Second Chebyshev Function is Big-Theta of x

Theorem
We have:


 * $\map \psi x = \map \Theta x$

where:
 * $\Theta$ is big-$\Theta$ notation
 * $\psi$ is the second Chebyshev function.

Proof
We show that:


 * $\map \psi x = \map \OO x$

and:


 * $x = \map \OO {\map \psi x}$

Note that:

Note that:

Clearly we have:


 * $\ds \sum_{n \le x} \map \psi {\frac x n} = \sum_{m \le x, \, m \text { odd} } \map \psi {\frac x m} + \sum_{m \le x, \, m \text { even} } \map \psi {\frac x m}$

So:

From Second Chebyshev Function is Increasing:


 * $\ds \map \psi {\frac x n} - \map \psi {\frac x m} \ge 0$

when $n < m$.

Suppose that $\floor x$ is an odd integer.

Then we have:

Similarly, if $\floor x$ is a even integer, we have:

We now show that:


 * $\ds \sum_{m \le x, \, m \text { odd} } \map \psi {\frac x m} - \sum_{m \le x, \, m \text { even} } \map \psi {\frac x m} = \map \OO x$

From the definition of big-O notation, there exists some $x_1 \in \R$ and positive real number $C$ such that:


 * $\ds \sum_{m \le x, \, m \text { odd} } \map \psi {\frac x m} - \sum_{m \le x, \, m \text { even} } \map \psi {\frac x m} \le x \ln 2 + C \map \ln {x + 1}$

for $x \ge x_1$.

Then we have:

As shown in Order of Natural Logarithm Function, for $x \ge 1$ we have:


 * $C \ln x \le C x$

Let:


 * $x_0 = \max \set {x_1, 1}$

So for $x \ge x_0$, we have:

Let:


 * $A = C + \paren {C + 1} \ln 2$

So, for $x \ge x_0$ we have:


 * $0 \le \map \psi x - \map \psi {x/2} \le A x$

So, for $x \ge 2^{k - 1} x_0$, we have:


 * $\ds 0 \le \map \psi {\frac x {2^{k - 1} } } - \map \psi {\frac x {2^k} } \le \frac {A x} {2^{k - 1} }$

Note that for $x < 2$, we have:


 * $\map \psi x = 0$

so for:


 * $k \ge \dfrac {\ln x} {\ln 2}$

we have:


 * $\ds \map \psi {\frac x {2^{k - 1} } } - \map \psi {\frac x {2^k} } = 0$

Set:


 * $\ds N = \floor {\frac {\ln x} {\ln 2} } + 1$

So we have, for $x \ge 2^{N - 1} x_0$:

So by the definition of big-O notation, we have:


 * $\map \psi x = \map \OO x$

Suppose that $\floor x$ is an odd integer.

Then:

Similarly if $\floor x$ is an even integer, we have:

Since:


 * $\ds \sum_{m \le x, \, m \text { odd} } \map \psi {\frac x m} - \sum_{m \le x, \, m \text { even} } \map \psi {\frac x m} = x \ln 2 + \map \OO {\map \ln {x + 1} }$

From the definition of big-O notation, there exists a positive real number $C$ and $x_2 \in \R$ such that:


 * $\ds \sum_{m \le x, \, m \text { odd} } \map \psi {\frac x m} - \sum_{m \le x, \, m \text { even} } \map \psi {\frac x m} \ge x \ln 2 - C \map \ln {x + 1}$

for $x \ge x_2$.

assume that $C > 1$.

We then have for $x \ge \max \set {x_2, 1}$:

We show that for sufficiently large $x$ we have:


 * $\ds x \ln 2 - x^{1/C} \ge \frac {\ln 2} 2 x$

This inequality holds :


 * $\ds \frac {\ln 2} 2 \ge x^{\frac 1 C - 1}$

That is:


 * $\ds x^{1 - \frac 1 C} \ge \frac 2 {\ln 2}$

Since $C > 1$, we have:


 * $1 - \dfrac 1 C > 0$

and so:


 * $\ds \paren {1 - \frac 1 C} \ln x \ge \map \ln {\frac 2 {\ln 2} }$

That is:


 * $\ds x \ge \map \exp {\frac {\map \ln {\frac 2 {\ln 2} } } {1 - \frac 1 C} } = x_3$

Then for $x \ge \max \set {x_3, x_2, 1}$, we have:


 * $\ds x \ln 2 - C \ln 2 - x^{1/C} \ge x \frac {\ln 2} 2 - C \ln 2$

If also $x \ge 4 C$, we have:


 * $\ds x \frac {\ln 2} 2 - C \ln 2 \ge x \frac {\ln 2} 4$

Let:


 * $x_4 = \max \set {x_3, x_2, 4 C}$

Then for $x \ge x_4$, we have:


 * $\ds \sum_{m \le x, \, m \text { odd} } \map \psi {\frac x m} - \sum_{m \le x, \, m \text { even} } \map \psi {\frac x m} \ge x \frac {\ln 2} 4$

So:


 * $\ds \map \psi x \ge x \frac {\ln 2} 4$

for $x \ge x_4$.

That is:


 * $\ds \frac 4 {\ln 2} \map \psi x \ge x$

From the definition of big-O notation, we have:


 * $x = \map \OO {\map \psi x}$

Since also:


 * $\map \psi x = \map \OO x$

we have:


 * $\map \psi x = \map \Theta x$