User gabriel benamy - MathOverflow

Can infinite polynomials be expressed as a product of its linear factors?

2013-01-10T16:39:56Z

Background: In the 1700s, Euler solved the Basel Problem, which was to solve $\sum_{n=1}^\infty\frac{1}{n^2}$ in closed-form. Euler showed that it was equal to $\frac{\pi^2}{6}$ by first expressing $\frac{\sin(x)}{x}$ in Taylor series form, and then writing it as the normalized product of the linear factors given by its roots.

$\frac{\sin(x)}{x} = (1 - \frac{x}{\pi})(1 - \frac{x}{-\pi})(1 - \frac{x}{2\pi})(1 - \frac{x}{-2\pi})...$

When multiplied out and the x^2 terms are collected, they come out to $-\frac{1}{\pi^2}(\sum_{n=1}^∞\frac{1}{n^2})$, which corresponds to x^2's coefficient of -1/6 in the Taylor Series expansion. Thus, the infinite sum comes out to $\frac{\pi^2}{6}$.

My question I have a function with an infinite number of complex zeroes, is continuous and differentiable everywhere, and has no infinities, complex or otherwise, for finite input (I believe this is called a holomorphic function, but I have yet to take a complex analysis class, so I'm not 100% sure). For reference, my function is $(x-1)\zeta(x)$. At x=1, this function is equal to 1, according to Mathematica.

Please note At x = 0, $(x-1)\zeta(x) = \frac{1}{2}$

If z is a non-trivial zero of $\zeta(x)$, so is 1-z

The trivial zeroes of $\zeta(x)$ are the negative even numbers

In the same fashion that Euler described sin(x)/x as the normalized product of its linear factors, can this function also be expressed as the normalized product of its linear factors? Namely, if $z_k$ is the kth non-trivial zero of the Riemann zeta function with "positive" imaginary component, then is this true:

$(x-1)\zeta(x) = (1/2)\prod_{n=1}^\infty(1-\frac{x}{z_n})(1-\frac{x}{1-z_n})(1+\frac{x}{2n})$

If not, why not?

Additional question In general, the only two classes of functions I can think of without any complex zeroes are non-zero constants and exponentials. So, can all functions be expressed as a product of a constant, an exponential, and its normalized linear factors?

If I need to make anything clearer, please let me know. I'm hastily posting this from a school computer and I might not be as clear as I would like.

Prove a parametrization function is surjective

2011-05-23T21:42:19Z

As a starting note, I would like to say that I haven't (yet) taken courses in Set Theory, so some higher-level notation may be lost on me (and I may not write everything conventionally), but I'll do my best.

My question is as follows: I have sets of numbers which satisfy a particular Diophantine equation. Take, for example, the Pythagorean Quadruplets:
$a^2 + b^2 + c^2 = d^2; a,b,c,d\in\mathbb{R}$
Wolfram Mathworld gives the following (unscaled) parametrization for this equation:
$a = 2mp$
$b = 2np$
$c = p^2-(m^2+n^2)$
$d = p^2+(m^2+n^2)$
However, this set of equations is not surjective, as it's missing the solution 36² + 8² + 3² = 37²

Wikipedia offers the following (unscaled) solution:
$a = m^2+n^2-p^2-q^2$
$b = 2(mq+np)$
$c = 2(nq-mp)$
$d = m^2+n^2+p^2+q^2$
This solves the above identity ({a,b,c,d} = {3,36,8,37}) with {m,n,p,q} = {2,4,-1,4}. This solution also happens to be surjective.

Here's the question: Given a Diophantine equation (e.g. Pythagorean Quadruplets), and a particular parametrization for this equation (e.g. Wikipedia's solution), how do you prove that the parametrization is surjective? That is, how do you prove that every set of integers that satisfies the Diophantine equation is generated by integer values of the parametrization?

Note that this is just an example; I want to know how it can be done in a general case.
Thanks for the help!
-Gabriel Benamy

Answer by Gabriel Benamy for Most memorable titles

2010-11-17T23:00:29Z

"What is infinity factorial (and why might we care)?"

The only downside is that it isn't actually typed up, but rather is hand-written and scanned, but the result of $\infty! = \sqrt{2\pi}$ is still rather intriguing.

Answer by Gabriel Benamy for Playing an (invertible) matrix game with two players

2010-11-11T21:31:23Z

Player A wins the trivial n=1 case by playing any non-zero number in (1,1). For all even n, player B wins by using symmetry a la a horizontal mirror. As Ben points out in the comments, if n = 3, player B can force a win. I had a long demonstration written out, but I decided against it (if you want, I can put it in later). Anyway, as for the general case, after a little searching, I found a paper called "A determinantal version of the Frobenius - König Theorem" by D. J. Hartfiel and Raphael Loewy, which can be purchased here.

The abstract, at least, says that given an n by n matrix A of, say, rational numbers, if the determinant is zero, then A must contain an r by s submatrix B such that r + s = n + p, and rank(B) ≤ p - 1 (no more than p - 1 linearly independent rows), for some positive integer p. This means that if we have, say, a 5x5 matrix whose determinant is zero, then there exists a submatrix B in A such that B is:

a 1x5, 2x4, 3x3, 4x2, or 5x1 matrix of 0s
a 2x5, 3x4, 4x3, or 5x2 matrix whose rows are all scalar multiples of each other
a 3x5, 4x4, or 5x3 matrix with no more than two linearly independent rows
a 4x5 or 5x4 matrix with no more than three linearly independent rows
a 5x5 matrix with no more than four linearly independent rows (duh)

While it doesn't say so explicitly, I think that it's a biconditional, so if player B manages to get one of these in the matrix, then she will win. However, even if it isn't biconditional, if player A can prevent any of those forming, he will win.

Of these two, I believe it would be easier for player A to prevent any of these forming than it would be for player B to force one of these, but I haven't given that in particular a great deal of thought. I hope this is helpful.

Answer by Gabriel Benamy for Checking if a positive integer is a power other than a first power

2010-07-01T18:11:10Z

I don't know how to do it in cubic time, but I suppose that, to use Newton's Method, you could do the following:

Find floor(log₂n), and this is the largest "power" that it can be. Then, define: $f_k(x) = x^k - n$ where n is your number and k is the floor value, and iterate Newton's Method until you get a number whose floor is m that fits any of the following:
1) m^k < n && (m+1)^k > n
2) m^k > n && (m-1)^k > n
3) m^k == n
If the third is true, congrats, you've found yourself a root! Otherwise, reduce k by 1 and try again.

The problem with this is that I don't know how long it'll take for such a value of m to be found. Wikipedia says Newton's Method has a quadratic convergence, and you're making a fixed number of operations each iteration of the Method, so I guess its running-time would be pretty fast.

Answer by Gabriel Benamy for Chance of something being fixed

2010-05-07T16:10:23Z

According to my statistics final which I took yesterday, the answer should be
$m=\lceil 2\left(1-\frac{1}{n}\right)\text{InverseErf}^2[1-p]\rceil$ where InverseErf[x] is the Inverse Error Function.

Answer by Gabriel Benamy for a^b = b^a when a is not equal to b.

2010-04-23T00:17:09Z

Given your original function, $f(x) = x^{x+1} - (x+1)^x = 0$, you can define the Taylor Series expansion around some value v to give an expansion such that f( x) = a for some a, which in itself isn't helpful. However, there exists an inversion, using the Lagrange Inversion Theorem, which allows you to specify the a which your f( x) will return and gives you the corresponding x.
The first few terms of the series are:
$a+\frac{x-f[a]}{f'[a]}-\frac{f''[a] (x-f[a])^2}{2 f'[a]^3}+\frac{\left(3 f''[a]^2-f'[a] f^{(3)}[a]\right) (x-f[a])^3}{6 f'[a]^5}+\frac{\left(-15 f''[a]^3+10 f'[a] f''[a] f^{(3)}[a]-f'[a]^2 f^{(4)}[a]\right) (x-f[a])^4}{24 f'[a]^7}$
Remember, though: this is HIGHLY volatile when you're not near the actual value. Using 3 makes the value explode upwards, and using 2 makes it explode in the negative direction. However, a value of 2.2 will converge to the value you're looking for. You can use the Inversion Theorem to calculate more terms in the inverse series and use it to calculate your constant to arbitrary precision, given enough computational time.

Of course, this isn't what you're looking for (i.e. closed form or 'nice' series), but it's the best that I've found for this particular problem. Unfortunately, it converges so slowly with respect to computational time that it probably becomes more convenient for you to use the secant method to get any sort of good approximation.

--Gabriel Benamy

Answer by Gabriel Benamy for What is your favorite "strange" function?

2010-04-22T15:32:06Z

The Weierstrass function is particularly intriguing, as it's a function that's everywhere continuous, but nowhere differentiable.
$f(x)= \sum_{n=0} ^\infty a^n \cos(b^n \pi x)$
where 0<a<1, and b is a positive odd integer such that $ab > 1 + \frac{3\pi}{2}$.
It challenges the notion that, just because a function is continuous, it must also be differentiable in most places, which I think is pretty cool.

Explicitly constructing an infinite set with particular size

2010-03-29T14:53:43Z

I would like to preface by saying that I have no significant experience working with set theory, so I'm probably making an intuitive mistake. I have figured out where the mistake probably is, but I can't figure out why it IS a mistake. I figured that this was the best outlet to ask my question.

I was reading about the Continuum Hypothesis on Wikipedia recently, particularly about the fact that it's undecidable in ZFC -- which means that it is undecidable whether or not there is an infinite set whose size is strictly between that of the natural numbers and that of the real numbers. Now, intuitively, a proof that it cannot be disproved would immediately give way to the fact that no counterexample could be constructed (for such a counterexample would disprove it, which is impossible), and thus it must therefore be true. But that's not where I'm going with this.

Cantor proved that the rational numbers are countable -- that there exists a counting method such that, given any integer, you could determine the unique rational number which corresponds to it, and given any rational number, you could determine the unique integer which corresponds to it. The proof is fairly cool, but that's not where I'm going, either. Essentially, this demonstrated that $\aleph_0^2 = \aleph_0$. Then, he went on to prove that all algebraic numbers were countable, which proved the stronger statement that for any finite n, $\aleph_0^n = \aleph_0$. But yet, the cardinality of the real numbers is still strictly greater.

He explicitly determined the cardinality of the real numbers as $2^{\aleph_0}$, or, strictly speaking, for any $\alpha > 1, \mathfrak c = \alpha^{\aleph_0}$, because, no matter which base you're in, the number of real numbers doesn't change. This means that the cardinality of the real numbers is strictly exponential, whereas the cardinality of the countable numbers is strictly polynomial.

This is where my confusion arises. If I construct a set whose size after an infinite number of steps is bounded by any polynomial, it is countable, whereas a set whose size is greater than every polynomial would not be countable.

The Adleman–Pomerance–Rumely_primality_test has running time, for a given n, of $n^{O\log(\log(n))}$, which is of super-polynomial running time -- there exists no polynomial that is strictly greater than that function. However, it is also sub-exponential -- there exists no exponential function that is strictly less than it, either. Therefore, it exists between the polynomials and exponentials.

Using this, I can construct a set of numbers whose size after n steps is equal to $f(n) = n^{O\log(\log(n))}$ by appending approximately f'(n) unique values to the end of the set. I have now explicitly constructed an infinite set whose size is $\aleph_0^{\log(\log(\aleph_0))}$, have I not? And, as I said before, it is, eventually, larger than any set whose size grows polynomially. But it is also smaller than every set whose size grows exponentially.

Of course, it also turns out that this set is countable -- I can give you the numbers in the set, if you so wish. But that's only part of the problem.

Using the same method, I can also construct a set whose size grows exponentially, if I only change my function to $f(n) = 2^n$. I then append f'(n) unique values to the end of my set and, voila, as I take a countably infinite number of steps, namely $\aleph_0$, the size of my set becomes $2^{\aleph_0}$, the cardinality of the continuum -- but, as before, I can tell the n^th item in my set, and if you give me any item in my set, I can tell you exactly where it lies.

Thus is my question: What did I do wrong? Either I have shown that $2^{\aleph_0} = \aleph_0$, which is exceedingly unlikely, or my assumption that my new set's size is equal to $2^{\aleph_0}$ is incorrect, and I cannot understand why.

Any help would be appreciated, thanks!

--Gabriel Benamy

Tschirnhaus Transformation

2009-12-01T18:16:48Z

Recently in my Intro to Proofs class, we've been talking about the fundamental theorem of algebra, which states that all polynomials of degree n always have n, not necessarily distinct, not necessarily real, solutions. Every high school student is taught the formula $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for solving quadratic equations, and there exist solutions to the general cubic (Cardano's formula) and general quartic (Ferrari's solution). But what about higher functions?
The Abel-Ruffini theorem states that general functions of degree 5 or higher cannot be solved using a finite number of additions, subtractions, multiplications, divisions, and root extractions. Now, that isn't to say that no function of 5 or higher can be solved -- quite the contrary, actually: x⁵ = 1 has solutions $x = \sqrt[5]{1}\in{1,-\frac{1\pm_1\sqrt{5}}{4}\pm_2 i\sqrt{\frac{5\pm_1\sqrt{5}}{8}}}$ where ±₁ and ±₂ are independent of the other but related to itself (thus both ±₁ are the same sign always), producing five solutions, four of which are complex.
But there also exist equations with no exact solutions, such as x⁵ - x - 1 = 0. Sure, it has a real solution at x≈1.1673, but that's only an approximation, good to 4 decimal places. However, this is not really what my question is about.

While they cannot be solved generally, quintic functions can be reduced significantly. Take, for example, the function x⁵ + a₄x⁴ + a₃x³ a₂x² + a₁x + a₀ = 0. Making the substitution x = v - a₄/4 produces the new equation v⁵ + b₃v³ + b₂v² + b₁v + b₀ = 0, where the b coefficients are in terms of the a coefficients. Realize that this is just a linear shift to the side of the previous equation, but it becomes simpler to manage, as it's missing the 4^th power term.
Tschirnhaus took it a step further, though. He used a method, which is now called the Tschirnhaus Transformation (the subject of my question) to solve the general cubic in a manner separate to Cardano's solution, and proposed that it could be used to solve any polynomial (and was mistaken). My question lies in what exactly it was that he did. Here is a paper which explains his transformation. I've followed it up to (7), and I understand that we're then solving for alpha in order to reduce the equation to a simple y³ + g, but then I am completely lost as to where the resulting equation came from. (7) is $res_{x}(P_{3},T) = y^3 + (p\alpha^{2}-\frac{1}{3}p^2 +3\alpha q)y + \frac{2}{3}\alpha^{2}p^{2}-\alpha^{3}q+q^{2}+\alpha pq+\frac{2}{27}p^3$, and the result is $y^{3} = 9q^{3}\alpha /p^{2}+\frac{4}{3}\alpha pq-\frac{8}{27}p^{3}-2q^{2}$
. Additionally, I followed the separate method for reducing (but not solving) the quintic through to (12), but I don't understand how (13) and (14) are derived.
If anybody here would be kind enough to explain where they came from, then thank you very much. I realize that there may be typos here, but I can't for the life of me understand what's going on anymore.

Thank you in advance.

Gabriel Benamy

PS. If any syntax is off, it's because the preview, auto preview, and tag look-ahead prompt are all down and I had to write everything on Wikipedia (which has somewhat different syntax, anyway).
PPS. Apparently, my notify email address is invalid, but I didn't have the option of changing it, so I had to uncheck "notify me" in order to post. Am I doing something wrong?

EDIT: Because apparently, I am unable to comment (as Mathoverflow functionality has completely died for me), I have to add this as an edit. Qiaochu Yuan said that I've misquoted both theorems. From my understanding, having "n complex roots" is the same as n, not necessarily real, roots. After all, 1 + i is "not necessarily real" but is a root of x³ - 3x² + 4x - 2 = 0. 1 is also a real root of this (and is also complex, because all real values are complex).
Additionally, he says that the Abel-Ruffini theorem says that "polynomials of degree 5 or higher cannot be solved using the arithmetic operations and root extractions." This is not true, as demonstrated with x^5 - 1 = 0, which is a polynomial of degree 5 and clearly can be solved. The AR theorem states that not all can be solved in such a way, not that none can be solved in such a way.

Answer by Gabriel Benamy for Unit in a number field with same absolute value at a real and a complex place

2010-03-11T17:24:03Z

The big problem is that Abs(a) = b implies that a = b x 1^c. Whether or not c is a rational number is another question, however. If I gave you $x^3 - 11x^2 + 55x -125$, its roots are x = 5, x = 3+4*i*, x = 3-4*i* In this case, |3+4*i*| = 5, and 5*(-1)^{ArcTan(3/4)/Pi} = 3+4*i*, and ArcTan(3/4)/Pi evaluates to ~ .295... Granted, this doesn't satisfy the requirement that the constant is equal to 1, but the point still stands that any two numbers which share an absolute value only have a factor difference of some power of unity.

Answer by Gabriel Benamy for Fundamental Examples

2010-03-02T21:18:25Z

The discovery of Transcendental numbers, or numbers that are not the root of any finite polynomial with rational coefficients.
Also, the proof that e and π were transcendental, the latter via the proof that e^a is only algebraic for transcendental values of a (and e^*i*π = -1 is algebraic, as is i, so therefore π must be transcendental). Their discovery, as well as the first explicitly created example, the Liouville number, sparked what's called "Transcendence theory".
And as it turns out, any randomly chosen real number is "almost surely" transcendental. In other words, the density of transcendental numbers among the real numbers is 1!!

Answer by Gabriel Benamy for Equality of the sum of powers

2010-03-01T18:09:17Z

One set of solutions for t = 3 is the class of numbers known as Taxicab Numbers, named after the number of a taxicab G. H. Hardy took, 1729, that Ramanujan mentioned was equal to 1³ + 12³ = 9³ + 10³. This particular example fails, as |10 - 9| = 1 < 2, but there are other Taxicab numbers, such as:
167³ + 436³ = 228³ + 423³ = 255³ + 414³.

This might be a helpful site for your question.

-Gabriel Benamy

Answer by Gabriel Benamy for On Polynomials dividing Exponentials

2010-02-25T19:30:10Z

The only m I've found that works up to 10,000 is 3, but I can't prove that it's the only one.
While I don't know how solve it directly, the exponential equation can be transformed into: $5^m + 3^m = 5^m\left(1+m!\sum _{k=0}^m \frac{(-2)^k}{k!(m-k)!5^k}\right) $, so you're looking for integer results to $\frac{5^m}{m^2-1}+m\frac{(m-2)!}{m+1}\left(\sum _{k=0}^m \frac{(-2)^k5^m}{k!(m-k)!5^k}\right)$.
The general form of the first equation here is: $a^m + b^m = a^m\left(1+m!\sum _{k=0}^m \frac{(b-a)^k}{k!(m-k)!a^k}\right)$, assuming a ≠ b (if a = b, then the numerator for k = 0, the numerator of the sum would be 0⁰ which should turn into 1). I would think that this might be a bit easier to solve, but I can't be sure.

Hope this helps!
-Gabriel Benamy

Answer by Gabriel Benamy for bounding roots of a polynomial with Rouche's Theorem

2010-02-24T15:34:28Z

I don't know about any further boundings, but n = 3 and k = 1/4, or polynomial $4z^3 - z^2-z-1 = 0$ has a solution (1/12 + 1/12 (235 - 6 Sqrt[1473])^(1/3) + 1/12 (235 + 6 Sqrt[1473])^(1/3)), whose absolute value is ~ 0.868877, which is greater than 1-k. Other {n,k} pairs are {2,3}, {4,6}, and {5,6}.

EDIT I noticed that if the roots are multiplied by nk, then as k goes from 0 to 1/n, the largest root in absolute value (which happens to be the largest root) goes from 0 to about 1. So I suppose that the roots are bound in the range (0, 1/n).

Can a limit to zero of a limit to zero assume they're both going at the same rate?

2010-02-23T20:47:23Z

My title can be a bit confusing, so here's a bit of background.
The corollary to the Fundamental Theorem of Calculus says that $\int_a^bf(x)dx = F(b)-F(a)$, assuming that F'(x) = f (x), or that the area under the curve f (x) from x = a to x = b is equal to the difference of values of the antiderivative of f (x) at a and b.

The following is my attempt to prove it.
1: The area under the curve of f (x) from x = a to x = b is equal to the area of the rectangles under the curve as you take more and more rectangles. See this image:
Mathematically speaking, it's $\int_a^bf(x)dx = \lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot f(a+hn)$
2: Let us replace our measly f (x) with its definition, in terms of the derivative of F (x), namely $f(x) = \lim_{j\to 0}\frac{F(x+j)-F(x)}{j}$. Thus, our first equation becomes

$\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot \lim_{j\to 0}\frac{F(a+hn+j)-F(a+hn)}{j}$

Now, my question is, since both h and j are going to zero via a limit, can I assume that they are effectively the same? Can I simply replace all instances of j with an h and rid myself of an unnecessary second limit? If I could, my proof would continue as follows:

3: Replacing all j's with h's yields:
$\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot \frac{F(a+h(n+1))-F(a+hn)}{h}$, and the *h*s can cancel out: $\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}F(a+h(n+1))-F(a+hn)$.
4: Thankfully, this becomes a telescoping series, as seen here:
$F(a+h(1))-F(a+0h) + F(a+h(2))-F(a+1h) + F(a+h(3))-F(a+2h) + ... = -F(a-h) + F(b-h)$
$ + F(a+h(\frac{b-a}{h}))-F(a+h(\frac{b-a}{h}-1) = F(b) - F(b-h)$
which, together, yields -F (a - h) + F (b) as the sum.
Putting this back in, we get $ \lim_{h \to 0} -F(a - h) + F(b) = F(b) - F(a) = \int_a^bf(x)dx = F(b)-F(a) $

However, steps 3 and 4 require the ability for me to assume that h and j are the same thing. My teacher (who admittedly doesn't deal with this too often), whom I asked first, said that perhaps h and j are going to 0 at different rates. However, I do not think that the concept of a limit to 0 at a rate actually means anything.
So the question I bring to you is: Is the operation that I performed to go from step 2 to step 3 a valid operation? If so, why? If not, why not?

Thanks for your help!
-Gabriel Benamy

Answer by Gabriel Benamy for hard diophantine equation

2010-02-23T21:04:13Z

I may be misunderstanding the question, but I do not believe that it has any integer solutions. At the very least, none are known to exist at the moment. Any solutions would be counterexamples to the Fermat-Catalan conjecture with {m,n,k} = {3,5,7} (since 1/3 + 1/5 + 1/7 = 71/105 < 1). The most I can tell you is that, for coprime {x,y,z}, there are finitely many solutions to your equation. I think your (x,y) = 1 means that they're coprime, anyway, so it follows that z must be coprime. Therefore, any solution at all would disprove the related Beal's conjecture.

Answer by Gabriel Benamy for Side-Angle-Side Congruence and the Parallel Postulate

2009-12-03T19:00:36Z

Ironically, I had intentions of working the other way around. As Hugh Thomas said, Euclid used a superposition of one figure onto another to demonstrate SAS congruence. My idea was to use a superposition of one figure onto another to demonstrate the parallel postulate, as well, but it never seemed to work well enough. However, regarding what Kristal Cantwell says, using the superposition "axiom" is contrary to hyperbolic geometry, and thus should be able to restrict it to our geometry.

Answer by Gabriel Benamy for Algorithm for finding the volume of a convex polytope

2009-12-03T18:45:16Z

Here's a fairly straightforward solution for polyhedra (3 dimensions), with running time O(v+ve), where v is the number of vertices and e is the number of edges. I suppose it could be extended to higher dimensions, but it would probably have much worse running time (I fear roughly exponential as in O(vⁿ), where n is the number of dimensions).

Let our polyhedron have n vertices, defined by their x,y,z coordinates: v₁, v₂, ..., v_n and let the lowest point be v₁ and the origin (modify the values for the others accordingly), and let it have e edges, defined by the vertices which they connect. Then, since we have coordinates for the vertices (as that is how we defined them), there must be a "ground level" plane p₀ running through the x and z axes (the y-axis being height, and the ground never having an elevation). Then, let v₂ be the point closest to the ground plane (shortest line perpendicular to the plane), and let v₃ be the next closest, etc, through v_n.

Through each of the points v₂ through v_n, draw a plane perpendicular to the ground, and let them be numbered p_m, where m is the subscript of the vertex through which it was drawn. Then, the volume of our polyhedron is equal to the sum of the volumes of the figures between the planes. We should have something resembling this:

Let the heights between the segments be h₁ through h_n-1, where height h_j is the height between planes p_j and p_j+1.

Now, through each plane, we have a polygon (or more, if the figure is concave), whose vertices' coordinates can be calculated easily as follows:
Let the edge that runs through the plane p_j have endpoints v_a and v_b. Then, the displacement vector is v_b - v_a (assuming the coordinates of v are in vector-form), and the percentage travelled up is $\frac{h_j-h_a}{h_{b-1}-h_a}$. Multiply this by v_b - v_a and add to v_a to calculate the new point of intersection for that edge:
Intersection point = $(v_b-v_a)\frac{h_j-h_a}{h_{b-1}-h_a}+v_a$
The area of these polygons can be determined using triangles, or a simplification of this very process in just 2 dimensions.

PlanetMath says that the volume of a prismatoid (which is the type of figure contained between sequential planes) is $h\frac{B_1 + B_2 + 4M}{6}$, where the Bs are the areas of the parallel polygons and M is the area of the midway polygon, which is exactly halfway between them (and parallel to them). Since we already know the area of each of the end polygons, and we can easily calculate the vertices of the midway polygon (using the previous paragraph's method), we can calculate the volume of the resulting prismatoids. Adding them up yields the total volume of the polyhedron.

I suppose that the only real issue in this case, then, is, via code, determining which edges run through any particular plane, but if we were to actually look at it, we could tell very easily.

A simpler version of this can be used to figure out the area of any polygon; simply draw lines through the vertices parallel to the x-axis and calculate the area of the resulting trapezoids as (b₁+b₂)/2

Answer by Gabriel Benamy for Famous mathematical quotes

2009-12-02T05:31:48Z

Dunno if it's appropriate, but: "Now, I've often thought of writing a mathematics textbook someday, because I have a title that I know will sell a million copies. I'm going to call it: Tropic of Calculus" -- Tom Lehrer, New Math

Answer by Gabriel Benamy for Integer division: the length of the repetitive sequence after the decimal point

2009-11-24T18:29:36Z

The length of the period of a fraction is indifferent to the numerator unless the numerator and the denominator are not in lowest terms (in which case, reduce them to apply the following method).
However, when they are in lowest terms, take the denominator and factor out all factors of the base you're in, as they change nothing (like 2s and 5s in base 10, 3s and 7s in base 21, etc). Then, the length of the period is equal to n, where n is the smallest integer such that your denominator divides bⁿ-1 (where b is your base). So, for example, 1/7 = 0.142857..., and the smallest 10ⁿ-1 such that 7 is a factor is 7 is 10⁶-1 = 999999, and 999999/7 = 142857, which is, incidentally, your repeating portion. The same thing even works for, say, 1/2 -- the smallest n is 0, because 10⁰ - 1 = 0, and all non-zero integers divide 0. This is one of the methods for factoring repunits. Note that n will always be less than the denominator.
Similarly, 1/14₇ (which is 1/11 in base 10) = 0.0431162355..., and the smallest n such that 7ⁿ-1 is a multiple of 11 is 10 (7¹⁰ - 1 = 282475248, and 282475248 / 11 = 25679568).

Multiplying by a constant does nothing to change the period, though it will change the numbers themselves. Interestingly, though, when the period n is equal to 1 less than the denominator (called "max" periodic length), multiplying the fraction by a constant (other than a multiple of the denominator itself) will actually produce a cycle of the period. 2/7 = 0.285714..., 3/7 = 0.428571..., 4/7 = 0.571428..., etc.

Edit: When you divide by factors of the base, let the whole number by which you divide be f. Then, $\lceil \log_{base}f\rceil$ (ceiling function) is equal to the number of non-repeating places before the repeating portion of the decimal. Thus, 1/6, divided by 2, yields 1/3 (1/6 has the same repeating length as 1/3), and we divided by 2. $\lceil\log_{10}2\rceil$ = 1, and indeed, 1/6 = 1.1(6...), with 1 non-repeating decimal.
Furthermore, for fraction m/n, $\lfloor\log_{base}n\rfloor$ is the number of leading zeroes in the repeating portion of the decimal.

Answer by Gabriel Benamy for On Euclid's proof of the infinitude of primes and generating primes

2009-11-24T14:35:25Z

Going with what Qiaochu Yuan said about f(x), it follows that we will never get those primes unless we start, even if we don't include multiplicities. Since we're starting with 'n', then we're taking the prime factors of 'n+1', then we're taking the prime factors of f(n+1), then f(f(n+1)), then f(f(f(n+1))) etc, even if we get, say, a 7² in there, our number is [f(f(f(n+1)))] / 7, which then goes into f(x). So no, unless you start with the infinite product $p_1 = \prod_{n=1}^\infty 6n-1$, you will never get any of those numbers.

It's funny, though; I'd had a whole demonstration started to show that you'll never get a multiplicity when taking f(f(...f(2)...)), but this is simpler. As for the Euclid-Mullin sequence, I have no idea.

Answer by Gabriel Benamy for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

2009-11-20T21:36:29Z

I suppose that yes, it does give some insight into the nature of the primes; or rather, their distribution amongst the natural numbers. Realize that Zeta( a ) is finite for all a > 1, and this means that given any positive real a greater than 1 will produce a finite value of $\sum_n^\infty=\frac{1}{n^a}$ (the closer to 1 it gets, the larger it will get). However, the partial sums of the reciprocals of the primes do not converge . What does this mean? Well, consider that there exists some exponent j and coefficient k such that, for every integer n , $\frac{k}{n^j}\geq\frac{1}{p(n)}$, where p ( n ) is the n ^th . If we flip the reciprocals, we also flip the direction of the inequality, and arrive at $\frac{n^j}{k}\leq p(n)$, or $n^j\leq p(n)\cdot k$. This means that, given j,k such that the above conditions hold, the function p ( n ) grows at least as fast as this function with exponent j . Looking back at our $\frac{k}{n^j}$, we examine $\sum_{n=1}^{\infty}\frac{k}{n^j}=k\cdot \zeta(n)\lt\infty$, as per the finite nature of k and the Zeta function at values greater than 1. But, the n ^th term in the series is always greater than the n ^th prime number, and therefore the whole sum is therefore greater than the sum of the reciprocals of the primes, which, as we know, is infinite. This creates a contradiction, in that the sum, which we know is finite, is greater than an infinite series. Therefore, there can't exist a j and k that satisfy $\frac{k}{n^j}\geq\frac{1}{p(n)}$ for all n , which means that the prime numbers themselves grow slower than any power.

All this does is it says the primes grow slower than any power, but it doesn't say how they grow, which the Prime number theorem does (and supports this conclusion). I suppose this could be generalized to say that any function which can be shown to, for all sufficiently large terms, grow slower than any power greater than 1, it is divergent (the primes being an example, and the evens being another).

Answer by Gabriel Benamy for A single paper everyone should read?

2009-11-20T19:52:49Z

One paper that I've read a few times and always loved was Who Can Name the Bigger Number? (also available in Spanish and French, for those who prefer to read in those). It discusses how our concept of "big numbers" has evolved over time, and talks about Turing machines and the "busy beaver" numbers, which represent a non-computable function.

Comment by Gabriel Benamy

2013-01-21T22:16:12Z

Thank you very much! That is indeed what I was looking for, but I didn't know what it was called. I'll take a look at the articles and try and decipher it. From what it looks like, I had the general idea down, but I was missing the pi exponential and some Gamma function stuff.

Comment by Gabriel Benamy

2010-12-17T19:34:58Z

I do not have any more insight into a solution at the moment. However, I gave this problem to one of my friends a while back, who in turn gave the problem to one of his professors at Columbia, who (again, in turn) is putting it as an extra credit question on her final next week. If anything turns up, I will be sure to let you know.

Comment by Gabriel Benamy

2010-11-18T02:43:57Z

Just to clarify, you are attempting to show that there exist an infinite number of non-overlapping ranges (N, N+1), where N is greater than, say, 2, such that h(x) has no zero in (N, N+1)?

Comment by Gabriel Benamy

2010-11-12T20:44:27Z

No problem. It's a very interesting problem, and after my Linear Algebra class today, I spoke with my professor about it for a few minutes, but he had another class to teach. His thinking is that it should be simple for player A to prevent submatrix <i>B</i> from forming, but didn't really give a constructive strategy. I will certainly try and work on this problem some more over the next few days and come up with something. I personally believe that player B always has a winning strategy when n > 1.

Comment by Gabriel Benamy

2010-11-11T23:00:34Z

True, but in the case of the 3x3, a 2x2 submatrix of zeros guarantees the determinant is zero, and I think that's what he meant.

Comment by Gabriel Benamy

2010-11-11T22:59:21Z

Thanks; I didn't catch that mistake in my read-through. The article's abstract mentions that it's n + p, so I mistyped it.

Comment by Gabriel Benamy

2010-07-04T15:39:08Z

I can't remember where (probably tvtropes), but when reading something about the Ackermann numbers (1 ^ 1, 2 ^^ 2, 3 ^^^ 3, etc), which are related to the Ackermann function, the joke was "it's always weird when looking at a sequence of numbers that goes: 1, 4, too big to count."

Comment by Gabriel Benamy

2010-05-18T22:27:04Z

What would happen if I gave the algorithm, say, .19, which is NOT generated by any a/b where a,b < 10?

Comment by Gabriel Benamy

2010-05-17T17:10:53Z

Ah, now I fully understand what the question is. So you're saying that every reduced a/b is unique up to the first 2n digits, where n = ceiling(log10(max(a,b))) (number of base-10 digits of the larger of a,b)? That's an interesting question...

Comment by Gabriel Benamy

2010-05-17T16:59:21Z

Is this a terminating or a repeating decimal? If it's a terminating decimal, it's a trivial solution: just multiply by 10^n for your numerator and stick the 10^n in the denominator and reduce. If it's a repeating decimal, just multiply p/q by 10^n and subtract p/q to get your repeating portion, then divide by (10^n - 1) to get your fraction. Then, reduce. Example: 0.123456789... Multiply by 10^9 to get 123456789.123456789... Subtract repeating portion to get 123456789 Divide by (10^9 - 1) to get 123456789/999999999 Reduce to 13717421/111111111

Comment by Gabriel Benamy

2010-05-17T16:50:55Z

I got the 15 arrangements worked out, but I can't figure out how to go from lots of variables to an actual numeric answer. I have "the answer is probably greater than a quarter" from trial and error, but past that, I have no idea.

Comment by Gabriel Benamy

2010-05-07T16:22:46Z

I was assuming that the number of times before the error occurs is roughly normally distributed - if the test is done enough times, a binomial distribution can be well-approximated by a normal distribution. $1-\text{Erf}\left[\sqrt{\frac{t\left(\frac{1}{n}-0\right)}{2 \frac{1}{n} \left(1-\frac{1}{n}\right)}}\right]$ is the equation for the probability of making a type-1 error (mistakenly assuming that we've fixed it).

Comment by Gabriel Benamy

2010-05-05T14:40:35Z

I'm unfamiliar with this subject, but a quick wiki on random walks gave me enough to write a simple program. Assuming that the walk can simply go back on itself (which would cause it to intersect itself), and that, say, {0,0}, {0,1}, {0,0} would constitute 2 steps, then assuming a normal distribution for the number of steps, there is about a 95% chance that the average number of steps is between 4.5652 and 4.59948, using a sample size of 100,000 random walks. Sorry it's not an analytic approach, though; I have a class in a few minutes and I didn't want to miss it working on this neat problem.

Comment by Gabriel Benamy

2010-04-30T03:56:29Z

I read his comment, but since I can't post there, I'll post it here; In his solution, he gets a "+1" at the end of his some in equation 1, and I'm not sure where that came from. If it's from the lambda, then I'm not quite sure what the lambda actually represents, but nowhere else in the sum could that +1 have arisen. And if that's the case, then I apologize; I have no idea. But if not, then actually, a can only exist if the sum of the sequence T converges, specifically to 1/2. And if it does, then a can be anything.

Comment by Gabriel Benamy

2010-04-29T13:58:11Z

He did say gamma > 0. I'm almost positive it converges to a real value when alpha = gamma = 1, as the imaginary part cancels itself out (limit of the imaginary component's integral from -a to a as a goes to infinity is zero, as it's an odd function). I wouldn't be surprised if this was the case for all alpha and gamma. I'll do some more experiments.