Question

Find the degree 3 taylor polynomial approximation to the function
f(x)=5 ln(sec(x)) about the point a=0 .
What is 5 ln(sec(x))?

Answer 1

This sounds like a question from Webwork. :P

Referring to the formula of the Taylor Series, you know you have to get the following values:
\[f(x) = 5\ln(\sec(x))\]
\[f'(x) = 5 \times \frac{1}{\sec(x)} \times \sec(x)\tan(x)\]
\[f''(x) = -\sec(x)\tan^2(x) + \sec^3(x) \]

Answer 2

It's given that the function is centred at x=0 (or in this case, a=0).
Note: cos(0)=1, tan(0)=0

So when f(0) = 0.
f'(0)=0
f''(0) = 1