Question

What is the Maclaurin Series for `tanax`?

Answer 1

`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 `x=0`

`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 + ...`