1. **Problem statement:** Evaluate the Taylor series for the function $f(x) = \cos x$ centered at $x=0$. 2. **Formula:** The Taylor series of a function $f(x)$ at $x=a$ is given by: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x - a)^n$$ where $f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at $x=a$. 3. **Apply to $f(x) = \cos x$ at $a=0$:** - $f(x) = \cos x$ - $f'(x) = -\sin x$ - $f''(x) = -\cos x$ - $f^{(3)}(x) = \sin x$ - $f^{(4)}(x) = \cos x$ 4. **Evaluate derivatives at $x=0$:** - $f(0) = \cos 0 = 1$ - $f'(0) = -\sin 0 = 0$ - $f''(0) = -\cos 0 = -1$ - $f^{(3)}(0) = \sin 0 = 0$ - $f^{(4)}(0) = \cos 0 = 1$ 5. **Write the Taylor series:** Only even derivatives are nonzero, so the series is: $$\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ 6. **Explanation:** The Taylor series for $\cos x$ at $0$ consists of even powers of $x$ with alternating signs. This series converges to $\cos x$ for all real $x$. **Final answer:** $$\boxed{\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}}$$