Theorem 3.1
Let
\(a, b\geq0\). Then the double inequality
$$ \frac{1}{\sqrt{x+a}}< \sqrt{\frac{2}{\pi}}e^{x}K_{0}(x)< \frac{1}{\sqrt{x+b}} $$
holds for all
\(x>0\)
if and only if
\(a\geq1/4\)
and
\(b=0\).
Proof
Let \(x>0\), \(f(x)\) be defined by Lemma 2.2, and \(f_{1}(x)\), \(f_{2}(x)\) and \(F(x)\) be respectively defined by
$$ f_{1}(x)=\frac{\pi}{2}-xe^{2x}K_{0}^{2}(x), \qquad f_{2}(x)=e^{2x}K_{0}^{2}(x) $$
(3.1)
and
$$ F(x)=\frac{\frac{\pi}{2}-xe^{2x}K_{0}^{2}(x)}{e^{2x}K_{0}^{2}(x)}=\frac {f_{1}(x)}{f_{2}(x)}. $$
(3.2)
Then from (1.5), (1.7) and (3.1) we clearly see that
$$\begin{aligned}& \lim_{x\rightarrow\infty}f_{1}(x)=\lim_{x\rightarrow\infty}f_{2}(x)=0, \end{aligned}$$
(3.3)
$$\begin{aligned}& \frac{f^{\prime}_{1}(x)}{f^{\prime}_{2}(x)}=\frac {-e^{2x}K^{2}_{0}(x)+2xe^{2x}K_{0}(x) [K_{1}(x)-K_{0}(x) ]}{ -2e^{2x}K_{0}(x) [K_{1}(x)-K_{0}(x) ]}=f(x). \end{aligned}$$
(3.4)
It follows from (3.2)-(3.4), Lemmas 2.1 and 2.2 together with L’Hôpital’s rule that the function \(F(x)\) is strictly increasing on \((0, \infty)\) and
$$ \lim_{x\rightarrow\infty}F(x)=\frac{1}{4}. $$
(3.5)
Note that (1.3) and (3.2) lead to
$$ \lim_{x\rightarrow0}F(x)=\lim_{x\rightarrow0} \biggl[\frac{\pi }{2e^{2x}K^{2}_{0}(x)}-x \biggr]=0. $$
(3.6)
Therefore, Theorem 3.1 follows easily from (3.2), (3.5), (3.6) and the monotonicity of \(F(x)\). □
Remark 3.2
From Lemma 2.2 we clearly see that the double inequality
$$ p< \frac{K_{0}(x)}{2 [K_{1}(x)-K_{0}(x) ]}-x< q $$
holds for all \(x>0\) if and only if \(p\leq0\) and \(q\geq1/4\).
From (1.7) and Remark 3.2 we get Corollary 3.3 immediately.
Corollary 3.3
Let
\(p, q\geq0\). Then the double inequalities
$$ 1+\frac{1}{2(x+p)}< \frac{K_{1}(x)}{K_{0}(x)}< 1+\frac{1}{2(x+q)} $$
and
$$ \bigl[\log \bigl(e^{x}\sqrt{x+p} \bigr) \bigr]^{\prime}< -\bigl[\log K_{0}(x)\bigr]^{\prime}< \bigl[ \log \bigl(e^{x}\sqrt{x+q} \bigr) \bigr]^{\prime} $$
(3.7)
hold for all
\(x>0\)
if and only if
\(p\geq1/4\)
and
\(q=0\).
Remark 3.4
Let \(p\geq0\). Then from inequality (3.7) we know that the function \(x\mapsto\sqrt{x+p}e^{x}K_{0}(x)\) is strictly increasing (decreasing) on \((0, \infty)\) if and only if \(p=0\) (\(p\geq1/4\)). We clearly see that the bounds for \(K_{1}(x)/K_{0}(x)\) given in Corollary 3.3 are better than the bounds given in (1.10) for \(\nu=0\).
From (1.3), (1.5) and Remark 3.4 we get Corollary 3.5 immediately.
Corollary 3.5
The double inequality
$$ \sqrt{\frac{\pi}{2}}=\lim_{x\rightarrow\infty} \bigl[\sqrt {x+p}e^{x}K_{0}(x) \bigr] < \bigl[\sqrt{x+p}e^{x}K_{0}(x) \bigr]< \lim_{x\rightarrow0} \bigl[\sqrt {x+p}e^{x}K_{0}(x) \bigr]=\infty $$
(3.8)
holds for all
\(x>0\)
if
\(p\geq1/4\), and inequality (3.8) is reversed if
\(p=0\).
Remark 3.4 also leads to Corollary 3.6.
Corollary 3.6
Let
\(p, q\geq0\). Then the double inequality
$$ \sqrt{\frac{y+p}{x+p}}e^{y-x}< \frac{K_{0}(x)}{K_{0}(y)}< \sqrt{ \frac {y+q}{x+q}}e^{y-x} $$
holds for all
\(0< x< y\)
if and only if
\(p\geq1/4\)
and
\(q=0\).
Remark 3.7
We clearly see that the lower bound for \(K_{0}(x)/K_{0}(y)\) in Corollary 3.6 is better than the bounds given in (1.11) and (1.12) for \(\nu=0\).
Remark 3.8
From the inequality
$$ \frac{\Gamma (x+\frac{1}{2} )}{\Gamma(x+1)}< \frac{1}{\sqrt {x+\frac{1}{4}}} $$
given in [10], (2.20), and the fact that
$$ \frac{1}{\sqrt{x+\frac{1}{4}}}>\frac{8\sqrt{x}}{8x+1} $$
for all \(x>0\) we clearly see that the lower bound given in Theorem 3.1 for \(\sqrt{2/\pi}e^{x}K_{0}(x)\) is better than that given in (1.8) and (1.9). But the upper bound given in Theorem 3.1 is weaker than that given in (1.8).
Remark 3.9
From Theorem 3.1 and Corollary 3.3 we clearly see that there exist \(\theta_{1}=\theta_{1}(x)\in(0, 1/4)\) and \(\theta_{2}=\theta_{2}(x)\in (0, 1/4)\) such that
$$ K_{0}(x)=\sqrt{\frac{\pi}{2(x+\theta_{1})}}e^{-x},\qquad K_{1}(x)= \biggl[1+\frac{1}{2(x+\theta_{2})} \biggr]\sqrt{ \frac{\pi}{2(x+\theta_{1})}}e^{-x} $$
for all \(x>0\).
Theorem 3.10
Let
\(x>0\), \(n\in\mathbb{N}\), \(R_{n}(x)=K_{n+1}(x)/K_{n}(x)\), \(u_{0}(x)=1+1/(2x)\), \(v_{0}(x)=1+1/(2x+1/2)\), and
\(u_{n}(x)\)
and
\(v_{n}(x)\)
be defined by
$$ u_{n}(x)=\frac{1}{v_{n-1}(x)}+\frac{2n}{x},\qquad v_{n}(x)=\frac {1}{u_{n-1}(x)}+\frac{2n}{x}\quad (n\geq1). $$
(3.9)
Then the double inequality
$$ v_{n}(x)< R_{n}(x)=\frac{K_{n+1}(x)}{K_{n}(x)}< u_{n}(x) $$
(3.10)
holds for all
\(x>0\)
and
\(n\in\mathbb{N}\).
Proof
We use mathematical induction to prove inequality (3.10). From Corollary 3.3 we clearly see that inequality (3.10) holds for all \(x>0\) and \(n=0\).
Suppose that inequality (3.10) holds for \(n=k-1\) (\(k\geq1\)), that is,
$$ v_{k-1}(x)< R_{k-1}(x)< u_{k-1}(x). $$
(3.11)
Then it follows from (3.9) and (3.11) together with the formula
$$ \frac{K^{\prime}_{\nu}(x)}{K_{\nu}(x)}=-\frac{K_{\nu-1}(x)}{K_{\nu }(x)}-\frac{\nu}{x}=-\frac{K_{\nu+1}(x)}{K_{\nu}(x)}+ \frac{\nu}{x} $$
given in [11] that
$$\begin{aligned}& R_{k}(x)=\frac{1}{R_{k-1}(x)}+\frac{2k}{x}, \\& v_{k}(x)=\frac{1}{u_{k-1}(x)}+\frac{2k}{x}< R_{k}(x)< \frac {1}{v_{k-1}(x)}+\frac{2k}{x}=u_{k}(x). \end{aligned}$$
(3.12)
Inequality (3.12) implies that inequality (3.10) holds for \(n=k\), and the proof of Theorem 3.10 is completed. □
Remark 3.11
Let \(n=1, 2, 3\). Then Theorem 3.10 leads to
$$\begin{aligned}& \frac{2(x+1)^{2}}{x(2x+1)}< \frac{K_{2}(x)}{K_{1}(x)}< \frac {4x^{2}+9x+6}{x(4x+3)}, \\& \frac{4x^{3}+19x^{2}+36x+24}{x (4x^{2}+9x+6 )}< \frac {K_{3}(x)}{K_{2}(x)}< \frac{2x^{3}+9x^{2}+16x+8}{2x(x+1)^{2}}, \\& \frac{2 (x^{4}+8x^{3}+28x^{2}+48x+24 )}{x (2x^{3}+9x^{2}+16x+8 )}< \frac{K_{4}(x)}{K_{3}(x)} < \frac{4x^{4}+33x^{3}+120x^{2}+216x+144}{x (4x^{3}+19x^{2}+36x+24 )} \end{aligned}$$
for all \(x>0\).
Remark 3.12
It is not difficult to verify that
$$\begin{aligned}& \frac{2(x+1)^{2}}{x(2x+1)}>\frac{1+\sqrt{x^{2}+1}}{x}, \qquad \frac {4x^{2}+9x+6}{x(4x+3)}< \frac{\frac{3}{2}+\sqrt{x^{2}+\frac{9}{4}}}{x}, \\& \frac{4x^{3}+19x^{2}+36x+24}{x (4x^{2}+9x+6 )}>\frac {2+\sqrt{x^{2}+4}}{x},\qquad \frac{2x^{3}+9x^{2}+16x+8}{2x(x+1)^{2}}< \frac{\frac {5}{2}+\sqrt{x^{2}+\frac{25}{4}}}{x}, \\& \frac{2 (x^{4}+8x^{3}+28x^{2}+48x+24 )}{x (2x^{3}+9x^{2}+16x+8 )}>\frac{3+\sqrt{x^{2}+9}}{x}, \\& \frac{4x^{4}+33x^{3}+120x^{2}+216x+144}{x (4x^{3}+19x^{2}+36x+24 )}< \frac{\frac{7}{2}+\sqrt{x^{2}+\frac{49}{4}}}{x} \end{aligned}$$
for \(x>0\). Therefore, the bounds given in Remark 3.11 are better than the bounds given in (1.10) for \(\nu=1, 2, 3\).