upper bound on $\vartheta (n)$
Theorem.
Let $\vartheta \mathit{}\mathrm{(}n\mathrm{)}$ be the Chebyshev function^{}
$$\vartheta (n)=\sum _{\begin{array}{c}p\le n\\ p\mathrm{prime}\end{array}}\mathrm{log}p.$$ |
Then $\vartheta \mathit{}\mathrm{(}n\mathrm{)}\mathrm{\le}n\mathit{}\mathrm{log}\mathit{}\mathrm{4}$ for all $n\mathrm{\ge}\mathrm{1}$.
Proof.
By induction^{}.
The cases for $n=1$ and $n=2$ follow by inspection.
For even $n>2$, the case follows immediately from the case for $n-1$ since $n$ is not prime.
So let $n=2m+1$ with $m>0$ and consider ${(1+1)}^{2m+1}$ and its binomial expansion (http://planetmath.org/BinomialTheorem). Since $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m+1}}\right)$ and each term occurs exactly once, it follows that $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m}}\right)\le {4}^{m}$. Each prime $p$ with $$ divides $\left(\genfrac{}{}{0pt}{}{2m+1}{m}\right)$, implying that their product^{} also divides $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m}}\right)$. Hence
$$\vartheta (2m+1)-\vartheta (m+1)\le \mathrm{log}\left(\genfrac{}{}{0pt}{}{2m+1}{m}\right)\le m\mathrm{log}4.$$ |
By the induction hypothesis, $\vartheta (m+1)\le (m+1)\mathrm{log}4$ and so $\vartheta (2m+1)\le (2m+1)\mathrm{log}4$. ∎
References
- 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 1938.
Title | upper bound on $\vartheta (n)$ |
---|---|
Canonical name | UpperBoundOnvarthetan |
Date of creation | 2013-03-22 16:09:47 |
Last modified on | 2013-03-22 16:09:47 |
Owner | mps (409) |
Last modified by | mps (409) |
Numerical id | 5 |
Author | mps (409) |
Entry type | Theorem^{} |
Classification | msc 11A41 |