% EXERCISE SEVEN
          
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\begin{document}
             
\begin{center}
	\textbf{Exercise Seven: Math Miscellany}
\end{center}  

\begin{dfn}
The \emph{inverse tangent} or \emph{arctangent} is denoted $\tan^{-1}x$ 
or $\arctan x$ and is the angle between $-\pi/2$ and $\pi/2$ whose tangent is 
equal to $x$.
\end{dfn}

\begin{ex}
Since $\tan(\pi/4)=1$, we also have $\arctan 1 = \pi/4$.
\end{ex}

\begin{ex}
However, $\tan(9\pi/4)=1$, but $\arctan 1 = \pi/4$.
\end{ex}

\begin{qtp}
Should we really be anthropomorphizing an angle by using the word ``whose'' here?
\end{qtp}

Now we use the arctangent and some calculus to derive a wonderful series.
\begin{align*}
\frac{\pi}{4}&=\arctan 1 \\
&=\int_0^1\frac{1}{1+x^2}\, dx && 
\text{Surely you remember that $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$.}\\
&=\int_0^1\frac{1}{1-(-x^2)}\, dx && 
\text{Prepare to use the series $\frac{1}{1-u}=1+u+u^2+u^3+\dots$}\\
&=\int_0^1[1-x^2+x^4-\dots]\, dx 
&& \text{Substitute $u=-x^2$ into the series above. Duh.}\\
&=\left[x-\frac{x^3}{3}+\frac{x^5}{5}-\dots\right] _0^1 
&& \text{Ever heard of the FTC?} \\
&=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots 
&&\text{A moment of silence, please.} 
\end{align*}

\begin{qtp}
What on earth do the reciprocals of the odd natural numbers have to do with $\pi$?!?
\end{qtp}
\end{document}