Mathematics - 2017-04-25 (page 2 of 4)

Daminark

07:00

Well

At least those are the most common exponents I've seen

Topologicalife

and 0, and 0.

Eric

I mean sometimes you do need $3$

Topologicalife

Good night guys, Have a nice day.

Daminark

See you @Topologicalife!

Eric

farewell

Daminark

07:01

What do you have in mind for 3 @Eric?

Balarka Sen

@blue No. Product rule says if you're product of two differentiable functions (at specific point say $a$) then you have $h' = fg' + f'g$. The functions are not differentiable at $a$, so you cannot apply it.

I think you need to go limits again.

Indeed, it's not too hard via the limit definition.

Anonymous

@BalarkaSen Yes, I agree. But in this case even if you apply the limit method of derivative you'll end up with basically the same thing. fg'+gf'. Of course it won't work when either of them is not derivable because f' and g' won't exist in that case!

Eric

@Daminark Gauss equations

Anonymous

You'll actually get g' and f' in the limit form. This is shorter detour to basically the same thing.

Daminark

Ah

Balarka Sen

07:05

@blue Yes. You can write the limit form for f' as f' because f is differentiable at a; but you cannot write the limit for for g' as g' because by hypothesis g is not differentiable at a. You're trying to contradict that.

This might seem pedantic but you should be careful with these things.

Eric

I seem to recall seeing a requirement that something was $C^{4}$ but I absolutely can't remember what it was

Daminark

Lol on the subject of equations, Neves said he'd be taking us to the diffeomorphism shop, meaning we'll do a bit of flow/vector fields

Anonymous

@BalarkaSen I should have mentioned that by $f'(a)$ I meant $\lim_{h \rightarrow 0} (f(a+h)-f(a))/h$

Anonymous

And not the value of $f'(a)$

Balarka Sen

Yep. But anyway, you're done.

Note that the hypothesis is that "$f$ is differentiable at everywhere except $x = a$ and $g$ is differentiable at everywhere except $x = b$ where $a \neq b$" and not "$f$ is not differentiable at $a$, $g$ is not differentiable at $b$, $a \neq b$"

Daminark

07:08

This, in order to prove that given a point on a manifold, you can find some ball around it for which any point in the ball has a diffeomorphism homotopic to the identity which swaps it with the center

For degree mod 2 stuff

Eric

fun fun

Anonymous

@BalarkaSen Yeah. I get it :)

Balarka Sen

@Daminark In fact for any two points on a manifold you can produce a diffeomorphism which takes one to the another. But that's a special case of what you said, of course; join the two points by a path, "thicken it up to a ball", and do your thing

The cool way to say it is that $\text{Diffeo}(M)$ acts transitively on $M$.

Daminark

Ah, that's a nice way of putting it!

And yeah we extended to that result afterwards

Balarka Sen

Modulo tubular neighborhood theorem.

Eric

07:11

@Balarka connected manifold

Daminark

Somehow thinking about things as actions makes life so much better

Balarka Sen

@Eric Thanks. I just keep forgetting adjectives.

Daminark

@Balarka we seem to have similar habits. This has caused a lot of confusion

Most often when things are compact

Balarka Sen

haha yes

Eric

oh @Daminark something even bigger than $C^{3}$, I think there's a result that every closed surface of regularity $C^{3 + \alpha}$ has at least two umbilic points. + some convexity requirements

so that's pretty high

Daminark

07:13

Actually yesterday I had one of my worst "I forgot to tell you a bunch of assumptions" moments yesterday

Oh snap @Eric

Balarka Sen

did you forget to tell literally the entirety of a problem statement?

thats happened to me before

Eric

lmao

Daminark

I think I might've done that before? Though I've often talked about problems with people who also had the problems so we would already decide what we'd work on

Though it has happened that I've switched problems without warning

Balarka Sen

"Hey man, did you know that it's not possible to have a nonvanishing vector field? [after 10 minutes of completely addled discussion] Oh, &%^# I meant on a 2-sphere"

Daminark

Oh that kind of thing

I do think I might have at one point been trying to explain Heine-Borel to someone

And I said it as "Every set is compact"

Balarka Sen

07:17

lol

Daminark

But yeah what happened yesterday was that I was doing this problem that I (wrongfully) thought would come up on our midterm

Which was that the typical continuous function mapped a first category set into a first category set

Turns out, in proving that it mapped to a measure zero set, 99% of the work was done, but I only figured that out as I was about to go to sleep that night

But anyway, I forgot to mention to people that we were assuming $A$ was closed and nowhere dense

Which led to confusion once I said "OK so the image of $A$ is $F_{\sigma}$ so being cat1 is the same as having empty interior

Jeez it's 2:30 I should actually sleep now

See you chat!

Anonymous

07:46

@BalarkaSen What is the reason for $f(x)^{g(x)}$ to be not defined for $x$ such that $f(x)<0$ ? At places where $g(x)$ is an integer should be defined isn't it? But desmos doesn't show those points :/

Anonymous

(I'm talking about real valued functions here)

Anonymous

Even wolfram alpha doesn't show the point $x=-1$ for $x^x$

Anonymous

@BalarkaSen I just realized that Desmos shows those points but they are practically invisible :P

Anonymous

One needs to hover their mouse over the exact point

Balarka Sen

08:19

@blue Desmos isn't showing them because they are not real.

Eg (-1)^(1/2). The cartesian plot cannot possibly show that point.

Anonymous

@BalarkaSen How is ${(-2)}^{-2}$ not real?

Balarka Sen

@blue It is but eg (-1/2)^(-1/2) isn't.

Anonymous

@BalarkaSen Yeah. I was talking about the integer points

Anonymous

Desmos doesn't show them

Balarka Sen

@blue Maybe it's not showing up because there are only isolated point which are real? Dunno

Anonymous

08:26

@BalarkaSen I think so. Even wolfram doesn't show

Anonymous

A bug perhaps

Anonymous

The limit exists at all integer points though

Balarka Sen

Limit of what?

Anonymous

Say limit x->-2 for x^x

Balarka Sen

Be very careful with saying stuff like that. That limit is not a real limit anymore; it's a complex limit.

Anonymous

08:28

Wolfram gives the limit correctly

Anonymous

@BalarkaSen In real analysis limit can be taken to be the value of the function at a point in case the left and right limits don't exist (not in real domain)

Anonymous

AFAIK

Balarka Sen

I have never in my life seen that convention.

But if that's so, a'right!

Anonymous

I can't find a source to backup my claim :P But we were taught so. For example if the left part of a function (at a point) doesn't exist then the rhl serves as the limit

SteamyRoot

Wait what?

Balarka Sen

08:37

@blue OK. It's reasonable now that I rethink this; this follows from the general definition of continuity of functions on arbitrary sets.

Eg f(x) = 1 if x = 0 and f(x) = 0 elsewhere is continuous restricted to the set {0, 2}. That's continuous on the discrete set so you'd want to say lim f(x) as x -> 0 is 1 on the domain {0, 2}

SteamyRoot

Well, if the function isn't defined for $x < x_0$, then the limit from the right is just the usual limit... kind of clear from the epsilon-delta definition of the limit

Balarka Sen

@Steamy He's saying the limit (from both sides) is the same as limit from the right, even though it doesn't make sense on the left.

This is a rather bad way to parse continuity of functions restricted to various domains.

SteamyRoot

Actually, I don't think "limit from the left" or "limit from the right" still makes sense in that case, unless you redefine it

Balarka Sen

Yeah, he's redefining it. None of this actually makes sense.

Anonymous

It's just a convention we follow in our calculus course...We've haven't started complex analysis yet (and neither the epsilon delta definition of limits). So I won't be able to comment on the formal definitions.

SteamyRoot

08:46

The usual definition to "limit from the left" would be the $L \in \mathbb{R}$ such that $$\forall \varepsilon > 0: \exists \delta > 0: \forall x \in \mathbb{R}: (0 < a - x < \delta \implies |f(x) - L|< \varepsilon)$$

But the left-hand side in that implication is never satisfied, so the implication is always true.

Unless I'm having a brainfart here, this would mean that any $L \in \mathbb{R}$ is the limit from the left....

Balarka Sen

I hate how high school calculus says limit without defining limit.

Anonymous

@BalarkaSen Me too. I wish I could go into the details. I just don't have enough time.

Balarka Sen

Ya understandably. It's just a garbage course.

Anonymous

I'll relearn all of this after a few months

Balarka Sen

In any case, @blue, it's interesting to try to extend the plot of x^x to the left as much as you can. Eg (-1/3)^(-1/3) is real.

(-p/q)^(-p/q) is real for all odd q.

Anonymous

08:54

@BalarkaSen That's interesting (-1/3)^(-1/3) is real?

SteamyRoot

Uhhh

Anonymous

Hmm, yes

Anonymous

it is

Anonymous

I didn't notice that earlier

Balarka Sen

@blue For an appropriate choice of branch, yes, it is.

SteamyRoot

08:58

Hmmm... those situations are a bit tricky, though.

Yeah, exactly. You'd have to take the branch of the $q$-th root that gives you a real value

Anonymous

$z=(-1/3)^{-1/3}$ Then $z^3=(-1/3)^{-1}$ Which is basically $-3$. So $z^3=-3$ has a real root.

SteamyRoot

However, I think that in the case of $x^x$, the expansion is something like $\sum_{i=0}^\infty \frac{(x \log(x) )^i}{i!}$

Balarka Sen

@blue It has a real root, but not all roots are real.

Anonymous

Of course. De moivre's

Anonymous

Gives one real root.

Balarka Sen

09:01

That real root is the "appropriate choice of branch".

Anonymous

@BalarkaSen I get it :)

SteamyRoot

So the principal branch of the logarithm won't give $(-p/q)^{-p/q}$ as real unless $p/q \in \mathbb{Z}$

I think...

meh

Anonymous

Ah, so basically $\text{"real number"}^{\frac{1}{\text{odd number}}$ will have a real root.

Balarka Sen

Will have a real value.

These multivalued functions are all very confuzzling.

Anonymous

@BalarkaSen ah, yes :P

SteamyRoot

09:08

There's no branch of $x^x$ that allows all of those $p/q$ powers to be simultaneously real... shame

Balarka Sen

That sounds right to me.

Anonymous

So if we are told to find the domain of $f(x)^{g(x)}$ the conditions should be $f(x)\geq0$ or/and $g(x) \in I$ or/and $g(x)=p/q$ where $q$ is an odd number. Am I missing something?

Anonymous

(f,g are real functions)

Anonymous

I'm trying to generalize it for f^g...

Alessandro Codenotti

Hi chat

Anonymous

09:14

Hi

Anonymous

@BalarkaSen Any ideas how to generalize the domain for $f^g$ ^^^^

Anonymous

I've tried to list down the points to check for

Balarka Sen

It's not that easy. To find the domain of $f^g$ you need an appropriate choice of branch for it.

I think only negative integers are the real values of $x^x$ for the standard branch.

Anonymous

@BalarkaSen Wolfram alpha takes into account only points for which f(x)>=0

Anonymous

I'm confused

Balarka Sen

09:18

I don't care about what W|A does, it mostly says garbage.

SteamyRoot

For example, you can ask WA to plot $x^x$ for negative integers

Balarka Sen

I am saying the question "Find the domain of $f^g$" is ill defined.

SteamyRoot

What it'll do then, is use the series expansion of the function, which has logarithms, and take the principal branch for the logarithm

Anonymous

@BalarkaSen I'm talking w.r.t real analysis

Anonymous

I think there should be a convention

Balarka Sen

09:19

@blue It is ill defined everywhere, in real or complex.

SteamyRoot

Resulting in a negative part that is mostly complex, but real on the negative integers

Balarka Sen

$f^g$ is not a function, it is a multivalued function.

You need a branch to ask the question.

Anonymous

Good question! One might even want to ask whether $4$ is in the domain of $(-1)^x$. The judgement would probably be that if we are talking about functions of a real variable, it isn't, even though of course $(-1)^4=1$. Not the only inconsistency in mathematical terminology. — André Nicolas Dec 12 '15 at 5:38

Anonymous

This sums it up ^ :P

Imago

Are there functions f, g such that f o g is strictly increasing, g o f strictly decreasing? By my intuition I would say no, I lack a rigorous proof however.

Balarka Sen

09:28

@blue It's all a complicated business. if you choose the standard branch of log, $f^g = \exp(g \log f)$ which is real iff the imaginary part of $\log f$ is $n\pi$ I think

Anonymous

10:03

Anonymous

In this problem we get two values for a

Anonymous

i.e. 0 and 2

Anonymous

But 2 is discarded

Anonymous

Because the base must be positive :/

Anonymous

I'm wondering why base can't be negative

Anonymous

10:17

Hmm, I think it is because the power isn't perfectly an integer as we are taking a limit

Anonymous

@BalarkaSen Yes, I checked out my calculus book. That's it. In real analysis by convention $u^b$ is $\exp(b \log u)$ whenever $b$ is not an integer. So only those points are taken in domain for which $u>0$.

Balarka Sen

@blue I see. Well, that settles it. However note exp(b log u) can be real even if u < 0

barto

If f is entire of finite order, is there a name for the degree $h$ of the polynomial $P$ in Hadamard's factorization $f(z)=z^me^{P(z)}\prod E_p(z/a_n)$? There's its genus $=\max(h,p)$ (and $p$ is called rank) but there does not seem to be a name for $\deg P$.
Maybe "Hadamard degree"? :p

Anonymous

@BalarkaSen Uh, what? How will log be defined then?

Balarka Sen

As a complex function.

Anonymous

10:24

@BalarkaSen Ah, right

Anonymous

BTW the problem I pasted above is based on exactly the same concept. We need to take the base>0 or we'll get an incorrect result.

Balarka Sen

a fair convention

Anonymous

I thought only physics has so many conventions. Now its maths :P

Balarka Sen

high school math

Anonymous

Agreed ^

Anonymous

10:27

:P

Anonymous

You are in high school aren't you? :D How do you get time to learn all this extra stuff like topology and all? @BalarkaSen

Balarka Sen

I just like what I do, and keep doing it

Anonymous

@BalarkaSen Good to know. Personally, I don't get sufficient time to complete even what is in my syllabus. Did you appear for the RMO/INPHO/IMO ?

Balarka Sen

I don't actually spent as much time on school as I should. No, I didn't.

user193319

10:42

I am trying to prove the following, though it may not be stated precisely: If f(x) => 0 for all x in its domain and L = lim_{x \to a} f(x) exists, where a is in f's domain, then L => 0. Here is my proof: By way of contradiction, suppose that L <= 0. Then -L => 0, and there exists a d (taking this to symbolize delta) such that |x-a| < d implies |f(x) - L| < - L. But this implies |f(x) - L + L| <= |f(x) - L| + L < 0 (triangle inequality), which means |f(x)| < 0 for every x such that |x-a| < d.

Contradiction....Would this work as a proof?

user340082710

Suppose that I have a binary operator $\oplus$. Normally, one uses the notation $x \oplus y$ to denote the result of apply $\oplus$ to $x$ and $y$. Suppose that I had a set of elements $S$ and a fixed element $y$, is there standard notation to denote the set $\{ x \oplus y : x \in S\}$ (i.e the set obtained by applying the operator to each element in $S$ with $y$)?

barto

11:08

@user340082710 $S\oplus y$ is standard. Or, if you like, $S\oplus\{y\}$

user193319

@barto Hello. Would you mind critiquing the short proof I gave above?

barto

@user193319 The triangle inequality gives |f(x) - L + L| <= |f(x) - L| + |L|, not |f(x) - L + L| <= |f(x) - L| + L

user193319

Whoops...Is there a way of fixing the proof?

Or at least know of a link to this problem on stackexchange? I am sure it has been asked before; I just couldn't find it through a google search.

barto

|f(x) - L| < - L implies f(x) - L < - L, contradiction

no triangle inequality needed

Essentially what you do is |f(x) - L| < e implies L - e < f(x) < L +e, and apply this for small e>0 (but e=-L suffices)

user193319

I just realized this: what if L = 0? Do we need to modify the statement slightly?

...Wait...If we have L=0, then we are done. So we can just ignore this case (duh!)

@barto I see now. Thank you very much!

Felix.C

11:55

I have a continuous function $f:M \subset \mathbb{R}^2 \mapsto \mathbb{R}$ with M connected. So $f(M)$ is obviously an interval. I also know from assumption that $f(M\{0}) = I_1 \cup I_2 \cup I_3 \cup I_4$ with $I_j$ intervals in $\mathbb{R}$ and $I_1 \cap I_2 \cap I_3 \cap I_4 = \emptyset$ this is obviously not possible. Since there at least three points missing and I removed only one but how could I write this down in a mathematic way...some stylistic tips highly appreciated :D

Ahh messed up with Latex, hope you understand it anyway

$f(M \setminus {0}) = I_1 \cup I_2 \cup I_3 \cup I_4$

Anonymous

12:11

Suppose f(x) is defined such that it is x+1 for all rational x and x+2 for all irrational x then is the limit x->0 defined/exists?

user193319

@blue No, I don't believe so. Consider the sequence sqrt{2}/n, which approaches 0 but every term is irrational.

@blue Oh, wait...You aren't asking about continuity. I don't think what I said applies.

Anonymous

@user193319 yes, continuity not needed

user193319

@Blue On second thought, that sequence will be helpful. If the limit exists, call it L, then if a_n is a sequence approaching 0, f(a_n) approaches L. This implies L =2. Now, can you find a rational sequence that approaches 0 and show that L is something different than 2, which would be a contradiction?

@Blue Don't overthink, if you are. It's probably a sequence you've seen a million times, and it isn't too different from the irrational sequence that I gave you.

Erik M

13:52

Can someone help me understand the answer to this question? math.stackexchange.com/questions/1771525/…

I don't see where $D$ comes from, or why $C_n$ can be written in the form stated in the answer.

Mary Star

14:36

Hello!!

Let f be an endomorphismus of a K-vector space V and v,w ∈ V ae eigenvectors of f for different eigenvalues. Let a, b ∈ K different than 0.
I want to show that av + bw is not an eigenvector of f.

I have done the following:

Since f is an endomorphismus of a K-vector space V, we ave that $f(av + bw)=af(v)+bf(w)$.
Let $\lambda, \mu$ are eigenvalues such that $f(v)=\lambda v$ and $f(w)=\mu w$.
So, we get $f(av + bw)=\lambda av+\mu bw$.
Since av+bw is not mapped to a multiple of itself it is not an eigenvector.

Astyx

14:50

@MaryStar How do you know $\lambda av+\mu bw$ is not a multiple of $av+bw$ ?

Mary Star

I thought so since $\lambda $ and $\mu$ are different eigenvalues. So, can it be that $\lambda av+\mu bw$ is a multiple of $av+bw$ ? @Astyx

Astyx

No, but you need to argue that

Mary Star

So, we have to add just that $\lambda av+\mu bw$ is not a multiple of $av+bw$ because $\lambda \neq \mu$ ? Or also something else? @Astyx

Astyx

You have to argue that $a, b \ne 0$ and $v$ and $w$ are linearily independant

Mary Star

How do we know that $v$ and $w$ are linearily independent? And how does it imply from there that $\lambda av+\mu bw$ is not a multiple of $av+bw$ ? @Astyx

Astyx

14:56

Unicity of decomposition over a linearly independant family

And try and figure out the first part by yourself

Simply Beautiful Art

^ I wonder how large of a countable ordinal this makes

Mary Star

@Astyx Ok!! Thank you!! :-)

Astyx

Glad to help

Balarka Sen

Morning @MikeMiller

Mike Miller

15:18

morning

Sha Vuklia

15:34

Hey @Steamy het klopt toch dat het overbodig is om $X^{-1}=Y$ te bepalen? We hadden toch meteen kunnen zeggen dat $X=(y_{ij})$, zonder met die inverses te hoeven werken?

SteamyRoot

Hmmm... eens kijken

Wat bedoel je met $X = (y_{ij})$ ?

En, wel, voor zover ik zie, is de gevraagde basis $\gamma$ gedefinieerd met behulp van de $y_{ij}$, dus om die basis te definiëren heb je de inverse nodig?

Sha Vuklia

ja dus ipv $Y=(y_{ij})$, vergeten we die $Y$ helemaal, en zetten we $X=(y_{ij})$ (beetje verwarrend, maar op deze manier hoef je niks aan te passen in de rest vh bewijs)

je kunt ook stellen: $X=(x_{ij})$

en dan in die kolom $y$ vervangen door $x$

hm, maar misschien gaat er dan toch iets mis

Dus we krijgen:
$$
co_\beta(w_j)=\begin{bmatrix}x_{1j}\\ \vdots\\x_{nj}\end{bmatrix}=Xco_\gamma(w_j),
$$
maar dan hebben we dat $X=id_\gamma^\beta$

SteamyRoot

Yup...

En het moet de basisverandering in de andere richting zijn

Sha Vuklia

oh, right. met die methode ga je altijd van $\gamma$ naar $\beta$, dus ze kiezen de inverse

SteamyRoot

Precies :)

Sha Vuklia

15:46

oke, thx!

Daminark

Hey everyone!

Sha Vuklia

hii @Daminark!

Daminark

How's it going?

Sha Vuklia

pretty good, kinda like always, you?

lol, and who upvoted Balarka Sen's "Heya" XD this chat really is random

Balarka Sen

16:02

i don't hope to get the starboard

but hey, i am on it

Daminark

So it seems

And lol today was fun, we started transversality

Balarka Sen

kewl

Daminark

And proved that if $F(X)$ intersects $Z$ transversely, then $F^{-1}(Z)$ is a submanifold of $X$

Balarka Sen

ofc, this is a generalization of the preimage theorem when Z = pt

Daminark

Yeah

Balarka Sen

16:10

what else? :)

Daminark

That was it, we started today by finishing the proof of the thing I told you yesterday, going to the diffeomorphism shop

Balarka Sen

Ah ok

Sha Vuklia

16:39

Guys, my book writes that a number $a$ is called algebraic if $a$ is a root of a polynomial $p\in\mathbb Z[X]$. However, as far as I know, algebraic numbers have to do with rational polynomials - so why don't they write $p\in\mathbb Q[X]$?

SoumyoB

@ShaVuklia why would you write something in your book that you do not know of? :O

Sha Vuklia

i don't understand your remark:d @SoumyoB

SoumyoB

unless some devil possessed me I wouldn't be able to write something that I do not know of in a book that I'm writing

SteamyRoot

Multiplying a polynomial with a nonzero constant doesn't change the set of roots

Alessandro Codenotti

@ShaVuklia it's the same, multiply by the lcm of the denominators of the coefficents to go from a polynomial in $\Bbb Q[x]$ to one with the same roots in $\Bbb Z[x]$

SteamyRoot

16:43

And a polynomial only has finitely many coefficients

So just multiply a rational polynomial by the lcm of the denominators

Sha Vuklia

oh, of course! thank you @Alessandro & @Steamy

Alessandro Codenotti

(Note that you can talk about stuff which is algebraic over rings/fields different from $\Bbb Q$, but that's not the case here)

Topologicalife

Hi guys.

SoumyoB

hi there

user123733

1

Q: Probability of original signal

Signal which can be green or red with probability 4/5 and 1/5 respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly 3/4. If the Signal received at station B is green, then the probability that the ongmal signal was...

BAYMAX

16:57

actually when does the solution of an ODE by Euler method converges ?

$\frac{dy}{dx} = f(x,y) , f(x_{0} = y_{0})$

$\frac{dy}{dx} = f(x,y) , y(x_{0})=y_{0}$

SoumyoB

@user123733 on it :)

user123733

@SoumyoB Thanks

Problem Solving Strategies

General chat for high school physics. For MathJax see meta.sta...

2

@SoumyoB

SoumyoB

I have an exam tomorrow, sorry won't be able to join that right now

but I'm answering your question

user123733

no problem

Little Rookie

17:18

What does the sup $x\in E$ notation means in the question?

Farhad Rouhbakhsh

x,y,z are real numbers and x^2+2y^2+5z^2=44.

what is the max of xy+yz+xz?

Sha Vuklia

@LittleRookie it means you take the supremum of the set $\{\vert f_n(x)-f(x)\vert: x\in E\}$.

Farhad Rouhbakhsh

any solution?

Little Rookie

yes i think so too

Farhad Rouhbakhsh

x,y,z are real numbers and x^2+2y^2+5z^2=44.
what is the max of xy+yz+xz?

Gunelle

17:38

Hey guys!

user123733

yo

Gunelle

I have a graph theory related question...

Is a graph containing vertices of degree 1 and 2 still considered to be a subcubic graph?

user123733

Sorry , I am just an high school student

Farhad Rouhbakhsh

x,y,z are real numbers and x^2+2y^2+5z^2=44.
what is the max of xy+yz+xz?

no one able to solve this f..... problem?

Gunelle

It isn't an easy one.

SoumyoB

17:54

@FarhadRouhbakhsh I remember doing these kinds of problems back in high school in preparation for JEE but those skills are now rusty as I never needed them here in college lol

I think there is a trick to do it without destroying the symmetry between x, y and z but I just can't remember it off hand

Daminark

How's it going?

Balarka Sen

@Farhad Do you know Lagrange maximization?

Ted Shifrin

Yeah, I was about to ask the same thing. But we can get it out of Cauchy-Schwarz, too.

Daminark

Oh hey @Ted!

Ted Shifrin

Aren't you supposed to be taking an exam, Demonark? Oh, no, that's later.

Daminark

17:58

That's Thursday :P

SoumyoB

hi @TedShifrin

Ted Shifrin

hi @SoumyoB.

Eric

hi chat

SoumyoB

were you talking about applying CS inequality to farhad's problem?

Ted Shifrin

I mumbled that, yes.

Problem Solving Strategies

Transcript for

$Mathematics$ Mathematics