Using this idea, I think we can even get a sequence which has $\mathbb Q$ as the set of all its limit points. Let $q_i$'s be an enumeration of all the rationals (this is possible due to countability of the rationals). Then the sequence: $q_1, q_1,q_2,q_1,q_2,q_3,...$ has the set of all limit points as $\mathbb Q$.
hmm, nope. My reasoning was wrong because if Q is the set of all limit points of such a sequence then Q should have been a closed set in R, which Q is not.
as for explaining what topology is in plain English, well, I can't do that, but I can explain the motivation (at least what I think the motivation is) behind the definition
So I think to give motivation for topology, you need to first look at metric spaces and open balls. We say that two metric spaces are equivalent if open balls from one space look like (are made up from) open balls from the other space
And this condition can be extended even further to say that if we take a metric space, and consider unions of all the open balls in it, then this family of subsets, which we call topology, for both of the metric spaces they are equal
Now, looking at open balls, they form a family which we can call a basis. From my little experience, it's very useful to define topology of a space using a basis rather than to give it directly
@robjohn Yeah, that's handy aswell. Also you can note that if $3p(x)+p(3x)=3p(2x)$ then equating the $n$-power terms we have $3+3^n=3\cdot 2^n$ which only has solutions for $n=1$ and $n=2$
Shame. I was hoping for something a bit more substantial, but fair enough.
I think it would be much easier to prove whether or not a number is transcendental if we can formulate it into a conjunction of two predicates A and B, where A is $\text{irrational}$ and B is some other predicate such that $\text{transcendental} := A\land B$.
we have rational numbers, and polynomials over those, and it's a bit like variables over rational numbers
So if say, $x, y\in \mathbb{R}$ are algebraically independent over $\mathbb{Q}$, then it basically means that expressions including $x, y$ and rational numbers act like polynomials in two variables over rational numbers.
and that basically means that any expression of the form $a_n\pi^n+...+a_0$ with $a_i$ rational numbers acts exactly like a polynomial over the rational numbers in the way that two are equal iff they have equal coefficients
so here $\pi$ is kind of like a variable of a polynomial, where by polynomial I mean an abstract expression of the form $a_nx^n+...+a_0$ and not a function
@robjohn Pretty much. Anders discusses various orbits around a torus on his site & in some Physic.SE posts. Sure, a full Dyson shell is astable, so you just need some (relatively) minor propulsion to compensate for drift, unlike a Ringworld, which has a nasty tendency to collide with its star. ;) On a related note, with a partial shell, you can make a Shkadov thruster. en.wikipedia.org/wiki/Stellar_engine It's not very fast, but it can travel a fair distance over the lifetime of a star.
@AMDG A couple of weeks ago I mentioned Carlson's modified AGM algorithm for the natural log. You asked if it works with a complex arg, but I somehow missed that question at the time. Yes, it does work on complex args, assuming your sqrt algorithm handles complex args, but it gets a bit sluggish near the branch cut.
In the usual convention, both sqrt & log have a branch cut on the negative real axis. For computation purposes, that's not actually a problem because you can just negate the arg & transform the result.
Yeah, I found Carlson's modified AGM algorithm actually. It can use division but doesn't necessarily require it.
Thanks.
I still have yet to see how well my approximation of the logarithm and arctan are for by integrating the infinite sum I have.
Who knows. Depending on circumstances, it could be pretty convenient. I also found a way to speed up convergence for it as well, but it seems to require more division capacity, so you could get pretty fast convergence if you can divide integers in $[0, 2^n]$. I found a solution that works just fine for floored quotients using n=8.
You can add an arbitrary constant to the $f_{lp2}(x)$ function to speed up convergence. Using 8 got me to just beyond 64-bit precision using only 6 terms of the series.
$f_{lp2}(x)$ is one of a family of functions which satisfies $2^{f_{lp2}(x)} \geq y$ so adding a constant for which this inequality still holds preserves its validity.
The original design is to find the smallest power of two when multiplied by a coefficient less than y yields a number that is greater than or equal to y.
The coefficient can effectively be dropped to get the above inequality.
@PM2Ring Do you know of anything analogous to $\int \frac{1}{x} dx = \ln(x)$ for the exponential function? I've found an OK approximation through repeated integration (shown in the desmos link), but I'm hoping to find something whereby I can just convert the sum I found into a function of the exponential.
@AMDG Well, the exponential function is its own integral, so you can't make much progress in that direction. ;) There's a reasonably fast simple algorithm for computing e to arbitrary precision, using its representation in factorial base: e=2.1111... but I guess that's not much use for arbitrary powers of e.
Yeah that's the issue with the exponential function, however, there are functions of the exponential which give something other than a constant in the derivative/antiderivative, e.g. $\operatorname{Ei}(x)$.
That's also the basis for the repeated integration approximation as well.
Since the exponential is its own derivative and antiderivative.
Carlson's log is so fast that it's been used to calculate exp(x) via Newton's method, but I haven't played with that much. And obviously if you don't want to use Halley's method, you're even less likely to want to use Newton's.
pmaddwd is a SIMD instruction available in the MMX subset of x86 for a number of processors. iirc the first version came out in around 2011. It computes $ax + ay$ on 16-bit operands and gives 32-bit operands as an output. SSE extends this to allow for larger word sizes like 32 and 64 bits.
While you do that, I'm going to finish my redstone elevator design.
Ok. He uses the identity $\exp(x) = 2^(x/\ln(2))$. With suitable range reduction, that resolves to $2^(n+f)$, for some integer n, and -0.5 <= f <= 0.5. The $2^n$ part is a simple bitshift. He does the $2^f$ part with a ratio of polynomials that he got from Cody & Waite ( a standard reference). It might be a Padé approximation, I guess.
I mean it opens up a whole world of approximations previously unusable.
Nested fractions... pade approximations... etc.
Yeah pade is ya boi when it comes to good rational approximations I've found, but lagrange surprisingly has come in clutch at least once for a low-precision logarithm approximation.
https://math.stackexchange.com/questions/4364501/is-a-clopen-set-a-paraconsistent-object I feel like this answer doesn't really answer anything, and the downvotes are a bit unfair here
In finding $n$th term for $1,1,2,1,2,3,1,2,3,4,...$ I tried to proceed as follows: Suppose that $m$th term of the sequence is $n$ then, the following inequality holds: $1+2+...+n\le m$ i.e., $m\ge \frac{n(n+1)}2$.
Hi Leslie!
In fact, stronger inequality would be: $\frac{n(n+1)}2+n\le m$
@Koro What about it? The Sphereworld robjohn mentioned is a hollow sphere with a star at the centre. It's astable because of the shell theorem (& Birkhoff's theorem in GR).
I am convinced that $a_k$ is the general term. It's just that I'm trying to understand how anyone would come up with that term? Is there an approach? Or was it guessed?
Here, I tried to come up with that systematically.
If k=2, then the beginning of 2+1st group should start at $(2^2+2)/2$th term that is at 3rd term. But clearly, the index of the third group begins at 4.
Firstly, we have this sequence : $1,1,2,1,2,3,1,2,3,4,...$ which is the sequence of integers $1$ to $k$ followed by integers $1$ to $k+1$. We could say a fractal sequence.
Secondly, we have this formula : $$a_n=\frac{1}{2}(2n+\lfloor\sqrt{2n}+\frac{1}{2}\rfloor-\lfloor\sqrt {2n}+\frac{1}{2}\rfloo...
yeah, but the idea is same I think. Now, I understand why the difference in $a_k=k\color{red}{-}\frac{r(k)^2+r(k)}2$
But even in that answer, I have one question, which I'll comment there as well, I think: what is $m$? I understand that $m$ is the nth term of the sequence. If so, then I think the first line in the answer is wrong.
We have the sequence: 1,(1,2),(1,2,3),(1,2,3,4),... Here $2$ starts to appear from the second 'group'.
So clearly if m=2 and n is large then the first inequality in the post is wrong.
Suppose that $n$ is the $m$th term of the sequence. Then the following holds: $$(1+2+3+...+n)\le m\implies \frac{n(n+1)}2\le m\implies n^2+n\le 2m$$
It follows that $(n+\frac 12)^2\le 2m+\frac 12\implies n\leq \sqrt{2m+\frac 12}-\frac 12$.
But if I choose $n=\left\lfloor \sqrt{2m+\frac 12}-\frac 12 \right\rfloor$, then it doesn't work.
Nice exercise from metric spaces: Let $(X, d)$ be a metric space and $U\subseteq X$ be a non-empty open set. Prove that the map $$f:y\mapsto \max\{r\in (0, 1] : B(y, r)\subseteq U\}$$ is continuous on $U$.
I am struggling with a bit of complex analysis. Is it true that the integral over an arc around a (single) pole of a meromorphic function is iangle of arcresidue at pole ?
@FliiFe the residue theorem does not cover partial angles. The contour should completely encircle the pole. If the pole is simple, then you can do what you are claiming from an arc that gets infinitely close to the pole.
If I have a Borel $\mathbb R^m$-valued measure $\mu$ and a function $f\colon \mathbb R^n\to\mathbb R$. What is might they mean by $\int_{\mathbb R^n}f(x)\,d\|\mu\|(x)$?
@robjohn I recently dealt with a bizarre PV question where this same sort of thing arose. People seem not to understand that higher-order poles can mess up arcs of circles.
@CroCo I have a straightforwad resolution. The idea is that non zero vectors $u,v$ lie on the same line iff there is some pair $\alpha,\beta$ not both zero such that $\alpha u_k +\beta v_k = 0$ for each index.
In the robot example, the second pair shows that the two rows are on the same line iff $c_{\theta_1} u_k +s_{\theta_1} v_k = 0$ (in particular, the angle formed by the two points must be in $\theta_1 + \{0,\pi\}$). If you work through the two other pairs, you get the equations $c_{\theta_1} s_{\theta_1+\theta_2} +s_{\theta_1} c_{\theta_1+\theta_2} = 0$ and $c_{\theta_1} s_{\theta_1+\theta_2+\theta_3} +s_{\theta_1} c_{\theta_1+\theta_2+\theta_3} = 0$.
This tells us that the rank drops iff both $\theta_2, \theta_3 \in \theta_1 + \{0,\pi\}$. This is consistent with the corresponding physical configurations.
