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Chebyshev theta function: Difference between revisions

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m →Estimates of the function: I didn't know if the slight refinement was due to Chebyshev
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'''Theorem (Chebyshev).''' If <math>x \ge 0</math>, then <math>\vartheta(x) \le
'''Theorem (Chebyshev).''' If <math>x \ge 0</math>, then <math>\vartheta(x) \le
2x</math>.
2x \log 2</math>.


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


Now suppose that <math>x \ge 2</math>.  Let <math>n = \lfloor x \rfloor</math>.  Then
Now suppose that <math>x \ge 2</math>.  Let <math>n = \lfloor x \rfloor</math>.  Then
Line 26: Line 26:
\prod_{\lfloor n/2 \rfloor < p \le n} p , </cmath>
\prod_{\lfloor n/2 \rfloor < p \le n} p , </cmath>
so
so
<cmath> x \ge 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 \ge \vartheta{x} - x , </cmath>
\ge \vartheta{x} - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta{x} - x \log 2x , </cmath>
by inductive hypothesis.  Therefore
by inductive hypothesis.  Therefore
<cmath> 2x \ge \vartheta(x), </cmath>
<cmath> 2x \log 2 \ge \vartheta(x), </cmath>
as desired.  <math>\blacksquare</math>
as desired.  <math>\blacksquare</math>


Revision as of 22:43, 29 March 2009

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

Estimates of the function

The function $\vartheta(x)$ is asymptotically equivalent to $\pi(x)$ (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 \[x \log 2 \ge \sum_{\lfloor n/2 \rfloor < p \le n} \log p = \vartheta{x} - \vartheta{\lfloor n/2 \rfloor} \ge \vartheta{x} - 2\lfloor n/2 \rfloor \log 2 \ge \vartheta{x} - x \log 2x ,\] by inductive hypothesis. Therefore \[2x \log 2 \ge \vartheta(x),\] as desired. $\blacksquare$

See also

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