sin(t)] (\mathcal{L}[\sin(t-\pi)] + \mathcal{L}[\sin(t)]).$$

Tug' Tegin

15:30

Hi there! Could somebody give me hint on evaluating $\cos(\alpha)\cos(\alpha/2)\cos(\alpha/4)...\cos(\alpha/124)$? I see that there is a geometric series, but can't find an approach to solve it.

there should be 128 not 124 in the argument of the last cos.

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