An equation of prime counting function

In summary, an equation of prime counting function is a mathematical expression used to determine the number of prime numbers less than or equal to a given number. This function is denoted as π(x) and is also known as the prime number theorem. It is calculated using the equation π(x) = Li(x) + O(x^(1/2)log(x)) and has been significant in understanding the distribution of prime numbers and their relationship to other numbers. However, it cannot be used to find the value of a specific prime number and has limitations in accuracy for smaller values and being an asymptotic function.
  • #1
secondprime
49
0
I have encountered the below problem-
Given,
##z(z-1)## has all prime < ##\sqrt{z} <n## , Prove(or disprove)-
## π(z)-w(z-1)-A= π(2z-1)- π(z) ## where A={0 ,1}, π (z) is the prime counting function, π(2z-1)- π(z) is the number of primes in between z and (2z-1), ##\omega(z-1)## is the number of distinct prime factors of (z-1).

I think the eq: is true and the proof I propose is -

Proof:
1.a. Consider (z+k) and (z-k) where 0<k<z, if both are prime or composite at the same time then, ##π(z)= π(2z-1)- π(z)##
is true .

If one is prime and another is not, then the equation is not true but I claim both have to be prime when one of them is prime and k | (z-1) .

let, the assumption is false , so (z-k) is prime and (z+k) is composite.

then (z+k)=cd where (z,k)=1(if (z,k)>1 then (z-k) is not prime.)

c,d does not divide z,k since k<z and (z,k)=1

so, at least c | (z-1) implies c| k(means c divides z since z=cd+k ) and leads to a contradiction since all primes (consecutive) of n are in z, (z-1) .So, assumption is true.

1.b. Consider (z+k) and (z-k)

where 0<k<z,

if both are prime or composite at the same time then, then the equation is true, if one is prime and another is not, then the equation is not true.

Say, (z+k) is prime an (z-k) is not, then z-k=(z-1)-(k-1) is also composite, this implies (z-1)+(k-1) is also composite.

Mod note: Removed link to external site.

Thus it can be shown, that there is a “one –to -one” relation between every composite number before z and a composite after z. So, the equation π(z)= π(2z-1)- π(z) holds.

2. But the primes of (z-1) are less than z, unbalance the equation (using above argument). So, w(z-1) is subtracted.

##π(z)-w(z-1)= π(2z-1)- π(z)##

3. The prime 2 cannot be considered above way so it is ignored in the numbers of prime so,

##π(z)-w(z-1)-A= π(2z-1)- π(z) ##, where A={0,1}

"Could you verify my proof, and/or point out any errors or ways I can improve it?"
 
Last edited by a moderator:
Mathematics news on Phys.org
  • #2

Thank you for sharing your proof with us. I always appreciate seeing others' approaches to solving problems. After examining your proof, I have a few comments and suggestions for you.

Firstly, I would like to commend you on your use of mathematical notation and symbols. It is clear that you have a strong understanding of the subject matter. However, I would suggest that you define your variables and symbols at the beginning of your proof, as it can help readers understand your reasoning more easily.

Secondly, I would like to address your first argument (1.a). While it is true that if both (z+k) and (z-k) are prime or composite, then the equation is true, I am not convinced by your reasoning for the case where one is prime and the other is not. Your claim that k | (z-1) does not necessarily lead to a contradiction. For example, let z=5 and k=2. Then (z+k)=7 is prime and (z-k)=3 is not, but k | (z-1) is still true. I suggest revisiting your reasoning for this case.

Additionally, I am not sure I understand your argument in 1.b. Could you clarify what you mean by a "one-to-one" relation between composite numbers before and after z? Also, I am unsure how this argument supports your overall proof.

Finally, I would like to suggest that you provide more detailed explanations for your steps in the proof. While your use of symbols and notation is clear, it would be helpful for readers to have a better understanding of your thought process and reasoning behind each step.

Overall, I think your proof has potential, but it could benefit from further clarification and explanation. I would also suggest seeking feedback from other scientists or mathematicians to help strengthen your argument.

Best of luck with your future research and problem-solving.
 

1. What is an equation of prime counting function?

An equation of prime counting function is a mathematical expression used to determine the number of prime numbers that are less than or equal to a given number. This function is denoted as π(x), where x is the given number. It is also known as the prime counting function or the prime number theorem.

2. How is the prime counting function calculated?

The prime counting function is calculated using the following equation: π(x) = Li(x) + O(x^(1/2)log(x)), where Li(x) is the logarithmic integral function and O(x^(1/2)log(x)) is the error term. This equation was first derived by mathematician Bernhard Riemann in 1859.

3. What is the significance of the prime counting function?

The prime counting function is significant because it helps us understand the distribution of prime numbers and their relationship to other numbers. It also plays a crucial role in various areas of mathematics, such as number theory, cryptography, and coding theory.

4. Can the prime counting function be used to find the value of a specific prime number?

No, the prime counting function cannot be used to find the value of a specific prime number. It only gives the total number of prime numbers up to a given number. To find the value of a specific prime number, techniques such as trial division or sieving algorithms are used.

5. Are there any limitations to the prime counting function?

Yes, there are some limitations to the prime counting function. It is accurate for large values of x, but for smaller values, it may not give an exact count of prime numbers. Also, it is an asymptotic function, which means it gives an approximation and not the exact number of prime numbers.

Similar threads

Replies
1
Views
745
  • General Math
Replies
15
Views
3K
Replies
8
Views
1K
Replies
4
Views
929
Replies
1
Views
903
Replies
3
Views
274
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
278
  • General Math
Replies
1
Views
2K
Back
Top