Mathematics - 2016-02-09 (page 1 of 3)

Nova

12:00 AM

It says: "By solving Equations 2 for and we obtain: X= xcos(theta) + ysin(theta)", but I cant seem to solve them

Ted Shifrin

12:13 AM

@aaadddaaa: Apply the Mean Value Theorem (or Rolle's Theorem) twice.

Hi @Alessandro

@Nova: I assume you don't know basic linear algebra? (That's the best way to do this.)

Nova

First step is to isolate X and Y in their respective equations. SO, X = [x+Ysin(theta)]/cos(theta) AND Y = [y - Xsin(theta)]/cos(theta)

is the only way to do this is using LA? I havn't taken it yet

@TedShifrin

Ted Shifrin

Well, I would recommend you multiply the first equation by $\cos\theta$ and the second equation by $\sin\theta$ and add. [No, you do not need linear algebra. It's just powerful.]

Nova

what is this? $\cos\theta$

Ted Shifrin

Oh, we use LaTeX here to make equations readable.

It's cos(theta) to you.

Nova

ah gotchya

Ted Shifrin

12:17 AM

Note that the point is going to be to use the one super-important identity, namely cos^2 + sin^2 = 1.

Nova

huh okay I thought that but I only got cos^2 + sin

so I couldnt plug 1 in

I ended up geting -[x+Ysin][sin/cos^2] + y = Y

Ted Shifrin

Do you see that what I told you to do works?

Nova

but I dont see how I can then combine the two big Y's

Ted Shifrin

They will cancel out !

Nova

No they wont...

Ted Shifrin

12:20 AM

They will if you do what I said !

Nova

ok you told me to use cos^2 + sin^2 = 1

how do I use it?

Ted Shifrin

That will be the coefficient of X. The coefficient of Y will be 0, because terms exactly cancel.

Nova

Ted, there is no way because look at the PDF I sent in the beginning

X and Y both have anwers in terms of sin and cos

Ted Shifrin

I did look. Yes, I know.

Nova

so how can Y have a coefficient of 0?

Ted Shifrin

12:22 AM

You're going to solve for X with one set of algebra, and then do a different set of algebra to get Y.

I'll type it out here (you can ignore $'s but I'm putting them).

Nova

ok cool

wait let me isolate Y first

Ted Shifrin

$x = X\cos\theta - Y\sin\theta$ and $y = X\sin\theta+Y\cos\theta$. So

Nova

and tell me if you get the same thing as me

Ted Shifrin

$ x \cos\theta = X \cos^2\theta - Y\sin\theta\cos\theta$ and $ y\sin\theta = X\sin^2\theta + Y\sin\theta\cos\theta$. Now we add and get $X = x\cos\theta +y\sin\theta$.

Heya @Clarinet

Clarinetist

Hi @Ted

@Ted Perhaps a basic linear algebra question... can a linear combination of an inner product of two vectors be written as the product of a square matrix and a column vector?

Ted Shifrin

12:26 AM

Huh? But the sum of two inner products is a scalar, right?

Clarinetist

Indeed

0

Q: Simplification of a product of three matrices

Define $$\mathbf{c}_t = \begin{bmatrix} x_{1t} \\ x_{2t} \\ \vdots \\ x_{Nt} \end{bmatrix}\in \mathbb{R}^N$$ where all entries are in $\mathbb{R}$, $t = 1, 2, \dots, p+1$. I am trying to simplify $$\sum_{j=0}^{p}\beta_j\mathbf{c}^{T}_{j+1}\mathbf{c}_{k+1} $$ where $k = 0, 1, \dots, p$ and $$\bol...

Ted Shifrin

So to get that you'd need a row vector times a column vector, not a square matrix times a column vector.

Clarinetist

Yes, but (assuming my simplifying is right), the above should result in $\mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}$. Let me get the notation out

Ted Shifrin

I don't want to look at all that yuck.

Clarinetist

$$\mathbf{X} = \begin{bmatrix}
\mathbf{c}_1 & \mathbf{c}_2 & \cdots & \mathbf{c}_{p+1}
\end{bmatrix}\in \mathbb{R}^{N \times (p+1)}\text{.}$$

Nova

12:28 AM

hey guys how do I enable LaTeX?

Ted Shifrin

So when you said linear combination you didn't mean linear combination.

@Nova: Click on LaTeX in chat over there >>>> ^^^

Clarinetist

Isn't $$\sum_{j=0}^{p}\beta_j\mathbf{c}^{T}_{j+1}\mathbf{c}_{k+1} $$ a linear combination of inner products?

Okay, wait

Ted Shifrin

The $\beta_j$ are scalars, @Clarinet?

Clarinetist

I'm confusing you

Yes, they are scalars

Ted Shifrin

Yes, then that's a linear combination of inner products.

Clarinetist

12:29 AM

This is what I'm trying to prove

$$\mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta} = \begin{bmatrix}
\sum_{j=0}^{p}\beta_j\mathbf{c}^{T}_{j+1}\mathbf{c}_{1} \\
\sum_{j=0}^{p}\beta_j\mathbf{c}^{T}_{j+1}\mathbf{c}_{2} \\
\vdots \\
\sum_{j=0}^{p}\beta_j\mathbf{c}^{T}_{j+1}\mathbf{c}_{p+1}
\end{bmatrix}$$

assuming that the expression is actually correct

Ted Shifrin

And a linear combination can be written as a matrix times the vector of $\beta$s.

Clarinetist

Yes, how do I see that?

Ted Shifrin

That's on page 20 of my book or something :P

Clarinetist

Dang, your book is at my work. -_-

Ted Shifrin

There are two fundamental ways to think about $A\mathbf x$.

You can use rows and get dot products or you can use columns and get the linear combination of the columns with the $x_j$ as coefficients.

Clarinetist

12:31 AM

Linear combinations of the columns of $A$... if I recall

Ted Shifrin

Right.

Clarinetist

So you're telling me that I could literally multiply those entries up there by a $\boldsymbol{\beta}$

instead of writing the $\beta_j$ sum

Ted Shifrin

But ordering is all wrong in what you're doing.

Clarinetist

Yeah, that's why I'm pretty sure I did something wrong

$$\begin{align*}
\sum_{i=1}^{N}\sum_{j=0}^{p}x_{ij}x_{ik}\beta_{j} &= \sum_{j=0}^{p}\sum_{i=1}^{N}x_{ik}x_{ij}\beta_{j} \\
&= \sum_{j=0}^{p}\beta_{j}\left(\sum_{i=1}^{N}x_{ij}x_{ik}\right) \\
&=\sum_{j=0}^{p}x_{ik}\begin{bmatrix}
x_{1j} & x_{2j} & \cdots & x_{Nj}
\end{bmatrix}
\begin{bmatrix}
x_{1k} \\
x_{2k} \\
\vdots \\
x_{Nk}
\end{bmatrix} \\
&= \sum_{j=0}^{p}\beta_j\mathbf{c}^{T}_{j+1}\mathbf{c}_{k+1} \tag{1}
\end{align*}$$

Ted Shifrin

The $j$th column of $\mathbf X^\top\mathbf X$ is $\mathbf X^\top\mathbf c_j$.

Clarinetist

12:35 AM

Hmm

Ted Shifrin

You've got things partially transposed, i think.

Clarinetist

I wonder if my initial expression is wrong

Ted Shifrin

Put the $\mathbf\beta$ on the left with a transpose?

Clarinetist

@TedShifrin This is a pretty famous equation in stats: $\mathbf{X}^{T}\mathbf{y} = \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}$

Here's how it's derived

You minimize the following expression

Ted Shifrin

Yes, I am fully aware of the normal equations. That's in chapter 5 of my book (with two different derivations).

It's about projections, or you can use calculus, yes.

Clarinetist

12:42 AM

@TedShifrin Dang, why didn't I bring your book home today? ugh

Ted Shifrin

Or you could just watch the appropriate video.

Clarinetist

$$\text{RSS}(\boldsymbol{\beta}) = \sum_{i=1}^{N}\left(y_i - \sum_{j=0}^{p}x_{ij}\beta_{j}\right)^2$$

where $x_{i0} = 0$ for all $i$

Ted Shifrin

I prefer the linear algebra derivation, projecting on the column space of $\mathbf X$.

Nova

holy moly Ted you are a genius

Clarinetist

$$\dfrac{\partial\text{RSS}}{\partial\beta_k}(\boldsymbol{\beta}) = \sum_{i=1}^{N}2\left(y_i - \sum_{j=0}^{p}x_{ij}\beta_{j}\right)(-x_{ik}) = -2\sum_{i=1}^{N}\left(y_ix_{ik} - \sum_{j=0}^{p}x_{ij}x_{ik}\beta_{j}\right)\text{.}$$

Ted Shifrin

12:42 AM

LOL, @Nova, not exactly :)

Nova

all you did was multiply throigh with cos and add the equations

Ted Shifrin

@Clarinet: You're writing yuck again.

Nova

ive been trying to do it by substituing one equation into anmother

Ted Shifrin

cos and sin respectively, @Nova.

That's what I told you to do ages ago :D

Nova

why does it work your way

but now my way

not*

Ted Shifrin

12:43 AM

Your way will work if you are careful; it's just not very pleasant.

Nova

I believe you, I just cant seem to figure it out ahaha

Thanks for your help though

that was awesome

Ted Shifrin

Sure, @Nova.

Nova

wish I could see elegant solutions like that

Ted Shifrin

With practice you will.

Clarinetist

@TedShifrin From a geometric perspective, the idea is you want to take your observed data given by a "dependent variable," whose values are given by $\mathbf{y}$ (the vector of the $y_i$), and minimize the squared difference between this and the regression equation $\sum_{i=0}^{N}x_{ij}\beta_j$). RSS above is merely this difference, and the second line involves finding the critical point by taking the partial with respect to one of the coefficients $\beta_k$ for some $k \in \{0, 1, \dots, p\}$.

If this is too much to take in, I understand

Ted Shifrin

12:46 AM

@Clarinet: I know all this stuff.

As I said, you can find expositions of different approaches in my book or my lectures. :)

Clarinetist

Did I mess up in the Calculus, I wonder

K, I will just look through the vids

Ted Shifrin

I am not going to check.

Clement C.

Hi

Clarinetist

What would happen if I purposefully put the $\beta$ on the right...

Ted Shifrin

Hi @Clement.

Clement C.

12:48 AM

Just in case: if anyone has time, can (s)he tell me what the commenter meants there? math.stackexchange.com/questions/1646837/… I don't see where (if) I'm wrong, but wouldn't want to be.

Ted Shifrin

It belongs on the right, @Clarinet.

Clarinetist

[you can ignore what I'm doing here, I just don't have paper on me]

Ted Shifrin

@Clement: There's a lot of context to that question and your answer that most of us just don't know. But I think you asked the commenter a perfectly reasonable question.

Clement C.

Thanks... well, I hope (s)he will answer. I genuinely can't understand the comment.

Ted Shifrin

Me either.

Clarinetist

12:52 AM

$$\begin{align*}
\sum_{i=1}^{N}\sum_{j=0}^{p}x_{ij}x_{ik}\beta_{j}
&= \sum_{i=1}^{N}x_{ik}\left(\sum_{j=0}^{p}x_{ij}\beta_{j}\right) \\
&= \sum_{i=1}^{N}x_{ik}\begin{bmatrix}
x_{i0} & x_{i1} & \cdots & x_{ip}
\end{bmatrix}\boldsymbol{\beta}
\end{align*}$$

Ted Shifrin

That looks better, @Clarinet.

I'm an artist

1

Q: Another way of doing integration

What's your option for calculating this integral? No full solution is necessary, it's optional as usual. Calculate $$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...

calculus real-analysis integration definite-integrals polylogarithm

Clarinetist

and $\begin{bmatrix}
x_{i0} & x_{i1} & \cdots & x_{ip}
\end{bmatrix} = \mathbf{r}_{i}^{T}$, the $i$th row of $\mathbf{X}$

So then that sum ends up being

$$\sum_{i=1}^{N}x_{ik}\mathbf{r}^{T}_i\boldsymbol{\beta}$$

Now "stacking" these, we have

Ted Shifrin

I still recommend that you look at how mathematicians do this so much more conceptually. :)

I'm an artist

Upvote

Clarinetist

12:57 AM

$$\begin{bmatrix}
\sum_{i=1}^{N}x_{i0}\mathbf{r}^{T}_i\boldsymbol{\beta} \\
\sum_{i=1}^{N}x_{i1}\mathbf{r}^{T}_i\boldsymbol{\beta} \\
\vdots \\
\sum_{i=1}^{N}x_{ip}\mathbf{r}^{T}_i\boldsymbol{\beta}
\end{bmatrix} = \begin{bmatrix}
\sum_{i=1}^{N}x_{i0}\mathbf{r}^{T}_i \\
\sum_{i=1}^{N}x_{i1}\mathbf{r}^{T}_i \\
\vdots \\
\sum_{i=1}^{N}x_{ip}\mathbf{r}^{T}_i
\end{bmatrix}\boldsymbol{\beta}$$

@TedShifrin I definitely will. This was given as an exercise without much explanation in a previous class I had. I can see why...

The prof's notes specifically say to solve the partials. It's disgusting

Ted Shifrin

It is, but again you'll see a better way to do it.

I'm an artist

$$\int_0^1 \frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-x)(\text{Li}_2(1-x){})^2 }{x}-\frac{ \log (x) \log ^4(1-x)\text{Li}_3(1-x)}{x}$$ $$-\frac{2 \log ^3(1-x)\text{Li}_2(1-x) \text{Li}_3(1-x) }{x}+\frac{\log ^2(1-x)(\text{Li}_3(1-x){})^2 }{x} \textrm{d} x$$

^^^

Clement C.

1:18 AM

I think I'll give up on what (s)he meant, then.

Clarinetist

2:09 AM

Hi @TedShifrin: btw, I figured it out. It so happens that the sum is equal to $\mathbf{c}^{T}_{k+1}\mathbf{X}\boldsymbol{\beta}$. Stack those on top of each other for $k = 0, 1, \dots, p$, and then you get $\mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}$. How nasty.

I look forward to reading your book's explanation tomorrow

Michael

@Clarinetist Where can I get his book? I am following his lectures and I feel like the practice questions in it are super helpful

Clarinetist

@Michael I got it on Amazon

Michael

@Clarinetist How much was it

Clarinetist

@Michael I don't remember.

Michael

@Clarinetist less than 30 bucks maybe?

Clarinetist

2:11 AM

@Michael Maybe if you can find it used

@TedShifrin It also required a trick I haven't seen in a long time - the (somewhat obvious) fact that an inner product commutes

OneRaynyDay

2:43 AM

Hi there! :) Could anyone redirect me to a tutorial about entropy? I'm using a softmax function to feed into a logistic regression for classification here and my stanford class said that the "cross entropy is equivalent to this expression of entropy and the Kullback-Leibler divergence". I understand that the divergence is the amount of information lost due to the mathematical model prediction of the "true" labels, but I can't find a place where entropy is clearly/intuitively described.

I looked on google and wasn't sure if my intuitions were correct - is the entropy just the possible number of states the function can take up?

Michael

3:14 AM

Guys, how do you explain Matrices to a layman?

zed111

@Michael

Matrix (mathematics)

In mathematics, a matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The dimensions of matrix (1) are 2 × 3 (read "two by three"), because there are two rows and three columns. The individual items in a matrix are called its elements or entries. Provided that they are the same size (have the same number of rows and the same number of columns), two matrices can be added or subtracted element by element. The rule for matrix multiplication, however, is that two matrices can be multiplied only when the number of columns in the first equals...

This serves a layman

for the beginning part

The Substitute

off topic. Anyone here familiar with systems of ODE? How can $ay''+by'+cy=f(x)$ be expressed as a 1st order system of ODE? I don't know how to deal with $y''$, and integrating doesn't seem the way to go.

usukidoll

anything with y'' is a 2nd order ode

Michael

3:29 AM

Is there something similar to periapsis(physics) in mathematics?

Simeon

Is a coordinate patch always a diffeomorphism?

@TheSubstitute Can you obtain an expression for the derivative of y'? How about the derivative of y?

usukidoll

How do I prove that x is an accumulation point in S
?
I got
Suppose that x doesn't belong to S and $x= supS$. Then, there is an $a \in S$ and $ \epsilon > 0$ such that $x-\epsilon < a \leq x$. The accumulation point x must be in between those intervals :/

The Substitute

Doh

thanks

usukidoll

sigh I think I'm missing the accumulation point definition all up in here

ah! so it has to be
$(x- \epsilon, x) \subseteq S$ that's a neighborhood
so by the accumulation point definition,A is an accumulation point of S if and only if each neighborhood of A contains a member of S different from A.
but who is the epsilon on here?

yay it's ted

Ted Shifrin

3:46 AM

Nah, it's not

usukidoll

awww

Ted Shifrin

Epsilon is any positive number it wants to be.

usukidoll

neglected to put that S is a nonempty set of real numbers... weee
but wait a sec do I let epsilon be 1 or something?

Ted Shifrin

No, epsilon can be any positive number ... particularly smaller and smaller and smaller.

usukidoll

oh gawd like 1/n or something?

Ted Shifrin

3:48 AM

yes, for $n$ as large as wanted

zed111

Hello everyone, do you know of a way to solve for F in this matrix equation: F^T F = C where T stands for transpose

@TedShifrin

usukidoll

so what do I write, since epsilon is any positive it wants to be let in be 1/n??

Ted Shifrin

@zed111, no, you can't ...

usukidoll

*any positive #

Ted Shifrin

You just say "let $\epsilon>0$ be arbitrary."

zed111

3:50 AM

You mean there is no way to solve this equation?

The Substitute

off topic again. What's the quickest proof of the fact that $\sum_{i=1}^n\frac{1}{i}-\log n$ is always between $0,1$?

usukidoll

then we have an arbitrary large epsilon ?

Ted Shifrin

I don't know what shapes the matrices have, but even with $1\times 1$, note that you're asking to solve $f^2 = c$, so if $c>0$, there are two solutions. With larger matrices, ... more options.

no, @usukidoll, the point is that you can have $\epsilon$ arbitrarily small.

@TheSubstitute: Do you know the integral test?

usukidoll

so epsilon has to be super small so that it goes closer to the accumulation point

Ted Shifrin

That's the right idea, @usukidoll ... the neighborhood shrinks closer and closer.

usukidoll

3:52 AM

I've read a pdf about accumulation points that made it way easier to understand than the textbook I have

Ted Shifrin

Well, then you don't need us.

usukidoll

WHAT! whoa I had writer's block

The Substitute

@TedShifrin thanks!

Ted Shifrin

That was enough hint, @TheSubstitute?

The Substitute

@TedShifrin I think so. Isn't there a formula for the error?

Ted Shifrin

3:55 AM

Just draw the picture with inscribed rectangles for the area under the curve ... and slide them over.

Well, slide the errors over.

@TheSubstitute, did you figure out your question about turning a second-order ODE into a first-order system? That should be in your book/course.

usukidoll

don't we need a min though like
min($x- \epsilon, x$)
so we can select an interval that's less than the min, so that interval is contained in ($x- \epsilon, x$) . So the neighborhood ($x- \epsilon, x$) and the interval contain infinite points, so x is an accumulation point of S ?!
this writing is so repetitive

The Substitute

Yea I did :)

Ted Shifrin

What min are you talking about, @usukidoll?

There isn't one (other than $0$).

usukidoll

ok scratch that, so there could be an interval contained in $( x - \epsilon,x)$. The neighbor and the interval may contain infinite number of points...

Ted Shifrin

not may ... must

usukidoll

4:00 AM

OH!

wait a sec... isn't there a theorem that tackles this?
like Bolzano-Weierstrass theorem

Ted Shifrin

no, we're talking about definitions of accumulation points

usukidoll

oh yeah.. O_o

it's a lemma in my book...mini proof x.x

every neighborhood of a contains points in A different from a
a real number a ... the Set A
I like the pdf version better

Ted Shifrin

Generally, I like using little letters and capital letters that are aligned. I like to have $x\in X$, $y\in Y$, $a\in A$, etc.

However, a limit point (accumulation point) of $A$ needn't be a point of $A$, so I would use a different letter.

usukidoll

like B

Ted Shifrin

no, little letter for elements, capital letter for sets.

usukidoll

4:11 AM

ohhh

so did I finish the proof or is there still more to be done?

Ted Shifrin

LOL, proof of what?

usukidoll

:/

so that was jacks...

wow x.x

proof of the x is the accumulation point of S. I think there's a bit more to be done

Ted Shifrin

I have no idea what your problem/exercise is.

usukidoll

Let S be a nonempty set of real numbers that is bounded from above (below) and let $x = sup S (inf S)$. Prove that either x belongs to S or x is an accumulation point of S

Ted Shifrin

Ah, ok. So what's your proof?

usukidoll

4:17 AM

k.. ummm
Suppose that x doesn't belong to S and $x= supS$. Then, there is an $a \in S$ and $ \epsilon > 0$ such that $x-\epsilon < a \leq x$.\\
As $ \epsilon > 0$ becomes arbitrarily smaller, the neighborhood shrinks closer to the accumulation point. \\
Therefore, there could be an interval contained in $(x- \epsilon,x)$. The neighborhood and the interval must contain an infinite number of points. Hence, x is an accumulation point of S. \\

actually that interval on the a in S part is from an exercise I did

Ted Shifrin

Oh, you were doing this 4 days ago.

usukidoll

and then I had to use that

yeah . this is linked

Ted Shifrin

OK, so if $x\in S$, we're done. If not, by definition of $\sup S$, for every $\epsilon>0$ there is $a\in S$ with $x-\epsilon<a<x$. Why does this mean $x$ is an accumulation point?

What's the definition of accumulation point?

usukidoll

Let S be a set of real numbers. Then A is an accumulation point of S
iff each neighborhood of A contains a member of S different from A. that's the definiton

by definition of sup s would be the x = least upper bound

Ted Shifrin

So what does a neighborhood of $x$ look like? Why have you shown every neighborhood contains a point of $S$ different from $x$?

Don't use $A$ as an element. Use little letters.

usukidoll

4:20 AM

we have an interval (x-e, e) right?

fffffff

Ted Shifrin

No.

usukidoll

(x-e,x)

Ted Shifrin

No, that's not a neighborhood of $x$.

usukidoll

what the hell O_O

Ted Shifrin

$x$ isn't even in that set!

usukidoll

4:21 AM

so x... x is the least upper bound, so that's outside the set

Ted Shifrin

What does a neighborhood of $x$ look like?

usukidoll

well this is the neighborhood definition
A set Q of real numbers is a neighborhood of a real number x
iff Q contains an interval of positive length centered at x-that is, iff there is
e > 0 such that (x-e,x+e) C Q.

Ted Shifrin

OK, so a neighborhood looks like $(x-\epsilon,x+\epsilon)$. Why does such a thing contain a point of $S$ other than $x$?

usukidoll

so Set S contains an interval of positive length center at x....... wait how can the x be centered when it's the least upper bound

Ted Shifrin

You really need to sit down and concentrate.

usukidoll

4:24 AM

sighhhhhh

Ted Shifrin

This interval does not live in $S$. It lives in $\Bbb R$.

usukidoll

this sucks more than Cauchy and convergence proofs. At least I know what I'm doing. Oh the interval is in R, not in the set S

interval lives in real numbers . . .

OneRaynyDay

Hey guys? I have a question about what this syntax means: P(yi | xi;W). namely the | and the ;

where P is probability and yi is a scalar and xi is a vector I believe

and W is a matrix

usukidoll

hmmm when x = sup(S) , there exists a in S such that $x-\epsilon < a \leq x$. When x = inf(S) , there exists a in S such that $x \leq a < x+ \epsilon$. So we have the interval $(x - \epsilon, x+ \epsilon)$ when x is both sup(s) and inf(s)

L33ter

4:39 AM

hey @TedShifrin

can you remind me something in vector calculus

I just forgot some stuff

if we have the following

lim (x,) --> (2,-1) e^{x^2 - 2y}cos(x + 2y)

this limit is exist and is e^6

since this function is continous

so you don't have to check it against different path etc

@TedShifrin here

usukidoll

I think I chased him away :(

Michael

Complex algebraic geometry

I think I just annihilated my mind. The only word I understand the word "the"

usukidoll

I'm in real analysis...book sucks

Michael

It looks cool though

usukidoll

but at least I did most of the homework...the converges exercise isn't that bad

Michael

4:53 AM

but I am only in elemntary. I did convergence too, the proofs can get nasty

usukidoll

I just have to do the a_n converges to a iff converges to 0

Michael

I am doing linear algebra and real analysis at the same time. Hopefully I can get through a few books in two years

$a_n$ converges to $a$ $iff$ converges to 0

usukidoll

so.. did you read my question above?

I think I'm stuck.. that's the one with the sups and infs

Michael

oh I'm no better

usukidoll

f........

Michael

4:55 AM

;)

usukidoll

so for the convergence proof I just have to use the definition

but when I do this
$\mid a_n - A \mid < \epsilon$

Michael

Triangle inequality?

I think it would work here

usukidoll

$\mid a_n - A \mid < \epsilon$
$\mid A + 0 \mid < \epsilon$
like this?

Michael

Uhm

usukidoll

$ \mid a_n -A +A +0 \mid$
$ \mid a_n +0 \mid$
$ \mid a_n \mid $
oh umm xx.xx

Michael

4:59 AM

Don't ask me for advice, u are alot better

usukidoll

how am I a lot better?

Michael

I am in highschool

usukidoll

O_O oh

Michael

im jking

First year uni

but im in health sciences

so its not mathematically

focused

I just picked up on linear algebra and real analysis because I thought they were interesting.

Are you a mathematics major?

usukidoll

holy shit. linear algebra is way easier than this

yeah I finish my BA this semester, but I need "analysis" crap courses to enter the MA fml

Michael

5:01 AM

did u do abstract algebra?

usukidoll

no

Michael

have fun

don't worry alot of people hate real analysis

usukidoll

I ignored those courses up until next semester... let me guess it's hard as hell

WOOHOO!

Michael

depends on who you are

some people think real analysis is easy

they just "get it"

then they see abstract algebra and groups

and blankly stare

usukidoll

back then I thought that I wouldn't be able to finish the ba in math... but it's so cool that this semester is my lasttttttttttttttttt...........

Michael

5:02 AM

so what do you want to specialize in

usukidoll

yeah it's not like discrete math where everything is so predictable

teaching post secondary uniz

Michael

Maybe if you like discrete

number theory? ;)

usukidoll

I took elementary number theory. it was easy

took all the courses that discrete math covers so last semester was super easy

hahahahha

Michael

discrete topology?

jks

usukidoll

oh hell no

Michael

5:05 AM

Intro to topology?

usukidoll

no

Michael

I am using this one

maths.ed.ac.uk/~aar/papers/munkres2.pdf

Munkres

I need another one though

Its super usukidoll friendly

usukidoll

awww

ok I just winged the last proof and it sounds like shit

maybe after I eat dinner I can tweak it but right now... it needs to go to the sideline

Michael

shoves proof

oh crap I just realized I have an assigment due tmrow

ok I am off, who needs sleep anyways

usukidoll

...ya

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$Mathematics$ Mathematics