Difference between revisions of "Chebyshev theta function"
m (→Estimates of the function: I didn't know if the slight refinement was due to Chebyshev) |
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<math>\theta</math>, is a function of use in [[analytic number theory]]. | <math>\theta</math>, is a function of use in [[analytic number theory]]. | ||
It is defined thus, for real <math>x</math>: | It is defined thus, for real <math>x</math>: | ||
| − | <cmath> \vartheta(x) = \sum_{p \le x} \log | + | <cmath> \vartheta(x) = \sum_{p \le x} \log p , </cmath> |
where the sum ranges over all [[prime number | primes]] less than | where the sum ranges over all [[prime number | primes]] less than | ||
<math>x</math>. | <math>x</math>. | ||
| Line 28: | Line 28: | ||
<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 | + | \ge \vartheta{x} - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta{x} - x \log 2 , </cmath> |
by inductive hypothesis. Therefore | by inductive hypothesis. Therefore | ||
<cmath> 2x \log 2 \ge \vartheta(x), </cmath> | <cmath> 2x \log 2 \ge \vartheta(x), </cmath> | ||
Revision as of 18:56, 18 September 2012
Chebyshev's theta function, denoted 
or sometimes
, is a function of use in analytic number theory.
It is defined thus, for real 
:
![\[\vartheta(x) = \sum_{p \le x} \log p ,\]](http://latex.artofproblemsolving.com/8/8/0/880766fe272dc6b1de54f36a29ec5e5582e069a8.png)
where the sum ranges over all primes less than
.
Estimates of the function
The function 
is asymptotically equivalent to
(the prime counting function) and . This result
is the Prime Number Theorem, and all known proofs are rather
involved.
However, we can obtain a simpler bound on .
Theorem (Chebyshev). If 
, then 
.
Proof. We induct on 
. For our base
cases, we note that for 
, we have 
.
Now suppose that 
. Let 
. Then
![\[2^x \ge 2^n \ge \binom{n}{\lfloor n/2 \rfloor} \ge \prod_{\lfloor n/2 \rfloor < p \le n} p ,\]](http://latex.artofproblemsolving.com/4/c/0/4c02f1710b662c21347a8a98f0c1c9970e1e0983.png)
so
![\[x \log 2 \ge \sum_{\lfloor n/2 \rfloor < p \le n} \log p = \vartheta{x} - \vartheta{\lfloor n/2 \rfloor} \ge \vartheta{x} - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta{x} - x \log 2 ,\]](http://latex.artofproblemsolving.com/b/a/5/ba522d895470039d2baa6606ee6ef316141951ae.png)
by inductive hypothesis. Therefore
![\[2x \log 2 \ge \vartheta(x),\]](http://latex.artofproblemsolving.com/f/6/d/f6d7c8e8dc96d0c3d8487b3b068223986a0fcc31.png)
as desired. 
