Thread for Non-CA Academic Questions

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Re: Thread for Non-CA Academic Questions

Post by Rhombic » November 12th, 2017, 7:57 pm

Can the resonance of a room (via what frequencies are perceived compared to the emitted ones) be used, after Fourier transform, to understand the shape of the room?
i.e. no echo or that kind of stuff, but the resonance of certain harmonics (which is known to happen but has never been used in this way)
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Re: Thread for Non-CA Academic Questions

Post by muzik » November 15th, 2017, 3:37 pm

Has any research been done into zero-sided polygons?
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Re: Thread for Non-CA Academic Questions

Post by Cclee » November 15th, 2017, 3:40 pm

muzik wrote:Has any research been done into zero-sided polygons?
Only to prove them impossible
^
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Re: Thread for Non-CA Academic Questions

Post by A for awesome » November 22nd, 2017, 5:28 pm

Me, a while ago, wrote:I'm currently trying to figure out how a universe in which Newton's second law is F=mj would behave. So far, I've determined that particles travel in parabolas (possibly degenerate) when not acted upon by any force, objects falling under a constant force with zero horizontal acceleration descend cubic curves, and that colliding objects attracted by a force rebound either superelastically or inelastic with a probability of 1 depending on how the force varies with distance — no kind of energy is in any way conserved, at least if my intuition is correct. I'm wondering what kind of work has been done on this problem.
I wrote some Python code to model this. It clearly demonstrates the lack of conservation of energy of a system consisting of a body orbiting a fixed point under an inverse square law. Run in a 100x51 terminal window for best results.

Code: Select all

from __future__ import print_function
from time import sleep
import operator
ARRAY_SIZE = (100, 50)
XRES_OVER_YRES = 0.5
class V:
    def __init__(self, *args):
        if len(args) == 1 and hasattr(args[0], '__iter__'):
            self._t = tuple(map(float, args[0]))
            self._d = len(self._t)
        else:
            self._t = tuple(map(float, args))
            self._d = len(args)
    def __str__(self):
        return "<" + str(self._t)[1:-1] + ">"
    def __repr__(self):
        return "V" + str(self._t)
    def __eq__(self, other):
        return self._t == other._t
    def __ne__(self, other):
        return self._t != other._t
    __hash__ = None
    def __len__(self):
        return self._d
    def __getitem__(self, ind):
        return self._t[ind]
    def __setitem__(self, ind, val):
        foo = self._t[ind] #To raise the correct exceptions
        l = list(self._t)
        l[ind] = val
        self._t = tuple(l)
    def __iter__(self):
        return (i for i in self._t)
    def __add__(self, other):
        return V(map(operator.add, self._t, other._t))
    def __sub__(self, other):
        return V(map(operator.sub, self._t, other._t))
    def __mul__(self, other):
        if type(other) == type(self):
            return sum(map(operator.mul, self._t, other._t))
        else:
            return V(map(operator.mul, self._t, (other,)*self._d))
    def __div__(self, other):
        return V(map(operator.div, self._t, (other,)*self._d))
    def __truediv__(self, other):
        return V(map(operator.truediv, self._t, (other,)*self._d))
    def __mod__(self, other):
        return V(map(operator.mod, self._t, (other,)*self._d))
    def __neg__(self):
        return V(map(operator.neg, self._t))
    def __concat__(self, other):
        return V(self._t+other._t)
    def __abs__(self):
        return sum(map(operator.mul, self._t, self._t))**0.5
class Obj:
    def __init__(self, x, v, a, disp=None):
        self.x = x
        self.v = v
        self.a = a
        self.visible = disp is not None
        if self.visible:
            self.disp = disp
    def tick(self, dt, n):
        for i in xrange(n):
            self.v += self.a*dt
            self.x += self.v*dt
        if self.visible:
            self.disp.objsToDisp.append(self)
    def jerk(self, j, dt):
        self.a += j*dt
    def getSymbol(self):
        return "?"
class Disp:
    def __init__(self, x_ul, xres):
        self.objsToDisp = []
        self.x = x_ul
        self.xres = xres
        self.yres = xres/XRES_OVER_YRES
    def display(self):
        disp = [[" " for i in xrange(ARRAY_SIZE[0])] for j in xrange(ARRAY_SIZE[1])]
        for obj in self.objsToDisp:
            x_obj = obj.x
            s = obj.getSymbol()
            disp[int(x_obj[1]/self.yres)][int(x_obj[0]/self.xres)] = s
        print("\n".join(["".join(i) for i in disp]), end="\r"+"\x1b[A"*(len(disp)-1))
        self.objsToDisp = []
d = Disp(V(0,0),1)
q = Obj(V(0,25),V(0,-3),V(0,0),d)
z = Obj(V(15,25),V(0,0),V(0,0),d)
t = 0.1
for i in xrange(int(15/t)):
    q.jerk((z.x-q.x)/abs(z.x-q.x)**3*60, t)
    q.tick(t, not not i)
    z.tick(t, not not i)
    d.display()
    sleep(t)
I plan to eventually make an ASCII platformer game based on this code, but who knows if it'll actually ever happen.
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)

$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$

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Re: Thread for Non-CA Academic Questions

Post by Majestas32 » November 22nd, 2017, 11:17 pm

gameoflifemaniac wrote:
calcyman wrote:
gameoflifemaniac wrote:Is it possible to emulate a normal computer in a quantum computer?
Yes, there's a standard trick for this. Given a function f : {0, 1}^n --> {0, 1}^m expressed as a circuit C of classical logic gates, you can replace the gates in the circuit with reversible equivalents to yield a reversible circuit C' (which may have lots of ancillary '0' inputs and some arbitrary messy outputs):

[n input bits][k ancillary '0' bits] --> [m output bits][n+k-m unwanted garbage bits]

Suppose we have another m ancillary '0' bits at this stage. Then we can CNOT the output bits with these ancillary bits to produce another copy of the output, like so:

[n input bits][k ancillary '0' bits][m extra '0' bits] --> [m output bits][n+k-m unwanted garbage bits][m output bits]

Now apply C' in reverse to the first n+k bits to clean up the mess we created:

[n input bits][k ancillary '0' bits][m extra '0' bits] --> [n input bits][k ancillary '0' bits][m output bits]

The upshot of this is that the ancillary '0' bits can be reused in a future computation. This combined circuit, ignoring the k ancillary '0' bits, actually computes the reversible function:

f : {0, 1}^(n+m) --> {0, 1}^(n+m)
(x, y) --> (x, f(x) XOR y)

Now, every reversible classical circuit is a quantum circuit (permutation matrices are unitary matrices), so this can be built out of quantum gates.
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Re: Thread for Non-CA Academic Questions

Post by muzik » November 23rd, 2017, 9:22 am

The degenerate star polygon {6/3} can be visualised as something along the lines of this (a compound of three digons):

Code: Select all

x = 122, y = 107, rule = B3/S23
102bo$102bo$101b2o$36bo64bo$36bo63bo$37bo62bo$37bo61bo$37bo61bo$38bo
59bo$38bo58bo$38bo58bo$39bo56bo$39bo55bo$40bo54bo$40bo53bo$40bo52bo$
41bo51bo$41bo50bo$42bo48bo$42bo47bo$43bo45bo$43bo44bo$44bo42bo$44bo42b
o$45bo40bo$45bo39bo$45bo37b2o$45bo36bo$46bo34bo$46bo33bo$46bo33bo$47bo
31bo$47bo30bo$48bo28bo$48bo28bo$49bo26bo$49bo25bo$50bo23bo$50bo23bo$
50bo22bo$51bo20bo$51bo19bo$52bo17bo$52bo16bo$53bo14bo$53bo14bo$54bo12b
o$54bo11bo$55bo9bo$55bo8bo$55bo7bo$56bo5bo$56bo3b2o$56bo2bo$8o49b2o$8b
21o28bo$29b19o8bobo$48b74o$54bo3bo$53bo4bo$52bo5bo$51bo7bo$50bo8bo$49b
o9bo$48bo10bo$47bo11bo$46bo13bo$46bo13bo$45bo14bo$44bo16bo$43bo17bo$
43bo17bo$42bo19bo$41bo20bo$40bo22bo$40bo22bo$39bo23bo$38bo24bo$37bo26b
o$36bo27bo$35bo28bo$34bo29bo$33bo31bo$33bo31bo$32bo33bo$31bo34bo$30bo
35bo$30bo36bo$29bo37bo$28bo38bo$27bo40bo$26bo41bo$25bo42bo$24bo43bo$
23bo44bo$23bo45bo$22bo46bo$22bo46bo$20b2o47bo$19bo49bo$18bo50bo$17bo
51bo$16bo$14b2o$12b2o$10b2o$8b2o!
Could this be seen as a compound of two {3/1.5} polygons, which would look something like this?

Code: Select all

x = 63, y = 86, rule = B3/S23
28bo$28bo$28bo$28bo$28bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$
29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$
29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$29bo$
29bo$29bo$29bo$28b2o$29bo$28b2o$28bobo$27bo2bo$26bo4bo$26bo5bo$25bo7bo
$24bo9bo$24bo10bo$23bo11bo$23bo12bo$22bo14b2o$21bo17bo$21bo18bo$20bo
20b2o$19bo23b2o$18bo26bo$18bo27bo$17bo29b2o$16bo32bo$15bo34bo$15bo35bo
$14bo37bo$13bo39bo$12bo41bo$12bo42bo$11bo44bo$10bo46bo$10bo47bo$9bo49b
o$8bo51bo$8bo52bo$7bo54bo$6bo$5bo$5bo$4bo$3bo$2bo$2o!

Has any research been done into "fractional stellation", for that matter?
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Re: Thread for Non-CA Academic Questions

Post by gameoflifemaniac » November 23rd, 2017, 9:42 am

How are bits converted into music?
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Re: Thread for Non-CA Academic Questions

Post by calcyman » November 23rd, 2017, 5:28 pm

gameoflifemaniac wrote:How are bits converted into music?
In a .wav file, you simply specify the amplitude of the sound wave (using 16-bit fixed-precision reals) every 1/44100th of a second. By Nyquist's sampling theorem, this is sufficient to fully recover any frequencies up to 22050 Hz, which is the upper range of human hearing. Then, you just pass it through a digital-to-analogue converter and into a loudspeaker, which therefore vibrates according to that sound wave.

(Other formats such as .mp3 are more complicated, but they are ultimately decompressed into .wav files.)

It's actually thanks to Fourier who had the idea that an orchestra could be decomposed into sinusoidal waves.
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Re: Thread for Non-CA Academic Questions

Post by Apple Bottom » November 24th, 2017, 12:30 pm

calcyman wrote:In a .wav file, you simply specify the amplitude of the sound wave (using 16-bit fixed-precision reals) every 1/44100th of a second. By Nyquist's sampling theorem, this is sufficient to fully recover any frequencies up to 22050 Hz, which is the upper range of human hearing. Then, you just pass it through a digital-to-analogue converter and into a loudspeaker, which therefore vibrates according to that sound wave.
Xiph.org's Monty has a good write-up on this sort of thing, BTW, in the context of 24-bit/192-KHz sampling as applied to digital music.
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Re: Thread for Non-CA Academic Questions

Post by muzik » November 24th, 2017, 4:35 pm

Why do the small stellated dodecahedron and great dodecahedron fit onto a sphere so well if they have a Euler characteristic of -6 - wouldn't it make more sense for them to tesselate a quadruple torus instead?
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Re: Thread for Non-CA Academic Questions

Post by Macbi » November 24th, 2017, 5:04 pm

muzik wrote:Why do the small stellated dodecahedron and great dodecahedron fit onto a sphere so well if they have a Euler characteristic of -6 - wouldn't it make more sense for them to tesselate a quadruple torus instead?
If you just look at a small stellated dodecahedron it looks like it's made out of 60 isosceles triangles with five stuck on each face of a dodecahedron. But this shape has an Euler characteristic of 2, just like the sphere. The small stellated dodecahedron only has an Euler characteristic of -6 if you consider it to be made of 12 stars which intersect each other. Since its faces intersect each other, the small stellated dodecahedron can't really be considered a valid polyhedron embedded in 3D space. So the idea of fitting it into a sphere doesn't really make sense.

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Re: Thread for Non-CA Academic Questions

Post by muzik » November 24th, 2017, 5:26 pm

Macbi wrote: Since its faces intersect each other, the small stellated dodecahedron can't really be considered a valid polyhedron embedded in 3D space.
By this logic wouldn't that invalidate every star polygon in 2D space?
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Re: Thread for Non-CA Academic Questions

Post by Majestas32 » November 25th, 2017, 3:58 pm

muzik wrote:
Macbi wrote: Since its faces intersect each other, the small stellated dodecahedron can't really be considered a valid polyhedron embedded in 3D space.
By this logic wouldn't that invalidate every star polygon in 2D space?
Only if they are drawn with intersecting lines. In that case they technically are not polygons.
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Re: Thread for Non-CA Academic Questions

Post by muzik » November 27th, 2017, 4:20 am

I don't remember hearing anything saying that self-intersecting polygons can't be regular...
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Re: Thread for Non-CA Academic Questions

Post by calcyman » November 27th, 2017, 9:39 am

I've heard 'polytope' used to mean any of the following four (increasingly restrictive) conditions:
  1. Abstract polytope (a partially ordered set of faces obeying certain properties);
  2. Polytope which can be immersed in R^n;
  3. Polytope which can be embedded in R^n;
  4. Convex polytope.
So the regular (flag-transitive) polyhedra (3-dimensional polytopes) by definitions (iv) and (iii) are just the 5 Platonic solids, whereas the regular polyhedra by definition (ii) include the Kepler-Poinsot self-intersecting polyhedra. Definition (i) also allows much more complicated things such as the Klein quartic, and square tessellations of a torus, and hemi-dodecahedra, etc.

But anyway, I think you're arguing over whether the definition (ii) or (iii) is correct. I've seen mathematicians use any of these four definitions, so you're both right.
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Re: Thread for Non-CA Academic Questions

Post by muzik » November 27th, 2017, 1:13 pm

How do i calculate the Euler characteristics of regular hyperbolic polygonal tilings, and is it possible to tessellate a certain order torus with these?
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Re: Thread for Non-CA Academic Questions

Post by calcyman » November 27th, 2017, 3:04 pm

muzik wrote:How do i calculate the Euler characteristics of regular hyperbolic polygonal tilings, and is it possible to tessellate a certain order torus with these?
Do you mean the infinite tessellation of the hyperbolic plane (such as the order-3 heptagonal tiling) or finite quotients thereof (such as the Klein quartic's tiling with 24 heptagons)?

Infinite: Read 'The Symmetries of Things' by John Conway, Heidi Burgiel, and Chaim Goodman-Strauss. If you calculate the Euler characteristic of a hyperbolic tessellation, you should end up with +infinity. But I seem to recall that the authors had a natural way to associate a finite negative number, such that it behaves correctly with respect to finite-index subgroups.

Finite: If you know the numbers of each type of face, you can calculate the number of edges. Moreover, if you know how many edges meet at a point, you can then obtain the number of vertices and calculate chi := V - E + F. If you want to ask more generally which polyhedra are possible, then this is a difficult question; even for the order-3 heptagonal case, it is essentially the study of Hurwitz surfaces:

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

This is deeply related to the studies of number fields, algebraic curves, and group theory.
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Re: Thread for Non-CA Academic Questions

Post by muzik » November 27th, 2017, 8:43 pm

So for polyhedra (and improper tilings?) it's 2, for Euclidean tilings it's 0, and hyperbolic tilings are infinity. Guess I won't be needing a torus for this one.

Do there exist any polyhedra which are isohedral, isogonal and isotoxal, but the individual faces aren't regular?
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Re: Thread for Non-CA Academic Questions

Post by muzik » December 1st, 2017, 4:06 pm

What happens if the pentagrams in the great stellated dodecahedron are made into pentagons following the same vertices?
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Re: Thread for Non-CA Academic Questions

Post by gameoflifemaniac » December 1st, 2017, 4:53 pm

There exists a formula for the reduced quartic equation, but is there a formula for the full quartic equation?
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Re: Thread for Non-CA Academic Questions

Post by Macbi » December 1st, 2017, 5:53 pm

gameoflifemaniac wrote:There exists a formula for the reduced quartic equation, but is there a formula for the full quartic equation?
Yes. It's easy to express the coefficients of the reduced quartic in terms of the original quartic, and then you can just substitute them in. But it's hilariously awful https://suhaimiramly.files.wordpress.co ... uartic.jpg

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Re: Thread for Non-CA Academic Questions

Post by gameoflifemaniac » December 2nd, 2017, 5:18 am

Macbi wrote:
gameoflifemaniac wrote:There exists a formula for the reduced quartic equation, but is there a formula for the full quartic equation?
Yes. It's easy to express the coefficients of the reduced quartic in terms of the original quartic, and then you can just substitute them in. But it's hilariously awful https://suhaimiramly.files.wordpress.co ... uartic.jpg
No. This is the formula for the reduced quartic equation, x^4+ax^3+bx^2+cx+d. But not for ax^4+bx^3+cx^2+dx+e.
And I knew about this formula, but I didn't see a formula for the full quartic equation.
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Re: Thread for Non-CA Academic Questions

Post by Macbi » December 2nd, 2017, 5:22 am

gameoflifemaniac wrote:
Macbi wrote:
gameoflifemaniac wrote:There exists a formula for the reduced quartic equation, but is there a formula for the full quartic equation?
Yes. It's easy to express the coefficients of the reduced quartic in terms of the original quartic, and then you can just substitute them in. But it's hilariously awful https://suhaimiramly.files.wordpress.co ... uartic.jpg
No. This is the formula for the reduced quartic equation, x^4+ax^3+bx^2+cx+d. But not for ax^4+bx^3+cx^2+dx+e.
And I knew about this formula, but I didn't see a formula for the full quartic equation.
Yeah, I read the linked page wrong. But everything else I wrote is still true, and the quartic formula is even worse than in that picture.

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Re: Thread for Non-CA Academic Questions

Post by calcyman » December 2nd, 2017, 7:30 am

The four solutions, according to Mathematica, are:

Code: Select all

x = -b/(4*a) - Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 - 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) - (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2

Code: Select all

x = -b/(4*a) - Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 + 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) - (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2

Code: Select all

x = -b/(4*a) + Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 - 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) + (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2

Code: Select all

x = -b/(4*a) + Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 + 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) + (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2
What do you do with ill crystallographers? Take them to the mono-clinic!

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gameoflifemaniac
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Re: Thread for Non-CA Academic Questions

Post by gameoflifemaniac » December 2nd, 2017, 10:27 am

calcyman wrote:The four solutions, according to Mathematica, are:

Code: Select all

x = -b/(4*a) - Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 - 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) - (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2

Code: Select all

x = -b/(4*a) - Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 + 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) - (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2

Code: Select all

x = -b/(4*a) + Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 - 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) + (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2

Code: Select all

x = -b/(4*a) + Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
        (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^
              2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
            (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)]/2 + 
    Sqrt[b^2/(2*a^2) - (4*c)/(3*a) - (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/(3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + 
           Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)) - 
       (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^
         (1/3)/(3*2^(1/3)*a) + (-(b^3/a^3) + (4*b*c)/a^2 - (8*d)/a)/(4*Sqrt[b^2/(4*a^2) - (2*c)/(3*a) + (2^(1/3)*(c^2 - 3*b*d + 12*a*e))/
            (3*a*(2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 
                   72*a*c*e)^2])^(1/3)) + (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e + Sqrt[-4*(c^2 - 3*b*d + 12*a*e)^3 + 
                (2*c^3 - 9*b*c*d + 27*a*d^2 + 27*b^2*e - 72*a*c*e)^2])^(1/3)/(3*2^(1/3)*a)])]/2
Thanks!
https://www.youtube.com/watch?v=q6EoRBvdVPQ
One big dirty Oro. Yeeeeeeeeee...

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