1 Introduction
The Stirling formula
has many important applications in statistical physics, probability theory and number theory. Due to its practical importance, it has attracted much interest of many mathematicians and has also motivated a large number of research papers concerning various generalizations and improvements; see for example, Burnside’s [1], Gosper [2], Batir [3], Mortici [4].
$$ n!\thicksim \sqrt{2\pi n}n^{n}e^{-n} $$
(1.1)
The gamma function \(\Gamma ( x ) =\int_{0}^{\infty }t^{x-1}e ^{-t}\,dt \) for \(x>0\) is closely related to the Stirling formula, since \(\Gamma (n+1)=n!\) for all \(n\in \mathbb{N}\). This inspired some authors to also pay attention to find various better approximations for the gamma function; see, for instance, Ramanujan [5, p. 339], Windschitl (see Nemes [6, Corollary 4.1]), Yang and Chu [7], Chen [8].
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More results involving the approximation formulas for the factorial or gamma function can be found in [9‐23] and the references cited therein.
In this note, we are interested in Smith’s approximation formula (see [24, equation (42)]):
It is easy to check that
which shows that the rate of \(S ( x ) \) converging to \(\Gamma ( x+1/2 ) \) as \(x\rightarrow \infty \) is like \(x^{-5}\). According to the comment in [8, (3.5)–(3.10)], it is well known that Smith’s approximation is an accurate but simple one for gamma function.
$$ \Gamma \biggl( x+\frac{1}{2} \biggr) \thicksim \sqrt{2\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2}:=S ( x ), \quad \text{as }x\rightarrow \infty . $$
(1.2)
$$ \Gamma \biggl( x+\frac{1}{2} \biggr) =\sqrt{2\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2} \biggl( 1+O \biggl( \frac{1}{x^{5}} \biggr) \biggr) , $$
The aim of this short note is to further prove the Smith approximation \(S ( x ) \) is an upper one. Our main result is stated as follows.
Theorem 1
The function
is strictly increasing and concave from
\(( 0,\infty ) \)
onto
\(( -\ln \sqrt{2},0 ) \).
$$ f ( x ) =\ln \Gamma \biggl( x+\frac{1}{2} \biggr) -\ln \sqrt{2 \pi }-x\ln x+x-\frac{x}{2}\ln \biggl( 2x\tanh \frac{1}{2x} \biggr) $$
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2 Proof of Theorem 1
To prove Theorem 1 we need the following two lemmas.
Lemma 1
The inequality
holds for all
\(x>0\).
$$ \psi^{\prime } \biggl( x+\frac{1}{2} \biggr) < \frac{4}{3} \frac{15x^{2}+4}{x ( 20x^{2}+7 ) } $$
(2.1)
Proof
Let
Using the recurrence formula [25, pp. 258–260]:
we have
It then follows that
which proves the desired inequality (2.1). □
$$ f_{1} ( x ) =\psi^{\prime } \biggl( x+\frac{1}{2} \biggr) - \frac{4}{3}\frac{15x^{2}+4}{x ( 20x^{2}+7 ) }. $$
(2.2)
$$ \psi^{ ( n ) }(x+1)-\psi^{ ( n ) }(x)=\frac{ ( -1 ) ^{n}n!}{x^{n+1}}, $$
$$\begin{aligned} f_{1} ( x+1 ) -f_{1} ( x ) =&\psi^{\prime } \biggl( x+ \frac{3}{2} \biggr) -\frac{4}{3 ( x+1 ) }\frac{15x^{2}+30x+19}{20x ^{2}+40x+27} \\ &{}-\psi^{\prime } \biggl( x+\frac{1}{2} \biggr) +\frac{4}{3} \frac{15x ^{2}+4}{x ( 20x^{2}+7 ) } \\ &{}-\frac{1}{ ( x+1/2 ) ^{2}}-\frac{4}{3 ( x+1 ) }\frac{15x ^{2}+30x+19}{20x^{2}+40x+27}+\frac{4}{3} \frac{15x^{2}+4}{x ( 20x ^{2}+7 ) } \\ =&\frac{144}{x ( x+1 ) ( 2x+1 ) ^{2} ( 20x ^{2}+7 ) ( 20x^{2}+40x+27 ) }>0. \end{aligned}$$
$$ f_{1} ( x ) < f_{1} ( x+1 ) < \cdots < \lim _{n\rightarrow \infty }f_{1} ( x+n ) =0, $$
Lemma 2
The inequality
holds for all
\(t>0\).
$$ \frac{\sinh^{2}t}{\cosh t}>\frac{t^{2} ( 21t^{2}+60 ) }{31t ^{2}+60} $$
(2.3)
Proof
It is obvious that the inequality what we consider is equivalent to
Simplifying and expanding it in power series lead us to
where
It is easy to check that \(a_{2}=a_{3}=0\) and \(a_{4}=49\,184>0\). It remains to prove \(a_{n}>0\) for \(n\geq 5\).
$$ f_{2} ( t ) = \bigl( 31t^{2}+60 \bigr) ( \sinh t ) ^{2}-t^{2} \bigl( 21t^{2}+60 \bigr) \cosh t>0. $$
$$\begin{aligned} 2f_{2} ( t ) =&60\cosh 2t+31t^{2}\cosh 2t-120t^{2}\cosh t-42t ^{4}\cosh t-31t^{2}-60 \\ =&60\sum_{n=0}^{\infty }\frac{2^{2n}}{ ( 2n ) !}t^{2n}+31 \sum_{n=1}^{\infty }\frac{2^{2n-2}}{ ( 2n-2 ) !}t^{2n} \\ &{}-120\sum_{n=1}^{\infty }\frac{1}{ ( 2n-2 ) !}t^{2n}-42 \sum_{n=2}^{\infty }\frac{1}{ ( 2n-4 ) !}t^{2n}-31t^{2}-60 \\ :=&\sum_{n=2}^{\infty }\frac{a_{n}}{ ( 2n ) !}t^{2n}, \end{aligned}$$
$$ a_{n}= \bigl( 62n^{2}-31n+120 \bigr) 2^{2n-1}-24n ( 2n-1 ) \bigl( 14n^{2}-35n+31 \bigr) . $$
To this end, it suffices to prove \(b_{n}=2^{2n-1}-6n ( 2n-1 ) >0\) for \(n\geq 5\), because the inequality
is clearly valid for \(n\geq 5\). We easily obtain
for \(n\geq 5\), which in combination with \(b_{5}=242>0\) yields \(b_{n}>0\) for \(n\geq 5\). This completes the proof. □
$$ \bigl( 62n^{2}-31n+120 \bigr) >4 \bigl( 14n^{2}-35n+31 \bigr) $$
$$ b_{n+1}-4b_{n}=6 \bigl( 6n^{2}-7n-1 \bigr) >0 $$
Now we are in a position to prove Theorem 1.
Theorem 1
Differentiating and simplifying yields
As an application of inequalities (2.1) and (2.3) it gives
Then it is deduced that
which in turn implies that
This completes the proof. □
$$\begin{aligned}& f^{\prime } ( x ) = \psi \biggl( x+\frac{1}{2} \biggr) - \ln x- \frac{1}{2}\ln \biggl( 2x\tanh \frac{1}{2x} \biggr) + \frac{1}{2x \sinh ( 1/x ) }-\frac{1}{2}, \\& f^{\prime \prime } ( x ) = \psi^{\prime } \biggl( x+ \frac{1}{2} \biggr) +\frac{1}{2x^{3}}\frac{\cosh ( 1/x ) }{ \sinh^{2} ( 1/x ) }-\frac{3}{2x}. \end{aligned}$$
$$\begin{aligned}& \begin{aligned} f^{\prime \prime } ( x ) &< \frac{4}{3}\frac{15x^{2}+4}{x ( 20x^{2}+7 ) }+\frac{1}{2x^{3}} \frac{\cosh ( 1/x ) }{\sinh^{2} ( 1/x ) }-\frac{3}{2x} \\ &=\frac{1}{2x^{3}}\frac{\cosh ( 1/x ) }{\sinh^{2} ( 1/x ) }-\frac{1}{6}\frac{60x^{2}+31}{x ( 20x^{2}+7 ) } \\ & \mathop{=}^{x=1/t}\frac{t^{3}}{2} \biggl( \frac{\cosh t}{ \sinh^{2}t}- \frac{31t^{2}+60}{t^{2} ( 21t^{2}+60 ) } \biggr) < 0. \end{aligned} \end{aligned}$$
$$ f^{\prime } ( x ) >\lim_{x\rightarrow \infty }f^{\prime } ( x ) =0, $$
$$ -\frac{1}{2}\ln 2=\lim_{x\rightarrow 0^{+}}f ( x ) < f ( x ) < \lim _{x\rightarrow \infty }f ( x ) =0. $$
3 Corollaries and remarks
Using the increasing property of \(f ( x+1/2 ) \) given in Theorem 1 and noting that
we have the corollaries.
$$ f \biggl( \frac{1}{2} \biggr) =\ln \frac{\sqrt{e}}{\sqrt{\pi } ( \tanh 1 ) ^{1/4}}\quad \text{and}\quad f \biggl( \frac{3}{2} \biggr) = \ln \biggl( \frac{2e\sqrt{e}3^{3/4}}{27\sqrt{\pi }\tanh^{3/4} ( 1/3 ) } \biggr) , $$
Corollary 1
The double inequality
holds for all
\(x>0\)
with the best constants 1 and
\(\alpha_{1}=\sqrt{e/ \pi }/ ( \tanh 1 ) ^{1/4}\approx 0.99573\).
$$ \alpha_{1}< \frac{e^{x+1/2}\Gamma ( x+1 ) }{\sqrt{2\pi } ( x+1/2 ) ^{x+1/2} [ ( 2x+1 ) \tanh ( 1/ ( 2x+1 ) ) ] ^{ ( 2x+1 ) /4}}< 1 $$
Corollary 2
The double inequality
holds for all
\(n\in \mathbb{N}\)
with the best constants 1 and
$$ \alpha_{2}< \frac{n!}{\sqrt{2\pi } ( ( n+1/2 ) /e ) ^{n+1/2} [ ( 2n+1 ) \tanh ( 1/ ( 2n+1 ) ) ] ^{ ( 2n+1 ) /4}}< 1 $$
$$ \alpha_{2}=\frac{2e\sqrt{e}3^{3/4}}{27\sqrt{\pi }\tanh^{3/4} ( 1/3 ) }\approx 0.99994. $$
By the decreasing property of \(f^{\prime } ( x+1/2 ) \) given in Theorem 1 and the facts that
the following corollaries are immediate.
$$\begin{aligned}& f^{\prime } \biggl( \frac{1}{2} \biggr) =\frac{1}{\sinh 2}- \frac{1}{2} \ln ( \tanh 1 ) +\ln 2-\frac{1}{2}-\gamma \approx 0.027823, \\& f^{\prime } \biggl( \frac{3}{2} \biggr) =\frac{1}{3\sinh ( 2/3 ) }- \frac{1}{2}\ln \biggl( 3\tanh \frac{1}{3} \biggr) -\ln \frac{3}{2}+\psi ( 1 ) +\frac{1}{2}\approx 0.00016946, \end{aligned}$$
Corollary 3
For
\(x>0\), the inequalities
hold, where the constants
\(1/2\)
and
are the best possible.
$$\begin{aligned}& \frac{1}{2}+\frac{1}{2}\ln \biggl( ( 2x+1 ) \tanh \frac{1}{2x+1} \biggr) -\frac{1}{ ( 2x+1 ) \sinh ( 2/ ( 2x+1 ) ) } \\& \quad < \psi ( x+1 ) -\ln \biggl( x+\frac{1}{2} \biggr) \\& \quad < \beta_{1}+\frac{1}{2}\ln \biggl( ( 2x+1 ) \tanh \frac{1}{2x+1} \biggr) -\frac{1}{ ( 2x+1 ) \sinh ( 2/ ( 2x+1 ) ) } \end{aligned}$$
$$ \beta_{1}=\ln 2-\frac{1}{2}\ln ( \tanh 1 ) +\frac{1}{\sinh 2}- \gamma \approx 0.52782 $$
Corollary 4
Let
\(H_{n}=\sum_{k=1}^{n}\)
for
\(n\in \mathbb{N}\). The inequalities
hold, where
\(1/2+\gamma \approx 1.0772\)
and
are the best possible constants.
$$\begin{aligned}& \biggl( \frac{1}{2}+\gamma \biggr) +\frac{1}{2}\ln \biggl( ( 2n+1 ) \tanh \frac{1}{2n+1} \biggr) -\frac{1}{ ( 2n+1 ) \sinh ( 2/ ( 2x+1 ) ) } \\& \quad < H_{n}-\ln \biggl( n+\frac{1}{2} \biggr) \\& \quad < \beta_{2}+\frac{1}{2}\ln \biggl( ( 2n+1 ) \tanh \frac{1}{2n+1} \biggr) -\frac{1}{ ( 2n+1 ) \sinh ( 2/ ( 2n+1 ) ) } \end{aligned}$$
$$ \beta_{2}=\frac{1}{3\sinh ( 2/3 ) }-\frac{1}{2}\ln \biggl( 3 \tanh \frac{1}{3} \biggr) -\ln \frac{3}{2}+1\approx 1.0774 $$
Finally, as a by-product of Lemma 1, we draw the following conclusion.
Theorem 2
Let
g
be defined on
\(( 0,\infty ) \)
by
Then
g
is strictly increasing and concave on
\(( 0,\infty ) \).
$$ g ( x ) =\ln \Gamma \biggl( x+\frac{1}{2} \biggr) - \biggl[ \frac{1}{2}\ln 2\pi +\frac{16}{21}x\ln x+\frac{5x}{42}\ln \biggl( x ^{2}+\frac{7}{20} \biggr) -x-\frac{\sqrt{35}}{42} \operatorname{arccot} \biggl( \sqrt{\frac{20}{7}}x \biggr) \biggr] . $$
Proof
Differentiation yields
where the inequality holds due to Lemma 1. This completes the proof. □
$$\begin{aligned}& g^{\prime } ( x ) =\psi \biggl( x+\frac{1}{2} \biggr) - \biggl[ \frac{5}{42}\ln \biggl( x^{2}+\frac{7}{20} \biggr) + \frac{16}{21}\ln x \biggr] , \\& g^{\prime \prime } ( x ) =\psi^{\prime } \biggl( x+ \frac{1}{2} \biggr) -\frac{4}{3}\frac{15x^{2}+4}{x ( 20x^{2}+7 ) }=f_{1} ( x ) < 0, \end{aligned}$$
Remark 1
Theorem 2 gives a new approximation for the gamma function
as \(x\rightarrow \infty \), which satisfies
$$ \Gamma \biggl( x+\frac{1}{2} \biggr) \thicksim \sqrt{2\pi }x^{26x/21} \biggl( x^{2}+\frac{7}{20} \biggr) ^{5x/42}\exp \biggl[ -x-\frac{ \sqrt{35}}{42}\operatorname{arccot} \biggl( \sqrt{\frac{20}{7}}x \biggr) \biggr] , $$
$$ \Gamma \biggl( x+\frac{1}{2} \biggr) =\sqrt{2\pi }x^{26x/21} \biggl( x ^{2}+\frac{7}{20} \biggr) ^{5x/42}\exp \biggl[ -x- \frac{\sqrt{35}}{42}\operatorname{arccot} \biggl( \sqrt{\frac{20}{7}}x \biggr) \biggr] \bigl( 1+O \bigl( x^{-5} \bigr) \bigr) . $$
Remark 2
Theorem 2 also offers an asymptotic formula for the psi function
Furthermore, by replacing x with \(x+1/2\), we have the following sharp inequalities:
for \(x>0\) with the best constant
for \(n\in \mathbb{N}\) with the best constant
Inequalities (3.1) first appeared in [26, Corollary 3.4].
$$ \psi \biggl( x+\frac{1}{2} \biggr) \thicksim \frac{5}{42}\ln \biggl( x ^{2}+\frac{7}{20} \biggr) +\frac{16}{21}\ln x\quad \text{as }x \rightarrow \infty . $$
$$\begin{aligned}& \frac{5}{42}\ln \biggl( x^{2}+x+\frac{3}{5} \biggr) + \frac{16}{21} \ln \biggl( x+\frac{1}{2} \biggr) \\& \quad < \psi ( x+1 ) < \lambda_{0}+\frac{5}{42}\ln \biggl( x^{2}+x+ \frac{3}{5} \biggr) +\frac{16}{21}\ln \biggl( x+ \frac{1}{2} \biggr) \end{aligned}$$
(3.1)
$$\begin{aligned}& \lambda_{0}=\frac{16}{21}\ln 2-\frac{5}{42}\ln \frac{3}{5}-\gamma \approx 0.011709; \\& \gamma +\frac{5}{42}\ln \biggl( n^{2}+n+\frac{3}{5} \biggr) + \frac{16}{21}\ln \biggl( n+\frac{1}{2} \biggr) \\& \quad < H_{n}< \lambda_{0}+\gamma +\frac{5}{42}\ln \biggl( n^{2}+n+ \frac{3}{5} \biggr) +\frac{16}{21}\ln \biggl( n+ \frac{1}{2} \biggr) \end{aligned}$$
$$ \lambda_{1}=1-\frac{16}{21}\ln \frac{3}{2}- \frac{5}{42}\ln \frac{13}{5}-\gamma \approx 0.00010718. $$
4 Conclusions
In this note, we mainly presented an upper bound of Smith’s approximation in accordance with the fact that the function \(x\mapsto \ln \Gamma ( x+1/2 ) - \ln S ( x ) \) is strictly increasing and concave on \(( 0,\infty ) \). As a consequence, we get some new sharp estimates to various classical inequalities concerning the gamma function and hyperbolic functions.
Acknowledgements
This paper is supported by the National Science Foundation of China grant No. 11371050.
Competing interests
The authors declare that they have no competing interests.
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