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Science Forum Index » Cognitive Science Forum » Uniqueness Demonstration with Prime Counting
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Message |
| James Harris |
 Posted: Thu Dec 04, 2003 11:16 pm |
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Guest
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I'm a guy who over 18 months ago made a major math discovery in the
area of prime numbers, but unfortunately have had difficulty being
fully heard.
I've been posting this partial difference equation, but the math is
kind of dry, so I thought I'd post *values* from that partial
difference equation to prove uniqueness.
If someone wishes to dispute my uniqueness claim, they can just give
their own math formula and post values from it, showing that it
matches mine.
Nothing like a demonstration, eh?
The formula I have is
dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,
sqrt(y-1))],
and here are some values from the calulation of p(100,10):
dS(100,2)=49
dS(100,3)=16
dS(100,4)=0
dS(100,5)=6
dS(100,6)=0
dS(100,7)=3
dS(100, =0
dS(100,9)=0
dS(100,10)=0
So there's the formula and some values. The sum of all those values
is 74, which subtracts from 100 to give 26, and subtracting 1 from
that gives you 25, which is the count of primes up to 100. You see,
the dS values are composite counts, which is how the formula can be
used to count prime numbers.
The fact that it's a partial difference equation is why you can
evaluate it for all those values, which is a key difference from
anything else in the math world.
Here the challenge to posters is a simple one: give a formula and
values that can match my formula feature for feature.
Let's see what happens.
Want more information?
See more at my blog http://mathforprofit.blogspot.com/
Go to my November 16, 2003 posting and you can get a Java program to
make your own dS calculations.
James Harris
"My math discoveries, found for profit"
http://mathforprofit.blogspot.com/ |
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| Sam Wormley |
| Posted: Thu Dec 04, 2003 11:28 pm |
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| Virgil |
| Posted: Fri Dec 05, 2003 12:52 am |
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In article <3c65f87.0312042016.447dddf6@posting.google.com>,
jstevh@msn.com (James Harris) wrote:
Quote: If someone wishes to dispute my uniqueness claim
Not at all. You are unique and we are all glad of it. |
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| John C. Randolph |
| Posted: Fri Dec 05, 2003 3:16 am |
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James Harris wrote:
Quote:
I'm a guy who over 18 months ago made a major math discovery in the
area of prime numbers, but unfortunately have had difficulty being
fully heard.
Nonsense. You've been heard ad naseum.
-jcr |
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| Wolf Kirchmeir |
| Posted: Fri Dec 05, 2003 4:45 am |
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On 4 Dec 2003 20:16:44 -0800, James Harris wrote:
Quote: 'm a guy who over 18 months ago made a major math discovery
Stop using the word "major." You are not competent to judge the value of your
discovery.
--
Wolf Kirchmeir, Blind River ON Canada
"Nature does not deal in rewards or punishments, but only in consequences."
(Robert Ingersoll) |
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| C. Bond |
| Posted: Fri Dec 05, 2003 10:18 am |
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James Harris wrote:
Quote: I'm a guy who over 18 months ago made a major math discovery in the
area of prime numbers, but unfortunately have had difficulty being
fully heard.
Puleez! You come through loud and clear (emphasis on loud, deemphasis on
clear), and often.
Quote: I've been posting this partial difference equation, but the math is
kind of dry, so I thought I'd post *values* from that partial
difference equation to prove uniqueness.
If someone wishes to dispute my uniqueness claim, they can just give
their own math formula and post values from it, showing that it
matches mine.
Nothing like a demonstration, eh?
Yeah, but nobody asked you to demonstrate your algorithm (which you
cryptically call a formula). They *did* ask you to demonstrate the
results from your so-called 'partial differential equation' which you
claim is of cosmic importance and inestimable value. To date, you have
done no such thing. No results, no values, no demonstration. Just rants,
harangues, diatribes, grandiose claims, diversions, dissembling and
insults. Stand up like a man and put up or SHUT UP.
--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com |
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| Randy Poe |
| Posted: Fri Dec 05, 2003 12:44 pm |
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jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312042016.447dddf6@posting.google.com>...
Quote: I'm a guy who over 18 months ago made a major math discovery in the
area of prime numbers, but unfortunately have had difficulty being
fully heard.
I've been posting this partial difference equation, but the math is
kind of dry, so I thought I'd post *values* from that partial
difference equation to prove uniqueness.
If someone wishes to dispute my uniqueness claim, they can just give
their own math formula and post values from it, showing that it
matches mine.
Nothing like a demonstration, eh?
The formula I have is
dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,
sqrt(y-1))],
and here are some values from the calulation of p(100,10):
dS(100,2)=49
dS(100,3)=16
dS(100,4)=0
dS(100,5)=6
dS(100,6)=0
dS(100,7)=3
dS(100, =0
dS(100,9)=0
dS(100,10)=0
http://mathworld.wolfram.com/LegendresFormula.html
"Legendre's Formula [phi(x,a)] counts the number of positive
integers less than or equal to a number x which are not
divisible by any of the first a primes"
phi(100,0) = 100
Difference between these values = 49 + 1
phi(100,1) = 50
Difference between these values = 16 + 1
phi(100,2) = 33
Difference between these values = 6 + 1
phi(100,3) = 26
Difference between these values = 3 + 1
phi(100,4) = 22
Do you notice any slight resemblance between the
sequence of values Legendre's formula generates
and your sequence of values? How can you say you're
not evaluating the same values?
As many people have pointed out, you are doing
Legendre's method with a notational difference, but
no saving in computation. Indeed if implemented as
written, you are adding a massive amount of computation
to Legendre's due to the number of zero terms which
you also evaluating and the number of things which
get evaluated multiple times by your recursion.
- Randy |
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| Uncle Al |
| Posted: Fri Dec 05, 2003 2:28 pm |
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| matt grime |
| Posted: Sun Dec 07, 2003 1:00 pm |
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On Thu, 04 Dec 2003 20:16:44 -0800, James Harris wrote:
Quote: I'm a guy who over 18 months ago made a major math discovery in the
area of prime numbers, but unfortunately have had difficulty being
fully heard.
I've been posting this partial difference equation, but the math is
kind of dry, so I thought I'd post *values* from that partial
difference equation to prove uniqueness.
If someone wishes to dispute my uniqueness claim, they can just give
their own math formula and post values from it, showing that it
matches mine.
Nothing like a demonstration, eh?
The formula I have is
dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,
sqrt(y-1))],
and here are some values from the calulation of p(100,10):
dS(100,2)=49
dS(100,3)=16
dS(100,4)=0
dS(100,5)=6
dS(100,6)=0
dS(100,7)=3
dS(100, =0
dS(100,9)=0
dS(100,10)=0
So there's the formula and some values. The sum of all those values
is 74, which subtracts from 100 to give 26, and subtracting 1 from
that gives you 25, which is the count of primes up to 100. You see,
the dS values are composite counts, which is how the formula can be
used to count prime numbers.
The fact that it's a partial difference equation is why you can
evaluate it for all those values, which is a key difference from
anything else in the math world.
Here the challenge to posters is a simple one: give a formula and
values that can match my formula feature for feature.
Let's see what happens.
Want more information?
See more at my blog http://mathforprofit.blogspot.com/
Go to my November 16, 2003 posting and you can get a Java program to
make your own dS calculations.
James Harris
"My math discoveries, found for profit"
http://mathforprofit.blogspot.com/
ok, i've got one.
let p(x) denote the number of primes less than or eqaul to n.
p(x) satisfies the following difference equation
dp(x) = p(x) -p(x-1)
and dp is either 0 or 1. Now the amazing thing here is that unlike your
one, James, I can give a finite set in which dp takes values. Your dS's
(or dSs if you prefer) are not given by some nice formula. Or at least you
haven't given one that doesn't involve S, which is kind of cheating a
little. So if you want to claim yours is better than mine then please
give me the nice closed expression for dS, that is one not in terms of S,
and from which S can be found.
For instance.
f(n) - f(n-1) = 0 ; n \in N, f(0)=1
has a somewhat different solution than
f(n) - f(n-1) = n ; n \in N, f(0)=0
and as for the challenge to produce numbers:
dp(0)=0
dp(1)=0
dp(2)=1
dp(3)=1
dp(4)=0
dp(5)=1
dp(6)=0
dp(7)=1
from which i conclude p(7)=4, let me check....2,3,5,7. yep it works.
m |
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