@Zach: In general, you should be doing more interesting ones, just so long as you make sure you have the computational skills down. (You seem to have skipped my favorite geometric ones at the end of section 2.)
There are some cool geometric things in section 5, too, even though you know the content there.
@TedShifrin I'm a bit confused about the difference between Euler characteristic of the total space (which is what it seems I'm computing if I compute e.g. $\chi$ of a cone or cylinder) as opposed to the Euler number of a bundle
Alternatively the beautiful theorem (why it's beautiful is a perhaps nontrivial exercise) that if $X$ has an $S^1$ action, $\chi(X)=\chi(X^{S^1})$, the latter being the fixed point set
@Ted I just did $\cos \theta = \frac{\vec{x} \cdot (\vec{x} + \vec{y})}{||x||||\vec{x+y}||}$ and for the other angle $\cos \theta = \frac{\vec{y} \cdot (\vec{x} + \vec{y})}{||y||||\vec{x+y}||}$ then i substituted $||x||$ for $||y||$ in the denominator, and distributed the dot products in the numerator
pretty good =) I did that proof you gave me the other day - I figured out it didn't matter that there were parallel lines, but if it was an isosceles triangle, the opposite angles must be congruent, and if a scalene triangle, the opposite angles can't be; therefore in a square the hypotenuse bisects, and otherwise it doesn't. @TedShifrin
It's not entirely clear from the problem statement, but assume the astronaut's weight vs. altitude is $W=W_0\left( {c \over c+x}\right)^2$ where $W_0=150$ and $c=6400$. Then
$$
{dW \over dt} = {dW \over dx}{dx \over dt} = vW_0c^2{d \over dx}(c-x)^{-2} = {2vW_0c^2 \over (c-x)^3}
$$
For $v=6$ and $...
I hope someone has time for this simple algebra question; Let $n\in\mathbb Z_{>0}$ and $a\in\mathbb Z$. 1. If $n$ is odd, then $\overline a=\overline{-a}\iff\overline a=\overline0.$ 2. If $n$ is even, then $\overline a=\overline{-a}\iff\overline a=\overline0$ or $\overline a=\overline{n/2}.$ It's easy to show the "$\impliedby$" direction for both statements. I'm having difficulties with the other direction though. Consider 1.
We have $a=q_1n+r$ and $-a=q_2n+r$, for $q_1,q_2,r\in\mathbb Z$ and $0\leq r<n$. We need to show that $0<r<n$ is not possible. Assume $r\in\{1,\dots,n-1\}$. We know that $a=q_1n+r=-q_2n-r$. So we can write $r=-(\frac{1}{2}q_1+\frac{1}{2}q_2)n$, since $n$ is odd. We need that $q_1$ and $q_2$ are even; is this somehow impossible? I need a contradiction.
It's not entirely clear from the problem statement, but assume the astronaut's weight vs. altitude is $W=W_0\left( {c \over c+x}\right)^2$ where $W_0=150$ and $c=6400$. Then
$$
{dW \over dt} = {dW \over dx}{dx \over dt} = vW_0c^2{d \over dx}(c-x)^{-2} = {2vW_0c^2 \over (c-x)^3}
$$
For $v=6$ and $...
Since you mod out the free part, the result is still a nice closed, oriented manifold (no singularities) and therefore $\chi\equiv 0\mod 2$ but on the other hand $\chi=\pm 1$
@Danu What's the question? A space with a free S^1-action has Euler characteristic zero because Z/n is a subgroup of S^1 for all n; and free Z/n-action means $\chi$ is div by $n$ for all $n$ - only zero is possible. Now if you have a fixed point just remove it. $X - p$ has $\chi = 0$. Assume $X$ is a CW complex with a vertex $p$. Then $\chi(X) = \chi(X - p) + \chi(p) = 1$.
@MikeMiller Yeah... It's a bit weird. We never proved orientability, but then at some point when it came to relaxing the orientability assumption he was like "yeah well it's oriented after all so we won't drop this assumption"
@BalarkaSen The divisibility by $n$ thing is nice. You just view it as an $n$-sheeted covering I guess and then the argument is sort of the same as Ted's :P
Thanks... I wish I could share the source code, but I'd have to go plundering my hard drive (And by that I mean shut down and boot Linux instead of Win10)
I secretly want to learn to be proficient at spectral sequences but I haven't had a great motivation to do that yet. Maybe I'd want to look for a nearby 1st year grad workshop on homotopy theory which covers Serre's theorem.
Let $n\in\mathbb Z_{>0}$ and $a\in\mathbb Z$. a) Prove $\overline a=\overline{-a}\iff \overline a=\overline0.$ b) Prove that 1. if $n$ is odd, then $\overline a=\overline{-a}\iff\overline a=\overline0,$ 2. if $n$ is even, then $\overline a=\overline{-a}\iff\overline a=\overline0$ or $\overline a=\overline{n/2}.$ c) Prove that if $n>2$, then $\phi(n)$ is even. I've shown a) and b). Here $\phi$ is the Euler function, defined by $\phi(n)=\#\{a\in\mathbb Z\mid1\leq a\leq n, \gcd(a,n)=1\}$. However, I have no clue how to use a) and b) to prove this. Could someone give me a hint?
Consider $\mathbb Nm\mathbb N$. Assume that for some $n\in\mathbb N$, it holds that $\phi(n)$ is even. Now consider $n+1$. We know that $\phi(n+1)=\phi(n)$ if $\gcd(n+1,m)\neq 1$, and $\phi(n+1)=\phi(n)+1$ if $\gcd(n+1,m)=1$. But I should be able to show that we can only add 2, and not 1. But I wouldn't know how to proceed:l
Hm, I'll post it in the forum, because I gotta leave now anyways
It also doesn't really suggest any obvious 'predictions'. For instance, suppose I construct the purported model of the hydrogen atom. Can I deduce the usual hydrogen spectrum in a sensible way?
Can I say something useful about the helium atom (the simplest two-electron atom) using it?
Now, those might be the wrong questions: they seem less interested in the electrons than in the nucleons.
If it says less than quantum mechanics already does, then I find it hard to believe I should be interested.
When a space shuttle is launched into space, an astronaut's body weight decreases until a state of weightlessness is achieved. The weight $W$ of a $150$-lb astronaut at an altitude of $x$ kilometers above sea level is given by $W=(\frac{6400}{6400+x})^2$. If the space shuttle is moving away fr...
@Krijn I don't know how to handle these questions. I stick to it, maybe? I do not place importance over knowing as much as over being able to solve things
2
Even simple things like a good combinatorics problem, not just high-tech stuff
And I am bad at that. I have improved a bit. Trying to get better
When a space shuttle is launched into space, an astronaut's body weight decreases until a state of weightlessness is achieved. The weight $W$ of a $150$-lb astronaut at an altitude of $x$ kilometers above sea level is given by $W=(\frac{6400}{6400+x})^2$. If the space shuttle is moving away fr...
I am interested in answers to this question particularly as it pertains to mathematics.
