## Math: Fleeting Thoughts

For the discussion of math. Duh.

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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.
nachomancer

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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.

rat4000
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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, \int \langle v_0,v_1,v_2 \rangle dt = \langle\int v_0 dt,\int v_1 dt,\int v_2 dt\rangle.

Eebster the Great

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### Re: Math: Fleeting Thoughts

Thanks again

rat4000
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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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### Re: Math: Fleeting Thoughts

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

humectant

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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

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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.
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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.
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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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### 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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### 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.

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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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### 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;

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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.

PerchloricAcid

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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.

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### Re: Math: Fleeting Thoughts

I thought this reddit thread was pretty funny.
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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!
PerchloricAcid

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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.
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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.
PerchloricAcid

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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.

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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 )
PerchloricAcid

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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.

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