I am trying to teach a teenage person math, but he doesn't seem to be able to grasp the concept of negative numbers and 0.
Again and again he finds -4 greater than -3 because he has spent several years seeing 4 greater than 3. Similarly he has never experienced adding 0 to stuff and stuff to 0 ...
@ಠ_ಠ I don't even think we've got one-star hotels. But there are places called "guest houses" that are more like homes. They're not expensive and the food is a lot like homemade food, which again, is clean.
You just don't want to eat at cheaper restaurants and on the street.
@MattN. I'm not sure what the question is. If $\{e_1,\dotsc,e_n\}$ is a basis of a vector space $E$, then every $x\in E$ can be written in a unique way as a linear combination $x = \sum\limits_{i=1}^n \alpha_i(x)\cdot e_i$ of the basis vectors (by definition of basis). It's an easy verification that the coefficient functionals are linear, I don't think there's a (named) theorem for that.
@DanielFischer I don't get it. Is it that obvious that given any $x$ then for any $\lambda \in \mathbb C$ there exists a linear functional $\alpha$ with $\alpha (x) = \lambda$?
@MattN. No. You need $x\neq 0$ to get any $\lambda\neq 0$. But that's not the point here. By definition of a basis, we have $n$ coefficient functionals $\alpha_i \colon E\to \mathbb{C}$ such that $x = \sum_{i=1}^n \alpha_i(x)\cdot e_i$ for every $x\in E$. Then observe that these functionals are linear.
@Hippalectryon honnêtement, pas la peine de se prendre la tête avec la démonstration de la théorie de la dimension qui ne servent strictement à rien par la suite
@DanielFischer Okay, I don't get it after all. I don't understand how the definition of a basis implies the existence of coefficient functionals. I only see that this is true in an inner product space where the coefficient functionals are the inner product with the basis vector $e_i$.
In fact I have never heard of coefficient functional before.
Maybe they are defined to be the maps that map the vector to the $i$-th coefficient.
@MattN. The definition of basis says that for every $x$, there is a unique $n$-tuple of coefficients. That gives you maps, of which you don't a priori know that they are linear. Then you prove linearity.
@JasperLoy I have used my textbook and it has been finished!In addition, some problems in my textbook don't have anything to write home about.Furthermore, I'm looking for a organized problem set in multivariable calculus to help me review stuff I've learned if I face lack of time due to the fact of focusing on the other areas!
@MrWho What I mean is, different places will use different books on calculus and will cover different topics and use different notation, so your question is vague =)
@JasperLoy So you're telling me people know different types of calculus and knowledge has been become private and people develop different notations, what a funny remark!
@JasperLoy Calculus is calculus, problem set is a problem set, nothing differs when you're doing a multiple integral and mathematical perspective do NOT change.
@JasperLoy I'm thinking of it, however, I don't want to become stagnant after these tons of practicing that I've made.Shortly, I want to still practice and learn new things from basics.
If anybody lurking has thoughts: I am wondering how to argue $L|K$ separable iff $L|M$ & $M|K$ separable for any $L|M|K$. If $L|K$ is finite then it fits into a tower of simple extensions and by counting embeddings we obtain the lemma that $L|K$ separable iff $\#\hom_K(L,\overline{K})=[L:K]$. This lemma doesn't apply to infinite extensions though. My definition of separability is $L|K$ separable if $L\otimes_K\overline{K}$ reduced.
And one more thing: for per parts for Riemann integral, we require the existence of both $\int_a^b f'g$ and $\int_a^b fg'$, whereas for improper integrals, the theorem says that if one exists, also does the other... Why?
In combinatorial game theory, the Sprague–Grundy theorem states that every impartial game under the normal play convention is equivalent to a nimber. The Grundy value or nim-value of an impartial game is then defined as the unique nimber that the game is equivalent to. In the case of a game whose positions (or summands of positions) are indexed by the natural numbers (for example the possible heap sizes in nim-like games), the sequence of nimbers for successive heap sizes is called the nim-sequence of the game.
The theorem was discovered independently by R. P. Sprague (1935) and P...
I don't know the argument, everywhere I look assumes the extensions are finite, but a few places mention that separability can be defined for infinite extensions too but don't extend the facts for finite extensions to infinite extensions.
@EnjoysMath Not quite. You have no natural ordering on the indices, so one usually takes summability, which in this setting is most likely the same as absolute convergence.
@seaturtles Consider the transcendental galois extension $\Bbb Q(x_1, x_2, \cdots, x_5)/\Bbb Q(e_1, e_2, \cdots, e_5)$, where $e_i$ are the elementary symmetric polynomials of $x_i$. take another similar extension $\Bbb Q(y_1, y_2, \cdots, y_5)/\Bbb Q(e'_1, e'_2, \cdots, e'_5)$ and let the galois groups be isomorphic. Is it possible to show that the pair $\{x_i, y_i\}$ is $\Bbb Q(e_1, e'_1, \cdots, e_5, e'_5)$-algebraically dependent?
@MikeMiller hmm, the argument I know uses the fact that $\#\hom_K(L,\overline{K})=[L:K]$ iff $L|K$ separable in order to prove transitivity, but this doesn't extend to infinite extensions.
@BalarkaSen when you speak of the pair $\{x_i,y_i\}$, the index $i$ is fixed right?
@seaturtles IIRC separability for algebraic extensions is equivalent to every element having separable minimal polynomial. are those only equivalent for finite extensions?
I don't get how he got that result for the repeating integral at the end. From the first integral he gets $-\sqrt{1-r^2}$, if he places in the required r value, he gets... ah. wait..
@seaturtles In fact, I'd guess it's even true in general with base field $\Bbb Q$ replaced through some algebraic number field. Do you have any idea how to approach this one?
@JasperLoy If you think about all the things that happened before you came into existence and all the things that will happen after you disappear then this is not really all that important.
There is some thing I saw on the interwebs and it went something like this: There is a bit of truth in every joke and a bit of fuck you in every 'whatever'. : )