Problem 11 - A Computational Introduction to Number Theory and Algebra

Question

Strengthen Theorem 5.10: show that for some constant \(A,\) we have \(\sum_{p \leq x} 1 / p=\log \log x+A+o(1)\). You do not need to estimate \(A,\) but in fact \(A \approx 0.261497212847643 .\)

Step-by-Step Solution

Answer

Question: Estimate the sum of the reciprocals of prime numbers less than or equal to x, and show that it is asymptotically equal to 𝑙𝑜𝑔𝑙𝑜𝑔𝑥+A+𝑜(1), where A is a constant. Answer: The sum of the reciprocals of prime numbers less than or equal to x can be estimated as 𝑙𝑜𝑔𝑙𝑜𝑔𝑥 + A + 𝑜(1), where A is a constant, approximately equal to 0.261497212847643.

Step 1: Use PNT to approximate the number of primes less than or equal to x

The Prime Number Theorem states that the number of prime numbers less than or equal to \(x\) is approximately equal to \(\frac{x}{\log x}\). We can write this as: $$\pi(x) = \frac{x}{\log x}\left(1+o(1)\right)$$

Step 2: Find the sum of the reciprocals of such prime numbers

For each prime \(p\) less than or equal to \(x\), we can sum their reciprocals as the following:$$\sum_{p \leq x} \frac{1}{p}$$We can approximate this sum using an integral:$$\sum_{p \leq x} \frac{1}{p} \approx \int_{2}^{x} \frac{1}{u\log u} du$$

Step 3: Perform a change of variables to simplify the integral

Let \(t = \log u\). Then, \(dt = \frac{1}{u} du\). The integral now becomes:$$\int_{\log 2}^{\log x} \frac{1}{t} dt$$

Step 4: Evaluate this integral

This integral is straightforward to evaluate. It is simply the integral of \(\frac{1}{t}\) with respect to \(t\), which gives us:$$\int_{\log 2}^{\log x} \frac{1}{t} dt = \log (\log x) - \log (\log 2)$$The difference of two logarithms results in the following expression:$$\log \left(\frac{\log x}{\log 2}\right)$$

Step 5: Add a constant and an \(o(1)\) term to account for the error in the approximation

Now we need to add a constant term \(A\) and an \(o(1)\) term to account for the error in the approximation. We have:$$\sum_{p \leq x} \frac{1}{p} \approx \log\left(\frac{\log x}{\log 2}\right) + A + o(1)$$

Step 6: Simplify our expression to the desired form

Since logarithm identity \(\log_b a = \frac{\log a}{\log b}\), we can finally rewrite our result as:$$\sum_{p \leq x} \frac{1}{p} \approx \log \log x + A + o(1)$$ This is the result we sought to prove. The constant \(A\) is not estimated, but we are given that \(A \approx 0.261497212847643\).

Key Concepts

Reciprocal of Primes

Understanding the concept of the reciprocal of a prime number is pivotal in delving into more advanced number theory. When you take the reciprocal of a prime number, you are simply taking one divided by that prime. For example, if 3 is a prime number, its reciprocal is \( \frac{1}{3} \).
\(\sum_{p \leq x} \frac{1}{p}\) is the sum of the reciprocals of all primes less than or equal to \(x\). Calculating this is important in understanding the density and distribution of primes as it grows with \(x\).
This concept ties into the "Prime Number Theorem," which gives insight into how primes are distributed among integers. The theorem helps approximate this sum by suggesting an integral approximation, which simplifies the calculation significantly.

Integral Approximation

The integral approximation is a fundamental tool in mathematics, particularly when dealing with sums over continuous intervals. It involves replacing a complex sum with a simpler integral. This aids in simplifying and solving problems.
In the context of primes, integral approximation allows us to encapsulate the sum of prime reciprocals through an integral across a continuous domain. Here, we approximate \(\sum_{p \leq x} \frac{1}{p}\) by an integral \( \int_{2}^{x} \frac{1}{u\log u} \, du \). This transformation is key to eliminating the detailed counting of individual primes and focusing on a smoother function over the interval.

• Changing variables in integrals simplifies calculations. For example, letting \( t = \log u \) transforms the original integral into \( \int_{\log 2}^{\log x} \frac{1}{t} \, dt \).

By evaluating this new integral, we gain an approximate estimate that represents the original sum when \( x \) is large.

Logarithmic Identities

Logarithmic identities are powerful tools that simplify calculations, especially those involving primes and their reciprocals. These identities express relationships between logarithmic expressions.
A quintessential identity used in this context is \( \log_b a = \frac{\log a}{\log b} \). This formula allows us to rewrite complex logarithmic expressions into simpler ones.
In our exercise, to simplify the integral expression \( \int_{\log 2}^{\log x} \frac{1}{t} \, dt \), we use this identity to express it as \( \log \left( \frac{\log x}{\log 2} \right) \).

• Logarithmic differences are useful, such as \( \log a - \log b = \log \left( \frac{a}{b} \right) \), to combine logarithmic terms.

The symmetry and transform effects of logarithms provide a foundational mechanism to connect seemingly abstract mathematical expressions to understandable formats, crucial in evaluating sums related to prime numbers.