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12:29 AM
I am likely doing something wrong. Not sure where though. Writing down the law of cosines for the sides (in original notation) $a$ and $b$, we get two equations. Subtracting them and simplifying yields $a^2-b^2=c(a\sin(x)-b\sin(y))$. But the law of sines gives $b\sin(y)=a\sin(x)$ so im getting $a^2=b^2$.
12:39 AM
How did you get the sin in the law of cosines?
oh obviously, i forgot to go back to angles $x$ and $y$ after messing with $\pi/2-x$ and $\pi/2-y$ in the formulas for $E_A$ and $E_B$
my brain is dead :)
Oh, $x$ and $y$ are not given fixed quantities. I was confused about the problem
Oh don't say that, everyone makes small mistakes xd
but $c$ is a known fixed quantity?
12:45 AM
yeah $c$ is known as well as the quantities $E_A=\frac{I\sin(x)}{a^2}$ and $E_B=\frac{I\sin(y)}{b^2}$ but $I$ is not known.
12:56 AM
imgur.com/a/XTcPZEj Tricky problem
Is there even a tool to determine if this has solutions like in linear systems the rank of a matrix?
may i recommend a mechanical pencil
0.3 mm
i'd be tempted to try it out with various parameters plugged in as numbers and see if a computer can take it the rest of the way
1:12 AM
The highest degree should be 4, so we have now sin(x) and sin(x+y) but my god that has to be ugly
it seems like it should be easier than im finding it. i am trying to write down algebraic equations that the unknowns must satisfy and then maybe we would try a dimension argument/heuristic. mathematica probably could do this but im a little stubborn ha. Unfortunately, im getting degree 4 stuff (which is not so nice to work with).
What is a good way to numerically test if a distribution is a good fit for a data sample?
@Quin Should I make a post for this problem? Maybe someone will find a nicer solution. On the one hand now we now that at least this is doable, but on the other 4th degree roots are very nasty.
sure! (idk what should/shouldnt be asked) but i dont see the harm in it. just make sure to clearly explain the prob. and show what you've done
Will my notes suffice? Writing this into latex will take forever
1:23 AM
actually, hold on a sec, i think ive been overlooking something. also, typically you would want to write into latex (annoying, yes, but many people wont read it otherwise)
I think that we should be able to set $I=1$ or something and just assume that it is constant. Then we can write $a^4E_A^2=\sin^2(x)=1-\cos^2(x)$ just right off the bat. Using the law of cosines twice (as before) adding and subtracting the expressions obtained for $a^2$ and $b^2$ I think we should end up getting $\frac{\cos(x)}{a}=\frac{\cos(y)}{b}=\frac{1}{c}$. Using this, we can write $a^4E_A^2=1-\frac{a^2}{c^2}$ and use the quadratic formula on $a^2$ to determine $a$.
We should then be able to repeat for $b$.
Hmmm, I'm thinking about the validity of setting I to a known constant.
its a light source, right? i think its reasonable to assume that it doesnt change its luminosity as you move it around. i could def be wrong with this so worth the think. i would also check the above stuff i said as ive been known to make a mistake or two (or three)
either way, if we dont set it to a constant, the above should produce expressions for $a$ and $b$ in terms of $I$
Where do you see I in your $$a^4(E_A)^2$$ expression?
if you included $I$, it would become $\frac{a^4E_A^2}{I^2}$
But that's the problem, I isn't known :((
1:38 AM
yeah, im not sure then. unfortunately i have to run but what I would try next is finding $a$ and $b$ in this way and seeing if there was some way to maybe take a quotient or some other expression to try to eliminate $I$. (also my brain is still fried!). Best of luck!
Thank you for your time!
I'm trying to formulate the triangle problem in vectors to see if that helps at all. Prob not :)
cya quin
1:56 AM
there really needs to be some sort of white board thing here.
2:12 AM
Here's the triangle problem post I made for anyone curious math.stackexchange.com/questions/4130116/…
2:30 AM
@robjohn bah! I was so close, sorry for the slow follow up. Before I cheated I had noticed a binomial like expression pop up in my attemp. Sadly I didnt follow through.
@user379685 i dont have a solution but that looks like fun!
3:10 AM
@copper.hat You installing for us?
@TedShifrin installing?
The white boards?
Finally found it: the beard emoji
3:40 AM
Happy almost birthday, @robjohn.
Happy birthday!
3:59 AM
happy birthday or amost birthday!
1 hour later…
5:23 AM
i know this must have a good answer, if I have a bunch of relations $c_1 - c_2 - c_4 = 0$, $-c_1 + c_3 - c_5 = 0$, etc, where the $c_i$ are integers, and I want to recover a generating set of relations , how can I do so?
uh or more precisely speaking, these $c_i$ are coefficients of generators, $g_1,...,g_6 \in \mathbb{Z}^{r}$ , and I know that I need these relations on the coefficients to hold if I want $c_1g_1 + ... + c_6g_6$ to be sent to $0$ in $\mathbb{Z}^m$ by some homomorphism $\phi : \mathbb{Z}^r \rightarrow \mathbb{Z}^m$
so is there an easy way I can figure out what the kernel of this homomorphism is?
5:56 AM
@TedShifrin thanks!
hi guys!
how are your maths endeavors going?
porridge i remember seeing something like this in a topology class where the instructor called it 'mere linear algebra,' which it sorta wasn't but kinda was. there are normal forms you can do that start with integer matrices and produce a simpler matrix output. i forget their names. we did it in computing homology.
or maybe it was cohomology.
6:16 AM
yeah thats the same context im thinking it about it in, the problem is in my class we just sort of find a basis for the kernel by what appears to be guesswork, and i am sure there is a way to do it systematically via things like SNF but i dont have the background knowledge to know how to do it that way
well not guess work, but trial and error
the issue i have is i dont know what rank the kernel is supposed to be, so its not clear how one knows when to tp
@leslietownes Smith normal form
Do row & column ops over $\Bbb Z$.
but column ops change the generating set so i would have to account for a new basis
i.e. do linear algebra but forget everything that makes it cool.
6:34 AM
Just do row ops to do what you asked. What I said allows you to read off the homology.
7:17 AM
@porridgemathematics How is going in your neighbourhood?
Which cruel evil has chopped me off from the chat
2 hours later…
9:31 AM
Can someone give an en example of a function $f \in L^{2}(\mathbb{R}) \backslash L^{1}(\mathbb{R})$ such that $\hat{f} \in L^{1}(\mathbb{R})$. with explaining , cuz I am not able construct one
10:27 AM
Is the idea to take a function in L1 and L2 and then apply fourier inversion formula on L1 and hope that inverse is not in L1?
10:46 AM
Getting very agitated
math.stackexchange.com/questions/4130324 I am asked to explain how to apply DCT. I include all the details and yet OP wants more
@ParclyTaxel Hi!
hi, I may need some help
you may not know me, but I remember you bcs you answered my first ever stack exchange question 4 years ago
There's another chat discussion in the comments on my answer which you can go into
Which question s29?
i think its deleted now. Nothing significamt, I just remember your name
10:56 AM
4 hours later…
2:54 PM
A point R with the x-coordinate 6 lies on line segment joining the point P(2,-3,4) and Q(8,0,10). Then distance of R from origin is...
What's this ? Is it the Answer?
I mean, one of those is the answer, plus or minus 47.
Have you thought about how to solve the problem?
What tools do you have at your disposal?
I thought I would need equation of line, but not getting how to find , then I thought may be slopes are same, but it's not 2D
Personally, I kind of like the idea of parameterizing the line.
Let $v$ be the vector from $P$ to $Q$. Then the line is given by $\ell(t) = tv + P$.
You then have $R = tv + P$ for some $t$, where $R$, $v$, and $P$ are all known quantities. Solve for $t$.
3:01 PM
@Rover I think use this form $(x-x_1/(x_2-x_1))=\lambda$.Find lambda then repeat same procedure for y,z and then use distance formula..
(Well, $R$ isn't known, but the $x$-coordinate of $R$ is known, which is good enough).
@XanderHenderson Okay, but I am not yet comfortable with lines properly, ok I will try out .
Well, this is an exercise which is meant to get you comfortable with lines.
And if you are not used to parameterized lines, then perhaps you just want to work with an equation for the line in the form $ax + by + cz = d$.
Which is why I asked you what tools you have at your disposal---how are you expected to tackle the problem?
@XanderHenderson Ok, right!
@XanderHenderson quite at basic level, where we are not introduced to anything other than distance formula, section formula..
@Bhavay Ok
I try very hard not to know formulae. The distance formula is just the Pythagorean theorem in disguise (so that is one formula I know), but I am not familiar with the "section formula".
3:09 PM
[ section formula](en.m.wikipedia.org/wiki/…)
After some quick Googling, this seems to be what you want.
With $P = (x_1, y_1, z_1) = (2,-3,4)$ and $Q = (x_2, y_2, z_2) = (8, 0, 10)$.
Then plug and chug.
Hmm yes, then we get the ratio in which it's divided, then the point and then distance from Origin.
You can use the $x$-coordinate of $R$ to determine the relative position of $R$ on the segment (this gives you $k$ in the formula), which you can then use to determine the $y$- and $z$-coordinates of $R$. Then compute distance to the origin.
3:17 PM
I get $\sqrt{53}$
@XanderHenderson Oops. That looks a lot like a plane to me.
@TedShifrin Oh, shoot. Yeah.
The plane normal to $\langle a, b, c \rangle$.
A plane :)
Again, I don't remember formulae to save my life.
Parametric equations, if you eliminate the parameter, give you $$\frac{x-x_0}a = \frac{y-y_0}b = \frac{z-z_0}c.$$ Personally, I like working with the parameter.
3:26 PM
@TedShifrin Me, too. That was my first suggestion.
For Rover, it's a matter of similar triangles. Just have to think in 3D, which is probably hard for a newbie.
So, it will be $\frac{x-2}{6}= \frac{y+3}{3}=\frac{z-4}{6}$ =$\lambda$ ?
Have you learned vectors, @Rover?
@TedShifrin yes
So that's the way to think of it, yes. A point on the line is gotten by starting at $(x_0,y_0,z_0)$ and adding some multiple $\lambda$ of the direction vector $(6,3,6)$.
I recommend understanding that (which is very intuitive) and not trying to memorize more formulas.
3:32 PM
But, yeah, you have $\lambda$. Now find the point.
Looks right.
Yeah, so answer is $\sqrt{101}$, it's not in options
Previously I did something wrong to get that as answer, I got it now.
Q: Possible typo in Chapter XII Proposition 1.1 in Lang Algebra

love_sodamThis post is related to my previous post. In the proof of proposition 1.1, I found some possible error. Here is the proof given in the text. Claim: If $|x|_1<1\Rightarrow |x|_2<1$ then $|-|_1,|-|_2$ induces the same topology on $K$. Supppose the condition holds. Then $|x|_1>1$ implies $|x|_2>1$ ...

I found some possible typo in Lang algebra textbook. Could someone check?
4:14 PM
Does anyone have any ideas about this?

why do you act like $X$ is not equipped with the subspace topology
it states as such in the exercise statement
could someone help me double check something? if $X = T^2 \cup_{f} D^2$ where we identify $T^2 = S^1 \times S^1$, and $f : S^1 = \partial D^2 \rightarrow S^1 \times S^1$ is the map $x \rightarrow (x,1)$, then the fundamental group of $X$ should be $\mathbb{Z}$?
@Thorgott Maybe I made a typo
tomasz did it all
I know that projective maps of a product topology is continuous, but what about if it's the subspace topology?
I guess that it's a subspace of a product topology.
So if subspace topologies of product topologies have continuous projective maps, then we are done with the continuity part.
4:26 PM
a restriction of a continuous map to a subspace with the subspace topology is continuous
@BrookTaylor it is continuous because you are restricting a projection to a subspace of $\mathbb{R}^2$
that's the entire point of the subspace topology
You're right, of course
4:27 PM
Hey does anyone know any online course for differential geometry?
there are lectures on youtube that are good. depends on the level (but im not aware of any very advanced lectures).
@porridgemathematics yeah
@Thorgott thanks for the confirmation, the way I computed it was using van kampens theorem, is there perhaps an easy homotopy equivalence of this space to $S^1$ or something?
perhaps if we can get away with collapsing the disk, we get a pinched torus.. which is homotopy equivalent to a wedge of a sphere and a circle... but i guess this space is not homotopy equivalent to $S^1$ since a wedge of a sphere and circle is not
so I guess my question is, can we get away with collapsing the disk (up to homotopy equivalence)?
if the disk was just a subcomplex in some CW structure of this space then I can see why we can, but its not clear to me that it is
I haven't done the computation, but surely this space should have non-trivial second homology
right, in that case $S^1$ goes out the picture, but homotopy equivalence to $S^2 \lor S^1$ stays
mind if I pivot a little and ask another topology question while were here?
4:43 PM
@Quin I've got physics experience of GR but would like to understand it more from a mathematicians perspective. I felt it was rushed
those are rather rigorous and quite nice. i enjoyed them a lot.
oh doh, I just realized it really is a subcomplex in a CW structure of this thing, we can take a one skeleton that looks like ( )---( ) with another line connecting two other opposite points of circles (those () are supposed to be circles), and then we can attach two disks to be the curved surface area of the resulting cylinder, and attach the boundaries of the top two circles to our final disk!
so the final disk (along with its boundary) is a bonafide subcomplex, and is contractible, so there is a homotopy equivalence from $X$ to the pinched torus as we suspected
isnt this what is basically done in the computation with van kampen? a nbhd of the disk retracts to a point and the remainder is a cylinder? maybe im not understanding the statement
the result is as you say, but I don't understand your reasoning
the disk is a subcomplex by definition of how you're ataching
4:49 PM
how? don't we need a CW structure on $X$ to say the disk is a subcomplex
what im saying is ive assigned it a CW structure and it is a subcomplex, doesn'
doesn't being a subcomplex depend on a CW structure?
@Quin yes it is, im trying to show that $X$ is homotopy equivalent to the pinched torus though, and if I was only interested in showing the fundamental group of $X$ is $\mathbb{Z}$ I could just use van kampen
@Quin i was asking what $X$'s homotopy type is
ah okay
pinched torus i.e. $S^2 \lor S^1$
im pretty sure this is accurate but just wanted to double check, if I have a $d$-simplex $\Delta^{d}$, and I collapse all the subsimplices of dimension $k \leq d$ to a point, I now have a wedge of spheres of dimensions $\geq k+1$, right?
@Thorgott could you elaborate on what you mean by it is a subcomplex by virtue of how it is attached?
5:10 PM
I wonder if the following is true
Let $X:=\Bbb R^2-\{a,b\}$ $a\neq b$. Then there exists an embedded nonempty graph $G$ in $X$ so that for any path connected cover $p:\tilde{X}\to X$ the inverse image $p^{-1}(G)$ is also path-connected.
yeah, but like, $T^2$ has a standard CW structure and if you attach 2-cell, there's an obvious CW structure on the attaching space
and wrt that CW structure, the disk is a subcomplex, because you attached it along a circle that's already a 1-cell in the standard CW structure
@porridgemathematics are you talking about the standard simplex?
@Thorgott ah yeah, that is true - I thought you probably meant this
yes I was
right in that case since its a contractible subcomplex, the quotient map is a homotopy eq
yeah, the subsimplices are just the boundary
and the standard $d$-simplex is a $d$-ball
so you're looking at ball modulo sphere, which is a sphere
right, well a wedge of balls in general right?
uh of spheres
just a single one, why would it be a wedge
5:24 PM
well for instance in $\Delta^3$, if I collapse the all simplices of dimension one or less
but i guess that isn't quite a wedge ..
oh, I misread, I thought you were collapsing all proper subsimplices
to be clear, do you want to collapse all subsimplices of a fixed dimension or all subsimplices of at dimension most that fixed dimension
in the former case, you will get non-Hausdorff spaces
in terms of CW complexes, i want to collapse a $k$ skeleton
so at most
yeah, I don't think it works in codimensions greater than 1
yeah it wont be a wedge in that case for sure
5:41 PM
I don't get why the suggested sets in the answer to the question I posted are not open and closed, respectively. I'm sure it's a stupid question, but I just want to understand.

what does the image look like?
It's the real line, right?
But I'm a bit unsure exactly how each set maps to it with this map
If it were $\pi:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$, then it would just the identity map for one coordinate.
I mean the image of the set he describes as counterexample under $\pi$
I don't know
Any clue for how to figure it out?
are you clear about which set he is describing?
5:48 PM
A an open ball laying on the left side of the open rectangle $(-1,1)\times(-2,2)$. Right?
perhaps this isn't clear, but the center of the ball is supposed to lie on the boundary of the rectangle
Then half of that ball is mapped to the image, right?
The other half is not in the subspace topology.
I'm not sure which image you're referring to
but yes, we take the intersection of this ball with $X$
which is an open subset of $X$, yes?
5:51 PM
ok, what does its image under $\pi$ look like?
But where does it get mapped to?
draw a picture if it helps
I have drawn the picture
A who?
5:52 PM
It's the plane with a closed and an open region removed
so if you have a picture of the open subset and a visual understanding of what $\pi$ does, namely projecting onto the $x$-axis, you can tell me what the image is
I guess it squeezes the space into a line.
So that we get an interval on the real line.
what kind of interval is the important quesiton
It can't be open
Because then the projective map would be open
But, that's not an explanation
It is open for plenty of open balls!
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