Exponentials bearbeitet, rotes Bullet für item

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}