Yeah, see, I was trying to install that but I got all tangled up in the virtual box setup. It kept telling me that the compressed image was corrupted and I couldn't import the .ova file
I gave up after around 40 minutes of not finding the answer online
Just in case anyone is wondering; I found something online by Sage called SageCloud; it's online cloud-saving Sage programming! Try it out if you're interested - it works for me.
Hey everyone, just a quick question. I'm reading through Rudin and I noticed that when Rudin discusses continuity of some function $f : X \to Y$, he notes that $X$ needs to be a compact metric space for $f$ to be continuous, but he doesn't mention the compactness of $Y$. Surely if $Y$ is non-compact then $f^{-1}$ doesn't exist?
As there would be elements of $Y$ that we couldn't map to $X$
I've been self-studying Information Theory, mainly from Ash and Shannon-Weaver, Roman. Why we assume for the development of the theory that the probability of error for a single random variable trasmitted is $p < 1/2$?
A reason I think may be is that the entropy $H(X)$ for the random variables $X_1,X_2, ..,X_n$ has a bound $H(X) \leq logn$, where equality holds for $p_i = 1/n, ~ \forall i=1,2,..,n$ e.g when tossing a fair coin. If the trial of error is $p=1/2$ we get maximum entropy, so there's no reason to work for $p> 1/2$
> A topologist is drinking tea from a cup when suddenly the handle drops off. The topologist is amazed: the new shape is different but he can still drink tea from it. And so he does until the bottom of the cup drops off. Now he is totally befuddled: the shape is equivalent to the original one but how can he drink his tea now?
Let $f(x) = \sum_{i=0}^n a_ix^i$ be a polynomial with nonnegative integer coeffiecients $a_i$, i.e. $f \in \mathbb{N}_0[x]$. How many times must you evaluate $f$ in some $x$ to find all coefficients?
Not tautological bundle, sorry. The dual of that ($O(1)$). Because $T\Bbb{CP}^n$ is stably direct sum of a bunch of those over $\Bbb{CP}^n$, and if I know it for $n = 1$ I can pullback.
Hi everyone :-) I have a quick question. My textbook has given that three lines $a_1x+b_1y+c_1$, $a_2x+b_2y+c_2$, and $a_3x+b_3y+c_3$ are concurrent, iff, for three constants $A, B$ and $C$, not all three zero $A(a_1x+b_1y+c_1)+B(a_2x+b_2y+c_2)+C(a_3x+b_3y+c_3)=0$. I was wondering why they have explicitly mentioned that all 3 constants can't be zero..?
@DHMO Yeah, no, I get that. Hang on, I thought I had thought this question through! I mean, of course all 3 can't be zero at the same time but...damn! I completely forgot what my original question was! Ugh. OK, I'll ask again if it comes back. Sorry :/ >.<
@MikeMiller I think I did it for tangent bundles. But if $w_2(O(1))$ was $0$, then $w(T\Bbb{CP}^2) = w(O(1) \oplus O(1) \oplus O(1)) = 1 \cdot 1 \cdot 1$ would have been $1$, clearly false as the top SW class of CP^2 is $\chi(\Bbb{CP}^2)$ mod 2 = $1$, right?
So $w(T\Bbb{CP}^n) = (1 + \alpha)^{n+1}$ where $\alpha$ generates $H^2(\Bbb{CP}^2)$. Doesn't that have the same S-W numbers as $\Bbb{RP}^n \times \Bbb{RP}^n$ (in which case the polynomial is $(1 + 2\alpha + \alpha^2)^{n+1} = (1 + \alpha^2)^{n+1}$ - $\alpha$ generator of $H^1(\Bbb{RP}^1)$)?
For thirty different reasons. The product of manifolds is orientable iff both factors are orientable. The Steifel-Whitney class of a product of manifolds is the sum of the pullbacks of the SW classes of each factor. For $w_1$, this says it's the sum of the $w_1$s from before, which belong to different factors of the direct sum decomposition of $H_1$. Etc.
Have a good joke? Share.
I know this is subjective, but the principle "should be of interest to mathematicians" trumps. (I hope.)
