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2:11 AM
a lot of things about me need to be fixed in the second printing.
 
2:27 AM
Makes sense I also guess at that level the "exercises" can actually fall into the category of the sorts of questions you'd be asking in one's research to an extent, not to mention the level of sophistication the questions would need to have.
 
@copper.hat I have explored myself as much as I can today. I think I will explore some math instead.
 
2:56 AM
@robjohn Enjoy the exploration! I'm browsing Ted's Abstract Algebra and wondering how come I can forget so much in such a short time...
 
3:41 AM
@copper.hat did you get a copy?
 
@robjohn Yes. In my lifelong question to make some progress with abstract algebra
 
very admirable.
 
EM4
3:57 AM
hello
 
 
2 hours later…
5:54 AM
@Ted: good evening.
 
6:30 AM
Good morning
 
good midnight(ish)
 
 
1 hour later…
7:45 AM
@robjohn So I thought more about the problem we discussed last night
I have written down a full solution, could you see if it has any mistakes?
 
ok
 
You did point out some mistakes/comments yesterday
but I'm not sure I fully understood them. So here it is:
First, we show the result for open intervals in $[0,2\pi]$. Let $A = (a,b)$.
$$\left|\int_a^b \cos nx\, dx\right| = \left|\frac{1}{n}(\sin bx - \sin ax)\right| \le \frac{2}{n} \xrightarrow{n\to\infty} 0$$
Now, suppose $A$ is a measurable set in $[0,2\pi]$. Fix some $\epsilon > 0$. There exists an open set $V$ such that $A \subset V$ and $m(V\setminus A) < \epsilon/2$. Since $V$ is open, it is a countable union of disjoint open intervals in $[0,2\pi]$, say $V = \bigcup_{i=1}^\infty A_i$. Clearly, $m(V) = \sum_{i=1}^\infty m(A_i)$. Since $m(V)$ is finite, we can find some $M\in \mathbb N$ suc
 
First problem: $\int_A \cos nx\, dx \le \int_V \cos nx\, dx$. That would be true if $\cos(nx)\ge0$, but it is not.
The next inequality is true, but since $\cos(nx)$ is not necessarily positive, the inequality is of limited use.
 
Aha
I think I can fix this. One moment.
@robjohn How's this? mathb.in/59328
 
Have you shown anything about $\int_a^b|\!\sin(nx)|\,\mathrm{d}x$?
that tends to $\frac2\pi(b-a)$, not $0$.
 
8:07 AM
@robjohn Whoa, that's interesting.
Where do we need that in this proof though? Are you saying that what I've done works for \cos nx and not for \sin nx?
 
8:31 AM
@epsilon-emperor No, I am saying that by putting $|\!\sin(x)|$ into the inequalities, you lose all the cancellation.
you will need to have $|\!\sin(x)|$ in the inequalities, but only in very limited places
The same is true for $\int_a^b|\!\cos(nx)|\,\mathrm{d}x$; it tends to $\frac2\pi(b-a)$
 
8:44 AM
@epsilon-emperor: think about the symmetric difference of the finite union of open intervals and the measurable set.
it is on that set that we need to control the integral (make it as small as we want).
oh, I guess I am talking to myself again.
 
9:02 AM
Small doubt on matrices and determinants... If I multiply a matrix by k, all the elements of the matrix are multiplied by k but if I multiply a determinant by k, only elements of one of the rows/columns are multiplied by k right? @robjohn
 
@AdilMohammed No. If you multiply a determinant by k, you get the determinant multiplied by k. Why should the matrix change at all?
 
@AdilMohammed if you multiply a column by $k$, then the determinant of the resultant matrix is multiplied by $k$.
I think that is what you were trying to say.
 
@robjohn Oh yes i meant like that, just one moment let me share a picture
 
If you multiply the matrix by $k$, then all the columns are multiplied by $k$, thus the determinant is multiplied by $k$ raised to the number of columns of the matrix.
 
9:12 AM
@AdilMohammed Hi, look up scalar triple product
you can interpret the determinant in 2-d and 3-d as some combination of dot and cross products between vectors
in this interpretation, the properties of the\ determinant is very simple
 
@Buraian that only applies in 3 dimensions
 
yah
 
thanks @burain
 
np
 
oh my keyboard is glitching now, so if i got a matrix 3*3 with all elements 3 then can i write it as 3 times a matrix with all elements 1?
ughh my keyboard is now goinf right to left now? how??
Nope reloaded now its fine
 
9:29 AM
@AdilMohammed yes
 
... @robjohn but in a determinant 3*3 with all elements 8, i can write only 8^3 determinant of all elements=1 right, like you told earlier? thankss got it
 
Yes, of course with a matrix with all elements $1$, the determinant is $0$.
 
@robjohn Thanks, thinking
 
Thanks once again
 
10:25 AM
Rob, I tried thinking of the symmetric difference, didn't get much yet.
Maybe I'll have to give it some more thought. It'd also be helpful if you could throw some light
 
10:53 AM
@robjohn mathb.in/59337
I'm taking a different approach now.
1. Prove for intervals
2. Prove for finite disjoint union of intervals
3. Prove for open sets = countable disjoint union of intervals
4. Try to prove for measurable sets, but get stuck
Any thoughts now? At least my proof works for open sets, I think. I have eliminated the problem (the inequality issue) you mentioned a few hours ago.
 
11:24 AM
I proved the twin prime conjecture. Again
Actually this is the first time I think the proof works
 
hi all
0
Q: Fermat's Last Theorem for Eisenstein Integers (excluding $\mathbb{Z}$ or $w\mathbb{Z}$ )

mickConsider the Eisenstein integers. Let p be an odd prime. $$x^p+y^p=z^p$$ with $x,y,z$ in the Eisenstein integers, excluding solutions in $\mathbb{Z}$ or $w\mathbb{Z}$ for a unit $w$. Are there any nontrivial solutions ? By trivial I mean excluding units and zero's for $x,y,z$. What has been found...

 
@mick
0
Q: An elementary proof of the twin prime conjecture using set and elementwise operations.

ExercisingMathematicianAn Elementary Proof of the Twin Prime Conjecture The set of prime number can be written: $$\bigcap_{c \notin \Bbb{P}} \Bbb{Z}\setminus c\Bbb{Z} = \Bbb{P} \tag{0}$$ We start with an elementwise addition that commutes with set complement and arbitrary intersection. For any family of sets $A_i \sub...

@mick I upvoted yours nice question
Took 5 hours to write that proof
But have tried many things before
I will give a cut of the millenium prize money to whoever can help me publish since I'm not accademic yet
 
12:32 PM
2 hours ago, by epsilon-emperor
Any thoughts now? At least my proof works for open sets, I think. I have eliminated the problem (the inequality issue) you mentioned a few hours ago.
Update: mathb.in/59338 (Corrections made using limsup, liminf)
 
12:44 PM
I think I've decided I'm just going to implement the complex matrix exponential and logarithm.
 
@epsilon-emperor taking a look...
 
So that leaves me with three questions:
1. What is the fastest converging series for $\exp(x)$ and $\ln(x)$.
2. How do I use Chebyshev polynomials and how do I compute the coefficients.
3. How can I combine minimax with Chebyshev to further improve the error?
 
@epsilon-emperor In your attempts, you fail to work with the fact that the integrands are not always positive. The inequalities don't really show anything; saying $x\lt\frac1{1000}$ does not guarantee that $|x|\lt1000$.
You have three sets to consider: the measurable set, the finite collection of open sets which cover most of the measurable set, and the remainder of the open sets which cover the rest of the measurable set.
 
So what do you all think of the Twin prime proof?
It is not a spoof
Pleae don't be aloof
Since the proof is not a spoof
Waves crashing restful
Haiku
^_^
 
1:31 PM
@robjohn I'm very sure we don't need to bother about that. I've shown $\limsup_{n\to\infty}\int_A \cos nx\, dx \le 0$ and $\liminf_{n\to\infty}\int_A \cos nx\, dx \ge 0$, which imply $\lim_{n\to\infty} \int_A \cos nx\, dx = 0$.
Also, there is no inequality now that depends on the integrand being positive. If you think there is one, could you tell me which?
Earlier there was one -- I've gotten rid of it now in the latest edit 59338
@robjohn Alright, let me think about this. Until then, it'd be great if you can point out where I'm using positivity of the integrand (I think I'm not using it)
 
@epsilon-emperor So this is trying to prove that the integral over the open set $A$ is the sum of the integrals over the $A_i$? That seems to be the case.
 
@robjohn This is trying to prove the result for an open set.
 
okay, for that, I think it is okay.
 
To elaborate, (1), (2) and (3) in http://mathb.in/59338 basically show:
(1) The result for open intervals
(2) The result for finite disjoint union of intervals
(3) The result for countable disjoint union of intervals (open sets)
Now I have to take (3) and work with measurable sets
 
1:46 PM
yes
 
2:13 PM
Why did they delete my post?
So rude
I rewrote it to meet the guidlines then it suddenly got deleted
That reflects poorly on the site
as a whole
 
@robjohn: Hi professor Rob, how are you?
 
The site doesn't like:
1. Creativitive ideas
2. Attempts at open problems
3. Positivity
 
Can you please direct me to some problems on recursive sequences as I want to solve them using a concept I recently learned on contraction mappings?
 
@Koro I will have to look for some. Right now I have to go for a while.
 
okay :-)
 
2:28 PM
@ExercisingMathematician Well the issue is that you seem to have tried to just get your idea out there without the intent of asking a question. StackExchange is meant to be Q&A style of problem solving, not a mathematics journal to publish your results. I honestly can't say that I know what the alternative is as I myself am looking for somewhere that I get more of my fundamental questions concerning maths answered.
All in all, this misunderstanding of SE leads to the misrepresentation of SE that you have mentioned as being contrary to those three things somehow and feeling incredibly unwelcoming to "outsiders".
Someone had to say it. Anyways...
Pretty simple logic: if one does not understand a group and how it operates and acts contrary to it, then he shouldn't wonder why he is rejected by that group.
So does anyone know what the fastest converging series is for the exponential and logarithm functions?
 
2:49 PM
lol my bad
 
ahaha. anyways i have deleted my off-topic comment
 
No worries
 
3:03 PM
-2
Q: What class of rings does this proof generalize to?

ExercisingMathematicianHere is a proof. Based upon all the operations used within it, I would like to know how to generalize it to other rings than $\Bbb{Z}$. Let $\Bbb{P}$ denote the set of $\pm$ the prime numbers but including $\pm1$. Define the set of composite numbers to be $C = \Bbb{Z} \setminus \Bbb{P}$. The set...

Can I get some upvotes? I've checked the proof 10 times. My question is actually now within the guidelines and is truely the next question a mathematician usually asks about a proof. Can it generalize and to what...
 
3:24 PM
@ExercisingMathematician There has been quite a lot of discussion about proof of open problems posted in MSE over the years. There is no general policy on that and, as you can see, opinion changes from time to time. But the most recent such discussion would be the most relevant.
 
3:59 PM
@AMDG The best known algorithms are not series. If you insist on using Taylor series, then argument reduction can get you to O(n^2.5) time for n-bit precision exp using schoolbook multiplication, and then a careful Newton-Raphson can get you the same for ln. But the best known is using AGM as wikipedia already cites:
The following tables list the computational complexity of various algorithms for common mathematical operations. Here, complexity refers to the time complexity of performing computations on a multitape Turing machine. See big O notation for an explanation of the notation used. Note: Due to the variety of multiplication algorithms, M ( n ) {\displaystyle M(n)} below stands in for the complexity of the chosen multiplication algorithm. == Arithmetic functions == == Algebraic functions == == Special functions == Many of the methods...
 
Well it doesn't strictly have to be a series. Just some approximation that quickly converges to e for an input of 1 (as a measure of error) that has sufficient digits of mantissa for scientific use. I figure having as many digits of mantissa as there are in the smallest distance we can use ($10^{-45}$ plus some abundance of digits) should be sufficient for anything practical.
 
@AMDG If you just need a constant then dump all the digits you want in.
 
I don't want a constant. I want a function.
 
It makes no sense to compute something when you can just hard-code it.
If you want exp, then I don't know why you say "e".
 
Because $\exp(1) = e$
 
4:04 PM
Oh so you mean you're only giving an example.
 
Yes. I said "as a measure of error".
 
Well you should be more precise. If you want to compute exp to 45-digit precision, just use argument reduction.
No point using a complicated algorithm that doesn't pay for such miniscule numbers.
 
Any arbitrary value as a target for convergence would work, obviously, but $e$ makes sense for the exponential function.
Argument reduction?
 
It's the standard technique.
3
A: Natural occurrences of a to the (b to the c)?

user21820It is easy to square a number. So instead of computing $\exp(z)$ you can compute $\exp(z/2^k)^{2^k}$ for some suitably large positive integer $k$. This is a very simple way of accelerating the convergence of the Taylor series for $\exp$, also known as "argument reduction". The double-exponential ...

 
4:07 PM
I have not
There are in fact a number of things I do not know about these things.
 
I posted the wrong url initially
 
I'll check that out in a moment.
 
@LadiesandGents That article is useless for the precision AMDG wants. 45-digit is much more than 64-bit double-precision.
 
And argument reduction is trivial to code. It can't really get any simpler.
 
4:09 PM
Can you help me understand argument reduction, please?
 
Do you first of all understand the identity stated in that linked post?
 
The one you linked, or the one Ladies and Gents linked?
 
Of course the one I linked. I said the other one is useless.
 
both seem to talk about argument reduction :'(
 
@ExercisingMathematician As a matter of fact, only reflects poorly on you.
 
4:12 PM
If you're referring to yours, then no, I do not understand how it works, though if I messed around a bit, I might understand it better. I understand the relationship of the exponential function to all exponential functions and vice versa, however, I was unaware that you could use $\exp(\frac{z}{2^k})^{2^k}$ for perfect powers of $k$ to compute $\exp(z)$, and I've never heard of the double exponential.
 
@ExercisingMathematician I flag users who beg for upvotes on very poor questions.
@AMDG Please "reply to" the user you are addressing?
 
@LadiesandGents Ok I missed that halfway down the article it started talking about range reduction, which is the same. However, the polynomial approximation is pointless. Simply argument reduction is sufficient.
@AMDG "double exponential" just means something with exponent in the exponent. It's not a technical term.
 
Oh ok, I see.
 
All you need to know is that exp(z) = exp(z/2^k)^(2^k), and then choose k carefully.
The point is that if k is large enough, then z/2^k is small enough so you need less terms of the series. But you cannot pick too small k, obviously, because you still need to square exp(z/2^k) k times.
 
@ExercisingMathematician Please don't even think about taking up Haiku or any form of poetry.
 
4:15 PM
@user21820 Can you give me an example or two for how to use this properly?
 
If you want to calculate e^113.8 you'll get something better by taking your approximation of e^{113.8/1024} and then raising that to the 1024
 
Right, but that's a chicken and egg problem, then. What approximation of $e^x$ do I choose to get this approximation using argument reduction?
 
I think the idea is that you can take the taylor series at $0$
 
Oh
 
Taylor series! The point is that you can compute how many terms you need based on how small z/2^k is.
If |z/2^k| < 1/2^c, then after m terms the error is bounded by at most 1/2^(m·c), at most off by a factor of two.
 
4:20 PM
Can I use a Chebyshev polynomial and use minimax to compute the optimal coefficients for it and use that as an approximation for this?
 
At the same time, every squaring you do will introduce error, so you need to bound that.
 
You can also read the link I posted but apparently it's not good
unfortunately it can't be downvoted
We should probably make internet providers disconnect users to prevent crankery.
 
@AMDG Using some pre-computed approximations may make it a bit faster on certain ranges, but what for? In the end, if you want your method to apply to arbitrary input, you might as well just code a version that works for all, and to arbitrary precision.
Just do the error analysis, and you'll see that it's not hard at all.
 
Well, I've never done this before, so it is quite a bit to take in all at once.
 
First understand how I bounded the error of the Taylor series if you use only the first m terms.
We will decide c > 0 later.
To repeat, we chose k such that |z/2^k| < 1/2^c and use Taylor series approximation with m terms, which yields t ≈ exp(z/2^k) where t is the approximate. |z|<2^k, which implies |exp(z/2^k)| = exp(Re(z)/2^k) ∈ [1/e,e]. This means fractional error of at most 2e/2^(m·c), since the real value is in [1/e,e] and the error is at most 2/2^(m·c).
Now compute the error upon squaring. Each time you compute x^2 from x, the fractional error of x^2 is roughly twice the fractional error of x.
When you actually do the code, you have to do proper analysis, but I'm just giving the rough sketch here of the error analysis.
 
4:32 PM
Well what is $c$ related to here?
I understand k and m, but not c
 
c is the precision we want to compute the series to. If you pick too large c, you will compute too many terms. If you pick too small c, then you either cannot achieve the desired precision at the end, or you need too much work.
The thing is that on every computation you also incur an error. For example if you multiply two q-bit numbers you introduce a fractional error of roughly 1/2^q.
 
So $c \ge 45$ digits of mantissa. What should the value of $c$ actually be in this context based on n-digits of mantissa? Is it 1:1 correspondence?
 
As I said, you have to wait and choose c later. You cannot choose c now because you haven't finished the analysis. It will not be obviously related to the desired precision.
For now let's ignore such errors in computation of the series, because it's not important. We will do all computations at q-bit precision. Of course q > p where p is the desired bits of precision. Assume you can get a q-bit precise sum of the first m terms, which as I said above would have a fractional of roughly 2e/2^(m·c). Now each squaring not only doubles the fractional error but also adds a fractional error of 1/2^q.
 
By bit, we're referring to base two digits, yes?
 
Yes.
So after k squarings, you would get a fractional error of roughly ( 2e/2^(m·c) + 1/2^q ) · 2^k. I leave you to check this by setting up the recursive relation (at each step double and add 1/2^q).
 
4:40 PM
Because if so, the funny thing is that I've had an idea in the back of my mind for symbolic computation of exponents by allowing bits to represent rational portions of bits, but I've been mostly ignoring it because I don't know of a convenient way to convert it to a usable approximation, but it would obviously speed up computation, and you could easily represent any irrationals with it and only limited by the number of finite bits required to represent the irrational.
 
@AMDG No. If you want to talk about funny things, I don't want to continue.
 
??
 
Do you want to listen or not?
 
I do, but since you mentioned arbitrary precision... and we're talking about bits... I thought I might try to contribute something, but I guess I'll just keep it to myself for now.
I'm only suggesting an encoding for numbers that allows one to represent irrationals exactly in finite form after all and reduces exponentiation to a rational number of bit shifts...
 
There is no such thing, period.
 
4:45 PM
Well that's like saying I can't represent the square root of two with the symbols $\sqrt{2}$.
 
@AMDG That's a syntactic representation, and has no relation with whether you can compute it easily.
 
Alright, but we still perform, as humans, algebraic operations on syntactic representations instead of infinite decimal expansions, mostly because the latter is impossible to represent otherwise using a finite number of symbols.
Is that not how we do algebra?
Floating point itself is already an abstraction over pure binary numbers since it requires some metadata in a header. What class of representation is floating point?
 
@AMDG That's not really correct. Humans do proofs, and that's very different from merely algebraic manipulation of syntactic representations of values. With mere algebra you can do very little.
Why do you think it's so difficult for humans to do integration? Because there is no simple algorithm for symbolic integration. And there cannot be one, for various reasons.
 
Which I find curious given how easy it is to find the derivative using a single simple algorithm for symbolic differentiation.
Namely, I'm talking about $\frac{f(x + h) - f(x)}{h}$
 
Hold on. I ask again, do you want to listen to the explanation (which you asked for) of how to use argument reduction to compute the series faster? I wanted to help you because you were not getting an answer, but if you're not interested just say so clearly.
 
4:56 PM
Yes, I am, and I have many other questions about maths that I want to answer, and I apologize for getting distracted and going on a tangent. Please continue.
I'm going to get a coffee though real quick
 
If $X$ is a set, and $\ell^{\infty}(X)$ is the set of all bounded complex-valued functions on $X$, what are uniformly precompact sets of $\ell^{\infty}(X)$?
 
Alright, I'm back.
 
@AMDG: So now you want ( 2e/2^(m·c) + 1/2^q ) · 2^k < 1/2^p. First remember that if you compute m terms of the series at q-bit precision, you will need roughly O(m·q) time, assuming you do it the right way (for the i-th term from the back add z/2^k then divide by i). Since we already need q > p, we don't want to make m too big. I claim that m = sqrt(p) is enough.
We shall pick c = q/m+3, so that the 2e/2^(m·c) is less than 1/2^q.
Sorry I made a mistake. The series computation will need O(m·q^2) time, and I messed up the "right way"; each step should be (multiply by z/2^k/i and then add 1).
That's why you want m = sqrt(p), so that total of that phase is O(p^2.5). (Because q will not be that much bigger than p, as we shall see.)
Note that the final k squarings will take O(k·q^2) time.
Since we only need |z/2^k| < 1/2^c, we can pick k just a bit bigger than c. Roughly k = c+log2(z) where log2 is binary log.
Now put that back into the inequality we want, knowing that our choice of c ensured 2e/2^(m·c) < 1/2^q. This means it is enough if we have 2/2^q · 2^k < 1/2^p. We want k = q/m+3+log2(z), so we need q > p+q/m+4+log2(z).
Since we chose m = sqrt(p), you get that q > (p+4+log2(z))/(1+1/sqrt(p)) is enough.
And we're done. You have an algorithm to compute exp to p-bit precision using schoolbook multiplication and argument reduction on the series that takes O(p^2.5) time, more or less.
 
5:27 PM
Question: how do you derive the inequality $|\frac{z}{2^k}| < \frac{1}{2^c}$ exactly? I tend to understand math better by "reinventing the wheel". I'm getting bits and pieces of what you're saying here, unfortunately. I'm doing my best with what little I know. Part of it might be just because the lack of pretty syntax makes it difficult to read, but I keep forgetting what all the individual variables are referring to.
I didn't want to interrupt, so I didn't say anything.
In particular, I tend to understand and learn things through intuition and extrapolating universal principles as my means of progress based on a few examples.
e.g. that the area of a circle takes an infinitely thin circular segment and has it rotate over all directions. The number of these segments that fill a given space for a given radius is the area of that circle.
I suppose that doesn't necessarily matter. I suppose I'll figure this out later when my mind isn't half-baked. I'm honestly not doing so well in terms of health right now. Normally I would understand this quite easily.
Thanks anyways.
 
5:46 PM
@AMDG I didn't derive it. I said "If |z/2^k| < 1/2^c, then after m terms the error is bounded by at most 1/2^(m·c)".
 
Yeah, but where do you get that inequality from?
Surely it isn't arbitrary...
 
v = z/2^k is the term used in the series 1 + v + v^2/2! + v^3/3! + ...
If |v| < 1/2^c, then if you take only the first m terms then the remainder is bounded by 2·|v|^m/m! ≤ 2·|v|^m.
 
Alright, got it. And what about the RHS? Where does $\frac{1}{2^c}$ come from?
 
It's for later use. As I said, we pick c later based on other considerations as well.
 
Yeah, but what if I subsitute 2^-c for any other value? What changes?
 
5:49 PM
You could have used ε instead of 1/2^c; it would make no difference except make it harder to think through the rest of the reasoning.
 
Yeah, but it still eludes me where you obtain this as a boundary.
 
The claim about the remainder term is simply ∀c∈ℕ ( |v| < 1/2^c ⇒ | exp(v) − sum of first m terms of series | ≤ 2·|v|^m ≤ 2/2^(m·c) ).
You are free to prove something different; I proved what is easy for me to use later.
 
@AMDG not sure if this is what you were looking for, decades ago a chap Andreas Griewank did a lot of work on automatic differentiation.
 
@user21820 I tend to be able to reason through things better if things are given as they are rather than in a form that might be "easier" to understand. For example, my time in the hospital was quite curious. I couldn't read the patient data sheets for medications, only the stuff that the doctors read was readable to me because it used clear and concise terminology with distinct names for every unique object and related unique patterns, even though I'm not a doctor.
For every term I didn't understand, I could just look it up in a medical glossary.
Likewise with math.
How you related z/2^k there was very helpful, and ∀c∈ℕ and the rest I also find incredibly helpful as well.
Like I didn't realize c was supposed to be some natural number.
 
Actually c does not have to be a natural number. However, it is convenient in code to make it so.
 
5:57 PM
I can imagine...
Having c being something other than a natural would also defeat the purpose of this implementation in the first place since it would also be used to compute complex and real nth roots.
 
In what I wrote earlier, I said things like m = sqrt(p) and c = q/m+3. I didn't require them to be naturals, but in code you should round them, and check your error analysis.
 
Well it would be awfully convenient for me if you could coalesce all of your explanation of error analysis into a single error function for me. :D
The rest I should be able to handle myself.
 
@AMDG No, because I am busy, and also because as I said earlier I only gave a sketch that ignores the errors in approximating the series. If you really want to do it yourself, you should slowly redo the analysis.
I gave all the key ingredients, but I skipped so many details that it would take too much effort on my part to present everything.
I'll just remark that if you just used the series with no argument reduction then you would need O(p^3) time with schoolbook multiplication when z = 1, which shows that the above method is actually useful.
 
Ah alright. Well in that case, is there something I can preferably watch or secondarily read that explains how to compute the error of Taylor series in more detail?
And in particular, the error relating to this method of approximation?
 
@AMDG I'm afraid the best way is to do it yourself, because error analysis depends on exactly how you implemented the approximation. Every single operation you do will add an error. I hope you already know about fractional error, as the intuitive estimates using that are usually roughly correct. However, you don't need to know anything about error analysis to bound the remainder, namely v^m/m! + v^(m+1)/(m+1)! + ...
If |v| < 1 and m > 1, it can be bounded by v^m/m! + v^m/m!/2 + v^m/m!/2^2 + ... = 2·v^m/m!
That's where the factor of 2 came from.
 
6:11 PM
How do you mean that every single operation I do will add error? Are you referring to things like floating point error and outside of using arbitrary precision computations? If so, then using the idea I have for a symbolic encoding would reduce the number of operations and improve the error by several orders of magnitudes.
 
@AMDG Stop talking nonsense about symbolic computation; it's bogus and there's nothing else to say. Until you can actually write code that can compute exp to arbitrary precision, don't talk about things you can't do.
And arbitrary precision computation HAS errors. You have to stop at some number of bits in every single operation. All that will add up.
 
Well that isn't quite what I meant by that regarding error, but more so that if I have something stored as $\sqrt{2}$ and I square this value, I get $2$ exactly, not an approximation of it, and the result here is, you'll notice, a symbol representing the quantity two.
All it involves is allowing bits to be fractional instead of whole bits, so in base two, sqrt(10) would become 110 as an exact representation by doubling the number of bits required to represent 10 as 110, then shifting right one to get 110. If you think this is nonsense, then I don't know what to tell you.
This would be a shift to the right by "half" a bit since for integer logarithms and exponentiation, the result is equivalent to shifting left or right by some constant.
 
It is nonsense, and I don't care what you want to say. Anywya I've given you enough to work with. If you actually attempt to write code and do the hard work, you're free to ask me for help with it. Other than that, sorry I have no interest in your fancy ideas (that to me are clearly nonsense).
 
so who's winning England v Ukraine?
 
england
although maybe not if one considers their storied history of f---ing up
 
6:23 PM
half ukrainian here but I have to admit that england plays better
 
Amazing. Literally a rejection of reason to call this model nonsense. There is zero error in 110 as a representation of sqrt(2), and zero error is introduced by the symbolic squaring of this value either. I guess some people are too set in their ways to recognize the truth for themselves. Anyways, I'll be on my way now. That's enough of this place for me for now. I can only handle so much. I will implement this in C with my encoding and prove that his statement is what is nonsensical here.
 
i forgot that was today, thanks for the reminder
i once wrote an arbitrary precision arithmetic library that handled numbers as text strings. it was slooooow.
 
did you use fft for multiplication?
 
no, that would have been too fast.
 
6:32 PM
1
Q: Dice rolling procedure and taking highest three of four numbers: how to calculate this probability?

ClarinetistConsider the following procedure with four six-sided dice: Roll all four of them. If any of them result in a one, then re-roll those dice until they no longer are ones. Take the highest three numbers of the dice and sum them. Suppose we repeat this procedure six times and generate numbers $Y_1,...

 
you know you've asked a good question when people spend time in the comments fighting the hypotheses of the question.
 
Lol
 
this does seem like something amenable to a closed form solution although i would handle it as you appear to have already done via monte carlo methods.
 
Yeah, I'm curious to see if there's a closed-form solution.
I think it can be done. I probably just don't know the right vocabulary or tools for it.
 
there was a guy at my grad school where if you gave him something like this it would turn into computing a coefficient of a multivariable polynomial. he was always right.
 
6:35 PM
LOL
 
but lulu's comment makes a lot of sense
although perhaps it's better to think of it as a first step to get the solution
 
he quit math after getting his phd. i think he just manages his family's money now. nice job if you can get it
 
Let $w=\exp(2i\pi/10)$. Then we have that $[\mathbb{Q}(w):\mathbb{Q}]=4$, but I'm getting that $\mathbb{Q}=\{a_1+a_2w+a_3w^3+a_4w^7+a_5w^9|a_i\in\mathbb{Q}\}$. Where's the mistake?
 
@Clarinetist rerolling the ones essentially gives you a five sided die.
 
@robjohn I was wondering about that, and I think it makes sense now... because that's all that gets thrown into $Y_i$ at the end of the day.
 
6:40 PM
So we just need to calculate the polynomial with the probabilities for $Y_1$ and raise it to the sixth and we're done
 
derivative it feels to me like the degree ought to be 5 but i have not thought in depth about it
 
@Clarinetist: The highest three of four dice does not simplify as well as the sum of four dice does. As far as I know, you have to case on the lowest number rolled and the number of those.
 
@leslietownes it's $\varphi(10)$
 
i'm also an idiot. i should have mentioned that
 
@Derivative there is no mistake, technically
it's just that $1,\omega,\omega^3,\omega^7,\omega^9$ are not linearly independent like you're probably thinking
 
6:42 PM
which one do I have to get rid of?
 
you probably have a choice.
 
doesn't make a difference
 
okay so it would probably be prettiest to get rid of 1, so I get the numbers coprime to 10 in the exponent
but then how do I get 1 as a linear combination of those?
 
do you understand why $[\mathbb{Q}(w)\colon\mathbb{Q}]=4$?
 
I got it thanks
 
6:55 PM
@leslietownes Did you later try using 'digits' that could be stored in 32-bit ints at least?
And I seriously regret wasting my time on a crank today.
 
@user21820 the prime crank?
 
@robjohn Oh, that was an earlier one. So that's two cranks now.
 
so twin cranks.
 
user no i did not.
maybe you will resolve the twin crank conjecture.
 
@robjohn Nope, fraternal cranks.
@leslietownes What does it say? No matter how long you wait, there will be new cranks popping up within 2 hours of each other?
Not 1 hour, because they take turns to annoy?
 
6:58 PM
in its weak form it asserts only that there are infinitely many twin cranks.
 
@leslietownes That's captured here. TP says "no matter how big n is, there will be primes p and q after n that are within 2 of each other".
 
hey chat!
 
it strikes me as untenable because there are only finitely many people.
 
@robjohn: By the way, could you easily find the first error in the probable RH crankery I asked you about?
 
only some of whom are cranks.
 
7:02 PM
different people have different thresholds to define crankery
 
underwood dudley published a book in which he described someone as a crank, and was sued for defamation. there is seventh circuit law on this now. calling someone a crank is an expression of opinion and not actionable.
 
@leslietownes Well, this phrasing avoid infinity, and if we assume humans won't die out then it's equivalent.
 
i assume that humans will die out.
but your point is well taken
 
In fact, you have to phrase the TP conjecture this way to get it down as a Π2-sentence.
 
maybe humans will die out but what about boltzmann humans?
 
7:05 PM
After the third minute 1:0 england.
 
oh wow.
 
@vitamind Is it surprising? I thought it's a poisson process.
=P
The next goal can occur any minute now... =P
 
that was a great goal. i do think it's funny that england supporters sing a song that is entirely about england losing.
 
@user21820 mute button lookin real good today
 
@leslietownes I don't have that (big) book, but there's a free illuminating article "What to do when the trisector comes".
 
7:12 PM
i have all of his books. a budget of trisections (later republished as the trisectors) and mathematical cranks.
 
@vitamind Lol you want to mute me?
 
he's a very engaging writer.
 
muting is for weaklings
 
I should say that I don't even support either team. I was just thinking about the mathematics behind goal distribution over time.
I mean, we know there is psychology, but if we ignore that it ought to be poisson, right?
If so, it wouldn't be any more surprising regardless of how earlier or late in the game.
 
yes. most of human endeavor can be reduced to observing poisson processes.
 
7:14 PM
@user21820 Mmhh :) well played
 
@LadiesandGents Besides Dudley's article, there are others who have spent time on the problem of crankery, even including two Math SE moderators (one past and one present):
38
Q: Spotting crankery

EugeneUnderwood Dudley published a book called mathematical cranks that talks about faux proofs throughout history. While it seems to be mostly for entertainment than anything else, I feel it has become more relevant in modern mathematics. Especially with the advent of arXiv, you can obtain research pa...

John Baez's list in the question itself is really on target, even though it was meant for physics.
 
yes i think baez's list is spot on.
there was a guy who used to send us very expensively bound volumes in which he argued a proof of the riemann hypothesis. it was all nonsense and ticked most of these boxes.
i never figured out how he identified my home address.
 
lol
 
i own a house now, so you can find me from state property records, but i didn't, at the time. and i'd just moved. and somehow he found me.
 
most "cranks" on the site check a small number of those boxes
they just post long stuff that goes nowhere
and is full of computations
 
7:23 PM
there was a funny guy on main a while ago who claimed that Abel-Ruffini was wrong and nobody but him understood radicals
 
I wish the people who wrote false proofs of stuff could at least write what they are doing
Like someone who actually solves the problem
Like in actual theorems where they go, the main theorem of the article is bla bla bal. The proof consists of first analyzing such possible forms, and utilizing this technique to show it must satisfy bla bla.
 
cranks are usually horrible at explaining (and also at understanding)
 
But most false proofs are just "recall the collatz conjecture is this". Now look at these drawings and we now do a bunch of computations. We won't break our results into neat lemmas though.
 
@leslietownes Expensive? What a waste of money. Give the money to me and I will make sure it goes to needy people.
@Thorgott Radical.
@leslietownes @Thorgott: Do either of you have familiarity with RH? There is in fact one probable RH crankery that I saw many years ago but it actually looked quite non-cranky at first glance, and I didn't notice any obvious sign of crankery. But I hesitate to ask people because, well, I don't want to be associated with asking about crankery...
 
7:48 PM
i don't have familiarity with it. i do remember a professor in graduate school expressing nervousness when someone sent him something that did not use "s" at the variable in the zeta function. the inference was, maybe this person, completely unaware of the usual notation, has actually come up with something.
people with phds in math are the biggest cranks. i had a guy buttonhole me at a conference and talk about the riemann hypothesis. all nonsense as far as i could tell but he had proved something decent in 1970 so they still let him in the door.
and he was very intelligent. you couldn't write him off although i do think he was wrong in his approach to RH.
i don't have the mental horsepower to figure out famously unsolved conjectures. you need to be partially mad to do it.
 
@user21820 I looked at it, but I haven't yet.
 
@leslietownes Ahaha.. I don't want to go near, but I still have some curiosity.
 
Is there a dupe target for finding multiplicative order of a residue mod n?
 
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