I am trying to develop a simple derivation of Chebyshev's estimates for the prime counting function $\pi(x)$.
Stopple's "A Primer of Analytic Number Theory" gives a good start:
By considering the following
$${2n \choose n} = 2n \cdot (2n-1) \cdot (2n-2)\cdot \ldots \cdot (n+2) \cdot (n+1) $$
He explains that there are $\pi(2n)-\pi(n)$ primes in this quantity. He also shows without too much effort that:
$$ n^{\pi(2n)-\pi(n)} < {2n \choose n} < 2^{2n}$$
He then takes logs to give:
$$ \boxed{\pi(2n)-\pi(n)< \frac{2n\log(2)}{\log(n)}}$$
This is very promising. Intuitively this hints at $\pi(n)$ being of the form $n/\log(n)$.
Question: How does one solve the above for $\pi(n)$ ?
And in general, how does one solve for $f(x)$ when faced with the following? $$ f(2x) - f(x) = g(x) $$
If $f(2x)-f(x) = g(x)$ then $f(2x/2^k)-f(x/2^k) = g(x/2^k)$.
Summing,
$\begin{array}\\ f(x)-f(1) &=\sum_{1 \lt k \le \log_2(x)}f(2x/2^k)-f(x/2^k)\\ &= \sum_{1 \lt k \le \log_2(x)}g(x/2^k)\\ &\lt \sum_{1 \lt k \le \log_2(x)}\dfrac{2\log(2)(x/2^k)}{\log(x/2^k)}\\ &=2x\log(2) \sum_{1 \le k \lt \log_2(x)}\dfrac{1/2^k)}{\log(x)-k\log(2)}\\ &=\dfrac{2x\log(2)}{\log(x)} \sum_{1 \le k \lt \log_2(x)}\dfrac{1}{2^k(1-k\log(2)/\log(x))}\\ \end{array} $
and once we show that the sum is bounded, we are done.