Question

find the taylor polynomial of orders 0, 1, 2, and 3 generated by f at a
$f(x)=\cos x,a = \frac{2\pi}{3}$
$p_0(x)=\square$
(type an exact answer, using \\(\pi\\) as needed.)

Explanation:

Step1: Recall the formula for Taylor polynomial of order 0

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(x = a\) is given by \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). For \(n = 0\), \(P_0(x)=\frac{f(a)}{0!}(x - a)^0=f(a)\) (since \(0!=1\) and \((x - a)^0 = 1\) for \(x
eq a\) and \((x - a)^0=1\) when considering the polynomial form).

Step2: Calculate \(f(a)\)

Given \(f(x)=\cos x\) and \(a=\frac{2\pi}{3}\). Then \(f(a)=\cos(\frac{2\pi}{3})\). Using the unit - circle definition of cosine (\(\cos\theta=x\) - coordinate of the point on the unit circle corresponding to the angle \(\theta\)), we know that \(\cos(\frac{2\pi}{3})=-\frac{1}{2}\).

Answer:

\(-\frac{1}{2}\)