切比雪夫第一及第二函數彼此相關，要驗證這點，可先將切比雪夫第二函數寫成如下形式：
ψ
(
x
)
=
∑
p
≤
x
k
log
p
{\displaystyle \psi (x)=\sum _{p\leq x}k\log p}
其中k
p
k
≤
x
<
p
k
+
1
{\displaystyle p^{k}\leq x<p^{k+1}}
k A206722 。一個更直接的關係如下：
ψ
(
x
)
=
∑
n
=
1
∞
ϑ
(
x
1
n
)
.
{\displaystyle \psi (x)=\sum _{n=1}^{\infty }\vartheta {\big (}x^{\frac {1}{n}}{\big )}.}
注意的是和的後半段只有有限多個非零數值，而這是因為有下式之故：
ϑ
(
x
1
n
)
=
0
for
n
>
log
2
x
=
log
x
log
2
.
{\displaystyle \vartheta {\big (}x^{\frac {1}{n}}{\big )}=0\quad {\text{for}}\quad n>\log _{2}x={\frac {\log x}{\log 2}}.}
切比雪夫第二函數是從1到n 最小公倍數 的自然對數：
lcm
(
1
,
2
,
…
,
n
)
=
e
ψ
(
n
)
.
{\displaystyle \operatorname {lcm} (1,2,\dots ,n)=e^{\psi (n)}.}
對於n lcm(1, 2, ..., n ) A003418 。
以下定理 將
ψ
(
x
)
x
{\displaystyle {\frac {\psi (x)}{x}}}
ϑ
(
x
)
x
{\displaystyle {\frac {\vartheta (x)}{x}}}
[ 3]
定理： 若
x
>
0
{\displaystyle x>0}
0
≤
ψ
(
x
)
x
−
ϑ
(
x
)
x
≤
(
log
x
)
2
2
x
log
2
.
{\displaystyle 0\leq {\frac {\psi (x)}{x}}-{\frac {\vartheta (x)}{x}}\leq {\frac {(\log x)^{2}}{2{\sqrt {x}}\log 2}}.}
注意：從此不等式 可推出
lim
x
→
∞
(
ψ
(
x
)
x
−
ϑ
(
x
)
x
)
=
0.
{\displaystyle \lim _{x\to \infty }\!\left({\frac {\psi (x)}{x}}-{\frac {\vartheta (x)}{x}}\right)\!=0.}
換句話說，若
ψ
(
x
)
/
x
{\displaystyle \psi (x)/x}
ϑ
(
x
)
/
x
{\displaystyle \vartheta (x)/x}
極限 ，則另一個也是如此，也就是兩者的極限相等。
證明： 由於
ψ
(
x
)
=
∑
n
≤
log
2
x
ϑ
(
x
1
/
n
)
{\displaystyle \psi (x)=\sum _{n\leq \log _{2}x}\vartheta (x^{1/n})}
0
≤
ψ
(
x
)
−
ϑ
(
x
)
=
∑
2
≤
n
≤
log
2
x
ϑ
(
x
1
/
n
)
.
{\displaystyle 0\leq \psi (x)-\vartheta (x)=\sum _{2\leq n\leq \log _{2}x}\vartheta (x^{1/n}).}
而由
ϑ
(
x
)
{\displaystyle \vartheta (x)}
ϑ
(
x
)
≤
∑
p
≤
x
log
x
≤
x
log
x
{\displaystyle \vartheta (x)\leq \sum _{p\leq x}\log x\leq x\log x}
因此有
0
≤
ψ
(
x
)
−
ϑ
(
x
)
≤
∑
2
≤
n
≤
log
2
x
x
1
/
n
log
(
x
1
/
n
)
≤
(
log
2
x
)
x
log
x
=
log
x
log
2
x
2
log
x
=
x
(
log
x
)
2
2
log
2
.
{\displaystyle {\begin{aligned}0\leq \psi (x)-\vartheta (x)&\leq \sum _{2\leq n\leq \log _{2}x}x^{1/n}\log(x^{1/n})\\&\leq (\log _{2}x){\sqrt {x}}\log {\sqrt {x}}\\&={\frac {\log x}{\log 2}}{\frac {\sqrt {x}}{2}}\log x\\&={\frac {{\sqrt {x}}\,(\log x)^{2}}{2\log 2}}.\end{aligned}}}
最後，將此不等式兩邊除以
x
{\displaystyle x}
對於切比雪夫函數，有以下已知的界線。其中p k k p 1 = 2p 2 = 3[1] [2]
ϑ
(
p
k
)
≥
k
(
log
k
+
log
log
k
−
1
+
log
log
k
−
2.050735
log
k
)
for
k
≥
10
11
,
ϑ
(
p
k
)
≤
k
(
log
k
+
log
log
k
−
1
+
log
log
k
−
2
log
k
)
for
k
≥
198
,
|
ϑ
(
x
)
−
x
|
≤
0.006788
x
log
x
for
x
≥
10
544
111
,
|
ψ
(
x
)
−
x
|
≤
0.006409
x
log
x
for
x
≥
e
22
,
0.9999
x
<
ψ
(
x
)
−
ϑ
(
x
)
<
1.00007
x
+
1.78
x
3
for
x
≥
121.
{\displaystyle {\begin{aligned}\vartheta (p_{k})&\geq k\left(\log k+\log \log k-1+{\frac {\log \log k-2.050735}{\log k}}\right)&&{\text{for }}k\geq 10^{11},\\[8px]\vartheta (p_{k})&\leq k\left(\log k+\log \log k-1+{\frac {\log \log k-2}{\log k}}\right)&&{\text{for }}k\geq 198,\\[8px]|\vartheta (x)-x|&\leq 0.006788\,{\frac {x}{\log x}}&&{\text{for }}x\geq 10\,544\,111,\\[8px]|\psi (x)-x|&\leq 0.006409\,{\frac {x}{\log x}}&&{\text{for }}x\geq e^{22},\\[8px]0.9999{\sqrt {x}}&<\psi (x)-\vartheta (x)<1.00007{\sqrt {x}}+1.78{\sqrt[{3}]{x}}&&{\text{for }}x\geq 121.\end{aligned}}}
此外，若黎曼猜想成立，則對於任意的
ε
>
0
{\displaystyle \varepsilon >0}
|
ϑ
(
x
)
−
x
|
=
O
(
x
1
2
+
ε
)
|
ψ
(
x
)
−
x
|
=
O
(
x
1
2
+
ε
)
{\displaystyle {\begin{aligned}|\vartheta (x)-x|&=O{\Big (}x^{{\frac {1}{2}}+\varepsilon }{\Big )}\\|\psi (x)-x|&=O{\Big (}x^{{\frac {1}{2}}+\varepsilon }{\Big )}\end{aligned}}}
對任意的
x
>
0
{\displaystyle x>0}
ϑ
(
x
)
{\displaystyle \vartheta (x)}
ψ
(
x
)
{\displaystyle \psi (x)}
[ 4] [3]
ϑ
(
x
)
<
1.000028
x
ψ
(
x
)
<
1.03883
x
{\displaystyle {\begin{aligned}\vartheta (x)&<1.000028x\\\psi (x)&<1.03883x\end{aligned}}}
對於1.03883這常數的解釋，可見A206431 的說明。
1895年，漢斯·馮·曼戈爾特 證明了[4]
ψ
(
x
)
{\displaystyle \psi (x)}
黎曼ζ函數 非平凡零點和的解析解
ψ
0
(
x
)
=
x
−
∑
ρ
x
ρ
ρ
−
ζ
′
(
0
)
ζ
(
0
)
−
1
2
log
(
1
−
x
−
2
)
.
{\displaystyle \psi _{0}(x)=x-\sum _{\rho }{\frac {x^{\rho }}{\rho }}-{\frac {\zeta '(0)}{\zeta (0)}}-{\tfrac {1}{2}}\log(1-x^{-2}).}
其中ζ′ (0)/ ζ (0)log(2π) ρ ψ 0 ψ 不連續點 （質數的冪）的取值為其左邊與右邊值的中間：
ψ
0
(
x
)
=
1
2
(
∑
n
≤
x
Λ
(
n
)
+
∑
n
<
x
Λ
(
n
)
)
=
{
ψ
(
x
)
−
1
2
Λ
(
x
)
x
=
2
,
3
,
4
,
5
,
7
,
8
,
9
,
11
,
13
,
16
,
…
ψ
(
x
)
otherwise.
{\displaystyle \psi _{0}(x)={\frac {1}{2}}\!\left(\sum _{n\leq x}\Lambda (n)+\sum _{n<x}\Lambda (n)\right)={\begin{cases}\psi (x)-{\tfrac {1}{2}}\Lambda (x)&x=2,3,4,5,7,8,9,11,13,16,\dots \\[5px]\psi (x)&{\mbox{otherwise.}}\end{cases}}}
就自然對數 的泰勒展開式 而言，解析解的最後一項可理解為xω / ω ω = −2, −4, −6, ...
∑
k
=
1
∞
x
−
2
k
−
2
k
=
1
2
log
(
1
−
x
−
2
)
.
{\displaystyle \sum _{k=1}^{\infty }{\frac {x^{-2k}}{-2k}}={\tfrac {1}{2}}\log \left(1-x^{-2}\right).}
類似地，此公式第一項x = x 1 / 1 極點 。這部分作為極點而非零點的事實，說明了項的變號。
一個由埃哈德·施密特 證明的結果指稱，對於某個特定的正常數K 正整數 x
ψ
(
x
)
−
x
<
−
K
x
{\displaystyle \psi (x)-x<-K{\sqrt {x}}}
同時有無限多個正整數x
ψ
(
x
)
−
x
>
K
x
.
{\displaystyle \psi (x)-x>K{\sqrt {x}}.}
[5] [6] 使用小o ，可將上式重述為
ψ
(
x
)
−
x
≠
o
(
x
)
.
{\displaystyle \psi (x)-x\neq o\left({\sqrt {x}}\,\right).}
哈代 與李特爾伍德 [7]
ψ
(
x
)
−
x
≠
o
(
x
log
log
log
x
)
.
{\displaystyle \psi (x)-x\neq o\left({\sqrt {x}}\,\log \log \log x\right).}
也就是說有無限多的正整數x
ψ
(
x
)
{\displaystyle \psi (x)}
x
x
log
log
log
x
{\displaystyle {\sqrt {x}}\,\log \log \log x}
切比雪夫第一函數也是x 質數階乘 x #
ϑ
(
x
)
=
∑
p
≤
x
log
p
=
log
∏
p
≤
x
p
=
log
(
x
#
)
.
{\displaystyle \vartheta (x)=\sum _{p\leq x}\log p=\log \prod _{p\leq x}p=\log \left(x\#\right).}
這說明了質數階乘x #e (1 + o (1))x o o 大O符號 一文的說明），而這點與質數定理共同確立了p n
切比雪夫函數可透過下式與與質數計數函數 發生關係。定義
Π
(
x
)
=
∑
n
≤
x
Λ
(
n
)
log
n
.
{\displaystyle \Pi (x)=\sum _{n\leq x}{\frac {\Lambda (n)}{\log n}}.}
那麼有
Π
(
x
)
=
∑
n
≤
x
Λ
(
n
)
∫
n
x
d
t
t
log
2
t
+
1
log
x
∑
n
≤
x
Λ
(
n
)
=
∫
2
x
ψ
(
t
)
d
t
t
log
2
t
+
ψ
(
x
)
log
x
.
{\displaystyle \Pi (x)=\sum _{n\leq x}\Lambda (n)\int _{n}^{x}{\frac {dt}{t\log ^{2}t}}+{\frac {1}{\log x}}\sum _{n\leq x}\Lambda (n)=\int _{2}^{x}{\frac {\psi (t)\,dt}{t\log ^{2}t}}+{\frac {\psi (x)}{\log x}}.}
從Π 質數計數函數 π
Π
(
x
)
=
π
(
x
)
+
1
2
π
(
x
)
+
1
3
π
(
x
3
)
+
⋯
{\displaystyle \Pi (x)=\pi (x)+{\tfrac {1}{2}}\pi \left({\sqrt {x}}\,\right)+{\tfrac {1}{3}}\pi \left({\sqrt[{3}]{x}}\,\right)+\cdots }
由於很明顯地，有π (x ) ≤ x
π
(
x
)
=
Π
(
x
)
+
O
(
x
)
.
{\displaystyle \pi (x)=\Pi (x)+O\left({\sqrt {x}}\,\right).}
黎曼猜想 指稱說黎曼ζ函數任意的非顯著零點的實部的值為1 / 2 |x ρ √x
∑
ρ
x
ρ
ρ
=
O
(
x
log
2
x
)
.
{\displaystyle \sum _{\rho }{\frac {x^{\rho }}{\rho }}=O\!\left({\sqrt {x}}\,\log ^{2}x\right).}
由上式可推得
π
(
x
)
=
li
(
x
)
+
O
(
x
log
x
)
.
{\displaystyle \pi (x)=\operatorname {li} (x)+O\!\left({\sqrt {x}}\,\log x\right).}
平滑化切比雪夫函數 定義如下：
ψ
1
(
x
)
=
∫
0
x
ψ
(
t
)
d
t
.
{\displaystyle \psi _{1}(x)=\int _{0}^{x}\psi (t)\,dt.}
顯然有
ψ
1
(
x
)
∼
x
2
2
.
{\displaystyle \psi _{1}(x)\sim {\frac {x^{2}}{2}}.}
^ 1.0 1.1 Joshua Knowles. Multiobjective Optimization Concepts, Algorithms and Performance Measures (PDF) . The University of Manchester: 34. 2 May 2014 [2023-12-07 ] . （原始内容存档 (PDF) 于2022-12-09）. ^ Ho-Huu, V.; Hartjes, S.; Visser, H. G.; Curran, R. An improved MOEA/D algorithm for bi-objective optimization problems with complex Pareto fronts and its application to structural optimization (PDF) . Expert Systems with Applications (Delft University of Technology). 2018. Page 6 equation (2) [2023-12-07 ] . doi:10.1016/j.eswa.2017.09.051 存档 (PDF) 于2024-04-16）. ^ Apostol, Tom M. Introduction to Analytic Number Theory. Springer. 2010: 75–76. ^ Rosser, J. Barkley; Schoenfeld, Lowell. Approximate formulas for some functions of prime numbers. . Illinois J. Math. 1962, 6 : 64–94 [2023-12-07 ] . （原始内容存档 于2016-08-18）.
^ Pierre Dusart ^ ψ θ π p k k k (log k + log log k − 1)k ≥ 2Mathematics of Computation , Vol. 68, No. 225 (1999), pp. 411–415.^ Mathematische Annalen , 57 (1903), pp. 195–204.^ Acta Mathematica , 41 (1916) pp. 119–196.^ Davenport, Harold Multiplicative Number Theory （页面存档备份 ，存于互联网档案馆 ）ISBN 0-387-95097-4 . Google Book Search.
![{\displaystyle {\begin{aligned}\vartheta (p_{k})&\geq k\left(\log k+\log \log k-1+{\frac {\log \log k-2.050735}{\log k}}\right)&&{\text{for }}k\geq 10^{11},\\[8px]\vartheta (p_{k})&\leq k\left(\log k+\log \log k-1+{\frac {\log \log k-2}{\log k}}\right)&&{\text{for }}k\geq 198,\\[8px]|\vartheta (x)-x|&\leq 0.006788\,{\frac {x}{\log x}}&&{\text{for }}x\geq 10\,544\,111,\\[8px]|\psi (x)-x|&\leq 0.006409\,{\frac {x}{\log x}}&&{\text{for }}x\geq e^{22},\\[8px]0.9999{\sqrt {x}}&<\psi (x)-\vartheta (x)<1.00007{\sqrt {x}}+1.78{\sqrt[{3}]{x}}&&{\text{for }}x\geq 121.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7301be23d04a6fb8d72a32bdccccb1cc2f9596f)
![{\displaystyle \psi _{0}(x)={\frac {1}{2}}\!\left(\sum _{n\leq x}\Lambda (n)+\sum _{n<x}\Lambda (n)\right)={\begin{cases}\psi (x)-{\tfrac {1}{2}}\Lambda (x)&x=2,3,4,5,7,8,9,11,13,16,\dots \\[5px]\psi (x)&{\mbox{otherwise.}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9e799e6e3a8daf7e8f40607109b76aba621e985)
![{\displaystyle \Pi (x)=\pi (x)+{\tfrac {1}{2}}\pi \left({\sqrt {x}}\,\right)+{\tfrac {1}{3}}\pi \left({\sqrt[{3}]{x}}\,\right)+\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/366ea3365a925e84a1dd8555ba2d8d27aac1207e)
