About bounds for the elliptic integral
of the first kind
Pál A. Kupán\(^\ast \) Róbert Szász\(^\ast \)
February 6, 2012
\(^\ast \)Department of Mathematics, Sapientia University, Sighişoarei st., no. 1C, Tg. Mureş, Romania, e-mail: kupanp@ms.sapientia.ro, szasz_robert2001@yahoo.com.
We deduce an inequality using elementary methods which makes it possible to prove a conjecture regarding the upper bound of the elliptic integral of the first kind, furthermore we also improve the lower bound.
MSC. 33C05
Keywords. Hypergeometric function; Elliptic integral; Inequality; Bounds.
1 Introduction
Legendre’s complete elliptic integral of the first kind is defined for \(r\in (0,1)\) by
\[ \mathcal{K}(r)=\int _{0}^{\pi /2}\tfrac {1}{\sqrt{1-r^2\sin {t}}}{\rm d}t. \]
This integral is a special case of Gauss’s hypergeometric function
\[ _2F_1(a,b;c;x)=\sum _{n=0}^\infty \tfrac {(a,n)(b,n)}{(c,n)}\tfrac {x^n}{n!}, \ \ x\in (-1,1), \]
where \((a,n)=\prod _{k=0}^{n-1}(a+k).\) We have
\begin{equation} \label{2}\mathcal{K}(r)=\tfrac {\pi }{2} \ {_2F_1}(\tfrac {1}{2},\tfrac {1}{2};1;r^2)=\tfrac {\pi }{2}\Big[1+\sum _{n=1}^\infty \Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)} {2\cdot 4\cdot ...\cdot (2n)}\Big)^2r^{2n}\Big].\end{equation}
1
In
[
1
]
the authors posed the problem to determine the best values \(\alpha ^*\) and \(\beta ^*\) such that
\begin{align} \label{1} \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{3/4+\alpha ^*r}{\lt}K(r) {\lt} \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{3/4+\beta ^*r}, \ \ r\in (0,1). \end{align}
This problem is equivalent to the following: determine the best values \(\alpha ^*\) and \(\beta ^*\) such that
\[ \alpha ^*{\lt}\Big[G(r)-\tfrac {3}{4}\Big]/r{\lt}\beta ^*, \ \ r\in (0,1), \ \textrm{where} \ \ G(r)=\tfrac {\log (2\mathcal{K}(r)/\pi )}{\log ([{\rm arth}(r)]/r)}. \]
The first part of this problem had been solved by the authors in
[
1
]
showing that \(\alpha ^*=0.\) Concerning the second part they conjectured that the mapping \(G:(0,1)\rightarrow {\mathbb {R}}\) is strictly increasing and convex. Since \(\lim _{r\nearrow 1}G(r)=1,\) this conjecture would imply \(\beta ^*=1/4.\) The result \(\beta ^*=1/4\) has been proved recently in
[
4
]
. The basic tool used in their proof is Theorem 1.25 from
[
2
]
. It seems very difficult to prove the conjecture regarding the monotonicity of \(G.\) In the following we will show that a different elementary approach leads to a result, which improves the upper bound conjectured in
[
1
]
.
In order to prove our results, we need certain lemmas, which will be exposed in the next section.
2 Preliminaries
Lemma
1
If \(a_n=\Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)}{2\cdot 4\cdot ...\cdot (2n)}\Big)^2,\) \(b_n=\tfrac {4}{\pi (4n+1)},\) \(c_n=\tfrac {1}{\pi (n+\tfrac {4}{\pi }-1)},\) \(x_n=\tfrac {a_n}{b_n}, \ \ n\in {\mathbb {N}^*},\) \(y_n=\tfrac {a_n}{c_n}, \ \ n\in {\mathbb {N}^*},\) then the sequence \(\big(x_{n}\big)_{n\in \mathbb {N}^*}\) is strictly increasing, the sequence \(\big(y_{n}\big)_{n\in \mathbb {N}^*}\) is strictly decreasing for \(n\geq 2\) and \(\lim _{n\rightarrow \infty }x_n=\lim _{n\rightarrow \infty }y_n=1.\)
Since
\[ \tfrac {x_{n+1}}{x_n}=\tfrac {(4n+5)(2n+1)^2}{(4n+1)(2n+2)^2}=\tfrac {16n^3+36n^2+24n+5}{16n^3+36n^2+24n+4}{\gt}1 \]
it follows that \(x_{n+1}{\gt}x_n, \ n\in {\mathbb {N^*}}.\) On the other hand, we have \(a_n=\Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)}{2\cdot 4\cdot ...\cdot (2n)}\Big)^2{\lt}\Big(\tfrac {2\cdot 4\cdot 6\cdot ...\cdot (2n)}{3\cdot 5\cdot ...\cdot (2n+1)}\Big)^2.\) This implies \(a_n{\lt}\tfrac {1}{2n+1}\) and finally we get \(x_n{\lt}\tfrac {\pi (4n+1)}{4(2n+1)}{\lt}\tfrac {\pi }{2}, n\in {\mathbb {N^*}}.\) Consequently, \((x_n)_{n\in {\mathbb {N}^*}}\) is convergent. Wallis product formula implies that \(\lim _{n\rightarrow \infty }x_n=1.\) Thus we have
\[ 1{\gt}x_{n+1}{\gt}x_n\geq {x_1}=\tfrac {5\pi }{16}=0.981\dots . \]
An analogous calculation implies the assertion regarding \(\big(y_{n}\big)_{n\in \mathbb {N}^*}.\)
Lemma
2
For all real numbers \(r\in (0,1),\) we have
\begin{eqnarray} \label{6} \mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg\{ 1+\tfrac {1}{4}r^2+\tfrac {4}{\pi }\sum _{n=2}^\infty \tfrac {r^{2n}}{4n+1}\bigg\} . \end{eqnarray}
We use the notations and the results of Lemma 1
\[ \tfrac {2}{\pi }\mathcal{K}(r)=1+\sum _{n=1}^\infty {a}_nr^{2n}=1+\sum _{n=1}^\infty \Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)} {2\cdot 4\cdot ...\cdot (2n)}\Big)^2r^{2n}, \]
and let
\[ h(r)=1+a_1r^2+\sum _{n=2}^\infty {b}_n{r^{2n}}=1+\tfrac {1}{4}r^2+\tfrac {4}{\pi }\sum _{n=2}^\infty \tfrac {r^{2n}}{4n+1}. \]
We introduce the notations \(\tfrac {2}{\pi }\mathcal{K}(r)=1+u(r)\) and \(h(r)=1+v(r).\) Lemma 1 implies \(a_{n}{\lt}{b}_{n}, n\in {\mathbb {N}^*, \ n\geq 2},\) and consequently \(u(r){\lt}v(r)\) for all \(r\in (0,1).\) Thus, inequality (3) holds.
Lemma
3
For all real numbers \(r\in (0,1),\) we have
\begin{eqnarray} \label{68} \tfrac {\pi }{2}\bigg\{ 1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}\bigg\} {\lt}\mathcal{K}(r). \end{eqnarray}
We use the notations and the results of Lemma1 in our proof again. We recall that
\[ \tfrac {2}{\pi }\mathcal{K}(r)=1+\sum _{n=1}^\infty {a}_nr^{2n}=1+\sum _{n=1}^\infty \Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)} {2\cdot 4\cdot ...\cdot (2n)}\Big)^2r^{2n}, \]
and let
\[ k(r)=1+\tfrac {1}{\pi }\sum _{n=2}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}. \]
Lemma 1 implies \(c_n{\lt}a_{n}, n\in {\mathbb {N}^*, \ n\geq 2},\) and consequently \(k(r){\lt}\tfrac {2}{\pi }\mathcal{K}(r)\) for all \(r\in (0,1).\) Thus, inequality (4) holds.
Lemma
4
(Bernoulli’s inequality) If \(\alpha \geq 1\) and \(a{\gt}-1\), then
\begin{equation} \label{mng} (1+a)^\alpha \geq 1+a\alpha . \end{equation}
5
If \(b\in [0,1]\) and \(\alpha \in (1,2),\) then
\begin{equation} \label{qw2}(1+b)^\alpha \geq 1+\alpha {b}+\tfrac {\alpha (\alpha -1)}{2}b^2+\tfrac {\alpha (\alpha -1)(\alpha -2)}{6}b^3.\end{equation}
6
We prove the second inequality. Let \(g:[0,1]\rightarrow {\mathbb {R}}\) be the function defined by \(g(b)=(1+b)^\alpha -1-\alpha {b}-\tfrac {\alpha (\alpha -1)}{2}b^2-\tfrac {\alpha (\alpha -1)(\alpha -2)}{6}b^3.\) We have \(g'(b)=\alpha (1+b)^{\alpha -1}-\alpha -{\alpha (\alpha -1)}b-\tfrac {\alpha (\alpha -1)(\alpha -2)}{2}b^2,\) \(g''(b)=\alpha (\alpha -1)[(1+b)^{\alpha -2}-1-(\alpha -2)b]\) and \(g'''(b)=\alpha (\alpha -1)(\alpha -2)[(1+b)^{\alpha -3}-1].\) Thus \(g''(0)=0\) implies that \(g''(b){\gt}0, \ b\in (0,1).\) An analogous argumentation shows that \(g'\) and \(g\) are strictly increasing on \((0,1)\) and so \(g(0)=0\) implies inequality (
6).
Lemma
5
Let \(w:(0,1)\rightarrow {\mathbb {R}}\) be the function defined by \(\tfrac {{\rm arth}(r)}{r}=1+\sum ^\infty _{n=1}\tfrac {1}{2n+1}r^{2n}=1+w(r).\) If \(w(r)+w(r)v(r)=\sum ^\infty _{n=1}\delta _nr^{2n},\) then \(\delta _n\leq \tfrac {1}{3}, \ n\in \mathbb {N}, \ n\geq 1.\)
Indeed \(\delta _1=\tfrac {1}{3}, \ \delta _2=\tfrac {17}{60},\) and if \(n\geq 3,\) then
\begin{align*} \delta _n& =\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi }\sum ^{n-2}_{k=1}\tfrac {1}{(2k+1)(4(n-k)+1)}\\ & =\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi (4n+3)}\sum ^{n-2}_{k=1}\Big(\tfrac {1}{2k+1}+\tfrac {2}{4(n-k)+1}\Big)\\ & =\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi (4n+3)}\Big(\sum ^{n-2}_{k=1}\tfrac {1}{2k+1}+\sum ^{n-1}_{k=2}\tfrac {2}{4k+1}\Big) \end{align*}
\begin{align*} & {\lt}\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi (4n+3)}\tfrac {3(n-2)}{5}\\ & \leq \tfrac {1}{5}+\tfrac {1}{20}+\tfrac {4}{25\pi }{\lt}\tfrac {1}{3}. \end{align*}
Lemma
6
[
4
]
Let \((a_n)_{n\geq 1}\) and \((b_n)_{n\geq 1}\) be two sequences of real numbers, and let the power series
\[ u(x)=\sum _{n=1}^\infty {a}_nx^n \ \ \textrm{and} \ v(x)=\sum _{n=1}^\infty {b}_nx^n \]
be convergent for \(|x|{\lt}1.\) If \(b_n{\gt}0, \ \ n=1,2,3,...,\) and if the sequence \(\Big(\tfrac {a_n}{b_n}\Big)_{n\geq 1}\) is strictly increasing (resp. decreasing), then the function
\(\tfrac {u}{v}:(0,1)\rightarrow {\mathbb {R}}\) is strictly increasing (resp. decreasing).
3 The main result
Recall that
\begin{eqnarray} v(r)=\tfrac {1}{4}r^2+\tfrac {4}{\pi }\sum _{n=2}^\infty \tfrac {1}{4n+1}r^{2n},\nonumber \end{eqnarray}
and
\[ w(r)=\tfrac {{\rm arth}(r)}{r}-1=\sum ^\infty _{n=1}\tfrac {1}{2n+1}r^{2n}. \]
Theorem
7
If \(r\in (0,1),\) then the following inequality holds:
\begin{equation} \label{123}1+v(r)<\big(1+w(r)\big)^{\tfrac {3}{4}+\tfrac {1}{4}r^2}.\end{equation}
7
We begin with the remark that (
7) is equivalent to
\begin{equation} \label{32}\bigg(1+\tfrac {w(r)-v(r)}{1+v(r)}\bigg)^{\tfrac {4}{1-r^2}}>1+w(r), \ \ r\in (0,1).\end{equation}
8
The inequality (5) from Lemma 4 implies that
\[ \bigg(1+\tfrac {w(r)-v(r)}{1+v(r)}\bigg)^{\tfrac {4}{1-r^2}}{\gt}1+\tfrac {4}{1-r^2}\tfrac {w(r)-v(r)}{1+v(r)}, \ \ r\in (0,1). \]
Thus, in order to prove (8), we have to show that
\begin{equation} \label{54}\tfrac {4(w(r)-v(r))}{(1-r^2)(1+v(r))}>w(r), \ r\in (0,1).\end{equation}
9
We have \(w(r)-v(r){\gt}\tfrac {1}{12}r^2, \ r\in (0,1).\) Thus
\[ \tfrac {r^2}{3(1-r^2)(1+v(r))}{\gt}w(r), \ r\in (0,1) \]
implies (9). This inequality is equivalent to
\[ \tfrac {r^2}{3(1-r^2)}{\gt}w(r)+w(r)v(r), \ r\in (0,1). \]
According to Lemma 5, we have
\[ \tfrac {r^2}{3(1-r^2)}=\sum _{n=1}^\infty \tfrac {1}{3}r^{2n}{\gt}\sum ^\infty _{n=1}c_nr^{2n}=w(r)+w(r)v(r), \]
and the proof is completed.
Theorem 1 and Lemma 2 imply the following result.
Corollary
8
If \( \ r\in (0,1),\) then
\[ \mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {1}{4}r^2}. \]
Theorem
9
If \(r\in (0,1),\) then
\begin{equation} \label{8nm9}1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}>\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {r^4}{200}}.\end{equation}
10
We introduce the notations \(\mu _1=\tfrac {4}{\pi }-1\) and \(z(r)=\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}=\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\mu _1}.\) Using this notation, (
10) will be equivalent to
\begin{equation} \label{1qwn2}\big(1+z(r)\big)^{\tfrac {200}{r^4+150}>1+w(r)}. \end{equation}
11
We shall prove this inequality in three steps. First assume that \(r\in [0,\tfrac {1}{5}].\)
In this case we use the second inequality of Lemma 4 putting \(\alpha =\tfrac {200}{r^4+150}\) and \(b=z(r),\) and we obtain
\begin{equation} \label{qwn2}(1+z(r))^\alpha \geq 1+\alpha {z(r)}+\tfrac {\alpha (\alpha -1)}{2}(z(r))^2+\tfrac {\alpha (\alpha -1)(\alpha -2)}{6}(z(r))^3.\end{equation}
12
On the other hand we have \(\tfrac {\alpha (\alpha -1)(2-\alpha )}{6}{\lt}\tfrac {1}{20},\) \(\tfrac {\alpha (\alpha -1)}{2}{\gt}0.22,\) \((z(r))^3{\lt}\tfrac {r^6}{50}, r\in (0,\tfrac {1}{5}).\) Thus, inequality (12) implies
\[ (1+z(r))^\alpha \geq 1+\alpha {z(r)}+0.22(z(r))^2-\tfrac {r^6}{1000}, \ \ r\in [0,\tfrac {1}{5}], \]
and consequently, in order to prove (11) we have to show that
\[ 1+\alpha {z(r)}+0.22(z(r))^2-\tfrac {r^6}{1000}\geq 1+w(r), \ \ r\in [0,\tfrac {1}{5}]. \]
This inequality is equivalent to
\begin{equation} \label{dfg21}0.22\Big(\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\mu _1}\Big)^2>\tfrac {r^6}{1000}+\sum _{n=1}^\infty \Big(\tfrac {1}{2n+1}-\tfrac {\alpha }{\pi (n+\mu _1)}\Big)r^{2n}, \ \ r\in [0,\tfrac {1}{5}].\end{equation}
13
Let us denote the coefficient of \(r^{2n}\) in \(0.22\Big(\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\mu _1}\Big)^2\) by \(d_n, \ n\geq 2.\)
In order to prove inequality (13), we will show that
\begin{equation} \label{d1fg21}d_2r^4\geq \tfrac {r^6}{1000}+\Big(\tfrac {1}{3}-\tfrac {\alpha }{\pi (1+\mu _1)}\Big)r^{2}+\Big(\tfrac {1}{5}-\tfrac {\alpha }{\pi (2+\mu _1)}\Big)r^{4}, \ \ r\in [0,\tfrac {1}{5}],\end{equation}
14
and
\begin{equation} \label{df2g21}d_n\geq \tfrac {1}{2n+1}-\tfrac {\alpha }{\pi (n+\mu _1)}, \ \ n\geq 3.\end{equation}
15
The inequality (14) holds, because
\begin{align*} d_2r^4& =0.22\tfrac {1}{16}r^4\geq \Big[\tfrac {1}{25000}+\tfrac {1}{450\cdot 25}+\Big(\tfrac {1}{5}-\tfrac {200}{(4+\pi )(\tfrac {1}{625}+150)}\Big)\Big]r^4\nonumber \\ & \geq \tfrac {r^6}{1000}+\tfrac {1}{450+3r^4}r^6+\Big(\tfrac {1}{5}-\tfrac {200}{(4+\pi )(r^4+150)}\Big)r^4\nonumber \\ & =\tfrac {r^6}{1000}+(\tfrac {1}{3}-\tfrac {\alpha }{\pi (1+\mu _1)}\Big)r^{2}+\Big(\tfrac {1}{5}-\tfrac {\alpha }{\pi (2+\mu _1)}\Big)r^{4}, \ \ r\in [0,\tfrac {1}{5}].\nonumber \end{align*}
It is sufficient to prove (15) for \(r=\tfrac {1}{5}.\) We have
\begin{eqnarray} d_n=\tfrac {0.22}{\pi ^2}\sum _{k=1}^{n-1}\tfrac {1}{(n-k+\mu _1)(k+\mu _1)}=\tfrac {0.44}{\pi ^2}\tfrac {1}{n+2\mu _1}\sum _{k=1}^{n-1}\tfrac {1}{k+\mu _1}. \nonumber \end{eqnarray}
If \(r=\tfrac {1}{5},\) inequality (15) is equivalent to
\begin{equation} \label{qqq}t_n=\tfrac {2\cdot 62500(n+\tfrac {1}{2})}{46876\pi (n+\mu _1)}+\tfrac {0.88}{\pi ^2}\tfrac {n+\tfrac {1}{2}}{n+2\mu _1}\sum _{k=1}^{n-1}\tfrac {1}{k+\mu _1}>1, \ n\in {\mathbb {N}^*}, \ n\geq 3.\end{equation}
13
We prove now that the sequence \((t_n)_{n\geq 3}\) is strictly increasing.
\begin{align*} & t_{n+1}-t_n\nonumber \\ & {\gt}\tfrac {2\cdot 62500}{46876\pi }\Big(\tfrac {n+\tfrac {3}{2}}{n+\tfrac {4}{\pi }}-\tfrac {n+\tfrac {1}{2}} {n+\tfrac {4}{\pi }-1}\Big)+\tfrac {0.88}{\pi ^2}\tfrac {n+\tfrac {3}{2}}{n+\tfrac {8}{\pi }-1} \Big(\sum _{k=1}^{n}\tfrac {1}{k+\mu _1}-\sum _{k=1}^{n-1}\tfrac {1}{k+\mu _1}\Big)\nonumber \\ & =\tfrac {0.88}{\pi ^2} \tfrac {n+\tfrac {3}{2}}{(n+\tfrac {8}{\pi }-1)(n+\tfrac {4}{\pi }-1)}-\tfrac {2\cdot 62500}{46876} \tfrac {\tfrac {3}{2\pi }-\tfrac {4}{\pi ^2}}{(n+\tfrac {4}{\pi })(n+\tfrac {4}{\pi }-1)}\nonumber \\ & =\tfrac {1}{n+\tfrac {4}{\pi }-1}\Big(\tfrac {0.88}{\pi ^2}\tfrac {n+\tfrac {3}{2}}{n+\tfrac {8}{\pi }-1}-\tfrac {2\cdot 62500}{46876}\tfrac {\tfrac {3}{2\pi }-\tfrac {4}{\pi ^2}}{n+\tfrac {4}{\pi }}\Big){\gt}0, \ n\geq 3.\nonumber \end{align*}
Consequently, inequality (13) holds, and the proof of inequality (10) is done for \(r\in [0,\tfrac {1}{5}].\)
In the second step we will prove that inequality (10) holds if \(r\in [\tfrac {1}{5},\tfrac {97}{100}].\) Let \(r_k=\tfrac {1}{5}+\tfrac {k}{200000}, \ k=\overline{0,154000}.\) The functions \(1+z(r)\) and \(\big(1+w(r)\big)^{\tfrac {3}{4}+\tfrac {r^4}{200}}\) are strictly increasing on \([\tfrac {1}{5},\tfrac {97}{100}].\) Thus, if the inequalities
\begin{equation} \label{951}1+z(r_{k-1})\geq \big(1+w(r_k)\big)^{\tfrac {3}{4}+\tfrac {r_k^4}{200}}, \ \ k=\overline{1,154000}\end{equation}
10
hold, then the inequality-chains
\[ 1+z(r)\geq 1+z(r_{k-1})\geq \big(1+w(r_k)\big)^{\tfrac {3}{4}+\tfrac {r_k^4}{200}}\geq \big(1+w(r)\big)^{\tfrac {3}{4}+\tfrac {r^4}{200}}, \ \ r\in [r_{k-1},r_k], \]
imply (10) for \(r\in [\tfrac {1}{5},\tfrac {97}{100}].\) The inequalities (10) can be verified easily using a computer program.
The third case is \(r\in [\tfrac {97}{100},1).\) In this case we will prove the following inequality, which is stronger than (10):
\[ 1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}{\gt}\bigg(\tfrac {arth(r)}{r}\bigg)^{\tfrac {151}{200}}, \ \ \ r\in [\tfrac {97}{100},1). \]
We define the function \(m:[\tfrac {97}{100},1)\rightarrow \mathbb {R}\) by \(m(r)=1+z(r)-\big(1+w(r)\big)^{\tfrac {151}{200}}.\) We have
\[ m'(r)=w'(r)\Big(\tfrac {z’(r)}{w’(r)}-\tfrac {151}{200}\tfrac {1}{(1+w(r))^{\tfrac {49}{200}}}\Big). \]
According to Lemma 6 the function \(\tfrac {z’}{w’}:(0,1)\rightarrow \mathbb {R}\) is strictly decreasing, and \(\lim _{r\nearrow 1}\tfrac {z’(r)}{w’(r)}=\tfrac {2}{\pi }.\) Thus
\[ \tfrac {z’(r)}{w’(r)}{\gt}\tfrac {2}{\pi }{\gt}\tfrac {151}{200}\tfrac {1}{(1+w(\tfrac {97}{100}))^{\tfrac {49}{200}}}\geq \tfrac {151}{200}\tfrac {1}{(1+w(r))^{\tfrac {49}{200}}}, \ r\in [\tfrac {97}{100},1), \]
and it follows that the mapping \(m\) is strictly increasing. Consequently, the inequality \(m(\tfrac {97}{100}){\gt}0\) implies \(m(r){\gt}0, \ r\in [\tfrac {97}{100},1)\) and the proof is complete.
4 Final comments
Theorem 2 and Corollary 1 imply the inequalities
\[ \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {r^4}{200}}{\lt}\tfrac {\pi }{2}\Big(1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}\Big) {\lt}\mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {1}{4}r^2}, \]
for \(r\in (0,1).\) Since
\[ \tfrac {\pi }{2}\Big(1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}\Big)=\tfrac {\pi }{2}+\tfrac {1}{2\mu _1}[_2F_1(1,\mu _1,\mu _1+1,r^2)-1] \]
it follows that the first inequality implies
\[ \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{3/4}{\lt}\tfrac {\pi }{2}+\tfrac {1}{2\mu _1}[_2F_1(1,\mu _1,\mu _1+1,r^2)-1], \ r\in (0,1), \]
which was conjectured in
[
3
]
. The second inequality has been established for the first time in
[
3
]
. The third inequality implies a conjecture from
[
1
]
. The authors of
[
4
]
proved that the following inequalities hold
\[ \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\alpha ^*r}{\lt}\mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\beta ^*r}, \ r\in (0,1), \]
with the best possible constants \(\alpha ^*=0\) and \(\beta ^*=1/4.\) Our results are improvements of these inequalities.
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