How to Apply Chebyshev's Bounds for the Prime Counting Function

What We're Solving:

We want to understand how Chebyshev proved important bounds for the prime counting function π(x) (which counts how many primes are ≤ x) in relation to x/log(x). This was a crucial step toward proving the Prime Number Theorem!

The Approach:

Chebyshev's brilliant strategy was to avoid counting primes directly (which is really hard!) and instead work with products and sums involving primes. He used the binomial coefficient (2n choose n) as a clever tool because it contains information about many primes at once. The key insight: if we can bound this binomial coefficient from above and below in different ways, we can extract information about how primes are distributed!

Step-by-Step Solution:

Step 1: Set up the binomial coefficient Consider $\binom{2n}{n} = \frac{(2n)!}{n! \cdot n!}$. This innocent-looking number will be our window into prime distribution!

Step 2: Find a simple lower bound Since $\binom{2n}{n}$ is the largest binomial coefficient in the expansion of $(1+1)^{2n} = 2^{2n}$, and there are $2n+1$ terms total, we get: $$\binom{2n}{n} \geq \frac{2^{2n}}{2n+1} \geq \frac{4^n}{4n}$$

Step 3: Find an upper bound using prime factorization Here's where it gets clever! For any prime $p$, the highest power of $p$ dividing $\binom{2n}{n}$ is: $$\sum_{k=1}^{\infty} \left(\left\lfloor \frac{2n}{p^k} \right\rfloor - 2\left\lfloor \frac{n}{p^k} \right\rfloor\right)$$

Chebyshev showed this is at most $\lfloor \log_p(2n) \rfloor$, so: $$\binom{2n}{n} \leq \prod_{p \leq 2n} p^{\lfloor \log_p(2n) \rfloor}$$

Step 4: Split the product cleverly Chebyshev separated primes into two groups:

• Large primes: $n < p \leq 2n$ (these divide $\binom{2n}{n}$ exactly once)
• Small primes: $p \leq n$

This gives us: $\binom{2n}{n} \leq \prod_{n < p \leq 2n} p \cdot \prod_{p \leq n} p^{\lfloor \log_p(2n) \rfloor}$

Step 5: Bound each part

• For large primes: $\prod_{n < p \leq 2n} p$ is exactly what we want to estimate!
• For small primes: $\prod_{p \leq n} p^{\lfloor \log_p(2n) \rfloor} < (2n)^{\pi(n)}$

Step 6: Take logarithms and use the bounds From our bounds on $\binom{2n}{n}$: $$n \log 4 - \log(4n) \leq \log\binom{2n}{n} \leq \sum_{n < p \leq 2n} \log p + \pi(n) \log(2n)$$

Step 7: Apply this inequality cleverly Through careful analysis of this inequality for different values of $n$, Chebyshev derived:

• There exist constants $A$ and $B$ such that: $A \frac{x}{\log x} < \pi(x) < B \frac{x}{\log x}$
• Specifically, he showed $A$ can be taken as about 0.89 and $B$ as about 1.11

The Answer:

Chebyshev proved that π(x) is bounded above and below by constant multiples of x/log(x). Specifically: $$0.89 \cdot \frac{x}{\log x} < \pi(x) < 1.11 \cdot \frac{x}{\log x}$$

This was revolutionary because it showed that x/log(x) captures the right "order of growth" for the prime counting function, even though the constants weren't quite right yet (the Prime Number Theorem later showed the ratio approaches 1).

Memory Tip:

Remember Chebyshev's approach as "Binomial Bounds Beat Brute force" - instead of trying to count primes directly, he used the binomial coefficient as a clever intermediary that naturally encodes information about many primes at once! The key insight is that different ways of expressing the same number can reveal hidden information about prime distribution.