I would prefer to take $r = a_n/a_{n+1}$; then if $a_0 = a$, the series is $a_n = ar^n$. so because $a_n/a_{n+1} = -2/5$, we get $a_n = 10\left(\frac{-2}{5}\right)^n $
Sure thing. I was about to write that I was having serious difficulties with it (this is the third series formula I've ever had to come up with and it was sort of a brick wall). Are there any formulas or concrete tips/tricks I can use to develop general terms?
Assuming it's geometric, are there any formulas? You mentioned $r = a_n / a_{n+1}$ earlier. Is that part of a formula or common technique? I have no repertoire I can pull out to tackle these, so it's currently pure intuition (and intuition fails often in math).
For a sufficiently large absolute constant $C$, there exists for every $n$ a prime number $r$ satisfying $(2\log{n})^6 \leq r \leq (C \log{n})^6, \text{ord}_{r}(n) > r^{\frac{1}{3}},$ and $P(r-1) > r^{\frac{2}{3}}$ where $P(m)$ denotes the largest prime divisor of an integer $m$.
is it true that the number of primes $r < x = (C\log{n})^6$ with $P(r-1) > r^{\frac{2}{3}}$ is at least $c\pi(x) \sim c\frac{C^6 \log{n^6}}{6\log(Cn)}$ where $c$ is an absolute constant?
getting 1/n is easy. it's not immediately obvious that you can do so for arbitrary rational numbers, but the fact that Egyptian fraction representations always exist means that the answer is yes
they also show in their paper that, for instance, you can get $e$ as a groupoid cardinality
sure. also, that's cheating, because groupoid cardinality isn't actually defined for infinite groupoids unless you're lucky, and it's just a fluke of nature that it converges for FinSet
In my undergrad upper div QM course we barely did any wave mechanics and I liked that perspective. It makes more sense to consider matrices rather than abstract operators.
not at the level of groupoids. the only way i can recognize that each of them has unit cardinality is to say "hey, it's a disjoint union of groups, and their cardinalities happen to add to 1"
i feel like you're looking for something deep here, and i'm pretty skeptical that there's anything deep at all; for some stupid heuristic evidence, see the journal bergner-walker was published in
Fair enough. And I guess I'd answer that this question is somewhat a test of it: Can one use it to validate an arithmetic argument by non-arithmetic reasoning?
If it can, then it's got something to it. If it can't, then I'm back to being 'eh, groupoids w/e' about it.
@MikeMiller I cannot say anything about it as I'm not really into immersion theory, @MikeMiller. Maybe you'll enjoy math.stackexchange.com/questions/1807683 although you're probably aware of all things mentioned.
Say I have to calculate the expected value of the number of aces from a deck. Whereby I pick cards without replacement. The distribution of the number of the cards is hypergeometric.
E(X_j), where j is in {0,1,2,3,4,5}
E(X)=E(X_1+X_2+X_3+X_4+X_5)=E(X_1)+E(X_2)+E(X_3)+E(X_4)+E(X_5) by linearity
Now in the next step, E(X)=5E(X_1). I don't get this why is this True even though the trails are dependent.
Question:
Say, I have to calculate the expected value of the number of aces from a deck. I pick cards without replacement. Thus, the distribution of the number of the cards is hypergeometric.
Formally,
X=#aces
Let X_j be indicator of j^{th} card being an ace
E(X_j), where j is in {0,1,2,3,4,5...
@TheQuantumPhysicist If $f$ is a map from $X$ to $Y$ where we have an operation $+$ on $X$ and an operation $\cdot$ on $Y$ then we call it a homomorphism between these two tructures
@SteamyRoot that is not always a requirement of monoid homomorphisms (though it sometimes is)
Anyway, if this is a map from the reals to the reals, then one might call it exponential, though there are also maps satisfying it which are not in any way like the exponential maps
Which is why I'm looking for a proper name of this... it's an interesting property that's helpful in data modeling that I don't know what to call in simple terms
Probably if it is related to data modelling, it really will be an exponential map if it satisfies this (and is from the reals to the reals). As the other such maps are just too weird
I'm told to find a set of generators and relations for $S_3$. I can't imagine what kind of generator would produce a permutation that switches the order of a set. Like $\{1,2,3,4\} \to \{1,3,2,4\}$ isn't a simple 'move everything to the right' or 'flip the order backwards' like in dihedral groups. Any advice?
(not that dihedral groups had cycle notation but rotations and reflections are analogous?)
Beginners in calculus may enjoy the following problem:
Compute the following integral:
$$\int_0^1 \int_0^1 \frac{\displaystyle(1+y) \log\left(\frac{1+x+y-xy}{1-x+2y+y^2-xy^2}\right)}{(1+y)^3+2x(1+y)(1-y-y^2)+x^2(1-y(3-y-y^2))} \ dx \ dy$$
(Since it is intended for beginners, it is of c...
@Obliv My guess is this: after seeing pretty many textbooks, books, I realized that one is formed to think in a certain way, and especially while following a certain way taught in the education system. At the beginning you have a larger freedom degree to see things that later you probably won't see anymore because of the way you are taught to see at things.
@Obliv So, I suspect there is a higher probability to see simple things when you are a beginner than later after you are pushed to follow certain directions. It's about much freedom in thinking.
Hi, I have this question: "Show that the simple extension $E = \mathbb{Q}(\sqrt[6]{2}) \subseteqq \mathbb{R}$ of $\mathbb{Q}$ has intermediate fields $\mathbb{Q} \varsubsetneqq F \varsubsetneqq E$. My guess is use $f(\alpha) = 0$, then th $\text{Irr}(\alpha:\mathbb{Q})$ gives me $X^{6} - 2 = 0$, now $[E(\alpha):E] = \text{deg} q$
@user1618033 I could not copy paste that integral successfully into Mathematica. If I could I would have Mathematica evaluated it numerically, and look up the answer in the inverse symbolic calculator.
@user1618033 I guess if english isn't your first language I see why you would make the mistake of labeling a problem like that for 'beginners'. A beginner in anything will not be able to perform many tasks at all since they are new to the subject of study. I think it's not a big deal but it is kind of misleading if many actual beginners to calculus click on that and see a double integral :p
@Obliv That integral is pretty easy just to note something a beginner might note easier because of the freedom of thinking. A beginner might not mean a very beginner (with 2,3 days of practice on integrals).
@MatsGranvik Yeah, it is. Well, my integral is not about bragging, but more like an experiment, I wanna see that beginners may note something experienced users cannot.
My problems differ much from all you usually see in other parts, because I simply don't use the style of other mathematicians, I simply create and solve problems in my own style, based on my research, not on what other say in textbooks, books, papers.
I don't copy anyone, my aim is to be completely original in the creational process, to come up with a new breeze in mathematics.
So, $\{\vec{x} \in \Bbb R^6 : x_1 x_6 - x_2 x_5 + x_3 x_4 = 0, \|\vec{x}\|^2 = 1\}$ is diffeomorphic to $S^2 \times S^2$.
But wasn't $x_1 x_6 - x_2 x_5 + x_3 x_4= 0$ the equation of the Grassmannian Gr(2, 4) as a projective variety? So this should mean link of Gr(2, 4) around 0 is $S^2 \times S^2$. Is there's a direct way to see this? I guess not, but asking anyway.
@MikeMiller oh, Qiaochu put up an answer to that MO question of mine. But it's the kind of answer that makes me less hopeful that there's any 'natural' solution of the kind I'd hoped for.
My thought was: If $a_0 \neq 0$, then scale it to $1$ and pick $a_1$ up to $a_n$, which gives you $n$-times $q^r$ picks, so $q^{rn}$. When $a_0 = 0$, repeat for $a_1 \neq 0$ to get $q^{r(n-1)}$. When $a_1 = 0$ etc. etc etc
This would give you $q^{rn} + q^{r(n-1)} + \ldots + q^r + 1$ points, right?
@Krijn $\Bbb P^n = (\Bbb A^{n+1} \setminus 0) / k^\times$, so if $|k| = q$ (I don't like your notation; say $\Bbb F_{q^r}$ if you like, but not $\Bbb F_r$!), then the cardinality of $\Bbb P^n$ is $(q^{n+1}-q)/(q-1)$
Well, that part cleared itself in the end, as it turned into $\sum_{r = 1}^\infty \sum_{k=0}^{n-1} q^{rk} \frac{t^r}{r} = \sum_{k=0}^{n-1} \sum_{r = 1}^\infty \frac{(q^kt)^r}{r} = \sum_{k=0}^{n-1} -\log(1-q^kt).$
the question isn't whether the sums do what you say they do, but why there are logs and $t^r/r$ showing up in algebraic geometry :) especially since implicit here is that $t$ is a small number
by this: (2) for any field $F$ let $F^{\times} = F - \{0\}$. <- does this mean the set of multiplicative inverses of $F$ is equal to the set $F - \{0\}$ or does it simply define $F^{\times}$ as the set $F$ without the element $0$? i'm guessing the former , right?
@MikeMiller OK, I think I am more comfortable with the material in chapter 1 G-P now. Is there a particular exercise you'd want me to do, or shall I move on?
@Krijn Mike's method is cleaner, but it's worth noting what you did was - as I suggested - using $\Bbb P^n = \Bbb A^{n+1} \cup \Bbb A^n \cup \cdots \cup \Bbb A^0$ and then computing cardinality of both sets.
re:first three section - I thought inverse function theorem for 1-1 functions on compact submanifolds was interesting, although not very hard. I think that was in section 3. Similar idea is used to prove injective maps are stable.
Oh, yeah, I do agree. My natural approach was to look at the (r+1)x(r+1) minors, but I couldn't fix it (there are too many minors, so that can't possibly work).
5.9-11; 6.8-11; look to be plenty in 7, so pick your favorites from the latter half; 8.8, 8.11-8.15
There's a workable approach in what you're describing.
The only key point is that there is no global function that cuts out the matrices of rank r. (Do you see why?) So you need to do it locally, for matrices of a particular nice form: the open set of matrices s.t. the top left has rank r.
If you are not feeling challenged, you can read Hirsch either as a substitute or in parallel.
It is not immediately obvious to me why there is no global function that cuts out the matrices of rank r (by the I suppose you're asking why it's not preimage of a point by some submersion at that point). Should it be?
I once found a cute, although far fetched explanation to that. Torsion in diffgeo of curves measures how much "flat" a curve is: if torsion is 0, the curve is flat, i.e., planar. Torsion-free modules (or in particular abelian groups), on the other hand, are flat, and vice versa.
Sure, the fibers of the G-bundle are reference frames at each point (for the GL_n-bundle corresponding to a vector bundle, literally frames) so gauge invariance is just a statement that you're independent of reference frame, yes?
The relation between what physicists call a gauge transformation ("well ehh my field just transforms like $\psi\mapsto \rho(g)\psi$ or, if it's in the adjoint rep, like $A\mapsto g^{-1}A g+g\partial g$") and what the usual mathematical definition is
I was bothering about the regular value more than the preimage part, oops. Thanks for the clarification. So, yeah, it is obvious and passing to an open set makes sense.
Question:
Say, I have to calculate the expected value of the number of aces from a deck. I pick cards without replacement. Thus, the distribution of the number of the cards is hypergeometric.
Formally,
X=#aces
Let X_j be indicator of j^{th} card being an ace
E(X_j), where j is in {0,1,2,3,4,5...
Beginners in calculus may enjoy the following problem:
Compute the following integral:
$$\int_0^1 \int_0^1 \frac{\displaystyle(1+y) \log\left(\frac{1+x+y-xy}{1-x+2y+y^2-xy^2}\right)}{(1+y)^3+2x(1+y)(1-y-y^2)+x^2(1-y(3-y-y^2))} \ dx \ dy$$
(Since it is intended for beginners, it is of c...
