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\begin{center}
Math 132: Calculus for Physical and Life Sciences 2\\ 
Problem Set 9 \\ 
Due Friday, April 18, 2008, at the beginning of class.
\end{center}
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General Directions: You must show all work for credit on these problems. 

\begin{enumerate}
\item Determine whether each of the following series converges
absolutely, converges conditionally, or diverges, using some
combination of the Alternating Series Test, the Ratio Test, and
the Integral Test. 

\begin{enumerate}
\item[(a)]  ${\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{e^n}}$

\sol 
Using the Ratio Test, 
\[
\lim_{n\to\infty} \left|\frac{ \frac{(-1)^{n+1}}{e^{n+1}}}{\frac{(-1)^n}{e^n}}\right|
= \lim_{n\to\infty} \frac{1}{e} < 1.
\]
Therefore, the series \emph{converges absolutely}.  Note: the fact that the ratio
is a constant says this is actually a geometric series with ratio $\frac{-1}{e}$.
Since this is less than 1 in absolute value, this series, and the series of 
absolute values both converge.    

\item[(b)] ${\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (2n+1)}{4n-3}}$

\sol
The terms in this series are not tending to zero: 
${\displaystyle 
\lim_{n\to\infty} \frac{2n+1}{4n-3} 
= \frac{1}{2} \ne 0}$. Therefore this series is \emph{divergent}.

\item[(c)]  ${\displaystyle \sum_{n=1}^\infty \frac{1}{n 7^n}}$

\sol
Using the Ratio Test,
\[
\lim_{n\to\infty} 
\left|\frac{\frac{1}{(n+1)7^{n+1}}}{\frac{1}{n 7^n}}\right| 
= \lim_{n\to\infty} \frac{n}{7(n+1)} = \frac{1}{7} < 1.
\]
So this series is \emph{absolutely convergent}.  (Note that the series itself has all
positive terms, so absolute convergence and ``ordinary'' convergence are the same in
this example.)

\item[(d)] ${\displaystyle \sum_{n=2}^\infty \frac{(-1)^{n-1}}{\ln(n)}}$

\sol
The function $f(x) = \frac{1}{\ln(x)}$ has $f'(x) = \frac{-2}{x(\ln(x))^2} < 0$ for
all $x > 1$ and ${\displaystyle \lim_{x\to\infty} \frac{1}{\ln(x)} = 0}$.  Therefore,
the series with the alternating signs converges by the Alternating Series Test. 
On the other hand, for all $x \ge 2$, $\ln(x) < x\ln(x)$, so $\frac{1}{\ln(x)} > \frac{1}{x\ln(x)}$
This implies that $\int_2^\infty \frac{1}{\ln(x)}\ dx$ diverges since $\int_2^\infty \frac{1}{x\ln(x)}\ dx$
diverges.  Therefore by the Integral Test,
${\displaystyle \sum_{n=2}^\infty \frac{1}{\ln(n)}}$ diverges.  The given
series is \emph{conditionally convergent}.  

\item[(e)]  ${\displaystyle \sum_{n=1}^\infty \frac{n^n}{n!}}$
\item[](Hint for (e): Try the Ratio Test on this one because of the powers and factorials.
The limit you will need to compute is a $1^\infty$ indeterminate form.
These are covered in Section 4.5 of the text if you need to review.)

\sol
Using the Ratio Test, 
\[
\lim_{n\to\infty} \left|\frac{ \frac{(n+1)^{n+1}}{(n+1)!} }{\frac{n^n}{n!}}\right|
= \lim_{n\to\infty}  \left(\frac{n+1}{n}\right)^n = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n.
\]
This limit is indeterminate of the form $1^\infty$.  To evaluate it, we will apply L'Hopital's
Rule after taking logarithms:
\[
\ln\left(\left(1 + \frac{1}{n}\right)^n\right) = n\ln\left(1 + \frac{1}{n}\right) = 
\frac{\ln\left(1 + \frac{1}{n}\right)}{\frac{1}{n}}.
\]
The limit here as $n\to\infty$ is indeterminate $0/0$.  So by L'Hopital it should be the 
same as the limit
\[
\lim_{n\to\infty} 
\frac{\frac{1}{1 + \frac{1}{n}}\cdot \frac{-1}{n^2} }{\frac{-1}{n^2}} =
\lim_{n\to\infty} \frac{1}{1 + \frac{1}{n}} = 1.
\]
Therefore 
\[
\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = e^1 = e > 1,
\]
and the series \emph{diverges}.
\end{enumerate}

\item For each of the following power series, 
\begin{enumerate}
\item[(i)] Determine the radius of convergence $R$ using the Ratio Test.
\item[(ii)] If $0< R < \infty$, determine whether the series converges at $a \pm R$.
\item[(iii)] Based on (i) and (ii), give the interval of convergence.
\end{enumerate}

\begin{enumerate}
\item[(a)]  ${\displaystyle \sum_{n=0}^\infty \frac{(x - 2)^n}{n}}$

\sol
(i)  Using the Ratio Test,
\[
\left|\frac{ \frac{(x-2)^{n+1}}{n+1}}{\frac{(x-2)^n}{n}}\right| = \frac{n}{n+1}|x - 2|.
\]
Therefore, 
\[
\lim_{n\to\infty} \frac{n}{n+1}|x - 2| < 1 \Leftrightarrow |x - 2| < 1,
\]
or $1 < x < 3$.  That is, the radius of convergence is $R = 1$.  

(ii)  If $x = 1$, the series is an alternating harmonic series, which converges
by the Alternating Series Test.  If $x = 3$, the series is a divergent harmonic
series.  

(iii)  The interval of convergence is $[1,3)$.

\item[(b)]  ${\displaystyle \sum_{n=0}^\infty \frac{2^n x^n}{n!}}$

\sol
(i)  Using the Ratio Test,
\[
\left|\frac{ \frac{2^{n+1} x^{n+1}}{(n+1)!}}{\frac{2^n x^n}{n!}}\right| = \frac{2|x|}{n+1}.
\]
Therefore, 
\[
\lim_{n\to\infty}  \frac{2|x|}{n+1} = 0 < 1
\]
for all $x$.  That is, the radius of convergence is $R = \infty$.  

(ii)  There are no endpoints to check for this series(!)

(iii)  The interval of convergence is $(-\infty,\infty)$.
\item[(c)]  ${\displaystyle \sum_{n=0}^\infty n! (x-4)^n }$

\sol
(i)  Using the Ratio Test,
\[
\left|\frac{(n+1)!(x - 4)^{n+1}}{n!(x - 4)^n}\right| = (n+1)|x-4|.
\]
Therefore for any particular $x \ne 4$, 
\[
\lim_{n\to\infty}  (n+1)|x - 4| = +\infty.
\]
If $x = 4$, then the series converges since all terms after
the first are zero.  That is, the radius of convergence is $R = 0$.  

(ii)  There are no endpoints to check for this series(!)

(iii)  The interval of convergence consists of the single number $x = 4$.
\end{enumerate}

\item Using the definition, find the
Taylor series of each of the following functions $f(x)$
at the indicated $a$.  Write the series using summation notation and also
give the first four nonzero terms.

\begin{itemize}
\item[(a)]  $f(x) = e^{2x}$, $a = 0$.

\sol 
Each time we differentiate $e^{2x}$, we get an extra factor of $2$ by the Chain
Rule.  Therefore, $f^{(n)}(0) = 2^n$, and the Taylor Series is
\[
\sum_{n=0} \frac{f^{(n)}(0)}{n!}(x - a)^n = \sum_{n=0}^\infty \frac{2^x x^n}{n!}.
\]
The first four nonzero terms are:
\[
1 + 2x + 2x^2 + \frac{4}{3} x^3.
\]
(Note that this is the same series as in part (b) of question 2 above(!))
\item[(b)]  $f(x) = \cos(x)$, $a = \pi/2$.

\sol
We have
\begin{eqnarray*}
f(\pi/2) &=& 0\\
f'(x) = -\sin(x) &\Rightarrow& f'(\pi/2) = -1\\
f''(x) = -\cos(x) &\Rightarrow& f''(\pi/2) = 0\\
f^{(3)}(x) = \sin(x) &\Rightarrow& f^{(3)}(\pi/2) = 1\\
f^{(4)}(x) = \cos(x) &\Rightarrow& f^{(4)}(\pi/2) = 0,
\end{eqnarray*} 
and thereafter the derivatives repeat according to the same pattern.
The series has the form
\[
\sum_{k=0}^\infty (-1)^{k+1} \frac{(x - \pi/2)^{2k+1}}{(2k+1)!},
\]
with first four nonzero terms given by 
\[
-(x-\pi/2) + \frac{(x - \pi/2)^3}{3!} - \frac{(x-\pi/2)^5}{5!} + \frac{(x-\pi/2)^7}{7!}.
\]
(Note, in this series, we can see the trig identity $\cos(x) = -\sin(x - \pi/2)$; 
do you see why?)
\end{itemize}

\item 
\begin{enumerate}
\item[(a)] Starting from the geometric series expansion for 
${\displaystyle f(x) = \frac{1}{1+x}}$, determine a power series 
(with $a = 0$) representing the 
function $\ln(1 + x)$.

\sol
By the geometric series formula with $r = -x$, we have
\[
\frac{1}{1 + x} = \sum_{n=0}^\infty (-x)^n = 1 - x + x^2 - x^3 + x^4 - \cdots\ ,
\]
and the series converges absolutely for all $x$ satisfying  $|x| < 1$.  Therefore
\begin{eqnarray*}
\ln(1 + x) &=& \int \frac{dx}{x + 1}\\
&=& \int \sum_{n=0}^\infty (-x)^n\ dx = 1 - x + x^2 - x^3 + x^4 - \cdots\ dx\\
&=& x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots + C\\
&=& \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n} + C,
\end{eqnarray*}
for some constant of integration $C$.  The value of $C$ can be determined
by setting $x = 0$.  Since $\ln(1) = 0$, $C = 0$, and we have
\[
\ln(1 + x) = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n}
\]
for all $x$ with $|x| < 1$.  
\item[(b)] What is the radius of convergence of your series from part (a)?

\sol
\[
\left|\frac{(-1)^{n}x^{n+1}/(n+1)}{(-1)^{n-1}x^n/n}\right| = \frac{n}{n+1}|x|.
\]
Therefore
\[
\lim_{n\to\infty}  \frac{n}{n+1}|x| = |x| < 1
\]
if and only if $|x| < 1$.  That is, the radius of convergence is $R = 1$.  

\item[(c)] Compute the Taylor series of $\ln(1 + x)$ at $a = 0$.  How
do your answers to parts (a) and (c) compare?

\sol
We have for $f(x) = \ln(1 + x)$, 
\begin{eqnarray*}
f(0) = \ln(1) &=& 0\\
f'(x) = \frac{1}{1+x} &\Rightarrow& f'(0) = 1\\
f''(x) = \frac{-1}{(1+x)^2} &\Rightarrow& f''(0) = -1\\
f^{(3)}(x) = \frac{2}{(1+x)^3} &\Rightarrow& f^{(3)}(0) = 2\\
&\vdots & \\
f^{(n)}(x) = \frac{(-1)^{n-1} (n-1)!}{(1 + x)^n} &\Rightarrow& f^{(n)}(0) = (-1)^{n-1}(n-1)!.
\end{eqnarray*}
Therefore, since the first term is zero anyway, we can start the summation 
from $n = 1$.  Cancelling terms in the factorials, the Taylor series is
\[
\sum_{n=1}^\infty \frac{(-1)^{n-1} (n-1)!}{n!} x^n = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n}.
\]
This is the same series as in part (a).
\end{enumerate}

\item 
\begin{enumerate}
\item[(a)] Starting from the Taylor series for $f(t) = \sin(t)$, find a 
power series representation for the function 
\[
Si(x) = \int_0^x \frac{\sin(t)}{t}\, dt.
\]

\sol
The Taylor series for $\sin(t)$ is 
\[
\sum_{k=0}^\infty (-1)^k \frac{t^{2k+1}}{(k+1)!} = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \cdots\ .
\]
Hence, dividing each term, then integrating, we get:
\begin{eqnarray*}
\int_0^x \frac{\sin(t)}{t}\, dt &=& \int_0^x 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \cdots\ dt\\
&=& x - \frac{x^3}{3\cdot 3!} + \frac{x^5}{5\cdot 5!} - \cdots \\
&=& \sum_{k=0}^\infty (-1)^{2k+1} \frac{x^{2k+1}}{(2k+1)\cdot (2k+1)!}.
\end{eqnarray*}
\item[(b)]  Find a power series representation for the function 
\[
F(x) = \int_0^x \frac{1}{1 - t^2}\, dt.
\]

\sol
We expand in geometric series, and integrate:
\begin{eqnarray*}
\int_0^x \frac{1}{1 - t^2}\, dt &=& \int_0^x \sum_{n=0}^\infty (t^2)^n\, dt\\
&=& \sum_{n = 0}^\infty \frac{x^{2n+1}}{2n+1}.
\end{eqnarray*}
\end{enumerate}

\item  Compute the requested Taylor polynomials.
\begin{enumerate}
\item[(a)] The Taylor polynomial of degree $n = 3$ for
$f(x) = \tan(x)$ at $a = 0$.

\sol
The derivatives and values we need:
\begin{eqnarray*}
f(0) = \tan(0) &=& 0\\
f'(x) = \sec^2(x) &\Rightarrow& f'(0) = 1\\
f''(x) = 2\sec^2(x)\tan(x) &\Rightarrow& f''(0) = 0\\
f'''(x) = 2\sec^4(x) + 4\sec^2(x)\tan^2(x) &\Rightarrow& f'''(0) = 2.
\end{eqnarray*}
So the Taylor polynomial is 
\[
p_3(x) = x + \frac{x^3}{3}.
\]

\item[(b)] The Taylor polynomial of degree $n = 6$ for
$f(x) = \cos(4x)$ at $a = 0$.  

\sol
We have
\begin{eqnarray*}
f(0) &=& 1\\
f'(x) = -4\sin(4x) &\Rightarrow& f'(0) = 0\\
f''(x) = -16\cos(4x) &\Rightarrow& f''(0) = -16\\
f^{(3)}(x) = 64\sin(4x) &\Rightarrow& f^{(3)}(0) = 0\\
f^{(4)}(x) = 256\cos(4x) &\Rightarrow& f^{(4)}(0) = 256\\
f^{(5)}(x) = -1024\sin(4x) &\Rightarrow& f^{(5)}(0) = 0\\
f^{(4)}(x) = -4096\cos(4x) &\Rightarrow& f^{(6)}(0) = 4096
\end{eqnarray*} 
So the Taylor polynomial is
\[
p_6(x) = 1 - \frac{16}{2!} x^2 + \frac{256}{4!} x^4 - \frac{4096}{6!} x^6 = 1 - 8x^2 + \frac{32x^4}{3} - 
\frac{256 x^6}{45}.
\]

\item[(c)] The Taylor polynomial of degree $n = 4$ for
$f(x) = {\root 3\of {x - 1}}$ at $a = 2$.  

\sol
We have
\begin{eqnarray*}
f(2) &=& 1\\
f'(x) = \frac{1}{3}(x - 1)^{-2/3} &\Rightarrow& f'(2) = \frac{1}{3}\\
f''(x) = \frac{-2}{9}(x-1)^{-5/3} &\Rightarrow& f''(2) = \frac{-2}{9}\\
f^{(3)}(x) = \frac{10}{27}(x - 1)^{-8/3} &\Rightarrow& f^{(3)}(2) = \frac{10}{27}\\
f^{(4)}(x) = \frac{-80}{81}(x-1)^{-11/3} &\Rightarrow& f^{(4)}(2) = \frac{-80}{81}.
\end{eqnarray*} 
So the Taylor polynomial is
\[
p_4(x) = 1 + \frac{1}{3}(x - 2) - \frac{1}{9}(x - 2)^2 + \frac{5}{81}(x - 2)^3 - \frac{10}{243}(x - 2)^4.
\]
\end{enumerate}

\item
\begin{enumerate}
\item[(a)] Compute the Taylor polynomial of degree $n = 4$ for
$f(x) = \sqrt{1 + x}$ at $a = 0$.  

\sol
We have
\begin{eqnarray*}
f(0) &=& 1\\
f'(x) = \frac{1}{2}(1+x)^{-1/2} &\Rightarrow& f'(0) = \frac{1}{2}\\
f''(x) = \frac{-1}{4}(1+x)^{-3/2} &\Rightarrow& f''(0) = \frac{-1}{4}\\
f^{(3)}(x) = \frac{3}{8}(1+x)^{-5/2} &\Rightarrow& f^{(3)}(0) = \frac{3}{8}\\
f^{(4)}(x) = \frac{-15}{16}(1+x)^{-7/2} &\Rightarrow& f^{(4)}(0) = \frac{-15}{16}.
\end{eqnarray*} 
So the Taylor polynomial is
\[
p_4(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4.
\]
\item[(b)]  Use your polynomial to approximate $\sqrt{1.1}$.  (What
$x$ should you use to do this?)

\sol
Note that $\sqrt{1.1} = f(0.1) \doteq p_4(0.1)$.  So we substitute $x = 0.1$ into 
$p_4(x)$ to get our approximation:
\[
\sqrt{1.1} \doteq p_4(0.1) = 1  + 
\frac{1}{2}(0.1) - \frac{1}{8}(0.1)^2 + \frac{1}{16}(0.1)^3 - \frac{5}{128}(0.1)^4 = 1.048808594.
\]

\item[(c)]  What is the error of your approximation, taking a
calculator value for $\sqrt{1.1}$ as the exact value?

\sol Taking $\sqrt{1.1} = 1.048808848$ as the exact value, the error in the approximation 
from (b) is
\[
1.048808848 - 1.048808594 = 2.54 \times 10^{-7}.
\]
(Note that the first 6 decimal places agree!)
\end{enumerate}

\item  Use Taylor series, not L'Hopital's Rule, to evaluate
each of the following limits.
\begin{enumerate}
\item[(a)]  ${\displaystyle \lim_{x\to 0} \frac{1 - \cos(x)}{x(e^x - 1)}}$.

\sol
Writing $\cdots$ for a sum of terms of higher degree in $x$, 
we have $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots$ 
and $e^x = 1 + x + \frac{x^2}{2} + \cdots$,
so ,
\[
\frac{1 - \cos(x)}{x(e^x - 1)} = \frac{\frac{x^2}{2} - \frac{x^4}{24} + \cdots }{x^2 + \frac{x^3}{2} + \cdots } = 
\frac{\frac{1}{2} - \frac{x^2}{24} + \cdots }{1 + \frac{x}{2} + \cdots },
\]
and 
\[
\lim_{x\to 0} \frac{1 - \cos(x)}{x(e^x - 1)} = \lim_{x\to 0} \frac{\frac{1}{2} - \frac{x^2}{24}\cdots }{1 + \frac{x}{2} + \cdots } = \frac{1}{2}.
\]
\item[(b)]  ${\displaystyle \lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \tan^{-1}(x)}}$.

\sol
Proceeding as in part (a),
\begin{eqnarray*}
e^x - e^{-x} - 2x &=& (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots)
- (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} + \cdots) - 2x\\
&=& \frac{x^3}{3} + \frac{x^5}{60} + \cdots.
\end{eqnarray*}
By an example done in class,
\[
\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots,
\]
so 
\[
x - \tan^{-1}(x) = \frac{x^3}{3} - \frac{x^5}{5} + \cdots.
\]
Therefore,
\[
\lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \tan^{-1}(x)} = \lim_{x\to 0} 
\frac{\frac{x^3}{3} + \frac{x^5}{60} + \cdots }{\frac{x^3}{3} - \frac{x^5}{5} + \cdots} 
=\lim_{x\to 0} \frac{\frac{1}{3} + \frac{x^2}{60} + \cdots}{\frac{1}{3} - \frac{x^2}{5} + \cdots} = 1.
\]
\end{enumerate}

\end{enumerate}
\end{document}
