Question

Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\sec x, \quad c=0 \text { (first three nonzero terms) } $$

Short answer

The Taylor series for the function \(f(x) = \sec x\) centered at \(c=0\) for the first three nonzero terms is \(1 + x^2/2\).

Step-by-step solution

Calculate the function at the center point

Firstly, calculate the value of \(f(x)\) at the center point \(c=0\). Since \(f(x) = \sec x\), when \(x=0\) it equals 1, i.e. \(f(0) = 1\).

Calculate first three derivatives

The next step is to calculate the first three derivatives of the function, and evaluate these at \(c = 0\). The first derivative of \(f(x) = \sec x\) is \(f'(x) = \sec(x)\tan(x)\), substituting \(x = 0\), \(f'(0) = 0\). The second derivative \(f''(x) = \sec(x)\tan^2(x) + \sec^3(x)\), for \(x = 0\), \(f''(0) = 0 + 1 = 1\). The third derivative \(f'''(x) = 2\sec(x)\tan(x)\tan^2(x) + 5\sec^3(x)\tan(x)\), when \(x = 0\), \(f'''(0) = 0\).

Build the Taylor series

The Taylor series is built up according to the formula: \(f(c) + f'(c)(x-c)/1! + f''(c)(x-c)^2/2! + f'''(c)(x-c)^3/3!\). Substituting the values from the previous steps and keeping only non-zero terms, the Taylor series becomes: \(1 + 0 + 1*x^2/2! + 0 = 1 + x^2/2\)