Let $a$ be a non-zero real number. For each integer $n$, we define $S_n = a^n + a^{-n}$. Prove that if for some integer $k$, the sums $S_k$ and $S_{k+1}$ are integers, then the sums $S_n$ are integers for all integers $n$.
@arctictern I just want to convince myself that the definition given this way is associative. $A \times (B \times C) = \{ f : \{1,2\} \rightarrow A \cup (B \times C) : f(1) \in A, f(2) \in (B \times C)\}$ right ?
@user19405892 if you prove it in the case of $k,k+1=0,1$ by induction, you can replace $a$ with $a^k$ and get that $S_k$ integer implies $S_{nk}$ integer for all $n$. So you also get $S_{n(k+1)}$ integer for all $n$. You might be able to combine those with the relation $S_kS_\ell=S_{k+\ell}+S_{k-\ell}$.
@TedShifrin That's one of those abuses of language which I'm sure I fall into myself at times. But it's one thing to abuse it for convenience, and another thing to insist upon it.
When I used to teach abstract algebra, I insisted on the overbars, but eventually, when we got to general quotient rings, relaxed the notation because I needed it for another layer of equivalence class (e.g., $\Bbb Z_m[x]/I$).
Anyways, I have a quick novice question if anyone cares to entertain it. Working on some algebraic geometry and I was recalling some old material from linear algebra / abstract vector spaces. In general what is the difference between k[x1,...,x2] and the polynomial space P^(n)?
Is it just that in the first we have control over the field?
@TedShifrin Hi, would you please explain me the last paragraph or two in theorem 3.10? In particular, I don't understand what "has the same breadth" means, and why we know that we can say the first vector is $R$ times the second, as opposed to say $-R$ (we know they're parallel, and why the curve actually gives us a whole circle - I understand it as "the points of the curve are $R$ away from some $(0,y_0)$, it's closed and simple, so it has to be a circle - is that right?
@Jake: Breadth is the distance between parallel tangent lines. When you're doing Cauchy-Schwarz, the dot product is equal to the product of the lengths if and only if the vectors are positive scalar multiples of one another. The deduction that the curve is a circle comes actually from the differential equations and the arclength parametrization of the original curve.
Professor @TedShifrin have you finished the classes with the at risk students yet?
user147690
Not quite nice to see him, or nice to not quite see him? If the latter is it more nice that you can only not quite see him, or is it nice that you can not quite see him, but not quite as nice as if you could not not quite see him.
@Brody Oficially I'm self-made, some of my interest reduces to the calculation of integrals and series (not really the ones you find one sites, or in papers - well, an exception here, stuff like the one you see in Ramanujan's notebooks)
Officially
I'm not really around, I have to finish some very imp stuff.
1 hour later…
Anonymous
08:29
@AlexClark Hello
Anonymous
@AlexClark My nick hasn't changed dammit,I am Ashwin
Anonymous
@AlexClark Link me your private room if you have one,we can speak over there
Is there anything I can do with $\int_{f_{min}}^{f_{max}} {Sin^2(f \pi t)/(f^2 \pi^2) df}$ or is this really something I can only evaluate numerically?
np just bail whenevs. So as a practical shortcut, if n is prime, do we always take the kth powers from the set $U(n-1)$ mod n to find the other generators?
e.g. the prior example when $\langle 3 \rangle$ is a generator of $U(31)$, all I needed to do to find the other generators of $U(31)$ is calculate $3^k \ \text{mod} \ 31$ for all $k \in U(30)$
But yes, in general, if $U(n)$ is cyclic and $x$ is a generator then the other generators are those of the form $x^k$ with $k$ coprime to $\varphi(n)$.
We best don't. What "really" forms a group is a set of equivalence classes of numbers in this case.
@SimpleArt Hmm, looks like I've got an incentive to actually answer the question you referenced before (rather than getting distracted by simpler matters as in our conversation yesterday)
ok. But points where the function jumps from 1 to 2 and then back to 1 are not allowed if it is to be an entire function? Is differentiability the keyword?
@MatsGranvik A set is called open, if it is a subset, S, of the Euclidean n-space, $ \mathbb{R}^n $, such that every element has in S has a neighbourhood in $ \mathbb{R}^n $, contained in A.
I think the sharper question is probably how well-behaved of a function $f(s)$ is
and my guess would be 'not very well-behaved at all'
i mean, you start off with $$\frac{c \zeta (c) \zeta (s c)}{\zeta (s c+c-1)}-\frac{\zeta (c) \zeta (s c)}{\zeta (s c+c-1)}=\frac{c-1}{\zeta(c)}\frac{\zeta (s c)}{\zeta (s c+c-1)}$$
prior to taking $c\to 1$, that's already got a poles at $s=1/c$ and at any $s$ where $\zeta(sc+c-1)=0$
along with zeroes as well
ideally, stuff would just nicely cancel in that limit. but i have no idea if that's true in this case.
@AkivaWeinberger: Yes, without taking some care you run into the standard set theotetic difficulties. Frequently the category of small categories is not sufficient, but the really careful people start talking about Grothendieck universes. No category of categories should contain itself.
Let $H$ be the subgroup of the additive group of rational numbers with the property that $\frac{1}{x} \in H$ for every nonzero element in $H$ prove that $H = 0$ or $H = \mathbb{Q}$. Does this mean that $H$ is made up of elements $\frac{1}{x}$ ?
Well if it's an additive subgroup of $\mathbb{Q}$, it has to infinite. So it makes sense it's either $0$ or $\mathbb{Q}$. But I must prove this coherently with math..
Eh.. I see how all rationals between $0$ and $1$ are in $H$ but I don't immediately notice that $\frac{5}{2}$ is in $H$. Like you said @arctictern if $H = \mathbb{Z}$ then every element $\frac{1}{x}$ exists where $x \in \mathbb{Z}$ if $x$ can only be integers, how can we achieve fractions of integers greater than $1$?
@arctictern If you meant "from just that information alone" I don't see how you could assume infinitely many more elements. But, if $H$ is a subgroup of $\mathbb{Q}$ then yes
Okay. Now that we have nx for all integers n (and some nonzero rational x in H), we know that 1/(nx) is in H. Now do the same thing with 1/(nx); what elements can we generate from 1/(nx)?
@akiva $\{\frac{p}{q}, \frac{2p}{q}, \frac{3p}{q}, \frac{4p}{q}...\} + \{\frac{q}{p}, \frac{2q}{p} ,\frac{3q}{p} , ... \} + \{\frac{p}{2q}, \frac{2p}{2q}, \frac{3q}{2q} ,...\} + $ and so on you can generate all rational numbers this way.
I just know if the subgroup is any subset of the integers, it will produce fractions which add to make all of $\mathbb{Q}$ and if it is a subset of rationals, it produces all of them additively too D:
Do diagonally-opposite points of a rectangle have a specific name, relative to the other two points?
Maybe that doesn't make sense. ^^;
To put it another way, if I have a rectangle NESW (assigned clockwise, as you might expect), is there a term that describes points E and W relative to N,S?
