1. **Problem 1:** Evaluate the function $2\cos(2\pi t)$. 2. **Problem 2:** Evaluate the infinite series $$\sum_{n=0}^{\infty} \frac{(-1)^n s_0 \omega^{2n}}{(2n)!} t^{2n}$$ for $n=20$ and then compare it with the first function using $n=40$. --- ### Step 1: Understanding the first function The first function is given as: $$y_1 = 2\cos(2\pi t)$$ This is a cosine wave with amplitude 2 and angular frequency $2\pi$. ### Step 2: Understanding the infinite series The infinite series is: $$y_2 = \sum_{n=0}^{\infty} \frac{(-1)^n s_0 \omega^{2n}}{(2n)!} t^{2n}$$ This is the Taylor series expansion of $s_0 \cos(\omega t)$. ### Step 3: Evaluate the series at $n=20$ We approximate the series by summing terms from $n=0$ to $n=20$: $$y_2^{(20)} = \sum_{n=0}^{20} \frac{(-1)^n s_0 \omega^{2n}}{(2n)!} t^{2n}$$ ### Step 4: Evaluate the series at $n=40$ Similarly, approximate the series by summing terms from $n=0$ to $n=40$: $$y_2^{(40)} = \sum_{n=0}^{40} \frac{(-1)^n s_0 \omega^{2n}}{(2n)!} t^{2n}$$ ### Step 5: Compare $y_1$ and $y_2^{(40)}$ Since $y_2$ represents $s_0 \cos(\omega t)$, if we set $s_0=2$ and $\omega=2\pi$, then: $$y_2 = 2 \cos(2\pi t) = y_1$$ The series approximation $y_2^{(40)}$ should be very close to $y_1$. ### Step 6: Summary of evaluation - For $n=20$, the partial sum $y_2^{(20)}$ approximates $2\cos(2\pi t)$ but with less accuracy. - For $n=40$, the partial sum $y_2^{(40)}$ is a much better approximation of $2\cos(2\pi t)$. ### Additional notes - Factorials grow very fast, so terms for large $n$ become very small. - The series converges to the cosine function for all real $t$. --- **Final answers:** $$y_1 = 2\cos(2\pi t)$$ $$y_2^{(20)} = \sum_{n=0}^{20} \frac{(-1)^n 2 (2\pi)^{2n}}{(2n)!} t^{2n}$$ $$y_2^{(40)} = \sum_{n=0}^{40} \frac{(-1)^n 2 (2\pi)^{2n}}{(2n)!} t^{2n}$$ where $y_2^{(40)}$ closely matches $y_1$.