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Paul Mackenzie
#6
Feb27-12, 08:57 PM
P: 16
Quote Quote by SteveL27 View Post
Can you back up to the beginning and help me to understand your notation?

What's the sequence for the even numbers? What's the sequence for the primes? I think if you can simply show a couple of examples, we can help sort out the notational issues.

It's not possible for anyone (well, for me, anyway) to follow your argument since the definition of f[x] was so garbled. I get that to each term of a sequence you assign a pair of numbers, but I'm unclear on what those numbers are. Then you start talking about 2N, sort of out of the blue, without defining N. And if you're not sure whether 0 is an even number, let's talk about that.

In other words, let's nail down the Intro before going forward.
Hi Steve and all:

Thanks for your help regarding the notation.



To understand where I coming from, I will describe f[x]
with reference to the prime number sequence {2,3,5,7,11,13}.

In this example the sequence is limited to all primes less
than 16 (viz 2N = 16).

I attach a pdf file which shows the functions f(x), f(2N-x),
and f(x).f(2N-x) where 2N = 16 and
where f(x) = 1 when x is prime
f(x) = 0 when x is otherwise;
and similarly f(2N-x) = 1 when 2N-x is prime
f(2N-x) = 0 when 2N-x is otherwise

The function f(x).f(2N-x) is one only when both x and 2N-x are
both prime so this function illustrates the goldbach partitions
of the even number 16. viz (3,13)(5,11)(11,5)(13,3).

The number of Goldbach partitions for 2N = 16
is then the sum of f(x).f(2N-x) from x = 0 through to x = 15 (2N-1),
which in this case four.

Looking at this problem you can see that figs. 1 and 2
are mirror images of each other about N = 8.

Furthermore Fig. 3 shows where these functions shown in
Fig. 1 and 2 are symmetrical.

The value 2N comes about so as to limit the prime number sequence.
So if 2N = 24 then the prime number sequence includes all primes
upto 24. I choose an even 2N because we are concerned with mirror
symmetry. Furthermore the sum of f(x).f(2N-x) from x = 0 through to 2N-1
will give you the number of goldbach partitions for that even number 2N.


So as to compare a value which is representative of the
symmetry for different values of 2N, I normalised this
sum to get a symmetry value I. The normalising value is
the sum of f(x).f(x) from x = 0 through to x = 15 (2N-1)

So the symmetry value I is the sum of f(x).f(2N-x)/ sum of f(x).f(x) from x = 0
through to x = 15 (2N-1)

which in this example I = 4/6. [Note the demoninator equals pi[2N=16] = 6]

As mentioned in my post, this symmetry value I for prime number sequences upto 2n
is then equal to g(2n)/pi(2n). I plotted this value for even numbers upto 80,000
and it seemed as though g(2n)/pi(2n) approaches zero as 2n appeoaches infinity

I realised this problem can be generalised for many integer sequences to obtain a symmetry
value [ with some restrictions]. I realised for example you could not apply this technique
to Fibonacci sequence. i.e. {0,1,1,2,3,5,.....} because two elements of the sequence are
equal, viz the second and third element "1". It seems that's where the confusion arises,
when I tried to deal with this.

I also gave an example regarding even numbers which probably does not correctly describe
the function f[x]. But I hope my depiction of the functions in these diagrams will assist
in your understanding. You are correct that my statement that "zero is not a even number"
was erroneous. I have removed it.


I think this generalisation could be applied to non-integer values and maybe some
real world applications. For instance you could take a finite seqment of speech, sample it,
reverse it, and multiply the sampled signal and reversed sampled signal together and divide it by the
power of the signal to obtain a value for its symmetry. But this requires a lot more work.

I hope these diagrams help with your understanding

Kind Regards
Attached Files
File Type: pdf sample.pdf (277.0 KB, 5 views)