# Power Series Expansion for Exponential of Tangent of x

## Theorem

$e^{\tan x} = 1 + x + \dfrac {x^2} 2 + \dfrac {x^3} 2 - \dfrac {3 x^4} 8 + \cdots$

for all $x \in \R$ such that $\left\lvert{x}\right\rvert < \frac \pi 2$.

## Proof

Let $f \left({x}\right) = e^{\tan x}$.

Then:

 $\displaystyle \frac \d {\d x} f \left({x}\right)$ $=$ $\displaystyle \sec^2 x \, e^{\tan x}$ Chain Rule for Derivatives $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d^2} {\d x^2} f \left({x}\right)$ $=$ $\displaystyle \sec^2 x \frac \d {\d x} e^{\tan x} + e^{\tan x} \frac \d {\d x} \sec^2 x$ Product Rule $\displaystyle$ $=$ $\displaystyle \sec^4 x \, e^{\tan x} + e^{\tan x} 2 \sec^2 x \tan x$ $\displaystyle$ $=$ $\displaystyle \left({\sec^2 x + 2 \tan x}\right) \sec^2 x \, e^{\tan x}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d^3} {\d x^3} f \left({x}\right)$ $=$ $\displaystyle \left({\sec^2 x + 2 \tan x}\right) \frac \d {\d x} \sec^2 x \, e^{\tan x} + \sec^2 x \, e^{\tan x} \frac \d {\d x} \left({\sec^2 x + 2 \tan x}\right)$ Product Rule $\displaystyle$ $=$ $\displaystyle \left({\sec^2 x + 2 \tan x}\right) \left({\sec^2 x + 2 \tan x}\right) \sec^2 x \, e^{\tan x} + \sec^2 x \, e^{\tan x} \left({\sec^2 x \tan x + 2 \sec^2 x}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({\sec^2 x + 2 \tan x}\right)^2 + \sec^2 x \tan x + 2 \sec^2 x}\right) \sec^2 x \, e^{\tan x}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d^4} {\d x^4} f \left({x}\right)$ $=$ $\displaystyle \left({\left({\sec^2 x + 2 \tan x}\right)^2 + \sec^2 x \tan x + 2 \sec^2 x}\right) \frac \d {\d x} \sec^2 x \, e^{\tan x}$ Product Rule $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sec^2 x \, e^{\tan x} \frac \d {\d x} \left({\left({\sec^2 x + 2 \tan x}\right)^2 + \sec^2 x \tan x + 2 \sec^2 x}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({\sec^2 x + 2 \tan x}\right)^2 + \sec^2 x \tan x + 2 \sec^2 x}\right)\left({\sec^2 x + 2 \tan x}\right) \sec^2 x \, e^{\tan x}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sec^2 x \, e^{\tan x} \left({2 \left({\sec^2 x + 2 \tan x}\right) \left({\sec^2 x \tan x + 2 \sec^2 x}\right) + 2 \sec^2 x \tan x + \sec^4 x + 4 \sec^2 x \tan x}\right)$

By definition of Taylor series:

$f \left({x}\right) \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!} f^{\left({n}\right)} \left({\xi}\right)$

This is to be expanded about $\xi = 0$.

Note that $\tan 0 = 0$ and $\sec 0 = 1$.

Thus:

 $\displaystyle f \left({0}\right)$ $=$ $\displaystyle e^0$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle \left.{\frac \d {\d x} f \left({x}\right)}\right\vert_{x \mathop = 0}$ $=$ $\displaystyle \sec^2 0 \, e^{\tan 0}$ $\displaystyle$ $=$ $\displaystyle 1 \times e^0$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle \left.{\frac {\d^2} {\d x^2} f \left({x}\right)}\right\vert_{x \mathop = 0}$ $=$ $\displaystyle \left({\sec^2 0 + 2 \tan 0}\right) \sec^2 0 \, e^{\tan 0}$ $\displaystyle$ $=$ $\displaystyle \left({1 + 0}\right) 1 \times e^0$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle \left.{\frac {\d^3} {\d x^3} f \left({x}\right)}\right\vert_{x \mathop = 0}$ $=$ $\displaystyle \left({\left({\sec^2 0 + 2 \tan 0}\right)^2 + \sec^2 0 \tan 0 + 2 \sec^2 0}\right) \sec^2 0 \, e^{\tan 0}$ $\displaystyle$ $=$ $\displaystyle \left({\left({1 + 0}\right)^2 + 1 \times 0 + 2}\right) 1 \times e^0$ $\displaystyle$ $=$ $\displaystyle 3$ $\displaystyle \left.{\frac {\d^4} {\d x^4} f \left({x}\right)}\right\vert_{x \mathop = 0}$ $=$ $\displaystyle \left({\left({\sec^2 0 + 2 \tan 0}\right)^2 + \sec^2 0 \tan 0 + 2 \sec^2 0}\right)\left({\sec^2 0 + 2 \tan 0}\right) \sec^2 0 \, e^{\tan 0}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sec^2 0 \, e^{\tan 0} \left({2 \left({\sec^2 0 + 2 \tan 0}\right) \left({\sec^2 x \tan 0 + 2 \sec^2 0}\right) + 2 \sec^2 0 \tan 0 + \sec^4 0 + 4 \sec^2 0 \tan 0}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({1 + 0}\right)^2 + 0 + 2}\right) \left({1 + 0}\right) + \left({2 \left({1 + 0}\right) \left({0 + 2}\right) + 0 + 1 + 0}\right)$ $\displaystyle$ $=$ $\displaystyle 8$

from which the result.