Determine the prime integers p, such that $\ mathbb F_p ^ *$ have 2 subgroups, one of order 17, G and order 19, H, such that there exists g in G with g + 1 in H.
@user104729 It's interesting that tends to be how mathematicians learn advanced topics ("reading a little, finding something you lack, then correcting that - then returning to the text"). I think the biggest thing I need to unlearn is the expectation to go through a book linearly.
If you take $\Bbb Z \supset \Bbb Z/(83)$ and let $b = 10$ do you say that $b$ is "algebraic" over $\Bbb Z/(83)$, since it satisfies $X^2 - 17 \bmod 83$?
@NV-US: Right. I found that, too, but didn't want to hassle trying to get into JSTOR. I'd have to log in through the UGA library and it's too much hassle (if I can even still do it, not sure).
Sure, @NV-US, you can ask :) It looks like it's got two basic ideas — the limit definition of the derivative and the intermediate value property for continuous functions. [Actually, without assuming the derivative is continuous, one can still prove it has the intermediate value property. Cool result.]
Oh, they're NOT assuming the derivative is continuous, @NV-US. That's good. They're using continuity of the secant slope.
@LeakyNun From an algebraic point of view, homology is a measure of how a chain complex fails to be exact, if you can associate a chain complex to topological spaces (or other objects), then homology gives you invariants. There's something to be said about holes and triangles, but I'm not really good at explaining geometry
@MatheinBoulomenos consider $f$ the automorphism of $\Bbb T^2 = \Bbb S^1 \times \Bbb S^1$ by swapping the two coordinates. It is not null-homotopic. How large must $n$ be such that if I embed the two tori into $\Bbb R^n$, $f$ becomes ambient-isotopic to the identity?
@LeakyNun just imagine you want to somehow show that a space has hole, e.g. a solid torus. One way to show this is to show that there exists a tetrahedron embedded in that space (without interior) that is not the boundary of some tetrahedron with interior (because there's a hole in the middle). Does this make sense?
that's the same idea with the parametric case you cited. if I change b, i'll get a different graph but the only change is one of scale. the two curves will be similar
and since that's not that interest of a difference, you might as well pick b=1 (and so a=k) for convenience in plotting
things are getting clearer now, thanks a lot for your time from a programmer's perspective the notation really messes me up, sometimes they mention t sometimes don't, sometimes the actual dependency is expressed indirectly like the (k=a/b) case I mentioned at first
@Leaky: Here's a linear algebra question for you. What is the smallest $n\ge 3$ so that you can join the matrix $\begin{bmatrix} 1 & \\ & 1 \\ &&\ddots\end{bmatrix}$ to the matrix $\begin{bmatrix} & 1 \\ 1 & \\ && \ddots\end{bmatrix} in $SO(n)$? Presumably the first matrix is the identity. Can you fill in the second matrix and get a continuous path?
$ \displaystyle \frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left| \frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right| dt = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2}$ --- where $\displaystyle \left(\frac n7\right)$ is the Legendre symbol.
I thought it was strange that this is unsolved until I tried to rewrite the RHS via Eisenstein representation of the Legendre symbol. It's a monstrosity!
the proof made sense to me. What i am stuck at is, after the limit definition of derivative, how can they say that there exists interval (x1,y1) and (x2,y2) such that the next inequality follows. I tried using epsilon-delta to no avail. @TedShifrin
Yeah, @Mathei, of course. But I was thinking about embeddings of the flat torus in $\Bbb R^n$ for $n\ge 4$. Not sure I know what to do in $\Bbb R^3$. In $\Bbb R^3$ you would have to turn it inside-out, and I don't like that.
where I was going was that, if you work out what $M$ would have to be for that to work at linear order in $t$, you'd get something whose powers don't make the structure of that matrix
in order for that to work to first-order in $t$, you'd need $M=\begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0\end{pmatrix}$
Yes, @NV-US, but not exactly. By the limit definition, you can choose $x_1$ and $y_1$ close to $c_1$ so that that quotient is close to $f'(c_1)$, but, in particular, still close enough so that it's less than $f'(c)$.
So what, @Semiclassic? Am I being dopey?
You really need pure imaginary eigenvalues for this to work.
$|x-c|<\delta 1, |y-c|<\delta 2$, and $|\frac{f(x)-f(y)}{x-y} - f'(c1)| < \epsilon$, are the inequalities i have (if correct). How does this show the existence of $(x1,y1)$? @TedShifrin
So choose $\epsilon$ less than the distance from $f'(c_1)$ to $f'(c)$, first of all, @NV-US. Then pick any $x,y$ satisfying your inequalities. (You can actually have one single $\delta$, but you're fine.)
