\documentclass{amsart} \usepackage[top=1in,bottom=1.5in,left=1in,right=1in]{geometry} \newcommand{\solution}{\noindent\textbf{Solution: }} \usepackage{enumerate} \begin{document} These are a few review problems that I stole from previous midterms at UC Berkeley. \section{Problem 1} Find the sums of the following series \begin{enumerate}[(a)] \item $\sum_{n=1}^\infty \left[\left(\ln\frac{n+1}{n}\right)-\ln\left(\frac{n+2}{n+1}\right)\right]$\\ \item $\sum_{n=3}^\infty \left(3\left(\frac{3}{4}\right)^n - 4 \left(-\frac{1}{2}\right)^{n+1}\right)$ \end{enumerate} \solution The first one is a telescoping series, so you just need to check that terms go to zero, and then the sum will be equal to the first term.\\ In the second problem, split the series into two smaller series, that is \[\sum_{n=3}^\infty \left(3\left(\frac{3}{4}\right)^n - 4 \left(-\frac{1}{2}\right)^{n+1}\right)=\sum_{n=3}^\infty \left(3\left(\frac{3}{4}\right)^n\right) - 4\sum_{n=3}^\infty\left( \left(-\frac{1}{2}\right)^{n+1}\right)\] These are both geometric series, so you can find there sum. Be careful with, because the series start the sum at $n=3$, not $n=0$ \section{Problem 2} Find the Maclaurin Series for $f(x) = xe^x$. Approximate $f(1)$ using a third degree approximation, and use Taylor's inequality to bound the error of your approximation. \\ \solution You can show that the $n$th derivative of $f(x)$ is \[f^{(n)}=ne^x+xe^x\] Evaluating at $x=0$, we have that the Maclaurin series is \[x+\frac{x^2}{1!}+\frac{x^3}{2!}+\frac{x^4}{3!}+\cdots +\frac{x^n}{(n-1)!}+\cdots\] Using the third degree approximation gives $1+1+\frac{1}{2}$. Taylor's inequality says that the error should be bounded by the maximum of the fourth derivative on the interval $[-1, 1]$. The maximum of $f^{(4)}(x)=4e^x+xe^x$ on the interval $[-1, 1]$ occurs on the right endpoint. The value of the fourth derivative here is $f^{(4)}(1)=5e$. Letting $M=5e$, we have that the error for this approximation is less than $\frac{M}{4!}(x-a)^4=\frac{5e}{4!}$. \section{Problem 3} Solve the differential equation $y'\sqrt{x^2-2x+2}=y^2$ for the initial condition $y(0)=1$ \section{Problem 4} Let $P'=P(P-1)^2 $. Find the equilibria solutions for this differential equation, and the behavior of $\lim_{t\to \infty} P(t)$ for all initial values for $P(t)$.\\ \solution The equilibria solutions to this differential equation are where $P'=0$, which occurs when $P=0$ and $P=1$. Since $P'$ is non-negative, it means that $P$ will always be increasing. \begin{enumerate} \item Case 1: $P(0)\leq0$. Then $P$ will increase to $0$ and stop, so $\lim_{t\to\infty} P(t)=0$. \item Case 2: $0