Miyakoda, Tsuyako

Kyoto University Research Information Repository

Title$n$-th Derivatives of Some Functions in terms of N-FractionalCalculus (Study on Non-Analytic and Univalent Functions andApplications)Author(s)Miyakoda, TsuyakoCitation数理解析研究所講究録 (2009), 1626: 44-55Issue Date2009-01URL http://hdl.handle.net/2433/140321RightType Departmental Bulletin PaperTextversionpublisherKyoto University$n$-th Derivatives of Some Functions in terms ofN-Fractional CalculusTsuyako MiyakodaDepartment of Information and Physical ScienceGraduate School of Information Science and TechnologyOsaka University, Suita 565-0871, Osaka, JAPANAbstractIn this article, N-fractional calculus and $n(\in Z^{+})$-th derivatives of functions$f(z)= \frac{1}{(\sqrt{z-b}-c)^{2}-d}$ $((\sqrt{z-b}-c)^{2}-d\neq 0)$are discussed. That is, n-th derivatives of the function,$(f(z))_{n}=(-1)^{n}(z-b)^{-1-n} \sum^{\infty}\frac{[1]_{m}[2+2m]_{k}[zk+1+m]_{n}}{m!k!}S^{k}T^{m}$$m,\succ-0$where$S= \frac{c}{\sqrt{z-b}}$ , $T= \frac{d}{z-b}$ $(|S|<1, |T|<1)$ .is reported for example.1 Definition of N-Fractional CalculusIn order to treat the derivatives of arbitrary order, we descrive the $d\Phi i\dot{u}tion$of fractional calculus and some basical theorems and identities.(I) Definition. (by K. Nishimoto, [1] Vol. 1)Let $D=\{D_{-}, D_{+}\},$ $C=\{C_{-}, C_{+}\},$ $C_{-}$ be a curve along the cut joinin$g$two points $z$ and $\infty+$ iIm(z), $c_{+}$ be a curve along the cut joinin$g$ twopoints $z$ and $\infty+iIm(z),$ $D-$ be a domain surrounded by $C_{-}$ , $D_{+}$ be adomain surrounded by $c_{+}$ (Here $D$ contains the points over the curve $C$ ).1626 2009 44-55 44Moreover $f$ let $f=f(z)$ be a regular function in $D(z\in D)$ ,$f_{\nu}$ $=$ $(f)_{\nu}=c(f)_{\nu}$$=$ $\frac{\Gamma(\nu+1)}{2\pi i}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^{\nu+1}}$ $(\nu\not\in Z^{-})$ , (1)$(f)_{-m}= \lim_{\nuarrow-m}(f)_{\nu}(m\in Z^{+})$ , (2)where$-\pi\leq arg(\zeta-z)\leq\pi$ for $C_{-}$ , $0\leq arg(\zeta-z)\leq 2\pi$ for $c_{+}$ ,$\zeta\neq z$ , $z\in C$, $\nu\in R$ , $\Gamma_{i}$ Gamma fundion,then $(f)_{\nu}$ is the hactional differintegation of arbitrary order $\nu$ (derivativesof order $\nu$ for $\nu>0$ , and integrals of order $-\nu$ for $\nu<0$ ), with respect to$z$ , of the function $f$ , if 1 $(f)_{\nu}|<\infty$ .(II) On the &actional calculus operator $N^{\nu}$ , some fundamentaJ proper-ties have reported. ([3], [5])Theorem A. Let fractional calculus operator (Nishimoto’s Operator)$N^{\nu}$ be$N^{\nu}=( \frac{\Gamma(\nu+1)}{2\pi i}\int_{C}\frac{d\zeta}{(\zeta-z)^{l/+1}})$ $(\nu\not\in Z^{-})$ , (Refer to[l]) (3)with$N^{-m}= \lim_{\nuarrow-m}N’’$ $(m\in Z^{+})$ , (4)and define the binary operation $\circ$ as$N^{\beta}oN^{\alpha}f=N^{\beta}N^{\alpha}f=N^{\beta}(N^{\alpha}f)(\alpha,\beta\in R)$, (5)then the set$\{N^{\nu}\}=\{N^{\nu}|\nu\in R\}$ (6)is an Abelian product group (having continuous index $\nu$ ) which has theinverse transform operator $(N^{\nu})^{-1}=N^{-\nu}$ to the fractional calculus operator$N^{\nu}$ , for the function $f$ such that $f\in F=\{f;0\neq|f_{\nu}|\leq\infty, \nu\in R\}$ , where$f=f(z)$ and $z\in C$ . $($ vis. $-\infty<\nu<\infty)$ .(For our convenience, we call $N^{\beta}oN^{\alpha}$ as product of $N^{\beta}$ and $N^{\alpha}$ . )Theorem B. Mractional calculus operator group $\{N^{\nu}\}$ is an Actionproduct group which has continuous index $\nu$ for the set of $F$ .45Theorem C. Let$S:=\{\pm N^{\nu}\}\cup\{0\}=\{N^{\nu}\}\cup\{-N^{\nu}\}\cup\{0\}(\nu\in R)$ . (7)Then the set $S$ is a commutative ring for the function $f\in F$ , when theidentity$N^{\alpha}+N^{\beta}=N^{\gamma}$ $(N^{\alpha}, N^{\beta},N^{\gamma}\in S)$ (8)holds. ([5])(III) We have following results for some elementary functions. ([1])(i)$((z-c)^{\beta})_{\alpha}=e^{-i\pi\alpha} \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(z-c)^{\beta-\alpha}$ $(| \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}|<\infty)$(ti)$(log(z-c))_{a}=-e^{-i\pi\alpha}\Gamma(\alpha)(z-c)^{-\alpha}$ $(|\Gamma(\alpha)|<\infty)$(m)$((z-c)^{-\alpha})_{-\alpha}=-e^{1\pi\alpha} \frac{1}{\Gamma(\alpha)}\log(z-c)$ , $(|\Gamma(\alpha)|<\infty)$where $z-c\neq 0$ in (i), and $z-c\neq 0,1$ in (ii) and (iii),(iv)$(u \cdot v)_{\alpha}:=\sum_{k=0}^{\infty}\frac{\Gamma(\alpha+1)}{k!\Gamma(\alpha+1-k)}u_{\alpha-k}v_{k}$ . $(u=u(z),v=v(z))$2 Preliminary(I) The following theorem is reported by K. Nishimoto [12].Theorem D. We have(i)$(((z-b)^{\beta}-c)^{\alpha})_{\gamma}=e^{-i\pi\gamma}(z-b)^{\alpha\beta-\gamma} \sum_{k\triangleleft}^{\infty}\frac{[-\alpha]_{k}\Gamma(\beta k-\alpha\beta+\gamma)}{k!\Gamma(\beta k-\alpha\beta)}(\frac{c}{(z-b)^{\beta}})^{k}$(1)$(| \frac{\Gamma(\beta k-\alpha\beta+\gamma)}{\Gamma(\beta k-a\beta)}|<\infty)$ ,and46(ii)$(((z-b)^{\beta}-c.)^{\alpha})_{n}=(-1)^{n}(z-b)^{\alpha\beta-n} \sum_{k=0}^{\infty}\frac{[-\alpha]_{k}[\beta k-\alpha\beta]_{n}}{k!}(\frac{c}{(z-b)^{\beta}})^{k}$(2)$(n \in Z_{0}^{+}, |\frac{c}{(z-b)^{\beta}}|<1)$ ,where$[\lambda]_{k}=\lambda(\lambda+1)\cdots(\lambda+k-1)=\Gamma(\lambda+k)/\Gamma(\lambda)$ with $[\lambda]_{0}=1$ ,(Pochhammer’s Notation).(II) The following theorem is reported by K. Nishimoto already[13].Theorem E. We have(i)$((((z-b)^{\beta}-c)^{\alpha})-d)^{\delta})_{\gamma}=e^{-i\pi\gamma}(z-b)^{\alpha\beta\delta-\gamma}$$\cross\sum_{m,k=0}^{\infty}\frac{[-\delta]_{m}[-\alpha(\delta-m)]_{k}\Gamma(\beta k-\alpha\beta(\delta-m)+\gamma)}{m!k1\Gamma(\beta k-\alpha\beta(\delta-m))}(\frac{c}{(z-b)^{\beta}})^{k}(\frac{d}{(z-b)^{\alpha\beta}})^{m}$ ,(3)$(| \frac{\Gamma(\beta k-\alpha\beta(\delta-m)+\gamma)}{\Gamma(\beta k-\alpha\beta(\delta-m))}|<\infty)$ ,and(ii)$((((z-b)^{\beta}-c)^{a})-d)^{\delta})_{n}=(-1)^{n}(z-b)^{\alpha\beta\delta-n}$$\cross\sum_{m_{l}k=0}^{\infty}\frac{[-\delta]_{m}[-\alpha(\delta-m)]_{k}[\beta k-\alpha\beta(\delta-m)]_{n}}{m1k!}(\frac{c}{(z-b)^{\beta}})^{k}(\frac{d}{(z-b)^{\phi}})^{m}$ ,(4)$(n\in Z_{0}^{+})$where$((z-b)^{\beta}-c)^{\alpha}-d\neq 0$ , $| \frac{c}{(z-b)^{\beta}}|<1$ , $| \frac{d}{(z-b)^{\alpha}\beta}|<1)$ .We apply these theorems to obtain some theorems for the function$m^{1}(z--c)^{2}-d$ .473 N-Fractional Calculus of Functions $\frac{1}{(\sqrt{z-}-c)^{2}-d}$Theorem 1. We have(i)$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{\gamma}=e^{-i\pi\gamma}(z-b)^{-1-\gamma}$$x\sum_{m,k=0}^{\infty}\frac{[1]_{m}[2+2m]_{k}\Gamma(\frac{k}{+2}+1+m+\gamma)}{m!k!\Gamma(zk1+m)}(\frac{c}{\sqrt{z-b}})^{k}(\frac{d}{z-b})^{m}(1)$$(| \Gamma(\frac{k}{2}+1+m+\gamma)|<\infty)$and(li)$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{n}=(-1)^{n}(z-b)^{-1-n}$$x\sum^{\infty}\frac{[1]_{m}[2+2m]_{k}[2k+1+m]_{n}}{m1k!}(\frac{c}{\sqrt{z-b}})^{k}(\frac{d}{z-b})^{m}$ (2)$m.k=0$$(n\in Z_{0}^{+})$where $(\sqrt{z-b}-c)^{2}-d\neq 0$ , $| \frac{c}{\sqrt{z-b}}|<1$ , $| \frac{d}{z-b}|<1$ .and $\sum_{m_{r}k=0}$ $:= \sum_{m=0}^{\infty}\sum_{k=0}^{\infty}$ .Proof of (i). We operate $N^{\gamma}$ to the both sides of following relation,$\frac{1}{(\sqrt{z-b}-c)^{2}-d}=(((z-b)^{1/2}-c)^{2}-d)^{-1}$ (3)and by setting $\alpha=2,$ $\beta=1/2,$ $\delta=-1$ in Theorem $E(i)$ we obtain (1),under the conditions stated before.Proof of (ii). We have the result by setting $\gamma=n$ in the equation (1).Furthermore by setting $\alpha=0$ in Theorem 1, we have the followingcorolary immediately.Corollary 1. We have48(i)$( \frac{1}{z-2c\sqrt{z-b}-b+c^{2}})_{\gamma}=e^{-i\pi\gamma}(z-b)^{-1-\gamma}\sum_{k=0}^{\infty}\frac{[2]_{k}\Gamma(kz+1+\gamma)}{k!\Gamma(\frac{k}{2}+1)}(\frac{c}{\sqrt{z-b}})^{k}$(4)$(| \Gamma(\frac{k}{2}+1+\gamma)|\leq\infty)$and(ii)$( \frac{1}{z-2c\sqrt{z-b}-b+c^{2}})_{n}=(-1)^{n}(z-b)^{-1-n}\sum_{-}^{\infty}\frac{[2]_{k}[zk+1]}{k!}(\frac{c}{\sqrt{z-b}})^{k}\succ 0$(5)$(n\in Z_{0}^{+})$ ,where $\sqrt{z-b}-c\neq 0$ , $| \frac{c}{\sqrt{z-b}}|\leq 1$ .Theorem 2. We have the following identities,(i)$(z-b)^{rightarrow 1/2} \sum_{m,k=0}^{\infty}\frac{[1]_{m}[2+2m]_{k}\Gamma(\frac{k}{+2}+1+m+\gamma)}{m!k!\Gamma(\mathfrak{T}k1+m)}(\frac{c}{\sqrt{z-b}})^{k}(\frac{d}{z-b})^{m}$$= \frac{1}{2\sqrt{d}}\sum_{b0}^{\infty}\frac{[1]_{k}\Gamma(2k+\frac{1}{2}+\gamma)}{k!\Gamma(\frac{k}{2}+12)}\{(\frac{c+\sqrt{d}}{\sqrt{z-b}})^{k}-(\frac{c-\sqrt{d}}{\sqrt{z-b}})^{k}\}$ (6)$(| \Gamma(\frac{k}{2}+1+m+\gamma)|<\infty)(|\Gamma(\frac{k}{2}+\frac{1}{2}\gamma|<\infty)$and(ii) for $n\in Z_{0}^{+}$ ,$(z-b)^{-1/2} \sum_{m,k=0}^{\infty}\frac{[1]_{m}[2+2m]_{k}[kz+1+m]_{n}}{m!k!}(\frac{c}{\sqrt{z-b}})^{k}(\frac{d}{z-b})^{m}$$= \frac{1}{2\sqrt{d}}\sum_{k=0}^{\infty}\frac{[1]_{k}[\frac{k}{2}+\frac{1}{2}]_{n}}{k!}\{(\frac{c+\sqrt{d}}{\sqrt{z-b}})^{k}-(\frac{c-\sqrt{d}}{\sqrt{z-b}})^{k}\}$ (7)where $(\sqrt{z-b}-c)^{2}-d\neq 0,$ $d\neq 0$ ,$| \frac{c}{\sqrt{z-b}}|\neq 1,$ $| \frac{d}{z-b}|\neq 1,$ $| \frac{c+\sqrt{d}}{\sqrt{z-b}}|\neq 1,$ $| \frac{c-\sqrt{d}}{\sqrt{z-b}}|\neq 1$ .49Proof of (i). We have the following relation,$\frac{1}{(\sqrt{z-b}-c)^{2}-d}=\frac{1}{2\sqrt{d}}(\frac{1}{\sqrt{z-b}-c-\sqrt{d}}-\frac{1}{\sqrt{z-b}-c+\sqrt{d}})$ . (8)From Theorem $D,(i)$ , we have$( \frac{1}{\sqrt{z-b}-p})_{\gamma}=(((z-b)^{1/2}-p)^{-1})_{\gamma}$ (9)$=e^{-i\pi\gamma}(z-b)^{-1/2-\gamma} \sum_{k=0}^{\infty}\frac{[1]_{k}\Gamma(\frac{k}{2(}+\frac{1}{2}+\gamma)}{kI\Gamma_{2^{+}2)}^{k1}}(\frac{p}{\sqrt{z-b}})^{k}$ (10)$(|\Gamma(2m+2+\gamma-k)|<\infty)$Therefore, setting $p=c+\sqrt{d}$ or $c-\sqrt{d}$ , we have$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{\gamma}=\frac{1}{2\sqrt{d}}((\frac{1}{\sqrt{z-b}-c-\sqrt{d}})_{\gamma}-(\frac{1}{\sqrt{z-b}-c+\sqrt{d}})_{\gamma})$$=e^{-2\pi\gamma}(z-b)^{-1/2-\gamma} \sum_{\succ 0}^{\infty}\frac{[1]_{k}\Gamma(zk+\frac{1}{2}+\gamma)}{k!\Gamma(|frack2+\frac{1}{2})}\{(\frac{c+\sqrt{d}}{\sqrt{z-b}})^{k}-(\frac{c-\sqrt{d}}{\sqrt{z-b}})^{k}\}$ .(11)Proof of (ii). Set $\gamma=n$ in (6).4 Semi Derivatives and Integrals[I] We have(i)$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{1/2}=-i(z-b)^{-3/2}$$\cross\sum_{m,k=0}^{\infty}\frac{[1]_{m}[2+2m]_{k}\Gamma(zk+z3+m)}{m!k1\Gamma(zk+1+m)}(\frac{c}{\sqrt{z-b}})^{k}(\frac{d}{z-b})^{m}$ (1)(semi derivative)and50(ii)$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{-1/2}=i(z-b)^{-1/2}$$x\sum_{m,k=0}^{\infty}\frac{[1]_{m}[2+2m_{k}\Gamma(++m)}{m1k!\Gamma(\frac{k]}{2}+1+m)}(\frac{c}{\sqrt{z-b}})^{k}(\frac{d}{z-b})^{m}$ (2)(semi integral)where$(\sqrt{z-b}-c)^{2}-d\neq 0$ , $| \frac{c}{\sqrt{z-b}}|<1$ , $| \frac{d}{z-b}|<1$ ,&om Theorem 1 by setting $\gamma=1/2$ and $-1/2$, respectively.[II] We have(i)$( \frac{1}{z-2c\sqrt{z-b}-b+c^{2}})_{1/2}=-i(z-b)^{-3/2}$$x\sum_{k=0}^{\infty}\frac{[2]_{k}\Gamma(k\mathfrak{T}+\S)}{k!\Gamma(zk+1)}(\frac{c}{\sqrt{z-b}})^{k}$ (3)(semi derivative)and(ii)$( \frac{1}{z-2c\sqrt{z-b}-b+d})_{-1/2}=i(z-b)^{-1/2}$$x\sum_{k=0}^{\infty}\frac{[2]_{k}\Gamma(2k+1\pi)}{k!\Gamma(\frac{k}{2}+1)}(\frac{c}{\sqrt{z-b}})^{k}$ (4)(semi integral)where$| \frac{c}{\sqrt{z-b}}|<1$ ,&om Corollary 1 by setting $\gamma=1/2$ and $-1/2$, respectively.515 Some Special CasesWhen the order of differentiation is some integer, our results coincide theclassical calculus. So, we illustrate some examples in cases of $n=0,1$ .(I) When $n=0$, from Theorem l.(ii), we have the followings,$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{0}=(z-b)^{-1}$$x\sum_{m,\succ-0}^{\infty}\frac{[1]_{m}[2+2m]_{k}[zk+1+m]_{0}}{m!k!}S^{k}T^{m}$ (1)Here we set$S= \frac{c}{\sqrt{z-b}}$ , $T= \frac{d}{z-b}$ .Indeed we haveR.H.S. of (1) $=(z-b)^{-1} \sum_{m=0}^{\infty}\frac{[1]_{m}}{m!}\mathcal{I}^{\pi n}\sum_{k=0}^{\infty}\frac{[2+2m]_{k}}{k!}S^{k}$ (2)$=(z-b)^{-1} \sum_{m\Leftrightarrow 0}^{\infty}\frac{[1]_{m}}{m!}\mathcal{I}^{m}(1-S)^{-2-2m}$ (3)$= \frac{1}{z-b}\frac{1}{(1-S)^{2}}\sum_{m=0}^{\infty}\frac{[1]_{m}}{m!}(\frac{T}{(1-S)^{2}})^{m}$ (4)$= \frac{1}{z-b}\frac{1}{(1-S)^{2}}(1-\frac{T}{(1-S)^{2}})^{-1}$ (5)$= \frac{1}{z-b}\frac{1}{(1-S)^{2}-T}=\frac{1}{(\sqrt{z-b}-c)^{2}-d}$ . (6)(II) When $n=1$ , our result is written as follows.$( \frac{1}{(\sqrt{z-b}-c)^{2}-d})_{1}=-(z-b)^{-2}$$\cross\sum_{m,\succ 0}^{\infty}\frac{[1]_{m}[2+2m]_{k}[42+1+m]_{1}}{m!k!}S^{k}\mathcal{I}^{m}$ . (7)Indeed we haveR.H.S. of (7) $=-(z-b)^{-2} \sum\frac{[1]_{m}}{m!}T^{m}\infty$$m-\triangleleft$$x\sum_{k=0}^{\infty}\frac{[2+2m]_{k}(\frac{k}{2!}+1+m)}{k}S^{k}$. (8)52Now we take notice that$\sum_{k=0}^{\infty}\frac{[2+2m]_{k}(\frac{k}{2})}{k!}S^{k}=\frac{1}{2}\sum_{k=1}^{\infty}\frac{[2+2m]_{k}}{(k-1)!}S^{k}$ (9)$= \frac{1}{2}S\sum_{k=0}^{\infty}\frac{[2+2m]_{k+1}}{k!}S^{k}$ (10)$= \frac{1}{2}S(2+2m)\sum_{k=0}^{\infty}\frac{[3+2mJ_{k}}{k!}S^{k}$ (11)$=S(1+m)(1-S)^{-3-2m}$ (12)and$\sum\frac{[2+2m]_{k}(1+m)}{k!}S^{k}\infty=(1+m)\sum\frac{[2+2m]_{k}}{k!}S^{k}\infty$$k=0$ $\succ_{-}0$$=(1+m)(1-S)^{-2-2m}$ . (13)Therefore we obtainR.H.S. of (7) $=-(z-b)^{-2}(1-S)^{-3} \sum_{m=0}^{\infty}\frac{[1]_{m}}{m!}(1+m)(\frac{T}{(1-S)^{2}})^{m}$ (14)$=-(z-b)^{-2}(1-S)^{-3} \{\sum_{m=0}^{\infty}\frac{[1]_{m}}{m!}(\frac{T}{(1-S)^{2}})^{m}+\sum_{m=0}^{\infty}\frac{[1]_{m}}{m!}(\frac{T}{(1-S)^{2}})^{m}\}$$=-(z-b)^{-2}(1-S)^{-3} \{(1-\frac{T}{(1-S)^{2}})^{-1}+\frac{T}{(1-S)^{2}}(1-\frac{T}{(1-S)^{2}})^{-2}\}$$=-(z- b)^{-2}(1-S)^{-3}\frac{(1-S)^{4}}{((1-S)^{2}-T)^{2}}$$=-(z-b)^{-2} \frac{1-S}{((1-S)^{2}-T)^{2}}$$=- \frac{1}{\sqrt{z-b}}\frac{\sqrt{z-b}-c}{((\sqrt{z-b}-c)^{2}-d)^{2}}$ . (15)This result coincides with the one obtained from the classical calculus$\frac{d}{dz}(\frac{1}{(\sqrt{z-b}-c)^{2}-d})$$(nI)$ The cases of $n=2$ and 3 are somewhat complicated , we will reportthose cases at anothor time.53Note. In this section we use the following identities.$\sum_{k=0}^{\infty}\frac{[\lambda]_{k}}{k!}S^{k}=(1-S)^{-\lambda}$ , $(|S|<1)$ (16)$[\lambda]_{k+1}=\lambda[\lambda+1]_{k}$ (17)$\sum_{k=0}^{\infty}\frac{[\lambda]_{k}k}{k!}S^{k}=\sum_{k=1}^{\infty}\frac{[\lambda]_{k}}{(k-1)!}S^{k}=S\sum_{k-\triangleleft}^{\infty}\frac{[\lambda]_{k+1}}{k!}S^{k}$ (18)References[1] K. Nishimoto; Ractional Calculus, Vol. 1 (1984), Vol. 2 (1987), Vol. 3(1989), Vol. 4 (1991), Vol. 5, (1996),DRPress, Koriyama, Japan.[2] K. Nishimoto; An Essence of Nishimnoto’s Fractional Calculus (Cal-culus of the 21st Century); Integrals and Differentiations of ArbitrwyOrder (1991), Descartes Press, Koriyama, Japan.[3] K. Nishimoto ; On Nishimoto’s &actional calculus operator $N^{\nu}$ (Onan action goup), J. Frac. Calc. Vol. 4, Nov. (1993), 1-11.[4] K. Nishimoto; Unification of the integrals and derivatives (A serendip-ity in fractional calculus), J. Frac. Calc. Vol. 6, Nov. (1994), 1-14.[5] K. Nishimoto ; Ring and Field Produced from The Set of N- liYactionalCalculus Operator, J. Frac Calc. Vol. 24, Nov. $(2003),29- 36$.[6] K. Nishimoto ; On the fractional calculus of functions (a $-z)^{\beta}$ and$\log(a$ -z), J. Frac. Calc. Vol. 3, May (1993), 19 - 27.[7] K. Nishimoto and S. -T. h ; Fractional calculus of Psi functions (Generalized polygamma functions), J. IFlirac. Calc. Vol. 5, May (1994),27-34.[8] S. -T. Tu and K. Nishimoto ; On the fractional calculus of functions(cz $-a)^{\beta}$ and $\log(cz$ -a), J. Nac. Calc. Vol. 5, May (1994), 35-43.[9] K. Nishimoto ; N-Nractional Calculus of the Power and LogarithmicFunctions, and some Identities, J. $\mathbb{R}ac$. Calc. Vol. 21, May (2002), 1-6.[10] K. Nishimoto ; Some Theorems for N-Fractional Calculus of Loggith-mic Eznctions, J. Frac. Calc. Vol. 21, May (2002), 7-12.54[11] K. Nishimoto; $N$-Ractional Calculus of Products of Some Power Func-tions, J. Rac. Calc. Vol. 27, May (2005), $83rightarrow 88$ .[12] K. Nishimoto ; N-Fractional Calculus of Products of Some CompositeFunctions, J. IFtac. Calc. Vol. 29, May (2006), 35 -44.[13] K. Nishimoto ; N-Fractional Calculus of Products of Some CompositeAlgebraic Rinctions, J. Frac. Calc. Vol. 31, May (2007), 11 -23.55

$n$-th Derivatives of Some Functions in terms of N-Fractional Calculus (Study on Non-Analytic and Univalent Functions and Applications)

http://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/140321/1/1626-05.pdf

$n$ -th Derivatives of Some Functions in terms of N-Fractional Calculus (Study on Non-Analytic and Univalent Functions and Applications)

Abstract

Similar works

Full text

Available Versions

Kyoto University Research Information Repository

nnn-th Derivatives of Some Functions in terms of N-Fractional Calculus (Study on Non-Analytic and Univalent Functions and Applications)

Abstract

Similar works

Full text

Available Versions

Kyoto University Research Information Repository

$n$ -th Derivatives of Some Functions in terms of N-Fractional Calculus (Study on Non-Analytic and Univalent Functions and Applications)