10.设u(x,t)e2xt−t2t是复数试证∂nu(x,t)∂tn∣t0(−1)nex2dndxne−x2。证左侧∂nu∂tn∣t0n!2πi∮e2xt−t2tn1dt右侧(−1)nex2dndxne−x2(−1)nex2n!2πi∮e−t2(t−x)n1dt令t−x−w则tx−w上式(−1)nex2n!2πi∮e−(x−w)2(−1)n1wn1⋅(−1)⋅dw(−1)nex2n!2πi∮e−x22xw−w2(−1)n1wn1⋅(−1)dwn!2πi∮e2xw−w2wn1dw∂nu∂tn\begin{aligned} 10. 设u(x,t)e^{2xt-t^2}t是复数试证\left.\frac{\partial^n u(x,t)}{\partial t^n}\right|_{t0}(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}。\\ 证 \\ 左侧\left. \frac{\partial^n u}{\partial t^n} \right|_{t0} \frac{n!}{2\pi i} \oint \frac{e^{2xt - t^2}}{t^{n1}} dt\\ 右侧(-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} (-1)^n e^{x^2} \frac{n!}{2\pi i} \oint \frac{e^{-t^2}}{(t-x)^{n1}} dt\\ 令 t-x -w则 t x - w\\ 上式 (-1)^n e^{x^2} \frac{n!}{2\pi i} \oint \frac{e^{-(x-w)^2}}{(-1)^{n1} w^{n1}} \cdot (-1) \cdot dw\\ (-1)^n e^{x^2} \frac{n!}{2\pi i} \oint \frac{e^{-x^2 2xw - w^2}}{(-1)^{n1} w^{n1}} \cdot (-1) dw\\ \frac{n!}{2\pi i} \oint \frac{e^{2xw - w^2}}{w^{n1}} dw \frac{\partial^n u}{\partial t^n}\\ \end{aligned}​10.设u(x,t)e2xt−t2t是复数试证∂tn∂nu(x,t)​​t0​(−1)nex2dxndn​e−x2。证左侧∂tn∂nu​​t0​2πin!​∮tn1e2xt−t2​dt右侧(−1)nex2dxndn​e−x2(−1)nex22πin!​∮(t−x)n1e−t2​dt令t−x−w则tx−w上式(−1)nex22πin!​∮(−1)n1wn1e−(x−w)2​⋅(−1)⋅dw(−1)nex22πin!​∮(−1)n1wn1e−x22xw−w2​⋅(−1)dw2πin!​∮wn1e2xw−w2​dw∂tn∂nu​​这是埃尔米特多项式Hermite Polynomials的母函数生成函数核心公式属于特殊函数/数学物理方法的经典结论。关键说明母函数定义埃尔米特多项式的指数型母函数e2xt−t2∑n0∞Hn(x)n!tne^{2xt-t^2} \sum_{n0}^{\infty}\frac{H_n(x)}{n!}t^ne2xt−t2n0∑∞​n!Hn​(x)​tn其中Hn(x)H_n(x)Hn​(x)为埃尔米特多项式。与上式的关联对母函数两边关于ttt求nnn阶导并令t0t0t0∂n∂tne2xt−t2∣t0Hn(x)\left.\frac{\partial^n}{\partial t^n}e^{2xt-t^2}\right|_{t0}H_n(x)∂tn∂n​e2xt−t2​t0​Hn​(x)而埃尔米特多项式有罗德里格斯公式Hn(x)(−1)nex2dndxn(e−x2)H_n(x)(-1)^n e^{x^2}\frac{d^n}{dx^n}\left(e^{-x^2}\right)Hn​(x)(−1)nex2dxndn​(e−x2)应用领域量子力学谐振子波函数概率论正态分布的正交多项式数学物理热方程、波动方程的特殊解