分数阶导数October 13, 2023 · 4 min readPuQingAI, CVer, Pythoner, Half-stack Developer在高中就已经学过函数的 nnn 阶导数,其中的 nnn 是正整数,1,2,3,⋯ ,n1,2,3,\cdots,n1,2,3,⋯,n,能否能够将这里的 nnn 推广至整数,以及推广至有理数,实数。 负整数阶导数 很自然的认为,函数的负整数导数应该是求它的不定积分,并且相差参数 CCC。 我们记导数算子为 D\operatorname{D}D,即 Df(x)≜df dx,D2f(x)≜d2f dx2,⋯ ,Dnf(x)≜dnf dxn.\mathrm{D} f(x) \triangleq \frac{\mathrm{d} f}{\mathrm{~d} x}, \quad \mathrm{D}^{2} f(x) \triangleq \frac{\mathrm{d}^{2} f}{\mathrm{~d} x^{2}}, \quad \cdots, \quad \mathrm{D}^{n} f(x) \triangleq \frac{\mathrm{d}^{n} f}{\mathrm{~d} x^{n}} .Df(x)≜ dxdf,D2f(x)≜ dx2d2f,⋯,Dnf(x)≜ dxndnf. 记 aDx−1{ }_{a} \mathrm{D}_{x}^{-1}aDx−1 表示变上限积分,即 aDx−1f(x)≜∫axf(τ)dτ,aDx−2f(x)≜aDx−1(aDx−1f(x)),……aDx−nf(x)≜aDx−1(aDx−(n−1)f(x)),x>a.\begin{aligned} { }_{a} \mathrm{D}_{x}^{-1} f(x) & \triangleq \int_{a}^{x} f(\tau) \mathrm{d} \tau, \\ { }_{a} \mathrm{D}_{x}^{-2} f(x) & \triangleq{ }_{a} \mathrm{D}_{x}^{-1}\left({ }_{a} \mathrm{D}_{x}^{-1} f(x)\right), \\ & \ldots \ldots \\ { }_{a} \mathrm{D}_{x}^{-n} f(x) & \triangleq{ }_{a} \mathrm{D}_{x}^{-1}\left({ }_{a} \mathrm{D}_{x}^{-(n-1)} f(x)\right), \quad x>a . \end{aligned}a