Mathematics - 2017-02-08 (page 1 of 2)

12:00 AM

So I'm trying to prove that $SO(2)$ is connected, it does make sense that given two matrices $A,B \in SO(2)$, we express each as a rotation and change one angle into another, thereby getting a path between them, right?

Gridley Quayle

So in one we have (x-e,x+e) a subset of {x} but not the other?

Balarka Sen

My favorite way to do it is Gram-Schmidt

Ted Shifrin

Well, that notation doesn't make sense, @Gridley, in general, but basically right. In one case you have an open ball $B(x,\epsilon) = \{x\}$.

Akiva Weinberger

I feel like Ivan Niven's one-page proof of pi's irrationality might actually be more complicated than it needs to be. (My guess is that I'm wrong, though)

Daminark

@Balarka Which of us were you talking to?

Balarka Sen

12:01 AM

To you :)

Ted Shifrin

OK, I'm outta here. My back is killing me.

Gridley Quayle

Yeah sorry. I thought singleton's were always closed

Daminark

Hmm, I don't see just yet how that'd work

See you @Ted

Akiva Weinberger

@GridleyQuayle Things can be both open and closed at the same time

Gridley Quayle

Thanks for your help.

Akiva Weinberger

12:02 AM

Especially in disconnected spaces

(Are you doing metric spaces or general topological spaces?)

(In the former, singletons are always closed.)

Gridley Quayle

We started we metric's and have moved to the general case

Balarka Sen

@Daminark Pick a path in SL_n and then Gram-Schmidt it to fit in SO(n)

Gridley Quayle

Another question I had is: when is frechet differentiable the same as gateaux differentiable?

Simply Beautiful Art

Has anyone in here studied oodle theory?

mt3

i wish ams notices go back to the old webpage design

Daminark

12:04 AM

Ah, well, we don't yet have that $SL(n)$ is connected :/

Balarka Sen

You can definitely prove that!

Gridley Quayle

In the lecture where we proved the MVT for higher dimensions I think my lecturer said that it they were equivalent for the purposes of the theorem

OneRaynyDay

12:16 AM

Hey guys! Can someone explain a little bit of functional basics to me in this example?:

I don't understand why the integral just vanishes here.

is it because we are evaluating at one of the specific slices of the integral?

and the constants get dropped outside of the integral?

Xam

Hello guys, does anyone know about primal sets?

Daminark

@Balarka I will definitely attempt that at some point soon, it sounds neat

For now, though, is it true that everything in $SO(n)$ can be written as $\begin{pmatrix}\cos(\theta)&-\sin(\theta)&0\\ \sin(\theta)&\cos(\theta)&0 \\ 0&0&I_{n-2}\end{pmatrix}$?

Well I only need it for $SO(3)$

But is it true in general?

Simply Beautiful Art

Then rotate the rectangle and repeat infinitely

Akiva Weinberger

12:40 AM

@SimplyBeautifulArt Ha. The trouble is in assuming that the removed triangles are isosceles (they aren't).

Simply Beautiful Art

Hehehe

:D

Love the little brain teasers

Adeek

12:55 AM

hey

just a quick question.

verification I guess not a question

Zach Hauk

hi

everyone

Akiva Weinberger

@Adeek Yes?

Simply Beautiful Art

@Adeek Please ask

Zach Hauk

@Akiva what studies pertain to symmetric polynomial and that stuff

algebraic geometry?

Akiva Weinberger

Galois theory, I think

And perhaps ring theory

I'm not an expert in either of these, though.

Simply Beautiful Art

1:02 AM

:|

But oodle theory

Akiva Weinberger

Actually, Wikipedia offers this:

> Symmetric polynomials are important to linear algebra, representation theory, and Galois theory. They are also important in combinatorics, where they are mostly studied through the ring of symmetric functions, which avoids having to carry around a fixed number of variables all the time.

@ZachHauk

https://en.wikipedia.org/wiki/Symmetric_polynomial#Symmetric_polynomials_in_alg‌ebra

Zach Hauk

i didnt get above the AIME cutoff

which im glad for, because that means i can study for another year

Akiva Weinberger

I don't know yet if I did

I suppose we're probably not allowed to discuss the problems yet.

Which one did you take? AMC 12 or 10?

Zach Hauk

10

Akiva Weinberger

Ah

Zach Hauk

1:06 AM

the only reason i was able to take it is because i got the highest school grade for amc 8

and so they send the highest scorer from the middle school to the public high school near me

to participate in the amc 10

Akiva Weinberger

What grade are you in, again?

Zach Hauk

so i didnt have much choice in the matter

8th.

and i feel like wasted potential.

Akiva Weinberger

Ah. I thought you were in high school for some reason.

"Wasted potential"??

Zach Hauk

nope, almost.

yeah, i spend most of my time doing other stuff

when i could be studying a lot harder, as i did in past years

Akiva Weinberger

You have three more years to try at the AMC 10, you know

Zach Hauk

1:08 AM

yeah, i know

well two now

and then two more for AMC 12

Akiva Weinberger

Oh, right

Zach Hauk

AMC 12 cutoff is a bit more lenient, right?

Akiva Weinberger

I think so.

Zach Hauk

hm, if you dont mind me asking

Akiva Weinberger

Well, fewer points, but also a smaller percentage of those who take it (IIRC)

Zach Hauk

1:08 AM

what scores have you gotten in previous years?

Akiva Weinberger

Perfect score on the AMC 8. Enough to pass to the AIME when I took the AMC 10 in 9th grade.

In 10th grade they made me take the AMC 12 instead of the 10, and I went to the AIME again

(I didn't pass the AIME either time)

And then 11th grade AMC was this morning.

Zach Hauk

yeah, it was this afternoon for me

i didnt do so well, only started studying for this like 4 days ago lol

i think my maximum score for amc assuming the ones i put down are correct is 105

Simply Beautiful Art

How was it @AkivaWeinberger ?

Zach Hauk

which is pretty garbage

because im trash :v

Akiva Weinberger

@SimplyBeautifulArt I wasn't able to do five questions (#17 and #22-25) since I ran out of time

but I'm pretty confident about the others?

Simply Beautiful Art

1:12 AM

D:

what is the AMC?

Zach Hauk

american mathematics competition

Simply Beautiful Art

ah

Zach Hauk

qualifies you for a series of exams leading to the USA IMO team

Simply Beautiful Art

WHOA WHOA, WAIT A MINUTE! HOW DO I SIGN UP?

and what's it about?

Zach Hauk

well

howq old are you?

Akiva Weinberger

1:13 AM

@SimplyBeautifulArt Are you from the US?

He's high school, I think

Zach Hauk

AH everyones smarter than me

Simply Beautiful Art

Yup

17

Zach Hauk

anyways just ask your school to order copies

next year

Akiva Weinberger

@SimplyBeautifulArt So my age ^_^

Simply Beautiful Art

^_^

Woo hoo

:D

What kinds of questions @AkivaWeinberger

Zach Hauk

1:14 AM

geometry, algebra, probability

are the main ones

Akiva Weinberger

You can look up past ones on the aops website

They're multiple choice. Nothing needs calculus.

Zach Hauk

except you can use calculus of course

Akiva Weinberger

You can, you don't need to

Zach Hauk

i didnt get any of the #20-25

Simply Beautiful Art

ok

Zach Hauk

1:16 AM

its weird

last year the highest amc 10 score at the high school was like 87

but the highest amc 12 score was like 130

Simply Beautiful Art

Hm

I hate geometry and probability

except some geometry

but too many annoyingly unintuitive formulas IMO

Zach Hauk

most of it isnt formulas...

some common theorems are like

power of a point

angle bisector theorem

Akiva Weinberger

It's out of 150 points, if I remember correctly. 6 points for right answers, 1.5 for skipped, and 0 for wrong answers. (So the expected points obtained by guessing randomly from the five choices is the same as the points you get by skipping EDIT: no)

Wait

Zach Hauk

uhh

wouldnt it have to be

1.2 for skipped

Akiva Weinberger

I assumed it was true without double-checking that

Zach Hauk

1:19 AM

in order to be the same

Akiva Weinberger

Huh, so skipping is actually more than guessing randomly

Weird.

It would be the same if there were only four choices per question.

Simply Beautiful Art

ok

Fargle

@AkivaWeinberger I guess they'd prefer an admission of a lack of knowledge over an attempt to bold-faced guess.

Zach Hauk

@AkivaWeinberger does that mean we can talk about it at 12:00am?

Akiva Weinberger

I dunno

Simply Beautiful Art

1:23 AM

artofproblemsolving.com/wiki/index.php/…

Oh, not too bad

Akiva Weinberger

@SimplyBeautifulArt The first few problems are always ridiculously easy.

Simply Beautiful Art

lol

Zach Hauk

@AkivaWeinberger where do you go for AIME?

Akiva Weinberger

They just set me up in a random room in my school the last two times

I don't know what they usually do

Ted Shifrin

DogAteMy: You still inside-out and upside-down? :)

Simply Beautiful Art

1:27 AM

I assume no calculators?

Akiva Weinberger

@Simply: Yes. @Ted: What?

Uh, maybe?

Zach Hauk

hey @Ted

Ted Shifrin

hi Zach

yh05

Which of the following functions is a solution of
some differential equation of the form
y’=f(y),
where f is continuously differentiable for all real y?

Zach Hauk

i didnt do well on the AMC10 but i have 4 years

left

yh05

1:30 AM

Can i have some hints to this problem

Ted Shifrin

Zach: I've never been a very good tricky competitive test taker, either.

Turns out I can write them pretty well, though :P

Zach Hauk

Ted's Geometry Test

1. Prove that there are infinitely many inscribed triangles of maximal area in an ellipse

Akiva Weinberger

of greatest area

Ted Shifrin

Our high school competition tests were all multiple choice (or short answer for the ciphering questions).

Zach Hauk

yeah they are multiple choice

Ted Shifrin

1:33 AM

actually, Zach, we had questions like you were practicing yesterday with roots of polynomials

Akiva Weinberger

There was a polynomial question on this year's AMC 12 but I didn't know how to do it

Simply Beautiful Art

Goodness, these problems aren't necessarily hard, I just keep making stupid mistakes

(>.<) I need sleep probably

Akiva Weinberger

Someone have a tranq dart

Zach Hauk

me neither

it was like question 24

a polynomial question, that is

i didnt know how to do it cuz i cant math

Akiva Weinberger

You were taking the AMC 10, it was a different question

Zach Hauk

1:37 AM

i know

im just saying

maybe it was the same one?

because they copy questions

(did it have to do with some polynomial that has the same roots as another except one more?)

Akiva Weinberger

Oh. So maybe it was that one.

I guess we'll see.

Zach Hauk

i didnt know how to do that one

Simply Beautiful Art

Goodnight

Kasmir Khaan

hi @ted

Zach Hauk

2:00 AM

@Akiva well i guess well wait and see

yh05

2:33 AM

Hello all, i need help for the following ODE question,
Which of the following functions is a solution of
some differential equation of the form
y’=f(y),
where f is continuously differentiable for all real y?

I was given multiple choices to pick from. How do i approach such a question?

arctic tern

@yh05 y is the inverse function of the antiderivative of f(x)

yh05

for instance im given a choice of y = sin(x)

the inverse function would be arcsin(x)

arctic tern

wait, you're given y=sin(x) as an option, but you're not told what f(y) is?

that makes no sense

yh05

im only given 5 functions

i mean 5 solutions

arctic tern

why don't you just tell us the problem?

yh05

2:43 AM

Which of the following functions is a solution of some differential equation of the form
y’=f(y), where f is continuously differentiable for all real y?

a) y = sin(t)
b) y = t^2
c) y = t^3 - t
d) y = cosh(t)
e) y = tanh(t)
f) all of the above

arctic tern

aha

yh05

i find it really difficult because f(y) is unknown

arctic tern

so find y' for each option and see if you can get it in terms of y (in a way that is true for all t)

for instance if y=sin(t) then y'=cos(t). can you write cos(t) in terms of sin(t)?

you could try: cos(t)=sqrt(1-sin^2(t)), however because of sign changes this isn't always true

yh05

i see

arctic tern

indeed, cos(t)=f(sin(t)) is not possible, because you can pick two values s and t for which sin(s)=sin(t) but cos(s)=/=cos(t)

yh05

2:48 AM

thanks so much

im trying out the other choices now

Im curious what does "f is continuous differentiable for all real y" have to do with the problem?

arctic tern

when it asks you if there is an f with y'=f(y), you are specifically to see if there is an f which is continuous and differentiable everywhere

Simple

3:41 AM

anyone here know numerical methods?

Akiva Weinberger

4:11 AM

Is there a name for mapping the alphabet to the numbers 1 through 26 in an order-preserving manner?

Like, A is 1, B is 2, C is 3, etc.

Secret

4:25 AM

alphabetical order or total order?

Samuel Yusim

4:37 AM

I would say alphabetical order is exactly that

but when you're talking to your friends outside of math you aren't going to write down an embedding of the alphabet into the integers

PVAL-inactive

4:52 AM

@MikeMiller Apparently in his book Partial Differential Relations, Gromov constructs an isometric (real analytic) embedding of any smooth surface into $\Bbb R^5$. There's lots of related stuff here. mathoverflow.net/questions/37708/… (for instance I think the 1939 result shows no isometric immersion of the flat Klein bottle into $\Bbb R^3$)

Akiva Weinberger

@Secret @SamuelYusim Alphabetical order is an order. I'm thinking, what would you call the one-to-one correspondence, if there is such a name

Samuel Yusim

I would call the correspondence alphabetical order as well as the order

I feel like people do that sometimes

where like, X is Y, but X is naturally associated to Z, so we say Z is Y

Akiva Weinberger

OK. So, I'd say, for example, "All of the letters of the word 'geek' correspond to prime numbers in the alphabetic order"

or maybe "All of the letters of the word 'geek' are in prime-number positions"

or what

Samuel Yusim

either sounds good to me

Akiva Weinberger

(7, 5, 5, 11)

@SamuelYusim 20,8,1,14,11,19

Samuel Yusim

5:03 AM

I did the first letter there and was like, it probably says thanks

and then I noticed the third letter was a so I'm gonna go with that

Akiva Weinberger

25,21,16

Martin Sleziak

@DHMO I have added a bounty.

Samuel Yusim

a friend gave me this dumb problem a while ago: consider the linear diophantine system given by

$O+N+E = 1$, $T+W+O = 2$, $T+H+R+E+E = 3$, ..., $F+O+U+R+T+E+E+N = 14$. Can this system be solved? If so, how many equations can you have before there's no solution?

Akiva Weinberger

So $(F+O+U+R+T+E+E+N)-(F+O+U+R)=T+E+N$?

That means $E=0$.

Samuel Yusim

yep

teen = ten

=tn

Akiva Weinberger

5:09 AM

Can the variables be negative?

Samuel Yusim

sure

actually just say they're positive

Akiva Weinberger

No, you need negatives.

Samuel Yusim

well fine, say they can be negative then

Akiva Weinberger

$(O+N+E)+(T+W+O)=3$, but the left-hand side would be greater than $T+E+N$.

Samuel Yusim

true

Mike Miller

5:10 AM

@PVAL-inactive Interesting. That's a book I'd like to understand, ever.

Akiva Weinberger

We don't need to worry about $S+I+X$ and $E+I+G+H+T$ equalling the right numbers; we can just set $X$ and $G$ to what they need to be @SamuelYusim

Samuel Yusim

anyway, have fun with this bad boy. I'm gonna go to sleep on account of I have a class in like 8 hours

Akiva Weinberger

$I=0$ also

@SamuelYusim A solution does not exist.

$N+I+N+E=9$, but $E=0$ and — as I'll show in a bit — $I=0$. This makes $2N=9$, contradiction.

$I=0$ follows from $T+H+I+R+T+E+E+N=(T+H+R+E+E)+(T+E+N)$

We get $I=E$, and thus $I=0$.

QED.

Samuel Yusim

so can you do it up to 12?

Akiva Weinberger

Dunno

The "teen" stuff made it a lot easier

Samuel Yusim

5:19 AM

I just realized that for any language you can do this, and get a constant which is the maximum number of equations before you can't solve the system

Perturbative

8:50 AM

Hey everyone, quick question

I'm trying to prove $\bar{A} = \Int(A) \cup Bd(A)$

I've proved that $Int(A) \cup Bd(A) \subset \bar{A}$, which is quite trivial

But proving the converse doesn't seem as easy

I've seen a proof that says "take $x \in \bar{A}$ and assume $x$ to be an exterior point of $A$", but how can an exterior point of $A$ be contained in $\bar{A}$?

Because $\bar{A} = A \cup A'$ where $A'$ are the limit points of $A$, if $x \in A$ then every neighborhood of $x$ intersects $A$, the same applies if $x$ is a limit point of $A$

But if $x \in \bar{A}$ and $x$ is an exterior point, then no neighborhood of $x$ intersects $A$, a contradiction...

Here's a link to the proof for the converse : math.stackexchange.com/a/704402/266135

Astyx

9:36 AM

@Perturbative He's proving that if a point is an exterior point, then it's not in $\overline A$

And the uses the contrapositive

Thus if $x\in \overline A$, it's either in the boundary of the interior

Astyx

10:01 AM

or*

elavarasan

10:47 AM

please anyone guide answer for this question

0

Q: Mixture and Alligation question:

There is a Vessel holding 40 litres of milk.4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in .After this 5 litres of mixtures of Mixture is replaced with the six litres of water and finally six litres of Mixture is Replaced with the six litres of wat...

algebra-precalculus

@Ramanujan guide me

Mary Star

11:54 AM

Hello!!

How could we check if the integral $\int_0^{\infty}\sqrt{x}\cos (x^2)dx$ converges?

I tried to use the direct comparison test, but it wasn't helpful... Coud you give me a hint?

Simply Beautiful Art

12:30 PM

yesterday, by Simply Beautiful Art

If anyone is interested, I'm hosting a big number contest. Code your number in any language (no experience needed), maximum of 256 characters, not including spaces, and try to reach the largest number you can. First submissions due Saturday. :-) http://chat.stackexchange.com/rooms/51337/this-is-the-realm-of-simply-beautiful-‌art

Perturbative

@Astyx Okay thanks for pointing that out!

Simply Beautiful Art

@MaryStar I'm not entirely sure on convergence, but it's possible to solve that integral in terms of the Gamma function

Mary Star

12:45 PM

@SimplyBeautifulArt To check the convergence it is not neccesary to calculate the integral. Do you have an idea how we could check the convergence?

Simply Beautiful Art

@MaryStar make the substitution $x=u^{2/3}$, then Dirichlet test

Aha! I can test for convergence!

XD

Rafa Budría

Hiya

Mary Star

@SimplyBeautifulArt Using that substitution we get the integral $\frac{2}{3}\int_0^{\infty}u^{-1/6}\cos (u^{4/3})du$, right?

Simply Beautiful Art

@MaryStar the cosine should be the only thing remaining

Mary Star

1:03 PM

Ahh... I tried that again:
For $u=x^{3/2}$ we get $du=\frac{3}{2}\sqrt{x}dx$.
Therefore we get the integral $\frac{3}{2}\int_0^{\infty}\cos (u^{4/3)du$.
Is this correct now? @SimplyBeautifulArt

Simply Beautiful Art

Yup

Now split it via the period of cosine and use alternating test

Akiva Weinberger

So I guess the crux of Niven's proof is that $\int_0^\pi x^n(\pi-x)^n\sin(x)\operatorname d\!x$ is a multiple of $n!$ (in $\Bbb Z[\pi]$).

Mary Star

@SimplyBeautifulArt Do we write the integral $\frac{3}{2}\int_0^{\infty}\cos (u^{4/3})du$ in the form $\frac{3}{2}\sum_{n=0}^{\infty}\int_{n\pi}^{(n+2)\pi}\cos (u^{4/3})du$ or $\frac{3}{2}\sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi}\cos (u^{4/3})du$ ?

Simply Beautiful Art

Yeah basically. I'm reading the raw LaTeX, so yeah

Mary Star

@SimplyBeautifulArt In the sum the integral goes from $n\pi$ to $(n+1)\pi$ or to $(n+2)\pi$ ?

Simply Beautiful Art

1:15 PM

No. You need to choose the bounds so that $cos(u^{4/3})=0$ at the bounds

Mary Star

Ahh

Simply Beautiful Art

Then we have a positive hump

Then a smaller negative hump

Astyx

Hi chat

Simply Beautiful Art

And then slightly smaller positive hump, etc.

And thus we may apply the alternating test

@Astyx o/

Mary Star

Do we have to take the integral is the sum from $(n\pi)^{3/4}$ to $((n+2)\pi)^{3/4}$, where $n$ goes from $0$ to infinity? @SimplyBeautifulArt

Mary Star

1:36 PM

To get an alternating series do we use again a substitution? @SimplyBeautifulArt

Simply Beautiful Art

First, good

@MaryStar second step, just show each hump is smaller than the previous

And sum up all the humps

Sum = alternating test

Mary Star

What exactly do you mean by hump? @SimplyBeautifulArt

Simply Beautiful Art

@MaryStar quick sketch the graph of $\cos(u^{4/3}$

Mary Star

@SimplyBeautifulArt We have the following graph:

Simply Beautiful Art

So you see the humps?

Mary Star

1:50 PM

@SimplyBeautifulArt Yes

Simply Beautiful Art

Show that the area of each hump goes to zero and apply alternating test

Mary Star

How could we show that the area of each hump goes to zero? Could you give me a hint? @SimplyBeautifulArt

Simply Beautiful Art

Uh...I just have an intuitive understanding on how to do things. I try to avoid getting my hands dirty. Perhaps someone can help?

Astyx

What are we doing ?

Simply Beautiful Art

2:06 PM

Determining the convergence of an integral

Mary Star

I want to check if the integarl $\int_0^{\infty}\sqrt{x}\cos (x^2)dx$ converges.
Using the substitution $u=x^{3/2}$ we get $du=\frac{3}{2}\sqrt{x}dx$.

Therefore, $$\int_0^{\infty}\sqrt{x}\cos (x^2)dx=\frac{3}{2}\int_0^{\infty}\cos (u^{4/3})du$$

We can write the last integral as $$\sum_{n=0}^{\infty}\int_{(n\pi)^{3/4}}^{((n+2)\pi)^{3/4}}\cos (u^{4/3})du$$
How could we continue from here to check the convergence? Do you have an idea?

navinstudent

hello everyone can anyone guide me in the following question math.stackexchange.com/questions/2134041/…

Ramanujan

2:24 PM

$$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n = 1$$ why??

Nevermind

0

A: Computing $\lim\limits_{n\to\infty}(1+1/n^2)^n$

You can do this without knowing anything about $e$. Using Bernoulli's Inequality we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \geq 1 - \frac{1}{n^{3}}$$ and clearly we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}$$ so that $$1 - \frac{1}{n^{3}} \leq \left(1 - \frac{1}{n^...

Hey-men-whatsup

@Ramanujan, could you help me?

I want to ask a math question

Astyx

2:40 PM

@MaryStar Shouldn't it be $n+1$ ?

Mary Star

@Astyx The upper bound of the integral?

Astyx

Yes

Mary Star

@Astyx Ah ok

Astyx

And first equality should be $2\over3$ I think

Mary Star

@Astyx You mean the constant term or the power?

Astyx

2:45 PM

Constant term

@MaryStar Maybe doing the variable change $u = x^2$, then integrating by part is a better idea ?

Ramanujan

@Hey-men-whatsup ask man,if I could then I will help :)

Hey-men-whatsup

3:05 PM

thanks man

here, kinda dumb Question, but eh..

0

Q: Choosing lowest value of a class group

I'm studying frequency distribution, but I found that the lowest value taken is varying. For this data (Has minimum value = 71): after it gets arranged in a table : The lowest value taken is 70 While for this data Has minimum value = 7.1: after it gets arranged in a table: The lowe...

distributions mathematical-statistics

Ramanujan

3:27 PM

@Hey-men-whatsup you question got deleted?

yh05

4:25 PM

Im doing a project on modeling ODE for my first course in ODE.

Anyone has a cool suggestion on the topic?

I thought of modelling love between characters in a movie.

Hey-men-whatsup

@Ramanujan, I've moved it math se..

Ramanujan

@Hey-men-whatsup you can notice that in second case maximum value is 9.1

And in first case maximum value is less than 90

I don't think picking different lowest values will affect/effect anything

Hey-men-whatsup

@Ramanujan, Exactly that's in my head now, I don't think picking different lowest values will effect anything, maybe it's just matter of a choice

Ramanujan

@Hey-men-whatsup and how are you computer classes going?

Simply Beautiful Art

4:43 PM