Mathematics - 2014-08-26 (page 1 of 3)

Mike Miller

00:14

Hiya @mixedmath

Jose Antonio

01:46

Hi I have other question which is easy I think

are there someone here?

well I'll post it Hi I´m continue working in group theory and I´m again stuck, this problem says: suppose $N_i$ for $1\le i \le k$ normal subgroups of $G$ and define $N$ as the intersection of the normal subgroups. Then there is an isomorphism between $G/N$ to a subgruop of $\prod G/N_i$.

Vibhav Pant

Hello

Jose Antonio

FIrst is almost trivial to show that $N$ is a normal subgroup. And I define the map as follows: $gN\mapsto \prod G/N_i$ so is not too difficul to show that the map is well defined and is injective and also is an homomorphism, so is sufficient to say something like is an isomorphism to the image of the map?

@VibhavPant hi

Vibhav Pant

So I was trying to get c omplex plot of $e^{2/3z\pi i} -1$, the plot should be zero (black) at integer values of $z$, right?

@JoseAntonio hi

Mike Miller

02:10

@JoseAntonio Remember that an ismorphism is just a bijective homomorphism - and it's a bijective homomorphism onto its image, so yup

Jose Antonio

First is almost trivial to show that $N$ is a normal subgroup. I define the map $f$ as follows: $gN\mapsto (gN_1, \ldots gN_k)$.

$f$ is **well-defined**. For is $hN=kN\iff h^{-1}k \in N$ and so $h^{-1}k \in N_i$ for all $i$. Hence $(hN_1, \ldots hN_k)=(kN_1, \ldots kN_k)$ as desired.

$f$ is **one to one**. If $(hN_1, \ldots hN_k)=(kN_1, \ldots kN_k)$, it follows that $\forall i (h^{-1}k \in N_i)$ i.e. $h^{-1}k\in N$ as desired.

$f$ is **homomorphism** Clearly we have a structure of a group. Then $f(gN kN)=f(gkN)$, so $(gkN_1, \ldots gk_k)=(gN_1kN_1, \ldots gN_kkN_k)=f(gN)f(hN)$

@MikeMiller thank as usual

Mike Miller

You got it!

Jose Antonio

@MikeMiller Thanks :)

Mike Miller

04:26

@TedShifrin This answer is incorrect, isn't it? Null-homotopy of pairs generally means that, for any $f_t$ in the homotopy, the second term maps to the second term. But there's no reason AFAICT that $M-D$ should map to $x_0$ in some potential null-homotopy.

Alas, you are but a ghost

skullpatrol

afloat in the ether of the internet...

mixedmath

04:42

@MikeMiller hiya! (only a few hours late)

skullpatrol

hi @mixedmath

mixedmath

@skullpatrol hello

Mike Miller

@mixedmath and hello back

skullpatrol

Greetings @Chris'ssis

Ted Shifrin

@Mike: AFAICT, he also confuses open and closed disks. He also doesn't relate $H_*(X/A)$ to $H_*(X,A)$, which can be done. I dunno. Way past my bedtime. Generally, I don't trust him, but this approach might work. I personally prefer mod 2 degree.

skullpatrol

04:54

Hi Professor @TedShifrin

Mike Miller

@TedShifrin Yup. He answered a tangent bundle question coherently, just now (showed $S^1 \times S^n$ was parallelizable by using your approach... embedding $S^n \times [0,1]$ in $\Bbb R^{n+1}$, used the constant vector fields in each direction, and pushed forward to $S^n \times S^1$

Jose Antonio

05:11

Hi

skullpatrol

hi

Jose Antonio

Again I need some hint with other problem of algebra... the entire day was dedicated to solve a lot of problems, abusing of this place do you think is possible give me a hint?

I'm stuck with the following. Let $N$ a normal subgroup of $G$. Show that if $[G:N]=4$, exists a normal subgroup $M$ of $G$ s.t. $[G:N]=2$. Since $G/N$ has orden 4, either is isomorphic to $\mathbb{Z}/4$ or $\mathbb{Z}/2 \times\mathbb{Z}/2$ and in any case we have a subgroup of order $2$ and for instance the same applies to $G/N$ but from here I'm not sure how to get the normal subgroup $M$ of $G$

sometimes I feel so stupid, hehe

blue

05:40

lattice correspondence

subgroups between G and N correspond to subgroups of G/N

the correspondence preserves indices; a subgroup of G/N of index k corresponds to a subgroup of G containing N of the same index k

note that the correspondence is as follows: given a subgroup of G containing N, take its image under the projection G->G/N to get the subgroup of G/N; given a subgroup of G/N, take its preimage under G->G/N (equivalently, take the union of all of the cosets of N it contains) to get a subgroup of G containing N.

Vrouvrou

06:25

Hello, help please :math.stackexchange.com/questions/909005/… if you can

skullpatrol

@Vrouvrou have you tried the homotopy room?

Vrouvrou

there is a homotopy room ?

skullpatrol

@Vrouvrou MO

Jose Antonio

@blue thank I think I got it math.stackexchange.com/questions/909436/…

Chris's sis

06:46

Greetings @robjohn @r9m

Here is a cute limit $$\lim_{x\to1} (2-x) ^{\tan(\pi x/2)}$$

to be done elementarily

@skullpatrol Hi

skullpatrol

@Chris'ssis Hi

Chris's sis

07:07

@r9m @robjohn have you seen this one? $$\lim_{x\to\infty} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \log(x+k)$$

robjohn

@Chris'ssis $$\lim_{u\to0}(1+u)^{1/\tan(\pi u/2)}=e^{2/\pi}$$

Chris's sis

@robjohn Indeed.

robjohn

@Chris'ssis is that not $0$?

Chris's sis

@robjohn Sorry, I was away. No, it's not $0$.

robjohn

For $n=1$, it is $\lim\limits_{x\to\infty}\log\left(\frac{x}{x+1}\right)=0$ and I don't see it being different for higher forward differences.

Chris's sis

07:24

@robjohn Sorry, it's $0$. I mixed $2$ limits.

robjohn

@Chris'ssis If it were multiplied by $x^n$, it might yield something interesting.

Chris's sis

@robjohn Really? Let me see ...

robjohn

$-(n-1)!$ I think

FractalHand

Is it possible to take differential equations course by knowing calculus I and a little bit of calculus II ?

robjohn

@FractalHand I don't know how your calculus classes are divided up, but the more calculus, the better for DE.

Chris's sis

07:35

@robjohn How did you get that?

robjohn

@Chris'ssis The formula you have there is $(-1)^n$ times the $n^\text{th}$ forward difference, which goes like the $n^\text{th}$ derivative. The $n^\text{th}$ derivative is $\frac{(-1)^{n-1}(n-1)!}{x^n}$

Chris's sis

@robjohn Ah, I see now.

Chris's sis

08:00

Let $f$ be a continuous real-valued function on $[-1,1]$ such that

$$a_n=N(n)\int_{-1/n}^{1/n} f(x) \ dx $$
where $N(n)$ is the number of perfect squares between $n^3$ and $(n+1)^3$. Compute

$$\lim_{n\to\infty} \sqrt{n} a_n$$

Chris's sis

08:11

Initially I wanted to write "compute it without pen and paper" but then I realized this might be annoying.

Ussing my immagination, I see that near $x=0$, we have that the integral can be approximated by a rectangle $$\int_{-1/n}^{1/n} f(x) \ dx \approx \frac{2}{n}\cdot f(0)$$

The number of perfect squares is $$N(n)\approx n^{3/2}$$

Q.E.D.

Ups, I did a mistake ...

I wanted to say that $$N(n)\approx \frac{3}{2} \sqrt{n}$$

Hence, we get that $$\lim_{n\to\infty} \sqrt{n} a_n=3 f(0)$$

This is a challenge I received this morning from a kid. It was pretty fun to attend it.

How about this version? $$\lim_{n\to\infty}n\left( \sqrt{n}N(n)\int_{-1/n}^{1/n} f(x) \ dx-3f(0)\right)$$

I need to buy a new keyboard ... some letters remain blocked here ...

Balarka Sen

08:41

@Chris'ssis Right. $N(n)$ is nowhere near $n^{3/2}$

Chris's sis

@BalarkaSen I thought of the correct thing, but I wrote another thing.

Balarka Sen

haha, how that happen to us all.

Chris's sis

:D

Balarka Sen

@Chris'ssis Do you have any interesting series question? maybe Euler sums?

Chris's sis

@BalarkaSen Isn't my limit above interesting? :-)

Balarka Sen

08:48

@Chris'ssis series

Not limit!

Especially a limit with an integral inscribed! Ughh

Chris's sis

@BalarkaSen OKAY ...

Prove that

$$\sum_{n=2}^{\infty} \frac{n(1-x)+2}{n(n+1)(n+2)}x^{n+1} (n-\zeta(2)-\zeta(3)-\cdots -\zeta(n))$$ $$=\log\left(\Gamma(2-x)\right)+\frac{3-\log(2\pi)}{2}x +\left(\frac{\gamma}{2}-1\right)x^2+\frac{1}{6}x^3-x \log\left(\Gamma(2-x)\right)-\log(G(2-x))$$

Balarka Sen

No, no, that's too much for me.

=)

Chris's sis

:D

Balarka Sen

If you have something a little elementary, ping it to me.

Otherwise, leave it.

rehband

May I suggest the following: Calculate $$\sum_{k=1}^{\infty} \frac{1}{k^n(k+1)}$$

:)

Balarka Sen

08:52

@rehband Partial fractions?

rehband

@BalarkaSen I don't know yet

Balarka Sen

$$\frac1{k(k+1)} = \frac1{k} - \frac1{k+1}$$

rehband

Sorry there is a typo!

1 sec

Chris's sis

@BalarkaSen Good point!

Balarka Sen

Wait, what?

rehband

08:54

No, there was no typo, I'm confusing problems :D

Sorry haha

Balarka Sen

@rehband =P

Well, there you have it.

rehband

True!

Balarka Sen

@Chris'ssis I am getting used to all this. Just a beginner.

=)

Chris's sis

@BalarkaSen Well, there is always a beginning, isn't it?

Balarka Sen

Ah, true.

rehband

08:57

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

Chris's sis

So, you have $$\frac1{k(k+1)} = \frac1{k} - \frac1{k+1}$$ What's next? :-)

@rehband Right, this is the next step. :-)

Balarka Sen

@rehband Keep partial factorizing.

rehband

Yes, keep splitting it up

Balarka Sen

Bleh.

Chris's sis

@BalarkaSen What do you mean? It's correct.

Balarka Sen

08:59

You're right.

Wait, is it going to be $\zeta(n) - \zeta(n - 1) - \zeta(n - 2) - \cdots - \psi(1) - \gamma$?

rehband

@BalarkaSen What's the $\Psi$?

Balarka Sen

Digamma function

rehband

@BalarkaSen What's the Digamma function? :D

Balarka Sen

mathworld.wolfram.com/DigammaFunction.html

rehband

@BalarkaSen Thx

Balarka Sen

09:05

Well, it's $\zeta(n) - \zeta(n- 1) - \zeta(n - 2) - \cdots - \zeta(2) - 1$

@Chris'ssis

By the way, @rehband, you can handle your mistyped sum in a similar way

Chris's sis

@BalarkaSen I don't help you this time ... :-)

rehband

@BalarkaSen Ah okay

Balarka Sen

@Chris'ssis No need to help, it's $\zeta(n) - \zeta(n- 1) - \zeta(n - 2) - \cdots - \zeta(2) - 1$

Just check the solution.

Nah

Chris's sis

@BalarkaSen Well, consider the simple case, $n=1, n=2, n=3$.

Balarka Sen

Ah, right, I mistook.

It's alternating.

rehband

09:11

Yes, nice, I got that too now

Balarka Sen

$$\sum_{k = 1}^\infty \frac1{k^n(k + 1)} = \zeta(n) - \zeta(n - 1) + \zeta(n - 2) - \cdots + \zeta(2) - 1$$

Of course, signs are show-off. It depends on the parity.

Chris's sis

@BalarkaSen There is a problem with the signs ...

Balarka Sen

@Chris'ssis Nothing, only that the sign I wrote holds for $n = 0 \bmod 2$

For $n = 1 \bmod 2$, it'd be $\zeta(n) - \zeta(n - 1) + \zeta(n - 2) - \cdots - \zeta(2) + 1$

Chris's sis

@BalarkaSen For instance, look at the last term for first few cases.

Balarka Sen

@Chris'ssis Just read what I wrote

Chris's sis

09:18

@BalarkaSen OK

Indrajith Indraprastham

Hello I am trying to solve this problem: "For each function f(n) and time t in the following table, determine the largest size n of a problem that can be solved in time t, assuming that the algorithm to solve the problem takes f(n) microseconds." This is the chart with answer : clrs.skanev.com/01/problems/01.html Can any one tell me how answer is found for (n Log n) and n! ..

Chris's sis

@BalarkaSen I want a beautiful solution there ...

Balarka Sen

@Chris'ssis Of what?

Chris's sis

@BalarkaSen $$\sum_{k = 1}^\infty \frac1{k^n(k + 1)} $$

@BalarkaSen Can you show me one?

Balarka Sen

@Chris'ssis Ah! I know! I know!

Chris's sis

09:26

@BalarkaSen :D

Indrajith Indraprastham

@BalarkaSen Do you know to find that answer?

Balarka Sen

@IndrajithIndraprastham No.

@Chris'ssis Wait a sec. Let me think about it. I think I have an idea.

$$\text{Li}_n(x) = \sum_{k = 1}^\infty \frac{x^k}{k^n}$$

Integrating w.r.t. $x$. and then setting $x = 1$ gives the desired.

Chris's sis

@BalarkaSen Horrible ...

Balarka Sen

Snap. Well, apparently that's not what you had in mind.

skullpatrol

The horror, the horror,...

Balarka Sen

09:31

@Chris'ssis I'm out of ideas.

Show me what you have.

Chris's sis

@BalarkaSen use $$a_n=\sum_{k = 1}^\infty \frac1{k^n(k + 1)}$$

combined with the difference above.

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

Balarka Sen

@Chris'ssis That looks almost the same as repeating the partial fraction method to me.

rehband

$$\zeta(n) - a_{n-1}$$

Balarka Sen

But good solution nevertheess

@rehband yeah that

Or $a_n + a_{n-1}$

@Chris'ssis Your proof uses almost the same machinaries, same ideas, same identity, just writes up stuffs a bit tidier. Why should I believe that it's of interest?

No, @rehband

Chris's sis

@BalarkaSen It is the brick of some very hard series.

rehband

09:41

:) sry

Balarka Sen

$$a_n + a_{n - 1} = \zeta(n)$$

rehband

Yep

Balarka Sen

@Chris'ssis This series? Or the solution?

I am talking of the solution.

Chris's sis

@BalarkaSen Not solution, but the series.

Balarka Sen

Your solution and the one by partial fractions are almost the same.

Chris's sis

09:42

@BalarkaSen Well, writing things as a recurrence relation makes the whole thing so easy.

Balarka Sen

@Chris'ssis As I said, it writes stuff up a bit tidier, nothing more.

I need to run. Byes. Have fun with series.

Chris's sis

OK. Bye. :-)

rehband

Take care!

Jasper Loy

Nice bird @FractalHand

FractalHand

@JasperLoy yeah

Jasper Loy

09:50

@FractalHand What brings you to this chat?

FractalHand

@JasperLoy finding some people with same interest, particularly, learning.

Chris's sis

$$\sum_{k = 1}^\infty \frac{(-1)^{n+1}}{k^n(k + 1)}=1-\zeta(2)+\zeta(3)-\zeta(4)+\cdots+(-1)^{n+1} \zeta(n)$$

rehband

@Chris'ssis Nice

Chris's sis

@rehband Thanks :-)

Chris's sis

10:13

$$\lim_{n\to\infty} (1+1-\zeta(2)+1-\zeta(2)+\zeta(3)+1-\zeta(2)+\zeta(3)-\zeta(4)+\cdots +1-\zeta(2)+\zeta(3)-\zeta(4)+\cdots+(-1)^{n+1} \zeta(n))$$

rehband

@Chris'ssis Weird limit! :P

Chris's sis

@rehband YES!!! :-)

rehband

@Chris'ssis Reminds me a little bit of problem 1.31 in Furdui's book

Chris's sis

@rehband Yeah, I know it. That's a very nice limit.

Chris's sis

10:43

@rehband That one diverges.

rehband

@Chris'ssis Hmm okay, how did you find that? :)

Chris's sis

@rehband $$\sum_{n=1}^{\infty} (-1)^{n+1} \underbrace{\sum_{k = 1}^\infty \frac{1}{k^n(k + 1)}}_{\displaystyle a_n}$$ and this fails the alternating series test.

rehband

@Chris'ssis Hmm okay

Chris's sis

@rehband Do you think I'm wrong? Look at what happens with $a_n$ when $n$ is large, it's almost $1/2$. Since it's an alternating series, we have after a while something like $-1/2+1/2-1/2+ ...$.

rehband

@Chris'ssis No, I just don't fully understand. What do you mean when you say it fails the alternating series test?

Yes

Chris's sis

10:51

Alternating series test

The alternating series test is a method used to prove that infinite series of terms converge. It was discovered by Gottfried Leibniz and is sometimes known as Leibniz's test or the Leibniz criterion. == Formulation == A series of the form Or, where an are positive, is called an alternating series. The alternating series test then says if {an} decreases monotonically and goes to 0 in the limit then the alternating series converges. Moreover, let L denote the sum of the series, then the partial sum approximates L with error bounded by the next omitted term: == Proof == Suppose we are given a...

rehband

Right!

Chris's sis

@rehband Swapping the summation order, one gets a wrong result. Actually this is what I wanted to point out.

rehband

@Chris'ssis Oh ok, got it

Chris's sis

@rehband That would have been $$\zeta(2)-1$$

skullpatrol

@Chris'ssis have you ever thought about practicing your writing skills on Wikipedia?

Chris's sis

10:54

@skullpatrol No. I don't think I have writing skills. :-)

skullpatrol

It would be a great place to develop them.

Balarka Sen

@skullpatrol Wiki removes contributions coming from personal calculations.

skullpatrol

And, of course, learn from others

Just a suggestion :-)

Balarka Sen

Most of what @Chris'ssis do are her own discoveries. Wiki only takes stuffs with unbiased references.

The best thing would be to set up a blog or something.

@skull I just read Poe's Raven. Eerie.

Balarka Sen

11:11

@TedShifrin Hello!

Chris's sis

11:29

@rehband leaving out the first term of the inner series, one gets $$\zeta(2)-\frac{5}{4}$$

skullpatrol

I just figured out that I have to acknowledge pings on my iPhone by tapping on them :-)

rehband

@Chris'ssis Ah okay

Chris's sis

brb

Huy

Yay, finally through with all exams. Now I can finally buy some more furniture and plants and make this place look nice. ^_^'

skullpatrol

11:44

How many did you write?

rehband

@Huy Congrats!

Huy

@skullpatrol: I just did two this semester, quantum mechanics and functional analysis, because I got a nice opportunity to work 50%.

skullpatrol

cool

rehband

12:23

@Chris'ssis Crazy trick: $$\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i+j} = \sum_{i=1}^{n} \sum_{j=1}^{n} \int_0^1 x^{i+j-1}dx$$ I would've never thought of writing it that way :D

Chris's sis

@rehband For computing the limit, when $n\to\infty$, just apply Stolz-Cesaro theorem.

rehband

@Chris'ssis Okay :) This is for proving that $$\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i+j} = 1 + (2n+1)H_{2n+1}-(2n+2)H_{n+1}$$

Chris's sis

@rehband This is from Ovidiu's book as well.

anorton

@robjohn I don't know if this has already been discussed, but I think that these two users are attempting to impersonate this user.

rehband

@Chris'ssis Yep:)

anorton

12:34

(Based on the comments in this question (10k) and this question, it seems that they're trying to discredit those usernames, which are commonly associated the user now known as Thursday.)

Chris's sis

13:06

@robjohn have you seen this one? It's cool I think. $$\lim_{n\to\infty} (1/2+3/2-\zeta(2)+1/2-\zeta(2)+\zeta(3)+3/2-\zeta(2)+\zeta(3)-\zeta(4)+\cdots +(1+(-1)^{n}/2)-\zeta(2)+\zeta(3)-\zeta(4)+\cdots+(-1)^{n+1} \zeta(n))=\zeta(2)-\frac{5}{4}$$

lol, my limits don't fit the screen anymore. :-)

Jose Antonio

hi guys a stupid question. suppose $G$ is simple and let $f$ be an homomorphism between $G \to H$. If $\#G\ne2$, $A\lhd H$, and $[G:H]=2$. Then $\text{Im}f$ is a subgroup of $A$. I know that the kernel of the homomorphism is a normal subgroup of $G$ so either is the trivial subgroup or the entire group. The last case is trivial, because everything is send to $1$ in $H$, and clearly the trivial group is a subgroup of the $A$.

The case when the kernel is trivial is my problem. Here we can conclude that the orden of the image is not two, because is the same as $G$, and also divides the orden of $H$, i.e., $#H=[H:A]\#A=2\#A$ i.e., the image divides the even number. But from here I'm not sure of how proceed, any idea ? Thanks in advance

Chris's sis

I just took an interesting test.

Wait, to give you the link, only 10 questions.

Jasper Loy

@Chris'ssis I think these tests are very silly, lol.

Chris's sis

@JasperLoy No, they are not at all! Not this one!

Here is the test - bitecharge.com/play/career

I'm shocked by the precision of this result. I only heard that some time ago from a psychologist.

Balarka Sen

13:21

silly enough.

heya @blue

blue

heya

Jasper Loy

bitecharge.com/play/career?sess=r5#r52821795563887187 @Chris'ssis is mine

Mats Granvik

Chef

You're naturally creative, imaginative, and you can instantly come up with new ideas in your head. More than anything you want to create something beautiful and enjoyable that will satisfy and make people happy. You have an uncanny ability to make masterpieces out of sheer nothingness.

robjohn

@anorton Thanks. I've brought this up with the other mods earlier today. I hope no one is really fooled.

Balarka Sen

i won't even click it.

robjohn

13:24

@Chris'ssis I don't know if I've seen that one exactly. I've seen some things that I think are similar. Neat :-)

Jasper Loy

@BalarkaSen Better than using representation theory to discover the meaning of life

Chris's sis

@JasperLoy Well, I think there is much truth in what I read there (although I only know you a bit). Do you agree with that result?

Jasper Loy

@Chris'ssis Sort of. I once took another long test and the result is the same, lol.

Chris's sis

@MatsGranvik Nice result! Do you agree with the result?

Mats Granvik

@Chris'ssis Well I like bananas with peanut butter, if that is what chef means.

Chris's sis

13:29

@MatsGranvik lol :-)

Mats Granvik

First meaning is probably person in charge like a boss. That I don't know. At work I am lowest in rank.

Chris's sis

@MatsGranvik This doesn't mean that one that you won't be a boss. This is just the beginning.

Chris's sis

13:41

@BalarkaSen did you take the test?

Balarka Sen

nopes.

Jasper Loy

@Chris'ssis I showed the people in the Eng room the test too.

Balarka Sen

as i said, i didn't even click it.

Chris's sis

@JasperLoy Nice. Do they take it?

@BalarkaSen OK

Jasper Loy

@Chris'ssis Yes, some of them.

I just realised that @blue almost never talks to me, lol.

blue

13:43

:/

Chris's sis

@robjohn Glad you like it! :D

Balarka Sen

@JasperLoy blue talks mostly about mathematics. you almost never talk of mathematics.

Jasper Loy

@BalarkaSen Yes. I am more interested in building relationships in this chat.

Balarka Sen

@blue You never saw my explanation of gram phenomenon.

le sigh

Jasper Loy

13:58

I form attachments to people quite easily.

robjohn

@JasperLoy great, you build relationships so that we can mourn when you delete your account? :-p

Jasper Loy

@robjohn I think I have promised somewhere in this chat never to delete my account again!

robjohn

@JasperLoy yes, several times, if I recall :-)

Jasper Loy

@robjohn This time, it is real, lol.

robjohn

I'll believe it if I don't see it!

3

Zachiel

15:55

Hello, people of math.SE

leo

@Zachiel hello

Zachiel

I'm not really great at math but I've got a question opened on rpg.se that might use some research on my part, so if anybody here wants to talk about optimization problems I'd be happy to learn something from you people. I have some basics (knapsack problem, rectangular distances, system solving for maximums).
Of course, since I'm basically asking for free help, I don't really expect much, but as they say "never ask, never get".

I would link the question, but I guess there would be too many subculture assumptions for you to parse it.

Transcript for

$Mathematics$ Mathematics