Riemann Hypothesis

May 14th, 2012

So in dawdling with the Riemann hypothesis, first of all I just want to go ahead and get an idea of the values of the function. In order to do that, I have been through a variety of crash course training on Complex Logs and calculations of e^z. Here are my notes on the subject up to this point, whether it is legible for you or not who knows.

So
Re(2) =
1/1^2 + 1/2^2 + 1/3^2

Re(-2) =
1^2 + 2^2 + 3^2

So the values are
1, 5, 14, 30, 55
= 1, 1*5, 5*2.8, 14*2.2
Does this have any significance, where the slope is e, or where the slope is PI?
or is the slope at e, PI?
This is while calculating an individual value of Re(S)

also
Re(-2) =
log(e^1)*log(e^1) = 1
+
log(e^2)*log(e^2) = 4

or
log(e^1)^(log(e^S))
+
log(e^2)^log(e^S))

(1+(1/n))^n = e
(1-(1/n))^n = 1/e

This is significant.

Re(2+i) =
1^(2+i)
=
1^2*1^i
=
log(e^1)^log(e^2)*log(e^1)^log(e^i)

or just to give us an idea to start with use:
log(e^1)*log(e^1)*log(e^1)^log(e^i)
still get stuck with the ^log(e^i)
how do we break this down?
apparently it is done as follows:
http://math.fullerton.edu/mathews/c2003/ComplexFunLogarithmMod.html
and ends up being
log(x+iy) = ln(x) + i*Arg(x+iy)

Arg(x+iy)
2*arctan(y/(sqrt(x^2+y^2)+x)

Arg returns the angle:
then use that with the vector
r(cos(theta)+i*sin(theta))

So in the case of i
log(1+i) = ln(1) + i*Arg(1+i)
=
calculate theta: 2*arctan(1/(sqrt(1+1)+1) = 0.785398163 = PI/4
ln(1) + i*0.785398163

or is it just this:
log(1+i) = log(sqrt(1^2+1^2))+i*atan(1/1)
0.150514998 + 0.785398163 i

e^i = 0.540302306 + 0.841470985 i

accoring to google:
log(1 + i) = 0.150514998 + 0.341094088 i
PI/9.21034038 * i
log(i) = 0.682188177 i

not sure if this is correct on google?
2 places show log(1+i) as log(sqrt(2)) + PI/4 * i

So if we think of this as a vector with length 0.5403…. as described in Polar notation,
can we add up all of the vectors to get an estimate of the value of the SUM?

so Re(1+i)
would be
1/1^1+i
=
1/1^log(e^1+i)
=
1/1^log(e^(log(e^1+i))) ….
=

e^(x + iy) = e^x(cos y + i sin y).

This is how we would have to do things.
so
1/1^log(e^1*(cos(1) + i*sin(1))
=
1/1^log(e*cos(1)+i*e*sin(1))
=
x:(e*cos(1)), y:(e*sin(1))
1/
1^(
ln(sqrt(x^2+y^2))
+ i*2*arctan(y/(sqrt(y^2+x^2)+x)

ln(sqrt((e*cos(1))^2+(e*sin(1))^2)) +
i*2*arctan((e*sin(1))/(sqrt((e*sin(1))^2+(e*cos(1))^2)+(e*cos(1)))

So from here we should be able to calculate some things.
This looks like it works out ok.
All this turns out to be 1+i

so 1^y = log(e^1)^log(e^y)

so log(u) root of s1 = 1
log(u) root of s2 = 2

but we should be able to do exponentiation in terms of e

********************
so 1^y = e^log(1)*y. But does this help us.
u:ln(sqrt((e*cos(1))^2+(e*sin(1))^2)) +
i*2*arctan((e*sin(1))/(sqrt((e*sin(1))^2+(e*cos(1))^2)+(e*cos(1)))

So we have 1/e^log(1)*u
+ 1/e^log(2)*u

where can we go from here?
this can then be reduced again to
e^(x + iy) = e^x(cos y + i sin y).
So then we will have 1/(x+iy)
***********************************

Then we have to figure out how to deal with this:
I guess then we deal with the vectors, and see what we can do that way.
So we have the magnitude and the angle, and then add them all up.
But what do we do with the 1/(x+iy).

a+bi/c+di = (ac+bd)/(c^2+d^2) + (bc-ad)/(c^2+d^2)*i
So in our case:
1+0i/c+di = (1*c+0)/c^2+d^2) + (0-1*d)/(c^2+d^2)*i

Then we just have x+iy without the fraction.

So this is the very simple way of estimating the elements of Re(S).
Really the first 100 or so elements is going to give a fairly good idea of the actual value, or if it is close to 0 or not.
I would like to get a general idea of how the Zeta Function looks in the positive side of the real number line, not just the critical strip. There is a simple graph here, but that is not sufficient. Anyway, calculating the actual values is a good exercise seeing as I am not familiar with the workings of imaginary exponents or logarithms anyway.

After we have a better understanding of the function, perhaps it will become clearer why exactly it is related to the patterns in prime numbers. Will have to read this again.

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More on Probability

May 14th, 2012

When I think about the subject recently, I still end up thinking about the probability of the next number being prime.
And the sequential calculation of this.

1/2 * 2/3 * 4/5 * 6/7 = 8/35

i.e. if we do it falsely and consider every number to be prime we get:

1/2 * 2/3 * 3/4 * 4/5 * 5/6 * 6/7 = 1/7 = 5/35 which is obviously not correct.

if we consider no prime numbers then we have the probability of being prime as 1 at any point.

So how can we rule out those numbers which aren’t actually prime by using this simple calculation?
That is the question I believe should have an answer to.
Although I dont know what that answer is.

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patterns in p^x

January 20th, 2010

I do see an interesting pattern in p^x, although it is tough to test very far.

p^2 for p up to 1000 seems to be of the form 6n+1. I believe this will hold true. Eulers 6n+1 Theorem can be written in the form x^2+3*y^2.  This is not related here, but searching for 6n+1 I found it.

6 = 3*2

In this case N also appears to be divisible by 4.  So basically 6*(4n)+1. This appears to be an extension of all primes above 2 being of the form 6n+1 or 6n-1, so (6n+1)^2 = 36n^2+12n+1 and (6n-1)^2 = 36n^2-12n+1, both of which will be of the form 6n+1.

12n(3n-1)+1 or 12n(3n+1)+1, which are basically 12n*x+1, so it is of form 6n+1 by default.

p^3 is also 30n+p (p being in the set 1, 7, 11, 13, 17, 19, 23, 29). Maybe we can work that out too from (6n+/-1), but I won’t spend the time here.  This works for any multiple of 6 up to 30.

30=5*3*2

p^4 appears to be of the form 210n+(1,121,151).  This is also of form (5!)*n+1 or 2*(5!)+1.  I imagine we can show with a 6n+/-1 calculation.

210=7*5*3*2

This seems to show promise for a pattern to me, but p^5 does not show any great noticeable pattern.

Results appear to be of the form 2310n+p or 2310n+p*p1, but not enough data to really think this is a pattern. Same pattern if we substitute 720 or 6!*n

2310=11*7*5*3*2

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More Graphical representations

January 17th, 2010

I tried again to do some more representations.

This time the pattern that is very obvious is above 25 or 5^2, we can see that all primes are of the form p^2+6n-2 or p^2+6n.  I’m sure there is some interesting explanation for this, but I do not see it off-hand.

Here is the resulting image from that.

primesquare_1

In order to see it well, you will have to download it and view it about 800%.  But the mentioned pattern is obvious.

I also find it interesting how i.e. 11^2 and 19^2 fall on the same line, and 17^2 and 31^2 and 23^2 and 43^2.

Unfortunately we can not see beyond that if this pattern continues, but obviously 29^2 and 55^2 is not in this, since 55 is no longer prime.

Perhaps there is another pattern like this farther down the primes, or perhaps not I don’t know.

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Ulam’s Spiral

January 2nd, 2010

I decided to play around with Ulam’s spiral a little bit today.

Here is the resulting image.   I decided we should see how primes are related to squares.  So at each square, I start at the ZERO pixel again.  You can see easily the first few are primes 1, 2, 3, 4 (2^2) is red.  Then 5, 7, and 9 (3^2) is red.

ulam1

Here is the important part of the code so it is easier to understand.

for ($c=0; $c<264*264; $c++){ // $root = sqrt($c); $isprime = false; if ($i){ $root = pow($c, 0.5);//1/$i); } if (($isprime=$p->isprime($c))){
$temp = $primecolor;
$lastprime = $c;
}
else{
$temp = $background;
}

imageSetPixel ( $image, $i, $j, $temp);
if (abs($root-round($root)) < 0.0001){
if ($i%8 == 0){
imagestring($image, 1, $i/2, $j, intval($root), $textcolor);
}
$i++;
$j=0;
}
else{
$j++;
}
}

imagepng($image);
// clean up
imagedestroy($image);

The thing that interests me the most is the diagonal lines right above 2^2, 3^2, 5^2, 7^2 and 15^2. These are full lines, and I wonder what exactly causes them. I’m sure there is some fairly simple explanation, but I don’t see one off-hand.

Another pattern can be seen at 7^2, 13^2, 19^2 and 43^2.  That pattern is x^2-2, (x+1)^2-3, (x+2)^2-4.  There is a sequence at 7^2 (2), at 13^2 (2, 4), at 19^2 (2, 4, 6), at 43^2 (2, 4, 6, 8, 10). I am not sure what causes this either.  They are all of the form 7*n – (n-1).  There is always a gap at (x+2)^2-4.  This is always divisible by x and x+4.   Maybe we should be thinking about what is definitely not prime instead of what is the pattern for prime numbers.  This works for even numbers as well.  Basically x^2-4 is never prime.  x^2-y^2 is never prime.  This is a famous known fact that was investigated and used extensively by Euler I believe.  So what other patterns do we have?

I think the idea of visualizing the primes in a variety of different ways and how they relate to x^2 for instance or x^3, etc. is a good exercise. I keep thinking eventually something will become apparent.  Here’s to hoping.

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