Mathematics - 2017-12-08 (page 2 of 3)

000(q4)
001(q4)
0100(0->2->3->5->5)
0101(0->2->3->5->3)
0110(0->2->3->1->3)
0111(0->2->3->1->0)
1000(0->1->3->5->5)
1001(0->1->3->5->3)
101(0->1->3->1)
11(0->1->0)

paths to 3:
---
2->3
1->3
5->3
---
0->2->3
3->1->3
0->1->3
3->5->3
---
1->0->2->3
1->0->1->3
---
0->1->0->2->3
3->1->0->2->3

Ashwin Gupta

Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2*sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine"

Leaky Nun

it doesn't have a period

but you might want to say that it oscillates more frequently as time progresses

Ashwin Gupta

yes indeed. Does that have a formal name?

Leaky Nun

2:55 AM

no idea

paths to 3:
---
1: 2->3
0: 1->3
1: 5->3
---
01: 0->2->3
10: 3->1->3
10: 0->1->3
01: 3->5->3
---
101: 1->0->2->3
110: 1->0->1->3
---
1101: 0->1->0->2->3
1101: 3->1->0->2->3

===
01: 0->2->3
10: 0->1->3
1101: 0->1->0->2->3
---
11: 0->1->0
---
11: 1->0->1
---
---
10: 3->1->3
01: 3->5->3
1101: 3->1->0->2->3

Lucas Henrique

hey guys.

Leaky Nun

base form: 01|10|1101

with states indicated: [0](0[2]1|1[1]0|1[1]1[0]0[2]1)[3]

Lucas Henrique

so I have a very fundamental question on functions.

Leaky Nun

loops: [0](1[1]1)[0], [1](1[0]1)[1], [3](1[1]0|0[5]1|1[1]1[0]0[2]1)[3]

Lucas Henrique

let $f: \Bbb R \ton\Bbb R$ be an arbitrary line. how can I formally prove that its image is the reals?

Leaky Nun

3:04 AM

@LucasHenrique by "line" do you mean $y=mx+c$?

Lucas Henrique

@LeakyNun yeah.

Leaky Nun

loops: [0](1(11)*1)[0], [1](1(11)*1)[1], [3](1(11)*0|01|1(11)*1(11)*01)[3]

@LucasHenrique well, what is the definition of "its image is the reals"?

combined regex: (1(11)*1)*(01|1(1(11)*1)*0|1(1(11)*1)*1(1(11)*1)*01)(1(11)*0|01|1(11)*1(11)*01)‌*

Lucas Henrique

@LeakyNun $\forall y \in \Bbb R \exists x \in \Bbb R: f(x) = y$.

Leaky Nun

slightly simplified: (11)*(01|1(11)*0|11(11)*01)(1(11)*0|01|11(11)*01)*

Lucas Henrique

injectivity is trivial.

Leaky Nun

3:08 AM

@Theo and this is just half of the regex :P

Lucas Henrique

i couldnt advance much

Leaky Nun

@LucasHenrique and how do you prove a statement that starts with $\forall$?

Lucas Henrique

@LeakyNun you pick a random?

Theo

@LeakyNun wow.. thank you

Lucas Henrique

holy shit

sorry

Leaky Nun

3:09 AM

Now $y$ is a random number

then you need to prove $\exists x \in \Bbb R: f(x)=y$

how do you prove a statement that starts with $\exists$?

Lucas Henrique

@LeakyNun in this case you just solve the equation backwards

Leaky Nun

can you solve it backwards?

Lucas Henrique

yeah.

Leaky Nun

I haven't told you anything

you've solved all of the question yourself

literally all that I asked you is to unfold definitions

remember this skill, it's very useful

Lucas Henrique

@LeakyNun yeah, it was ridiculously easy

@LeakyNun actually that was my thought from the beggining

but for some reason I thought I was wrong

Leaky Nun

3:12 AM

now, the "proper" term for "random" is "arbitrary"

Lucas Henrique

dummie me.

@LeakyNun yeah ikr

Leaky Nun

the proof goes like, "let $y$ be arbitrary. Then, when $x=\dfrac{y-c}m$, we observe that $f(x)=y$. QED"

Lucas Henrique

since its surjective and injective then its a map

Leaky Nun

right

Lucas Henrique

i’m a high schooler. as a graduate, are epsilon-lambda proofs that beautiful?

(totally unrelated)

cuz I love them

constructions, inequalities, etc

Leaky Nun

3:14 AM

good for you

@LucasHenrique well, beauty is in the eye of the beholder

my answer isn't that useful anyway

@AkivaWeinberger hola

Lucas Henrique

@LeakyNun why tho?

Leaky Nun

@LucasHenrique is "beauty is in the eye of the beholder" a useful answer to the question "are epsilon-lambda proofs that beautiful"?

Lucas Henrique

nope

Leaky Nun

exactly

GFauxPas

3:40 AM

My professor spent at least 20 minutes today proving that a trefoil isnt an unknot, it was fun

David Reed

@LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$

I enjoyed developing the lebesgue integral though. I thought that was cool

Leaky Nun

@GFauxPas how did your professor prove it?

GFauxPas

3:55 AM

put the trefoil on a torus, put the torus in $\mathbb S^3 \cup \{\infty\}$, took its complement, broke it down into factors of spaces with the free product of amalgamation,

used van kampen's theorem to find $\pi_1$ of the trefoil

and then found a sub-group of its $\pi_1$ that wasn't abelian

which it would be, if it were the unknot, we showed

took the knot's complement, not the torus's complement

Leaky Nun

4:15 AM

hmm

Secret

6:30 AM

blogs.ams.org/blogonmathblogs/2017/12/06/…

Exposing open problems to high schoolars to solve

Secret

10:51 AM

I suspect $\{0\}$ is actually closed, since it is the complement of the union of all nonzero singletons, all being open sets

In addition, the sets $\Bbb{R} - \{a\}$ is also closed

The topology is almost discrete with a (suspected) limit point of $\{0\}$

As Leaky suspect, any set containing $\{0\}$ will be closed since the complement is a union of open sets

Vrouvrou

Hello

someone knows the Lagrange differential equation ?

Secret

In particular, this topology does not contain neither (open nor closed) sets

The topology also seemed to be second countable, with a countable base of the form all open intervals not containing zero

Alessandro Codenotti

This space doesn't look second countable, almost every singleton is open

Secret

But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons?

uh wait a sec...

... I need arbitrary intersection to produce singletons from open intervals...

ok nvm then

Alessandro Codenotti

11:07 AM

According to your argument the discrete topology on $\Bbb R$ should also be second countable while it's not

Anyhow this space is metric, so second countable iff separable and it shouldn't be hard to show that this space isn't separable

Secret

About first countability, is a constant sequence e.g. a,a,a,a,a,a,... a valid neighbourhood sequence for the point $\{a\}$ $a\neq 0$?

because it seems the neighbourhood of any nonzero singleton can only be itself in this topology

Alessandro Codenotti

You can just take $\{a\}$, if a point has a minimal nbhd you don't even need a sequence

Secret

I see

Alessandro Codenotti

@Secret $\{a\}\cup\{b\}$ is an open nbhd of both $a$ and $b$ if they are distinct and not the origin

Secret

also my mistake earlier: 0 is actually not a limit point since the neightbourhood $\{0\}$ contains only 0 and no other points

Alessandro Codenotti

11:22 AM

This is not a nbhd of $0$, it's not open

Secret

right, cause there are no open balls of 0 radius

hmm... 0 does not even have a nbhd, since any set containing 0 is closed

I have no idea how to deal with points having empty nbhd

o wait a sec...

the open set of any topology must contain the whole set itself

so I guess the nbhd of 0 is $\Bbb{R}$

Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…)

Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time

Alessandro Codenotti

11:48 AM

@Secret That's not a problem, the open balls centered in $0$ are open as well

Secret

hmm... something is not really adding up...:

$d(a,b)=|a|+|b|=|a|+|0|+|b|+|0|=d(a,0)+d(b,0)$

So going from a to b directly is no more efficient than go from a to 0 and then 0 to b

hmm...

$d(A \to B \to C) = d(A,B)+d(B,C) = |a|+|b|+|b|+|c|$

$d(A \to 0 \to C) = d(A,0)+d(0,C)=|a|+|c|$

so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination

But if the starting point is nonzero, then the distance increases quadratically

Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius)

only when travelling via the origin, will such qudratic penalty in travelling distance be not apply

More interesting things can be said about slight generalisations of this metric:

Jarek Duda

12:13 PM

Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ...

To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ...

2

Q: Analog of Vandermonde determinant for fitting a quadratic form?

1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d...

However, there remains the question when fitting ellipsoid is well defined (det != 0) - it seems basic for e.g. algebraic geometry ... ?

Secret

$d(x,y) = \begin{cases} |x|+|y| & x \ne y \\ 0 & \{(x,y) \in ([a,b] \cup [0,c] \cup [d,\infty),[a,b] \cup [0,c] \cup [d,\infty)) : x = y\} \end{cases}$

Here, the shortest route become dependent on whether the desintation in question and everything in between is isolated (have x,y coordinates lie on the null line)

Jarek Duda

Here describing two points as intersection of such ellipses adds only symmetric (-x) points - the question is when it is true (also in higher dimensions)?

Secret

Any alg geom guys on? I know zilch about alg geom to even start analysing this question

Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit

To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — balu Apr 13 at 18:24

grr, thought I can get some more intuition in SR by using open balls

Jarek Duda

1:02 PM

@Secret, if you want to understand SR, study sine-Gordon model: extremely simple (lattice of coupled pendula) but already has Lorentz contraction, SR scaling of mass/momentum, time dilation for breathers: en.wikipedia.org/wiki/Sine-Gordon_equation en.wikipedia.org/wiki/Breather dropbox.com/s/aj6tu93n04rcgra/soliton.pdf

Semiclassical

1:17 PM

“Theorem X: (1) and (2) are equivalent. Proof: Clearly, (1) iff (2).” Haaate youuu book author

TastyRomeo

At least omit the proof then. Don't get anyone's hopes up

Semiclassical

tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two.

The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$.

And all it has in the proof is the assertion that the above is clearly true.

Haaaaate

Leaky Nun

1:33 PM

@Semiclassical but it’s clearly true :p

Semiclassical

(The argument i see is that their claim amounts to the matrix having a Cholesky factorization, and that’s true whevever the matrix is PSD. But they say nadda )

Haaaate @LeakyNun

Daminark

Hey everyone!

Balarka Sen

Yo

Daminark

@Semi this reminds me, in my analysis book the differentiation chapter started with a review of differentiation in $\mathbb{R}$

And the first theorem it gave... The proof said "Look in a rigorous calculus textbook"

Semiclassical

Lol

Daminark

1:40 PM

In fairness it was basic stuff and by that point everyone knew Spivak but like, you can't write that and then proceed to still put the pirate face qed

Alessandro Codenotti

Hi chat

Daminark

Though in that book, the real annoying thing was that half the proofs were exercises, half were "this is trivial", and half had typos

Semiclassical

To fill in my parenthetical: if such vectors exist, the above matrix has elements of the form $v_j^Tv_k$

So we can write the above matrix as $M=(v_1,v_2,v_3)^T(v_1,v_2,v_3)=V^T V$

And this is psd since $x^T M x=x^T V^T V x = \|Vx\|^2\geq 0$

user8469759

0

Q: Numerical computation of divergence and curl on a discrete surface.

I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k...

numerical-methods computational-geometry

Secret

1:57 PM

Let $r_0,r_1,r_2,..$ be irrationals

Consider $a,b$ both irrational and the interval $[a,b]$

Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$

It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint

This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will

2

Q: How did Irrationals Stay in [0,1] with small open cover of Rationals removed?

If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals. One way for an irrational number $\alpha$ to be in this new set is b...

real-analysis lebesgue-measure

5

Q: Complement of a countable open cover of the rationals

Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals. Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat...

general-topology fake-proofs

user193319

2:12 PM

Suppose that $(X,d)$ is a metric space equipped with addition. Does it follow that this addition operator is continuous with respect to the metric topology?

Secret

2:32 PM

I felt like I need to think this through again. I don't think I understood the open cover of the rationals and irrationals

Secret

3:19 PM

hmm...

(For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs)
Let $\lambda^*$ be the Lebesgue outer measure
We previously proved that $\lambda^*(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero.

We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^*([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting,
or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi…

W also knew that one way to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^*(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^*(\Bbb{I}\cap [-1,1])$ is to use $\lambda^*(\Bbb{Q}\cap [-1,1])+\lambda^*(\Bbb{I}\cap [-1,1]) = \lambda^*([-1,1])$ and thus deducing $\lambda^*(\Bbb{I}\cap [-1,1]) = 2$

However, what I am interested here is to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below:

We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages:

First denote an enumeration of the rationals as follows:

Narcissus jewel

3:41 PM

The name was actually chosen via this chain of events:

1) Milne released a beautiful textbook: Algebraic groups (Theory of group schemes of finite type over a field)

2) He included an image of a butterfly and a quote from Grothendieck about viewing a reductive algebraic group

3) I took the picture and needed a name.

4) I googled a bunch of butterflies

5) Found some that looked good, and eliminated ones whose names were boring.

Secret

$\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short:

David Reed

@Narcissusjewel Nice. I clicked on "sign up via facebook" and my full name and profile were automatically migrated with very little in the way asking my permission

way of*

Secret

$\{p \in \Bbb{N}, q \in \Bbb{N} - \{0,1\} : \pm \frac{p}{q} \in \Bbb{Q} \land \text{lcm} (p,q)=pq\}$

typo: Forgot 0 in the enumeration, which begins before $\frac{1}{2}$

$C_1 = (-1,0) \cup (0,1)$

$C_2 = (-1,-\frac{1}{2}) \cup (-\frac{1}{2},0) \cup (0,\frac{1}{2}) \cup (\frac{1}{2},1)$

After this point, we are going to use an n-tuple as a shorthand of the union of intervals

$C_3 = (-1,-\frac{1}{2},-\frac{1}{3},0,\frac{1}{3},\frac{1}{2},1)$

$C_4 = (-1,-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},0,\frac{1}{3},\frac{1}{2},\frac{2}{3}‌,1)$

$C_5 = (-1,-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},0,\frac{1}{4},\frac{1}{3‌},\frac{1}{2},\frac{2}{3}‌,1)$

$C_6 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},0,\frac{1}{‌4},\frac{1}{3},\frac{1}{2},\frac{2}{3}‌,\frac{3}{4},1)$

$C_7 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},
-\frac{1}{5},0,\frac{1}{5},\frac{1}{‌4},\frac{1}{3},\frac{1}{2},\frac{2}{3}‌,\frac{3}{4},1)$

$C_8 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{2}{5}, -\frac{1}{4}, -\frac{1}{5},0,\frac{1}{5}, \frac{1}{‌4},\frac{2}{5},\frac{1}{3},\frac{1}{2}, \frac{2}{3}‌,\frac{3}{4},1)$

$C_9 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{3}{5},-\frac{1}{3},
-\frac{2}{5}, - \frac{1}{4‌},-\frac{1}{5},0,\frac{1}{5},\frac{1}{‌4},\frac{2}{5},\frac{1}{3}, \frac{3}{5},\frac{1}{2}, \frac{2}{3}‌,\frac{3}{4},1)$

$C_{10} = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{3}{5},-\frac{1}{3}, -\frac{2}{5}, - \frac{1}{4‌},-\frac{1}{5},-\frac{1}{6},0,\frac{1}{6},\frac{1}{5},\frac{1}{‌4},‌\frac{2}{5},\frac{1}{3}, \frac{3}{5},\frac{1}{2}, \frac{2}{3}‌,\frac{3}{4},1)$

Therefore: $C = \lim_{n \to \aleph_0} C_n$

Observe $C$ will eventually exclude all rationals. We now want to check the infinum of the intervals in the nth stage

actually... I am not very sure how to check whether all consecutive pair of intervals do have a length tends to zero as $n \to \aleph_0$

Eric Silva

4:28 PM

@Daminark whenever i hear someone here speak about Sally it's always "Sally was a great man but I have some problems with his pedagogy"

Secret

Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $|p-q| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero

However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works...

Lucas Henrique

hey there

I'm almost sure this is wrong

artofproblemsolving.com/school/mathjams-transcripts?id=438

In fact, Lotta visits client $N$ on day $2^{N-1}$

My approach was a recurrence:

Let's say that for $N$ clients, Lotta will take $d_N$ days to retire.

For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$

Where $d_1$ = 1.

Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.

Narcissus jewel

4:56 PM

@DavidReed That's certainly easier :P.

Lucas Henrique

My result is that $d_N = 2^N - 1$

Mary Star

Hi! How can we solve the equation: sinx+cosx=1,2 ?

Lucas Henrique

try squaring both sides

and using the identity $2 \mathrm{sin}x \, \mathrm{cos}x = \mathrm{sin} 2x$

Mary Star

5:12 PM

@LucasHenrique Thanks!! :-)

Lucas Henrique

@MaryStar np

Dragneel

Hi guys. Is this valid? I'm skeptical about the $(5^2)^{30}$ because it equals $5^{60}$, and it was $5^{30}$ before. Why would they double the exponent?

https://i.gyazo.com/a0b49487716b1adcf52ebb13ce96e2cd.png

nickgard

5:34 PM

The intermediate step should be $(5^2)^{15}\cdot 5$, but then the argument proceeds the same and the final answer is correct.

Dragneel

Okay, just what I thought. Thanks :)

Christopher 2EZ 4RTZ

6:38 PM

hello

I need some help with mathjax

I basically have a complicated equation for a question but I can't type it in mathjax

is there any way one of you nice people would be willing to take my equation and put it in mathjax?

@EricSilva could you help?

halp

David Reed

ok i just went through this yesterday

Christopher 2EZ 4RTZ

David Reed

top right of the screen click the link where it says latex in chat

Christopher 2EZ 4RTZ

I tried

David Reed

what happened

Christopher 2EZ 4RTZ

6:47 PM

ohh lol different link

i can view mathjax but I can't write it well

David Reed

in terms of you knowing the language or your computer not rendering it?

knowing the language

oh ok

2nd time i needed it

that I can't help you with other than to give you a link

Ted Shifrin

6:48 PM

@Christopher2EZ4RTZ: You should recognize the first integral as $-b/a$ times the area of a semicircle of radius $a$. The second term integrates to $2ab$, of course.

Christopher 2EZ 4RTZ

@TedShifrin I just need the latex to use in my question :P

David Reed

ok gotcha

Ted Shifrin

But I answered the question :) You want \int_{-a}^a \big(-\frac{b}{a}\sqrt{a^2-x^2}+b\big)dx ... That's the hard part.

David Reed

hold up

Christopher 2EZ 4RTZ

@DavidReed i got it (martin gave it)

Ted Shifrin

6:56 PM

Howdy, Demonark.

Daminark

How's it going?

Ted Shifrin

Going fine, and you?

Done with finals yet?

Antonios-Alexandros Robotis

hi @TedShifrin, @Daminark.

Ted Shifrin

hi @Antonios.

Christopher 2EZ 4RTZ

Hey would math.meta.stackexchange.com/questions/4666/… be ok to post?

or should I edit it?

Daminark

7:00 PM

Just did both my finals this morning

Ted Shifrin

You should have some parentheses inside the integral for clarity, as I put them. Do you want an answer to your question? I already gave you most of the answer.

Christopher 2EZ 4RTZ

nice

Daminark

Hey Antonios!

Ted Shifrin

Wait, two finals already? I guess they're not 3-hour finals.

Tug' Tegin

Hi everyone! What abot link text?

Daminark

7:01 PM

2 hour finals

Ted Shifrin

Aha.

I was used to 3-hour, both as a student and as a prof.

Christopher 2EZ 4RTZ

@TedShifrin I need to post it to main (added some edits btw) so I can show my old math teacher the proof

Ted Shifrin

Hmm, OK.

Daminark

I see. I think I prefer 2 hours since a day like today would've been draining if I had to test for 6 hours almost straight

Christopher 2EZ 4RTZ

@TedShifrin does the newest version look fine?

Ted Shifrin

7:03 PM

I prefer 3 hours to give people time to think and for the exam to be somewhat comprehensive.

Daminark

Yeah that's fair

Ted Shifrin

You don't need the big parentheses, unless you remove the fraction and just write 1/2 times the whole thing.

Christopher 2EZ 4RTZ

kk

Daminark

One of my professors gave us about 4 and a half hours for the test.

Christopher 2EZ 4RTZ

anything else?

Daminark

7:05 PM

Last year she gave them unlimited time and a harder test but someone took 8-9 hours, which she felt was cruel to the TA

Christopher 2EZ 4RTZ

XD

Ted Shifrin

Well, Christopher, it's good enough.

Christopher 2EZ 4RTZ

@TedShifrin what should i title it?

Ted Shifrin

I don't know. It's basically "What is an explanation for this equality?" You do not want to do the integral by some horrendous means. As I said, you want to observe how it's related to the area of a semicircle. And then it's immediate.

Antonios-Alexandros Robotis

One of my friends has extra time for his exams (well – did in undergrad) and they were 6 hours long.

Christopher 2EZ 4RTZ

7:09 PM

0

Q: Proof request for finding the area of a rectangle using an ellipse and an integral.

$$\frac{\int_{-a}^a (-\frac ba \sqrt{a^2-x^2} + b\;)\mathrm{d}x + \frac{\pi ab}2}2=ab$$ The above equation finds the area of a rectangle given A and B (both > 0). I wrote this myself as a challenge. Using desmos I verified that it was true (the desmos). I was wondering if someone could write a pr...