OMG Prime number theorem!!! I actually loved it! So today I solved a quite fancy NT problem with other people's help though!
Problem: Prove that there exists an interval of the $[n^2,(n+1)^2]$ which contains at least $1000$ primes.
Proof 1: By PNT, if $n$ is large enough $$\pi(n) ~ \frac{n}{\log(n)}$$
So $$f(n)=\pi( {n+1}^2)-\pi( n^2) = \frac{(n+1)^2}{\log({n+1}^2)}- \frac{n^2}{\log(n^2)}$$
$$= \frac{(n+1)^2}{2\log({n+1})}- \frac{n^2}{2\log(n)}$$
Then $$f'(n)= \frac{(2n+2)\log(n+1)-(n+1)}{2[\log(n+1)]^2}-\frac{2n\log(n)-n}{2[\log(n)]^2}$$
$$=\frac{n+1}{\log(n+1)}+\frac{n}{2[\log(n)]^2}- (\frac{n}{\log(n)}+\frac{n+1}{2[\log(n+1)]^2})$$
Take $$h(x)= \frac{(x)}{\log(x)}, g(x)=\frac{x}{2\log^2(x)}$$
we have basically $$(h(n+1)-h(n))>>>( g(n+1)-g(n)) $$ since $h$ grows faster than $g$ we have $$h(n+1)-h(n) > (g(n+1)-g(n)$$ for large enough $n$.
Infact $f'(n)$ is strictly increasing.
Now, by MVT, note that there exist $t_1$ such that $$f(2)-f(1)=f'(t_1), t\in [1,2]$$
Similarly there exist $t_2$ such that $$f(3)-f(2)=f'(t_2), t\in [2,3]$$
and so on, there exist $t_{n-1}$ such that $$f(n)-f(n)=f'(t_{n-1}), t\in [n-1,n]$$
So $f(n)-f(1)$ tends to infinity as $f'$ is increasing and positive..
So $f(n)$ is increasing and diverges!
Tonight I will revise O notation and make notes! Hopefully I can survive!
Sunaina
Ohh nicee fancy
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