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Author Topic: Mathematics Help Thread  (Read 192074 times)

bloop_bleep

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Re: Mathematics Help Thread
« Reply #2535 on: January 06, 2021, 03:01:03 pm »

A measure is just a method of assigning a "size" to a set which respects some intuitive properties. The one most commonly used on the real numbers is the Lebesgue measure. Basically, you start by assigning the size b - a to intervals of the form [a, b]. So [0, 1] has measure 1 and [2, 4] has measure 2. Which intuitively makes sense usually. Then you can get the measures for a whole bunch of other sets by applying measure properties. So if you have two disjoint sets A and B, the measure of A union B is the sum of the measures of A and B. So the measure of [0, 1] union [2, 4] is 3. This union rule applies to countable unions, so unions where you can list the elements of the union in order. So for example the union of all sets {r} where r is a rational number works, since rational numbers are countable. But the union of all {r} where r is real doesn't work. So from this you can get that the measure of the set of rational numbers is 0, since the measure of every set {r} where r is real is 0 (since it's the interval [r, r] which has measure r - r = 0), so since the rational numbers are a countable union of such sets the rational numbers have measure 0. More exotic sets that have measures are the classical Cantor set with measure 0 and the fat Cantor set which has measure 1/2.
« Last Edit: January 06, 2021, 03:02:56 pm by bloop_bleep »
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MaximumZero

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Re: Mathematics Help Thread
« Reply #2536 on: January 09, 2021, 10:22:59 am »

Y'all don't have to waste time explaining for the benefit of my dumb ass who won't get it. Just carry on and do your thing.
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Holy crap, why did I not start watching One Punch Man earlier? This is the best thing.
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bloop_bleep

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Re: Mathematics Help Thread
« Reply #2537 on: January 15, 2021, 11:52:10 am »

Y'all don't have to waste time explaining for the benefit of my dumb ass who won't get it. Just carry on and do your thing.

You sure? I think you can understand it. Let me know if you want another explanation.

Anyway, I think I have a proof for Van der Waerden's Theorem which can be used to prove that any subset of the positive integers which my professor calls "piecewise synthetic" "piecewise syndetic", which is a set that is the intersection of a set with bounded above gaps between consecutive elements and a set with arbitrarily long runs of consecutive integers, has arithmetic progressions of arbitrarily long length. That was an exercise he gave. I might write it up and I could show it to you guys if you want.

As an example:

Code: [Select]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Syndetic set (gaps bounded above by 3):
X X   X     X X       X        X  X
Thick set (arbitrarily long runs):
X   X X   X X X       X  X  X  X  X
Piecewise syndetic set (intersection of syndetic and thick set):
X     X     X X       X        X  X

EDIT: There is another exercise which is to find a positive upper density subset of the natural numbers that is not piecewise syndetic.
« Last Edit: January 15, 2021, 06:40:22 pm by bloop_bleep »
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MaximumZero

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Re: Mathematics Help Thread
« Reply #2538 on: January 15, 2021, 06:37:15 pm »

I realize that the post above may have come across as a lot more bitter than I meant it to. Sorry, all. I was trying to be funny and fell flat. But seriously, I'm fascinated with just how much I would have to learn just to be considered semi-literate at these levels. I relish watching the discussion.
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Holy crap, why did I not start watching One Punch Man earlier? This is the best thing.
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bloop_bleep

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Re: Mathematics Help Thread
« Reply #2539 on: January 17, 2021, 01:27:41 pm »

No, it's ok, I understood it was a joke.

Anyway, I've written up a proof of Steinhaus's Theorem using the Lebesgue Density Theorem, though a starting point was given in class. Also, a classmate told me that an example of a positive upper density set that is not piecewise syndetic is the set of square-free integers, which apparently was an example given in class.
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Folly

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Re: Mathematics Help Thread
« Reply #2540 on: February 10, 2021, 11:40:02 pm »

Hey.
I'm not good at mathing...or wording for that matter. But when it comes to mathing, I can usually muddle through just mashing buttons on a calculator and yelling at it until it gives me what what I want.
Since I started messing with Lua a while back(I'm also bad at scripting) I've frequently found myself in situations where I had an iterating value X and needed a correlating value Y which would start at one value and approach another value without ever quite reaching it. Today I finally coaxed a free graphing calculator out of Google and poked at it until it gave me this:



where 'x' is my iterative value, 'b' is the baseline that y is equal to when x is 0, 't' is the upper range that y approaches, and 's' is a scale that shifts the arc up or down as needed.

If anyone has a cleaner or more capable method for what I'm trying to do, I'd love to hear about it. If not, hopefully someone else can get some use out of what I came up with~
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WealthyRadish

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Re: Mathematics Help Thread
« Reply #2541 on: February 11, 2021, 12:23:16 am »


Bear in mind that for small values of 's' with that function you may end up with errors due to floating point precision.

You might find these functions interesting:

https://en.wikipedia.org/wiki/Sigmoid_function

A function based on (1-e-x) or (2/pi)atan(x) is probably what I would go for.
« Last Edit: February 11, 2021, 12:35:59 am by WealthyRadish »
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Folly

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Re: Mathematics Help Thread
« Reply #2542 on: February 11, 2021, 01:52:58 am »

Thanks!
With that wiki page I was able to guess at which buttons to mash until my graph got rid of the hole in the middle. I'm not sure how exactly, but that hole probably would have caused problems with something eventually.
Much appreciated!

« Last Edit: February 11, 2021, 02:09:14 am by Folly »
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methylatedspirit

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Re: Mathematics Help Thread
« Reply #2543 on: February 15, 2021, 11:28:16 pm »

In my Calculus exam, I completely forgot how to integrate anything that wasn't individual polynomial terms. The exam's over, but I'd figure I'd share these questions because I don't have a damn clue how to solve these and I fear these might eat away at my soul if I continue to not know.


I don't know how to solve these. I have some vague ideas about replacing the complex terms with u, but I forgot the steps that would make that method work. I want a step-by-step solution.
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Vector

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Re: Mathematics Help Thread
« Reply #2544 on: February 16, 2021, 01:15:54 am »

3cosx - sin^3xcosx

I can integrate 3cosx! it's 3sinx

I can integrate -sin^3xcosx! it was chain ruled. Undo!! -> power was 3! rewind to 4, cosx is chainruling the base sinx, so antiderivative is 1/4*sinx^4

answer: 3sinx-0.25sinx^4 + C


Other one: when I take the derivative of, say, sqrt(x), it becomes (1/2)x^(-1/2). This is a clue for what happens here.

What if the original function was sqrt(3x^2+1)? Then the derivative is (6x/sqrt(3x^2+1))*(1/2).

That is pretty close to what we got in the integrand, it's just that the constant was wrong. We got 3 and we want 2.

So multiply by 2/3 -> the correct integral/antiderivative is (2/3)sqrt(3x^2+1)+C.
« Last Edit: February 16, 2021, 01:18:27 am by Vector »
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bloop_bleep

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Re: Mathematics Help Thread
« Reply #2545 on: February 16, 2021, 06:48:06 pm »

Both of those can be done with u-subs. The first one make cos(x) dx into dsin(x), second make 2x dx into dx^2, then that into 1/3 d(3x^2 + 1). Then they're essentially polynomial integrals.
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methylatedspirit

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Re: Mathematics Help Thread
« Reply #2546 on: April 22, 2021, 12:59:35 am »

Suppose if you had this function, f(x), and a sequence of random real numbers, R, such that its values are in the range (0,1], and f's definition looks like:
f(x) = sin(2pi * (x / R[0])) when 0 < x < R[0]
f(x) = sin(2pi * ((x - R[0]) / R[1])) when R[0] < x < R[0] + R[1]
f(x) = sin(2pi * ((x - R[0] - R[1]) / R[2])) when R[0] + R[1] < x < R[0] + R[1] + R[2]
...

And the intended idea here is that every time f(x) completes a full cycle, its period changes to a random value for the next cycle. And when that cycle ends, it changes again. And over and over across the non-negative real number line. It's basically a sequence of 1-cycle sine waves of varying periods just glued together at the zero crossings in succession.

Does a function like this, with a period that changes randomly every cycle, still considered to be a periodic function? Is there a name for this class of function?
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da_nang

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Re: Mathematics Help Thread
« Reply #2547 on: April 22, 2021, 05:13:06 am »

A periodic function is a function for which there exists a non-zero constant T such that for all x in the domain, f(x+T) = f(x).

A necessary condition for this to be true for f(x) is for all Rs to be the same, the probability of which is zero.

So I'd say that f(x) is an almost-certainly aperiodic function.
« Last Edit: April 22, 2021, 05:17:38 am by da_nang »
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #2548 on: April 22, 2021, 08:22:46 am »

It's not necessary for all Rs to be the same, it's enough if the sequence R itself is periodic. But again, the probability of that is zero.
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methylatedspirit

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Re: Mathematics Help Thread
« Reply #2549 on: June 26, 2021, 01:32:50 am »

I just realized that the derivative of f(x) = ln(k*x), a is a constant, a >0, is always 1/x. It's because it goes:
f(x) = ln(k*x)
f'(x) = k/x * k
f'(x) = 1/x.

But why? What's the intuition behind this fact? The idea that's in my mind is: "No matter how fast you make ln(x) increase by multiplying x with a nonnegative constant k, its slope is always 1/x.", and that feels counterintuitive to me, even if the math checks out.
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