A reformulated series expansion of the arctangent function

TL;DR

This paper presents a new real rational function representation of a previously complex-number-based series expansion of the arctangent function, simplifying its computation and analysis.

Contribution

It reformulates a complex series expansion of arctangent into a real rational function, enhancing understanding and potential applications.

Findings

The arctangent series can be expressed as a real rational function.

Simplification from complex to real form facilitates computational efficiency.

The new representation may improve numerical methods involving arctangent.

Abstract

In our recent publication we obtained a series expansion of the arctangent function involving complex numbers. In this work we show that this formula can also be expressed as a real rational function.

Equations

$$\frac{-i}{{\left(x+i\right)}^m},$$

$$\frac{-i}{{\left(x+i\right)}^1}=-\frac{1}{1+x^2}-i\frac{x}{1+x^2}$$

$$\frac{-i}{{\left(x+i\right)}^m}$$

$$=i^1\left(\frac{-i}{{\left(x+i\right)}^{m-1}}\right){\left(-\frac{1}{1+x^2}-i\frac{x}{1+x^2}\right)}^1$$

$$=i^2\left(\frac{-i}{{\left(x+i\right)}^{m-2}}\right){\left(-\frac{1}{1+x^2}-i\frac{x}{1+x^2}\right)}^2=\cdots$$

$$=i^n\left(\frac{-i}{{\left(x+i\right)}^{m-n}}\right){\left(-\frac{1}{1+x^2}-i\frac{x}{1+x^2}\right)}^n=\cdots$$

$$=-i^{m+1}{\left(-\frac{1}{1+x^2}-i\frac{x}{1+x^2}\right)}^m,m\ge n.$$

$$-i^{m+1}{\left(-\frac{1}{1+x^2}-i\frac{x}{1+x^2}\right)}^m=-i^{m+1}\sum _{n=0}^m{\left(-\frac{1}{1+x^2}\right)}^{m-n}{\left(-\frac{ix}{1+x^2}\right)}^n\left(\begin{array}{rl}& m\\ & n\end{array}\right)$$

$$\frac{-i}{{\left(x+i\right)}^m}=\sum _{n=1}^m\frac{{\left(-1\right)}^n{x}^{m-\left(2n-1\right)}}{{\left(1+x^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right)+i\frac{{\left(-1\right)}^n{x}^{m-2\left(n-1\right)}}{{\left(1+x^2\right)}^m}\left(\begin{array}{c}m\\ 2\left(n-1\right)\end{array}\right).$$

$$\mathrm{Re}\left[\frac{-i}{{\left(x+i\right)}^m}\right]=-\mathrm{Re}\left[\frac{-i}{{\left(x-i\right)}^m}\right]$$

$$\mathrm{Im}\left[\frac{-i}{{\left(x+i\right)}^m}\right]=\mathrm{Im}\left[\frac{-i}{{\left(x-i\right)}^m}\right]$$

$$-\frac{-i}{{\left(x-i\right)}^m}=\sum _{n=1}^m\frac{{\left(-1\right)}^n{x}^{m-\left(2n-1\right)}}{{\left(1+x^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right)-i\frac{{\left(-1\right)}^n{x}^{m-2\left(n-1\right)}}{{\left(1+x^2\right)}^m}\left(\begin{array}{c}m\\ 2\left(n-1\right)\end{array}\right).$$

$$\left(\frac{-i}{{\left(x+i\right)}^m}-\frac{-i}{{\left(x-i\right)}^m}\right)$$

$$=2\sum _{n=1}^m\frac{{\left(-1\right)}^n{x}^{m-\left(2n-1\right)}}{{\left(1+x^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right).$$

$$\frac{d^m}{d{x}^m}\arctan \left(x\right)=\frac{{\left(-1\right)}^m\left(m-1\right)!}{2i}\left(\frac{1}{{\left(x+i\right)}^m}-\frac{1}{{\left(x-i\right)}^m}\right).$$

$$\frac{d^m}{d{x}^m}\arctan \left(x\right)=\left(m-1\right)!\sum _{n=1}^m\frac{{\left(-1\right)}^{m+n}{x}^{m-\left(2n-1\right)}}{{\left(1+x^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right).$$

$$\frac{d^m}{d{t}^m}\arctan \left(x\right)={\mathrm{sgn}}^{m-1}\left(-x\right)\frac{\left(m-1\right)!}{{\left(1+x^2\right)}^{m/2}}\sin \left(m\arcsin \left(\frac{1}{\sqrt{1+x^2}}\right)\right),$$

$$\mathrm{sgn}\left(x\right)=\{\begin{array}{rl}1,& x\ge 0\\ -1,& x<0,\end{array}$$

$$\arctan \left(x\right)=i\underset{M\to \infty }{\lim }\sum _{m=1}^{\lfloor \frac{M}{2}\rfloor +1}\frac{1}{2m-1}\left(\frac{1}{{\left(1+2i/x\right)}^{2m-1}}-\frac{1}{{\left(1-2i/x\right)}^{2m-1}}\right).$$

$$\arctan \left(x\right)=$$

$$\times \left(\frac{1}{{\left(\left(2\ell -1\right)+2i/x\right)}^{2m-1}}-\frac{1}{{\left(\left(2\ell -1\right)-2i/x\right)}^{2m-1}}\right).$$

$$\arctan \left(x\right)=i\sum _{m=1}^{\infty }\frac{1}{2m-1}\left(\frac{1}{{\left(1+2i/x\right)}^{2m-1}}-\frac{1}{{\left(1-2i/x\right)}^{2m-1}}\right).$$

$$\frac{1}{{\left(1+2i/x\right)}^m}=\frac{{\left(x/2\right)}^m}{-i}\frac{-i}{{\left(x/2+i\right)}^m}.$$

$$\frac{-i}{{\left(x/2+i\right)}^m}$$

$$=\sum _{n=1}^m\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-\left(2n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right)+i\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-2\left(n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2\left(n-1\right)\end{array}\right).$$

$$\frac{1}{{\left(1+2i/x\right)}^m}$$

$$=i{\left(\frac{x}{2}\right)}^m\left(\sum _{n=1}^m\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-\left(2n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right)+i\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-2\left(n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2\left(n-1\right)\end{array}\right)\right).$$

$$\frac{1}{{\left(1-2i/x\right)}^m}=\frac{{\left(x/2\right)}^m}{-i}\frac{-i}{{\left(x/2-i\right)}^m}$$

$$-\frac{1}{{\left(1-2i/x\right)}^m}$$

$$=i{\left(\frac{x}{2}\right)}^m\left(\sum _{n=1}^m\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-\left(2n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right)-i\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-2\left(n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2\left(n-1\right)\end{array}\right)\right).$$

$$\frac{1}{{\left(1+2i/x\right)}^m}-\frac{1}{{\left(1-2i/x\right)}^m}=2i{\left(\frac{x}{2}\right)}^m\sum _{n=1}^m\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-\left(2n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right)$$

$$\frac{1}{m}(\frac{1}{{\left(1+2i/x\right)}^m}$$

$$=\frac{2i}{m}{\left(\frac{x}{2}\right)}^m\sum _{n=1}^m\frac{{\left(-1\right)}^n{\left(\frac{x}{2}\right)}^{m-\left(2n-1\right)}}{{\left(1+{\left(\frac{x}{2}\right)}^2\right)}^m}\left(\begin{array}{c}m\\ 2n-1\end{array}\right).$$

Full text

Keywords: arctangent function, rational function

1 Derivation

1.1 Derivatives of the arctangent function

Consider the following function

[TABLE]

where $x$ is a real variable and $m$ is a non-negative integer. Since

[TABLE]

it is not difficult to show by induction that

[TABLE]

Applying the binomial formula to this identity

[TABLE]

and then separating its right side into the real and imaginary parts, after some trivial rearrangements we obtain

[TABLE]

Taking into consideration that

[TABLE]

and

[TABLE]

we can also find

[TABLE]

Consequently, taking sum of equations (1) and (2) provides

[TABLE]

The ${{m}^{\text{th}}}$ derivative of the arctangent function can be represented as [1, 2]

[TABLE]

Comparing equations (3) and (4) immediately yields

[TABLE]

Notably, in contrast to the equation (4), the series expansion (5) involves no complex numbers.

It interesting to note that the equation (4) is also directly related to the identity

[TABLE]

where the signum function is defined as

[TABLE]

by de Moivre’s formula (see [1] for derivation procedure). However, in the work [1] this identity is shown without signum function that, according to Lampret [3], is necessary to account for $x\in\mathbb{R^{-}}$ in derivatives of the acrtangent function.

1.2 The arctangent function

In our recent publication we have derived a series expansion of the arctangent function [4]

[TABLE]

We used the notation $\left\lfloor M/2\right\rfloor+1$ in equation (6) only for chronological reason to keep consistency with its previously published variation (see [2] for details)

[TABLE]

Since at $M\to\infty$ the upper limit in summation $\left\lfloor M/2\right\rfloor+1$ also tend to infinity, the series expansion (6) can be rewritten as

[TABLE]

In order to exclude the complex numbers in equation (7) we can apply a similar approach as we have made already for derivation of the identity (5). In particular, we note that

[TABLE]

Consequently, for the function

[TABLE]

we can simply replace the variable $x\to x/2$ in the identity (1) to obtain

[TABLE]

Comparing now equations (8) and (9) we get

[TABLE]

[TABLE]

Similarly, writing

[TABLE]

and replacing $x\to x/2$ in equation (2) results in

[TABLE]

[TABLE]

Therefore, taking sum of equations (10) and (11) leads to

[TABLE]

or

[TABLE]

Replacing in equation (12) the index $m\to 2m-1$ provides

[TABLE]

Consequently, applying the equation (13) to the formula (7) we have

[TABLE]

or

[TABLE]

or

[TABLE]

[TABLE]

As we can see, the equation (14) is represented in form of a real rational function.

Although performing computation by truncating equation (14) is not optimal due to double summation, its application may be more convenient in theoretical analysis. In particular, reformulation as the real rational function (14) can help understand, for example, a behavior of equation (7) at small $x\to 0$ in computing pi (see [4] for details) and estimate its error in truncation.

2 Conclusion

We derived an equivalent (14) to series expansion (7) of the arctangent function in form a real rational function. Since equation (14) involves no complex numbers, its application may be more convenient in theoretical analysis.

Acknowledgments

This work is supported by National Research Council Canada, Thoth Technology Inc. and York University. The authors wish to thank Dr. J. Guillera and Dr. M. Shao for discussions and useful information.