@PM2Ring What parts are you having trouble approximating well, and for what? I think speculatively you could do something fast using exp and extracting cosh out of that, and $\frac{1}{x}$... well, I'm still kind of working on that one, but consider taking a look at oeis.org/A204983 and use A007733 instead of A002326 in the corresponding place in the expression.
Note: it's listed as conjecture only because I'm not really competent to formally prove it despite the fact that it's readily demonstrable to me.
@copper.hat My cosh function is well-behaved. It's a bijection, in fact it's monotonic. But that exponential growth makes it hard to get rough approximations.
@Obliv I'm not quite sure what you mean, but if $AX=A$ then $X=I$. We show that by left- multiplying by $A^{-1}$. Hopefully, you can show that a matrix always commutes with its inverse.
@PM2Ring Well let's say $A$ is a system of equations, and $I$ is also the coefficient matrix of a system of equations (does this make sense), then what is this adjoining process? It's not adding/multiplying the systems
You haven't said much about your computing domain anyways. Assuming it's binary, this is providing linear, crude approximation to the expression you want computed.
Blasted thing. I can't seem to escape the quote. /shrug
Ah I see. Well yeah, I get there's a necessity for cheap and high precision Newton seeds at least, though, but I mean... if you could just manage to look at the stuff in the function folder, well, let's just say it speaks for itself on the cheap side of things relatively speaking.
@copper.hat That's not bad, although it goes negative for small $u$. I guess I could handle small & large $u$ as separate cases. But instead I've been doing $x=y \sinh^{-1}(u)$, with $2\ge y\ge 1$. And then the trick is to find good estimates of $y$.
@AMDG Not in this context. I've used it when working with Mercator projection. But it's (probably) not so useful in this context, since I don't need to involve circular trig functions.
I don't need to compute cosh, Sage can do that to any precision. It can also, simplify symbolic expressions involving hyperbolic functions, to an extent.
Arbitrary precision function evaluation in Sage is quite fast. I'm just trying to simplify my algorithm for calculating x from u. It's easy to compute u from x.
My early versions needed >9 loops of Newton's method for large x. Now it does 1 or 2 rounds of $x\leftarrow \cosh^{-1}(1+ux)$ followed by 3 rounds of Newton, for 120 bits of precision.
I've noticed in some of these problems the inverse of a 2x2 matrix can be $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \to \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ but not always, what's the condition?
Each round of Newton has to do a cosh & a sinh. I suppose I could replace the sinh with a sqrt(cosh^2-1). My initial approx also needs asinh, tanh and an exponentiation (to a non-integer power).
@PM2Ring What if you tried approximating the integral of the expression, approximated that with something having a known inverse in closed form, then used the difference quotient to approximate the inverse?
The shape seems relatively simple. Possibly even something that could be approximated by cosh or smth idk
We need more fundamental operations for isolating variables.
idk, find a geometric object that relates to the equation itself. That usually helps
I mean alternatively, we're saying that $\cosh(\phi) = u\phi + 1$ and $\cosh(\phi)$ is necessarily a point on a hyperbola, and $u\phi + 1$ is a parabolic shape. Haven't thought enough nor done enough research nor know enough about hyperbolic geometry to say much. Or just conic sections in general. Anyways, I'm going to retire to bed. Good night!
Well there might be something worth looking at with $ux = 2^p (2^q - 1)$. idk. It relates a reciprocal to the infinite repeating integer value in its binary expansion. Presumably extends to all bases. Also, I'm working on solving $2^n x = y$ given $y$, so something might come out of that as an operation. Like I said: more fundamental operations. That's why I'm working on an isomorphism between boolean logic operations and numerical operations. Anyways, night!
A ship is accelerating with constant proper acceleration $a$, i.e., that's the gravity that the passengers feel. Then $x=a\tau$ and $u=d/\tau$, where $d$ is the distance in the rest frame, and $\tau$ is the proper time (i.e., the ship's clock time), in units where the speed of light $c=1$. In those units, standard gravity is ~1.0323 light-years/year^2.
@copper.hat Well, done properly, it's way better than no grading at all (which is the trend in today's academia). I don't know the context of your remark, but I am a big fan of well-written and -executed computational homework on WeBWork with proofs, of course, still graded by a competent TA or professor.
@TedShifrin it was a bit of a rant based on a call i had with a friend (prof) yesterday. i have never used WeBWork. the idea of learning with no grading is like taking a driving test on one's own.
@copper.hat cool. It would be interesting to visit Europe some day
@PM2Ring I remember bringing that up to Dr William Kaufman, who was the director of the Griffith Observatory, and he poo-pooed me because I was in 7th grade.
Written exams were only invented a few centuries ago. Partly to deal with the subjectivity of oral examination, but also to enable the professors to handle a larger number of students in a timely fashion.
@robjohn Well, "everyone knows" you can't go faster than c. :) But (mean) celerity is a useful quantity if you want to zip around the galaxy, and you aren't concerned that centuries will pass on Earth during your absence.
Of course, the amount of energy required to do that is insane, even if you have an ideal antimatter-powered engine.
It would be nice if some kind of hyperspace / sponge space travel is possible... It'd be a bit sad if the only way to settle the galaxy is in generation ships, or in ships that require the entire current energy budget of Earth just to go a few measly light years in a reasonable timeframe.
I'm sure I read a book (or 2) by Kaufman. I'm sure he understood about celerity, but maybe he thought you were claiming that the coordinate velocity $d/t$ would exceed $c$. It's so easy to have misunderstandings when discussing relativity, unless all parties are using the same terminology.
@copper.hat Well, the fact remains that I have had colleagues who literally couldn't be bothered to grade homework in upper-level courses (even), and graded based solely on exams. I asked some of these colleagues how our students were supposed to learn to write mathematics and improve ... and he said, "Well, look at me. I made it." He was horrid.
Because it is, @Obliv. That's the beauty of the fact.
That's called a "cyclic permutation" (the order is preserved if you go around in a cycle/circle).
So I could theoretically have a 1000x1000 matrix and jumble it up with row swapping and as long as I keep the count of how many I do, I will have the det..
Oh okay
Doesn't seem convincing but I'll take your word for it.
@Obliv You can divide the permutations for a given matrix into two sets, known as the odd permutations and the even permutations. The odd ones flip the sign of the determinant, the even ones preserve the sign. Play with permutations of small sequences (length 3 and 4) to see how this works. It's a pretty fundamental thing, and it keeps coming up all over the place in physics, so it's a Good Idea to try & get a solid understanding of it.
Suppose that I have a topological vector space (TVS) X. Let A be a convex open nbd. of 0.Suppose that $x_0\notin A$. Then there exists a continuous linear functional f on X such that f(x_0)=1 and f(a)<1 for all a in A.
Let $p_A$ be the Minkowski functional of A. Then, it turns out that $p_A$ is subadditive, positively homogeneous and that $A=\{x\in X: p_A(x)<1\}$. Using which we define a function f($\lambda x_0)=\lambda$, which extends to f (using the same notation) satisfying $f( x)\le p_A(x)$.
@Obliv There's no need to mess with the stickers. You can be evil & dismantle the cube & reassemble it into an odd permutation, making it impossible to solve.
@PM2Ring Also not entirely sure what you mean by dividing the permutations into even and odd. Swapping R1 and R2 gives you a -, but doing it again makes it positive again? so how would you say one is an odd or even permutation
There's a classic old Web page that teaches introductory group theory, with the later sections focused on the Rubik's cube. It's quite good, IMHO. It's a shame that the author didn't continue it. dogschool.tripod.com
@Obliv If you start from an even permutation swapping one row takes you to an odd permutation. And if you start from an odd permutation, swapping 1 row takes you to an even permutation.
@Koro maybe I'm misunderstanding the issue, but if x is in the unit ball of X/ker T, then q(x) is in the unit ball of x, so $x=q^{-1}(q(x))$ is in the image through $q^{-1}$ of an element of the unit ball of Y. But x was arbitrary so we're done
Suppose that D is a countable dense subset of R. Then there can not exist a function f:R->R which is continuous exactly on D.
Hint 1 (showing that the set of points C at which f is continuous is a G_\delta set.): Take $U_n=$ union of all open sets U, diam U <1/n. Then $C=\cap U_n$.
This is okay.
Hint 2 (show that D is not a G_\delta set): Suppose on the contrary that it is so. Then $D=\cap W_n$, where $W_n$ is open in R. Define $V_d= R-\{d\}$. Then show that W_n and V_d are dense in R.
Suppose that W_n and V_d are dense in R. How does this contradict that D is not a G_\delta set?
W_n is dense because it contains a dense set. V_d is dense by definition.
Prove the following theorem:
Theorem: If $D$ is a countable dense subset of $\Bbb{R}$, there is no function $f : \Bbb{R} \to \Bbb{R}$ that is continuous precisely at the points of $D$.
(a) Show that if $f : \Bbb{R} \to \Bbb{R}$, then the set $C$ of points at which $f$ is continuous is a $G_{\del...
Line is passing through point $(0,a)$ making with $OY$ axis angle $\phi$. Find pdf and cdf of intersection point of line with $OX$ axis. a)$\phi \in [0,\pi/2],b)\phi \in [-\pi/2,\pi/2].$
I stared on this problem a lot but can't get any idea how to solve this. Can you give some hints?
My work.
$y=...
Let $E ⊂ (0, 1)$ be the set of all real numbers with decimal representation using only the digits $1$ and $0$: $E := \{ x ∈ (0, 1) : ∀j ∈ N, ∃d_{−j} ∈ \{1, 0\} \text{ such that } x = 0.d_{−1}d{−2} . . .\}$. $f : E → ℘(N) \text{ s.t if } x ∈ E, x = 0.d_{−1}d_{−2} . . ., f(x) = \{j ∈ N : d_{−j} = 1\}$. I am trying to prove that f is injective. This is my reasoning. Set $f(x)$ tells us about the position of $1$ and $N-f(x)$ tells about position of $0$. From this it implies that if $f(x)=f(y)$ then $x=y$.
Okay in the first link they defined the space but in the second link they didn't
Here it means continuous functions which vanish at infinity because fourier transform of $L^1$ function is uniformly continuous and vanishes at infinity (by Riemann lebesgue lemma).
yeah, the exercise involved a construction involving numbers having only the digits 1 or 2 (and in particular, not 9 and not 0) in their decimal expansions, which do not raise the concern
Why does the Principe of induction Not Work in gödels incompleteness, Like OK there ist a Statement that ist true but Not provable by some theory, then add that and all next Statements too that occur by the Same logic
@leslietownes Oh, I was just looking at the standard diagonalization argument, and didn't realize we were discussing something else. I need to read back more.
not to exaggerate too much, but reading back more on that particular problem/family could take the better part of several days. i recommend against it.
im having a bit of trouble seeing why the $b_j$ in the LHS of the inequality needs to be complex conjugated D:
i see that the proof provided by Rudin wouldn't work (at least in the same way), but I don't see an actual reason for it having to be complex conjugated
the result would be true without the conjugates, so in some sense this inequality doesn't "require" the conjugates. the conjugates do make the thing inside | |^2 on the LHS a complex inner product on C^n, which it would not be without any conjugates.
if that's rudin, however, i'm not sure he ever uses the complex inner product on C^n that much. it's just a common way of writing it, for those who do use it.
physicists sometimes put the conjugate in the first variable (over the a's) instead of the second
:P i wish there were some footnotes or something which made the precise connection between the complex modulus and so on with metric spaces, inner product spaces and so on
oh gosh i misunderstood the proof. the same proof works for non conjugated
that is strange :P i would think that such a nontrivial detail would actually matter in the proof but i guess maybe like you mentioned one is to think about innerproducts
@SAJW If $x\ge y$, why not factor it out as $x!-y! = y!\big(x(x-1)\cdots (x-y+1) - 1\big)$?
@anak This is more a question of which part of the world you come from. The European school (including South America) tends to use $\wedge$ for $\times$. It really confuses people, ultimately, because a bivector is not a vector. (Of course, physicists will be careful to call the cross-product a pseudovector rather than a vector.)
So I have two series $\{1\}= H_0\subset H_1\subset H_2 \subset ...\subset H_n= H, \{1\}=K_0\subset K_1\subset...\subset K_m= K$, such that the factor groups are abelian.
Do you know the proof for the following statement or where I can find it?
Semidirect product of solvable groups is solvable?
I thought that this property is so general that it should somewhere on the Net, but unfortunately I didn't find the answer.
Thanks
Suppose that $H$ and $K$ are solvable. Given $\phi: K\to Aut (H)$, a homomorphism, then their semi-direct product $H\rtimes K$ is also solvable.
I came to know that I need to use the following theorem here: A group $G$ is solvable iff there exists a solvable subgroup $N$ such that $G/N$ is solvab...
Does anyone remember the theorem which goes something like this: If G is a finite group, H is a subgroup. Let i be index of H in G. If i! (i factorial) is not divisible by something, then something happens?
Suppose that $H$ and $K$ are solvable. Given $\phi: K\to Aut (H)$, a homomorphism, then their semi-direct product $H\rtimes K$ is also solvable.
I came to know that I need to use the following theorem here: A group $G$ is solvable iff there exists a solvable subgroup $N$ such that $G/N$ is solvab...
Ok. I have a following task: Let the rank of the Jakobi matrix of the correspondence $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is $n$ for each point. Prove that f translates regular curves into regular curves
Suppose that o(G)=n, where G is finite subgroup of C*. If g is any non identity element, then $g^k=1$ for some $k|n$. Suppose that k<n. Then there is g' in G not in <g>.
a lot of the most popular proofs use a teensy bit o' field theory.
the set of elements of order d in a subgroup of the multiplicative group of a field is contained in the set of solutions of x^d = 1. and if you toss out the roots corresponding to elements with orders properly dividing d there are at most phi(d) elements of order exactly d. but sum_{d | n} phi(d) = n, so if n is the order of your group, you have to have exactly phi(d) elements of order d for each d dividing n. in particular phi(n) >= 1 element of order n.
i'm not sure if i'd even call that field theory. just counting solutions to a polynomial equation.
in C* maybe you can do something more concrete. these things lie on the unit circle and one of them makes a smallest positive angle with the positive x axis?
Mmh i think of gödels incompleteness vs consistency Like the halting Problem, because the solution i ready where very similar. But ist the Numbers of "different" unprovable Statements Infinite? Like different thoughtbranches of a Axiom based system. That there are infinitely many of the Same branch i understand by now.
Do you know the proof for the following statement or where I can find it?
Semidirect product of solvable groups is solvable?
I thought that this property is so general that it should somewhere on the Net, but unfortunately I didn't find the answer.
Thanks
