Find the Taylor polynomial of degree two approximating the given function centered at the given point. \[ f(x)=\cos (4 x) \text { at } a=\pi \] \[ p_{2}(x)= \]
Solution
Step 1 ：Given the function \(f(x) = \cos(4x)\) and the point \(a = \pi\), we are asked to find the Taylor polynomial of degree two approximating the function centered at the point \(a\).
Step 2 ：The Taylor polynomial of degree two for a function \(f(x)\) centered at a point \(a\) is given by the formula: \(p_{2}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2\).
Step 3 ：First, we need to find the first and second derivatives of the function \(f(x) = \cos(4x)\). The first derivative \(f'(x)\) is \(-4\sin(4x)\) and the second derivative \(f''(x)\) is \(-16\cos(4x)\).
Step 4 ：Next, we evaluate these derivatives at the point \(a = \pi\). We find that \(f(a) = 1\), \(f'(a) = 0\), and \(f''(a) = -16\).
Step 5 ：Substituting these values into the Taylor polynomial formula, we get \(p_{2}(x)=1 - 8*(x - \pi)^2\).
Step 6 ：\(\boxed{p_{2}(x)=1 - 8*(x - \pi)^2}\) is the Taylor polynomial of degree two approximating the function \(f(x)=\cos (4 x)\) centered at \(a=\pi\).
