if $k$ is a field, $K/k$ is algebraic, and $\overline{k}$ is an algebraic closure, then $K/k$ is separable $\Leftrightarrow$ $K\otimes_k\overline{k}$ is reduced.
you have to remember that a topological space is it's own thing - in general, there's no "ambient space" it lives in. this space doesn't live in $\Bbb R^2$, it's its own thing.
I get the feeling that connectedness conveys- but something like [1,2] [3,4] is disconnected because it's the union of two disjoint open subsets of [1,2] [3,4]?
Please help with this question: Find a general term (as a function of the variable n) for the sequence {a1,a2,a3,a4,......}={8/3,64/9,512/27,4096/81,.......}.
Off topic: $k$ is an integer. If $f(\pi kx )$ has domain $[0,1]$ on the real line and $f$ (when defined on the real line) has zeros at all integer multiples of $\pi$, how many roots does f(\pi kx) have on $[0,1]$? My response: $\pi kx$ is an integer multiple of $\pi when $kx=j$ is an integer. Wouldn't the answer be infinity since there are infinitely many rationals between 0 and 1?
From the category of answers deserving more upvotes, consider this hint by Gerry Myerson with its awesome density of useful hints in just a few sentences.
@TheSubstitute Remember that $k$ is given, thus fixed.
@TheSubstitute The second caveat: Both endpoints $0$ and $1$ are included, which gives one extra repetition that needs to be taken into account.
@TheSubstitute Third caveat: You may need to also consider $k=0$ and $k<0$. The case $k=0$ may seem strangely out of place, but it is instructive insofar as it hints to the rationale of characteristic 0 in algebra.
Are there any issues with considering the 'class of all cateogries' or the 'class of all functors'? (Note that I don -note- speak of the category of all categories or the category of all functors)
@Mike: are you speaking with me? And why does it matter why I want to? The motivation is to have a mathematical object $Cl$ and and make "let $F$ be a functor" be written as let $F\in Cl$. But the question is a mathematical one.
@NiftyKitty95 Besides size, classes may have inclusion issues. I am definitely not an expert in these things, but I suppose that things are fine as long as everything can be reformulated in terms of the defining properties. That is, if the notion of class can be eliminated completely. But then, it's a non-issue. That might explain Mike's initial response.
Well, if you formulate a cateogry, e.g. the category of sets Set, then Ob_Set is not a set. Set isn't an object in ZFC of course. But if it's possible to formalize Ob_Set, then it must also be possible to formalize classes.
Sure, there are set theories in which one talks about classes. I know one is abbreviated as NGB but I don't know its name. You can also do it with Grothendieck Universes without leaving ZFC, IIRC.
In the former case you probably do run into issues with talking about a "class of all classes" just like you do with sets in ZFC. Russel's paradox is hard to do away with. In the latter you just don't mention classes.
All sets are classes. And from all the classes: A class is a set if it's in a another class
Implying there can be classes which just happen not to be member of another class and then these aren't sets
I want to put functors in a class - mostly because I don't know if they fit in a set, and because I don't quite know what a functor is, if I don't model it via set theory
I look for good definitions in the nLab, Wikipedia, I know Joy of Cats, Goldblatt, Simmons (informal)
it's for my own notes, I don't quite know under which "type" to introduce functors etc.
I don't mind forming the category of all functors, even if 'all functors' is larger than any the class of all my sets, but I just don't want to be formally inconsistent
or make a defintion so that what I naively set out to be 'the class of all…' or 'the category of all..' is in fact not a class resp. category.
I've proved that every (non-elemantry) tree has at least two maximal independent sets, with equality only for stars. I'm trying to extend that proof now for general graphs, with the assumption that diamater>=2 holds the same 'status'. Any suggestions/hints on how to approach this?
I've considered heading to the u,v vertexes and the path between them that forms the diameter, but not sure what to do with it actually..
If I have equation $y'(x)=y(x)+c e^{y(x)}$ and $y(0)=y'(0)=2$, can I just plug in that to get $2=2+c e^2 \implies c=0$, and then that my particular solution has to satisfied equation $y'=y \implies y=ae^x$ and from $y(0)=2$ I get $y=2e^x$. I think that this way isn't right?
I think I may try to prove this via construction. Every connected graph G with $diam(G)\ge 2$ has a maximum path (the diam path) which is also a maximal path...
Which is also bipartite
And I can construct the two independent sets on the path: All the even vertices on one hand, and all the odd vertices on the other hand.
Too clumsy to try adding to these sets the vertices outside the path. Gah, I'm lost..
In real analysis, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x.
The inequality states that
:(1 + x)^r \geq 1 + rx\!
for every integer r ≥ 0 and every real number x ≥ −1. If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads
:(1 + x)^r > 1 + rx\!
for every integer r ≥ 2 and every real number x ≥ −1 with x ≠0.
Bernoulli's inequality is often used as the crucial step in the proof o...
I've proved that every nontrivial tree has at least two maximal indepndent sets, with equality only for stars via the bipartition of the trees.
I am trying to extend that proof to general graphs, and I think the following is correct:
Every connected graph G with $diam(G)\ge 2$ has at least t...
@DanielFischer I see, thanks! In general, I've almost never figured out when he isn't on-line. My impression has been that he's around all the time. :-)
@PedroTamaroff Do you know the following lemma which says that If $ f\in C^2 , Df(x_0) =0, D^2f(x_0 \neq 0)$ then there exists a local diffeomorphism $\phi$ such that $f \circ \phi^{-1} (y) = \sum_{i=1}^n \pm y_i^2$ .
Define $y=F(x)$ as $y = (f(x) - f(x_0) , x_2- x_{20} , ..... x_n - x_{n0}$ where each entry is the diagonal entry of $n \times n$ , $x_0 = x_{10} , ...... , x_{n0}$ is a particular vector in $\mathbb R^n$ . my target is to show that if $det DF(x_0) \neq 0$ then $f\circ F^{-1} = f(x_0) + y_1$ . @DanielFischer I couldn't verify the last relation with the inverse i have computed .
@Complexanalysis Sorry, I don't get what is what. Start with $f$, that is a (continuously?) differentiable function on some open subset $U$ of $\mathbb{R}^n$, right?
@Complexanalysis Since $F$ doesn't change the last coordinates, $F^{-1}(y) = (g(y),y')$ for some $g$. So $f(F^{-1}(y)) = f(g(y),y')$. On the other hand, $(f(F^{-1}(y)),y') = F(F^{-1}(y)) = y$, so $f(F^{-1}(y)) = y_1$.
@Vrouvrou look, all I know about the $M_k$ and the $\beta_k$ are what you give in the question. I don't know what the Morse numbers are apart from that. From what you give in the question, I cannot answer the question.
When using the Cauchy Riemann equations for determining whether a function is differentiable, most of the books I read tell you to split the function into that $f(z) = u(x,y) + iv(x,y)$ thing, but is there a good reason for this? Can't I just take the partial derivatives of $f(z)$ along the two axes and get two complex numbers?
no it's juste that i dont understand in the answer given what it means by " To compute these Taylor coefficients use the know expansion $(1+t)^{-1}=\sum_{q=0}^\infty(-1)^q t^q.$"