# What is the Maclaurin Series for #tanax#?

##### 1 Answer

# tan ax = ax + 1/3a^3x^3 +2/15a^5x^5 + ...#

#### 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 + ...#

We start with the function

# f^((0))(x) = f(x) = tanax #

Then, we compute the first few derivatives:

# f^((1))(x) = (sec^2ax(a) #

# \ \ \ \ \ \ \ \ \ \ \ \ = a sec^2(ax) #

# f^((2))(x) = (2a sec^2ax)(secax tanax)(a) #

# \ \ \ \ \ \ \ \ \ \ \ \ = 2a^2 sec^2ax tanax #

# \ \ \ \ \ \ \ \ \ \ \ \ = 2a^2 (1+tan^2ax) tanax #

# \ \ \ \ \ \ \ \ \ \ \ \ = 2a^2 (tanax+tan^3ax) #

# f^((3))(x) = 2a^2{asec^2ax+3atan^2ax sec^2ax} #

# \ \ \ \ \ \ \ \ \ \ \ \ = 2a^3sec^2ax{1+3tan^2ax} #

# \ \ \ \ \ \ \ \ \ \ \ \ = 2a^3sec^2ax{1+3(sec^2ax-1)} #

# \ \ \ \ \ \ \ \ \ \ \ \ = 2a^3sec^2ax{1+3sec^2ax-3} #

# \ \ \ \ \ \ \ \ \ \ \ \ = 6a^3sec^4ax-4a^3sec^2ax #

# vdots #

Now we have the derivatives, we can compute their values when

# f^((0))(x) = 0 #

# f^((1))(x) = a #

# f^((2))(x) = 0 #

# f^((3))(x) = 2a^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 + ...#

# \ \ \ \ \ \ \ = ax + 1/3a^3x^3 + 2/15a^5x^5 + ... #