Two questions: Do any of the two sites hosting the Inverse Symbolic Calculator work for anyone ? Does anyone know of a downloadable version of this very helpful tool ?
@r9m Cody asked me about that integral about a week ago. But I haven't thought much about it. Did you see those two integrals I posted 5 days ago? They're really not that difficult to evaluate (especially using contour integration), but I thought they looked cool.
@Lucian, you can do a little yourself with the PSLQ algorithm and later variants, gp-pari has it. However, the value of ISC is the extremely large list of comparison numbers, maintained by Carma at Newcastle. Not something we could do ourselves.
your notation doesnt work in chat. have no idea what you're typing. but my point stands, for degrees, wolfram alpha yields approximately .9984 for cos(pi) and approximately i0.05477 for sin(x). Adding these two yields 1.053275
For latex look at the starboard . By that I mean to say that you will see that at the right hand corner of the screen there are starred messages. Click on the first one. It says Chat guidelines....
pi is in radians Pi radians = 180 degrees. Obviously it won't be true for pi degrees . pi is approximately 3.14 . so you are just saying 3.14 degrees which will not give you the desired result.
@BalarkaSen Well I guess my question is two-fold, what does $x^2+x+1$ look like as a product of linear factors over its splitting field as an extension of $\Bbb F_2$ and what does the splitting field as an extension of $\Bbb F_2$ look like as an $\Bbb F_2$-vectorspace. Hopefully that makes sense, I am significantly more fatigued than yesterday
@Chris'ssistheartist looks fun! I'll give it a try :-) I am trying to figure out Cody's integral (the one I asked you about the other day) .. I have a tedious approach in mind, I'm looking for short/slick approaches :)
@Rememberme The ability to see beyond the world of the living ?! :P you can't call that a disease :P
@PeterTamaroff Yes, some examples are $\Bbb R$ with basis open sets $[a, b)$, the long line, and $\Bbb R$ with basis open sets of the form $U - \text{countable set}$, where $U$ is open in the standard topology
In Spivak's Different Geometry volume 1 in the appendix we see that requiring a manifold be Hausdorff and metrisisable are equivalent conditions, There are other such conditions.
So yeah screw you @GaloisintheField - as you can see I didn't lie. Don't accuse me of doing that again.
I wouldn't mislead when it comes to something serious @GaloisintheField - you were really offensive.
@GaloisintheField I'm not, I can take a picture of the page in the book if you like.
@GaloisintheField no it doesn't, it says the "manifolds we'll deal with" satisfy that, notice it uses metric ones too, not topologies. If you look in the appendix at the back ....
I haven't got the book to hand, and I don't recall the exact definition of closed (other than it returns to where it starts) so if that isn't it. I'll need the definition and 3 minutes.
I vaguely recall existence of something like "ExerciseWiki", where goal was to collect various mathematical exercises (and solutions). Do I remember it wrong? Or does anybody remembers seeing such thing (and, ideally, has the link)?
maths.kisogo.com/index.php?title=D-system I try and do stuff like this. I found 2 definitions, they're listed, with references and proved equiv. The search is now at the point where it is actually worth using though. Which I am quite proud of.
@Danu I would either go with some space derived from $\omega_1$ or with some weak or weak* topology. Whatever you feel more comfortable working with. You can find some posts on the main with some examples. I listed a few of them here.
Lastly, maths.kisogo.com/index.php?title=Topological_property_theorems I've given an index (as best as one can) to theorems, so say you know something about your space and you're looking to show something else, you can look up in these (user-sortable) tables given in dictionary order (dictionary as in on the keys, not name)
@Chris'ssistheartist that last form can be simplified to $\displaystyle \frac{22}{9}\zeta(3)-\frac{4\pi\sqrt{3}}{27}\psi_{1}(1/3)+\frac{8\pi^{3}\sqrt{3}}{81}$
@r9m Maybe he has a nice way he doesn't share, just test it.
Even affected by my condition I crawled myself anf got a lot of new identitites and other stuff, but I need more time to recover before being like an A-bomb for each integral, series, limit I meet. :-)
(especially some clever ones involving the root of unity)
I've been working on a set of problems for the past 6 hours and I'm about to explode from frustration! Either my book has errors, wolfram alpha is broken or integration by substitution does not work! Help!!
I'm trying to solve the following ODE:
$$
\frac{dh}{dt}\; =\; \frac{1\; -\; 20kh\; +\; kh...
@r9m I've been waiting a bit for some stuff to be published that is reference in my book too. It's awul how long you might expect for something to be published.
@r9m No, next year. With 300 problems all would have been already finished, just waited for the publishing moment. Even if you're done with your book, you need to wait until all your book is revised.
The process of publishing is not that fast, think that for some articles I waited for more than 6 months, and it was just an article. ;)
@r9m lol, but who cares that? I still have no published book. ;) Well, patience, patience, patience ...
Why? For a simple reason: I work hard ceaselessly. While others talk about known problems in mathematics, or about philosophy related to mathematics and other stuff like that, I work extremely hard and get the results I also talk here once in while.
@BalarkaSen Were you willing to continue our discussion on the splitting field of $x^6+1$ for $\Bbb F_2$ whenever you are online? I'll be in the Linear & Abstract algebra room
Initially I wanted to write something like "While others talk about my mathematics ... and the rest ...", but I just don't wanna seem wrongly perceived. ;)
@GaloisintheField Well, if $\alpha$ is a root of $x^2 + x + 1$ in the splitting field of the polynomial, then the splitting field looks like $\Bbb F_2(\alpha)$. What else can be said? The polynomial factorizes as $(x + \alpha)(x + \alpha^2)$. You can provide an explicit isomorphism between the splitting field and $\Bbb F_4$, of course.
If the square root of non-perfect squares are all irrational, then does this definition continue to the cube root of all non-perfect cubes are irrational?
@r9m My point is, if you have really thought about what life is about, you will have found that there isn't really a point, and you are left with assigning your own point, which the common choice is enjoyment. If enjoyment is your life-value-metric, then from my observation, dumb people are usually winning.
@GaloisintheField In terms of enjoyment I'm hard to beat when a attend some integrals, series and limits and make some discoveries ... :-) It's about music, dancing, jumping around, shouting, screaming, drinking champaigne ...
that was a quote from an onion video about a football team's pre game coin toss that makes them realize "the randomness of life and the triviality of their own existence"
I hate the bar, because bar is used for so many other things in mathematics (e.g., closure). But some intro to higher math texts use it for complement, for example. I certainly would never, never use it.
@Ted BTW I talked to anon the other day because of identifying $H^2$ with upper triangular matrices
@Ted I looked up QR. Now if I want to identify SL(2,R)/SO(2,R) with upper triangular matrices with positive diagonal entries, wouldn't I have to show that if I take a matrix A in SL(2,R) with A = QR, and multiply it by a matrix B in SO(2,R), then AB = Q'R for the same R?
In the QR decomposition, the orthogonal matrix goes on the left, so you are actually identifying SO(2)\SL(2,R) with upper triangulars with positive entries. Anyway, if G=HK for trivially intersecting subgroups H and K, to show H\G can be identified with K do you think you have to show for all g=hk that multiplying on the right by some h' yields h'k for the same k? Why would you think that?
To show SO(2)\SL(2,R) can be identified with upper triangular matrices, it suffices to show that SL(2,R)=SO(2)*B (where B=upper triangular matrices with positive diagonals) and SO(2) intersects B trivially. This is true more generally: whenever a (lie) group G acts (smoothly) on a set (space) X, if H is a point-stabilizer and K is a subgroup that acts transitively, you can prove G=HK (or G=KH, they're equivalent) and H,K intersect trivially.
I'm thinking of a map $SL2/SO2 \to U2$. The version $SL2 \to U2$ is already well defined by the QR decomposition and requiring positive diagonal. now if I look at an equivalence class, the map shouldn't depend on the representative and quotienting groups is done by right translation, no?
It looks like the only problem is you're multiplying by B on the right instead of the left, whereas like I said QR puts the orthogonal matrix on the left so we're talking about the right coset space SO(2)\SL(2,R).
Say X and Y are in the same right coset of SO(2). Then X=AY for some A in SO(2). Write the QR decompositions X=QR and Y=Q'R'. Then QR=(AQ')R'. Since QR decompositions are unique, Q=AQ' and R=R'.
Try to prove what I said about group actions in general though: if G acts on X, H is a point-stabilizer and K a transitive subgroup, then G=HK and H,K intersect trivially. For example, let E(2) be the group of rigid motions acting on R^2. Then H=SO(2) is the point-stabilizer of 0, and K=(R^2,+) acts by translations. Then the decomposition E(2)=HK means that every rigid motion is uniquely a rotation around 0 followed by a translation.
err, actually a translation follows by a rotation around 0 sorry
err, I want more than transitive
a regular group action
so that K can be identified with X
another example: $G=S_n$ acting on $\{1,\cdots,n\}$. We can regard $H=S_{n-1}$ as the point-stabilizer of $n$, and $K=\langle(12\cdots n)\rangle$ a subgroup which acts regularly on $\{1,\cdots,n\}$. Then $G=HK$, so $K$ is a right transversal for $H$.
you can look it up, but the way I think about it is: $K\curvearrowright X$ regularly if $K\cong X$ as $K$-sets, ore equivalently if for all $x\in X$ the map $K\to X$ defined by $k\mapsto kx$ is a bijection. thus, $K$ can be "identified" with $X$. For instance, the cyclic group $C_n$ acting on $\{1,\cdots,n\}$, or a vector space acting on itself by translation.
@Krijn keep some problems at the back of the mind, start off doing it in between other works, as anon said. most of the time, subconsciousness helps you to do mathematics. this is a well-known theory proposed by Poincare.
some people make serious progress by sleeping with a problem on the mind, letting the subconsciousness do the thing. this might not work universally. e.g., i have never done mathematics in my sleep. but i can make nontrivial progress by walking with a problem in the mind and never thinking about the problem at all
One of my peers is studying for the GRE and ran into the following problem:
When is $|x-4|$ equal to $4-x$?
I tried to attempt teaching this by developing the intuition for why
$$|x| = \begin{cases}
x, & x > 0 \\
-x, & x < 0 \\
0, & x = 0 \end{cases}$$
by taking positive numbers, negative n...
@robjohn as you mentioned ,, that is exactly the Hlawka Inequality :-) and as for the resemblance to IE formula ,. the inequality is not true for $4$ or more variables in general :)