The third map $\pi_*(E) \to \pi_*(Fp)$ in the middle should be just coming from the homotopy equivalnce $E \to Fp$, $e \mapsto (e, 0_e)$ like we discussed
Well I'm not sure if pullback is the right word but what I have in mind is like
An element of $\pi_q(E, p^{-1}(b))$ is like homotopy class rel $p^{-1}(b)$ of a map $\sigma : (D^q, S^{q-1}) \to (E, p^{-1}(b))$
compose that with the fiber projection $E \to B$
$p^{-1}(b)$, which contains the boundary of $\sigma$, pushes down to $b$
so I get a map $\pi \circ \sigma : (D^q, S^{q-1}) \to (B, b)$
where $S^{q-1}$ goes to $b$
So that's just homotopy class of a map $S^q = D^q/S^{q-1} \to (B, b)$, i.e, elt of $\pi_q(B)$ (by omitting the b)
@Daminark Does that make sense?
It's a good exercise to prove it's an isomorphism by the way
but once you have this, verify that these maps make the whole diagram commute. Now take five terms on both LES's so that the middle map is $p_*$. Then use the five lemma
Isn't mathematical induction basically the "easy way out"? Because we first substitute and check and then assume true for all m and then prove it to be true for $(m+1)$ also.
It is an easy way out. Why would you want a hard way out?
If induction gives a way to prove a mathematical proposition easily, that's good. It means we have a effective tool
It might not be the most enlightening to prove something by induction as opposed to doing the combinatorics or something, but it's certainly a way to get the job done
Consider the quadratic programming problem $$\min \frac{1}{2}\langle x, Qx \rangle + \langle c, x \rangle \\ \text{s.t.}\, Ax \geq b $$
where $Q$ is a positive definite matrix of dimension $n$, $A$ is a matrix of dimension $m \times n$, $c \in \mathbb{R}^{n}$, and $b \in \mathbb{R}^{m}$.
I need...
@ALannister I don’t know about quadratic programming in particular, but the main difference I see between your quadratic and linear problems is that <x,Qx> is always positive regardless of x due to the positive definiteness
Anyways. I think the difference between QP and LP should basically be the same as that between minimizing a quadratic function of one variable and a linear function
@tuki your deduction that the ode to be solved is u’+ru=0 is incorrect
The ode to be solved is the one they give you, and no other
That said, what you did would be effective if you substituted v=u’
That gives a first-order ODE in v(r) which you can solve as you did in your work. Integration then yields u(r). You should have two constants of integration, enough to satisfy the boundary conditions uniquely
@Daminark Actually the map here seems to not be too hard. $E$ is naturally a subspace of $Fb$, of pairs $(x, 0_x)$ where $x \in E$ and $0_x$ is the constant path at $x$. $\rho : Fp \to B$ restricts to $p : E \to B$ on this subspace. So there is a natural inclusion $p^{-1}(b) \hookrightarrow \rho^{-1}(b) = F$
I have the following problem:
I tried to solve this problem by creating a function $h$ which is a shifted and scaled version of $g$, such that $h$ has a period of $2\pi$ on the interval $[-\pi, \pi]$, so that I can apply the theorem above. However, I am stuck, because I don't know how to make ...
That said, if your goal from the start was to prove that$$-\left\|x-\frac{a+b}{2}\right\|^2+\|x-a\|^2= \left\|\frac{a-b}{2}\right\|^2,$$I'd have just done it with the other trick from the start
please @AkivaWeinberger after that they say that : but $\frac{a-b}{2}\in K$ because $K$ is convex , then $||\frac{a-b}{2}||^2\leq ||x-a||^2-||x-\frac{a+b}{2}||^2\leq0$
Consider the quadratic programming problem $$\min \frac{1}{2}\langle x, Qx \rangle + \langle c, x \rangle \\ \text{s.t.}\, Ax \geq b $$
where $Q$ is a positive definite matrix of dimension $n$, $A$ is a matrix of dimension $m \times n$, $c \in \mathbb{R}^{n}$, and $b \in \mathbb{R}^{m}$.
I need...
I have the following problem:
I tried to solve this problem by creating a function $h$ which is a shifted and scaled version of $g$, such that $h$ has a period of $2\pi$ on the interval $[-\pi, \pi]$, so that I can apply the theorem above. However, I am stuck, because I don't know how to make ...
