定理3
地图
$$t\mapsto I_{\mu}/I_{\nu}$$
(21)
是日志-凹入
\((-\infty,0)\)
和
\((0,\infty)\)
如果
\(0<\nu<\mu\).
不平等
保持正确
\((-\infty,0)\)
和
\((0,\infty)\),哪里
\(增量>0),\(\nu>-1\)
和
\(\nu-\delta+1\英寸(0,\infty)\).
证明
让\(\mu=\nu+\delta\)对一些人来说\(增量>0)和
$$R{\mu,\nu}(t):=\ln\biggl[\frac{I{\mu}(t)}{I{\nu}(c)}\biggr]=\ln\figl[\frac{t^{delta}\varGamma$$
(23)
利用引理1,我们有
$$R{\mu,\nu}^{\prime}(t)=\delta(2\nu+\delta)\frac{1}{t}\frac}\sum_{n=0}^{infty}a{n}t^{2n}}{\sum_{n=0.}^{infty}b_{n}t ^{2n}}$$
(24)
哪里
$$开始{对齐}&a{n}=\frac{1}{4^{n}}\frac}\varGamma$$
(25)
$$开始{aligned}&b_{n}=\frac{1}{4^{n}}\frac}\varGamma(nu+\delta+1)\varGamma(nu+1)\varGamma。\结束{对齐}$$
(26)
比率\(1/t\)正在严格减少\((0,\infty)\).由于系数的比率\(a{n}\)和\(b{n}\)由提供
$$r_{n}=\frac{a{n}}{b_{n{}=\frac{1}{2n+2\nu+\delta}\longrightarrow0\quad\text{作为}n\longrightarrow\infty$$
(27)
函数\(F(t):=(sum{n=0}^{infty}a{n}t^{2n})/正在上减少\((0,\infty)\).自\(R_{\mu,\nu}^{\prime}(t)\)是正递减函数的乘积\(1/t\)和\(F(t)\)和常数\(\增量(2\nu+\增量)\),\(R_{\mu,\nu}^{\prime}(t)\)正在严格减少\((0,\infty)\).因此\(R_{\mu,\nu}^{\prime}(t)>0\)、和\(R_{\mu,\nu}^{\prime\prime}(t)<0\),\(所有t>0\)。此外,\(R_{\mu,\nu}^{\prime}(t)=-\vert R_{\ mu,\nu}^{\ prime}(t)\vert\)哪里\(\vert R_{\mu,\nu}^{\prime}(t)\vert\)正在上增加\((-\infty,0)\),所以很明显\(R_{\mu,\nu}^{\prime}(t)\)在负实数域上减小。因此,\(映射到I_{mu}/I_{nu}\)启用了日志压缩\((-\infty,0)\)和\((0,\infty)\)。只有当\(\mu\neq\nu\).
为了证明定理的后一部分,我们声明以下属性:
$$\frac{\varGamma^{2}(n+\nu+1)}{\var伽玛(n+\nu+\delta+1)\varGamma$$
(28)
对于任何整数\(第1页),前提是\(n+\nu-\delta+1>0\),\(\nu>-1\),\(增量>0)和\(\nu+1>\delta\).何时\(n+\nu-\delta+1>0\),我们有\(\varGamma(n+\nu+\delta+1)\varGamma(n+\nu-\delta/1)>0\)因此,以下是正确的:
$$开始{对齐}和\frac{\varGamma^{2}(n+\nu+1)}{\var伽玛(n+\nu+\delta+1)\varGamma(n+-nu-\delta+1)}\\&\quad=\frac}\varGadma^{2{(n+/\nu)}{\ varGamma(n++\delta)\varGamma(n+\nu-\delta^{2}-\增量^{2}}\\&\quad>\frac{\varGamma^{2{(n+\nu)}{\varGamma(n+/\nu+\delta)\varGamma(n+\nu-\delta”)}。\结束{对齐}$$
(29)
通过归纳,方程式(29)暗示
$$开始{aligned}&\frac{\varGamma^{2}(n+\nu+1)}{\var伽玛(n+\nu+\delta+1)\varGamma(n+-nu-\delta/1)}\\&\quad>\frac}\varGarma^{2{$$
(30)
$$\开始{aligned}&\quad>\frac{\varGamma^{2}(\nu+1)}{\var伽玛(\nu+\delta+1)\varGamma(\nu-\delta/1)},\end{aligned}$$
(31)
对于任何整数\(第1页).根据引理1,的产品\(I_{\nu+\delta}\)和\(I_{\nu-\delta}\)成为
$$开始{对齐}I_{nu+\delta}(t)I_{nu-\delta{(t$$
(32)
$$开始{对齐}&=\sum_{n=0}^{\infty}\frac{\varGamma\结束{对齐}$$
(33)
$$开始{对齐}&=\sum_{n=0}^{\infty}\biggl(\frac{\varGamma^{2}(n+\nu+1)}{\varGamma(n+/\nu+\delta+1)\varGamma(n+-nu-\delta+1)}\bigr)\\&\quad{}\cdot\frac{\var伽马(2n+2\nu+1 \nu+1)}\分形{t^{2n+2\nu}}{2^{2n-2\nu}n!}。\结束{对齐}$$
(34)
通过应用(28),我们有
$$开始{对齐}I_{\nu+\delta}(t)I_{\nu-\delta}}\frac{\varGamma(2n+2\nu+1)}{\varGamma(n+2\nu+1)\varGamma^{2}(n+\nu/1)}\frac{t^{2n}}{n!}。\结束{对齐}$$
(35)
自
$$I_{\nu}^{2}(t)=t^{2\nu}\sum_{n=0}^{\infty}\frac{1}{2^{2n-2\nu{}\frac{\varGamma(2n+2\nu+1$$
(36)
结果是
$$\frac{I{nu}^{2}(t)}{I{nu+delta}$$
(37)
为了证明\(1),我们需要声明
$$g(\nu,\delta):=\frac{\varGamma^{2}(n+\nu+1)}{\varGamma(n+/\nu+\delta+1)\varGamma(n+\nu-\delta/1)}<1$$
(38)
通过出租
$$G(\nu,\delta):=\ln\biggl[\frac{\varGamma^{2}(n+\nu+1)}{\var伽玛(n+\nu+\delta+1)\varGamma(n+-nu-\delta/1)}\biggr]$$
(39)
拿\({\partial}/{\partial \delta}\)在\(G(\nu,\delta)\)产量
$$\frac{\partial}{\paratil\delta}G(\nu,\delta)=-\psi(n+\nu+\delta+1)+\psi(n+\nu-\delta/1)$$
(40)
哪里ψ表示digamma函数。它可以展开为Euler–Mascheroni常数之和[10],即。\(伽马{E})和谐波级数
$$开始{对齐}\frac{\partial}{\parial\delta}G(\nu,\delta)&=-\Biggl(-\gamma_{E}+\sum_{k=0}^{\infty}\Biggl(\frac}1}{k+1}-\frac[1}{k+n+\nu+\delta+1}\bigr)infty}\Biggl(\frac{1}{k+1}-\frac{1'{k+n+nu-\delta+1}\bigr)\biggr)\\&=-2\sum_{k=0}^{\infty{\biggl(\frac{\delta}{(k+n+\nu+1)^{2}-\delta^{2}}\biggr)<0,\end{对齐}$$
(41)
所以\(g(\nu,\delta)\)正在严格减少\(\增量\ in(0,\ infty)\)鉴于n个和ν.自\(g(\nu,\delta)\longrightarrow 1)作为\(\delta\longrightarrow 0\),\(g(\nu,\delta)<1),其中\(\增量\ in(0,\ infty)\)从不平等中(37)和(38),我们得出结论
$$开始{aligned}和I_{nu}^{2}(t)-I_{nu+r}}{varGamma^{2}(n+\nu+\delta+1)}\\&\qquad{}\cdot\biggl\varGamma(n+\nu+2\delta+1)}\biggr)\biggr]\\&&quad>0。\结束{对齐}$$
(42)
因此,\(1). □
不平等(42)等于中报告的Turán型不等式[8]. 这种不平等有很多种说法。Thiruvenkatachar等人最初使用系数比较法证明了不等式[9]. 后来,Joshi等人[6]和Baricz[三]利用产品扩展来证明属性,但它们的证明在细节上有所不同。不等式用于证明中发现的其他属性[11,12]. 我们的研究使用了Thiruvenkatachar论文中的级数展开的Cauchy积,但我们改进了该性质以覆盖非整数δ条件。
定理4
鉴于
\(\nu>-{1}/{2}\),\(t映射到t^{mu}I_{nu}(t)\)
严格来说是日志-凹入
\((0,\infty)\)
对于
$$\开始{aligned}&\textstyle\begin{cases}\mu>-\nu;&-1/2<\nu<1/2,\\mu>1/2;&\nu\ge 1/2,\end{cases}\displaystyle \\&\textstyle\begin{cases}\mu>-\nu;&-1/2<\nu<1/2,\\mu>1/2;&\nu\geq 1/2,\end{cases}\displaystyle\end{aligned}$$
严格来说是日志-凸的
\((0,\infty)\)
对于
\(\mu<-\nu\).
证明
根据(2),MBF-I可以根据\(\gamma _{\nu}(t^{2})\)作为
$$t^{\mu}I{\nu}(t)=\frac{t^{\ mu+\nu}\gamma_{\nu{(t^{2})}{2^{\nuneneneep \varGamma(\nu+1)}$$
(43)
然后是
$$\ln t^{\mu}I_{\nu}(t)=(\mu+\nu)\ln t+\ln\gamma _{\nu}\bigl(t^{2}\bigr)+\ln 2^{\nu}\varGamma(\nu+1)$$
(44)
关于吨描述如下:
$$\begin{aligned}和\frac{d}{dt}\bigl[\ln t^{mu}I{nu}(t)\bigr]=\frac{mu+nu}{t}+2t\frac}\gamma_{nu}^{prime}(t^{2})}{gamma_}\nu}$$
(45)
$$开始{aligned}和\frac{d^{2}}{dt^{2{}\bigl[\lnt^{mu}I{nu}(t)\bigr]\\&\quad=-\frac}\mu+\nu}{t^{2neneneep}+2\frac[\gamma_{nu}^{prime}\nu}^{\prime\prime}(t^{2})\gamma_{\nu}$$
(46)
$$\begin{aligned}&\quad=\Biggl\lbrace\sum_{n=0}^{\infty}\frac{\varGamma^{2}(\nu+1)\varGarma(2n+2\nu-1)}{4^{n}\varGamma(n+2\nu+1)(\varGamm(n+\nu+1))^{2{}\frac{t^{2n}}{n!}\\&\qquad{}\cdot\Biggl[\Biggl(压裂{2\nu^{2}-\nu-n}{1-2\nu-2n}\biggr)-\mu\biggr]\biggr\rbrace\cdot\bigl[t^{2}\gamma_{\nu}^{2{\bigl(t^{2\bigr)\bigr]^{-1}。\结束{对齐}$$
(47)
发件人(46),我们考虑这个术语
$$S_{n}:=\biggl(\frac{2\nu^{2}-\nu-n}{1-2\nu-2n}\biggr)-\mu$$
(48)
条件\(菜单>1/2)暗示\(1-2\nu-2n\le 1-2\nu<0\)对于\(n\le 0),所以我们有
$$\开始{aligned}&\inf\biggl\{\frac{2\nu^{2}-\nu-n}{1-2\nu-2n}\biggr\}=\textstyle\begin{cases}1/2;&-1/2<\nu<1/2,\\-\nu;&\nu\geq 1/2,\end{cases}\displaystyle\end{aligned}$$
(49)
$$\开始{aligned}&\sup\biggl\{\frac{2\nu^{2}-\nu-n}{1-2\nu-2n}\biggr\}=\textstyle\begin{cases}-\nu;&-1/2<\nu<1/2,\\1/2;&\数字1/2。\结束{cases}\displaystyle\end{aligned}$$
(50)
对于\(\mu>\sup\{(2\nu^{2}-\nu-n)/(1-2\nu-2n)\}\),这意味着\(S_{n}<0\),和中非正系数系列的乘积(47)和积极作用\([t\gamma_{nu}(t^{2})]^{-2}\)为非负。因此,\(({d^{2}}/{dt^{2{}})(lnt^{mu}I{nu}(t))<0\)这使得\(t^{\mu}I_{\nu}\)log-concave打开\((0,\infty)\)注意,除非\(\mu=\nu=1/2\),\(S_{n}<0\)对一些人来说\(n\in\mathbb{n}\)这意味着\(({d^{2}}/{dt^{2{}})(lnt^{mu}I{nu}(t))严格来说是否定的。因此,\(t映射到t^{mu}I_{nu}(t)\)严格为log-convave on\((0,\infty)\).
对于对数-凸函数\(\mu<\inf\{(2\nu^{2}-\nu-n)/(1-2\nu-2n)\}\),\(S_{n}>0\),这意味着在条件\(\nu>-{1}/{2}\),\(\mu>{1}/{2}\)因此,我们看到\(({d^{2}}/{dt^{2{}})(lnt^{mu}I{nu}(t))是负系数级数和非负函数的乘积\([t\gamma_{\nu}^{2}(t^{2{)]^{-2}\)因此,\(t^{\mu}I_{\nu}\)启用了日志压缩\((0,\infty)\). □
定理4给出了另一种证明映射的对数压缩性的方法\(u\mapsto\sqrt{u} 我_{\nu}(u)\)在里面\((0,\infty)\)已在中报告[13]. 什么时候?\(\mu+\nu=0\),术语\(-(\mu+\nu)/t^{2}\gamma_{\nu}^{2{(t^{2neneneep)\)消失了。然后\(D^{2}(\t^{mu}I_{nu}(t))对所有人都存在\(t>0),因此我们可以将对数凸性的域扩展到\(\mathbb{R}^{+}\)结果产生了另一种方法来证明\(t\mapsto\gamma_{nu}(t^{2})=2^{nu}\varGamma(\nu+1)t^{-\nu}I_{nu{(t)\)在\(\mathbb{R}^{+}\)由Neumann演示[三,5]. 在本文中,我们将对数压缩条件推广到\(单位:(-1/2,1/2))。根据(46), (49)和(50),人们可以看到\(t映射到t^{1/2}I_{nu}(t)\),\(t映射到t^{-\nu}I_{\nu}(t)\),\(菜单>1/2),似乎分别是对数凸性和对数凸性的最佳条件,形式如下\(t ^{\mu}I_{\nu}(t)\).
定理5
地图
\((x,y)\mapsto(xy)^{\mu}\gamma_{\nu}(xy)\)
是日志-凹进
\(\mathbb{R}^{+}\times\mathbb{R}^{+{)
前提是
\(\mu=\nu+{1}/{2}\),\(菜单>1/2).等效,
$$(x,y)\mapsto(xy)^{\nu/2+1/2}I_{\nu}(\sqrt{xy})$$
(51)
是日志-第一象限内的凹面,前提是
\(菜单>1/2).
证明
对于\(l(x,y)=-\ln[(xy)^{mu}\gamma_{nu}(xy,其导数表示如下:
$$\开始{aligned}&\frac{\partial l}{\partic x}=-\frac}\mu}{x} -年\压裂{\gamma{\nu}^{\prime}(xy)}{\gama{\nu{(xy)},\end{aligned}$$
(52)
$$\开始{aligned}&\frac{\partial^{2} 我}{\partial x^{2}}=\frac{\mu}{x^{2}}-y^{2{压裂{\gamma_{nu}^{\prime\prime}$$
(53)
$$\开始{aligned}&\frac{\partial^{2} 我}{\partial y^{2}}=\frac{\mu}{y^{2}}-x^{2{压裂{\gamma_{nu}^{\prime\prime}$$
(54)
$$\开始{aligned}&\frac{\partial^{2} 我}{\partial xy}=-\frac{\gamma_{\nu}^{\prime}(xy)}{\gama_{\nu}(xy)}-xy\frac}\gamma_{\nu}^{\ prime\prime{(xy)\gamma{\nu{(xy)-[\gamma_a{\nuneneneep ^{\prime}。\结束{对齐}$$
(55)
然后是的Hessian矩阵\(l(x,y)\)是
$$\开始{bmatrix}\frac{\mu}{x^{2}}-y^{2} 克_{nu}^{prime\prime}(xy)&-g{nu}^{prime}^{2} 克_{\nu}^{\prime\prime}(xy)\end{bmatrix}$$
(56)
哪里
$$\begin{aligned}&g_{\nu}(t)=\ln\gamma{\nu}(t),\\&g_{\nu}^{\prime}(t)=\gamma{\nu}^{\prime}(t)>0,\\&g_{\nu}^{\prime}(t)=\gamma{\nu}^{\prime}(t)/\gamma{\nu}(t)-\bigl(\gamma{\nu}^{\prime}(t)/\gamma _{\nu}(t)\bigr)^{2}<0。\结束{对齐}$$
二阶导数为负\(t\mapsto\gamma_{\nu}\)在中是log-concave\((0,\infty)\)前提是\(\nu>-1\).我们必须证明Hessian矩阵是正定的,方法是
$$\begin{aligned}E&=\begin{bmatrix}u\\v\end{bmatrix}^{T}\bbegin{bmatrix}\frac{\mu}{x^{2}}-y^{2} 克_{\nu}^{\prime\prime}(xy)&-g{\nu{^{\prime}^{2} 克_{\nu}^{\prime\prime}(xy)\end{bmatrix}\begin{bmatricx}u\\v\end{bmatrix}\\&=\biggl(\frac{u^{2} 年^{2} +伏^{2} x个^{2} }{x^{2} 年^{2} }\biggr)\mu-(uy+vx)^{2} 克_{\nu}^{\prime\prime}(xy)-2uvg{\nu}^{\trime}。\结束{对齐}$$
(57)
案例1:\(紫外线<0)很明显
$$\开始{aligned}E(u,v)&=\biggl(\frac{u^{2} 年^{2} +伏^{2} x个^{2} }{x^{2} 年^{2} }\biggr)q-(uy+vx)^{2}g{nu}^{prime\prime}。\结束{对齐}$$
(58)
案例2:\(紫外线>0).
$$\开始{aligned}E(u,v)&=\biggl(\frac{u^{2} 年^{2} +伏^{2} x个^{2} }{x^{2} 年^{2} }\biggr)\mu-(uy+vx)^{2} 克_{\nu}^{\prime\prime}(xy)-2uvg_{\nu{^{质数}(xy)\\&=\frac{(uy-vx)^{2}}{x^{2} 年^{2} }\mu-(uy-vx)^{2} 克_{\nu}^{\prime\prime}(xy)\\&\quad{}+\frac{2uv}{xy}\mu-4uvxyg{\nu}^{\ prime\prime}uv}{xy\gamma{nu}(xy(xy)-4x^{2} 年^{2} \gamma_{nu}^{prime\prime}(xy)^{2} 年^{2} \bigl[\gamma_{\nu}^{\prime}(xy)\bigr]^{2} -2倍\gamma_{\nu}^{\prime}(xy)\gamma_}\nu}(xy)\biger)\\&=\frac{uvH(xy$$
(59)
哪里
$$\begin{aligned}&H(xy)=\sum_{n=0}^{\infty}\frac{\varGamma^{2}(\nu+1)\varGamma(2n+2\nu-1)}{4^{n}\varGamm(n+2\nu+1)\varGamma^}(n+\nu+1$$
(60)
$$\begin{aligned}&Q(n,\nu,\mu)=2(n+\nu)\bigl(n+2n\nu+4\nu^{2}-2\较大)。\结束{对齐}$$
(61)
接下来,我们找到了隐含的条件\(Q(n,\nu,\mu)>0).如果\(n+\nu>0\)为所有人\(n \geq 0),我们有\(\nu>0\).通过考虑\(n+2n\nu+4^{2}-2>0\),我们有\(\nu<-1/2\)或\(菜单>1/2-n/4)。对于\(菜单>1/2-n/4),我们有\(第1/2页>第1/2页至第4页)为所有人\(n \geq 0)因此,\(Q(n,\nu,\mu)>0)什么时候\(第1/2页).
那么我们假设\(\mu=\nu+{1}/{2}\),\(菜单>1/2)根据定义\(-\ln(xy)^{\mu}\gamma_{\nu}(xy))是正定矩阵。映射\((x,y)\mapsto(xy)^{\mu}\gamma_{\nu}(xy)\)在以下条件下为对数曲线\(\mu=\nu+{1}/{2}\)哪里\(菜单>1/2)此外,
$$开始{对齐}(xy)^{\mu}\gamma_{\nu}(xy)&=2^{\nuneneneep \varGamma(\nu+1)(xy$$
(62)
意味着\((x,y)\mapsto(xy)^{\nu/2+1/2}I_{\nu}(\sqrt{xy})\)为原木凹面。□
定理5可以应用于统计学,因为它证明了Kibble双变量γ分布的对数似然函数的对数相合性[14,15,16]其概率密度函数为
$$\begin{aligned}f(x,y|\nu,\lambda{1},\lambda{2},\rho)&=\frac{(\lambda{1}\lambda{2})^{\nu}{(1-\rho)\varGamma(\nu)}\biggl(\frac{xy}_{1} x个+\λ_{2} 年}{1-\rho}\biggr)I_{\nu-1}\bigl(\frac{2\sqrt{\rho\lambda_{1}\lambda_{2} xy公司}}{1-\rho}\biggr)。\结束{对齐}$$
(63)
在自由度ν和形状参数ρ给出了分布的极大似然估计,并将其归结为凸优化问题。证明详情如下:
$$开始{对齐}f(x,y|\nu,\lambda_{1},\λ_2},\ρ)&=\frac{\varGamma(\nu)}\\&\quad{}\cdot\exp\biggl(-\frac{\lambda_{1} x个+\λ_{2} 年}{1-\rho}\biggr)I_{\nu-1}\bigl(\frac{2\sqrt{\rho\lambda_{1}\lambda_{2} xy公司}}{1-\rho}\biggr)\end{对齐}$$
(64)
$$\begin{aligned}&=\biggl[\frac{(xy/\rho)^{\frac}\nu-1}{2}}{(1-\rho_{1} x个+\λ_{2} 年}{1-\rho}\biggr)\biggr]\\&\quad{}\cdot\biggl[(\lambda_{1}\lambda{2})_{2} xy公司}}{1-\rho}\biggr)\biggr]。\结束{对齐}$$
(65)
让\(κ=2\sqrt{\rho-xy}/(1-\rho)\)我们证明了在域的仿射变换下,对数压缩是守恒的。因此,以下情况是等效的:
$$开始{aligned}&(\lambda_{1},\lambda _{2})\mapsto(\lampda_{1\lambda_2})^{\frac{\nu+1}{2}}I{\nu-1}$$
(66)
$$\begin{aligned}和\bigl(\kappa^{2}\lambda_{1},\lambda{2}\bigr)\mapsto\bigle(\kappa^}\lampda{1}\lambeda{2{\biger)^{\frac{\nu+1}{2}}I{\nu-1}}$$
(67)
$$开始{对齐}&(\lambda_{1},\lambda _{2})\mapsto\bigl(\kappa^{2}\lambada_{1{,\lambda_2}\bigr)\\&\hphantom{(\lampda_{1',\lampda _{2])}\mapsto\ bigl{nu-1}(\kappa\sqrt{\lambda{1}\lambda{2}})\quad\text{是log-concave},\end{aligned}$$
(68)
$$\begin{aligned}&(\lambda_{1},\lambda{2})\mapsto(\lampda_{1\lambda_2})^{\frac{\nu+1}{2}}I{\nu-1}(\kappa\sqrt{\lambada_{1{\lampda{2{})\squad\text{islog-concave}。\结束{对齐}$$
(69)
通过考虑以下领域\(λ{1})和\(λ{2}),\(f(\lambda _{1},\lambda _{2})\)是由3个组件组成的产品:
$$\开始{对齐}和(xy)^{\frac{\nu-1}{2}}/\bigl[\rho^{\frac{\nu-1-}{2{}}(1-\rho)\varGamma(\nu)\bigr]\quad\text{(常量)},\end{aligned}$$
(70)
$$\开始{aligned}&\exp\bigl[-(\lambda_{1} x个+\λ_{2} 年)/(1-\rho)\bigr]\quad\text{(对数线性函数)},\end{aligned}$$
(71)
$$\begin{aligned}&(\lambda_{1}\lambda{2})^{\frac{\nu+1}{2}}I_{\nu-1}(\kappa\sqrt{\lambd_{1}\ lambda_2}})\quad\text{(log-concave函数)}。\结束{对齐}$$
(72)
因此,地图\((\lambda{1},\lambda{2})\mapstof)原木凹入\(\mathbb{R}^{+}\times\mathbb{R}^{+{).
定理6
如果
\(-1/2<\nu<0\),\(I_{nu}\)
是日志-凸的
\((0,\infty)\)
如果
\(菜单>0),\(I_{nu}\)
是日志-凹函数开
$$\bigle(0,\bigl(16\nu^{4}-16\努^{3}-24\nu^{2}+4\nu+5\biger)^{1/2}/2\biger$$
2007年,诺依曼推测[4]第一类修正贝塞尔函数在\((0,\infty)\)据我们所知,这似乎仍是一个悬而未决的问题[13]. 在我们的研究中,该函数仅在由参数确定的特定区间上是对数压缩的ν.
证明
的二阶导数\(在I_{nu}\中)关于吨是\(-\t^{2}+2\gamma{2})/\gamma_{nu}(t^{2{)+4t^{2](\gamma{nu}^{prime\prime}{2})\)。该术语可以简化为\(\varOmega(t)\)/\([t\gamma_{nu}(t^{2})]^{2{)哪里\(\varOmega(t)=-\nu\gamma_{nu}(t^{2})\gamma_2}_{\nu}^{\素数}(t^{2})\gamma_{\nu}^{\prime}(t ^{2{)\).的二阶导数的符号\(在I_{nu}\中)取决于\(\varOmega(t)\)因为分母是非负的。通过采用(8),\(\varOmega(t)\)以系列的形式重写
$$\varOmega(t)=\sum_{n=0}^{\infty}a{n}\frac{t^{2n}}{n!}$$
(73)
哪里
$$a_{n}=\frac{\varGamma^{2}(\nu+1)\varGarma(2n+2\nu-1)}{4^{n}\varGamma(n+2\nu+1)\varGamma^}(n+\nu+1$$
(74)
然后我们考虑
$$\varOmega(t)=a_{0}+\sum_{n=1}^{\infty}a_{n}\frac{t^{2n}}{n!}=-\nu+\sum_{n=1}^{infty{a_{n}\frac{t^[2n}}}{n!}$$
(75)
条款\(\varGamma^{2}(\nu+1)\),\(\varGamma(2n+2\nu-1)),\(\varGamma(n+2\nu+1)),\(\varGamma(n+\nu+1)\)、和\(n+\nu\)属于\({a{n})是积极的\(\nu>-1/2\)和\(第1页)因此\({a{n})仅由确定\(n-2\nu^{2}+\nu\).给定\(-1/2<\nu<0\),\(n-2\nu^{2}+\nu\)是积极的,所以我们得出结论\(a{n}\)对任何情况都是积极的\(n \geq 0).任意\(第0页),\(\varOmega(t)\)是一个正函数,因此\(ln I_{\nu}\)是积极的。这证明了\(I_{nu}\)在第一个条件下。对于以下情况\(\nu>0\),\(a{n}<0\)等于\(n-2\nu^{2}+\nu<0\)和\(n<2个^{2}-\数字\),因此我们可以推断存在\(n_{0}\)这样的话\(a{m}\leq0)对于任何\(m\leqn_{0}\)和\(a_{n}>0\)对于任何\(n>n{0}\).通过引理2,我们的结论是\(t>0)这样的话
$$开始{对齐}\varOmega(t)&=\sum_{n=0}^{\infty}\frac{\varGamma^{2}(nu+1)\varGamma(2n+2\nu-1)}{4^{n}\varGamm(n+2\nu+1)\valGamma^}(n+\nu+1)}\\&\quad{}\cdot\bigl[2(n+/nu)\bigl(n-\nu(2\nu-1)\bigr)\bigr]\frac{t^{2n}}{n!}\\&=\sum_{n=0}^{infty}\frac}\varGamma^{2}(nu+1)\varGamma(2n+2\nu+1)}{4^{n}\varGamma(n+2\nu+1)\varGamma^{2}(n+\nu+1)}\\&\quad{}\cdot\biggl(\frac{1}{2}-\压裂{(\nu-\frac{1}{2})(\nu+\frac}{2{)}{n+\nu-\压裂{1}}}\biggr)\压裂{t^{2n}}{n!}。\结束{对齐}$$
(76)
按引理1,我们有
$$开始{对齐}\varOmega(t)&=\frac{1}{2}\gamma_{nu}\bigl=0}^{\infty}\frac{\varGamma(\nu+1)\varGamma(\nu+2)\varGamma(2n+2\nu+2)}{4^{n}\varGamm(n+2)\varGamma(n+\nu+2)\varGamma(n+\nu+1)}\\&\quad{}\cdot\frac{。\结束{对齐}$$
(77)
自\(适用于所有n in mathbb{无}_{0}\)和\(\nu>{1}/{2}\),我们有
$$\开始{对齐}&(n+\nu)^{2}+(\nu+2)(n+\nu)+(\nu+1)>(n+\tu)^{2}-\压裂{1}{4},结束{对齐}$$
(78)
$$开始{aligned}&\frac{(n+\nu+1)(n+2\nu+1。\结束{对齐}$$
(79)
因此,我们有了不等式
$$开始{aligned}\varOmega(t)&<\frac{1}{2}\gamma_{nu}\bigl(t^{2}\bigr)\gamma_2}\bigle(tqu{2}\figr)-\Biggl\lbrace\frac{n=0}^{infty}\frac{\varGamma(nu+1)\varGarma(nu+2)}{4^{n}\varGamma(n+2\nu+2){\varGamma(n+\nu+1)\varGamma(n+\nu+2)}\frac{t^{2n}}{n!}\Biggr]\Biggr\rbrace\&=\fracc{1}{2}\gamma_{nu}\bigl c{1}{2})}{(\nu+1)}\frac{\gamma{\nu+1}(t^{2})}{\gama{\nu}\bigr)\gamma_{\nu}\bigl(t^{2}\biger)\biggl[1-2\biggl。\结束{对齐}$$
(80)
由于Kokologiannaki的工作[17],我们可以利用不等式
$$-\frac{\nu+1}{t^{2}}+\sqrt{\frac}$$
(81)
不平等(80)成为
$$\varOmega(t)<\压裂{1}{2}\gamma_{nu}\bigl{F}(F)_{\nu}(t)$$
(82)
哪里
$$\开始{aligned}\mathcal{F}(F)_{纽}(t)&=1-2\biggl(纽-\frac{1}{2}\biggr)\biggl(纽+\frac}1}{2\biggr}\\&\quad{}-2\biggl(\nu-\frac{1}{2}\biggr)\biggl(\nu+\frac}{1}}{2{biggr(\nu+1)^{2}}{t^{4}}+\frac{1}{t^}}}}。\结束{对齐}$$
(83)
自\(\gamma_{\nu}>0\),函数\(\varOmega(t)<0)当且仅当\(\mathcal{F}(\nu,t)\leq 0\)因此,在区间上保证了第一类修正贝塞尔函数的对数压缩性\((0,({16\nu^{4}-16\努^{3}-24\nu^{2}+4\nu+5})^{1/2}/2)\). □
根据证明,MBF-I不应在\(\mathbb{R}^{+}\)正如人们猜测的那样。然而,我们怀疑确保MBF-I的对数压缩性的区间的实际上界至少应该是\(2\nu^{2}\).