Exponentials bearbeitet, rotes Bullet für item

exponential.tex

@@ -92,14 +92,14 @@ $$
 These formulas should be memorized, both in their long polynomial form and their more concise summation notation form.
 
 \subsection*{Example}
-Use Euler's formula to show that $e^{i \pi}=-1$. Answer
+Use Euler's formula to show that $e^{i \pi}=-1$. \hyperref[ExponentialsAnswer1]{Answer}
 \subsection*{Example}
-Compute $1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}-\cdots$. Answer
+Compute $1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}-\cdots$.  \hyperref[ExponentialsAnswer2]{Answer}
 \subsection*{Example}
-Check that taking the derivative of the long polynomial for $\sin x$ gives the long polynomial for $\cos x$ (hence, verify that $\frac{d}{d x} \sin x=\cos x$ ). Answer
+Check that taking the derivative of the long polynomial for $\sin x$ gives the long polynomial for $\cos x$ (hence, verify that $\frac{d}{d x} \sin x=\cos x$ ).  \hyperref[ExponentialsAnswer3]{Answer}
 
 \subsection*{Example}
-Show that the long polynomial for $e^x$ satisfies the first property above, namely $e^{x+y}=e^x e^y$. Hint: start with the long polynomials for $e^x$ and $e^y$ and multiply these together, and carefully collect like terms to show it equals the long polynomial for $e^{x+y}$. Answer
+Show that the long polynomial for $e^x$ satisfies the first property above, namely $e^{x+y}=e^x e^y$. Hint: start with the long polynomials for $e^x$ and $e^y$ and multiply these together, and carefully collect like terms to show it equals the long polynomial for $e^{x+y}$.  \hyperref[ExponentialsAnswer4]{Answer}
 
 
 \section{More on the long polynomial}
@@ -114,4 +114,27 @@ $$
 \end{aligned}
 $$
 
-Each polynomial in the sequence is, in a sense, the best approximation possible of that degree. Put another way, taking the first several terms of the long polynomial gives a good polynomial approximation of the function. The more terms included, the better the approximation. This is how calculators compute the exponential function (without having to add up infinitely many things).	
+Each polynomial in the sequence is, in a sense, the best approximation possible of that degree. Put another way, taking the first several terms of the long polynomial gives a good polynomial approximation of the function. The more terms included, the better the approximation. This is how calculators compute the exponential function (without having to add up infinitely many things).	
+
+\begin{figure}[h]
+	\centering
+	\includegraphics[width=0.7\linewidth]{ExponentialApproximants}
+%	\caption{}
+	\label{fig:exponentialapproximants}
+\end{figure}
+
+\section{EXERCISES}
+\begin{itemize}
+	
+	\item[\textcolor{red}{\tikzcircle{3pt}}] So, how good of an approximation is a polynomial truncation of $e^{\text {? }}$. Use a calculator to compare how close $e$ is to the linear, quadratic, cubic, quartic, and quintic approximations. How many digits of accuracy do you seem to be gaining with each additional term in the series?
+\end{itemize}
+%- 
+%- Now, do the same thing with $1 / e$ by plugging in $x=-1$ into the series. Do you have the same results? Are you surprised?
+%- Calculate the following sum using what you know:
+%$$
+%\sum_{n=0}^{\infty}(-1)^n \frac{(\ln 3)^n}{n!}
+%$$
+%- Write out the first four terms of the following series
+%$$
+%\sum_{n=0}^{\infty}(-1)^n \frac{\pi^{2 n}}{2^n n!}
+%$$