If you do it the way Robert did, you'll have the same number of forms as complex dimension, so Nirenberg tells you that you get a new holomorphic structure on your space making all the $1$-forms type $(1,0)$.
I now understand Nirenberg's Theorem. How can it be that it doesn't appear anywhere modern in neat print?
Hi I want to compute cubic equation. the first thing I think about it is to convert it to quadratic equation. Now, I need to find at least one factor for the equation, it seems that the equation has no integer value. I wonder what will you do if you have this case, the equation is as following: x^3 - x^2 - 1 = 0
Hey all! Is there an appropriate room to discuss: https://math.stackexchange.com/questions/2625467/algorithm-to-tell-if-an-infinite-sequence-is-random-or-not
I'm having troubles to prove that the following space is not a topological manifold:
Let $r:S^1\to S^1$ be a rotation of $\frac{2\pi}{3}$, i. e., $r(\cos\theta,\sin\theta)=\left(\cos\theta+\frac{2\pi}{3},\sin\theta+\frac{2\pi}{3}\right)$, such that $X=B^2/\sim$, where the partition is given by
$...
Is a sufficient proof under the axioms of integers that $a = (-(-a))$? By additive inverse, $a + (-a) = 0$. Let $\alpha = (-a)$. Then $\alpha$ $\in$ $\mathbb{Z}$. Then by additive inverse, $$\alpha + (-\alpha) = 0$$ Which is identically $$(-a) + (-(-a)) = 0 = a + (-a)$$ Subtracting $(-a)$ from both sides gives us $$(-(-a)) = a$$.
Intuitively, an algorithmically random sequence (or random sequence) is an infinite sequence of binary digits that appears random to any algorithm. The notion can be applied analogously to sequences on any finite alphabet (e.g. decimal digits). Random sequences are key objects of study in algorithmic information theory.
As different types of algorithms are sometimes considered, ranging from algorithms with specific bounds on their running time to algorithms which may ask questions of an oracle machine, there are different notions of randomness. The most common of these is known as Martin-Lö...
@user777 you could look at the derivative, which is positive except on a small intervall and then use the monotonicity on suitable intervals and the intermediate value theorem
Sorry for re tagging but they are in seperate lines ... also id prefer to chat here before i respond to ur comment: @TheGreatDuck So all possible infinite sequences can be expressed as a functions ?
@ted so I'm trying to prove that $ab = (-a)(-b)$. I was struggling to prove $(-1)a = -a$, so I was thinking I would circumvent doing so by letting $-a = x$ and $-b = y$. Then $(-a)(-b) = xy$ Then if we take $(-x)(-y)$, we have $ (-x)(-y) = (-(-a))(-(-b)) = ab$.
You say Voltaire, I say Pushkin, you say Sartre, I scream Dostoyevsky, you say French revolution, I retort, Bolshevism!!! you say Paris, I say Stalingrad, you say guillotine I say: GULAAAAG
@Ted hmm...Something along the lines of expanding to $(a+0)$ or something. Yeah this is from class. We did the $(-1)a = (-a)$ at the end of class, and I stupidly didn't copy everything down.
you need to formalise pattern somehow, probably as some kind of computable relation that can be obtained from the sequene after performing the row algorithm you did there
Oh no kidding? It's just a 1 credit "Intro to Proofs" seminar for the first five or six friday afternoons. Its ~ an hour drive to another school, but it breaks up the monotony and it's part of the "honors" program there, so it looks good on a transcript.
I'm having troubles to prove that the following space is not a topological manifold:
Let $r:S^1\to S^1$ be a rotation of $\frac{2\pi}{3}$, i. e., $r(\cos\theta,\sin\theta)=\left(\cos\theta+\frac{2\pi}{3},\sin\theta+\frac{2\pi}{3}\right)$, such that $X=B^2/\sim$, where the partition is given by
$...
I'll use cohomology with coefficients $\mathbb{Z}/2$ everywhere.
Suppose that the space $P=\mathbb{R}P^{n-1}$ embeds in $S^{n}$ (where $n>2$). Recall that
$$ H^*(P)=(\mathbb{Z}/2)[x]/x^{n} = (\mathbb{Z}/2)\{1,x,\dotsc,x^{n-1}\} $$
By examining the top end of the long exact sequence of the pai...
We didn't do semidirect products in the lectures. We were given the definition on an exercise sheet and we had to work out the details, like the splitting criterion etc., but were expected to know them after that
@Ted Ok I think I have it! My prof actually played with the $0\cdot a = 0 \forall a$ proof, so assuming that, I have $(-1)a$$(-1)a + 0a$ $= (-1)a + 0(-a)$ $= (-1)a + (1 + (-1))(-a)$ $(-1)a + 1(-a) + (-1)(-a)$ $(-1)(a + (-a)) + (-1)a$ $(-1)0 + (-1)a$ $=(-1)a$
@Ted I didn't succeed at trying to formalize my idea about Darboux theorem which I was ranting about earlier. Do you want to hear the idea/already know it?
There are definitions of "fractal" for which the rationals might be considered fractal, though I think that most people who work in the field would dismiss those definitions
@XanderHenderson I see, so under the modern setting, for people working in the field, what properties do fractals usually have. Both you and balarka said they are not necessary scale invariant?
The two most useful tricks in all of analysis: add zero (often in the form of $x + (-x)$ or something similarly silly), and multiply by one (for example, $\dfrac{x}{x}$)
And consider some homomorphism $\phi:K\to \text{Aut}(H)$. We can construct a group operation on the Cartesian product $H\times K$ by $(h,k)(h',k') = (h\phi(k)(h'),kk')$
Now, turns out these two things are the same. In the abstract semi-direct product of $H$ and $K$, the group is the internal semi direct product of the subgroups $H\times \{0\}$ and $K\times \{0\}$
And if you know $G = HK$ where $H$ is normal and $K\cap H = \{e\}$, then $G$ is isomorphic to the abstract semi-direct product of $H$ and $K$ where the homomorphism is $k\mapsto k^{-1}(\cdot )k$
@ted Ok, so $0 = 1 + (-1)$. Multiply by $(-a)$ to get $0(-a) = (-a)(1 + (-1))$. Distribute to get $0 = (-a) + (-1)(-a)$. Then $(-a)$ must be the additive inverse of $-(-a)$, so then $(-1)(-a)$ must be identically equal to $a$?
algebraic number theory is really beautiful stuff. It combines many different aspects of mathematics and some of the result are just to good to be true
algebraic number theory isn't that demanding, you just need abstract algebra, commutative algebra, non-commutative algebra, group cohomology, representation theory, abstract harmonic analysis, K-theory, complex analysis and scheme theory
It's a thing from this call of duty game, never played that part at least but one of your comrades dies and you have to press F on your controller to pay respects. I guess the absurdity of it all made it a meme
@XanderHenderson it takes a lot of work to get to the frontiers of research in most subjects (if it's not obscure). A friend of mine proved new stuff in his bachelor thesis on ANT, so it's not as hard as you make it out to be
Clockwise from top left 1. It's some drawing made by someone, that guy did not include a generation gormula, but it seemed to be based on the mandrobrot set 2. Binary tree 3. Kosch snowfake, skeriski triangle and other related objects 4. Graph of Thomae function
@Secret I know what a binary tree is, I'm just saying that I don't think that there are many notions of fractal for which a binary tree would be considered fractal
@XanderHenderson Hmm, based on what you said, it seems that what most people in the field think are fractals does not necessary need to be self similar nor generated from a recursive process
@Mathein but yeah the GRE that you have to do for entry into grad school seems to be just a bunch of computational calculus and linear algebra, /maybe/ with some fancier topics thrown around
@Daminark when I took the math subject GRE, it was about 50% stupid integration problems that required knowing a trick (or were otherwise quite computationally intensive), about 25% linear algebra, with the rest of the exam filled out with a hodgepodge of other more interesting questions
Like I can do single variable stuff alright (at least the standard bits, but all the really sneaky stuff... maybe not), but the geometric business and parametrization from multi just goes way over my head
Can a nonclosed open subset of a $T_1$ topological space be compact?
I mean an open compact set which is not clopen.
