Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\sec x, \quad n=2 $$

Short answer

Maclaurin polynomial of degree 2 for \(f(x) = \sec x\) is \(1 + \frac{x^2}{2}\).

Step-by-step solution

Calculation First and Second Derivatives at 0

First, the first and second derivatives of the function \(f(x) = \sec x\) at \(x=0\) need to be calculated. Using the chain rule and the definition of the derivative, we start by finding \(f'(x) = \sec x \tan x\) and \(f''(x) = \sec x \tan^2 x + \sec^3 x\). Now, substituting \(x=0\) into these, we have: \(f'(0) = \sec(0) \tan(0) = 1*0 = 0\) and \(f''(0) = \sec(0) \tan^2(0) + \sec^3(0) = 1*0 + 1 = 1\).

Maclaurin Polynomial Formulation

Next we use these derivatives to form our Maclaurin polynomial. A Maclaurin series is a Taylor series expansion of a function about 0, in the form of: \( f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \cdots\). Using our calculated derivatives and \(f(0)\), our polynomial with degree \(n = 2\) becomes: \(1 + 0 \cdot x + \frac{1}{2} \cdot x^2\).