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 $a^2=b^2$.
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).
@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.
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 $a$.
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 $I$
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!
@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!
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?
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.
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
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
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".
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.
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.
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.
This 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 $|x|_2>1$ ...
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 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
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
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
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?
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.
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.
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
