Mathematics - 2018-03-21 (page 2 of 5)

Google books ? PDFs? It’s hard work

@Cows, have you tried u-substitution?

lol

ok, that may have been a little mean :P

1:02 AM

You have better look with physics chat , math people don’t really do calculus much

I did 4 semesters of calculus, 2 semesters of differential equations... all for naught

I have done integrals in the past

not sure why

still dumb as shit

@orbit u-substitution ? What are you talking about? this technique is not enough

What about integration by parts?

1:03 AM

@user525966 I discussed this a while back with @Shobhit

I'm trolling

here

Jan 29 at 13:39, by Shobhit

hi. I was able to prove that $[a,b] = \cap_{n \in N} (a - \frac{1}{n}, b + \frac{1}{n})$ and $(a,b) = \cup_{n \in N} [a + \frac{1}{n}, b - \frac{1}{n}]$, but i cannot picture this. Like how is this happening?. Why union on first gets a closed set and the opposite but same issue with second one. Please help.

trig substitution

where is that insane guy who used to do integrals here

Use FTC

wow

1:04 AM

amazon.com/Handbook-Mathematical-Functions-Formulas-Graphs/dp/…

I know that after factoring you will have a sum of elliptic integrals

amazon.com/Table-Integrals-Products-Sixth-Gradshteyn/dp/…

Here’s a couple of them

you can then do u sub for and get moduli and so forth but it is not easy to get to that point

@zee wow they look old

@AkivaWeinberger "The boundary map $\partial: H_n(X,A) \to H_{n-1}(A)$ has a very simple description: If a class $[\alpha] \in H_n(X,A)$ is represented by a relative cycle $\alpha$, then $\partial[\alpha]$ is the class of the cycle $\partial \alpha$ in $H_{n-1}(A)$."

why do u need to solve this?

turns out solving this integral is equivalent to proving the Riemann Hypothesis... tough luck

1:06 AM

@AkivaWeinberger I don't understand why $\partial\alpha$ would be in $C_{n-1}(A)$. Am I missing something obvious?

I asked him that a dozen times!

@0celo7 is that to me?

No

$\partial :C_n\to C_{n-1}$

@Zee hah, I’ve got an old copy of G&R on my desk

what is the issue

1:07 AM

@Semiclassical for some reason am not surprised

And I used to have a copy of A&S

Lol

@0celo7 it’s from $H_n(X,A)$ to $H_{n-1}(A)$

the long boi in relative homology

@LeakyNun It's the restriction of a chain map tho

I managed to get G&R for free during an office move

There was s bunch of free books

I got 200 book from a philosophy office move , am still waiting for the day a math office move happens

1:08 AM

I thought $H_n(X,A)$ is built from $C_n(X)/C_n(A)$ @0celo7

I was literally walking back and fourth from the office to the trunk of my car with my bag full of books and just dumping them in

@Zee this was a Physics office move. So good opportunity

anakhronizein

Did you throw those philosophy books in the dump, Zee?

@LeakyNun Yeah, and the boundary operator descends to the quotient to give you the map in the relative long boi

@Zee, The problem I am having to be precise is reducing elliptic integrals to standard form. The polynomial I am dealing with is degree 4 . I was going to follow mathworld.wolfram.com/EllipticIntegral.html , but it is not straight forward for degree 3 and 4

At this stage funny things happen

1:10 AM

why r u trying to solve this

@orbit-stabilizer give it up

@0celo7 how does the boundary operator give you something in A?

No, I have them at my house , took them when I stopped reading philosophy, idk what to do with them ,but I would never throw a book in trash

@0celo7 you never give up

I also got a copy of Arnold because of that which is nice

1:11 AM

you persist and persevere

until the job is done

what am i talking about

@Semiclassical I need to get in my advisor's will

@orbit-stabilizer I thought i mentioned earlier? I want to compute error. If I am making an approximation, I need to know enough to appropriate the right bounds if that makes sense

He has $20k in books that I need to have

Nice , I like Arnold

@0celo7 haha

1:12 AM

@Zee did you see the link I mentioned?

Sorry Cow , gotta leave

It's also not fair. He leaves receipts in books and the \$100 books today were \$15 when they were released.

@0celo7 “those were the days”

@LeakyNun Thinking

ok

1:13 AM

@LeakyNun Page 115 of Hatcher is probably better than anything I'll say

@0celo7 who was the guy you mentioned who solved all integrals on here? Can I contact him?

what a coincidence @0celo7

my quote is from 117

See the first bullet point near the middle of 115 then.

I think it answers your question.

Please tell people I need help with this. I will be available to learn about this at anytime.

:o

1:16 AM

Does it?

How do I do this limit

yes, but now I need to think about why it is true

$n[e-(1+1/n)^n]$

as n -> inf

@LeakyNun Good luck.

@Zee I think the reference might help. Let me grab a copy real quick

1:21 AM

@orbit-stabilizer i’d write $(1+1/n)^n =e^{n \ln (1+1/n)}$

Then you can use the Taylor series of log to get the higher order terms

@Semiclassical, thanks I'll try that

@Semiclassical Hi ! Got a question.

@Semiclassical Given two points in 2-d , I can calculate the equation of line on which they lie using two point slope form. How do I find the equation of line in 3-d when I'm given two sets of points like $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$

@LeakyNun Do you understand it?

thinking

@LeakyNun For a regular $[\alpha]\in H^n(X)$, what is $\partial\alpha$

1:28 AM

I’m not focusing now,sorry

@LeakyNun Describe $H_n(X,A)$ to me.

user525966

is there a certain order i should read things in if i want to prove everything as i work up

@AkivaWeinberger some quotients by some kernels and images of some quotients

@LeakyNun look, for what I said above, $\partial\alpha=0\in C_{n-1}(X)$

here you have $\partial\alpha=0\in C_{n-1}(X,A)$

what do you know about zero in this guy

@user525966, chapter 1 - 8 in Rudin

Or use Tao's analysis

Just pick an Analysis book and go with it

user525966

1:38 AM

any othres?

Abbot's understanding analysis

@LeakyNun It's basically the set of chains in $X$ whose boundary is in $A$

Abbot's would probably be the easiest of the 3

@AkivaWeinberger that’s what everyone says

Tao's does more work on fundamental stuff, like constructing the reals etc

1:40 AM

and I just told you how to prove that @LeakyNun !

@LeakyNun Work through an actual example. Like $X$ being the torus and $A$ being a parallel union a meridian.

Note that $X/A$ is a sphere.

okok, sorry, I’ll think about this later, many thanks

sorry

(^Not obvious)

Is homotopy theory just studying spheres ?

And wedges of them...

seems like that

Sounds silly but I think 1/2 of math is studying some generalization of circles

1:55 AM

@Zee most of algebraic geometry is just studying $S^1\times\cdots\times S^1$

nah

that wasn't serious.

But I was , am serious

It really seems to me that homotopy is realy about studying circles and wedges of them

homotopy theory is about studying one parameter families of continuous maps.

Lots of math is about understanding one parameter families.

Yes sure , but I mean in a geometrical sense , that’s what it boils down to , circles

1:58 AM

Definitely at least 1/2 of AG is about understand one (\Bbb C not \Bbb R) parameter families.

What does one mean by homotopy theory? Like, the study $\pi_m(S^n)$?

(<- not a homotopy theorist)

The fundamental group of loops , that’s what I think of

(^I actually need to use an up arrow to point at my name, don't I)

Not on desktop

Oh right

<^>v Spin attack

2:01 AM

That looks more like an SO(2) attack tbh.

Anyone interested in discussing markov chains

idk much about them myself

@Zee okay

@hungryWolf Once there were two Marks, which was unacceptable, so they challenged each other to a game of ping-pong to see who could keep their name and who would have to change theirs for the summer

It was a Mark-off

(I think Mark won)

@AkivaWeinberger I thought you were going to start an example with a story

2:07 AM

Has anybody read Milnor Morse theory ?

I think everyone does at some point

Have you ?

does anybody have a good reference for a topological dynamics text?

I am contained in the set of everyone

Have you at this point ?

2:08 AM

yes

Well the set you defined contained future people...

Anyway

fair enough

Was the middle two sections good preparation for riemanian geometry ?

@hungryWolf you start

I knew Riemannian geometry long before I read it

2:10 AM

Do you think it makes a good intro ?

I always thought it explains RG really badly but most people think it's great

Faust

blarg i escaped!

Was there something that you didn’t like ?

too short, no examples, no exercises

True

was The first chapter relevant to anything aside from Morse theory itself ?

2:12 AM

the section with focal points is useful for all math

all good math anyway

Lol good math

does it have any homotopy theory ?

there is homotopy theory involved, yes

Cool, I like homotopy theory

Was there anything estoric or was it all useful for “good math “?

like I said, everyone (should) know that book

if you don't use it at some point, you're probably doing something wrong

Ya , but I can’t afford to explore around at this time

But that book seems to fit the bill

It’s annoying that it has no problems though

I sat next to Milnor once actully, I was so nervous

2:18 AM

@JoeShmo ok

I've been reading about markov chains

I'll try to summarise what I've read

correct me if I'm wrong

you probably know more than me if you've actually been reading :-). but go ahead

We can calculate the probability of a thing to happen

like say the probability of tommorow being cloudy

if we assume there are just 3 types of weather: cloudy, rainy and sunny

that'd be 1/3rd probability

If we're able to find a way to observe the weather of a day based off weather of previous day

I mean if we can find any previous data

Let us say we're able to calculate the probability of today being sunny if yesterday was cloudy

etc.. and all 9 combinations sunny - rainy .. etc

Guys , can someone help me with something , how do I convert a line in cartesian form to vector form ?

$$\dfrac{x-1}{4}=\dfrac{y-4}{0}=\dfrac{z-6}{-2}$$

we using this to calculate the weather of say 20 days later, I'll call it a markov chain

as it depends on previous probabilities

Tanuj can you write it without Latex ?

2:26 AM

@Zee why ? Just enable the mathjax on your browser

Am I the only one who can't see latex in chat

Idk how...

facepalm

Am not good with technology

See , its in the description of the room

2:26 AM

@Tanuj you're too nice

okay, @Zee can't see too. thank god :b

I did go there but I was still clueless

in general, one could have a finite state machine, with edges labeled according to their probabilities, where the probability of arriving in any one state is the sum of the product of the edges that form a path from that state to an initial state.

@0celo7 Nah , in fact you are to think like that.

@Zee tinyurl.com/cfqcvpc

@Tanuj what

2:28 AM

@0celo7 nothing man. Do you have any idea about this though ?

indeed, the stable state of the markov process is the limit of A^n as n -> infinity; where A is the matrix representation of the markov process.

3 mins ago, by Tanuj

$$\dfrac{x-1}{4}=\dfrac{y-4}{0}=\dfrac{z-6}{-2}$$

^if such a limit exists.

I don't know how to divide by zero.

i.e. A converges

2:28 AM

Some people can divide by zero?

Oh , I guess on the Riemann sphere

@0celo7 nvm , it's just a ratio , we aren't infact dividing by 0

@Zee Can you help me with that ?

Okay, so in my example A = matrix(all 9 possibilities) I guess?

@0celo7 @AkivaWeinberger thanks, I see why it is true algebraically, but I’m not sure how to interpret it visually

I can’t help you if I can’t read it

I must be getting old

sorta. A_(i,j) = p(i,j) where p is the labeling function of probabilities on the edges

its not uncommon that A is a sparse matrix. I.e., most entries are 0's

2:31 AM

@Zee I just told you how to enable mathjax , or if you don't wanna do that for some reason , you can always copy the latex and paste it in the "ask question" section of any SE website

@LeakyNun Just to check: when $\partial\alpha=0$ in $C_{n-1}(X,A)$, then in fact $\partial\alpha$ is in the kernel of $C_{n-1}(X)\to C_{n-1}(X)/C_{n-1}(A)$, i.e. $C_{n-1}(A)$.

one decomposes A its Jordan Normal Form, A = PXP^-1, and thus A^n = PX^nP^-1

@0celo7 right

where X is quasi-diagonal, and therefore easier to iterate

@JoeShmo I think I may need to read more to understand what you are saying, these edges of state machine and stuff. Can you represent them visually, like a drawing or something?

2:32 AM

Tanuj that’s too much work

@Zee Just go here okay ? tinyurl.com/cfqcvpc

Dude I did

im just using fancy language. the one and only way to represent a finite markov chain is pictorially

Am on mobile maybe that’s why

are you familiar with a graph?

2:32 AM

@Zee drag the first link to your bookmarks bar , or just bookmark i

@Zee It's also given how to do that on mobile

there -- upload.wikimedia.org/wikipedia/commons/thumb/9/95/…

@JoeShmo NOT. I am reading markov chains examples as a part of Linear algebra course

yeah that makes sense. but are you familiar with a discrete graph?

@JoeShmo I am a total noob to math, sorry

I am not familiar with discrete graph

Just write it like we used to do back in the days

2:34 AM

that picture was helpful by the way

a graph is a bubble chart with arrows going between bubbles

Why are you making me do iT work ?

@Zee I've probably got a better idea.

just like that wikipedia picture?

2:35 AM

yes. just like that

that wikipedia picture happens to be a "directed" graph. because the arrows have direction

owh got it

What’s with the 0?

@Zee It's just a direction ratio , right ? We aren't actually dividing by $0$ , it's just a ratio.

@Semiclassical Can you help on this one

I don’t know what’s a direction ratio...

Well, the denominators of the two on the outside are nonzero in general

2:37 AM

@JoeShmo I am going to go read a few more examples. bye

bye all

so the left and right terms are finite. then so there's only one way that middle term can possibly make sense

TC

Bye

enjoy. come back with questions

Do you see what it is?

2:38 AM

owh yeah

bye @Zee

@Semiclassical yea , but how do I convert this to the vector form ? Like ax+by+cz+k=0 ?

How many free variables are in that expression you just wrote?

Leave it to semiclassical to solve a division by zero problem

@Semiclassical I don't get what you mean by free variables , 3 ? x, y and z ?

Can you pick x,y,z independently of one another in that equation?

2:40 AM

@Semiclassical No

Indeed not. What about x and y---can you pick those independently of one another? (I'm saying nothing about z in this case.)

@Semiclassical I'm not sure.I can I guess.

Correct.

there's no restriction on x and y, but once you choose them you know what z is

hence you can solve for z in terms of x,y. z is then a function of the two free variables x,y.

And that shouldn't surprise you, because that's the equation of a 2D object (a plane)

By comparison, a line is a 1D object. For that reason your options are more constrained.

specifically, there can only be one free variable here. for reasons that'll become obvious in your example, let's just call this free variable t rather than picking one of x,y,z

the point is now that, since you're working with a line, the coordinates x,y,z should be linear functions of t.

Can you write the equations for that? You'll need to include some coefficients; I don't really care how you label them.

well , can you show it how ? I'm really confused about the whole thing

I'm wililng to point you in the right direction, but I won't write the equations.

But start with this: What's the equation for x as a linear function of t?

2:48 AM

@Semiclassical yup , done

Okay. What is it?

$x=4t+1$

$y=4$ and $z=-2t+6$

Okay.

Which if you write your position vector as $\vec{r}(t)=(x(t),y(t),z(t))$, gives you a vector equation

so, what's $\vec{r}(t)$ here?

hmm , idk , do I just add $x$ , $y$ and $z$ ?

No.

A vector is not a sum of components.

A vector is a list of components.

2:52 AM

oh $\vec{r}(t)=(4t+1)i+4j+(-2t+6)k$ ?

That works, yeah. I'd also have accepted $\vec{r}(t)=(4t+1,4,-2t+6)=(4,0,-2)t+(1,4,6)$

And the thing is, that's the best you can do as far as the equation of a line.

In particular, you have to write $y=4$ in some form or another. the value of $y$ is fixed in this problem and doesn't change along the line

But $y=4$ is still linear insofar as it's of the form $y=mt+b$ albeit with $m=0$.

@Semiclassical So can I write this as ax+by+cz+d =0 ?

That would have two free variables. How many free variables does $\vec{r}(t)$ have?

I don't understand

Why?

How many numbers can I actually choose in $\vec{r}(t)=(4,0,-2)t+(1,4,6)$ (typo above)

Can I pick the x component independently of the y component or the z component?

2:59 AM

yes

nitsua60

@Semiclassical (fixed it)

So if I say x=4t+1=13 and y=-2t+6=3, I can do that?

nah

Correct. (Fun fact: I almost wrote x=4t+1=13 and y=-2t+6=0, in which case I'd have looked foolish since t=3 would work!)

So I can't pick x independently of z.

@Semiclassical I've had some cookies in my freezer for 5 months. Still safe to eat?

3:02 AM

yea

To put the point more sharply: Once I pick t, I know what x,y,z all are.

So there's only one free variable in this problem.

okay yes

For that reason alone, I should not expect to have a representation like ax+by+cz+d=0.

The closest you can get is to solve for t.

The closest you can get is to solve for t, i.e. $t=\frac{x-4}{1}=\frac{z-6}{-2}$

You'll note the absence of $y$ from that. $y$ doesn't depend on $t$, so you have to write separately that $y=4$.

got it.

That said, note that the ratios I just wrote can further be rearranged to $-2(x-4)=1(z-6)$ or just 2x+z=14

And that is the equation of a plane. But your line is not that plane, because your line lies both on that plane AND on the plane y=4

3:06 AM

@Semiclassical Let's suppose I'm also given a point on this line as $(5,4,4)$ can I now write this as $ax+by+cz+d=0$ ?

That is, your plane is the intersection of the plates 2x+z=14 and y=4.

In all of what I've written, how many free variables has a line had?

1

Right. ax+by+cz+d=0 has two free variables, so no: You can never express a line using a single plane.

At best, you can express it as the intersection of two planes.

@Semiclassical okay ! Thanks a ton :)

Backpropagation calculus | Appendix to deep learning chapter 3

one final point

when you wrote earlier $\frac{x-1}{4}=\frac{y-4}{0}=\frac{z-6}{-2}$

what I was attempting to hint is that, regardless of what x and z are, the values on the left and right are perfectly sound quantities.

If I know that x=5, for instance, then (x-4)/4=1 and I can deduce z=-4. In both cases, the fractions are fine.

That's not the case for (y-4)/0, since dividing a finite quantity by zero is undefined.

The only case in which this sorta works is if you have y=4, in which case you have 0/0 which is indeterminate.

That's still problematic and means you shouldn't write that those fractions are equal. But it does correctly suggest that y=4 is what you need.

(If y had any t dependence, then you'd have had (y-a)/b with b nonzero and this wouldn't have been a problem to begin with.)

Faust

3:28 AM

rock fairy farey

which is a real mathimagical series?

Niing

5:54 AM

After I dive into the math behind the neural-network, I found that there no so called "back-propogation", it's just that those weights and bias in each layer would influence the final cost of a this one single input.

Here is the link:

►

7:03 AM

Please have a look here

7:32 AM

why with this pde $u_{tt}-c^2u_{xx}=0$ and with change of variables $$x=\epsilon+n \\t=\epsilon-n\\\implies \epsilon=(x+t)/2,n=(x-t)/2$$ one gets $$$$

$\frac{\partial}{\partial x}(\frac{1}{2})\frac{\partial u}{\partial\epsilon}=\frac{1}{2}(\frac{\partial^2u}{\partial\epsilon\partial n})$?

please someone explain the equality between the 2 partials

are you sure you typed that correctly?

7:48 AM

yes

why?

@anon

just it has x,epsilon on the left and epsilon,n on the right

yes, they can also commute

since $\partial/\partial\epsilon = \partial/\partial x +\partial/\partial t$ and $\partial/\partial n=\partial/\partial x-\partial/\partial t$ that seems to say $\partial^2 u/\partial x\partial t=-\partial^2u/\partial t^2$

8:04 AM

I don't understand

what don't you understand in my message?

wait a sec. I'm trying to understand again

$\partial u/\partial x=\partial u/\partial\epsilon+\partial u/\partial n$?

8:22 AM

that is true

well, missing a 1/2 tho

$\partial/\partial x=\frac{1}{2}(\partial/\partial\epsilon+\partial/\partial n)$

it's the chain rule $\partial f/\partial x=(\partial \epsilon/\partial x)\partial f/\partial\epsilon+(\partial n/\partial x)\partial f/\partial n$

with $\partial\epsilon/\partial x=1/2$ and $\partial n/\partial x=1/2$

8:50 AM

:) got it

finally...

thanks @anon

Marks Polakovs

10:00 AM

I'm completely stumped on a simple probability/combinatorics question.

`A class consists of $6$ students from City A and $8$ students from City B. A committee of 5 students is chosen at random from the class. Find the probability that there are $3$ students from City A and $2$ students from City B.`

My original reasoning was $\frac{{6\choose 3}*{8\choose 2}}{{11\choose 5}}$, but this gives an answer $> 1$

6+8=14 not 11

Marks Polakovs

facepalms Thank you!

Balarka Sen

@anon quickmaths

PrithiviRaj

10:17 AM

in CSIR-TIFR-ISI-NBHM, 21 mins ago, by PrithiviRaj

17 hours ago, by PrithiviRaj

May we ask questions like "interview experience and questions asked at ISI(M.MATH) / CMI(M.Sc. Mathematics) / IISERs and IISc for Integrated Ph.D admission" ??

on MSE

@BalarkaSen

Balarka Sen

That probably would not be on-topic.

PrithiviRaj

okay !

Is there any other way to know this?

10:45 AM

How to write transposition $(1,3)$ as product of $(1,2)$ and $(2,3)$?

try

Shobhit

If $f$ is a non-negative function in $M(X$,X), then there exist sequence $(\phi_{n})$ in $M(X$,X) such that, $(\phi_{n})$ is a monotonically increasing sequence converging to $f$, and each $\phi_{n}$ has only a finite number of real values, where a function in M(X,X) means it is real valued X-measurable function from $X$.

The proof given in my book for this uses the construction of sets $E_{kn}=\{x \in X:k2^{-n} \le f(x) <(k+1)2^{-n}\}$ for $k=0,1,..,n2^n-1$ and $E_{kn}=\{x \in X : f(x) \ge n\}$ for $k=n2^{n}$. I dont understand the idea behind this, why this? how did they come up with this?

$M(X,{\bf x})$

Shobhit

too late to edit the message

$(1,2)(2,3)(1,2)$

10:49 AM

right

heh