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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 $1$?

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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

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  • $\begingroup$ Where did checking the two cases with $x-2n$ come from? $\endgroup$ – Xtrfyable Mar 26 at 23:04
  • $\begingroup$ the floor function of x/2 goes up by one every two of x, hence there is two unit intervals of x to consider. $\endgroup$ – Paul Childs Mar 27 at 2:16
  • $\begingroup$ Got it, thank you $\endgroup$ – Xtrfyable Mar 28 at 2:55
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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 $\lfloor 2n/p^k \rfloor - 2\lfloor n/p^k \rfloor \in \{0,1\}$

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