Statement from the Text:-
Suppose the given curve is $$x=f_1(t),y=f_2(t),z=f_3(t).$$ On
eliminating $t$ between $f_1$ and $f_2$, we obtain a relation
$\phi(x,y)=0.$ Similarly, between $f_1$ and $f_3$, we get
$\psi(x,z)=0$.
Reference:- An Elementary Course in Partial Differential Equations, T. A...
I responded to you before. They are being sloppy, as applied math people sometimes are. Consider $f_1$ and $f_2$ smooth and both zero on an interval. Their statement is just wrong without assuming non-vanishing derivative for (at least) one of the functions.
I also had constantly plant lilypad on the pool lane so that I have an easier time planting chompers there and I would have an easier time micromanaging. Lilypads don't get hit by peashooters so we can plant as many as we want without any worries.
the point of having immitator-sunflowers is that, a lot of sunflowers will die under the fire of peashooters and gatling pea zombies. So we have to somehow replenish them as fast as possible. We can also use them as a temporary shield.
this sounds like a very axiomatic game. you could replace 'sunflowers' with 'elements of g in G' and 'night stage' with 'abelian group' and it would be more or less the same conversation.
Hi, I have a doubt in the solution of the given problem: >The value (s) of $a$ for which the area of the triangle included between the axes and any tangent to the curve $x^ay=\lambda^a$ is constant is/are?
solution^
What does $\frac{dA}{da}=0$ mean? Afaik when $\frac{dy}{dx}=0$ this implies that $y$ is constant wrt $x$, so if that is the case then $A$ is constant wrt $a$ and also we are asked to find $a$? I am confused
Even though I don't get to play video games that much due to studying, your discussion with me reminded me of a lot of childhood memories. Especially the Gargantuar boss music and the ending credit song, they have a special place in my heart. Thanks for having this discussion with me :D
A right $R$-module $B$ is flat if and only if $0\to B\otimes_R I\to B\otimes_R R$ is exact for every finitely generated left ideal $I$.
The proof of this statement is given in Rotman Advanced Modern Algebra Part 1. My question is on if direction so I only state if part of the proof: Conversely,...
@BalarkaSen i've been thinking about why time exists at all, like why do things have internal clocks, and why do the different internal clocks seem to have the same speed everywhere (at least if you ignore special relativity)
like one second is defined in relations to the vibrations of an 138-Cs atom iirc
Nothing really significant happens to the geometry of the spacetime as you cross the event horizon though. If you have a supermassive blackhole with a huge Schwarzschild radius, you won't feel anything while crossing the event horizon.
It's an artifact of the coordinate system. Tells you how artificial time is.
first of all i dont really understand what "continuous time" means. what is "discrete time"? the assumption is spacetime is a pseudoriemannian 4-manifold
time dilation is completely consistent with this assumption, right?
its just a local computation with the Minkowski metric
If $X$ is a right $G$-space and $Y$ is a left $G$-space then $X\times_G Y$ for $X\times Y/(xg,y)\sim(x,gy)$ something formal? The name of this space is the 'balanced product' of $X$ and $Y$. The notation is for pullback but I can't see any pullback here
Very generally, if $A$ is a monoid object in a monoidal category, we can define a "tensor product over $A$" $X\otimes_A Y$ between a right $A$-module $X$ and a left $A$-module $Y$ to be a coequalizer of the two maps $X\otimes A\otimes Y\to X\otimes Y$ (where the first map uses the multiplication ...
There is a diagonal (left) $G$-action on $X\times Y$ given by $g.(x,y)=(xg^{-1},gy)$ and $X\times_GY$ is the orbit space of this action
the inverse $g^{-1}$ is an artifact of transforming a right into a left action, as usual
equivalently, it is the orbit space of the diagonal (right) $G$-action on $X\times Y$ given by $(x,y).g=(xg,g^{-1}y)$
the way you have written the quotient is basically just a symmetrized way of describing this quotient
this has an obvious universal property with respect to balanced $G$-maps out of $X\times Y$, but that's probably only of questionable usefulness
you can also note that $X\times_GY$ naturally surjects onto the product of the orbit spaces $G\X\times Y/G$
an example where this construction is used is in equivariant topology, where for a $G$-space $X$, one considers $X//G=X\times_GEG$ ($EG$ being the total space of the universal $G$-bundle). this is called a homotopy quotient (for reasons I don't understand). if the action is free, it is homotopy-equivalent to $X/G$, but in general it is not. the point is that this constructions plays a lot better with homotopy-theoretic constructions than the ordinary quotient $X/G$.
@onepotatotwopotato I don't think it is a tensor product
a tensor product is more like the space $X\times Y$ with (one of) the diagonal actions
of which this is the orbit space
even putting variance aside, this is not a bifunctor from a category to itself (like a monoidal structure), but rather it is a pairing of categories right $G$-spaces $\times$ left $G$-spaces $\rightarrow\mathbf{Top}$
let's see what this looks like algebraically. i think we're better off assuming our spaces CW complexes and our $G$-actions cellular. this ensures our orbit spaces to carry a natural CW structure s.t. the projections become cellular maps. to the (left, forthcoming) $G$-space $X$, we associate the cellular chain complex $C_{\bullet}(X)$, which carries an induced $G$-action by chain maps. the cellular projection $X\rightarrow X/G$ induces an isomorphism $C_{\bullet}(X)_G\rightarrow C_{\bullet}(X/G)$ (the former being the $G$-coinvariants).
and the previous observations tell us that the operations of tensor product and taking coinvariants do not commute
that's where group homology can enter the picture
taking coinvariants is right-exact, but not exact, which is the fundamental complication of equivariant algebraic topology
if you use cohomology instead of homology and replace coinvariants by invariants, the same general picture holds (though I'm not sure if you might need some finiteness assumptions)
Trying to understand the following lemma by Gromov: Let $M = S^1 \times D^2$ be a solid torus and suppose $M$ is equipped with a hyperbolic metric near $\partial M$ such that $\partial M$ is a quotient of a horosphere in $\Bbb H^3$. Then the metric extends all the way inside to a negatively curved metric on $M$ if the length of the geodesic representative of the meridian is at least $2\pi$
Seems like it's at least necessary because of the following. Take a meridianal disk $D$ in $M$. By Gauss-Bonnet, $\int_D K dA + \int_{\partial D} k_g ds = 2\pi$. Now $K < 0$ and $k_g = 1$ because a collar of $\partial D$ in $D$ looks like a hyperbolic annulus, whose outer boundary has geodesic curvature $+1$. So $\ell(\partial D) = 2\pi + \int_D (-K) dA > 2\pi$.
Well, you're given a hyperbolic metric only near $\partial M$ at first. Then, given hypothesis, you can extend all the way to $M$ keeping it negatively curved (perhaps not hyperbolic)
Correct. Note that $\partial M$ is a quotient of a horosphere, which is flat. So really, it's a flat torus. Any meridian will be a geodesic on $\partial M$, and all of them are the same length.
You're right that that's superfluous. Some people call meridian the homology class instead of the curve. Any Euclidean meridian is indeed a geodesic on $\partial M$.
I didn't think of it that way. That's true because it's completely orthogonal to the family of geodesics emanating from the point at infinity on the horocycle.
That wouldn't nail $k_g=1$, though, I don't think. It would just say constant. But I have the computation in my hyperbolic geometry section, along with an exercise about curves of $k_g=c$ for different $c$.
I have the following question: Show that the subspaces of R3 are precisely {0}, R3, all the lines through the origin and all the planes in R3 through the origin. I can show that the first 3 are subspaces however I don't quite understand how to show the last condition.
What I've done is the following: I broke the question down into 4 cases and I know that when dim U = 2, it will be a plane. Thus the basis will have a length of 2. Say (x1,x2,x3),(y1,y2,y3) is my basis. Thus my subspace will be the span of the list. However I can't draw the line that this shows all the planes through the origin
I feel like I'm missing something and/or misunderstanding something
Maybe I am not understanding your question... any subspace must include zero (the zero vector, i.e. the origin), which means that all subspaces must pass through the origin. So the only planes which can be subspaces are planes though the origin.
Nah, its on me, I realize I phrased myself poorly. I know that every subspace must contain 0. I think my confusion arises more from showing that the span of the list is a plane
Maybe I am confused about what you are trying to show. Are you trying to show that planes through the origin are subspaces?
Or are you trying to show that if $V$ is a subspace, then $V$ must be one of (a) the trivial space $\{0\}$, (b) a line through the origin, (c) a plane through the origin, or (d) all of $\mathbb{R}^3$?
Okay, so what you would need to do is show that if you have two linearly independent vectors (say, $x$ and $y$), the set of all linear combinations of those vectors form a plane, i.e. they satisfy an equation of the form you gave.
If you are familiar with the dot product (and some basic results about the dot product), then there is a pretty slick proof.
Otherwise, you might have to get your hands dirty with some computation.
The key observation is that if $\langle x, y, z\rangle$ is in the plane defined by the equation $ax + by + cz = 0$, then $\langle x, y, z \rangle \cdot \langle a, b, c \rangle = 0$.
This means that every element of the plane is perpendicular to $\langle a, b, c\rangle$, no?
@TedShifrin I suspect it was the cheapest flight. Dubrovnik-Frankfurt-Boston-SFO. I was so sure that he would miss the connection that I had lined some some places to stay and found an early morning flight. All for naught thankfully.
[Edited]
Show that there are no real solutions of
$$1 + \sum_{n = 1}^{\infty} \frac{x^n}{\prod_{k=1}^{n} H_k} = 0$$
where
$$H_k = \sum_{i=1}^{k} \frac{1}{i}.$$
I managed to prove this, and I want to know if my proof is correct, and if there are any other (better) ways of proving it. (This is my f...
At the beginning part of the proof, why $1+\sum_{n=1}^\infty {x^n\over \prod_{k=1}^n H_k} = 1+\sum_{n=1}^\infty \left({x^{2n-1}\over\prod_{k=1}^{2n-1}H_k}+{x^{2n}\over\prod_{k=1}^{2n}H_k}\right)$? We don't know the convergence of the given power series
Statement from the Text:-
Suppose the given curve is $$x=f_1(t),y=f_2(t),z=f_3(t).$$ On
eliminating $t$ between $f_1$ and $f_2$, we obtain a relation
$\phi(x,y)=0.$ Similarly, between $f_1$ and $f_3$, we get
$\psi(x,z)=0$.
Reference:- An Elementary Course in Partial Differential Equations, T. A...
