Series expansion for $\arctan(1-x)$

Question

Series expansion for $\arctan(1-x)$

I try to expand this function into its Taylor series by means of differentiating it and then integrating it terms by terms but I fail to obtain the correct result.

The derivative of $\arctan(1-x)=-\dfrac{1}{x^2-2x+2}$. By using long division, I can obtain the series expansion for $-\dfrac{1}{x^2-2x+2}$ and it is

$-\dfrac{1}{2}-\dfrac{x}{2}-\dfrac{x^2}{4}+\dfrac{x^4}{8}+\dfrac{x^5}{8}+\dfrac{x^6}{16}...$

I integrate this series terms by terms and I obtain: $0-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$

The series according to Wolfram is: $\dfrac{\pi}{4}-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$.

I notice that if I use the indefinite integral, there is going to be an arbitrary constant left after the integrating process. How do I obtain this $\dfrac{\pi}{4}$? Without it, is my series wrong?

Is there any other methods to expand this function into Maclaurin series without directly employing the Maclaurin formula

Answer 1

When you integrate it, you get a $+C$. To get it's value, you can substitute in $x=0$ to get that $C=\arctan(1)$.

Answer 2

For $n\ge1$, we have \begin{align*} [\arctan(1-x)]^{(n)} &=-\biggl(\frac{1}{x^2-2x+2}\biggr)^{(n-1)}\\ &=-\sum_{k=0}^{n-1} \frac{(-1)^kk!}{(x^2-2x+2)^{k+1}} B_{n-1,k}(2x-2,2,0,\dotsc,0)\\ &\to-\sum_{k=0}^{n-1} \frac{(-1)^kk!}{2^{k+1}} B_{n-1,k}(-2,2,0,\dotsc,0), \quad x\to0\\ &=-\sum_{k=0}^{n-1} \frac{(-1)^kk!}{2^{k+1}} 2^k B_{n-1,k}(-1,1,0,\dotsc,0)\\ &=-\sum_{k=0}^{n-1} \frac{(-1)^kk!}{2^{k+1}} 2^k \frac{1}{2^{n-k-1}}\frac{(n-1)!}{k!}\binom{k}{n-k-1}(-1)^{2k-n+1}\\ &=(-1)^{n}\frac{(n-1)!}{2^n}\sum_{k=0}^{n-1} (-1)^k2^{k} \binom{k}{n-k-1} \end{align*} where we used the formula \begin{equation}\label{Bell-x-1-0-eq} B_{n,k}(x,1,0,\dotsc,0) =\frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}x^{2k-n}. \end{equation} Consequently, we arrive at \begin{equation*} \arctan(1-x)=\frac\pi4+\sum_{n=1}^\infty\frac{(-1)^{n}}{2^n}\Biggl[\sum_{k=0}^{n-1} (-1)^k2^{k} \binom{k}{n-k-1}\Biggr]\frac{x^n}{n}. \end{equation*}

References

1. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
2. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.
3. Feng Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstratio Mathematica 55 (2022), no. 1, 710--736; available online at https://doi.org/10.1515/dema-2022-0157.

Answer 3

For $n\in\mathbb{N}$, the derivative of $\arctan z$ is \begin{equation*} (\arctan z)^{(n)} =\frac{(n-1)!}{(2z)^{n-1}}\sum_{k=0}^{n-1}(-1)^k\binom{k}{n-k-1}\frac{(2z)^{2k}}{(1+z^2)^{k+1}}. \end{equation*} The function $\frac{\arctan z}{z}$ has Taylor's series expansion \begin{equation}\label{arctan-pi-4-ser-eq} \frac{\arctan z}{z}=\sum_{n=0}^{+\infty}(-1)^n\Bigl[\frac{\pi}{4}+T(n)\Bigr](z-1)^n,\quad |z-1|<\sqrt{2}\,, \end{equation} where \begin{equation}\label{T(n)-dfn-notation} T(n)= \begin{cases} 0, & n=0;\\ \displaystyle \sum_{k=1}^{n}\frac{(-1)^{k}}{2^{k/2}k}\sin\frac{3k\pi}{4}, & n\in\mathbb{N}. \end{cases} \end{equation} Hence, we derive \begin{equation}\label{arctan(1-z)-pi-4-ser-eq} \frac{\arctan (1-z)}{1-z}=\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^n,\quad |z|<\sqrt{2}\,, \end{equation} which can be rearranged as \begin{align} \arctan (1-z)&=\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^n -\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^{n+1}\\ &=\sum_{n=0}^{\infty}\Bigl[\frac{\pi}{4}+T(n)\Bigr]z^n -\sum_{n=1}^{\infty}\Bigl[\frac{\pi}{4}+T(n-1)\Bigr]z^{n}\\ &=\frac{\pi}{4}+\sum_{n=1}^{\infty}[T(n)-T(n-1)]z^n\\ &=\frac{\pi}{4}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2^{n/2}n}\sin\frac{3n\pi}{4}z^n\\ &=\frac{\pi}{4}-\frac{z}{2}-\frac{z^2}{4}-\frac{z^3}{12}+\frac{z^5}{40}+\frac{z^6}{48}+\frac{z^7}{112}-\frac{z^9}{288}-\dotsm. \end{align}

1. Feng Qi and Mark Daniel Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function, arXiv (2021), available online at https://arxiv.org/abs/2110.08576v1.