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12:00 AM
sorry, I forgot to edit one thing, $\phi = x$ and $\theta = y$
 
 
2 hours later…
1:42 AM
@nbro in the first line, 4 variables map to 3 variables?
even if so, the first entry in your matrix is $-\sin\phi\cos\theta = \dfrac{\partial}{\partial \phi} \cos\phi\cos\theta$, why?
if so is the case, then the second entry is $\dfrac{\partial}{\partial \theta} \cos\phi\cos\theta$
then what is the third and the fourth entry?
 
 
1 hour later…
3:11 AM
so is $1 = \sqrt{1}$
?
 
@BAYMAX yes
 
then this must be true $e^{i2\pi} = e^{i\pi}$
?
@LeakyNun
 
@BAYMAX no it isn't
the property that fails for complex power is $(e^b)^c = e^{bc}$
$\sqrt 1 = \sqrt{e^{2i\pi}} = \left(e^{2i\pi}\right)^{0.5} {\Huge \color{red}\ne} e^{i\pi} = -1$
 
ok
1
Q: What's wrong in this proof of $i^2=1$

Umesh shankarComplex number $i=\sqrt{-1}$ Now i consider $$\frac{1}{i}=\frac{1}{\sqrt{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i$$ so $$i^2=1$$

I was thinking of the answer here
?
If 1 =$\sqrt{1}$ then ...
 
Square root has two branches
$-2$ and $2$ are both square roots of $4$
$i$ and $-i$ are both square roots of $-1$
and if you multiply the two statements above
$\{-2i, -2(-i), 2i, 2(-i)\} = \{2i, -2i\}$ are both square roots of $-4$
you would get back the two square roots by multiplying, but you never know which gives which
1 min ago, by Leaky Nun
$i$ and $-i$ are both square roots of $-1$
If you multiply this by itself:
$\{(i)(i), (i)(-i), (-i)(i), (-i)(-i)\} = \{-1,1\}$ are both square roots of $1$
 
3:19 AM
multiplying the square roots of 4 with those of -1 to get those of -4 is odd to me
 
but there is no guarantee that the principle value multiplied by the principle value will give you back the principle value
@arctictern isn't that how we prove $\sqrt{ab} = \sqrt a \sqrt b$?
@BAYMAX however, $\sqrt \cdot$ just takes the principle value
 
you just said $\sqrt{}$ takes the principal value, right after asking me if we use both positives and negative square roots to prove $\sqrt{ab}=\sqrt{a}\sqrt{b}$ (I presume you're talking about $a,b>0$)
 
oops, sorry for my inconsistency
how do you denote the square roots of $x$?
 
$\{u:u^2=x\}$
 
So, the theorem I intended to prove is $\{u:u^2=x\} \{u:u^2=y\} = \{u:u^2=xy\}$
@BAYMAX do you understand?
 
3:30 AM
any simple counter argument wecan give @LeakyNun
 
@BAYMAX to what?
@arctictern can we find a larger domain in which the statement holds?
 
in which $\sqrt{ab}=\sqrt{a}\sqrt{b}$? sure, as long as $a,b$ are reals and not both negative, for example. or you could specify $a,b$ both have real part $>0$. not sure off the top of my head what the full solution set in $\Bbb C^2$ is.
 
like $(a^{x})^{y} \neq a^{xy} ?$
In case of complex numbers
 
@BAYMAX counterexample: $((-1)^2)^{1/2}\ne (-1)^{2(1/2)}$.
(-1, 2, 1/2 are complex numbers!)
 
@arctictern he said complex number
for complex numbers $z^{1/2}$ is multivalued
 
3:35 AM
I use principal branch for a single-valued function, which is the default interpretation of $z^w:=\exp(w\ln z)$ for complex $z,w$
 
in which case the $\ne$ would mean the possible values are not the same
so your statement still holds
@arctictern alright.
 
ok
 
hi @Baymax, @Leaky, tern
 
hi
 
hello@TedShifrin
How is your oral test
 
3:47 AM
Huh? Surgery? That was a week ago. Everything seems to be fine. Thanks :)
 
nice
Buddhabrot
nice one!
 
Hi DogAteMy ... you're burning midnight oil these days.
 
4:04 AM
to determine the number of parrots in a sparse population
an ecologist captures 30 parrots
and puts ring around their necks and releases them
After a week he captures
40 parrots
and finds that 8 of them have rings on their necks
what is the approximate parrot population
I think he found 32 new parrots
so there can minimum of 32 + 40 parrots
72 parrots minimum
any help on this?
 
I don't know if the question is really clear.
 
like which part?
 
But you would expect the ratio of parrots with rings to parrots without rings that you captured
to be the same as the ratio of parrots with rings to parrots without rings in the total population
So the ratio is 5:1 and they're are 30 parrots with rings
in the total population.
 
so we are claiming that total population is 120 ?
 
approximately
I guess the ratio is 4:1
whoops
actually no
we would expect 120 parrots without rings
If the sampling was truly random.
 
4:14 AM
so 120 parrots without rings
and 30 rings
so 150 in total
 
yeah but I imagine any sampling in reality like that would have to be much bigger to hope for accuracy.
 
yes
there must be a large number of elements to be sampled
Let $f,g$ be two entire functions $\forall z \in \mathbb{C}$
$g(z) \neq 0 \forall z \in \mathbb{C}$
and if $|f(z)| \leq |g(z)|$
 
You haven't finished your question but the hint is to use Liouville.
 
:)
yeah
entire and bounded
just I want to make some observations
now $f(z) \neq 0 \forall z \in C$
I think this is correct!
now from Liouville's theorem $f$ is a constant function
 
No that's not right
f isn't bounded necessairly.
For instance let g(z)=e^z and f(z)=(1/10)*e^z
 
4:27 AM
ok that depends on $g(z)$
Now if it is bounded then $f$ is constant right1
Next an we say that $f(0) = 0$
?
 
Well what can you conclude if $g$ is not bounded?
In the example I gave $f$ is a constant multiple of $g$.
Is that always the case?
 
hmm
yes
Like we can always choose $c \in C$ st $|c| < 1$
such that $f(z) = c.g(z)$
and it satisfies $|f(z)| < |g(z)|$
 
we are given that $|f(z)| \leq |g(z)|$
then $|f(z)|/|g(z)| \leq 1$
and we can write this always as $|f(z)|/|g(z)| \leq c$ for some $c \in \mathbb{C}$
where $|c | \leq 1$
 
Yeah but you claimed f(z)=cg(z) for some |c| \leq 1
how do you get that?
Hi @ted
 
4:38 AM
hi @PVAL. @Baymax, you'd better use real numbers in inequalities, not complex numbers.
 
Hey guys!
 
oh no — it's Demonark
 
actually how about dealing with $||$ of complex numbers @TedShifrin
modulus of complex numbers?
 
I would write $|f(z)|\le c$ with $c>0$ real.
There is absolutely no point of having a complex number on the RHS.
Haven't we had this discussion more than a month ago?
 
oh
I am sorry
 
4:45 AM
What'd I do?
 
You're just your evil self, Demonark.
 
I thought I was a benevolent demon!
 
I thought of writing $|f(z)|/|g(z)| \leq |c|$ where $c \in C$ such that $|c| < 1$
 
Here is a good one for you, Demonark.
No, you're not benevolent.
@BAYMAX: Make things as simple and clear as possible. Hence my way.
 
Yes I will follow that!
 
4:47 AM
I gave him the hint. I would hate for you to post a complete solution.
It's not quite dominated convergence without something else.
 
I actually don't like your hint that much.
I'd rather use subadditivity of the measure.
and see it directly.
 
He already said he'd been told to use monotone convergence.
 
or in other words additivity of the integral on disjoint sets.
 
So you have to prove that $L^1$ means the tail integrals go to $0$.
I gave my hint because of his wording.
Offhand, I'm not seeing immediately how to prove the tail integrals go to $0$ without something similar.
 
Well $\Bbb R = [-1,1] U [-2,-1) U (1,2] U [-3,-2) U (2,3] ...$
 
4:53 AM
hi chat
 
what happened to \cup?
heya @Eric
oh, so you're going to convert to a series, @PVAL?
 
gotcha
 
and by additivity of the measure the series converges in particular the tail dies.
 
right ... I might have thought of that. As I said, I read what the OP said. I usually try to be responsive.
 
4:55 AM
That kind of technique shows up a lot in similar problems
 
So does the chopping :)
 
I'm not sure you can always use the dominated convergence theorem
 
And looking at $\min(f,n)$. :P
 
er monotone.
 
So wait the naive approach would be to just consider $f = \lim_{n\to\infty} f\chi_{[a-n,b+n]}$ and then do dominated convergence right then
 
4:57 AM
anyone watching Designated Survivor
Kiefer's a better pres than Trump.
 
The reason is that since $f\in L^1$, you can give some large interval outside of which $f$ becomes small, and then stitch some constant function and $|f|$
 
Hell, I'd be better. Ugh.
 
What is designated survivor
 
Demonark, he has $a\to -\infty$ and $b\to\infty$. So ... meh.
I see, you're trying to make a particular sequence? But who are $a$ and $b$?
 
Some starting point, I guess arbitrarily chosen?
 
5:00 AM
can't you apply dominated convergence twice
first with a sequence of indicators that becomes a ray in one direction and then another
 
Well, officially, for a double limit you can't sequence the limits without justification.
 
right right that's true
 
Well, does it work to just take your starting points and build the sequence to increment each by $1$, so that if you are taking some limit, you'll eventually "outpace" this sequence?
 
Anonymous
Is there any way to quickly derive the algebraic expressions for cyclic determinants like |{1,a,bc},{1,b,ca},{1,c,ab}|
 
Anonymous
(Other than expanding it wholly)
 
5:03 AM
@Daminark What happens if you look at say a/b
 
Demonark: The limits cannot be coordinated like you did.
You need two unrelated "rates" of going to infinity.
 
so we wrote $\mathbb{R}=....\cup(-2,-1]\cup(-1,0]\cup(0,1]....$
 
yeah ok so you really do want to show that $f \in L^{1}$ means it has small tails, useful idea anyway
 
Anonymous
For that determinant I can see that (a-b)(b-c) and (c-a) should be factors. But how can I say if that will have a positive sign or negative sign?
 
Anonymous
Ideas, anybody?
 
5:09 AM
You can test the sign with one set of numbers, @blue.
 
Anonymous
Ohkay, I can test with 1,2,3 or something like that then
 
Right. Do you know it's not something interesting times $(a-b)(b-c)(c-a)$? Just a constant? Why?
 
Anonymous
@TedShifrin Yep, it could be some constant times that too. I see
 
Why not a random polynomial times that?
 
Anonymous
@TedShifrin Can't be more than 3 degree
 
Anonymous
5:12 AM
It's a 3*3 determinant
 
What you wrote wasn't linear!!
 
Anonymous
Oh yeah, but even then I can't see how it can be more than 3 degree
 
no, you're right — it is degree 3.
OK, so you need to determine the constant. Plugging in one set of points does it
 
Anonymous
@TedShifrin Right. Thanks! (That was a good idea btw)
 
LOL, @blue, thanks. I've been around a while :P
 
Anonymous
5:15 AM
I know I know :P
 
Are you trying to say something, Demonark?
 
It was probably not a good idea
 
ugh im having trouble focusing on math :(
 
Imagine Soug's face lighting up when he sees you working in Ryerson on a Friday night as inspiration :P
 
Occasionally you need to take a break, @Eric.
 
5:22 AM
that makes it so much worse @Daminark
 
This happened to us at one point since we were working on his Buck pset, he sorta notices us, is all like "What are you doing here?" and with all this joy, explains something to us that I don't remember offhand. Now, I wasn't quite working with the others since I was dysfunctional after an all nighter and had to work on compsci.
 
yeah the trouble is I have a galois theory pset and a presentation for neves in a couple days and if I put it off tonight future me is gonna hate current me for it :/
 
Soug just kinda looks at my screen with all the code and is like "Ehhh"
 
Well, an hour break is good.
 
yeah I'll stop thinking about this stuff for a bit
clear my head
 
5:24 AM
Go for a brisk walk in the cool air.
 
it is quite cold outside :P ill play some music for a bit
 
plays the sound of Hagoromo chalk being used on the boards in the manifolds room (= your algebra classroom @Eric)
 
music - like Fade by Alan Walker
or Hall of fame by Script
oh all have their own taste!
 
@Daminark by music i meant id play like a musical instrument (Hagoromo is musical though)
 
Oh I thought you were going for "play $\simeq$ put on" thing
 
5:35 AM
bye
 
See you @Baymax!
 
see ya@Daminark
 
@Eric what do you play?
 
heya @Alessandro
 
Mandolin mainly @Alessandro
 
5:42 AM
Hi @Ted
I thought Italians were supposed to play the mandolin :P
 
it's a very popular instrument in Brazilian popular music actually
 
6:00 AM
Interesting, I didn't know
 
yup it's one of the central instruments in choro music
 
 
2 hours later…
8:11 AM
What does it mean for a chain map to be "natural"?
 
that... is a good question
any context for that?
 
yes, one minute
In my lecture there was a Lemma: Any two natural chain maps $C_*(X)\otimes C_*(Y)\to C_*(X)\otimes C_*(Y)$ that are identity in degree zero are chain homotopy equivalent
The proof was not done so I could not guess the definition of natural from the proof
 
So, grammatically I can make sense of that
erm... loading...
 
It might be that it is induced by a map $X\to Y$?
 
oh... wait
this might be something stupid about the axiomatic definition of homology
 
8:19 AM
context is Eilenberg Zilber lemma, the wiki page of that says "the maps are natural" and links to the definition of a natural transformation
 
perhaps it just means 'functorial'?
 
In what sense?
 
sth like: that if you have a continuous map $X\to X'$ then this induces a map of chain complexes $C_*(X')\otimes C_*(Y)\to C_*(X')\otimes C_*(Y)$
the wiki page says 'natural in X and Y' which is what makes me think something like this is happening.
 
0
Q: A question regarding formulating a PDE for a minimization problem.

Rajesh DachirajuQuestion : I'd like to formulate a pde for the following minimization problem. Let $\Omega$ be a convex, closed, compact set in $\mathbb{R}^d$ with a smooth boundary. Given a data $(x_i,d_i)$, $x_i \in \Omega^{\mathrm{o}} $ ,$d_i \in \mathbb{R}$, $i = 1,2,3...N$, $N>d$ and $\sum\limits_{i=1}^...

 
I dont get it, a map $X\to X'$ already induces a map $C_*(X)\to C_*(X')$?
 
8:30 AM
yeah so the wiki page sense doesn't transfer over quite, but I do feel like I'm getting a grip on it somehow
So the EZ theorem is a blackbox that says X and Y give you some F and G
If you map X continously to X', the F and G are supposed to change by a morphism as well, if this can be understood to mean anything.
 
I think I get what you mean
 
so it seems like 'natural chain maps' in the lemma is supposed to mean 'natural constructions for chain maps'. I'm not totally sure I can parse this either but the grammar at least makes more sense ;P
 
But this doesnt make sense in the first context I described above
In that Lemma it really is "you've got a chain map, it is natural, then for this map you've got"
so its like a property that a chain map can have, and not a property of a construction of chain maps has
 
Yeah, I see the concern
Hmm but actually I still think my interpretation could work, in that:
hmm, okay, it would be the same as your 'maps on spaces' guess.
um... wait... $C_*$ is a functor, right?
 
yes
googling some more I found the following
 
8:44 AM
So that's the end of the story right?
Maps on spaces is what you want.
 
'natural chain map C_*(A)\to C_*(B)' means 'a natural transformation $C_*\to C_*$' perhaps?
 
which functors?
a natural chain map between $C_*(X)\otimes C_*(Y)$ and $C_*(X)\otimes C_*(Y)$ does not give you a map between $C_*(Z)\otimes C_*(A)$ and itself
(Of course the book from which the picture comes doesnt appear to define $Q$...)
 
I think I'll ask the prof in the seminar today
 
8:50 AM
probably for the best
 
the definition appears to be in masseys book, but key parts are not in the google preview so I cannot see it from there
but it looks related to commutativity of some diagram like a natural transformation
 
How do I do this question:
If cube root a + cube root b + cube root c = 0 , then (a+b+c)^3 = ?
 
you cant do a lot with it
 
The options are:
A) 27abc
 
mainly because im sure $a+b+c$ can take any value
 
8:56 AM
B) 3abc
C) 9abc
D) abc
@s.harp Is it possible to solve this question?
 
if anything it has to be A) by looking at examples
 
@s.harp How did you come to this solution?
 
by looking at examples
you can probably show it by doing things like $(a+b+c)^3=(a+b+c -(\sqrt[3] a+\sqrt[3] b+\sqrt[3] c)^3 )^3$
 
I don't think it has to be done that way :/
 
It's basically just substitution...?
 
9:06 AM
Let us denote $x=\sqrt[3]a$, $y=\sqrt[3]b$, $z=\sqrt[3]c$.
 
It's indeed A) - just work out what $a+b+c$ is, and ignore the cube around it
 
You have $x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$ from the well known factorization of $x^3+y^3+z^3-3xyz$
Since $x+y+z=0$, you get $a+b+c=x^3+y^3+z^3=3xyz = 3\sqrt[3]{abc}$.
 
I did something similar, but I used $x^3 + y^3 + z^3 = x^3 + y^3 - (x+y)^3 = -3x^2y - 3 xy^2 = -3xy(x+y) = 3xyz$
 
Well-known at least to some people, Eric.
 
9:12 AM
Hi chat
 
momo asty
 
ohi
 
What's up ?
 
@Astyx le chandelier
 
Wow
 
9:17 AM
sorry if it's not funny
 
@LeakyNun I can't find anymore what I posted before, but there are 4 input variables, but the vector-valued function is defined only in terms of two variables, that's why you have many zeros
 
@nbro what are you referring to?
 
Celebrating Star Wars day by watching the original in ASCII
 
Oh yeah it's may the fourth isn't it
 
@LeakyNun regarding my last writings here, i.e. the jacobian of a vector-valued function
and stay calm
 
9:18 AM
Yup :D
 
first of all
 
@nbro alright
 
I'm disappointed that I haven't seen a single student wear any Star Wars-related clothing today
If I had to teach a class today, I'd definitely sneak a "I find your lack of faith disturbing" in there.
 
Haha :p
 
@LeakyNun I found it, if you're still interested in discussing about it: chat.stackexchange.com/transcript/message/37151371#37151371
 
9:20 AM
@nbro nah, I don't know much
 
Thanks, got the answer.
 
@Abcd hi
 
Hello :)
 
@LeakyNun ok
So, guys, anyone wants to (dis)confirm that the Jacobian of $f(x, y, w, z) = [cos(x)cos(y), cos(x)sin(y), sin(y)]$ is $\begin{bmatrix} -sin(x) * cos(y) & -cos(x) * sin(y) & 0 & 0 \\ -sin(x) * sin(y) & cos(x) * cos(y) & 0 & 0 \\ cos(x) & 0 & 0 & 0 \end{bmatrix}$?
 
I disconfirm
No $\cos x$ at the bottom
And a $\cos y$ at the bottom second collumn
Hi Balarka
 
9:25 AM
Shouldn't that be a $\cos (y)$ in the second?
Damnit, sniped
 
@Astyx Sorry, it should be cos(y), it's just a typo
 
Hi @Astyx
 
@nbro then it's also on the wrong collumn
 
wait
let me see if I also copied that wrongly, indeed, it looks to me that should be in the second column
@Astyx No, sorry, the third element of the function is actually sin(x), not sin(y)...
 
Then it's right
Unless I'm not
 
9:30 AM
But you are.
 
So it is
I think therefore I am right
Anyway I'll go now
Catch ya later
 
Bye
@Astyx If only exams were that easy
 
Nice to know that I was right at the beginning and that the only thing where I f***** up was in copying it from my program to this chat
lol
thanks!
 
@Astyx I'm gonna use and abuse this.
 
How do I do this question: The sum of n terms of a series is An^2 + Bn, then the nth term is: A) A(2n-1)B B) A(1-2n) + B C ) A(1-2n)- B D)A(2n-1)+B
A) A(2n-1)B
B) A(1-2n) + B
C ) A(1-2n)- B
D)A(2n-1)+B
 
9:48 AM
That problem is definitely not worded very well.
(it's D, though)
 
How?
 
Either use some reasoning, or just calculate the four sums?
 
I am not able to do this one using the formulae for AP.,
@SteamyRoot How?
 
You can use AP.
 
How??
 
10:01 AM
All you're doing is taking the sum of something that consists of constants and $(2n-1)$ or $(1-2n)$
 
10:13 AM
@SteamyRoot But we don't know the first term of the series
 
Take $n = 1$... ?
 
Assumption?
i can't understand you clearly, sorry :(
 
That's why I said the problem wasn't worded clearly.
What they're asking, is pretty much: Suppose that $\sum_{n=1}^N a_n = AN^2 + BN$. Which of the following can be $a_n$?
 
∑Nn=1an
Can you explain what this means?
the LHS of your equation.
 
Have my lunch break now, sorry.
 
10:17 AM
okay
 
ok
 
 
1 hour later…
11:32 AM
The bisectors of the angles of an acute angled triangle ABC meets BC, CA and AB at X, Y and Z respectively then:
A) BX.CY.AZ = XC.AY.BZ
B) BX.AY.AZ = XC.CY.ZB
C) BX.ZB.AZ = XC.YA.CY
D) NONE OF THESE
How to do this question?
 
11:50 AM
Okay, I finally think I can conclude that $\sqrt{1+\sqrt{2+\sqrt3}}$ cannot be reduced any further than $\frac12\sqrt{4+2\sqrt2+2\sqrt6}$
 

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