In February 2013, Sadegh Nazardonyavi and Semyon Yakubovich posted on arxiv: Sharper estimates for Chebyshev's functions $\vartheta$ and $\psi$.

I have a question about Theorem 2.27 on page 22.

My question regards the argument for this:

$$\vartheta(x) < 1.000027651, \;\;(x > 0)$$

I can follow the beginning of the proof:


$$8\cdot10^{11} \le x < 10^{16} $$

From Theorem 2.2.5 on p21 (which is taken from N. Costa Pereira. Estimates for the Chebyshev function $\psi(x) - \theta(x)$. Math. Comp.,44(169):211-221,1985.):

$$\psi(x) - \vartheta(x) > \sqrt{x} + \frac{6}{7}\sqrt[3]{x}, \;\;\;(2,036,329 \le x \le 10^{16})$$


$$\vartheta(x) < \psi(x) - \sqrt(x) - \frac{6}{7}\sqrt[3]{x}$$

I am unclear on the next step:

$$ \psi(x) - \sqrt(x) - \frac{6}{7}\sqrt[3]{x} < \left\{1.0000284888 - \frac{1}{\sqrt{x}} - \frac{6}{7}\frac{1}{\sqrt[3]{x^2}}\right\}x $$

I can see that:

$$- \sqrt(x) - \frac{6}{7}\sqrt[3]{x} < \left\{- \frac{1}{\sqrt{x}} - \frac{6}{7}\frac{1}{\sqrt[3]{x^2}}\right\}x $$

I am unclear how it is established that:

$$\psi(x) < 1.0000284888x$$

If anyone can help me to understand this, that will be very helpful.




It seems that $2.84888\times 10^{-5}$ is the $\varepsilon$ you compute using Theorem 2.12 for $x\ge e^{27.407\cdots}=8\times 10^{11}$. For some reason this appears as a fragment on page 36.

  • $\begingroup$ Thanks very much! I thought that was the case but I wasn't clear. $\endgroup$ – Larry Freeman May 17 '13 at 19:10

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