diff --git a/ExponentialApproximants.png b/ExponentialApproximants.png
new file mode 100644
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diff --git a/exponential.tex b/exponential.tex
index dc137d2..853c104 100644
--- a/exponential.tex
+++ b/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).	
\ No newline at end of file
+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
+%$$
+%\sum_{n=0}^{\infty}(-1)^n \frac{\pi^{2 n}}{2^n n!}
+%$$
\ No newline at end of file
diff --git a/solutions.tex b/solutions.tex
index dc65399..bedcbae 100644
--- a/solutions.tex
+++ b/solutions.tex
@@ -91,9 +91,39 @@
 		\chapter*{EXPONENTIALS}
 		\section*{Euler's formula}
 		\subsection*{Answer 1}
+		
 		\label{ExponentialsAnswer1}
 		Setting $x=\pi$ in Euler's formula gives $e^{i \pi}=\cos \pi+i \sin \pi=-1$.
-%	
+		
+		\subsection*{Answer 2}
+		\label{ExponentialsAnswer2}
+		Note that this is the long polynomial for $\cos x$, evaluated at $x=\pi$. So the value is $\cos \pi=-1$.
+		
+		\subsection*{Answer 3}
+		\label{ExponentialsAnswer3}
+		Computing the derivative term by term gives
+		$$
+		\begin{aligned}
+			\frac{d}{d x} \sin (x) & =\frac{d}{d x}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\right) \\
+			& =1-3 \frac{x^2}{3!}+5 \frac{x^4}{5!}-\ldots \\
+			& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots
+		\end{aligned}
+		$$
+		which is the long polynomial for $\cos x$, as desired.
+
+		\subsection*{Answer 4}
+		\label{ExponentialsAnswer4}
+		Beginning with $e^x \cdot e^y$, we find
+		$$
+		\begin{aligned}
+			e^x \cdot e^y & =\left(1+x+\frac{x^2}{2!}+\cdots\right)\left(1+y+\frac{y^2}{2!}+\cdots\right) \\
+			& =1+(x+y)+\left(\frac{x^2}{2!}+x y+\frac{y^2}{2!}\right)+\cdots \\
+			& =1+(x+y)+\frac{x^2+2 x y+y^2}{2!}+\cdots \\
+			& =1+(x+y)+\frac{(x+y)^2}{2!}+\cdots,
+		\end{aligned}
+		$$
+		which is the long polynomial for $e^{x+y}$, as desired.
+		
 %		\subsection{Euler's Formula}
 %		
 %		\subsection{Additional Examples}
diff --git a/upenn.pdf b/upenn.pdf
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diff --git a/upenn.tex b/upenn.tex
index 52746d4..26ec66c 100644
--- a/upenn.tex
+++ b/upenn.tex
@@ -28,6 +28,14 @@
 \setkomafont{disposition}{\normalcolor\bfseries} % alle überschriften fett und serifenlos
 %\setkomafont{disposition}{\normalcolor}
 
+
+\usepackage{tikz}
+
+\newcommand{\tikzcircle}[2][red,fill=red]{\tikz[baseline=-0.5ex]\draw[#1,radius=#2] (0,0) circle ;}%
+
+
+
+
 \setlength\parindent{0pt}
 
 \everymath{\displaystyle}