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Recently I am working on a project. It requires the nth derivative of the function $e^{i\frac{1}{\sqrt{1+\lambda}}z}$ with respect to $\lambda$. Or equivalently, finding the nth order Taylor series near $\lambda=0$ to nth order. Using mathematica to expand it to 3rd order, we can get \begin{align} e^{i z}-\frac{1}{2} i \lambda e^{i z} z-\frac{1}{8} \lambda ^2 \left(e^{i z} z (z-3 i)\right)+\frac{1}{48} \lambda ^3 e^{i z} z \left(i z^2+9 z-15 i\right)+O\left(\lambda ^4\right) \end{align} If we type the following to mathematica

SeriesCoefficient[E^(I 1/Sqrt[1 + [Lambda]] z) , {[Lambda], 0, n}]

It will also give some complicated expression. Is it possible to get a closed form of the nth derivative?

Siyi Zhou
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1 Answers1

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My answer is as follows: \begin{align*} \frac{\operatorname{d}^n}{\operatorname{d}x^n} \operatorname{e}^{\operatorname{i}\frac1{\sqrt{1+\lambda}\,}z} &=\frac{\operatorname{d}^n}{\operatorname{d}x^n} \operatorname{e}^{\operatorname{i}z(1+\lambda)^{-1/2}}\\ &=\sum_{k=0}^n \operatorname{e}^{\operatorname{i}z(1+\lambda)^{-1/2}} B_{n,k}\biggl(\operatorname{i}z(1+\lambda)^{-1/2-1}\biggl\langle-\frac12\biggr\rangle_1, \operatorname{i}z(1+\lambda)^{-1/2-2}\biggl\langle-\frac12\biggr\rangle_2,\\ &\quad\operatorname{i}z(1+\lambda)^{-1/2-3}\biggl\langle-\frac12\biggr\rangle_3, \dotsc, \operatorname{i}z(1+\lambda)^{-1/2-(n-k+1)}\biggl\langle-\frac12\biggr\rangle_{n-k+1}\biggr)\\ &=\operatorname{e}^{\operatorname{i}z(1+\lambda)^{-1/2}} \sum_{k=0}^n (\operatorname{i}z)^k (1+\lambda)^{-k/2-n} B_{n,k}\biggl(\biggl\langle-\frac12\biggr\rangle_1, \biggl\langle-\frac12\biggr\rangle_2, \dotsc, \biggl\langle-\frac12\biggr\rangle_{n-k+1}\biggr)\\ &=\operatorname{e}^{\operatorname{i}z(1+\lambda)^{-1/2}} \sum_{k=0}^n (\operatorname{i}z)^k (1+\lambda)^{-k/2-n} \sum _{\ell=k}^n s(n,\ell) \biggl(-\frac12\biggr)^\ell S(\ell,k)\\ &\to \operatorname{e}^{\operatorname{i}z} \sum_{k=0}^n (\operatorname{i}z)^k \sum _{\ell=k}^n s(n,\ell) \biggl(-\frac12\biggr)^\ell S(\ell,k), \quad \lambda\to0, \end{align*} where $s(n,\ell)$ and $S(\ell,k)$ denote the Stirling numbers of the first and second kinds respectively. Therefore, we obtain the power series expansion \begin{equation*} \operatorname{e}^{\operatorname{i}\frac1{\sqrt{1+\lambda}\,}z} =\operatorname{e}^{\operatorname{i}z} \sum_{n=0}^\infty\Biggl[\sum_{k=0}^n (\operatorname{i}z)^k \sum _{\ell=k}^n s(n,\ell) \biggl(-\frac12\biggr)^\ell S(\ell,k)\Biggr]\frac{\lambda^n}{n!}, \quad |\lambda|<1. \end{equation*}

References

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