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Consider Wilson's Theorem and consider what will happen to \((j-2)!\) modulo primes and composites (this is Exercise 7.7.6). Use this to prove the bizarre formula in Section 21.1.
Consider Wilson's Theorem and consider what will happen to \((j-2)!\) modulo primes and composites (this is Exercise 7.7.6). Use this to prove the bizarre formula in Section 21.1.
Come up with two functions \(f(x)\) and \(g(x)\) that both go to infinity as \(x\to\infty\), such that \(f(x)\) is always ahead of \(g(x)\), but \(f\) and \(g\) are asymptotic (to each other).
Come up with two functions \(f(x)\) and \(g(x)\) that both go to infinity as \(x\to\infty\), but that switch the lead infinitely often and \(f\) and \(g\) are asymptotic.
Show that the two limits in the Prime Number Theorem are really equivalent. That is, show that if \(\lim \pi(x)/Li(x)=1\), then the other limit is 1, and vice versa.
Find an arbitrarily long sequence of consecutive composite numbers using factorials.
Come up with two functions \(f(x)\) and \(g(x)\) such that \(f(x)\) is \(O(g(x))\) and \(g(x)\) is \(O(f(x))\), but are not asymptotic.
Use Proposition 21.3.5 to show that \(\lim_{x\to\infty}\pi(x)/x=0\).
Show that if \(n>1000\) then \begin{equation*}\frac{\log(2)+1}{\log(n)}<\frac{2}{\log(2)+\log(n)}=\frac{2}{\log(2n)}\end{equation*} To do this, you should compare \(2\log(n)\) and \(\log(2)(\log(2)+\log(n))\) and their derivatives for \(n=1000\) and up, then divide the two expressions appropriately. You will need to show that if \(f(x_0)> g(x_0)\) and \(f'> g'\) for \(x\geq x_0\), then \(f>g\) as well.
Verify that \(3.394\frac{n}{\log(n)}<\frac{2n}{\log(2n+1)}\) for \(n>1000\). You will need to verify that the derivative of \(\frac{\log(n)}{\log(2n+1)}\) is positive there.