What is the $n^\text{th}$ derivative of $f(x)=\frac{1}{1+x^2}$

Question

I want the taylor series expansion around some value $a$ of the function $f(x)=\frac{1}{1+x^2}$. I used the general formula \begin{eqnarray} f(x) = f(a) + \sum_{n=1}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n \end{eqnarray} But unfortunately, I cannot compute any general formula for $f^{(n)}(a)$. The first derivative is $$ f^{(1)}(x)= -\frac{2x}{(1+x^2)^2}.$$The second derivative is $$ f^{(2)}(x)= \frac{6x^2-2}{(1+x^2)^3}.$$The third derivative is $$ f^{(3)}(x)= \frac{24x(x^2-1)}{(1+x^2)^4}.$$The fourth derivative is $$ f^{(4)}(x)= -\frac{24(5x^4-10x^2+12)}{(1+x^2)^5}$$. The fifth derivative is $$ f^{(5)}(x)= \frac{240x(3x^4-10x^2+3)}{(1+x^2)^5}$$.

What is the $n$-th derivative of the function for working with the above taylor series which I want to use to prove something?

Answer 1

$$2f(x)=\frac1{1+ix}+\frac1{1-ix}.$$ Therefore $$2f^{(n)}(x)=\frac{(-i)^nn!}{(1+ix)^{n+1}}+\frac{i^nn!}{(1-ix)^{n+1}}.$$

Answer 2

Another form of the $n$-th derivative: $$f^{(n)}(x)=(-1)^n n!\frac{\sin((n+1)\cot^{-1} x)}{(1+x^2)^{(n+1)/2}}.$$ It's easy to prove by induction.

Answer 3

For $0\le k\le n$, the Bell polynomials of the second kind $B_{n,k}(x_1,x_2,\dotsc,x_{n-k+1})$ satisfy \begin{equation}\label{Bell-x-1-0-eq} B_{n,k}(x,1,0,\dotsc,0) =\frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}x^{2k-n}. \end{equation} See the papers [2,4] and Section 1.4 in [3] below. By the Faa di Bruno formula \begin{equation}\label{Bruno-Bell-Polynomial} \frac{\text{d}^n}{\text{d}x^n}f\circ g(x)=\sum_{k=0}^nf^{(k)}(g(x)) B_{n,k}\bigl(g'(x),g''(x),\dotsc,g^{(n-k+1)}(x)\bigr) \end{equation} and some properties of the Bell polynomials of the second kind, we find \begin{align*} \biggl(\frac1{1+x^2}\biggr)^{(n)} &=\sum_{k=0}^{n}\frac{(-1)^kk!}{(1+x^2)^{k+1}}B_{n,k}(2x,2,0,\dotsc,0)\\ &=\sum_{k=0}^{n}\frac{(-1)^kk!}{(1+x^2)^{k+1}}B_{n,k}(2x,2,0,\dotsc,0)\\ &=\sum_{k=0}^{n}\frac{(-1)^kk!}{(1+x^2)^{k+1}}2^kB_{n,k}(x,1,0,\dotsc,0)\\ &=\sum_{k=0}^{n}\frac{(-1)^kk!}{(1+x^2)^{k+1}}2^k \frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}x^{2k-n}\\ &=\frac{n!}{(2x)^{n}}\sum_{k=0}^{n}(-1)^k\binom{k}{n-k}\frac{(2x)^{2k}}{(1+x^2)^{k+1}}, \quad n\ge0. \end{align*}

By the method used in the paper [1] below, we can straightforwardly obtain another formula \begin{equation*} \biggl(\frac1{1+x^2}\biggr)^{(n)} =\frac{(-1)^{n}}{(1+x^2)^{n+1}} \begin{vmatrix} 2x & 1+x^2 & 0 & \dotsm&0& 0& 0\\ 2 & 4x & 1+x^2 & \dotsm& 0&0& 0\\ 0 & 6 & 6x & \dotsm& 0& 0&0\\ \dotsm & \dotsm & \dotsm & \ddots&\dotsm& \dotsm& \dotsm\\ 0 & 0 & 0 & \dotsm&2(n-2)x& 1+x^2& 0\\ 0 & 0 & 0 & \dotsm&(n-2)(n-1)&2(n-1)x & 1+x^2\\ 0 & 0 & 0 & \dotsm&0& (n-1)n& 2nx \end{vmatrix},\quad n\ge0. \end{equation*}

References

1. Feng Qi, Determinantal expressions and recursive relations of Delannoy polynomials and generalized Fibonacci polynomials, Journal of Nonlinear and Convex Analysis 22 (2021), no. 7, 1225--1239.
2. Feng Qi and Bai-Ni Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterranean Journal of Mathematics 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
3. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.
4. Feng Qi and Miao-Miao Zheng, Explicit expressions for a family of the Bell polynomials and applications, Applied Mathematics and Computation 258 (2015), 597--607; available online at https://doi.org/10.1016/j.amc.2015.02.027.