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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))?
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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: \[f(x) = 5\ln(\sec(x))\] \[f'(x) = 5 \times \frac{1}{\sec(x)} \times \sec(x)\tan(x)\] \[f''(x) = -\sec(x)\tan^2(x) + \sec^3(x) \]
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
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