Question

What is the taylor series expansion for the tangent function (tanx)?

Answer 1

`tan x = x + 1/3x^3 +2/15x^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) = tanx`

Then, we compute the first few derivatives:

`\ \ \ \ \ \ \ \ \ \ \ \ = sec^2(x)`

`f^((2))(x) = (2 sec^2x)(secx tanx)()`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2 sec^2x tanx`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2 (1+tan^2x) tanx`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2 (tanx+tan^3x)`

`f^((3))(x) = 2{sec^2x+3tan^2x sec^2x}`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2sec^2x{1+3tan^2x}`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2sec^2x{1+3(sec^2x-1)}`
`\ \ \ \ \ \ \ \ \ \ \ \ = 2sec^2x{1+3sec^2x-3}`
`\ \ \ \ \ \ \ \ \ \ \ \ = 6sec^4x-4sec^2x`

`vdots`

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

`f^((0))(x) = 0`
`f^((1))(x) = 1`
`f^((2))(x) = 0`
`f^((3))(x) = 2`
`vdots`

Which permits us to form the Maclaurin serie:

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

`\ \ \ \ \ \ \ = x + 1/3x^3 + 2/15^5x^5 + ...`