Can This 4-Dimensional Integral Provide a New Way to Calculate Pi(x)?

[matt grime]

but you'd already said that your metohd was the best so you must have calcualted this> or are you admitting to making intellectually fradulent claims?

 

[eljose]

sorry matt perhpas i exagerated a bit..i thought it was the best because (at least i thought) it provided Pi(x) without calculating a sum or without knowing all the primes but i don,t know how to calculate the time involved running it.

i have given an integral for Pi(e^1/t) so if we want to know the value of Pi for high values we need to calculate the integral for small t

you needn,t make me fun of my poor engilsih or my scarce knowledge of numerical methods, in fact i am a physicist (surprised?) by i like math and someday i would like to contribute to math and physics...

 

[keebs]

i have given an integral for Pi(e^1/t) so if we want to know the value of Pi for high values we need to calculate the integral for small t

t may be small, but e^1/t is large.

you needn,t make me fun of my poor engilsih or my scarce knowledge of numerical methods, in fact i am a physicist (surprised?) by i like math and someday i would like to contribute to math and physics...

What do you do in physics then? Where do you work?

 

[matt grime]

I know you're a physics student, you have said so before. No one is making fun of your poor English, though I have in the past explained to you that the "papers" you have written are poorly presented and thus that will count against you attempting to submit them to anyone. Nor is anyone making fun of the fact that you know no numerical methods. we are pointing out that your claims of speed simplicity uniqueness and originality are unfounded and that making grand claims such as "this probably deserves a field's medal but you're too snobbish to accept that someone not famous could have come up with it" are not winning you any supporters. attacking those who point out the mathematical flaws with such non-mathematical attacks also only serves to underline that you are not prepared to listen and that you probably cannot defend oyur idea on mathematical grounds. you have been given a lot of attention on this forum for your ideas.

 

[Hurkyl]

A run-time analysis is probably best demonstrated by a (simple) example:

Your algorithm is to check if the number n is prime by trial dividing n by every number up through √n.

In pseudocode, it looks like:

for m in the range 2, 3, 4, ..., [√n]
   r := the remainder of n / m
   if r == 0
      print "n is composite"
      quit
end for
print "n is prime"

In the worst-case, we have to go through this loop [√n] - 1 times. Each time through the loop, we must compute one remainder. So, we say that, in the worst-case, this algorithm tests for primality using O(√n) remainder operations. (We can say exactly how many, but that level of detail is generally not of theoretical importance)

(We must also increment m, test if we're done looping, test if the remainder is zero, and other minor things, but these are considered insignificant compared to the rest of the work being done)

Now, if we assume a remainder operation consumes one "unit" of time, we would say that this algorithm runs in O(√n) time, or in "square-root time". For example, that would mean if we quadruple the size of the input, the running time is only doubled.


However, that assumption is not really valid.

I don't recall precisely how much time it takes to compute the remainder of a / b, so I'll make an underestimate, and say it takes O(log a) time to compute the remainder of a / b. (Because it takes at least log a time to simply read all the digits of a)

So, this primality testing algorithm really runs in O(√n log n) time.

Compare this to, say, the AKS primality test, which is said to run in O((log n)^{4 + \epsilon}) time. (ε, here, is your favorite small positive number. Basically, it's saying that using the exponent 4 is an underestimate, but using anything bigger than 4 is an overestimate)

By calculus, we know that √n log n grows much faster than (log n)^(4 + ε), so we say that the latter algorithm is asymptotically faster than the former.

In practice, that means when n is sufficiently large, the AKS primality test will generally take (much) less time than trial division.

 

[eljose]

let,s suppose we know a method to calculate pi(e^t) by means of a double complex integral..then we will need a time O(t^a) a>0 so to know the value of pi(u) we will need time O(ln^a(u)) so this method will be better than other that goes like O(t^a).

 

[matt grime]

and...?

 

[eljose]

the final equation is:

\pi(e^u)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}t^p\frac{\int_0^{\infty}ln\zeta(x)x^{p-1}/x}{\int_0^{\infty}x^{p-1}/(exp(x)-1)}

now the time to calculate goes like O(ln(t)^a) with a>0

 

[eljose]

sorry it was the integral

\pi(2^u)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}t^p\frac{\int_0^{\infty}ln\zet a(x)x^{p-1}/x}{\int_0^{\infty}x^{p-1}/(exp(x)-1)}

 

[eljose]

\pi(e^t)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}t^p\frac{\int_0^{\infty}ln\zet a(x)x^{p-1}/x}{\int_0^{\infty}x^{p-1}/(exp(x)-1)}dp

there is and error where it puts lna(x) is ln\zeta(x)

i have obtained this integral from solving the integral eqauation for \pi(e^t) that,s it:

ln\zeta(s)/s=\int_0^\infty}dt\pi(e^t)/(exp(st)-1) now i think i,m the first in solving this integral equation