quick follow up to my posts on prime numbers and cosine waves

just wanted to post a quick follow-up to the entry I had about cosine waves...  taking the cosine value to a higher power just helps in math terms to guarantee the values are in an acceptable range...

when actually trying to program these functions, it is best to make use of the "floor" function to scrap any values that are between 0 and 1... this helps retain only the 1 values from the waves in the SUM, and is less taxing on the computation.

so rather than SUM(Cosine(pi*n/k)^(2^n), for k = {1,n}), utilizing SUM(FLOOR(Cosine(pi*n/k))) will yield the same results as the sigma/divisor functions.

that's all, just wanted to throw that out there in case anyone has read this and actually wants to implement it!


more details on the FLOOR function and its applications can be found here:  http://en.wikipedia.org/wiki/Floor_and_ceiling_functions

scribbles from 2009 - sigma and divisor functions

More prime stuff

Took photos of my notes rather than typing out the formulas.  Playing around with the cosine waves led me to this summation in 2009, Using cosine squared keeps the function even and continuous (the absolute value of cosine has non-differentiable points, or cusps). 

Description from Wikipedia (http://en.wikipedia.org/wiki/Divisor_function):

 "...

The sum of positive divisors function σ_x(n), for a real or complex number x, is defined as the sum of the xth powers of the positive divisors of n, or

..."

Here are the first two of these sigma functions as determined by my cosine representation, the divisor function δ(n), and the sum of divisors function σ(n):

These equations should probably have a '~' over the equals signs... By way of construction, these waves only equal 1 when N is divisible by the K value.  I took the cosine squared function to the Nth power because it would be positive, even (and symmetric), retain all 1 values, and minimize any values that were less than 1 (making use of the fact:  if 0 < x < 1, then lim [x^n] = 0 as n à ∞). 

Here is another clip from my notes where I've represented sigma function with a slightly different ordering:

The sigma and divisor functions have so many applications throughout mathematics and they are related to many unsolved problems.  I'm still looking for a closed-form equation, but haven't found the right path to go down... Lots of "rabbit trails" that seem promising at first, but end up winding in circles.  I keep thinking there might just be some trig identity that helps collapse sums of waves, but haven't found it yet. 

Kind of related, from 2008:

I also found a conjecture of mine concerning the amount of prime numbers between powers of 2... No proof yet, but computer tests look promising as the "i" value increases to infinity, the error value decreases to 0...

Simply put, let π(n) be the number of prime numbers less than or equal to n. 

If...

C = π(2^n) = the number of primes up to 2^n

A = π(2^(n-1))      and...

B = π(2^(n-2))      ...

Then C < 3A-2B

The number of primes up to 2^n is less than three times the primes up to 2^n-1 minus twice the primes up to 2^n-2.

From my notes, I had written, "the number of primes in between 2^i and 2^(i+1) is less than 2 times the number of primes between 2^(i-1) and 2^i." 

Equation form:



π(2^(i+1)) - π(2^i) < 2 * [ π(2^(i)) - π(2^(i-1)) ]

==>

π(2^(i+1)) - π(2^i) < [ 2*π(2^(i)) - 2*π(2^(i-1)) ]

==>

π(2^(i+1)) < 3*π(2^(i)) - 2*π(2^(i-1))

==>

π(2^i) < 3*π(2^(i-1)) - 2*π(2^(i-2))

So after a little manipulation, and shifting the index, you get the same inequality as shown in the conjecture... I've tried studying recursion with these kinds of inequalities, but found that the accuracy decreases as you expand further..

I've found that the conjecture fails when i=5, but couldn't compute high enough to find another exception to the rule.  Here is that example:

π(2^5) = π(32) = 11  prime numbers up to 32:  (2,3,5,7,11,13,17,19,23,29,31)

but

π(2^4) = π(16) = 6  prime numbers up to 16: (2,3,5,7,11,13)

π(2^3) = π(8) = 4  prime numbers up to 8:  (2,3,5,7)

3 * π(2^4) - 2 * π(2&3) = 3*6 - 2*4 = 10  < 11...

So π(32) is not less than 3*π(16) - 2*π(8) as the conjecture suggests it should be, it's off by 1...

That's the only hiccup that I've found in testing this conjecture.  I have yet to find another value for 'i' which causes the inequality to fail, but that's only as far as my computer can count for now - better programming will yield better results. It's very possible that way down the number line, some huge power of 2 exists that has more primes than the estimate... but that's why it's still a conjecture!

Thanks for reading - that's more than enough prime numbers for today :)

Trig Waves and Divisibility

Finally uncovered some old matlab/octave generated graphs from my prime number research (2008).  I don't want to take the time to fully detail each formula/function, but the images are interesting to study, and you can probably figure them out if you really want to.

This first graph is the one that guitar harmonics reminded me of in an earlier post...  It's the absolute value of sine/cosine waves increasing in period and amplitude by integer amounts.

As you can see, taking the absolute value makes it much easier to manage, because you can cut the period in half, and you don't have to mess with those annoying negative numbers.

This graph is increasing sine waves, with the asymptotes showing how the trend would continue if more and more waves were plotted. In this graph, N=3.

You can see that the waves all equal zero whenever they reach a number that is divisible by each waves period.  More waves coincide at integers that are highly divisible, and an obvious pattern emerges, and repeats.

Note how the asymptotes alternate +/- (positive/negative).

This next graph is essentially the same function, but this time, N=50.  Although this looks much more chaotic, the same symmtry is still underlying. 

(the first three asymptotes are identical to the graph above... note how much more they've "filled-in" with the resulting waves...)

I haven't decided which I like better, sine or cosine - they each have their advantages when dealing with the prime numbers and divisibility.  Taking the absolute value of the waves chosen makes all of the asymptotes positive, as you can see in this graph:

This graph shows the exact same information as before, but now we are restricted to positive numbers.

I only used the increasing amplitude in my graphs to make the the symmetry easier to see.  Modifying the formula to give each wave equal amplitude results in this representation of the symmetry:

It's tougher to discern where the waves of each period start and stop.  I've filled in black dots for all of the prime numbers up to 60, and included the circles to show the symmetry of the spacing of the waves.  Remember, when the waves all coincide at the axis, the number is composite (divisible by other numbers - not prime).  This is a visual representation of the Seive of Eratosthenes made much quicker with symmetric waves.  Wherever the waves do not touch (up to a limit) will be the prime numbers.

So an easy way to see where the waves all line up is to take the SUM of the values of each wave at each integer.  Using the equal amplitude waves and taking the SUM results in this crazy graph:

This is the sum of equal amplitude cosine waves up to 50 partial sums (kind of reminds me of the stock market, but that's another post for later). The points in the graph where it seems to dip or have peaks and valleys is just another representation of the prime number symmtry.  Each major dip happens at a highly divisible number (e.g. 24, 30, 36, 48, 60).  A cool thing happens when you take sums of symmetric functions:  the sums also retain a similar symmetry.

I guess that's enough on this topic for now.  All of these graphs and functions are from research I did in 2008, so I've got plenty more to cover.  I'll bring on another installment soon enough!

The Goldbach Conjecture (an unsolved problem)

The Goldbach conjecture (GBC) has been around in one form or another since the mid-1700s. 
It is easily stated, but has yet to be proven...

The statement:
"Every even number is the sum of two prime numbers"

Is what is known as the "strong Goldbach conjecture".  This was noted in correspondence between two very well known mathematicians of the time, Leonhard Euler and Christian Goldbach.

I am always drawn to study unsolved problems in mathematics because in the past, solutions to such problems have given insight to many other fields, and even require invention of new techniques and models to solve, which advances the science of mathematics even further. 

So here is an introduction to my research on the unsolved Goldbach Conjecture.

From the beginning,
"Every even number is the sum of two prime numbers"

So letting each even number be represented by 2n, we can rewrite this statement as:

2n = p + q    (where p and q are prime numbers less than 2n)

now, it turns out that you can do a little algebraic manipulation to this equation to obtain:

n + n = p + q

which then yields:

n - p = q - n

Now, closely consider this equation...  What this means is that if GBC is true, there is an underlying "symmetry" in the primes.  How?

This arrangement of the equation lets one see that p and q are equidistant from N (which is half of 2n, obviously).  This picture makes it easy to imagine:

So now, what's the big deal?  Why should somebody try rearranging 2n = p + q at all?

IF the GBC is true, this means that for every N, there exist 2 primes, p and q, such that:

1. p < n
2. n < q < 2n    ------>  (see below how this can be n < q < 2n-2 by Bertrand's postulate)
3. neither p or q divide 2n
4. d(p,n) = d(q,n)  (distance function d(x,y)...)

So my goal is to try to force a contradiction in the following manner:

We know that p exists, (any arbitrary prime less than N).
According to Bertrand's postulate,
(if n > 3 is an integer, then there always exists at least one prime p with n < p < 2n − 2)
We know that there must exist some prime q such that n < q < 2n-2
So we can close our bounds on condition 2 above just a little further as indicated above.

It is only left to prove that there exists one prime Q greater than n, less than 2n-2, such that Q-N = N-P, and P does not divide 2n.

IF it can be shown that no such Q can exist, we have contradicted Bertrand's postulate, and therefore GBC must be correct.

But that's the hard part.  I still have yet to find a successful route of attack to force this contradiction. 

Some things to consider if you're going to try to attack it:

1.  Non-prime numbers can be equidistant from prime numbers too...  i.e. 20 = 11+9   but 9 is not prime... How can one guarantee that at least one prime Q (where n < Q < 2n-2) is the same distance from N as a prime less than N?

2.  For each prime P < N, there exists a number N + (N - p) = 2N - P = Q' 
Is there an effective way to check primality of this Q' in each case where P is prime and P < N?  We know that at least one prime number Q must exist in (n, 2n-2), but how do we know which P' < N it is associated with?

From this line of attack I can see two possible ways of proving GBC:
Firstly, by proving the existence of Q for every N and 2N (induction? pigeon-hole principle?).
Or secondly, by assuming Q does not exist and forcing a contradiction to Bertrand's postulate saying that there must be a prime in (n, 2n-2). 

So that's a good introduction so far.  Still no solution as of yet, but I'll keep working on it and see where it goes!