and so, a line in projective $2$-space is a point in dual projective $2$-space, therefore a pencil (set of lines) corresponds to a line (Set of points)
i understand, its just sometime confusing when points are lines and lines are points :P
Let $\{e_k\}$ be a sequence in X(normed space). Then it is said to be shauder basis iff for every x in X there exists unique scalar $\alpha_k$ s.t $x = \Sigma_{k = 1}^{\infty} \alpha_k e_k$
every hilbert space admits an orthonormal basis; if $K$ is the minimum cardinality of a dense subset of the hilbert space, there's an orthonormal basis of cardinality $K$
would the answer on this problem be considered an example or counter-example? I have seen it before on a similar question, but am having a hard time understanding how that answers the question...
More specifically than the title, let $\alpha : V \to V$, where $V$ is a finite-dimensional inner product space. If $ \|\ \alpha(v) |\| = \|\ v \|\ $ for all $v$ in some orthonormal basis for $V$, must $\alpha$ be unitary?
I believe $\alpha$ must be unitary with these conditions. However, I'm n...
@MikeMiller just saw a really cool theorem. For every functional on a hilbert space H such that f isn't 0 functional. There exists unique y such that f(x) = <x,y> for all x.
that is crazy.
That means that $kernel(f) = \{0\}$ for all $f \in dual(H)$ where f isn't 0 functional.
@AkivaWeinberger I should also note that the method I did for this one is rather different than I did for the squares; with that one I cheated by using GraphicsGrid, whereas here I used Translate to put each hex tile in the right place.
There's probably a clever way to make the chunk rectangular rather than a parallelogram, but I didn't want to think about that :P
Define $f:\mathbb C^2\setminus\{(0,0)\}\to\mathbb P^1$ by $(z_1,z_2)\mapsto[z_1^2,z_1z_2,z_2^2]$. Then $f$ cannot be extended holomorphically to $\mathbb C^2$.
So lift $f$ to its "blow up", i.e., $f:\tilde{\mathbb C}^2\setminus E\to\mathbb P^1$, where $\tilde{\mathbb C}^2=\{((z_1,z_2),[l_1,l_1])\in\mathbb C^2\times\mathbb P^1:z_1l_2=z_2l_1\}$ and $E=\pi^{-1}(\{(0,0)\})$, where $\pi:\tilde{\mathbb C}^2\to\mathbb C^2$ is the first projection map.
Can I now extend $f$ across $E$?
I observed that $\tilde{\mathbb C}^2=\{((z_1,z_2),[z_1,z_2])\}$ and $E=\{(0,0)\}\times\mathbb P^1$.
Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.
There is something i do not quite understand when it comes to learning math. There are problems were you just have to find connections and i can't understand how to do all of those connections. For example look at this question: http://math.stackexchange.com/questions/294213/prove-that-13-23-n3-1-2-n2
You can use the $(a+b^2)$ relation or the mean value of the first and last item in the sum.
This semester's culminated in deciding to begin studying for a given exam at 2 a.m. the night before. Working so far, but it's probably not the optimal choice...
What is the common definition, if I say "let $\pi \colon \widetilde{X} \to X$ be a universal covering", is $X$ already path-connected?
I would say universal coverings are defined as coverings where $\widetilde{X}$ is simply-connected, i.e. path-connected and $\pi_1(\widetilde{X}) = \{e\}$.
Okay, think I got it. If covering maps are assumed to be surjective, then a space $X$ admitting a universal, i.e. simply-connected, covering space $\widetilde{X}$, is automatically path-connected. Does this make sense?
@AntonioVargas But it feels like if i already know how to solve it it is not a problem. I dont like that i have to memorize it all if you know what im saying
@AntonioVargas For example in induction there is formulas to solve arithmetic sums and geometric sums, is it really about memorizing theese kind of things?
@AntonioVargas Im talking about $\frac{n(n+1)}{2}$ for arithmetic sum for example
@Rakso, I don't remember those formulas exactly, but I do know how I would derive them. It's not about memorizing the formulas, it's about learning what they mean and where they come from. The goal is to give you ideas to solve different but related problems.
This is an attempt to generalize this question. $$x^n-2y^n=1\implies \frac{x}{y}=\left(2+\frac{1}{y^n}\right)^{\frac 1 n}=2^\frac1n\left(1+\frac{1}{2y^n}\right)^\frac1n<2^\frac 1n\left(1+\frac{1}{2ny^n}\right)$$
$$0<\frac{x}{y}-2^\frac1n<\frac{2^\frac1n}{2ny^n}$$ but since the irrationality meas...
[Some interesting stuff (?)] $$\frac{d}{dx}x^n=nx^{n-1}$$ $$\int x^n dx=\frac{1}{n+1}x^{n+1}$$ Now $$\int x^{-1} dx=\frac{1}{-1+1}x^{-1+1}=\frac{1}{0}x^{0}=\color{red} {\textrm{undefined}}$$ $$\therefore\textrm{Define: } \ln x=\int_a^x t^{-1}dt$$ Where is the derivative analogue of this $x\neq -1$ 'gap' in the integral power rule? Actually it is right here: $$\frac{d}{dx}x^0=0x^{0-1}=0x^{-1}=0 \color{red}{\textrm{ (Often skipped as it is straightforward: $x\neq 0)$}}$$ NB $0^0=\textrm{undefined, but in most applications, it is natural to set as 1}$
@Secret because when you do $\int x^ndx=\frac{x^{n+1}}{n+1}$ you're looking for a function whose derivative is $x^n$, because of the fundamental theorem of calculus. What happens when $n=-1$ is simply that you find out that there's no function of the form $ax^b$ whose derivative is $\frac{1}{x}$
Yeah I know but in the past I am always being annoyed of this apparent lack of symmetry in the power rules of integral and derivatives, and then some time later I realise the symmetry is still there: it is being hid by zero in the derivative case
Perhaps a better way to phrase it is: We all knew the following:
> What happens when $n=−1$ is simply that you find out that there's no function of the form $ax^b$ whose derivative is $\frac{1}{x}$
but the above give a reason on why it is the case
I am pretty sure most people are aware of the why but for me, it just clicks recently, shwoing that I still need to work hard on my calculus knowledge
A similar kind of logic applies to the case on why there is no elementary antiderivative of the gaussian integral
One can say it is impossible to find an elemntary function whoose derivative is the gaussian function
but the more fundamental reason is because it is an outcome of Risch algorithm
This line of thinking is one reason I like universal algebra: It provide a pathway to explain why the axioms of the various algebraic system work the way they do and how at the most abstract level they interact with each other
The answer is: The derivative power rule does somewhat "break" at the n=0 case, it does not look like it because for $x\neq 0$ absorbption property of zero hid the signs of the "damage"
two variable limts give as another perspective on how it breaks as $\lim_{(x,y)\rightarrow (0,0)}x^y$ does not exists because you can get any value between 0 and 1 depending on how you approach the origin
More specifically than the title, let $\alpha : V \to V$, where $V$ is a finite-dimensional inner product space. If $ \|\ \alpha(v) |\| = \|\ v \|\ $ for all $v$ in some orthonormal basis for $V$, must $\alpha$ be unitary?
I believe $\alpha$ must be unitary with these conditions. However, I'm n...
