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As the quartic equation Ax^4 + Bx^3 + Cx^2 + Dx + E = 0 can be reduced to x^4 + (B/A)x^3 + (C/A)x^2 + (D/A)x + (E/A) = 0 without loss of information (unless A is 0), you can just do a trivial substitution on the formulas Macbi gave to find its solutions: Expanding the formula into a standard form, of course, is another matter. I used to amuse myself doing these by hand when I was 13, but now I am too busy for it.

So even without consulting Mathematica, it is trivial to see that exists such a formula--but when you asked "is there a formula," maybe you meant "has the formula ever actually been written, and simplified." My answer is "probably; I could easily imagine someone like me doing this." Calcyman's formulas don't seem to be simplified thoroughly, though; they still contain square roots and cube roots of non-polynomial rational expressions, as well as sums of fractions having different denominators. I think that if the formula were to be entirely simplified, no term but A would be taken to a power higher than the fourth. (In the expanded cubic formula, no term but A has an exponent higher than 3.) If provable, this would work as a good test to see whether an evaluation by hand was correct.

Can somebody give me an overview of data types in Python?

AforAmpere wrote:

muzik wrote:Is 11 truly the only number where n, n+2, n+6, n+8, n+90, n+92, 9+96, n+98, n+180, n+182, n+186 and n+188 are all prime? If there are any more such numbers out there I will be surprised.

Probably not, as calcyman said, but none that I could find up to 10^8.

No others < 304075581810 = 29# * 47 .

AbhpzTa wrote: No others < 304075581810= 29# * 47 .

What did you use to check?

AforAmpere wrote:

AbhpzTa wrote: No others < 304075581810= 29# * 47 .

What did you use to check?

Excel and Notepad (and this page).
No others < 433469446410 = 29# * 67 .

I've finally found a way to accurately graph them.

Are there any well-known names for these shapes - the "bubble polygons", as I like to call them? They definitely seem to resemble different types of torus knot, without the whole over/under thing, since we're dealing with extending the definition of a star polygon here.








In order: {3/2}, {5/3}, {5/4}, {6/5}, {7/4}, {7/5}, assuming we're going with regular star polygon notation.

What are signed trees? http://oeis.org/A000060

I have a conjecture, but I don't know how to prove it:
[∑ₙ̊₌₁ aₙ×sin(bₙx+cₙ)]≠k for all real values of x and a finite positive integer o, where each aₙ, bₙ, and k are constant, non-zero real numbers, and each cₙ is any constant real number.

In other words, I conjecture that you can't use add sine waves alone to get a constant, non-zero value.

I base this upon the nature of the universe: If we could do this, then we could probably use waves to produce a permanent higher-energy state. Such a state would limit the effects of entropy, thereby testing the bounds of the Second Law of Thermodynamics.

EDIT: Thanks calcyman! I now have a proof:
Consider the following infinite integral:

Upon dividing by bₙ then subtracting cₙ from the top and bottom parts of the equation, we have an equivalent form that works for all terms. Zero times any real number is zero, and zero plus zero is zero, so the entire sum is also zero. The function, when used on zero, also produces 0.
Let f~(f(x))=x.
Let f(x) be the integral described above.
Since f(0)=0, f[f(0)]=0.
f~{f[∑ₙ̊₌₁ aₙ×sin(bₙx+cₙ)]}=f~{f[f(0)]}
[∑ₙ̊₌₁ aₙ×sin(bₙx+cₙ)]=[f(0)]=0
Since k≠0, this proves the proposition.

Also, as corollaries, sin² x and cos² x cannot be produced under the same conditions, either.

Gamedziner wrote:I have a conjecture, but I don't know how to prove it:
[∑ₙ̊₌₁ aₙ×sin(bₙx+cₙ)]≠k for all real values of x and a finite positive integer o, where each aₙ, bₙ, and k are constant, non-zero real numbers, and each cₙ is any constant real number.

In other words, I conjecture that you can't use add sine waves alone to get a constant, non-zero value.

I base this upon the nature of the universe: If we could do this, then we could probably use waves to produce a permanent higher-energy state. Such a state would limit the effects of entropy, thereby testing the bounds of the Second Law of Thermodynamics.

Integrate both sides of your equation. The left-hand side is bounded, and the right-hand side is linearly increasing.

Are there any ways to infinitely tesselate space in a way that would normally close over and create a polyhedron?

Still waiting on answers to both questions.

muzik wrote:Do there exist any polyhedra which are isohedral, isogonal and isotoxal, but the individual faces aren't regular?

I don't think so, but for Euclidean tilings the following construction counts:

http://mathstat.slu.edu/escher/upload/e ... nd-net.png

After further consideration, I believe that the requirements cannot be satisfied by a finite polyhedron. Any finite polyhedron obeying these requirements would have all its vertices lying on the same sphere (as a consequence of isogonality), and all its edges would be the same length (as a consequence of isotoxality). Since each face of the polyhedron must fulfill the same requirements, the faces would all have edges the same length, and vertices lying on the same circle. Thus, all angles on a single face must be congruent. Since the edges are also the same length, it follows that each face must be a regular polygon, contradicting the requirement that the faces aren't regular.

muzik wrote:What happens if the pentagrams in the great stellated dodecahedron are made into pentagons following the same vertices?

They become the pentagons of the ditrigonal dodecadodecahedron, or of the great ditrigonal icosidodecahedron, and do not meet at edges anymore, although they still come three to a vertex.

Is it possible for a tessellation to have a negative amount of polytopes or vertices?

Using the equation 720/(360-(V*A)), where V is the amount of polygons meeting to an edge and A is the internal angle of said polygon, to give the amount of vertices gives expected results for the platonic solids and infinity for what would make euclidean tilings, but hyperbolic tilings give negative results. The heptagonal tiling has -28 vertices whereas its dual has -12, the octagonal tiling has -16 vertices and its dual -6, and the infinity-gon tiling has -4 vertices with its dual theoretically having negative infinity.

The hyperoperarions are incrementation, addition, multiplication, exponentiation, tetration, pentation etc. But is there something past incrementation, or is it so elementary, that it can't be divided further?

gameoflifemaniac wrote:The hyperoperarions are incrementation, addition, multiplication, exponentiation, tetration, pentation etc. But is there something past incrementation, or is it so elementary, that it can't be divided further?

This is a great question. But I've always thought the incrementation operation is itself a bit dodgy. In particular 3 + 0 = 3 and 4 + 0 = 4. So 3 incremented with itself 0 times should be 3, and 4 incremented with itself 0 times should be 4, but a 0-ary function doesn't take any inputs so it should always take on the same value.

Macbi wrote:

gameoflifemaniac wrote:The hyperoperarions are incrementation, addition, multiplication, exponentiation, tetration, pentation etc. But is there something past incrementation, or is it so elementary, that it can't be divided further?

This is a great question. But I've always thought the incrementation operation is itself a bit dodgy. In particular 3 + 0 = 3 and 4 + 0 = 4. So 3 incremented with itself 0 times should be 3, and 4 incremented with itself 0 times should be 4, but a 0-ary function doesn't take any inputs so it should always take on the same value.

Oops, I meant zeration. Sorry!

gameoflifemaniac wrote:

Macbi wrote:

gameoflifemaniac wrote:The hyperoperarions are incrementation, addition, multiplication, exponentiation, tetration, pentation etc. But is there something past incrementation, or is it so elementary, that it can't be divided further?

This is a great question. But I've always thought the incrementation operation is itself a bit dodgy. In particular 3 + 0 = 3 and 4 + 0 = 4. So 3 incremented with itself 0 times should be 3, and 4 incremented with itself 0 times should be 4, but a 0-ary function doesn't take any inputs so it should always take on the same value.

Oops, I meant zeration. Sorry!

http://math.eretrandre.org/tetrationfor ... hp?tid=122

So, you randomly choose bits n times. You add up the bits and make a plot of how many times a number appeared. What distribution is this?

gameoflifemaniac wrote:So, you randomly choose bits n times. You add up the bits and make a plot of how many times a number appeared. What distribution is this?

https://en.wikipedia.org/wiki/Binomial_distribution?

Macbi wrote:

gameoflifemaniac wrote:So, you randomly choose bits n times. You add up the bits and make a plot of how many times a number appeared. What distribution is this?

https://en.wikipedia.org/wiki/Binomial_distribution?

Thank you! This makes sense!

How many different patterns are possible in those pattern lockers, those that are common in phones?

I remember a YouTube video that said it was like 110000

gameoflifemaniac wrote:How many different patterns are possible in those pattern lockers, those that are common in phones?

389,112.

Source

What is the optimal number system (fewest # of digits and data)?
How many patterns are there that have a bounding box x*y?

Ternary, and approximately 2^(xy-3)