## Math: Fleeting Thoughts

For the discussion of math. Duh.

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nachomancer
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### Re: Math: Fleeting Thoughts

dissonant wrote:I often wonder about Brouwer's fixed point theorem. I.e "Every continuous function f from a closed disk to itself has at least one fixed point."

Say, by means of an analogy we are stirring a cup of coffee. At any two points in time there is some point which has not moved. Do these fixed points trace out some kind of "path"? I would think so. Intuitively, if the derivative is 0 at a point and the derivatives change continuously over the surface then if you are going to be a fixed point at the next point in time you would need to be arbitrarily close to the original.

Although, I would be happy if fixed point jumped around a lot too...

The points definitely don't move continuously. Consider rotating the coffee first about one point, than about any other point.

rat4000
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### Re: Math: Fleeting Thoughts

Once more regarding simple integration and physics: if I have a velocity that is a three-dimensional vector and want to figure out what happens to the object, can I just take every element of the vector and integrate to get the speed in that particular... dimension? plane? direction? at a particular moment? It's what looks most intuitive, but maybe it's wrong, or maybe there's another way to do it.

By now it must be obvious that, while I like maths, I'm not all that knowledgeable about it; I apologize if the simple things that I'm flooding this thread with have become annoying.

Eebster the Great
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### Re: Math: Fleeting Thoughts

rat4000 wrote:Once more regarding simple integration and physics: if I have a velocity that is a three-dimensional vector and want to figure out what happens to the object, can I just take every element of the vector and integrate to get the speed in that particular... dimension? plane? direction? at a particular moment? It's what looks most intuitive, but maybe it's wrong, or maybe there's another way to do it.

By now it must be obvious that, while I like maths, I'm not all that knowledgeable about it; I apologize if the simple things that I'm flooding this thread with have become annoying.

Mostly you can just integrate vectors component-by-component. So for example, [imath]\int \langle v_0,v_1,v_2 \rangle dt = \langle\int v_0 dt,\int v_1 dt,\int v_2 dt\rangle[/imath].

rat4000
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### Re: Math: Fleeting Thoughts

Thanks again

doogly
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### Re: Math: Fleeting Thoughts

You only have to be careful if your equations of motion include coupling between different directions. For example, if you are working with a rotational moment of inertia that is not diagonalized, you have to do that first, and then you can proceed. In the Kepler problem, you're doomed!
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Magnanimous
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### Re: Math: Fleeting Thoughts

Egosearching for good measure. I'm sure I'll have a fleeting thought eventually.

humectant
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### Re: Math: Fleeting Thoughts

I've been taking a course on Riemann Surfaces and the Riemann-Hilbert Problem with applications to partial differential equations. So far we've only really seen the Riemann-Hilbert Problem arise in elliptic pde's, and even then only linear ones. It seems unlikely that hyperbolic or parabolic pde's would give rise to such methods, but I wonder sometimes if there are any known examples of fully nonlinear 2nd order (or more) pde's which can be solved via a boundary value problem in the complex plane, like a Riemann-Hilbert problem. Sadly, I've never had the chance to ask, and my research is far removed from the subject.

PM 2Ring
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### Re: Math: Fleeting Thoughts

I just noticed a cute coincidence: the cube root of 3/2 is approximately the natural logarithm of pi.

(3/2)**(1/3) ~= 1.14471424
log(pi) ~= 1.14472989

Also,
1/(log(pi)**3 - 1.5) ~= 16261.11692974
log(pi)**3 - 1/16261 ~= 1.49999999955779
log(pi)^3 ~= 1.50006149639
1.5+1/16261 ~= 1.50006149683

And
exp((3/2 + 1/16261) ** (1/3)) ~= 3.141592653943
pi ~= 3.14159265358979

fishfry
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### Re: Math: Fleeting Thoughts

Buttons wrote:Let's talk about mathematical eponymous adjectives! Pretty much all the ones I can think of are of the form name+ian. What exceptions do you know? My favorite I just learned yesterday: the eponymous adjective for MacMahon is Mahonian! Beats platonic any day, if you ask me.

Boolean.

the tree
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### Re: Math: Fleeting Thoughts

It's there a word to describe the smallest degree of polynomial to have a given number as it's root?

So if it was f(x) then when x is a rational number, f(x) would be 1. f(sqrt(2))=2, f(pi)=infinity.

I dunno if that would have any interesting properties or whatever but it feels like it should have a name.

doogly
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### Re: Math: Fleeting Thoughts

The degree of the minimal polynomial
http://en.wikipedia.org/wiki/Minimal_po ... _theory%29
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jestingrabbit
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### Re: Math: Fleeting Thoughts

Yeah, that's right, but you do need to specify that the coefficients of the polynomial are all rational numbers, or, what amounts to the same thing, integers.
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Talith
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### Re: Math: Fleeting Thoughts

It's also the dimension of the field that the polynomial is taken over augmented by the root. So for instance the dimension of Q(a) where Q is the field of rationals, a is the square root 2, then the dimension of Q(a) as a Q-vector space is 2, because a minimal polynomial (over Q) of sqrt(2) is given by x2-2. Further reading: http://en.wikipedia.org/wiki/Field_extension http://en.wikipedia.org/wiki/Galois_theory.

antonfire
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### Re: Math: Fleeting Thoughts

It's also called the degree of that (algebraic) number. Fun fact: the growth rate of this thing is an algebraic number of degree 71.
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Dason
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### Re: Math: Fleeting Thoughts

I love that the sampling distribution of the sample mean for independent and identically distributed standard Cauchy random variables is itself distributed as a standard Cauchy.
double epsilon = -.0000001;

PerchloricAcid
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### Re: Math: Fleeting Thoughts

PM 2Ring wrote:I just noticed a cute coincidence: the cube root of 3/2 is approximately the natural logarithm of pi.

(3/2)**(1/3) ~= 1.14471424
log(pi) ~= 1.14472989

Way cool, man
Reminded me of this.

PM 2Ring
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### Re: Math: Fleeting Thoughts

PerchloricAcid wrote:
PM 2Ring wrote:I just noticed a cute coincidence: the cube root of 3/2 is approximately the natural logarithm of pi.

(3/2)**(1/3) ~= 1.14471424
log(pi) ~= 1.14472989

Way cool, man
Reminded me of this.

Thanks, HClO4. I'm glad I'm not the only one here that thinks it's cool.

Yeah, comic #1047 did inspire me (slightly) to make that post, and also http://xkcd.com/217/, although I've been interested in rational / integer expressions that are approximately equal to irrational values for a long time. One of my favourites is Ramanujan's constant.
Also see http://en.wikipedia.org/wiki/Almost_integer

That thing with the polynomial roots is interesting! I guess that it's not that surprising that an expression for the interval containing the roots can be given in terms of the coefficients, but it is a little surprising (IMHO) that a simple quadratic expression gives such good bounds.

Magnanimous
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### Re: Math: Fleeting Thoughts

I thought this reddit thread was pretty funny.

PerchloricAcid
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### Re: Math: Fleeting Thoughts

Thanks, PM 2Ring. Interesting stuff

I have a growing fascination for Riemann's theorem, which states that you can rearrange the terms of conditionally convergent series so that it converges to ANY value or diverges.

I mean, you can take any conditionally convergent series and rearrange its terms so it converges to, say, Pi or Phi or e or 2 or -154 or whatever number you fucking wish. Wow!

Tirian
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### Re: Math: Fleeting Thoughts

I'm not out to harsh your squee, but it's pretty ordinary once you realize what they're up to. If you've got some conditionally convergent sequence a, then you can partition it into subsequences p and n such that p are all the positive terms and n are all the strictly negative terms. Since a is conditionally convergent, p and n both diverge. So let x be the number that we want the rearranged sequence to converge to. (WLOG, I'll assume that x is positive.) Start with enough terms of p to drive the sum over x, which you can do because p diverges. Then pick enough terms of n to drive the sum back under x (again, since n diverges). "What's left" of p and n are both divergent since the sum of the terms we used from each are finite, so we can do it again, and again and again and so on. Will we eventually use every term of a? Yes -- the i'th term of a is somewhere within the first i terms of either n or p and therefore will be used within the first i iterations of our drawing.

PerchloricAcid
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### Re: Math: Fleeting Thoughts

I know, I've learnt the proof together with the theorem, but it's very fun nevertheless.

PM 2Ring
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### Re: Math: Fleeting Thoughts

PerchloricAcid wrote:I know, I've learnt the proof together with the theorem, but it's very fun nevertheless.

What she said. Still, Tirian's explanation is a lot easier to read than the formal proof on the Wiki page. And although it's not hard for us to understand what's going on, we ought to bear in mind that this "paradox" was considered to be a major problem by great mathematicians like Euler before Riemann showed how to resolve it.

Note that it can also apply to infinite products, eg the Wallis product for pi/2, since an infinite product of positive terms can be turned into an infinite sum by taking logarithms.

PerchloricAcid
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### Re: Math: Fleeting Thoughts

Tirian's explanation is basically the proof formulated informally.
Even the formal proof is quite easy to understand. I mean, it's easy to get once you've heard the idea. However, Riemann was damn smart to come up with it in the first place

Btw, here's a fun and related quote by Niels H. Abel: “Divergent series are the invention of the devil, and it is shameful to base on them any demonstration whatsoever.”

(The Wallis product is new to me, so I'll be checking it out right now. I haven't actually worked with infinite products almost at all; we've used the converting-infinite-products-to-sums-by-ln-ing-them trick in some convergence test proofs. Math is fun, damn )

Sizik
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### Re: Math: Fleeting Thoughts

I noticed that if you consider the most significant bit as repeating infinitely leftward, then two's complement signed integers are essentially 2-adic numbers.
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### Re: Math: Fleeting Thoughts

Sizik wrote:I noticed that if you consider the most significant bit as repeating infinitely leftward, then two's complement signed integers are essentially 2-adic numbers.

Yes, though there are other 2-adic numbers as well, those corresponding to 'infinite integers', i.e. semi-infinitely long strings of bits.

phlip
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### Re: Math: Fleeting Thoughts

Ugh, I wish I could remember how to statistics properly...

OK, so I'm trying to test whether the probabilities quoted by some game are accurate. I have a dataset of claimed percent-chance-to-hit and then a yes/no value of whether or not it actually hit. I don't have enough data to try to do something like "just look at the points where it says 75%, and see if 3/4 of them actually hit", since that's only, like, 2 points... I need to analyse the whole dataset as one big thing.

Now, I figured out I can just run a linear-regression trend line through the data and get an on-average chance-to-hit value at each point, but I have no idea what analyses need to be done to figure out error bars for this thing. From inspection, I don't believe the trend line is statistically significantly different from y=x, but I have no idea how to check that. I've never really worked with this sort of binary data before.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

cyanyoshi
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### Re: Math: Fleeting Thoughts

It's funny how statistics knowledge just disappears into the aether when you stop using it! The last statistics class I took was in high school, way before I took linear algebra. I had to relearn everything about Bayesian statistics from scratch when I was doing basically the same analysis as you are doing! Why they don't require a statistics course for my major (aerospace engineering) is anyone's guess.

Tirian
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### Re: Math: Fleeting Thoughts

I'm just brainstorming here. But I recall that you can demonstrate that two variables are not independent by doing a linear regression and rejecting the null hypothesis that m=0. I wonder if you can do the same thing here except that you have two null hypotheses: m=1 and b=0. Perhaps if you can't reject either, then you can't reject the claim that the formula for hit success is valid.

cyanyoshi
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### Re: Math: Fleeting Thoughts

It's a little more complicated than that, since there is a range of hit percentages that go from 0% to 100%. The tricky part is determining what the likelihood is that the hit rate is any given percentage while taking into consideration the data you collect, i.e. the number of hits and misses for a predetermined sample size. One way to do that (or at least the way I did it with my very informal analysis) is to first assume that the true hit rate was randomly chosen from a set of possible values, such as every whole multiple of 1% up to 100%. For each candidate percentage, I calculated the probability of getting the collected data (from a binomial distribution). Then I was able to build a distribution of relative likelihoods of the true hit rate. If you are testing to see if the true hit percentage is 75%, you would see if "75%" falls within your confidence interval from the percentage suggested by the data. I am by no means an expert on statistics, so maybe there is an easier way to go about this problem under the assumption that the range of possible values is continuous, like with the Student's t-test. I just don't know how to properly incorporate the knowledge that the hit percentage must fall in the interval [0,1].

Dopefish
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### Re: Math: Fleeting Thoughts

I think the relevant wiki page is on logistic regression, but I was very tired and somewhat drunk when I looked into this last night, not to mention it's been ages since I did stats.

Dason
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### Re: Math: Fleeting Thoughts

You could probably use logistic regression if you converted the assumed hit probabilities into the logit of the those probabilities and used that as your predictor. This wouldn't work if 0 or 1 are values that are hypothesized for the hit probabilities. At this point you would want to simultaneously test if the coefficient in for the predictor you created was 1 and if the intercept was 0.

If you use R...

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# 20 observationsN <- 20# Simulate the 'true' hit probabiltiiesp <- sort(runif(N))# Simulate the number of observations we get from each of the hit probabiltiesn <- rpois(N, 2) + 1# Simulate the number of hitsy <- rbinom(N, n, p)# Create our predictorx <- log(p/(1-p))# Need this for running the logistic regression in Rn.fail <- n-y# Run logistic regression using x as the predictoro <- glm(cbind(y, n.fail) ~ x, family = binomial)# Create a reduced model where we force the intercept# to be 0 and the coefficient in front of x to be 1o.red <- glm(cbind(y, n.fail) ~ offset(x) - 1, family = binomial)# Test if the full model where the intercept and slope# can vary provides a significantly better fit than# the reduced model.  If it does then we have evidence# the true hit probabilities aren't as advertised.anova(o.red, o, test = "Chisq")

The output of the last line is

Code: Select all

> anova(o.red, o, test = "Chisq")Analysis of Deviance TableModel 1: cbind(y, n.fail) ~ offset(x) - 1Model 2: cbind(y, n.fail) ~ x  Resid. Df Resid. Dev Df Deviance Pr(>Chi)1        20     11.486                     2        18     10.730  2  0.75591   0.6853

Results may vary since I forgot to set a seed.

So our p-value is 0.6853. So we don't have significant evidence that the hit probabilities aren't as advertised. Note that this only tests a very specific type of departure from the advertised hit rate though. I toyed around with the idea of just a likelihood ratio test but after running some simulations the sampling distribution of the test statistic was kind of wonky so I either messed something up or the asymptotics hadn't kicked in yet. It wouldn't be *too* bad to use monte carlo methods to get a better estimate of the sampling distribution of the likelihood ratio statistic though.

But the logistic regression approach seems reasonable. The method of just using linear regression might be flawed due to the nature of the data but then again regression is relatively robust... I'd have to toy around with some simulations...
double epsilon = -.0000001;

WibblyWobbly
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### Re: Math: Fleeting Thoughts

The other day, while trying to fall asleep, I was doing some recreational math when I found that the only happy Mersenne primes I could come up with were 7 and 31, and further that the perfect numbers generated from those primes (28 and 496) were themselves happy (perfectly happy, if you will). Now, I also found that 8128 was happy, so there are other perfectly happy numbers. But are there any other happy Mersenne primes anyone knows of? I tried to calculate as much as I could before falling asleep (got to 231-1 before it happened), but I didn't seem to find any. Might have messed up the math, might have missed something, just wondered if anyone else had considered it.

LucasBrown
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### Re: Math: Fleeting Thoughts

The 48 known Mersenne primes have exponents 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, and 57885161.

Five of these produce happy Mersennes primes — 3, 5, 2281, 19937, 132049, and 42643801.

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#! /usr/bin/env pythonfrom sys import stdoutdef is_happy(n):    while n != 89 and n != 1: n = sum(int(i)**2 for i in str(n))    return n==1exps = [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161]happy, unhappy = [], []for n in exps:    print n,    stdout.flush()    if is_happy(2**n - 1):   happy.append(n)    else:                  unhappy.append(n)    print happyprint len(happy), len(unhappy)

cyanyoshi
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### Re: Math: Fleeting Thoughts

It's funny how things work out, but I was looking more in-depth into the very same problem I had glanced at previously: essentially determining the probability of "heads" for a biased coin where each flip is independent from previous flips and the probability can be any real number from 0 to 1 and blah blah blah. Doing a bit of research confirms that I reinvented the wheel, but it was a nice exercise to derive for myself the probability (likelihood?) density function for performing n trials, resulting in k number of heads. The distribution looks like l(p) = C(1-p)n-kpk, where C is a positive constant to make the integral from p = 0 to 1 equal unity.

It's neat how p = k/n is the most likely value for the coin's probability of heads (i.e. the mode of the distribution), but the mean probability is always a bit closer to 0.5 because that seems to be our best guess assuming complete ignorance, in a sense. That lingering uncertainty goes away as n gets really big, confirming my intuition about these kinds of tests. I also like how the function turns out to be a nice-and-easy polynomial of degree n rather than something more weird. Like I said, it's funny how things work out.

Eebster the Great
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### Re: Math: Fleeting Thoughts

Well sure, because we can't be "completely ignorant" of the probability of heads. We at least know it must be between 0 and 1.

Tirian
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### Re: Math: Fleeting Thoughts

It seems to me that one of the fundamental theorems of plane geometry that comes up over and over again is that if ABC is a right triangle with hypotenuse AB and D is the foot of the altitude from C to AB, that (AB/BC)=(BC/CD). Is there a common name for that theorem aside from "Euclid X.33"? If not, would you feel the urge to write more than "Since ACB is a right angle and CD is perpendicular to AB, we may conclude that (AB/BC)=(BC/CD)" in a paper?

jestingrabbit
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### Re: Math: Fleeting Thoughts

Tirian wrote:It seems to me that one of the fundamental theorems of plane geometry that comes up over and over again is that if ABC is a right triangle with hypotenuse AB and D is the foot of the altitude from C to AB, that (AB/BC)=(BC/CD). Is there a common name for that theorem aside from "Euclid X.33"? If not, would you feel the urge to write more than "Since ACB is a right angle and CD is perpendicular to AB, we may conclude that (AB/BC)=(BC/CD)" in a paper?

I'd probably want to slip the word "similarity" in there somewhere.
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cyanyoshi
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### Re: Math: Fleeting Thoughts

I kind of stumbled on an interesting property of analytic functions the other day, and I wondered if it had a proper name. Provided that the series converges,

f(x) + f'(x)*h/1! +f''(x)*h2/2! + f'''(x)*h3/3! + ... = f(x+h)

or alternatively, ehDf = f(x+h)
where h is a constant and D is the differential operator, and Dnf is interpreted as the nth derivative of the function f evaluated at x.

As someone who has never taken a course on real or complex analysis (other than vanilla calculus), I just thought this was really cool to discover on my own!

brenok
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### Re: Math: Fleeting Thoughts

This is Taylor Series. I thought it was taught in Calculus.

cyanyoshi
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### Re: Math: Fleeting Thoughts

And it is at this moment, I facepalmed harder than I ever facepalmed. All I did was reverse the constant and variable in Taylor's formula.