Mathematics - 2016-10-31 (page 1 of 3)

user3502615

00:11

hey guys i just found a breakthrough in mathematics on vixra vixra.org/pdf/1607.0499v1.pdf

0celo7

this better be good

> Theorem 1.1. Kothe conjecture is right.

user3502615

im laughing so hard right now

this website's a fucking gold mine

i just found a "proof" of fermat's last theorem

typed up using ms word

vixra.org/pdf/1609.0393v1.pdf

0celo7

> King’s Mongkut Institute of Technology Ladkrabang
Mechanical Engineering , Thailand

That guy probably has a PhD and everything

> Mr. Sattawat Suntisurat

Maybe not?

user3502615

he may not have a PhD

but he has a catchy name

sounds like "hakuna matata"

0celo7

Maybe that proof is correct. Will anyone check?

user3502615

00:16

I tried to check it

but i really got lost in all the pseudomathematics

ok uhh wtf

this guy posted 3 proofs of this on vixra

vixra.org/pdf/1609.0361v1.pdf

> Shortest proof in the world of The Fermat’s Last Theorem\

pretty impressive

Ted Shifrin

@Balarka: Not sure what I'm supposed to read and what not. :) You use planar to know $\tau = 0$. ... 16b is far more subtle. You're fine if no geodesics have $\kappa=0$ somewhere. But otherwise ...

Hope you got some good sleep finally.

0celo7

Ok let's check it

the proof I mean.

Eq. (1) is correct.

what the heck is (2) lol

Ted Shifrin

hello 0celo

0celo7

@TedShifrin Hi. Is a 55 average normal in a first year grad class?

For an exam, I mean.

Ted Shifrin

Sort of a meaningless question.

Different professors have different testing/grading styles.

My numbers were always a bit lower than most people were used to, but I didn't use the usual 90-80-70 grading scale.

0celo7

00:25

@user3502615 I think equation 2 in that paper is wrong

Ted Shifrin

It also depends whether you're talking upper undergraduate/MA level or PhD level.

0celo7

I should email the author

Ted Shifrin

"Think"ing the equation is wrong isn't good enough, 0celo. Do you have a counterexample?

0celo7

@TedShifrin No, I think it is equivalent to Fermat.

But he pulled it out of thin air

@TedShifrin First year quantum

Ted Shifrin

Honestly, there are a lot of frauds out there. I personally wouldn't waste my time on it.

0celo7

00:30

All the incoming grad students are taking

@TedShifrin I'm not wasting time. I got a new phone and am babysitting the backup process

Ted Shifrin

First year as in Ph.D. level or M.A. level?

LOL, oh.

0celo7

@TedShifrin MA. It's a 500 level class

@TedShifrin So I'm doing whatever is least boring but does not require lots of attention

Ted Shifrin

Then it's less surprising. Probably the professor would have been happier with an average 10-15 points higher.

But some professors write exams expecting a 40 average.

user3502615

i want to be a professor

0celo7

The grading scale is 90-80-70 I think

Ted Shifrin

00:32

Then I hope you learn to be a good teacher, @user3502615. That's not a priority for too many.

0celo: I doubt the professor will give half the graduate students Ds and Fs.

user3502615

im going to try and solve some IMO problems tonight

Ted Shifrin

I don't think I've ever solved one.

But I haven't tried many.

user3502615

what do i have to do if i want t get into an IMO?

do i have to be in high school?

Ted Shifrin

You have to be on your national team.

user3502615

honestly

Ted Shifrin

00:36

People train for years, first locally, then on national teams.

user3502615

i suck at number theory

0celo7

@TedShifrin He sent out a message telling anyone who got below a 50 to schedule a meeting with him

(I did not get below a 50)

Ted Shifrin

If the exam was one people should have gotten 80s on, then a serious graduate student is in trouble doing so poorly, so it's good of him to force a meeting. I think that shows caring.

user3502615

@0celo7 what class?

0celo7

oh he definitely cares

he's doing an optional problem session once a week now

usukidoll

00:44

Hi

0celo7

@user3502615 QM, first half of Sakurai

user3502615

sakurai? the man who made Super Smash Bros.?

0celo7

Yes

he's also a physicist

usukidoll

Is there an efficient way to grab all polynomials of less than or equal to degree 3 in mod 3? That's 3^3 =27 polynomials and all I got are 16 or 17. It's like all the polynomials is the form Ax^3+Bx^2+Cx+D, Ax^2+Bx+C, Ax+B and A

Ted Shifrin

you should appreciate him, 0celo ... atypical.

@usukidoll: You should be able to figure out for yourself what you're missing.

0celo7

00:47

@TedShifrin Hmm?

I appreciate the prof just fine, I'll be going the problem sessions even though I got the grade I wanted.

Ted Shifrin

I'm just saying a professor who cares is atypical.

user3502615

im disappointed in my 8th grade physics teacher; today, he told the class that acceleration is an "exponential increase in speed"

0celo7

lol

Ted Shifrin

Oh dear ...

user3502615

im no physicist but im pretty sure acceleration has linear speed.

Ted Shifrin

00:50

But 8th grade physics teachers aren't exactly super well-trained.

No, that's wrong, @user3502615.

user3502615

constant acceleration?

Ted Shifrin

Did you say constant acceleration?

user3502615

yes

thats what he meant

Ted Shifrin

Where?

Why did he mean that?

user3502615

sorry i thought it was implied

Ted Shifrin

00:51

NOOOOOOO

0celo7

You broke Ted, dammit

Ted Shifrin

You're as bad as he is.

user3502615

but.. wait

NOOO

usukidoll

X^3, X^3+X^2, X^3+X^2+X, X^3+X^2+X+1, X^3+X, X^3+X^2+1 , X^3+X+1, and X^3+1 is all I can grab for polynomials in the form Ax^3+Bx^2+Cx+D oh wait a minute... I think I need to redo this since each letter can be set as 0 1 or 2

user3502615

where did my life go wrong????

Ted Shifrin

00:51

Right @usukidoll.

user3502615

but $\vec{v}' = \vec{a}$, right???

0celo7

how do you know about derivatives in 8th grade...

user3502615

for velocity and acceleration vectors $\vec{v} and \vec{a}$

Ted Shifrin

But $\vec a(t)$ can (and usually is) be a function of $t$.

user3502615

yeah so

Ted Shifrin

00:52

For example, the acceleration due to the earth's gravitational field far above the surface of the earth is not constant.

0celo7

OK. My new phone knows when I'm looking at it and turns itself on. FREAKY

usukidoll

X. X when you sleep so late and can't even think lel

user3502615

$\vec{v} = \int \vec{a} dt$ right?

0celo7

I'm scared

user3502615

and of course the integral of a constant acceleration would be.... linear velocity??

0celo7

00:53

@user3502615 Not quite

usukidoll

Of?

0celo7

You're solving an ODE, you have to make sure the boundary conditions are correct

it's more than just an integral

Velocity is linear for constant acceleration, yes

user3502615

ooh, take that, TEDDY

usukidoll

So A= 0 1 2 B = 0 1 2 C = 0 1 2 D= 0 1 2 x. X

0celo7

But that's not what you said originally

user3502615

00:54

what did i say?

i need to get a username

one second

Ted Shifrin

@user3502615: I think you owe me a bit more respect than that.

0celo7

you didn't say constant acceleration...

Ted Shifrin

quickly loses patience

user3502615

@TedShifrin i kid you.

usukidoll

~_~

Yesterday there was a lot of rain

Ted Shifrin

00:57

@usukidoll: Of course they can't all be $0$. But, yeah, that's the count you did in the first place :P

user3502615

@0celo7 i shouldve mentioned it. in the situation, he was talking about constant acceleration

0celo7

huh? who are you

Ted Shifrin

the twit just changed his name.

meow-mix

i have name-switched

because

usukidoll

5 out of 8 x^3 polynomials were irreducible

0celo7

00:58

oh, I see

meow-mix

its hard to memorize my numbers

0celo7

@TedShifrin twit?

meow-mix

@TedShifrin i've never been called a "twit" before. interesting

usukidoll

Ahahahahaha

0celo7

@meow-mix Ted is old-fashioned like that

Ted Shifrin

00:59

@usukidoll: For degrees up to 3, deciding irreducibility is easy. But in higher degrees you want to think about a generalization of the sieve of Erasthosthenes (for making a lit of prime numbers).

0celo7

@TedShifrin Something must be working, my dad was in the mall all day today with my mom. That's not an easy task :)

Ted Shifrin

@0celo7 Huh?

0celo7

His back

Ted Shifrin

Ohhhh, good.

0celo7

We were talking about it last night

Ted Shifrin

01:00

I go to the chiro/massage therapist in the morning. I hope they fix me before my 2 weeks of long drives.

meow-mix

long drives are the worst

0celo7

Oh, I didn't think about that. He probably won't come to visit me now

I'm going home for Thanksgiving, that's a looooong drive to DC

meow-mix

a while back, when i visited my cousins in florida (keep in mind, i live in new jersey), i would have a 24 hour drive to them and back

Ted Shifrin

They're in DC? Yeah, that's not too close.

0celo7

I'm going to probably skip probability (ha) so I don't have to drive through 1AM...

meow-mix

01:04

as much as i love pens, doing math with pens is the worst

i dont even like pencils, i just use $\LaTeX$ instead :D

holy shit

i just saw this GIGANTIC spider

and so i get a shoe and try to kill it and the THING FUCKING FLINGS AT ME

THE THING CAN FLY

0celo7

you're gonna get suspended if you don't tone it down

meow-mix

i apologize

Andrew Li

01:44

Why are the solutions of the quadratic inequality x^2 + 8x + 16 >= 0 all real numbers? Is it because anything you plug in for x will make the inequality true?

arctic tern

@AndrewLi (x+4)^2 >= 0 is always true - a real number squared will be 0 or bigger

0celo7

@arctictern anything?

Andrew Li

@arctictern When x is any real right?

arctic tern

> a real number squared will be 0 or bigger

Andrew Li

Ok

I have one more question

So is it possible to say that if an equation has one zero or no zeros that the inequality has infinite solutions?

0celo7

01:56

@AndrewLi no

the solutions could be negative

but in that case your quadratic is a square

Andrew Li

I'm having a really hard time understanding how to find the solutions to an inequality

0celo7

so the solutions are automatically positive

Andrew Li

Well yeah

0celo7

@AndrewLi you draw the solution curve for =

you know how do do that, right?

Andrew Li

With the inequality 0 > -2x^2 + 4x + 4, the zeros are -0.73 and 2.73

@0celo7 Of course

How do you determine what the solutions are?

arctic tern

01:58

to solve quadratic inequalities, you need to complete the square, then know how to split u^2>k and u^2<k up into parts

Andrew Li

I need to?

I've been using the quadratic formula

Isn't it just to find the zeros?

arctic tern

quadratic formula, completing the square, same thing

Andrew Li

Yeah

I get that you have to find the zeros of the parabola

So you try to find all x that satisfy the inequality right?

arctic tern

BTW, "how do I solve inequalities" is far, far more general than "how do I solve quadratic inequalities"

it's not a good idea to leave important context out of questions

Andrew Li

Yeah, sorry

Why would an quadratic inequality such as -x^2 + 2x - 15 < 0 have all real numbers as solutions?

arctic tern

02:01

you can find the real zeros (assuming there are any), use the leading coefficient to determine if the parabola opens up or opens down, then use that to determine if you're doing an interval between the zeros or to the right and left of them

Andrew Li

But what if no (real) zeros?

arctic tern

@AndrewLi that's -(x-1)^2<14

if no real zeros, then either all real numbers or no solutions

Andrew Li

Ok. How would I tell?

arctic tern

by whether the graph opens up or down

e.g. if the graph opens up and there are no real zeros, then the whole parabola is above the x-axis, so all its values are positive

Andrew Li

Ok

So -x^2 + 2x - 15 < 0 will always be true. I see that, but the graph has no real zeros and it opens down

Ohhhh

Nevermind. I got it

So if two solutions, check for or or and.

Thanks @arctictern!

TheGreatDuck

02:34

@arctictern you were here last night, correct?

@arctictern I was curious if you knew of any functions other than constant or piece wise constant functions that always give a function in their set when composed together in any manner with each other.

0celo7

02:52

@TheGreatDuck the identity?

Secret

03:08

[Random] Can we have probability dependent probability distributions, that is, the random variable is itself a probability of another random variable?

0celo7

@Secret Yes.

Let $X:S\to\Bbb R$ be an r.v., then $W=g(X)$, where $g:\Bbb R\to\Bbb R$, is also an r.v.

It makes sense to talk about $f_W$, the pdf.

Secret

I see

0celo7

But maybe that's not what you're asking?

You can compute $f_W$ in terms of $f_X$ btw.

Secret

Well it sounds right, If we restrict the measure g such that its range lies between 0 to 1, and is additive, then g will be a probability map

Thus W is in effect, a r.v that is also a probability because it obeys the probability axioms

Well the question is inspired because of wondering about something more general than the dynamics of quanutm mechanics. Since in quantum wavefunctions have a deterministic dynamics, thus the probability distribution will also be deterministic, it is not unreasonable to ask for generalisations where the wavefunction evolve stochastically, and such question on the existence of such pdf will be a good starting point for such investigations

Adeek

03:26

hey @0celo7 here ?

0celo7

yes

Adeek

want to discuss something in differential geometry

so why is X a complex manifold ?

so I read somewhere that if an action is free if all the stabilizers acting on an element x is trivial.

0celo7

to be discussed later dude

Bob Baxley

I just added an analysis (meta.math.stackexchange.com/questions/21259/…) of an existing answer on the distrubition of reputation scores. Anyone care to give it a look to see if my conclusions are correct? I think I spotted an error in the previous, highly voted, answer, but I could be wrong.

0celo7

when you quotient a manifold by a free and properly discontinuous group action, you get a manifold on the other end

I can never remember all the criteria for such things though

Adeek

03:31

I see

0celo7

you could check Lee

but hopefully that book will explain it

Adeek

yeah I will check lee

0celo7

@Adeek do you have do Carmo?

Adeek

yeah

0celo7

page 22/23 contains what you want.

Adeek

03:34

okay awesome

0celo7

he doesn't mention it there but you need the action to be free so that your quotient is Hausdorff, iirc

Adeek

Differential Geometry of Curves and Surfaces ?

0celo7

Riemannian

Adeek

okay

I see

0celo7

@Adeek Page 548 in Lee.

and the following

Adeek

03:37

rhanks a lot @0celo7

0celo7

np

Adeek

okay that is exactly what I want nice.

hm

why is that true what he wrote [1/2,1]

0celo7

what?

Adeek

@0celo7 we are quotienting by $\mathbb{Z}$ right ?

so we have $\mathbb{C}^n / \mathbb{Z}$

then [v] = [v + n] for any integer n.

why $[v] = [v + \frac{1}{2}]$

0celo7

what claim exactly?

Mike Miller

03:47

Atiyah just posted a six page paper claiming to prove S^6 has no complex structure.

8

Adeek

@0celo7 To see this we observe bla bla is surjective with $(v,1)$ and $(v,1/2)$ mapping to same point

@MikeMiller that is amazing

I wonder how such proofs work to prove it can't have complex structure

Semiclassical

here's the paper, for convenience: arxiv.org/abs/1610.09366

Adeek

amazing

Semiclassical

@MikeMiller I don't have the tools to evaluate/understand that paper, but I'm struck by how short it is

Ted Shifrin

@MikeM: Wow, Atiyah must be more than 90 now. Amazing ...

0celo7

03:55

My first thought was, "He's alive?"

Adeek

I was gonna say he is 90 and proved unsolved problem

amazing

Semiclassical

born in 1929

so 87. not far off

Mike Miller

I mean, it's six pages. So when I get home I can probably just read it.

Semiclassical

yeah.

I mean, it uses terminology that I'm not in a position to grasp, e.g. K-theory and KR-theory

Ted Shifrin

Most is history. One page of my download is blacked out. I'll try again. I definitely don't know all the tools.

Adeek

04:56

hey @0celo7

this should be rank n right ?

it should be n not m right

Mike Miller

Rank is dimension of the image. The dimension of the image should be m.

Adeek

oh ok

here where what is the values does j goes from and k ? it is $1 <= j < m$ and $1<= k <= n$?

?

yeah that is correct that is the values and $j$ and $k$ takes.

Josué

05:30

I'm showing $0\rightarrow\mathbb C^*\rightarrow\mathcal O^*\xrightarrow{\alpha}\Omega\rightarrow0$ is exact.

But I'm having trouble showing $\alpha$ is surjective.

The underlying manifold is $1$-dimensional and complex.

Balarka Sen

06:10

@MikeMiller Pretty cool. Wish I could read that.

NaCl

07:21

$g^{-1}(f(g(a),g(b)))=f(a,b)$?

NaCl

07:59

ok doesnt hold

Tobias Kildetoft

08:19

@AlexClark Cool, I will send you an email when I have the code ready. The current version can be sped up by something like a factor 5 I think and it would be a waste of time to run it before doing that.

Anubhav

@MikeMiller Sounds interesting :)

Secret

09:16

Took me ~1hr to digest this paragraph

I am so bad at tracking the behaviour of morphisims

Tobias Kildetoft

09:28

@Secret What is this from? It seems terrible to read

Secret

https://i.sstatic.net/DOBFe.png
You need to ask MAFIA, it is something he is reading about group theory

abenthy

09:58

I want to understand why a diffeomorphism $f \colon M \to M$ with $f^p = id_M$ for an odd prime $p$ is orientation-preserving, what definition of orientation(preserving) is best for this?

Tobias Kildetoft

@abenthy Hmm, the only other possibility is that it is orientation reversing, isn't it?

abenthy

yes

Tobias Kildetoft

so this is really just the statement that $(-1)^n = -1$ when $n$ is odd.

abenthy

10:15

so the composition of orientation-reversing diffeomorphisms is orientation-reversing again?

Tobias Kildetoft

no, the composition of an odd number of them is

abenthy

10:41

Oh yes, I think this makes sense because a diffeomorphism $f \colon M \to M$ is orientation-reversing iff for every $p \in M$, the map $d_pf \colon T_pM \to T_{f(p)}M$ maps a positively oriented base to a negatively oriented base.

Is the set of fixed points $F = \{ x \in M : f(x) = x \}$ orientable if $f$ preserves the orientation?

Balarka Sen

11:07

@abenthy As Tobias said, look at the degree of the diffeomorphism. If it's orientation reversing, degree would have been $-1$, so $(-1)^p = -1$. That's not the degree of the identity map.

abenthy

@BalarkaSen Yeah I thought about this, but are degrees defined for non-compact manifolds?

Balarka Sen

I gave the counterexample to the wrong question, sorry.

@abenthy Maybe you can think about compactly supported cohomology.

abenthy

Oh yeah that seems to work :) Nice

Balarka Sen

Interesting question about the fixed point set. I don't know how to answer that, but I am skeptic.

abenthy

$F = \phi^{-1}(\Delta)$, where $\phi \colon M \to M \times M, x \mapsto (x,f(x))$ and $\Delta$ the diagonal.

Balarka Sen

11:23

Yeah, but how are you going to incorporate the information $f^p = \text{id}$ to that?

Note that $p$ being odd is important. I can give you counterexamples for $p = 2$.

abenthy

Maybe $\phi$ is orientation-preserving because $f$ is, and so $F$ is orientable somehow?...

In a paper I read "If $p$ is odd, then the $\mathbb{Z}/p\mathbb{Z}$-action is orientation-preserving, so that the fixed-point set is orientable." without further explanation.

Balarka Sen

Ah, yeah, maybe my involution counterexample is not orientation preserving.

Your logic sounds right. $\phi$ is transverse to $\Delta$, is orientation-preserving, so preimage should be orientable.

user228700

@DHMO: I have a quick homework-tsy question about parabolas.

user228700

I've been asked to find the equation of a parabola whose vertex is at the origin and the equation of its direction is $x+y=2$. I figured it out, but the answer isn't correct :/ I found $a=\sqrt{2}$. Then, using the fact that the focus of this parabola is at a distance of $\sqrt{2}$ units from the vertex and $2sqrt\2$ units from the directrix, I found two equations to find the coordinates of the focus but I'm not obtaining real roots :/

user228700

If I find the coordinates of the focus, since I already know the equation of the directrix, I can find out the equation of the parabola...

abenthy

11:34

@BalarkaSen which theorem are you refering to here about the transversality?

user228700

@DHMO: Also, I've done all this only because from the equation of the directrix, it is clear that this parabola is not a standard parabola and will be tilted w.r.t the coordinate axes.

user228700

I've also been asked to find the length of the latus rectum of this parabola, and this is given by $4a$, so I get $4\sqrt2$, which matches the answer given in my textbook.

user228700

So Ik the value of $a$ is correct...

user228700

Thoughts?

Balarka Sen

@abenthy If $f : M \to N$ is transverse to $W \subset N$, everything is orientable and $f$ is orientation preserving, then $f^{-1}(W)$ is orientable. I believe this is because pullback of the normal bundle of $W$ is the normal bundle of $f^{-1}(W)$ in $M$, and pulling back the orientation on the normal bundle of $W$ gives an orientation on the normal bundle of $f^{-1}(W)$.

Since $M$ is orientable, that gives a bundle-orientation on the tangent bundle of $f^{-1}(W)$.

abenthy

11:41

Wow, that looks nice. Thank you.

Balarka Sen

Not clear that $\phi$ is transverse to $\Delta$ though.

abenthy

I think it might have something to do with the other assumptions in the article, but then its not good written I think :/

Balarka Sen

I'll again advice you to ask it somewhere else, or wait for other people to come :)

I'm a bit brain-dead to think about it clearly.

abenthy

11:58

Thanks, I have asked a question on SE: math.stackexchange.com/questions/1992912/…

Balarka Sen

12:25

@abenthy By the way, the earlier approach is void. There is a involution on $\Bbb{CP}^2$ with fixed point set $\Bbb{RP}^2$: just complex conjugate coordinatewise. This is also orientation preserving.

So that your diffeomorphism has odd order is more important than it being orientation preserving.

Danu

@abenthy Check out Guillemin & Pollack's book for how to induce orientations on the preimage.

Balarka Sen

That's not his question

Danu

54 mins ago, by Balarka Sen

@abenthy If $f : M \to N$ is transverse to $W \subset N$, everything is orientable and $f$ is orientation preserving, then $f^{-1}(W)$ is orientable. I believe this is because pullback of the normal bundle of $W$ is the normal bundle of $f^{-1}(W)$ in $M$, and pulling back the orientation on the normal bundle of $W$ gives an orientation on the normal bundle of $f^{-1}(W)$.

abenthy replied "wow that looks nice"

Guillemin & Pollack do it

Balarka Sen

doesn't sound relevant if that approach doesn't work, which it probably doesn't.

Danu

It's still interesting.

What is your beef with me giving a reference for something that "looks nice"?

Balarka Sen

12:35

I am not going to waste time arguing

Mahmoud

12:51

Hello.

teadawg1337

Good morning, @Mahmoud

Actually, it's probably afternoon where you are

Mahmoud

Guys what's the name of the mathematical field that formalizes functions.

@teadawg1337 Thanks $:)$ It's 12:52

Good whatever time of day you're in @teadawg1337 $:)$

Tobias Kildetoft

@Mahmoud Set theory I would guess

Mahmoud

$f: E \to F$ Things like this @TobiasKildetoft

Tobias Kildetoft

@Mahmoud yes, formally that is a certain type of set

Mahmoud

12:55

Can you give me the formal name of this study ?

Tobias Kildetoft

It is just set theory

(naive set theory)

Mahmoud

Okay thank you @TobiasKildetoft but what do we call it ?

Tobias Kildetoft

what do you mean? There is no name for that particular little bit of set theory itself

Mahmoud

If want to learn more about this what do I google ? For example.

Tobias Kildetoft

google the definition of "function"

Mahmoud

13:00

Alright It wasn't that fancy to have a name anyway.

Thank you @TobiasKildetoft It was mentioned in a set theory book as you said (Naive Set Theory) By Halmos, Paul, R

Aldon

13:29

Hi I have a simple question, for a normal definite integral can you always reverse the limits of integration?

Balarka Sen

Upto a sign, yes

Aldon

Even if one of the limits is infinity right? You just multiply a negative one to the integral to reverse the limits?

Balarka Sen

Yes, whenever the indefinite integral makes sense.

Aldon

Thanks! @BalarkaSen if I could give you rep here in chat I would've already done so.

Danu

Atiyah has submitted a proof that $S^6$ is not complex!! See here.

Balarka Sen

13:42

Mike said so earlier this morning.

Danu

Oh, damn

Interesting!!

Balarka Sen

It's sorta exciting, but I don't understand any of the ideas involved in the proof, so...

Jessy Cat

0

Q: Having trouble getting continuous $u(x,t)$ for $2u_{x}+u_{t}=0$

I am trying to solve the following initial boundary-value problem for the constant-coefficient, first-order wave equation: $2u_{x}+u_{t} = 0$, $x>0$, $t>0$ I.C.: $u(x,0)= \begin{cases} x-1, & 0<x<1 \\ 1, & x\geq 1 \end{cases}$ B.C.: $u(0,t)=0$, $t>0$ Since $a = 2$, $f(x) = \begin{...

If anyone could help me, that would be great. But, hints and rhetorical questions will not be a whole lot of help right now

Danu

@BalarkaSen I'm excited anyways :D :P

Aksel'sRose

When writing proofs where there is a set of defined characteristics to meet, how do you chose an f and g? For instance, additive property for a function defined as p(x) mapsto p(a)-p(0).

Tobias Kildetoft

14:40

@Aksel'sRose I have no idea what you are trying to ask there.

Aksel'sRose

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A: How to determine if a function is a linear functional.

Consider your example, i.e. $V = C([0,1])$, and the functional $f\mapsto \max\limits_{x\in[0,1]}\{f(x)\}$. To be precise, the functional which we may call $L$, is a function $L:V\to \mathbb{R}$ and is defined by $$Lf = \max\limits_{x\in[0,1]}\{f(x)\}.$$ Consider the functions $f,g\in V$ defined ...

see the comment on the first answer

Axoren

This room is really full right now.

Aksel'sRose

@TobiasKildetoft sorry realized I didnt reply to you. math.stackexchange.com/questions/1992318/…

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