Problem 4 - A Computational Introduction to Number Theory and Algebra

Question

For each positive integer \(k,\) let \(P_{k}\) denote the product of the first \(k\) primes. Show that \(\varphi\left(P_{k}\right)=\Theta\left(P_{k} / \log \log P_{k}\right)\) (here, \(\varphi\) is Euler's phi function).

Step-by-Step Solution

Answer

Question: Show that \(\varphi(P_k) = \Theta\left(\frac{P_k}{\log \log P_k}\right)\), where \(P_k\) is the product of the first \(k\) prime numbers. Answer: We have shown that there exist positive constants \(c_1\) and \(c_2\) such that \(c_1 \cdot \frac{P_k}{\log \log P_k} \le \varphi(P_k) \le c_2 \cdot \frac{P_k}{\log \log P_k}\). Hence, \(\varphi(P_k) = \Theta\left(\frac{P_k}{\log \log P_k}\right)\).

Step 1: Calculate Euler's phi function for \(P_k\)

Euler's phi function can be found for any number by using the formula: $$\varphi(n) = n \cdot \prod_{p | n} (1 - \frac{1}{p})$$ where \(p | n\) are the distinct prime factors of \(n\). Here, \(P_k = p_1 p_2 \cdots p_k\), where \(p_i\) is the \(i\)-th prime number. Notice that \(P_k\) has exactly \(k\) distinct prime factors, namely the first \(k\) primes themselves. Therefore: $$\varphi(P_k) = P_k \cdot \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right)$$

Step 2: Proving the Lower Bound

To prove that the expression is a lower bound, i.e., \(\varphi(P_k) \ge c_1 \cdot \frac{P_k}{\log \log P_k}\), we can use the following inequality for the first \(k\) primes: $$p_i \le i(\log i + \log \log i) \quad \text{for } i \ge 6$$ It is easy to check that this inequality also holds for \(i=1,2,3,4,5\). Using this inequality, we have: $$\varphi(P_k) = P_k \cdot \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) \ge P_k \cdot \prod_{i=1}^k \left(1 - \frac{1}{i(\log i + \log \log i)}\right)$$ Further, we use the fact that: $$\left(1 - \frac{1}{i(\log i + \log \log i)}\right) \ge \frac{1}{(\log i + \log \log i) \log i (1 + \varepsilon)}$$ Now, taking the product over the first k terms for \(k\) sufficiently large: $$P_k \cdot \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) \ge P_k \cdot \prod_{i=1}^k \frac{1}{(\log i + \log \log i) \log i (1 + \varepsilon)} \ge c_1 \cdot P_k \cdot \frac{1}{\log k + \log \log k} = c_1 \cdot \frac{P_k}{\log \log P_k}$$

Step 3: Proving the Upper Bound

To prove that the expression is an upper bound, i.e., \(\varphi(P_k) \le c_2 \cdot \frac{P_k}{\log \log P_k}\), we will use the fact that the first k prime numbers have an approximately harmonic sum: $$\sum_{i=1}^k \frac{1}{p_i} \approx \log \log k$$ Since \(\varphi(P_k) = P_k \cdot \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right)\), we can rewrite it using the product of a sum identity: $$\prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) = \exp\left(\sum_{i=1}^k \log \left(1 - \frac{1}{p_i}\right)\right)$$ Now, we make use of the inequality \(\log(1-x) \le -x\) for \(0 < x < 1\) to get: $$\sum_{i=1}^k \log \left(1 - \frac{1}{p_i}\right) \le -\sum_{i=1}^k \frac{1}{p_i} \approx -\log \log k$$ Thus, we have: $$\varphi(P_k) = P_k \cdot \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) \le P_k \cdot e^{-\log \log k} = c_2 \cdot \frac{P_k}{\log \log P_k}$$

Step 4: Conclusion

With the bounds shown in steps 2 and 3, we have proved that \(\varphi(P_k) = \Theta\left(\frac{P_k}{\log \log P_k}\right)\).

Key Concepts

Prime Numbers

Prime numbers are the building blocks of all numbers. A number is considered prime if it is greater than one and has no divisors other than 1 and itself. Unlike composite numbers that can be divided evenly by other numbers, primes possess a unique property: they stand alone without extra divisors.
For example, the numbers 2, 3, 5, 7, and 11 are all primes because they cannot be divided evenly by any number other than 1 and themselves.
Understanding prime numbers is crucial for studying number theory and other mathematical disciplines because they are the fundamental components used in the multiplication to form any given integer. They are used extensively in many applications, including cryptography and algorithms.
In the context of the given exercise, we work with the product of the first \(k\) primes, represented as \(P_k = p_1 p_2 \cdots p_k\), where \(p_i\) is the \(i\)-th prime number. This enables us to explore broader mathematical functions, such as Euler's phi function and its boundaries in relation to prime products.

Euler's theorem

Euler's theorem is an essential concept in number theory. It establishes a connection between exponentiation and modular arithmetic, particularly dealing with numbers that are relatively prime.
In simple terms, the theorem states: for two numbers, say \(a\) and \(n\), if they are coprime (which means they have no common divisor other than 1), then:
\[ a^{oldsymbol{ ext{Euler's } oldsymbol{oldsymbol{ ext{phi}}} f function } (n) } \equiv 1 mod n \]
This means that when you raise \(a\) to the power of Euler’s phi function of \(n\) and then divide by \(n\), the remainder will be 1.

• The 'phi function', \(\varphi(n)\), counts the integers up to \(n\) that are coprime with \(n\).
• This theorem is vital in the computation of the phi function, especially in the exercise relating \(\varphi(P_k)\) to prime products.

Euler's theorem not only underpins the phi function but also lays the foundation for advanced cryptographic protocols like RSA, where security hinges on the difficulty of factoring large numbers into prime ones.

Asymptotic Notation

Asymptotic notation describes how functions behave as their input grows large. It provides a way to express the limiting behavior of sequences and functions, particularly focusing on growth rates.
In the context of our exercise, we use the notation \(\Theta(n)\), which specifically expresses a "tight bound" around a function's growth.

• \(\Theta\)-notation signifies that a function grows at the same rate from above and below with respect to the input size.
• For example, the statement \(f(n) = \Theta(g(n))\) implies that there exists constants \(c_1\) and \(c_2\) such that \(c_1 \cdot g(n) \leq f(n) \leq c_2 \cdot g(n)\) holds true for sufficiently large \(n\).

In our exercise, the statement \(\varphi(P_k) = \Theta\left(\frac{P_k}{\log \log P_k}\right)\) portrays how closely the behavior of \(\varphi(P_k)\) matches with \(\frac{P_k}{\log \log P_k}\). As \(k\) becomes large, this asymptotic notation accurately describes \(\varphi\)'s growth in relation to the product of prime numbers.