Power Series Expansion for Real Arctangent Function

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Theorem

The arctangent function has a Taylor series expansion:

$\arctan x = \begin{cases} \displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {2 n + 1} & : -1 \le x \le 1 \\ \displaystyle \frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} } & : x \ge 1 \\ \displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} } & : x \le -1 \end{cases}$


That is:

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


Proof

From Sum of Infinite Geometric Progression:

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

for $-1 < x < 1$.

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

\(\displaystyle \int_0^x \frac 1 {1 + t^2} \rd t\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \int_0^x \left({-t^2}\right)^n \rd t\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \arctan x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {2 n + 1}\) $\quad$ Primitive of Reciprocal of $\dfrac 1 {1 + t^2}$, Integral of Power $\quad$

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

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

$\Box$


Now consider the case $x \ge 1$:

\(\displaystyle \arctan x\) \(=\) \(\displaystyle \frac \pi 2 - \arctan \left({\frac 1 x}\right)\) $\quad$ Sum of Arctangent and Arccotangent $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {2 n + 1} \left({\frac 1 x}\right)^{2 n + 1}\) $\quad$ as $x \ge 1$, $0 < \dfrac 1 x \le 1$, so may apply the above result $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} }\) $\quad$ $\quad$

$\Box$


We also have:

\(\displaystyle \arctan \left({-x}\right)\) \(=\) \(\displaystyle -\arctan x\) $\quad$ Arctangent Function is Odd $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^{n + 1} \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} }\) $\quad$ per above result $\quad$

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

\(\displaystyle \arctan x\) \(=\) \(\displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^{n + 1} \frac 1 {\left({2 n + 1}\right) \left({-x}\right)^{2 n + 1} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^{n + 1} \frac 1 {-\left({2 n + 1}\right) x^{2 n + 1} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {\left({2 n + 1}\right) x^{2 n + 1} }\) $\quad$ $\quad$

$\blacksquare$


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