Hello. I'm doing a math phd (just finished quals). If one specializes in pure math (say dynamical systems with no particular application in mind) how difficult can it be to transition into industry after the PhD? I am not particularly interested in applied math, but I'm not too optimistic about the future of the pure mathematician (unless he/she solves some millenial problem!!) and I would like to think that I have the option of switching from academia to industry. Any input is appreciated.
I am a 2nd year math PhD student. If one specializes in pure math (say dynamical systems with no particular real-world application in mind), how difficult can it be to transition into industry after the PhD? What are measures one can take to make this a smooth transition? I am not particularly in...
Ah, well maybe someone here can help me out. Why does Hungerford add the "free group" hypothesis in this problem? This appears to be something true of all groups. My solution is extremely suspiciously short, but I honestly don't see anything wrong with it: i.imgur.com/FY7Rd3v.png
Yeah you could have a bunch of different members of the generating set, at different orders there @KajHansen - unless (random order of powers of different elements)=$x^n$ for some $x$ it isn't true. Hope that helps.
@AlecTeal, I'm trying a different route altogether
Suppose $F$ is the free group on a set $X$, and let $I$ be an indexing set where $|I| = |X|$. Then the group $\displaystyle \prod_{i \in I} \mathbb{Z}_n$ can be realized as the homomorphic image of $F$, and hopefully the kernel matches the supposed normal subgroup. Just thinking through my reasoning and making sure it's airtight.
@KajHansen You have only shown for the generating set $S$ of $G$ that $gSg^{-1}\subset S$. To show the subgroup is normal you're going to need to do that for any $g$, $gGg^{-1}\subset G$. These two statements are not equivalent.
Yeah, I noticed my mistake pretty quick in that original image @PVAL, but what of the idea right above your post? I think I've filled out the details, and it seems to work.
I take my generating set on that product to just be $\{ \mathbf{e}_i \}$, which denotes a tuple zeros except a $1$ in the $i^\text{th}$ position. Then $\phi$ is induced by the inclusion $X \hookrightarrow G$.
Your kernel is the normal closure of the desired group. You need to show the normal closure is actually the desired subgroup. In general if you have a quotient of a free group relators $r_i$, the kernel of the map is just the normal closure of the subgroup generated by the $r_i$. You are asked to show that in this case the normal closure is the same as the group. Start with a generic element in $G$ (not a generator) and show that after conjugating, the element is still in $G$.
I am TA'ing linear algebra at the moment, and I realized that I cannot come up with a satisfactory definition of a linear equation. I have some ideas, but it just never seems quite "right" (and it also gets rather complicated)
So now the question is whether an equation is linear if it is equivalent to an equation of the form $L(x) = y$, but this seems too weak somehow
since for example, over the reals $x^2 = 1$ would then be linear (being equivalent to $0=1$) and also over $\mathbb{F}_2$ (being equivalent to $x=1$), but not over $\mathbb{C}$.
which then bring in "is linear over any field in which it makes sense", but that just seems wrong somehow
@MikeMiller Mainly because there was an exercise asking them to state whether $2x_1 = 4x_2 - x_3$ was linear. And the book said "yes" while the answers from the previous lecturer said "no" (due to an imprecise definition of the term in the book)
I think what you want is precisely "Equations that can be derived from $Ax = y$ by elementary linear transformations", where by that I mean you can multiply either side by a scalar or add terms from one side to the other.
Singing while discovering tons of new amazing results ... (that is now I mean). I should have one or two assistants that record all my stuff I create every day, this would be very nice (in terms of time management).
@Chris'ssistheartist last night I was hearing sounds that seemed like Doppler effect :P .. it's amazing and hilarious at the same time what a delirious brain can make you experience
@MikeMiller $f : M \to N$ be a map between closed compact orientable manifolds of nonzero degree. I want to prove that it's surjective. Here's what I have done : assume it's not surjective. then $f$ factors as $f : M \to N - \{x\} \hookrightarrow N$. $H_nf$ factors as $H_n M \to H_n(N - \{x\}) \to H_nN$. $N - \{x\}$ is not compact so the middle group is zero, forcing $\deg \, f = 0$, and the whole world blows up.
why can't we do the following? $G$ acts on $M$ by orientation preserving action, thus orbit of a single point $x \in M$ all have the same local homology. Pushforwarding by covering map gives a point $y \in M/G$. Now $U$ be a small nbhd of $y$ which is evenly covered, and lift this to $M$ to get a bunch of disjoint neighborhoods of the orbits.
Now, $M$ is orientable, so any orbit s.t. every element lies in each of the disjoint nbhds above get the local orientation from the total nbhd
Right, what you mean is that the orientation of $y \in M/G$ is the pushforward of the orientation of one of the elements in its preimage. The choice doesn't matter.
And it's locally the same because one can pick a neighborhood as you say.
@anon Ah ok, honestly I've never working with this kind of problems. I think and immediately corollary, is that $\dot{x}=A(t)x$ has the same solution as the scalar case, right?
@NaCl Still a bit .. off. For instance, If $\sum t_k=m$ is $\le\sum n_k$ and $\sum n_k\le m$ then $m=\sum n_k$. You're aware of this right? Also, when you say "I want to generate tuples from that," do you mean you want the components/coordinates of the tuple to be taken from the multiset?
Also, what's the purpose of having repeated elements in the multiset?
ok, I am totally confused. the two converses I have in mind is 1) $M/G$ is orientable only if $G$ acts by orientation preserving homeos 2) $M/G$ is orientable only if $M$ is orientable.
@NaCl Would it be correct to say you have some list of positive whole numbers $n_1,n_2,\cdots$ and a chosen whole number $m$ and you want to enumerate the lists of nonnegative integers $t_1,t_2,\cdots$ such that $t_k\le n_k$ for each $k$ and $\sum t_k=m$?
So how about this: say "Given whole numbers $n_1,n_2,\cdots,n_f$ and $m$, let $T$ be the collection of all tuples $(t_1,\cdots,t_f)$ with $0\le t_k\le n_k$ for $k=1,\cdots,f$ and $\sum_{k=1}^f t_k=m$. Based on examples, it seems to me that $T$ depends only on $m$ and the sum $\sum_{k=1}^f n_k$. Is this correct?"
@Agawa001 Oui, le chiffre etait choisi volontairement, mais le fait de demander de trouver la distance au lieu du temps est bien une erreur involontaire.
Given I have to use X and Y amounts of A and B to reach a sum of 10,000, and given I must use a combination of exactly 100 X and Y to get this, I'm left with:
xa + yb = 10000
And x+y = 100 of course.
So I solved it to y = -200, x = 300
But this makes no sense practically. Because I can't have -200 cash.
So I'm trying to think of how I'd use another combination.
@DanielFischer That exercise doesn't even mention anything about the continuity of $f(x)$. It simply says, "Show that if $F(s) = \int_{0}^{\infty} e^{-sx} f(x) \, dx $ converges for $s=s_{0}$, then it converges uniformly on $[s_{0}, \infty)$."
You can't just put ask someone to give in three values for the price of three items, and output combinations of them that give you a precise amount for a precise price.
If you must spend $100 to get 100 items, and you input prices 0.1, 0.1, 0.1. There is no way you can spend exactly $100 to get 100 of them, you must get more.
@DanielFischer If it is indeed true, would it directly imply $\lim_{s \to 0^{+}} \int_{0}^{\infty} e^{-sx}f(x) \, dx = \int_{0}^{\infty} f(x) \, dx$ as long as $\int_{0}^{\infty} f(x) \, dx$ converges as an improper Riemann integral? Or is there something I'm missing here.