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Chebyshev theta function: Difference between revisions

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The function <math>\vartheta(x)</math> is [[asymptotically equivalent]] to
The function <math>\vartheta(x)</math> is [[asymptotically equivalent]] to
<math>\pi(x)</math> (the [[prime counting function]]) and <math>x</math>.  This result
<math>\pi(x)\ln x</math> (<math>\pi(x)</math> is the [[prime counting function]]) and <math>x</math>.  This result
is the [[Prime Number Theorem]], and all known proofs are rather
is the [[Prime Number Theorem]], and all known proofs are rather
involved.
involved.
Line 27: Line 27:
so
so
<cmath> x \log 2 \ge \sum_{\lfloor n/2 \rfloor < p \le n} \log p
<cmath> x \log 2 \ge \sum_{\lfloor n/2 \rfloor < p \le n} \log p
= \vartheta{x} - \vartheta{\lfloor n/2 \rfloor}
= \vartheta(x) - \vartheta(\lfloor n/2 \rfloor)
\ge \vartheta{x} - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta{x} - x \log 2 , </cmath>
\ge \vartheta(x) - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta(x) - x \log 2 , </cmath>
by inductive hypothesis.  Therefore
by the inductive hypothesis.  Therefore
<cmath> 2x \log 2 \ge \vartheta(x), </cmath>
<cmath> 2x \log 2 \ge \vartheta(x), </cmath>
as desired.  <math>\blacksquare</math>
as desired.  <math>\blacksquare</math>

Latest revision as of 12:28, 1 April 2014

Chebyshev's theta function, denoted $\vartheta$ or sometimes $\theta$, is a function of use in analytic number theory. It is defined thus, for real $x$: \[\vartheta(x) = \sum_{p \le x} \log p ,\] where the sum ranges over all primes less than $x$.

Estimates of the function

The function $\vartheta(x)$ is asymptotically equivalent to $\pi(x)\ln x$ ($\pi(x)$ is the prime counting function) and $x$. This result is the Prime Number Theorem, and all known proofs are rather involved.

However, we can obtain a simpler bound on $\vartheta(x)$.

Theorem (Chebyshev). If $x \ge 0$, then $\vartheta(x) \le 2x \log 2$.

Proof. We induct on $\lfloor x \rfloor$. For our base cases, we note that for $0 \le x < 2$, we have $\vartheta(x) = 0 \le 2x \log 2$.

Now suppose that $x \ge 2$. Let $n = \lfloor x \rfloor$. Then \[2^x \ge 2^n \ge \binom{n}{\lfloor n/2 \rfloor} \ge \prod_{\lfloor n/2 \rfloor < p \le n} p ,\] so \[x \log 2 \ge \sum_{\lfloor n/2 \rfloor < p \le n} \log p = \vartheta(x) - \vartheta(\lfloor n/2 \rfloor) \ge \vartheta(x) - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta(x) - x \log 2 ,\] by the inductive hypothesis. Therefore \[2x \log 2 \ge \vartheta(x),\] as desired. $\blacksquare$

See also

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