# Power Series Expansion for Real Arctangent Function

## Theorem

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

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

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

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 $x \le -1$:

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

$\blacksquare$