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\def\header{{\hfill Exam 1\hfill A. Miller \hfill Spring 2004
                                       \hfill Math 222\hfill}}
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    \string& \qquad \string\\ \string\hline}
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\begin{document}

\bigskip

Show all work.  
Simplify your answers.  
Circle your answer.

No notes, no books, no calculator, no cell phones, no pagers, no
electronic devices. 


\par\vskip .25in
\begin{center}
Name\makebox[3in]{\hrulefill} \\
\end{center}

\bigskip
Hand in to your TA.  Circle your TA's name:

{\large
\begin{tabbing}
xxxxxxxxx \=Ben Ellison xxxxxxxxxxxxxxx \=          Derek Moffitt \kill
\> Adam Berliner \>        Dan McGinn \\
\> Ben Ellison \>          Derek Moffitt \\
\> Jon Godshall \>         Richard Oberlin \\
\> E. Alec Johnson \> \\
\end{tabbing}
}



\bigskip

\begin{center} \Large
 \begin{tabular}{||c|c|c||} \hline\hline
  Problem & Points & Score \\  \hline\hline
   \input{exam.tbl}
  \hline \hline
 \end{tabular}
\end{center}

\immediate\openout\examTBLFile=exam.tbl

\bigskip
 
Solutions will be posted shortly after the exam:
www.math.wisc.edu/$\sim$miller/m222

\setcounter{page}{0}

\newpage 

\prob{4} % 8.1-37
Find the following integral:
$$\int {{e^{\arctan(x)}}\over{1+x^2}}\;\;dx$$
\answer Substitute $u=\arctan(x)$ and $du={1\over{1+x^2}}dx$ getting
$$\int {{e^{\arctan(x)}}\over{1+x^2}}\;\;dx=\int e^u\;du=e^u+C=
e^{\arctan(x)}+C$$

\prob{5}   % 10.8 -17
In the power series for the following function 
$$(1+x)^{5\over 2}=\sum_{n=0}^\infty a_nx^n$$
what is $a_3$?

\answer $f(x)=(1+x)^{5\over 2}$
\par\bigskip $f^\prime(x)={5\over 2}(1+x)^{3\over 2}$
\par\bigskip $f^{\prime\prime}(x)={15\over 4}(1+x)^{1\over 2}$
\par\bigskip $f^{\prime\prime\prime}(x)={15\over 8}(1+x)^{-{1\over 2}}$
\par\bigskip $f^{\prime\prime\prime}(0)={15\over 8}$
\par\bigskip $a_3={f^{\prime\prime\prime}(0)\over 3!}=
({15\over 8})({1\over 6})={5\over 16}$


\prob{5}   % 10.7 - 2
Starting with the geometric series, show how to obtain the
power series representation for 
$$f(x)={1\over(1+x)^2}$$
\answer \par ${1\over 1-x}=1+x+x^2+x^3+\cdots$ 
\par\bigskip substituting $-x$
\par ${1\over 1+x}=1-x+x^2-x^3+\cdots$ 
\par \bigskip differentiating both sides
\par  $-{1\over (1+x)^2}=-1+2x-3x^2+4x^3+\cdots$ 
\par\bigskip multiplying by $-1$
\par ${1\over (1+x)^2}=1-2x+3x^2-4x^3+\cdots=\sum_{n=0}^\infty(-1)^n(n+1)x^n$

\prob{5}   % 10.6 - 17
Determine all $x$ such that the following series converges: 
$${3^2x^2\over 3} + {3^3x^3\over 4} + {3^4x^4\over 5} +{3^5x^5\over 6}
+\cdots $$
\answer $\sum_{n=2}^\infty {3^n x^n\over n+1}$.  Apply the ratio test
$${{|{3^{n+1} x^{n+1}\over n+2}|}\over {|{3^n x^n\over n+1}|}}=
|x|\cdot{3^{n+1}\over 3^n}\cdot{n+1\over n+2}=
|x|\cdot{3}\cdot{n+1\over n+2}\to |x|\cdot 3
$$
So the series converges for $|x|\cdot 3< 1$ or equivalently $|x|<{1\over 3}$ 
and diverges for $|x|>{1\over 3}$.  For the two end points, 
$x={1\over 3}$ and $x=-{1\over 3}$, the
series is 
$$\sum_{n=2}^\infty {3^n x^n\over n+1}=
\sum_{n=2}^\infty {3^n ({1\over 3})^n\over n+1}=
\sum_{n=2}^\infty {1\over n+1}$$
which is a tail of the harmonic series and so diverges, and
$$\sum_{n=2}^\infty {3^n x^n\over n+1}=
\sum_{n=2}^\infty {3^n (-{1\over 3})^n\over n+1}=
\sum_{n=2}^\infty {(-1)^n\over n+1}$$
which is an alternating series and so converges.  Hence this power
series converges for exactly those $x$ which satisfy
$-{1\over 3}\leq x<{1\over 3}$.

\prob{5}  % 10.5 - 21
Determine whether the following series is absolutely convergent,
conditionally convergent, or divergent and explain why.
$$\sum_{n=1}^\infty (-1)^{n+1}{n\over {n^2+1}}$$
\answer \par 
$${{{n\over {n^2+1}}}\over{1\over n}}={n^2\over{n^2+1}}\to 1 \mbox{ as }
n\to\infty$$
The harmonic series $\sum{1\over n}$ diverges, so by the limit comparison
test $\sum_{n=1}^\infty {n\over {n^2+1}}$ diverges.  The original series
converges by the alternating series test and so the series is conditionally
convergent.

\prob{5}  % 10.3 - 12 
Use the integral test to
determine whether the following series converges or diverges.
$$\sum_{k=2}^\infty {1\over {k(\ln(k))^2}}$$

\answer $\int{1\over {x(\ln(x))^2}}dx =\int{1\over u^2}\;du={-1\over u}$
after substituting $u=\ln(x)$ and $du={1\over x}dx$.  Hence
$$\int_2^\infty{1\over {x(\ln(x))^2}}dx=-\left({1\over\ln(x)}
\right]^\infty_2=
{1\over\ln(2)}-{1\over\ln(\infty)}={1\over\ln(2)}$$
which is finite.  Since the function is positive and eventually
decreasing the associated series converges by the integral test.


\prob{5} % 8.2 - 17
Find
$$\int \tan^4(x)\;dx$$
\answer Using $1+\tan^2(x)=\sec^2(x)$ we have that
$$\int \tan^4(x)\;dx=\int (\sec^2(x)-1)\tan^2(x)\;dx=
\int \sec(x)^2\tan^2(x)\;dx- \int \tan^2(x)\;dx 
$$
The first integral is $\tan^3(x)\over 3$ after substituting
$u=\tan(x)$ and $du=\sec^2(x)dx$.  The second integral is
$$\int \tan^2(x)\;dx=\int \sec^2(x)-1\;dx=\tan(x)-x$$
giving us that answer
$${\tan^3(x)\over 3}-\tan(x)+x+C$$

\prob{5} % 8.3 - 21
Find
$$\int \sqrt{1-x^2}\;dx$$
\answer Substituting $x=\sin(\theta)$ and $dx=\cos(\theta)d\theta$ gives
us 
$$\int \sqrt{1-x^2}\;dx=\int\sqrt{1-\sin^2(\theta)}\cos(\theta)d\theta
=\int \cos^2(\theta)\;d\theta$$
using the formula $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$ and so
$$\cos(2A)=\cos^2(A)-\sin^2(A)=\cos^2(A)-(1-\cos^2(A))=2\cos^2(A)-1$$
and hence $\cos^2(\theta)={{1+\cos(2\theta)}\over 2}$ we have
that 
$$\int\cos^2(\theta)\;d\theta=\int{{1+\cos(2\theta)}\over 2}d\theta=
{1\over 2}\theta+{1\over 4}\sin(2\theta)+C$$
Since $\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$ we have
that $\sin(2\theta)=2\sin(\theta)\cos(\theta)$.  
Since $x=\sin(\theta)$ and
$\cos(\theta)=\sqrt{1-\sin^2(\theta)}=\sqrt{1-x^2}$ we have
that the
answer is
$${1\over 2}\arcsin(x)+{1\over 2}(x\sqrt{1-x^2})+C$$

\prob{5}  % 8.5 
Find
$$\int {{2x^2-1}\over{x^2(x-1)}}\;dx$$
\answer Let 
$${{2x^2-1}\over{x^2(x-1)}}={A\over x}+{B\over x^2}+{C\over (x-1)}$$
cross multiplying and equating numerators gives us
$$2x^2-1=Ax(x-1)+B(x-1)+Cx^2=(A+C)x^2+(-A+B)x-B$$
so $-B=-1$, $A+C=2$, $-A+B=0$.  Solving these gives $A=B=C=1$ and
therefore
$$\int {{2x^2-1}\over{x^2(x-1)}}\;dx=
\int {1\over x} + {1\over x^2} + {1\over(x-1)} \;dx
=\ln|x|-{1\over x}+\ln|x-1|+C$$

\prob{6}  % 9.3 - 13
Evaluate or show that it diverges:
$$\int_2^{\infty} {\ln(x)\over x^2}\;dx$$
\answer Using integration by parts 
let $u=\ln(x)$ and $dv={1\over x^2}\;dx$ we have
that $du={1\over x} \;dx$ and $v=-{1\over x}$ and so
$$\int {\ln(x)\over x^2}\;dx =
-{1\over x}\ln(x)+\int {1\over x^2}dx=
-{1\over x}\ln(x)- {1\over x}+C
$$
Hence 
$$\int_2^{\infty} {\ln(x)\over x^2}\;dx=\left. -{1\over x}\ln(x)- {1\over x}
\right]^\infty_2$$
$$\lim_{b\to\infty}{\ln(b)\over b}=\lim_{b\to\infty}{{1\over b}\over 1}=0$$
by L'Hopitol's Rule, so
our integral converges to
$${1\over 2}\ln(2)+{1\over 2}$$

\finishtbl

% \printanswers

\end{document}


