Recent Forum Posts
 From categories: All categories Ideas: Student Ideas Ideas: Help! Technical Questions: Working in the Wiki Technical Questions: When & Where
page 1123...next »
08 Dec 2008 01:15
in discussion Ideas / Student Ideas » Goldbach's twin primes

So…

Oddly enough, this sequence is in the OEIS… A054735 (Idon't know why I couldn't find it earlier!)

But it did mention that the formula to find any term is $p^q + q^p \mod pq$, where $p,q$ is the nth twin prime pair.

Re: Goldbach's twin primes by , 08 Dec 2008 01:15
08 Dec 2008 00:58
in discussion Ideas / Student Ideas » Pi and the ancient Egyptians

Well, after a little math…

(1)
\begin{align} \pi_{Egyptian} = 3 + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} = 3 + \frac{13}{81} = \frac{3*81+13}{81} = \frac{256}{81} = \frac {(4)^4}{(3)^4} = {\Big{(} \frac{4}{3} \Big{)}}^4 \end{align}

…which is what we gave in class.

But yes, it does come from the Rhind Papyrus.

Re: Pi and the ancient Egyptians by , 08 Dec 2008 00:58
08 Dec 2008 00:33
in discussion Ideas / Student Ideas » Merten's Conjecture

This post moved me to re-visit Goldbach's conjecture…

I am guessing that the expected value of the difference between consecutive primes, $x = p_{k+1} - p_k$, has something to do with the proof (or lack thereof).

It also may have to do with the fact that we can partition the prime numbers into groups of equal magnitude (in particular, if $m = 2^k$, we get $2^{k-1}$ groups) such that $p_i \equiv r \equiv 2j-1 \mod m, \; \forall \; 1 \leq j \leq 2^{k-1}$.
Thus, for a given $p_1, p_2$ we get $p_1 + p_2 \equiv (r_a + r_b) \mod m$. Also note that $r_a + r_b \equiv 2i \mod m \forall \; 0 \leq i \leq 2^{k-1}-2$ for all possible choice of $r_a, r_b$. These sums are sets even numbers. (Note that these groups should more than span $2 \mathbb{Z}$, although it would be nice to know if there is a reasonably calculable amount of overlap between the groups).

Re: Merten's Conjecture by , 08 Dec 2008 00:33

Here is something that I wanted to tell you about in class but didn't not had time:

Plato’s Formula:
Let m be contained the set of integers with m > 1. Then we can generate a Pythagorean triple by letting a = 2m, b = m^2 – 1, c = m^2 + 1.

Ex. Let m = 2, so we have that a = 2(2) = 4, b = 4 – 1 = 3, c = 4 + 1 = 5.
Then 4^2 + 3^2 = 5^2, which is true.
The formula generates finitely many triple but not all Pythagorean triples.

Formula for generating all Pythagorean triples

Consider the Pythagorean triple x – y – z where (x,y,z) = d. Let x = du,
y = dv, and z = dw where (u,v,w) = 1. So we have u^2 + v^2 = w^2, thus
u – v – w is a triple. Then we can say that every Pythagorean triple is a multiple of primitive triple.

Ex. Let x = 6, y = 8, z = 10. We have that 6^2 + 8^2 = 10^2. Then (6,8,10) = 2. We have that:
x = 2(3) = 6
y = 2(4) = 8
z = 2(5) = 10

This tells that 3 – 4 – 5 is a Pythagorean triple, which we know is true.

Using this idea in the equations for generating infinitely many primitive triples that Dan showed we can generate all triples by adding a parameter d. So we have:

a = d*(m^2 – n^2), b = d*(2mn), c = d*(m^2 + n^2) where m, n, d are contained in the set of integers with m>n>0 and d positive, n or m is odd and (m,n) = 1.

Conclusion:
The formula generates all Pythagorean triples but not uniquely.

There's also another way to generate primitive triples that I don't think was discussed in class, although it isn't linear algebra. If you let $a = 2k+1$ (any odd number), then $b = (a^2-1)/2$ and $c = (a^2+1)/2$. To prove it:

$a^2+b^2 = c^2$
$(2k+1)^2+(((2k+1)^2-1)/2)^2 = (((2k+1)^2+1)/2)^2$
$(2k+1)^2+((4k^2+4k+1-1)/2)^2 = ((4k^2+4k+1+1)/2)^2$
$(2k+1)^2+(2k^2+2k)^2 = (2k^2+2k+1)^2$
$4k^2+4k+1+4k^4+8k^3+4k^2 = 4k^4+8k^3+8k^2+4k+1$
$4k^4+8k^3+8k^2+4k+1 = 4k^4+8k^3+8k^2+4k+1$

Taken from the Wolfram website. I am pretty sure this method wasn't covered but I can't always be sure…

If $<a_0, b_0, c_0>$ is a Primitive Pythagorean Triple, then $<a_0, b_0, c_0>U_i$ generates a new primitive triple $<a_i, b_i, c_i>$ where

$U_1 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_2 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_3 = \left| \begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right|$.

Let's exam a simple example. We were told that $<3, 4, 5>$ is a primitive triple. So…

$<3 ,4, 5>U_1 = <5, 12, 13> ,$ $<3, 4, 5>U_2 = <21, 20, 29> ,$ $<3, 4, 5>U_3 = <15, 8, 17> ,$

…which can easily be verified to be valid!

01 Dec 2008 05:12
in discussion Ideas / Student Ideas » Nontotient numbers

I think that it's not that bad. But you also left out the fact that all odd numbers (excluding 1) are nontotient.
So remembering that $\phi(x)$ is multiplicative, we note the following: suppose $n = p_1^{a_1} * p_2^{a_2}$, with $p_1, p_2 \neq 2$ (there could be a 2 in there, but not more than one, since $\phi(2) = 1$ but that 4|$\phi(2^k)$ and 14 isn't divisible by 4). Then $\phi(n) = p_1^{a_1 -1}(p_1 -1) + p_2^{a_2 -1}(p_2 -1)$. But 2|$p_1 -1$ and 2|$p_2 -1$ and thus that implies that 4|$\phi(n)$ which would be impossible. And thus n cannot be a product of odd primes. So the only case left is that $n=p^{a}$ (again possibly times 2), and since p is odd let p = 2k+1, $k \in \mathbb{Z}$. But then $\phi(n) = p^{a-1}(p-1)$ which implies that $\frac{14}{2} = 7 = (2k+1)^{a-1}*k$. And then there's no k that makes that work (that would need to be more rigorous, but it's certainly a start).

Re: Nontotient numbers by , 01 Dec 2008 05:12
30 Nov 2008 21:19
in discussion Ideas / Student Ideas » Nontotient numbers

A nontotient number $n \in \mathbb{N}$ is a number such that there is no $x \in \mathbb{N}$ where $\phi(x) = n$.

The smallest such number is 14.

My question is how would one prove for example that 14 is nontotient?

Nontotient numbers by , 30 Nov 2008 21:19
28 Nov 2008 23:55

I was using StumbleUpon and I stumbled across a website that listed something special about a lot of the numbers between 0 and 9999. If anything, it seems like a great jumping off point for exploring unfamiliar topics in number theory.

25 Nov 2008 19:10
in discussion Ideas / Student Ideas » Fermat's Last Theorem Documentary

Here's something that I quickly stumbled over. It's interesting to see, if accurate, how the mathematical processes of theorem development are portrayed throughout this film.

45:21 Minutes

Fermat's Last Theorem Documentary by , 25 Nov 2008 19:10
21 Nov 2008 18:19
in discussion Ideas / Student Ideas » Fermat's last theorem for other n

here's an amazing link for a bunch of proofs of Fermat's last theorem for other n, such as 3 and 5. There's even a proof for n being a prime number.

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html

Fermat's last theorem for other n by , 21 Nov 2008 18:19
Other proofs that Pi is irrational by , 21 Nov 2008 01:42
19 Nov 2008 21:29
in discussion Ideas / Student Ideas » Pi and the ancient Egyptians

In class we learn that for the ancient Egyptians pi was approximately 3.16. Last semester I did a research paper of geometry before Euclid and learn that we know that Egyptians used pi approximately 3.16 from a geometry problem in the Rhind Papyrus. The problem asked, "A circular field has diameter 9 Khet. What is its area?” (1 khet is 100 cubits). From the solution of this problem is determined that the Egyptians used Π = 3+1/9+1/27+1/81~ 3.1605.

Pi and the ancient Egyptians by , 19 Nov 2008 21:29
19 Nov 2008 19:46
in discussion Ideas / Student Ideas » Merten's Conjecture

Define

(1)
\begin{align} M(n) = \sum_{1\le k \le n} \mu(k) \end{align}

where $\mu$ is the Moebius function

Merten's conjecture says that

(2)
\begin{align} \left| M(n) \right| < \sqrt {n} \end{align}

Now in 1985, this conjecture was proved false and also that a counter example exists somewhere between $10^{14}$ and $e^{1.59 \cdot 10^{40}}$.

So this result kind of kills peoples argument that "Goldbach's Conjecture is true because we've seen it to be true for any number tested".

But the fact that it's true for so many numbers makes me think it's more than just a coincidence. What do you guys think?

Merten's Conjecture by , 19 Nov 2008 19:46
19 Nov 2008 19:30
in discussion Ideas / Student Ideas » Reciprocal Fibonacci constant

Let $F_k$ be the $k$th Fibonacci number.

Define

(1)
\begin{align} \psi = \sum_{k=1}^{\infty} \frac{1}{F_k} \approx 3.359885666243177553172011302918927179688905133731 \ldots \end{align}

the sum of the reciprocals of every Fibonacci number

It has been proved to be irrational which looks like a daunting task to me

http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant

Reciprocal Fibonacci constant by , 19 Nov 2008 19:30
18 Nov 2008 23:29
in discussion Ideas / Student Ideas » Series Converging to Pi

Similarly, I have found other series converging to various multiples of $\pi$ in one of my other classes. Sadly, this of course requires methods outside the scope of this class but it's still pretty interesting that a large sum of rationals add to such a peculiar irrational.

$\frac{\pi^2}{8} = $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$$

This sum can be derived with the Fourier sine series of $f(x)=1$ on $(0,\pi)$.

$\frac{\pi^2}{6} = $\sum_{n=1}^{\infty} \frac{1}{n^2}$$

This sum can be derived with the Fourier sine series of $f(x)=x$ on $(0,l)$.

$\frac{\pi^4}{90} = $\sum_{n=1}^{\infty} \frac{1}{n^4}$$

This sum can be derived with the Fourier cosine series of $f(x)=x^2$ on $(0,l)$.

(Additionally, this required the use of Parseval's Equality).

Re: Series Converging to Pi by , 18 Nov 2008 23:29
18 Nov 2008 06:33
in discussion Ideas / Student Ideas » Series Converging to Pi

Chip and Josh in class today gave the series $\sum_{i=1}^{k} (-1)^{i-1}\frac{4} {2 i - 1} = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} ...$ as an example of a series that converges to $\pi$.

After doing some further research, I stumbled upon two other series that have convergence related to $\pi$:

Ramanujan's Formula

(1)
\begin{align} \pi = 2 \sqrt{3} \sum_{n=0}^{\infty}{\frac{(-1)^{n}}{(2n+1)3^{n}}} \end{align}

Chudnovsky Algorithm

(2)
\begin{align} \frac{1}{\pi} = 12\sum_{k=0}^{\infty}{\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+\frac{3}{2}}} \end{align}

The Chudnovsky Algorithm is based on a rapidly converging hypergeometric series. It was used to generate over a billion digits of $\pi$! Mathematica uses it today to calculate $\pi$.

Series Converging to Pi by , 18 Nov 2008 06:33
16 Nov 2008 10:37
in discussion Ideas / Student Ideas » Riemann Zeta Function

"We have known since the 1920s that the first two numbers are 1 and 2, but it wasn't until a few years ago that mathematicians conjectured that the third number in the sequence may be 42—a figure greatly significant to those well-versed in The Hitchhiker's Guide to the Galaxy."

Haha, that would be amazing if it really was 42. Then it really could be the answer to life, the universe, and everything!

Re: Riemann Zeta Function by , 16 Nov 2008 10:37
13 Nov 2008 05:34
in discussion Ideas / Student Ideas » Riemann Zeta Function

"In their search for patterns, mathematicians have uncovered unlikely connections between prime numbers and quantum physics. Will the subatomic world help reveal the elusive nature of the primes?"

My boyfriend sent me this article a couple of weeks ago, and I had been meaning to post it to the forum. It seems only appropriate now! I don't know how much has changed since this was published in 2006, but it's still interesting anyway.
Here you go: Riemann

Riemann Zeta Function by , 13 Nov 2008 05:34
03 Nov 2008 22:11
in discussion Ideas / Student Ideas » Primality Testing

So after lecture last week I was interested in the difference between primality tests that require a factorization and tests that don't. We saw one or two tests last week that didn't require a factorization, So I decided to look for more and have also included some that do require a factorization.

there are many many more if you're interested and wikipedia has many links.

Primality Testing by , 03 Nov 2008 22:11
page 1123...next »