If I wanted to solve a system of equations consisting of equations such as frac(1/x)+frac(2/y)+frac(4/z)=1 Would it be safe to assume that these are just 1/1(x) etc?
It is sad that this site is being overrun by spam. For the past few days I've been downvoting and flagging spam posts, but I need help from I think two other users. You can only flag five times per hour. I think that currently around 60% of the posts are spam. Thank you.
In how many ways can the number 1;000;000 (one million) be written as the product
of three positive integers $a, b, c,$ where $a \le b \le c$?
(A) 139
(B) 196
(C) 219
(D) 784
(E) None of the above
This is my working out so far:
$1000000 = 10^{6} = 2^{6} \cdot 5^{6}$ and then to consider th...
Remember the problem I asked you about regarding the convergence of characteristic functions. I discovered that the dyadic cubes are used in the standard proof of separability of Lp space.
@DanielFischer :) No that just offers hope, I still have to study it. I was just surprised that it was a standard idea used, it seemed very inventive. I found it online and in Brezis.
In here : "To the question "what is 2 + 3" a French primary school pupil replied: "3 + 2, since addition is commutative". He did not know what the sum was equal to and could not even understand what he was asked about! ". And here.
@DanielFischer To prove the last part of the MSE post of [Convergence of characteristic functions](http://math.stackexchange.com/questions/774552/convergence-of-characteristic-functions-on-hypercube), is the following fine: To show that $\int gf_{m} = \frac{1}{2}\int g$ for all m large enough. Would it be fine to simply note that by construction as you stated for $m$ large enough $O_{1,2}^{m} \cap A$ takes up exactly half of $A$.
Then taking $g \in A$ and assume that $g = \chi_{E}$ for some dyadic cube $E$, consider the sequence of characteristic functions $\{\chi_{O_{1}^{m}}\}_{m}$ then:
@LucioD Apart from the fact that you used $A$ to mean two different things, once a dyadic cube, and once the space of linear combinations of characteristic functions of such. Of course you need to say (and best give a reference) that the space of linear combinations of characteristic functions of dyadic cubes is dense in $L^1$.
@DanielFischer One of the $A$'s should be an $E$. I'm only trying to prove the last part not the whole question. I still have to study the density proof.
@DanielFischer Do you have a geometric idea of what the weak* convergence of these characteristic functions $\chi_{O_{1}^{m}}$ and $\chi_{O_{1}^{m}}$ mean, or is it just an abstract idea at this stage?
For the particular case of $\chi_{O_i^m}$, it means that $L^1$-functions aren't too irregular. Although putting that into words would tend to be either horribly imprecise, or effectively repeat the definitions ;) Generally, hmm, I go with abstract.
@Nick not true if were working with a magma of the set ${0,1,2,3,.....,120}$ with the magma operation "$+$" defined as $a+b=20ab \mod 121$, thus $2+3=120$
Please, tell me, what may mean notation $\tr F \wedge G$ for a Lie-algebra valued 2-forms $F$ and $G$? Should I take scalar product of coefficients with respect to the Killing form?
It is sad that this site is being overrun by spam. For the past few days I've been downvoting and flagging spam posts, but I need help from I think two other users. You can only flag five times per hour. I think that currently around 60% of the posts are spam. Thank you.
@Sush Because it uniquely determines the first and the second component of the pair. So it does what an ordered pair should do, hence we can take it as a definition. Note, however, that outside set theory nobody cares what definition is used, all one cares about are the properties ordered pairs have.
@Sush We have by the definition $(x,x) = \{ \{x\},\{x,x\}\}$. Now, since $x = x$, by definition of equality of sets it follows that $\{x,x\} = \{x\}$, and hence $(x,x) = \{ \{x\},\{x,x\}\} = \{\{x\},\{x\}\}$. Once more using equality of sets, $\{\{x\},\{x\}\} = \{\{x\}\}$, so finally $(x,x) = \{\{x\}\}$. We find the first component of the pair by looking at the element of the singleton set that $\{\{x\}\}$ contains. For the second component, either $\{\{a\},\{a,b\}\}$ contains two elements,
then the second component is the element of the non-singleton element of $\{\{a\},\{a,b\}\}$ that is different from the element of the singleton element (what a mouthful ;), or $\{\{a\},\{a,b\}\}$ contains only one element (that is the case here), and the second component is the same as the first.
So, to plot $\{\{x\}\}$ in the Cartesian plane, you go $x$ units to the right, and $x$ units up, and mark the point.
@Anthony Oy. You need a domain on which you can define a holomorphic function that does what you want (logarithm, roots, etc.). Now, the point of a branch-point (Ha!) is that you can't define a holomorphic function doing that in a punctured neighbourhood of the branch-point, or even in any annulus containing that branch-point in the inner disk, unless there is a corresponding other branch point (or several) that annihilate(s) the effect of the branch point also contained in the inner disk.
So you need something to prevent the existence of annuli surrounding the wrong number of branch-points. The simplest (geometrically) method is a branch-cut.
@Anthony Consider $\sqrt{z^2-1}$. You have two branch-points, at $1$ and at $-1$. Going around each of the branch points on a small circle (so small that it doesn't enclose the other branch point), has the effect of multiplying with $-1$, that is, if you start with one value and go around the branch point, when you complete the circle (the first time), you have the negative of the value you started with. Now, if you take a larger circle, surrounding both branch points, you get a factor of $-1$
from both branch-points, and the net-effect is that you come back to the value that you started with. The effects of the two branch-points cancelled.
So you can, for example define a holomorphic branch (two, exactly, differing by a factor of $-1$) of $\sqrt{z^2-1}$ on an annulus $1 < \lvert z\rvert$.
So for example, you can see that something the loops around the branch point at z = 0 jumps into the other sheet before passing through the cut (say) [0, -\infty]
@BalarkaSen I meant a nonzero ring homomorphism $\mathbb R \rightarrow \mathbb C$. If you can prove that there aren't any nonstandard ones, I'd be very interested, because that contradicts a comment I saw (and I've been unable of doing so)
@Khallil I've computed that integral in Mathematica and it's actually equal to: $$\ln \left(x^2+1\right)- x \arcsin\left(\frac{2 x}{x^2+1}\right)+\dfrac\pi2 -\ln 2.$$
@Hakim I'm a bit shaky on the values of $x$ for which the integral is valid. I used a Weierstrass substitution to get it into that form. The form you have was a direct result of integrating by parts, right?
(It seems as though the two are equal for $-1 \leqslant x \leqslant 1$)
So there can't be possibly any map $\psi : \Bbb {R} \to \Bbb {C}$ that is nontrivial, restricting to $\Bbb {R}$. Shouldn't that be enough or have I misunderstood your question? @Mike.
@BalarkaSen Now, you can prove that there are no automorphisms of $\Bbb R$, using roughly that idea (prove that if $x >0$, then $f(x)>0$ and you're done), but I don't see why the only embeddings should come from autos of $\Bbb R$.
Ah, I see what you mean. I just meant that if you can show that $\psi(\Bbb R) \subset \Bbb R$ (where by $\Bbb R$ I mean the standard copy of $\Bbb R$ in $\Bbb C$), then we're done, by restricting the codomain
$f$ wasn't specified at all. I'm guessing it can be any function mapping $\mathbb{R}$ to $\mathbb{R}$. There wasn't any context either. I found it on a random internet forum.
@Hakim I tried a counterexample of $f : x \mapsto x$ and showed that the inequality would suggest that $y(1+x)<0$ so I chose a $y<0$ and an $x$ such that $x+1<0$.
The poster of the question told me to read the question again.
In how many ways can the number 1;000;000 (one million) be written as the product
of three positive integers $a, b, c,$ where $a \le b \le c$?
(A) 139
(B) 196
(C) 219
(D) 784
(E) None of the above
This is my working out so far:
$1000000 = 10^{6} = 2^{6} \cdot 5^{6}$ and then to consider th...
