# Chebyshev function - show that $\psi(x)>(x-2)\log2-\log(x+1)$

The question I'm trying to do is this:

Assume $$x>2$$ and $$n=\lfloor x/2\rfloor$$. Show that $$\psi(x)>(x-2)\log2-\log(x+1)$$, given the inequality $$2n\log2-\log(2n+1)<\psi(2n)$$.

All I've really done is substitute $$n$$ in to get

$$2\lfloor x/2\rfloor\log2-\log(2\lfloor x/2\rfloor+1)<\psi(2\lfloor x/2\rfloor)$$

No idea what to do.. could it be a manipulation involving $$\lfloor 2x\rfloor- 2\lfloor x\rfloor=0$$ or $$1$$?

The Chebyshev function doesn't come into it much once you have what you are given (just $$\psi(x)=\psi(\lfloor x \rfloor)$$). Doing a check on the two case $$1 \ge x-2n > 0$$ and $$2 > x-2n \ge 1$$ gives

for the former: $$2 \lfloor \frac{x}{2} \rfloor = \lfloor x \rfloor$$,

for the second: $$2 \lfloor \frac{x}{2} \rfloor = \lfloor x \rfloor - 1$$.

In either case it is sufficient to prove:

$$(x-2) \log 2 - \log(x+1) < \lfloor x \rfloor \log 2 - \log ( \lfloor x \rfloor +1 )$$

$$\log(x+1)-\log(\lfloor x \rfloor +1) > (x-\lfloor x \rfloor -2)\log2$$

which is a given as the LHS >= 0 and the RHS < 0

• Where did checking the two cases with $x-2n$ come from? – Xtrfyable Mar 26 '19 at 23:04
• the floor function of x/2 goes up by one every two of x, hence there is two unit intervals of x to consider. – Paul Childs Mar 27 '19 at 2:16
• Got it, thank you – Xtrfyable Mar 28 '19 at 2:55

Also a bound $$ax >\psi(x) > b x$$ can be obtained from $$n \log 4 +O(\log n)= \log{2n \choose n} = \log(2n)! - 2 \log n! = \sum_{p^k \le 2n} (\lfloor 2n/p^k \rfloor - 2\lfloor n/p^k \rfloor) \log p$$ $$\in [\psi(2n)-\psi(n),\psi(2n)]$$ as $$\lfloor 2n/p^k \rfloor - 2\lfloor n/p^k \rfloor \in \{0,1\}$$