Power Series Expansion for Real Arctangent Function

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Theorem

The arctangent function has a Taylor series expansion:

$\arctan x = \begin {cases} \ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1} & : -1 \le x \le 1 \\ \ds \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \ge 1 \\ \ds -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \le -1 \end {cases}$


That is:

$\arctan x = \begin {cases} x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 - \cdots & : -1 \le x \le 1 \\ \dfrac \pi 2 - \dfrac 1 x + \dfrac 1 {3 x^3} - \dfrac 1 {5 x^5} + \cdots & : x \ge 1 \\ -\dfrac \pi 2 - \dfrac 1 x + \dfrac 1 {3 x^3} - \dfrac 1 {5 x^5} + \cdots & : x \le -1 \end {cases}$


Proof

From Sum of Infinite Geometric Sequence:

$(1): \quad \ds \sum_{n \mathop = 0}^\infty \paren {-x^2}^n = \frac 1 {1 + x^2}$

for $-1 < x < 1$.

From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

\(\ds \int_0^x \frac 1 {1 + t^2} \rd t\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t^2}^n \rd t\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \arctan x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1}\) Primitive of Reciprocal of $\dfrac 1 {1 + t^2}$, Integral of Power

For $-1 \le x \le 1$, the sequence $\sequence {\dfrac {x^{2 n + 1}} {2 n + 1} }$ is decreasing and converges to zero.

Therefore the series converges in the given range by the Alternating Series Test.

$\Box$


Now consider the case $x \ge 1$:

\(\ds \arctan x\) \(=\) \(\ds \frac \pi 2 - \map \arccot x\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \frac \pi 2 - \map \arctan {\frac 1 x}\) Arctangent of Reciprocal equals Arccotangent
\(\ds \) \(=\) \(\ds \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {2 n + 1} \paren {\frac 1 x}^{2 n + 1}\) as $x \ge 1$, $0 < \dfrac 1 x \le 1$, so $(2)$ may be applied
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} }\)

$\Box$


We also have:

\(\ds \map \arctan {-x}\) \(=\) \(\ds -\arctan x\) Arctangent Function is Odd
\(\ds \) \(=\) \(\ds -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {\paren {2 n + 1} x^{2 n + 1} }\) from $(3)$

Substituting $x$ for $-x$ gives us the expansion for $x \le -1$:

\(\ds \arctan x\) \(=\) \(\ds -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {\paren {2 n + 1} \paren {-x}^{2 n + 1} }\)
\(\ds \) \(=\) \(\ds -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {-\paren {2 n + 1} x^{2 n + 1} }\)
\(\ds \) \(=\) \(\ds -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} }\)

$\blacksquare$


Also see


Sources