# What is the Maclaurin Series for tanax?

Feb 5, 2018

$\tan a x = a x + \frac{1}{3} {a}^{3} {x}^{3} + \frac{2}{15} {a}^{5} {x}^{5} + \ldots$

#### Explanation:

The Maclaurin series is given by

 f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...

${f}^{\left(0\right)} \left(x\right) = f \left(x\right) = \tan a x$

Then, we compute the first few derivatives:

 f^((1))(x) = (sec^2ax(a)
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = a {\sec}^{2} \left(a x\right)$

${f}^{\left(2\right)} \left(x\right) = \left(2 a {\sec}^{2} a x\right) \left(\sec a x \tan a x\right) \left(a\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {a}^{2} {\sec}^{2} a x \tan a x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {a}^{2} \left(1 + {\tan}^{2} a x\right) \tan a x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {a}^{2} \left(\tan a x + {\tan}^{3} a x\right)$

${f}^{\left(3\right)} \left(x\right) = 2 {a}^{2} \left\{a {\sec}^{2} a x + 3 a {\tan}^{2} a x {\sec}^{2} a x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {a}^{3} {\sec}^{2} a x \left\{1 + 3 {\tan}^{2} a x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {a}^{3} {\sec}^{2} a x \left\{1 + 3 \left({\sec}^{2} a x - 1\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {a}^{3} {\sec}^{2} a x \left\{1 + 3 {\sec}^{2} a x - 3\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 6 {a}^{3} {\sec}^{4} a x - 4 {a}^{3} {\sec}^{2} a x$

$\vdots$

Now we have the derivatives, we can compute their values when $x = 0$

${f}^{\left(0\right)} \left(x\right) = 0$
${f}^{\left(1\right)} \left(x\right) = a$
${f}^{\left(2\right)} \left(x\right) = 0$
${f}^{\left(3\right)} \left(x\right) = 2 {a}^{3}$
$\vdots$

Which permits us to form the Maclaurin serie:

 f(x) = (0) + (a)/(1)x + (0)/(2)x^2 + (2a^3)/(6)x^3 + ... (f^((n))(0))/(n!)x^n + ...

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = a x + \frac{1}{3} {a}^{3} {x}^{3} + \frac{2}{15} {a}^{5} {x}^{5} + \ldots$