Mathematics - 2018-04-10 (page 3 of 4)

Leyla Alkan

13:00

Hi all! I didn't understand how the yellow parts are derived

Akiva Weinberger

Yeah 'cause it's symmetric

Oh I thought you were about to ask about the blue underlined bits

It's Pythagoras. $ds^2=dx^2+dy^2$.

Leyla Alkan

oh

what about this: ıf $\bar x$ is $0$ how can they calculate $l$?

Akiva Weinberger

I don't think you use $\bar x$ to find $l$

In that picture^, $dx$ is the width of that triangle, $dy$ is the height, and $ds$ is the solid pink line segment

'cause you're breaking up the curve into lots of little line segments and adding them

(Image source)

Leyla Alkan

Okay, I got that. Thanks@AkivaWeinberger

Btw, I got a Chegg account yesterday. But I can't get answers right away as I can here.

MSE is a heaven on earth, lol

Akiva Weinberger

What's Chegg?

Balarka Sen

13:11

a q&a site iirc

Akiva Weinberger

Looks like a tutoring site

> The name Chegg is a portmanteau of the words chicken and egg, based on the founders' experience after graduating from college; they could not land a job without experience, but could not get experience without a job. - Wikipedia

Leyla Alkan

It's also a Q&A site, but there is also numerous textbooks' solutions

Akiva Weinberger

I like it

Leyla Alkan

Hahaha yes

Balarka Sen

@AkivaWeinberger Learn and understand Anosov maps and then prove that spheres do not admit Anosov diffeomorphisms for me

Add a comment here when you're done :D

Sirmimer

13:14

Can someone help with the first few steps of isolating $\frac{x_1}{x_2}$ from the following equation $(\frac{a_1x_1}{a_2x_2})^{(p-1)}$

$(\frac{a_1x_1}{a_2x_2})^{(p-1)}=l$ *

I figured substituting the expression into $\frac{x_1}{x_2}=\apha$ and then isolating alpha. But I still couldn't quite figure it out from the new equation: $(\frac{a_1\alpha}{a_2})^{(p-1)}$

of course i ment: $(\frac{a_1\alpha}{a_2})^{(p-1)}=l$

Akiva Weinberger

Remember that $(xy)^n=x^ny^n$

In your case you have $(\frac{a_1}{a_2})^{p-1}(\frac{x_1}{x_2})^{p-1}=\ell^*$

Then $(\frac{x_1}{x_2})^{p-1}=\ell^*(\frac{a_2}{a_1})^{p-1}$

And finally take the $p-1$ root for

$\frac{x_1}{x_2}=\sqrt[p-1]{\ell^*}\frac{a_2}{a_1}$

ÍgjøgnumMeg

@AkivaWeinberger why not just take the $(p-1)$th root straight away?

Akiva Weinberger

You could have also taken the $p-1$-th root as the first step

Balarka Sen

pew pew

Akiva Weinberger

When I'm sniped should I even bother writing the rest of the sentence

ÍgjøgnumMeg

13:21

just to nitpick hahah

Akiva Weinberger

But yeah I kinda want to avoid roots as long as possible, you know? Kinda like avoiding fractions

Otherwise they might get annoying

Wouldn't have happened in this case but could happen in general

Balarka Sen

Yeah I'm paranoid about that too

Sirmimer

Alright I got it. Took me a little bit with the second line (before root). Makes sense thx

Balarka Sen

Eg I also always make my logs into exponentials

Even in unnecessary contexts

ÍgjøgnumMeg

it's cool

I type $3 - 4$ into a calculator

just in case it's not what I think it is

Akiva Weinberger

13:25

@BalarkaSen Yeah, otherwise you might have to simplify $\sqrt[\Large\log_xy]y$ or something

(It's $x$)

Balarka Sen

Lol I did the dot product of a vector with (1, 0, 0) using calculator once

In my defense it was an exam and I was nervous

@AkivaWeinberger Right, ugh

ÍgjøgnumMeg

It's okay to identify the polynomial ring $R[X]$ with $\bigoplus_{i \in \Bbb N} R$, right?

Akiva Weinberger

Seems to me the latter doesn't have an obvious multiplicative structure

Besides, would $R(X)$ be $\bigoplus_{i\in\Bbb Z}R$?

Balarka Sen

As R-modules it's fine

ÍgjøgnumMeg

I mean as $R$-modules of course

Balarka Sen

13:30

@AkivaWeinberger I think that would be R[X, 1/X]

Akiva Weinberger

Arright sure then

@BalarkaSen Oh, right

$\frac1{x+1}\notin R[X,1/X]$

Balarka Sen

Mhm

Akiva Weinberger

Laurent polynomials

Balarka Sen

Well, $\frac{1}{x+1} \notin R(X)$ either

:P

Akiva Weinberger

No?

Oh. $X$

Balarka Sen

13:32

Lol

Sorry

ÍgjøgnumMeg

lel

Akiva Weinberger

Big X, little x, sideways +, same thing

ÍgjøgnumMeg

$\frac{1}{\times + 1}$

Balarka Sen

Accurate

orbit-stabilizer

what is

Akiva Weinberger

13:35

The future of civil engineering, ladies and gentlemen

orbit-stabilizer

$f^{(99)}(z)$ where $f(z) = e^{iz}$? I thought it was $-ie^{iz}$

Balarka Sen

Looks correct.

$f^{(n)}(z) = i^n e^{iz}$.

orbit-stabilizer

Hmmm. Then the answer key says $\int \frac{100! e^{iz}}{(z+1)^{100}}dz $ along a circle of radius 1 at -1 is equal to $100\pi i e^{-i}$

Akiva Weinberger

Is that not $f^{(100)}$?

orbit-stabilizer

I don't agree with that $i$ in the answer. It should be $2\pi i Res(-1)$

Where $Res(z_0) = \lim_{z\rightarrow z_0} \frac{1}{m-1!} \frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^{m}f(z)]$

Akiva Weinberger

13:41

wikimedia.org/api/rest_v1/media/math/render/svg/…

orbit-stabilizer

Setting, $m = 100$, we get that we take the derivative of $e^{iz}$, 99 times.

@AkivaWeinberger, that's cauchy's integral formula. not residue theorem

Akiva Weinberger

Oh, I see, there's an $n+1$ in the denominator

@orbit-stabilizer Yeah but it applies here too

orbit-stabilizer

fair

So, it seems like Cauchy's formula gives me the right answer. But then, I must be messing up somewhere in the residue theorem

Akiva Weinberger

http://www.wolframalpha.com/input/?i=residue+of+(100!+e%5Eiz)%2F((z%2B1)%5E100)‌+at+-1

Seems like Wolfram Alpha agrees with you, in that the residue is $-100ie^{-i}$

orbit-stabilizer

itisamystery.com

Akiva Weinberger

13:46

and thus the integral should $200\pi e^{-i}$ I think

orbit-stabilizer

Yeah, that's what I get.

Wait, that's what I get for cauchy as well..

So.. they probably fked up?

Akiva Weinberger

Prolly

orbit-stabilizer

gah

wish me luck, I go to fail my complex variables final

Akiva Weinberger

Luck

I wish it upon thee

Xander Henderson

Finals? Man... my quarter just started.

Akiva Weinberger

13:51

It'll happen before you know it ^_^

Leyla Alkan

Do you also know about this @AkivaWeinberger?

Akiva Weinberger

$ds$ should equal $|dg|$

'cause $ds$ is the distance between $g(t)$ and $g(t+dt)$

Leaky Nun

$H^1(S_n,\Bbb Z)\cong\Bbb Z$

i'm high on group cohomology

Akiva Weinberger

I was very confused 'cause I thought that said $S^n$

Leaky Nun

you're high on topology

Akiva Weinberger

13:59

Why do we want to take the cohomology of groups

Leaky Nun

because it's dank

Leyla Alkan

Thanks :)

Akiva Weinberger

So my vague understanding is that we turn $G$ into spaces and take the cohomologies of those spaces? And there's also a native group-theoretic way as well @LeakyNun

Balarka Sen

Equivalent approaches yes

Leaky Nun

@AkivaWeinberger do you know Tor and Ext?

Akiva Weinberger

14:02

Vaguely. Tor measures torsion and Ext measures magic

Leaky Nun

do you know group ring?

Akiva Weinberger

The name rings a bell

Wahey, "rings"

Leaky Nun

wow

so if $R$ is a ring and $G$ is a group, then $R[G]$ is a ring that is, as an $R$-module, the free product of $G$ copies of $R$

Akiva Weinberger

Hey guys quick question, A or B

Leaky Nun

the multiplication is given in a way that "g" multiplied by "h" gives "gh"

Mike Miller

14:04

Group cohomology is an algebraic tool to capture all the information you can out of a group; just like if a map between CW complexes induces an iso on homotopy groups, it is an equivalence, you can check that a map of finite groups which induced an iso on group homology is an isomorphism of groups

Akiva Weinberger

I see.

Leaky Nun

@AkivaWeinberger A

Mike Miller

You can use it to prove things about those groups, and it often has group-theoretic information hidden in it

Akiva Weinberger

Can I get a concrete example?

Leaky Nun

@AkivaWeinberger so, e.g. $\Bbb Z[C_3] = \{a_0 + a_1 g + a_2 g^2 \mid a_0, a_1, a_2 \in \Bbb Z\}$

where e.g. $(2g) \times (3g^2) = 6$

Akiva Weinberger

14:05

Rmhrm.

Leaky Nun

$(1+2g) \times (g) = g + 2g^2$

Akiva Weinberger

I see.

Mike Miller

Eg, consider H^*(G;Z/p). This is a commutative ring (more or less). We can talk about its Krull dimension; in the simplest case where it's polynomial on n generators, the dimension is n.

As it turns out the dimension of this ring is equal to the rank of the largest (Z/p)^n in G.

Leaky Nun

now, giving an abelian group $M$ the structure of a $\Bbb Z[G]$ module is the same as giving it a linear $G$-action

i.e. other than $1x=x$ and $g(hx)=(gh)x$, you also require $g(x+y)=gx+gy$

so a $\Bbb Z[G]$-module is also called a $G$-module

Akiva Weinberger

@MikeMiller I see

Balarka Sen

14:07

The standard baby example is that $G$ has torsion implies $H_n(G; \Bbb Z)$ is nonzero in arbitrarily high dimensions.

Mike Miller

The cohomology of BSO(3) with mod 2 coefficients is Z/2 [w_2, w_3] those guys living in deg 2 and 3, no further relations

Leaky Nun

Then, $\Bbb Z$ is a $G$-module where the action is trivial, i.e. $gn=n$ for every $g \in G$ and $n \in \Bbb Z$

Mike Miller

SO(3) has a Klein 4-subgroup correspondingly: S(O(1) x O(1) x O(1))

Akiva Weinberger

BSO?

Leaky Nun

Then, the cohomology group $H^n(G,M)$ where $M$ is a $G$-module, is $\operatorname{Ext}^n_{\Bbb Z[G]}(\Bbb Z,M)$

Mike Miller

14:08

Ignore the B

I'm taking group cohomology now for compact Lie groups

And I was thinking of the classifying space approach

Balarka Sen

BG is the space corresponding to G that you take singular cohomology of to compute the group cohomology of G (so you can ignore it, like Mike said)

Akiva Weinberger

Right, and the Klein 4-subgroup is the symmetries of a rectangle

@BalarkaSen Ah, I see

Mike Miller

SU(2) has cohomology Z/p [p_1] with p_1 living in degree 4

It therefore can only have a Z/p living inside it

Via the inclusion inside the circle subgroup

If a group acts freely on a sphere, then its group cohomology is "periodic" (H_* is jsomorphic to H_{*+n} for all *)

So we see from the above that SO(3) does not act freely on a sphere: the rank of cohomology above was increasing linearly with degree

Leaky Nun

@MikeMiller are our approaches different?

Mike Miller

(the number of 2-variable monoids of degree n is n+1)

Akiva Weinberger

14:12

I gotta go for a bit

Mike Miller

Same answer, yours is slightly harder to generalized to compact Lie groups, mine is slightly harder to generalize to nontrivial coefficients

Leaky Nun

interesting

Akiva Weinberger

I'm back

Balarka Sen

The torsion fact has been useful in several different points in my life. Akiva once asked whether orientation double cover of a nonorientable (not necessarily closed) manifold can be contractible. The answer is no, because then the base has to be a K(Z/2, 1) which has cohomology in arbitrarily high dimensions

In particular can't be a manifold

Saved my ass that day because I was rambling garbage involving Mazur manifolds to PVAL prior to realizing this

Leaky Nun

@AkivaWeinberger after some calculations, $H^1(G,M)$ can be described as the free abelian group of "twisted homomorphisms" from $G$ to $M$, quotient by the functions that look like $g \mapsto gm-m$ for some $m$

now, $f:G \to M$ is a twisted homomorphism if it satisfies $f(gh) = gf(h)+f(g)$

Akiva Weinberger

14:18

Is it called "twisted" because there's a hidden connection to braid groups or is it called "twisted" 'cause it looks weird

Leaky Nun

well normally we would have $f(gh)=f(g)+f(h)$

so you kind of twist it a bit

Mike Miller

I would call them projective representations (you're allowed to scale, and compositions are defined up yo scaling)

Balarka Sen

twist by the G-action

Leaky Nun

right

Akiva Weinberger

@MikeMiller Free means that the actions have no fixed points, right?

Like $x\mapsto gx$ with $g\ne e$

Leaky Nun

14:20

up yo scaling

free: $\forall g (\forall x (gx=x \to g=e))$

faithful: $\forall g(\forall x(gx=x)\to g=e)$

you literally move a single parenthesis

Akiva Weinberger

Yo, Frankenstein

Leaky Nun

aqui viene frankenstein

aqui va frankenstein

Balarka Sen

10/10 will watch this movie

Akiva Weinberger

Turns out it was bad

Balarka Sen

I refuse to believe that

Leaky Nun

14:27

@MikeMiller so, uh, how do you, eh, go from, eh, cohomology group relative to some G-module, to, eh, "the" cohomology group of a group?

Akiva Weinberger

Whoa, someone want a gem for $15?

Diamonds aren't even the pretties gem, why put that on a ring when you can get a prettier one for cheap

Leaky Nun

why are people still falling for the diamond prank

@BalarkaSen do people use Tor and Ext?

orbit-stabilizer

14:54

It's in 30 minutes..eek

I used L'Hopital on $\lim_{z\rightarrow \frac{\pi}{2}} \frac{zcos(z)}{(2z-\pi)^2}$ to get a finite value. I definitely messed up somehow.

Szabolcs

Graph theory terminology question: Is there a term for the set of vertices that are not part of any cycle?

orbit-stabilizer

That would be equal to $\lim_{z\rightarrow \frac{\pi}{2}} \frac{cos(z) - zsin(z)}{4(2z-\pi)}$

Which should be equal to: $\lim_{z\rightarrow \frac{\pi}{2}} \frac{-sin(z)-(sin(z) +zcos(z))}{8}$

Mike Miller

@LeakyNun Use Z as the trivial G-module, or Z/p if you're interested in torsion information.

orbit-stabilizer

Which exists.

Mike Miller

Don't use Q.

Leaky Nun

14:57

ok

Abcd

Hi @LeakyNun ... ?

Leaky Nun

mental note: stay away from Q

what

Abcd

@LeakyNun $f(a+b)= f(ab), f(-1/2)= -1/2$, find $f(1005)$, any hints?

Leaky Nun

a=1, b=-1/2

Abcd

I know that

Mike Miller

14:59

H^*(G;Q) = 0, not hard to show

Leaky Nun

then a=1/2, b=-1/2

and then a=1,b=b, induction

Abcd

I know that too

@LeakyNun Oh, what?

Leaky Nun

what is f(0)?

Abcd

f(0)= f(everything/ anything)

Leaky Nun

what

Abcd

15:00

f(1)= f(0)

Leaky Nun

right

Abcd

f(-1/2)= f(0 )

Leaky Nun

so ignore the part where i said a=1/2,b=-1/2

let's see how we can reach an integer

oh wait what

oh right

just substitute a=0, b=b lol

i'm stupid

done, why ask me

substitute a=0, b=b, what do you get?

Abcd

Wait,

f(1005)= f(0)

f(1005)= -1/2

done

Leaky Nun

lol

orbit-stabilizer

15:11

If I have $e^{6i\theta} = e^{i\pi}$, I get $6\theta = \pi$, but there are more solutions, are there not?

ah, $+ 2\pi n / 6$

Abcd

15:40

$f(x).f(1/x)$ is satisfied by polynomial function $f(x)=\pm x^n +1$

How?

philmcole

Is it true, that if $A$ is diagonalizable and $AB=BA$ then also $B$ is diagonalizable? If found only proofs which required $A$ having $n$ distinct eigenvalues. But my assumption allows for repeated eigenvalues. Is it still true?

For example: math.stackexchange.com/questions/285546/…

Abcd

(@LeakyNun any idea?)

Leaky Nun

no

philmcole

I found a proof showing that $B$ is "block diagonal": https://math.stackexchange.com/questions/46544/why-does-a-diagonalization-of-a-matrix-b-with-the-basis-of-a-commuting-matrix-a?lq=1
This still doesn't say it's diagonalizable, right?

Alright, I found the solution. It's not true, see second example here: en.wikipedia.org/wiki/Commuting_matrices

Abcd

16:01

@LeakyNun Sorry made a mistake. $f(x).f(1/x)= f(x)+ f(1/x)$ is satisfied by polynomial function $f(x)=\pm x^n +1$

16 mins ago, by Abcd

(@LeakyNun any idea?)

Leaky Nun

hmm ok?

Abcd

@LeakyNun could you tell me how its derived?

Leaky Nun

no idea

Balarka Sen

16:21

Blockhead - Downtown Science (Full Album)

►

pretty damn ill

Szabolcs

16:43

Is there a standard term for components of a graph that look like these?

user193319

1

Q: Subsequence of Measurable Functions

Given a sequence $\{f_n\}$ of measurable functions, why does there exist a subsequence $\{f_{n_k}\}$ such that $\lim_{k \to \infty} \int_E f_{n_k} = \liminf \int_E f_n$? I need to use this in a theorem I am proving, but I don't see how to justify it. Just for your information, I am trying to prov...

Need help proving a subsequence exists. See my comments on B. Mehta's answer.

Zee

17:22

/wave

//wave

Alisha

@Zee Maybe you should consider looking up the manual

Curio

Consider these 2 infinite sets: 3, 3.3 , 3.33, 3.333 ... and 4 , 3.4 , 3.34 , 3.334. They tend to 10/3. Are they contiguous?

Leaky Nun

what is contiguous?

Abcd

17:55

Let $f: \mathbb{R^+} \to \mathbb R $ be a function which satisfies $f(x).f(y)= f(xy)+2 (\frac 1x + \frac 1y + 1)$ $\forall x,y >0$, then find $f(x)$

Attempt:
$$f(1)= 2 \text{ or } -3 \\
f(x).f(1/x)= 1/x+x \\
(f(x))^2= f(x^2)+2(2/x +1 )$$

Step 2 if f(1)= -2

Balarka Sen

$f(x)f(1) = f(x) + 2(1/x + 2)$ is probably useful.

Abcd

Correction* f(1) = -2 or 3

@BalarkaSen Oh right. Which value of $f(1)$ should I take, -2 or 3?

Oh its a quadratic

so f(x) should have two values

got it'

Balarka Sen

Yeah I don't think $f(1)$ is completely determined.

These functional equation questions kinda suck because most of the time they are vague.

user193319

18:11

0

Q: Lebesgue's Theorem for Riemann Integrability Overkill?

Let $f$ be a bounded function on $[a,b]$ whose set of discontinuities has measure zero. Show that $f$ is measurable. Then show that the same holds without the assumption of boundedness. This problem comes from the section where Lebesgue's theorem about Riemann integrable functions is presen...

real-analysis measure-theory proof-verification

Silent

@LeakyNun, here $z(H\int K)$ need not posses either $x$ or $y$, right?

Dattier

@Abcd $f(x)\times f(1)=f(x) +2(1/x+2)$ so $f(x)=\dfrac{2+4x}{x(f(1)-1)}$

Mike Miller

@BalarkaSen I kinda like them, just pushing around symbols until things work

$f(x) = 1/x + 2$ or $f(x) = \frac{-2}{3} \left(1/x + 2\right)$ I guess

Curio

@LeakyNun en.m.wikipedia.org/wiki/Contiguity

Balarka Sen

@MikeMiller It's fun to work out, yeah. I guess I just don't like the parsing as much.

Sometimes you'll see the domain/range unspecified

Semiclassical

18:26

@MikeMiller it's sorta closer to a modeling challenge, i.e. given constraints provide an appropriate function

Balarka Sen

That boils my blood a little

Semiclassical

The trouble with that analogy, of course, is that a model usually has some motivation behind it

MA-Maddin

Semiclassical

whereas a random question need not

MA-Maddin

can someone tell me what I have to google for, to find the formula that calculates the ID from the values of num1, num2, num3 ?
so 1,2,3 = 1
1,3,4 = 4
i guess there must exist one already :P
num3 > num2 > num1
sum[ID+1](num1, num2, num3) > sum[ID](num1, num2, num3)

Semiclassical

18:29

@MA-Maddin look at ID's 6 and 7

then 1+4+5=10 for ID 6 but 2+3+4=9 for ID 7

so that's actually gone down

MA-Maddin

oh, you're right. Sry, so the last rule doesn't apply :P

Semiclassical

similarly, note that ID 1 has 1+2+3=6 and ID 9 has 3+4+5=12. So if you were increasing by at least one with each ID, you'd need ID 9 to have at least 6+8=14

the real problem, though, is that all you have are a collection of numbers.

MA-Maddin

that are all sorted combinations of 3 numbers from a pool of 1-5.

Semiclassical

you can devise a function that will give those numbers in response to the inputs, but absent some specific pattern there's no reason to think said function is very interesting

In particular, there's no way at all from this data to guess what ID 10 would give

Silent

Is it possible that $z(H\cap K)$ has neither $x$ nor $y$ here?

MA-Maddin

18:34

yea, the Input is always 3 numbers. The Output the ID.

Semiclassical

ah, other way around then.

I see the pattern now, yeah

MA-Maddin

I guess there exist a function already, just don't know the name :P

Semiclassical

I think I see another way of looking at this, actually.

The Testosterone Fanatic

GUYS

I just made another meme of mine

does anyone want to see it

Semiclassical

you can encode the combination 135 using stars-and-bars as * | * | *

Zee

18:36

Hell yea

Semiclassical

so there's stars in positions 1,3,5 and bars in positions 2,4

The Testosterone Fanatic

@Zee it's offensive and not suited for this chat, so I'll post it in another separate room

actually, never mind... There's no way for me to post it here -_-

Semiclassical

@MA-Maddin I think you've missed an ID along the way: between 7 and 8 you should also have 2 3 5

MA-Maddin

oh, yes. good catch :D

Semiclassical

in which case you have 10 IDs possible, which corresponds to there being 5-choose-3=10 ways to pick the 3 stars among the 5 spots

what I think you can say is that you're doing some sort of lexicographic ordering

123 , 124 , 125 , 134 , 135 , 145 , 234 , 235 , 245, 345

where you order just as you would if you were doing an alphabetical sort

MA-Maddin

18:43

thats right

Abcd

@Dattier I know, then we'll have to substitute the two values of $f(1)$, and two answers are given ...

Semiclassical

The wiki page on lexicographical ordering may be worth looking at, particularly the section on finite subsets: en.wikipedia.org/wiki/Lexicographical_order#Finite_subsets

Balarka Sen

@TheTestosteroneFanatic who the hell makes a makes a meme of themselves

isn't that memetic narcissism

Dattier

@Abcd yes

Lozansky

Hi @Dattier

Dattier

18:47

Hi

Mary Star

Hello!! Does someone of you have an idea how we could show that the Gamma function is continuous for $x>0$ using the mean value theorem of differential calculus?

MA-Maddin

@Semiclassical yeah, that looks nice. Thank you.

Semiclassical

What I don't know is how one would do the following: Given a particular element (n1 n2 n3), determine where it appears land in the relevant lexicographic ordering

there may not be a nicer way than just "take all such elements, order them, and look where that one is"

Dattier

@Lozansky how did the exams go?

Semiclassical

But this is getting firmly into algorithms territory and I'm not an expert on that

@MA-Maddin oh, this looks even more useful: geeksforgeeks.org/lexicographic-rank-of-a-string

so "lexicographic rank" is the relevant search word

"The time complexity of the above solution is O(n^2). We can reduce the time complexity to O(n) by creating an auxiliary array of size 256."

ooook

Lozansky

18:54

@Dattier It went well, I did PDE before Fourier so I already knew like 60% of the material

MA-Maddin

@Semiclassical :D you helped me anway, thank you :)

Semiclassical

np

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$Mathematics$ Mathematics