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4:03 PM
multiply with sin(a/128)
 
Another hint: $\sin(2x)=2\sin(x)\cos(x)$
 
if i have $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $w:\mathbb{R}^3\rightarrow \mathbb{R}$ and for example if we have $f(x,y)=x^3+3xy^2$ $$ f(x,y)=x^3+3xy^2 $$ $$ x^3+3xy^2-f(x,y)=0 $$we can mark this as function of $w(x,y,f(x,y))$
$$ w(x,y,z)=x^3+3xy^2-f(x,y) $$ Is it true that:

$$ \nabla w(x,y,z)=\begin{bmatrix} \frac{\delta}{\delta x} \\ \frac{\delta}{\delta y} \\ \frac{\delta}{\delta z} \end{bmatrix} $$
in every situation $\nabla(x,y,z)$ is normal vector for tangent plane of $f(x,y)$ ??
 
4:30 PM
if someone wants to help
0
Q: Normal vector for tangent plane if function of $f:\mathbb{R}^2\rightarrow \mathbb{R} $

TukiProblem if i have $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $w:\mathbb{R}^3\rightarrow \mathbb{R}$ and for example if we have $f(x,y)=x^3+3xy^2$ $$ f(x,y)=x^3+3xy^2 $$ $$ x^3+3xy^2-f(x,y)=0 $$we can mark this as function of $w(x,y,f(x,y))$ $$ w(x,y,z)=x^3+3xy^2-f(x,y) $$ Is it true that: $$ \n...

 
4:47 PM
@leaky you're a logic man right?
 
@CookieToast hmm?
 
I was just curious if you would explain to me the fundamental differences between the logic courses offered in most philosophy departments and the courses offered in math departments, if you knew?
 
no idea
 
:P Fair enough
Thanks anyways
 
"Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives."
@CookieToast
 
4:52 PM
So what does that make the philosophy of mathematics?
2
 
Does MVT still hold if the function is continuous and differentiable on $[a,b] \cap E$ for some closed set $E$.
 
Can someone explain why a 4x4 rank 1 matrix with an eigenvalue of 0 multiplicity 3, has to be diagonalizable? I understand that rank 1 would mean that the dimension of the Eigenspace(0) is 3 but where does the other eigenvalue come from?
 
comes from the fact that the characteristic polynomial always has 4 solutions
 
Huy
depends on the field
 
@Huy come on
 
Huy
4:58 PM
:P
help me with my Laplace transform
I never did this kind of math before
 
@Huy just ask
 
2 hours ago, by Huy
hey, I'm looking at an exercise about solving a differential equation using the Laplace transform. I get the (correct) result, that
$$\mathcal{L}[y] = \frac{1}{(1+s^2)^2} e^{-\pi s} + \frac{1}{(1+s^2)^2}.$$
now, the standard solution suggests to "know" the value of
$$\mathcal{L}^{-1} \left[\frac{1}{(1+s^2)^2}\right]$$
which I don't. Instead, I rewrote my result as
$$\mathcal{L}[\sin(t)] \cdot \mathcal{L}[\sin(t-\pi)] + \mathcal{L}[\sin(t)]^2,$$
which is
$$\mathcal{L}[\sin(t)] (\mathcal{L}[\sin(t-\pi)] + \mathcal{L}[\sin(t)]).$$
 
Huy
must be something super trivial
 
> Since $\sin(t-\pi) = -\sin(t)$ and $\mathcal{L}$ is linear, the bracket is $0$, hence all of it must be. Where's the error in my reasoning?
not enough context
 
Huy
5:03 PM
what context is missing
 
well, of course, the differential equation
 
Huy
why
it is not required
this is just about Laplace transform and why my approach leads to 0, even though the $\mathcal{L}[y]$ expression is the same as the standard solution provides
 
I'm also not required to help you
 
Huy
you're not
 
@LeakyNun so a 4x4 matrix always has 4 solutions to the characteristic equation (even if these solutions can be the same)?
 
5:06 PM
@kourbou the characteristic equation is a quartic polynomial
it always has 4 (complex) solutions
 
Ohh... I forgot about the complex eigenvalues, thank you
 
@kourbou no, the remaining eigenvalue must be real
 
How come?
 
is your matrix real?
 
It doesn't specify, but probably
 
5:09 PM
then your characteristic polynomial would look like $x^3(x-\lambda)$
where $\lambda$ is a non-zero real number
because $0$ is an eigenvalue with multiplicity $3$, i.e. $(x-0)^3$ is a factor
do you know basic facts about polynomials?
 
Yes of course
But a polynomial of degree n doesn't necessarily have n real solutions
 
but it already has three 0's
that means $x^3$ is a factor
and the quotient must be linear
 
Oh okay that makes sense, thank you
 
In an experiment, n coins are tossed, with each one showing up heads with probability p independently of the others. Each of the coins which shows up heads is then tossed again. What is the probability of observing 5 heads in the second round of tosses, if we toss 15 coins in the first round and p = 0.4?
(Hint: First find the mass function of the number of heads observed in the second round.)
How do I start?
 
Huy
First find the mass function of the number of heads observed in the second round.
 
5:14 PM
It's awfully complicated.
I have a sigma inside a sigma.
I started by assuming that k coins show up heads in the first round.
Out of the k coins, 5 show up heads in the second round.
The product of those two gives the probability of having 5 heads in the second round if k heads had shown up in the first round.
I then sum it up from k = 5 to k = n.
It gets very complicated and after all that I got the wrong answer.
So I suspect that there is a simpler method which I am overlooking.
 
If $X$ is well-ordered set in the order topology, and $Y$ is a subspace of $X$, will the subspace topology on $Y$ be the order topology on $Y$?
 
Uh, I hope I'm not interrupting some important discussion, but can someone take a look at my question?
https://math.stackexchange.com/questions/2613617/a-recursive-divisor-function
 
5:35 PM
if you can evaluate a limit to two different values, does the limit exist?
 
@CausingUnderflowsEverywhere limit doesn't exist if you can evaluate limit to two different values when approaching different paths ?
if you think of $\epsilon \delta$ definition of limits this would mean you need to get arbitrary close to the limit in order it to exists. If not this would mean limit doesn't exists
 
@CausingUnderflowsEverywhere limit is not even well-defined in this case, hence it doesnt exist
 
In your case if you can evaluate two different values for limit this would mean you cannot get arbitrary close to the limit
 
thats why you need limsup and lim inf
 
@quallenjäger so i would be right about this ?
 
5:43 PM
More or less yes.
 
I got the following:

$f(x) = x$

$\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $
$= \lim_{h \to 0} \frac{x + h - x}{h}$
$= \lim_{h \to 0} \frac{h}{h}$
$= \lim_{h \to 0} 1$
$= 1$

While my textbook says:

$\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ) $
$= \lim_{h \to 0} \frac{k - k}{h}$
$= \lim_{h \to 0} 0$
$= 0$

because apparantly if $f(x) = k$ , then $f(x + h) = k$ for all h}

so how does the limit exist if you can get two answers?
oops I guess I didn't find the proper latex notation, sorry.
 
in the set A= {1/n st n in N} why is 0 the only limit point?
 
Why is $f(x+h)=k$?
Are you sure your function is right?
 
did I evaluate the limit wrong when I received 1 as an answer
 
No, just you are dealing with two completely different functions
Your textbooks function is constant
 
5:50 PM
$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ would be definition for derivative
 
and what you proved is a function with slope 1
 
the text book says (Since $f(x) = k$ and $f(x+h) = k$ for all $h$) as the comment next to their work
 
Then your function should be $f(x)=k$
and not $f(x)=x$
 
oh
I'm quite silly aren't I
Thank you Tuki and quallenjager
 
No problem
 
5:53 PM
if someone could take a look at this:
0
Q: Normal vector for tangent plane if we have function of $f:\mathbb{R}^2\rightarrow \mathbb{R} $

TukiProblem if i have $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $w:\mathbb{R}^3\rightarrow \mathbb{R}$ and for example if we have $f(x,y)=x^3+3xy^2$ $$ f(x,y)=x^3+3xy^2 $$ $$ x^3+3xy^2-f(x,y)=0 $$we can mark this as function of $w(x,y,f(x,y))$ $$ w(x,y,z)=x^3+3xy^2-f(x,y) $$ Is it true that: $$ \n...

 
@BalarkaSen So here's something that's probably well known but I found pretty interesting
Say you have a polygon, and to each face you attach a normal vector whose magnitude is the area of the face
Then the sum of those vectors should be 0
 
I am not sure about concave stuff
Wait, polygon or polyhedra?
 
(This came about when I was thinking about how people don't like to think about vectors as having "direction" and "magnitude", but here the direction and magnitude do represent different things)
 
(Correspondingly, face or edge?)
 
@BalarkaSen Er, polyhedron, sorry.
 
5:55 PM
is it 0 or a 0 vector ;)
 
Well, either, I guess. Though for polygons it's just $i$ times the sum of the edge vectors and so trivially zero. @BalarkaSen
 
@Akiva Normal outward vector?
 
@Cau Guess
@BalarkaSen Yeah
 
@CausingUnderflowsEverywhere is there any difference in zero vector and zero ?
 
Well, $0\in\Bbb R$, whereas $0\in\Bbb R^n$ (or even $0\in V$ more generally)
Not to be confused with the $0$ that's in rings $R$ or fields $K$ :P
or the $0$ matrix that's in $\Bbb R^{m\times n}$
 
5:57 PM
don't know I havent done the homework in vectors yet
 
@AkivaWeinberger What's a counterexample for polyhedrons? (Should I think about it?)
I am guessing it has to be very nonconvex
 
@BalarkaSen There is none, at least that's what I'm conjecturing
 
I see
 
I'm pretty sure if you prove it for right tetrahedra then you're good, to be honest
(Like, tetrahedra whose vertices are $0$, $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$)
And that led to my Pythagorean-like thing from earlier
 
Well. It just looks like you're integrating the area 2-form on a closed manifold, if you want to think smoothly.
Which is zero by Stokes' theorem
 
5:58 PM
10 hours ago, by Akiva Weinberger
So say you have a plane in 3-space, and a shape on that plane with area $A$
10 hours ago, by Akiva Weinberger
and say $A_{xy}$ is the area of its projection onto the $xy$-plane, and similarly for the other two planes
 
I think (area 2-form is closed, right?)
 
10 hours ago, by Akiva Weinberger
Do you have $A^2=A_{xy}^2+A_{yz}^2+A_{xz}^2$??
If so, then it's true for right tetrahedra, and thus all polyhedra by adding and subtracting those together
and I'm pretty sure it is true
 
My Stokes logic makes me think it is true
 
@BalarkaSen To be clear, adding 2-forms is isomorphic to adding the normal 1-forms?
 
I am sure you can cook up a PL version of it
 
6:01 PM
I guess that's why physicists can get away with representing rotations by vectors
and adding the vectors to add rotations
and justifying magic like gyroscopic precession and stuff
 
@Akiva Not sure what the normal 1-forms are. The 2-form is locally like $\omega = n_1 dx \wedge dy + n_2 dy \wedge dz + n_3 dz \wedge dx$ where $n = (n_1, n_2, n_3)$ is the outward pointing unit normal.
I got the order backwards. WHATEVER
 
Ah, I see. So then it's trivially the same as adding $n$s
 
Yup
 
and so your S'tokes thing should work
 
lmao
 
6:03 PM
(It's after the S innit)
 
I am going to star that just to anger Ted
 
You have incurred The Wrath of Ted
@BalarkaSen So, yeah, that Hatcher proof was pretty cool.
 
I like it
 
I guess I had a bit of a mental block 'cause, you know, "Fractals, therefore how can we know anything"
 
Hah yeah
TOP is weird
 
6:05 PM
Especially with the manifold that wasn't triangulatable
(with boundary)
E8 or whatever it was
 
Something like that
 
Hi, Purr Collection
 
is proving that the lower riemann sum is bounded sufficient enough to prove it's integrable on that interval?
hi blue ball
 
Test it on $1_{\Bbb Q}$
 
@MeowMix um
 
6:06 PM
(That's a horrible notation. Why is the important bit small?)
 
HEY he's a blue ball
 
don't be immature
 
I mean technically it's a plane
 
12 year old intensifies :3
 
6:07 PM
xD
 
still 12?
when's your birthday
 
HOROSPHERES BIKHES
 
I spent years of my life not knowing that "hairy ball theorem" had a double entendre
 
I'm already 32
and it's only been 5 months
 
I thought you were, like, 14 or something
(Meow, I mean)
(Zach)
 
6:08 PM
what
yeah im 14
however ive already invented my own physics with over 2000 symbols
2000 confirmed symbols
 
still need over 9000
 
That sounds like the start to a parody of the Navy SEAL copypasta
 
also what is $1_\Bbb{Q}$
exactly
 
blackberry bold
 
$1_{\Bbb Q}(x)=\begin{cases}1,&x\in\Bbb Q\\0,&x\notin\Bbb Q\end{cases}$
It's discontinuous everywhere, sends rationals to $1$ and irrationals to $0$
 
6:11 PM
oh
 
Similarly, one can define $1_A$ for any set. You'll also find notation like $I_A$ and $\chi_A$
 
the upper sum is always 1, lower sum is always 0, got it
 
Huy
how u doin @BalarkaSen
 
Yeah, and it's not Riemann integratable
 
@Huy
 
6:11 PM
so you need $\sup(L(f, P)) = \inf(U(f,P))$
 
We meet again
 
It is Lebesgue ("leBEG") integratable, though
 
Huy
of course
at least once a year
 
integratable?
 
How is life?
 
6:12 PM
@MeowMix Something's integratable if its integral exists?
How else would you say it
Able to be integrated
 
integrable
 
i thought i saw integrable somewhere
 
Huy
exhausting. too much work. but I decided to learn a bit about Fourier series, transform and Laplace transform, so I picked up some lecture notes from a course at my uni and started solving some exercises. pretty fun stuff.
 
I think Rudin used "integratable"
I'm wrong
It is "integrable"
@MeowMix Here's a fun thing. Say you have a right tetrahedron,
meaning its vertices are $(0,0,0)$, $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$
Like, they're the origin and points on the axes
 
Yeah i thought measure 0 discontinuity => integrable
 
6:15 PM
Given the areas of the faces on the $xy$-, $yz$-, and $xz$-planes, find the area of the slanted face
 
@BalarkaSen <=>
 
right
 
@BalarkaSen Not Riemann
unless it's bounded, maybe
 
I thought it was true in Riemann. For bounded, yeah
 
Otherwise take the function that's $0$ everywhere except for points of the form $1/n$ where it's $n$
($n\in\Bbb Z$)
 
6:16 PM
confusingly called lebesgue criterion
 
Defined on $[0,1]$
I think that was in Ted's book somewhere?
 
Not the other direction I do not think
=> probably
 
yeah riemann + bounded
is what you need for <=>
 
i read on a text that $cot(\pi /2 + i\pi y ) = tanh(\pi y)$ but im getting the inverse of $tanh$ , someoene can explain this equality so i will see my error?
 
Well, $\cos(\frac\pi2+x)=-\sin(x)$, and $\sin(\frac\pi2+x)=\cos(x)$
So $\cot(\frac\pi2+x)$ should be $-\tan(x)$
Thus $\cot(\frac\pi2+i\pi y)$ is $-\tan(i\pi y)$
Now, $\sin(ix)=i\sinh(x)$ and $\cos(ix)=\cosh(x)$, right?
So $\tan(ix)$ should be $i\tanh(x)$
And then $-\tan(i\pi y)$ should be $i\tanh(\pi y)$
So I'm getting $i$ times your answer
 
6:22 PM
not mine :P
 
And Wolfram seems to agree
@Liad ?
 
Oh. They're taking absolute values
 
so yea it is ok
 
$|i\tanh(\pi y)|=\lvert\tanh(\pi y)\rvert$
Yeah
'Cause it's $|i|\cdot\lvert\tanh(\pi y)\rvert$ and $|i|=1$
 
6:25 PM
when i write it by def. i get that the absolute value is less than $\dfrac{e \ ^ {y\pi}+e \ ^ {-y\pi}}{e \ ^ {y\pi}-e \ ^ {-y\pi}}$
 
And what is $\tanh$ when you write out the definition
 
1/my answer
 
@EricSilva Hi
I have been reading the book you have recommended, Frank Morgan
 
so is $0/0$ defined in math? I thought it was indeterminate :/ still a bit confused about how we could determine
$\lim_{h \to 0} \frac{k-k}{h}$ to be 0 if we couldn't factor out the h from the denominator
 
The proof of the Lipschitz Function is nearly almost everywhere $C^1$
1
Q: Mean-Value Theorem for set

quallenjägerSuppose $f:A\rightarrow \Bbb R $ is Lipschitz continuous and its derivative $Df$ is continuous on a closed subset $E$, $E\subset A$. Define $$\eta_\delta(a)=\text{sup}_{0<|x-a|<\delta \text{ and }x\in E}\frac{|f(x)-f(a)-Df(a)(x-a)|}{|x-a|}$$. I would like to proof the pointwise convergence for $\...

Any idea to this?
 
6:29 PM
@Cau When $h$ is not zero, $\frac{k-k}h$ equals $0$, right?
 
@CausingUnderflowsEverywhere It is a constant 0 sequence.
 
Now, when we look at $\lim_{h\to0}\frac{k-k}h$, we are not plugging $h=0$ into it.
 
Suppose that $X$ is a well-ordered set in the order topology, and that $A,B \subseteq X$ are nonempty sets satisfying $\overline{A} \cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$. Does it follow that $\overline{A} \cap \overline{B} =\emptyset$?
 
We look at the set of values you get when $h$ ranges over all nonzero values, and see if we can take the limit as $h$ approaches zero.
Since $\frac{k-k}h=0$ for all nonzero $h$, we have $\lim_{h\to0}\frac{k-k}h=\lim_{h\to0}0$.
And $\lim_{h\to0}0=0$.
@user193319 What is $\bar A$? Closure?
 
and h being 0 doesnt have to be defined for the limit to exist, got it thanks
 
6:31 PM
@AkivaWeinberger Yes.
 
@user193319 Let $X$ be $\{-1,-1/2,-1/3,-1/4,\dots\}\cup\{0\}$. This is well-ordered, and has order type $\omega+1$ (if you know what that means).
Let $A$ be $\{-1,-1/3,-1/5,\dots\}$, i.e. the negative inverse odd numbers, and let $B$ be $\{-1/2,-1/4,-1/6,\dots\}$, i.e. the negative inverse even numbers.
$\bar A=A\cup\{0\}$ and $\bar B=B\cup\{0\}$.
 
@user193319 it's wrong, the closure of $A$ $B$ can intersect.
 
$\bar A\cap B=A\cap\bar B=\emptyset$, but $\bar A\cap\bar B=\{0\}$.
(Another way to do this is to have $X$ be $\{0,1,2,3,4,\dots,\omega\}$ where you define $\omega$ to be greater than any other element. Then you can take $A$ and $B$ to be the even and odd numbers, and then $\bar A\cap\bar B$ would be $\{\omega\}$.)
 
Dang...All right. Thanks!
 
(It's essentially the same thing, with the elements relabeled.)
 
6:40 PM
@AkivaWeinberger BTW, I think your $A^2 = A_{xy}^2 + A_{yz}^2 + A_{xz}^2$ thing is a consequence of $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ where $\alpha, \beta, \gamma$ are the angles the normal to the plane $A$ belongs in makes with the $x, y, z$-axes.
 
sorry my internet shat itself
i forget how to do surface area calculations
 
Hm, I never heard of that before @BalarkaSen
 
oh hey direction cosines
 
So that works for any vector sticking out of the origin?
 
is it like $\int\int\sqrt{1 + \partial_x + \partial_y} dA$?
 
6:40 PM
(aka any vector)
 
That's a simple application of pythagoras
Yeah
 
Hm, I see it
Cool
Right, and those are the factors that determine how much the area $A$ changes when it gets projected onto the axis planes
Neat
 
Yeah.
 
@MeowMix The LaTeX is $\iint$ \iint, by the way
 
oh
 
6:48 PM
You want to square those as well, I think
 
square which
 
$(\partial_x f)^2$ etc
 
$\displaystyle\iint\sqrt{1+{f_x}^2+{f_y}^2}\operatorname d\!A$
 
oh right
what an ugly integral
does u-sub work ok on this?
 
depends on the integral
 
6:53 PM
@MeowMix Did you think about the right tetrahedron thing?
 
yes
im trying right ow
thats where that integral came from
 
Agh how do I make that small
Ah.
Right, so you have the areas of the right triangle faces and want the area of the slanty face
 
yes
so you can get the value of $a$, $b$ and $c$ rather easily no?
by just doing $\frac{ab}{ac}$
wait no
that gives $\frac{b}{c}$
 
Those being the lengths of the edges that go along the axes?
 
yes
 
6:58 PM
$\dfrac{(ab)(ac)(bc)}{(bc)^2}$ I guess if you want to go that way
 
right
but i mean
if you give me the side areas there has to be somethig nice with that
otherwise you would have given me just $a$, $b$, and $c$
 

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