### Sunday, November 23, 2003

## JSH: I WILL GET MY MONEY

If you fucking morons think that I will let you get away with not giving me credit for my fucking math discoveries then you have another fucking thing coming.

What the fuck??!!!

Where's "pure math" now, huh? Where's loving math for the fucking beauty of it now you fucking shits??!!!

LOOK AT IT!!!

Here is the partial difference equation and instructions for integrating.

dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))],

S(x,1) = 0.

And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,2) to dS(x,y).

http://mathforprofit.blogspot.com/

You fucking shits. I will get credit for my discovery and get fucking paid, and you best believe that I will not fucking let you stupid fucks get away with your fucking stupid bullshit--"pure math" my ass--without me coming at you fuckers with some fucking PURE fucking PURE AS FUCK math that you stupid shits have been shitting on for over a fucking YEAR!!!

You goddamn FUCKS!!!

What the fuck is wrong with you shits??!!! Don't you even believe in your own stupid shit? Where's "pure math" now?

Where is it?

What the fuck??!!!

Where's "pure math" now, huh? Where's loving math for the fucking beauty of it now you fucking shits??!!!

LOOK AT IT!!!

Here is the partial difference equation and instructions for integrating.

dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))],

S(x,1) = 0.

And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,2) to dS(x,y).

http://mathforprofit.blogspot.com/

You fucking shits. I will get credit for my discovery and get fucking paid, and you best believe that I will not fucking let you stupid fucks get away with your fucking stupid bullshit--"pure math" my ass--without me coming at you fuckers with some fucking PURE fucking PURE AS FUCK math that you stupid shits have been shitting on for over a fucking YEAR!!!

You goddamn FUCKS!!!

What the fuck is wrong with you shits??!!! Don't you even believe in your own stupid shit? Where's "pure math" now?

Where is it?

### Saturday, November 15, 2003

## Implementing the partial difference integration

I've talked about my find of a partial difference equation that you can integrate to get a count of prime numbers, and given some quick instructions, but I saw one post that leads me to consider that there are differences between how programmers tend to think, and how others do, so here's some additional help on the instructions.

Here's the difference equation and the instructions for the integration:

dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))],

S(x,1) = 0.

Note that it's a

Like if you're using Java or C++, you'll need to cast the sqrt() function, though I saved the cast for what follows in a C++ program I posted:

And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,2) to dS(x,y).

That is, you can get away with just casting x, to something discrete, while using non-discrete x and y, but it's just a waste of storage space.

For programmers "you get S as the sum of dS from dS(x,2) to dS(x,y)" means you sum up to and

for (i=2; i<y; i++;){}

which is WRONG, and if you do that you will probably get a wrong answer.

Note: p(x,sqrt(x)) here gives the same value as the traditional pi(x).

For faster calculations you need to use

dS(x,y) = [ p(x/y, sqrt(x/y)) - p(y-1, sqrt(y-1)][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))]

when sqrt(x/y) < y-1.

That's a BIG deal, as the pure math implementation is VERY SLOW, and even that quick speed-up won't push you very far.

However, the point is that the difference equation integration does work, which those of you who can program can verify for yourselves.

Then you should search to see if ANYONE has ever used a partial difference equation integration to get a count of prime numbers because then you can see that yes, I'm the only one in recorded history to present this method.

The above instructions are easy enough to program into a computer and if you follow them, you'll notice that you do get a correct count for prime numbers, and you should also notice that unless y is prime dS(x,y) = 0, so yes, it'll also tell you

The research is a shining example of the importance of

The independent researcher is important for the future.

Here's the difference equation and the instructions for the integration:

dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))],

S(x,1) = 0.

Note that it's a

**discrete**function, so for you programmers that means you need to use int's or long's or some discrete variable type.Like if you're using Java or C++, you'll need to cast the sqrt() function, though I saved the cast for what follows in a C++ program I posted:

And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,2) to dS(x,y).

That is, you can get away with just casting x, to something discrete, while using non-discrete x and y, but it's just a waste of storage space.

For programmers "you get S as the sum of dS from dS(x,2) to dS(x,y)" means you sum up to and

**including**y, as I don't doubt some of you may write something likefor (i=2; i<y; i++;){}

which is WRONG, and if you do that you will probably get a wrong answer.

Note: p(x,sqrt(x)) here gives the same value as the traditional pi(x).

For faster calculations you need to use

dS(x,y) = [ p(x/y, sqrt(x/y)) - p(y-1, sqrt(y-1)][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))]

when sqrt(x/y) < y-1.

That's a BIG deal, as the pure math implementation is VERY SLOW, and even that quick speed-up won't push you very far.

However, the point is that the difference equation integration does work, which those of you who can program can verify for yourselves.

Then you should search to see if ANYONE has ever used a partial difference equation integration to get a count of prime numbers because then you can see that yes, I'm the only one in recorded history to present this method.

The above instructions are easy enough to program into a computer and if you follow them, you'll notice that you do get a correct count for prime numbers, and you should also notice that unless y is prime dS(x,y) = 0, so yes, it'll also tell you

**which**numbers are prime.The research is a shining example of the importance of

**independent**researchers willing to check in areas that more staid academics think are fallow.The independent researcher is important for the future.

### Sunday, November 09, 2003

## Factorization example, end of argument

Here's an example that shows the problem with the ring of algebraic integers.

Consider

(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)

where the c's

Now then, you have as a zero of the factorization x = -7/c_1, so let x= -7/c_1, so you have

49(-7^3/c_1^3 + 5(49)/c_1^2 - 21/c_1 + 2) = 0

which is

2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.

But that is a non-monic primitive irreducible over Q, so c_1 and by symmetry c_2 cannot be algebraic integers. However they must be algebraic numbers, and it can be shown that any algebraic number can be written as the ratio of algebraic integers.

So then there must exist f, such that fc_1 is an algebraic integer, and letting g = fc_1 and multiplying both sides by f, I have

(gx + 7f)(c_2 x + 7)( c_3 x + 2) = 49f(x^3 + 5x^2 + 3x + 2)

so a zero is now x = -7f/g, which gives

49f(-7^3f^3/g^3 + 5(49)f^2/g^2 - 21f/g + 2) = 0

which is

2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0

which proves that f must have 2 itself as a factor for g to be an algebraic integer.

But looking back again at

(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)

that would mean that c_3 has a factor that is 2, which can distribute to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of 2 by symmetry.

But you see, you only have just that one 2.

Now then, are mathematicians and math groupies rational, or are you cranks?

Will you give up when you see the truth, or will you keep fighting as if mathematics is just a joke to you, as if you never really cared whether or not it was true, as long as everyone you cared about agreed with you?

What's important to you sci.math newsgroup?

Consider

(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)

where the c's

**should**be algebraic integers. Notice that only two of the c's can have 7 as a factor.Now then, you have as a zero of the factorization x = -7/c_1, so let x= -7/c_1, so you have

49(-7^3/c_1^3 + 5(49)/c_1^2 - 21/c_1 + 2) = 0

which is

2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.

But that is a non-monic primitive irreducible over Q, so c_1 and by symmetry c_2 cannot be algebraic integers. However they must be algebraic numbers, and it can be shown that any algebraic number can be written as the ratio of algebraic integers.

So then there must exist f, such that fc_1 is an algebraic integer, and letting g = fc_1 and multiplying both sides by f, I have

(gx + 7f)(c_2 x + 7)( c_3 x + 2) = 49f(x^3 + 5x^2 + 3x + 2)

so a zero is now x = -7f/g, which gives

49f(-7^3f^3/g^3 + 5(49)f^2/g^2 - 21f/g + 2) = 0

which is

2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0

which proves that f must have 2 itself as a factor for g to be an algebraic integer.

But looking back again at

(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)

that would mean that c_3 has a factor that is 2, which can distribute to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of 2 by symmetry.

But you see, you only have just that one 2.

Now then, are mathematicians and math groupies rational, or are you cranks?

Will you give up when you see the truth, or will you keep fighting as if mathematics is just a joke to you, as if you never really cared whether or not it was true, as long as everyone you cared about agreed with you?

What's important to you sci.math newsgroup?