\documentclass[11pt]{article}


\usepackage{amsfonts}
\usepackage{rotate}

\renewcommand{\labelenumi}{\underline{\underline{\bf\theenumi.}}\ }

\setlength{\hoffset}{-1in}
\setlength{\textwidth}{160mm}
\setlength{\parindent}{0pt}
\setlength{\voffset}{-1in}
\setlength{\textheight}{235mm}
\setlength{\parindent}{0pt}
\setlength{\parskip}{1ex}

\def\R{\mathbb{R}}
\def\abs#1{\mathopen|#1\mathclose|}

\let\Dst=\displaystyle




\begin{document}

% \centerline{\bf Homework }
% \centerline{\bf
% UTK -- M148 -- Honors Calculus II -- Spring 2015}
% \centerline{\bf
% Jochen Denzler}

\setcounter{page}{6}


\begin{enumerate} \relax\setcounter{enumi}{36}\relax

\item
The parabolas $y=(x-2)^2$ and $x=(y-2)^2$ intersect to form a
`quadrangle' with curved sides. Find the coordinates of the intersection
points. (Two can be guessed, the other two need calculation). 
Then calculate the area of this `quadrangle'. Make sure to split the
integral appropriately, depending on which formulas apply for the `upper' and
`lower' boundary curves. 

\item
A cognac glass has the shape of a ``sphere with a cap cut off''. See figure 
for a cross section. 
If the radius of the ball is $R$ and the glass takes up three fourths of the 
vertical diameter, what is the volume? (Yes I know these glasses are
not meant to be filled to the brim, but calculate the total volume
anyways.) 

\fbox{%
\begin{picture}(138,220)
\put(0,0){\special{psfile=hwk38fig.eps}}
\put(123,144){$\frac32R$}
\put(67,25){\bf Fig.\ for \#38}
 \end{picture}%
}
\hfill
\fbox{%
\begin{picture}(135,220)
\put(0,0){\special{psfile=hwk40fig.eps}}
\put(70,5){footprint}
\put(85,165){vertical} 
\put(85,152){cross} 
\put(85,139){section}
\put(85,126){(at $y=0$)}
\put(60,207){\bf Fig. for \#40}
 \end{picture}%
}


\item
Two cylinders, each with a circular cross section of $R$ intersect in
such a way that their axes meet at a right angle.  What is the volume
of the intersection body? I find it difficult to draw, but Rogawski
(Sec 6.2, Hwk 21 of the 2008 edition) has a figure of it, and also
some hints. 

\item 
A glass has the shape of a circular cylinder. It is initially full
with water. We pour out some water carefully by leaning the glass
sideways until the surface of the remaining water touches the bottom
of the glass along a diameter. What percentage of the water remains in
the glass? 

Assume the radius is $R$ and the height $h$, but express the remaining
volume as a fraction of the total volume $\pi R^2h$. 

Note/Hint: There are at least three ways of doing the problem. One in which
the slices are rectangles, one in which the slices are right
triangles, and one in which the slices are segments of a circle. 
Set up integrals for all three of them. 

Evaluate the integral obtained in at least two of the ways. 

The figure above shows various
sections of the glass, which is standing, with the water (grey) `frozen' in
place, centered at the origin of
the $x$--$y$--plane. The vertical axes is the $z$ axis.

Rectangles: 
slices parallel to the $y$--$z$--plane, different slices have
different $x$. 
\\
Triangles: 
slices parallel to the $x$--$z$--plane, different slices have
different $y$
\\
Circular segments:
slices parallel to the $x$--$z$--plane, different slices have
different $z$


\item
(a)
Given a torus that is obtained by rotating a circle of radius~$r$,
centered at $x=R$ on the $x$--axis,  about the $y$--axis, use the
shell method to calculate its volume. (The answer should of course
coincide with the answer $V=2\pi^2r^2R$ obtained by slicing in class.)

(b) 
Now assume $R<r$, so the circle that rotates about the $y$ axis
intersects the $y$--axis. We only cosider the part of the circle that
is in $x\ge0$, let it rotate about the $y$--axis. Use the shell method
to find the volume of the rotation body. [Note that in this case, the
  slicing method from class would be more cumbersome, since we would
  have washers for some $y$ and disks for other $y$.]

\item
The area below the curve $z= e^{-x^2}$ and above $z=0$ rotates about
the $z$--axis. Calculate its volume by means of shells. {\sf Note: 
You can expect to get a finite volume, even though the body extends
all the way to $\infty$. The volume is to be understood here as an
improper integral, i.e., you first consider only the area from $x=0$
to $x=R$ rotating, and the volume you obtain this way. Then you let
$R\to\infty$. }

\item
Let's go back to volume by slicing and do the previous problem that
way. We have the body above the $x$--$y$--plane and below the rotated
version of the curve $z=e^{-x^2}$, which is the surface $z=e^{-r^2}$,
where $r^2=x^2+y^2$. Slice the body by planes perpendicular to the
$y$--axis.
You will run into the difficulty that you cannot evaluate
$\int_{-\infty}^\infty e^{-x^2}\,dx$. Just call this unknown number
$I$. Its numerical value is $I\approx 1.8$.

Now express the volume of the said body as a simple expression
involving the quantity $I$.

Compare your result with the previous problem. From this comparison, 
conclude the {\em
  exact\/} value of $I=\int_{-\infty}^\infty e^{-x^2}\,dx$. 

{\sf [So here you have an example where you find the value of a
    definite integral without going through the process of finding an
    antiderivative, where an antiderivative is actually not available
    to you in this 
    example. --- There are some  advanced methods to find definite
    integrals without antiderivatives, like
    such as are based on complex-variable techniques (Math 443); but
    apart from those advanced methods, available 
    examples are unique and based on individual ad-hoc tricks.]

Knowledge of the value of this integral is desired, due to its frequent
occurrence in probability (`bell curve') and sometimes physics. }

\item
Calculate the perimeter of the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as an integral
$2\int_{-a}^a\ldots\,dx$. (You will not be able to evaluate this
integral, so leave it alone, except for algebraic simplification of
the expression under the integral.) 

Use the trig substitution $x=a\sin u$ to convert the obtained integral
into another integral (which you still won't be able to finish up, but
which may turn out to be more convenient in some cases, like for
numerical purposes). 

{\sf [Note: Integrals containing trigs under a square root (in not too
    complicated a manner), or 3rd or
    4th degree polynomials under a square root, or the ratio of
    two quadratics under a square root, are known under the label
    `elliptic integrals' for the very reason that they show up when
    calculating the perimeter of an ellipse. It is good to recognize
    them and to know that they will {\em not\/} be evaluated by Calc 2 methods
    at all, unless you can get rid of either the trig or the square
    root by simple algebra or a simple $u$-substitution. These
    integrals do show up in applications. If you recognize them, you
    save time by {\em not\/} trying your antiderivative prowess on
    them in vain; they are well-studied, and mathematics has a plethora of
    wisdom about them if you are desperate enough to ask for it.] }

\item
(a)
Calculate the length of the arc of the parabola $y=ax^2-1$ between $x=0$
and $x=b$. [The `$-1$' part is irrelevant here and just offered for
  convenience in part (b).]  You may borrow a good deal of work from either 27d or 29d,
after doing either a trig or hyp substitution to finish up the
integral.

{\sl The `limit'  part of b is voluntary; but if you skip it you have to
  do part c instead. Part c is much easier, but way less fun. }

(b) A downsized version of a Putnam competition problem in 2001 asks
whether it is possible to choose $a$ in such a way that the part of
the parabola inside the disc $x^2+y^2\le1$ is longer than 4. Assume
$a>\frac12$ and find the appropriate intersection point $(b,
ab^2-1)$. Calling $L(a)$ the length of that part of the parabola inside
the disk, as a function of $a$, calculated according to part (a), show
that $a(L(a)-4)$ goes to $\infty$ as $a\to\infty$. Conclude the
correct answer to the Putnam question. 

(c) Instead of calculating the limit as $a\to\infty$ in part (b), use a pocket
calculator to evaluate the length of the piece of the parabola inside
the disk in the case $a=100$. Answer the Putnam question. 
{\sf FYI: there are no calculators on the Putnam competition.}

\item
Take the ellipse from problem 44, let it rotate about the $x$-axis. 
Find the surface area of the body thus obtained. You may assume
$b<a$. 
{\sf Note: This time,
  evaluation of the integral is quite manageable. }

\item
For the torus from \#41a, calculate its surface area. 

\item
\rule{0pt}{2cm}The function $r(\varphi) = a (1-\cos\varphi)$ describes\\ 
a heart--shaped
(or apple--shaped?) curve \\
called the cardioid. \\
(a) Calculate the area $A$ 
it encloses;\\
(b) also calculate the length $L$ of this curve.\\
(c) As a simple consistency check, confirm \\
the isoperimetric inequality, namely that \\
$A/L^2< \pi/(2\pi)^2$.   
\hfill\smash{\fbox{%
\begin{picture}(163,160)
\put(0,0){\special{psfile=hwk48fig.eps}}
\end{picture}
}}

\item 
(a) 
That same cardioid now rotates about the $x$--axis. Calculate the
surface area~$S$ of the apple--shaped body thus obtained. 

(b) Also calculate the enclosed volume~$V$. {\bf Hint: }Slices in the form
of disks or washers may actually be easiest. Note that
$x=r(\varphi)\cos\varphi$ and $y=r(\varphi)\sin\varphi$ and use the
formula $\int_{x_{min}}^{x_{max}} \pi y^2\,dx + \int_{x_{max}}^0 
\pi y^2\,dx$ (explain why?), with $x,y,dx$ properly expressed in terms of
$r(\varphi)$ and $\varphi$.

(c) Confirm that your results are consistent with the isoperimetric
inequality 
$V^2/S^3 < (\frac43\pi)^2/(4\pi)^3$. 


\newpage

\item
You sip a drink from a cocktail glass with a straw. 
The cocktail glass is rotationally symmetric with a vertical cross
section given by $y= x^2$ (where $\abs{x}\le3$). The straw extends 6
unit lengths (meaning $6cm$) above the rim of the glass. Assuming the
density to be 1.01937 
(meaning 1.01937 g/cm$^3$, as even juice is mostly water, plus a bit
of sugar), and no ice
(horror, that must have \marginpar{\small \raggedright 
I made up the density in such a way
  that $1.01937\times9.81 = 10$ for numerical convenience}
happened on a study abroad, but it does simplify the math;-), what is
the work it takes yu to empty the glass by this method, using that 1
Joule = 1~Nm and N=m\,kg/sec$^2$? Use the
acceleration of gravity to be 9.81 m/sec$^2$.

\item
Here you will numerically evaluate the integral
$\Dst\int_0^1\sqrt{(1-x^2)\Bigl(1-\frac{x^2}{2}\Bigr)}\, dx$. (Recall the
comment made in assignment \#44 that this is an integral not amenable to the
algebraic techniques of Calc 2 --- albeit well-studied in advanced
mathematics). 

(a) Use each of trapezoidal rule, midpoint rule, and Simpson's rule
while splitting the interval of integration into four subintervals of
equal length, comparing the results with 6 digits behind the decimal
point. 

(b) Calculate the 2nd derivative of the integrand and show that it
is negative for all $x$ between 0 and 1. {\sl Hint: Combining all
  terms, you may get a numerator $-6 + 12 x^2 - 9 x^4 + 2 x^6$. Be a
  bit inventive with the algebra to determine its sign when
  $0<x<1$. If you don't see a good idea use that 
\rotate[f]{\mbox{$x$ to power six is always less than one, to get rid
    of that one term}}.}
Explain why the error estimates (relying on $M_2$ or $M_4$ such that
$\abs{f''(x)}\le M_2$ or $\abs{f''''(x)}\le M_4$) 
for each of the three rules given in the book are useless in this
case. Also give explicit upper and lower bounds for the integral based
on your knowledge that $f''(x)<0$. 

(c) Now for comparison, substitute $x=\sin u$ and use each of
trapezoidal, midpoint and Simpson rule on the new integral, again with
four subintervals of equal length, again giving 6 digits behind the
decimal point in each case. 
Give error margins for the results, based on the formulas; 
finding bounds $M_2$ and $M_4$ for the second and fourth derivatives
of the integrand is a bit clumsy: you may instead simply use the best 
values found by means of symbolic algebra and graphing software,
namely  $M_2=2.5$ and $M_4= 15.25$. 

(d) Observe the improvement in precision when comparing (a) and (c). 
Compare the result for trapezoidal and Simpson rules with 10
equidistant intervals (and the respective error margins
obtained). {\sl [Just to save labor, I am skipping midpoint in this
    part.]}



\item 
\begin{quote}\sl 
``How large is the number `1000!' ?'' Can we get a simple formula that allows us 
to give a good approximation for $n!$ for large $n$ that does not
require to go through all the multiplications one by one?
\end{quote}

Integrals can help in this task. We can compare 
$\ln n! = \ln 1 + \ln 2 + \ln 3 + \ln 4 + \ldots + \ln n$ with an
integral that we can actually calculate. So, just for a change, in
this problem we will use (easy) integrals to give a good approximation for
(messy) finite sums, rather than using easy finite sums to find an
approximation for messy integrals!

{\bf Task 1: } Use the midpoint rule with $\Delta x =1$ on the
integral $\int_{3/2}^{n+1/2} \ln x\,dx$ to show that 
$$\textstyle
\ln n! \ge (n+\frac12)\ln (n+\frac12) - (n+\frac12) - \mbox{some
  number}
$$
(of course you want to calculate this ``some number'' precisely!)
Sketch a graph and observe that the correct sign of the 2nd derivative of
the logarithm function tells you why you have `$\ge$' and not
`$\le$'. 

{\bf Task 2: } Use the trapezoidal rule (with $\Delta x=1$) on the
interval from 2 to $n$ to show that 
$$\textstyle
\ln n! \le \frac12\ln(2n) + n\ln n - n + \mbox{some number}
$$
(again you want to calculate this number exactly...)
Sketch a graph and observe, from the sign of the second derivative of
the logarithm function, why you get `$\le$', not `$\ge$'. 

{\bf Task 3: } The $\ln (n+\frac12)$ in the result of task 1 is just a
bit un-beautiful, we'll prefer $\ln n$: 
Explain with a 1-term Riemann-sum 
why $b\int_a^b \frac{dx}{x} >(b-a)$ when $b>a>0$. Use this to show
(by convenient choice of $a$, $b$) that $(n+\frac12)\ln(n+\frac12) >
(n+\frac12)(\ln n) + \frac12$.  

Combine this result with your result from Task 1 to show
$$
\ln n! \ge (n+\frac12)\ln n - n+ \mbox{some
  number}
$$
where again you want that number precisely. 


{\bf Task 4: } Combine your results and apply the exponential function
to them to find that 
$$
C_1 \sqrt{n} \,\frac{n^n}{e^n} \le n! \le C_2 \sqrt{n}\, \frac{n^n}{e^n}
$$
where we want the $C_1$ and the $C_2$ both as exact numbers and
numerically with 3 decimals. 


{\bf Task 5 } for you is only to watch in amazement, you don't have
work of your own here: Advanced Calculus methods that go beyond what
we can do here improve on the result from Task 4; 
they actually show that 
$$
\frac{n!}{\sqrt{n}\, n^n/e^n} \to \sqrt{2\pi}
\mbox{ \ \ as  \ \ } n\to\infty
$$
The numerical value of $\sqrt{2\pi}$ with 3 decimals is 2.507, which
should be neatly between the $C_1$ and $C_2$ obtained above. 

Yet another occurrence of $\pi$ in a context that has nothing to do
with circles!


\item
(a) Determine, possibly depending on~$x$, whether the series
  $\Dst\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}$ converges or diverges.   \\
(b) For which $x$ does the series $\Dst\sum_{n=0}^\infty
  \frac{x^{n^2}}{n!}$ converge? (While we are discussing only
  series with positive terms, you may pretend $x>0$.)\\
(c) Determine, depending on~$x$, whether the series $\Dst\sum_{n=0}^\infty
  \frac{n!^2 x^n}{(2n)!}$ converges. (Again, you may presume $x>0$ for
  now. And you may leave a borderline case, where ratio or root test is
  inconclusive, undecided.)
\item
Apply the ratio and the root tests to the $p$-series
$\Dst\sum_{n=1}^\infty\frac{1}{n^p}$. Do you get a useful answer? 
\item
Recall from a prior problem that there are certain positive constants
$C_1$ and $C_2$ (namely $C_1=2.43$ and $C_2=2.62$ would work) such
that $C_1\sqrt{n} \,(\frac ne)^n < n! < C_2 \sqrt{n}\,(\frac ne)^n$. 
This will help you, as the question here is to decide the convergence
of $\sum\frac{x^n}{n!}$  by means of the {\em root\/} rather than the
ratio test. (For the time being, as we study series with positive
terms only, you may pretend $x>0$.) 
\item
Does $\Dst\sum_{n=1}^\infty \frac{n^{1000}}{3^n}$ converge? Use ratio
test, root test and limit comparison (choosing an appropriate partner
series to compare with). 
\item
(a) Consider the series $1 + \frac{2}{3} + \frac{1}{3^2} + \frac{2}{3^3} +
  \frac{1}{3^4}+\frac{2}{3^5} + \ldots$. The denominators are powers
  of~3, but the numerators alternate between 1 and 2. In $\sum$
  notation, you could write this as
$$
\sum_{n=0}^\infty \frac{(3-(-1)^n)/2}{3^n} 
\qquad \mbox{ or } \qquad
\sum_{n=0}^\infty \frac{(3-\cos(\pi n))/2}{3^n} 
\;.
$$
Show that the ratio test is inconclusive, but use either the root test
or direct comparison to decide on whether the series converges
\\
(b) Nothing to do here except to read \& acknowledge: Is the series
$\frac12+\frac13+\frac15+\frac17+\frac1{11}+\ldots$ convergent or
divergent? (The terms are the reciprocals of the prime numbers.) None
of the methods seen so far or later in Calc 2 are good enough to
decide this question. This goes to showing that one can easily invent
series whose convergence is difficult to decide. But for the series
that are useful in calculus, the decision can usually be made routinely.

{\sf (In case you really want to know: The sum of the reciprocals of
  prime numbers is known to be a divergent series.)}
\item
Consider $\Dst\sum_{n=2}^\infty \frac{1}{n(\ln n)}$. Does this series
converge or diverge? (Hint: Comparison with an integral is the way to go.) ---
Next decide $\Dst\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ where $p>0$ is
any given real number. (The answer will depend on $p$.)
{\sf Note: You learn more by simply following the method I used in
  class when deciding the $p$-series by comparison with an
  integral. Explain the   situation with a figure using terms like
  `left' or `right endpoint' 
  Riemann sum and `decreasing'. --- Once you have done this, reading
  up precisely on the highlight box about `integral comparison' in the
  textbook  will make more sense than it would if you were to just
  plug quantities into the highlight box paradigm. }

\item 
(In continuation of \#53c:)
Use the result from Hwk~\#52 (requoted in \#55), together with an
appropriate comparison, to decide whether the series
$\sum_{n=0}^\infty \frac{4^n\,n!^2}{(2n)!}$ converges or
diverges. 

\item
We have seen how the ratio test can be used to show that
$1-\frac{x^2}{2!}+\frac{x^4}{4!} -\frac{x^6}{6!} +-\ldots =
\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$ converges absolutely for
every~$x$ (cf.~\#53a), and we have seen that the value of this series 
cannot be anything else but $\cos x$ because partial sums are
alternatingly above and below $\cos x$  (cf.~\#26). 

(a) Explain why
Leibniz' alternating series test applies to this series for $x=1$, but
does not apply to it for $x=2$. 
\\
(b) For which $x$ exactly does the alternating series test apply to the
cosine series?

\item 
Based on the knowledge that $e^x= \sum_{n=0}^\infty
  \frac{x^n}{n!}$, write a series for $e^{-x}$ and then conclude that
  the series in \#53a converges to $\cosh x$. {\sl Hint: Write out the
    sums with the `\ldots' notation rather than manipulating
    expressions involving $\sum_{n=0}^\infty$, as you may lack the
    experience to do the manipulation of $\sum_{n=0}^\infty$
    expressions correctly/expediently.}

\item 
(a) What is the radius of convergence of the power series
$\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{4^n\,n!^2}$~?
\\
(b) The series defines a function $J(x):= \sum_{n=0}^\infty (-1)^n
\frac{x^{2n}}{4^n\,n!^2}$. Since you won't be able to find a formula
for the value of the series, you know practically nothing about this
function. Calculate $J(1)$ with to 3 decimals accuracy. How many terms
did you need? How many terms would you need to calculate $J(2)$ with 3
decimals accuracy? 
Calculate $J(0)$, $J'(0)$, $J''(0)$, $J'''(0)$ and $J^{(4)}(0)$ exactly. 

\item
Using the geometric series $\frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots =
\sum_{n=0}^\infty t^n$ and the substitution $t=-x^2$ to find a power
series representation for $\frac{1}{1+x^2}$. 
Integrate it to obtain a power series representation of $\arctan x$. 
What is the radius of convergence of the obtained series? 
%
(In each step, write the power series both in a `\ldots' form giving enough
terms to see the general pattern and using the $\sum$ notation.)

Substitute $x=1$ into the arctan series (believing me that doing so is
permissible). Obtain a beautiful but useless series representing
$\pi$. How many terms would you need to calculate in order to obtain
$\pi$ up to an accuracy of $10^{-4}$ by this series? 

Calculate $\arctan \frac15$ to 5 decimals using this series. (Of course
in doing so you pretend your pocket calculator does not know arctan.)
How many terms do you need to do this?

Also calculate $\arctan\frac1{239}$ to 5 decimals. How many terms do
you have to use for this purpose? [We'll later see why, of all crazy
  numbers, I would specifically care for $1/239$.] 

\item
Write down the power series representing $\ln(1+x)$ and $-\ln(1-x)$
respectively (obtained by integrating a geometric series.) What is the
radius of convergence?

Obtain a power series for $\ln \frac{1+x}{1-x}$ from the preceding
two. Which is the radius of convergence for each of these series? 

Use the series for $\ln (1+x)$ and the one for $\ln\frac{1+x}{1-x}$, each with
a specific choice of $x$, to obtain a series with value $\ln 2$. 
[You need to take my word for it that the choice of $x$ in the first
  series is actually permissible.]

One of the two series is almost useless for practical calculation of
$\ln 2$, because it would take way too many terms for decent precision. 
In the other one, roughly how many terms do you estimate you'd have to
calculate to get 5 digits precision for $\ln 2$~?


\item
By multiplying the corresponding power series, obtain a power series
representing $e^x\,sin x$. In doing so calculate all powers up to
$x^7$; you are not expected to see a general formula for the $n$th
coefficient, and so you will not be able to use the $\sum$ notation; 
just calculate the first few terms in the series as indicated. 

\item
Write down power series for $1+\ln(1+2x)$ and $1+\ln(1+x)$ up to the
$x^4$ term. By long division, obtain a power series for 
$\Dst\frac{1+\ln(1+2x)}{1+\ln(1+x)}$, also up to the 4th term. 

\item
It is possible to obtain the well-known formula $e^x\,e^y = e^{x+y}$
directly by multiplying the power series, without using any prior
knowledge about the exponential function. By multiplying 
the power series $1+x+\frac{x^2}{2!}+\frac{x^3}{3!} +
\frac{x^4}{4!}+\ldots$ with the series $1+y+\frac{y^2}{2!}+\frac{y^3}{3!} +
\frac{y^4}{4!}+\ldots$ and applying the binomial formula, verify that
the product is indeed 
$1+(x+y)+\frac{(x+y)^2}{2!}+\frac{(x+y)^3}{3!} +
\frac{(x+y)^4}{4!}+\ldots$, at least up to power 4. 

{\sf Note: If you know the general binomial theorem in the form 
$(x+y)^n = \sum_{k=0}^n \frac{n!}{k! (n-k)!} x^k y^{n-k}$, you can
  obtain the analogous calculation for {\em all\/} terms, not just up
  to the 4th power. However, for this homework, we'll be content with
checking the result only up to power 4.}

\item
Recalling the power series of $\sin x$, and of $\cos y$, plug  the
series of $\sin x$ for $y$ into the series for $\cos y$, calculating
terms up to order $x^6$. This way you will obtain the beginning of the
power series of $\cos(\sin x)$. 

\item
Calculating the first five derivatives of $\tan x$ and using Taylor's
formula, find the first 3 nonvanishing terms (i.e., up to order $x^5$)
for the Taylor series of $\tan x$, and compare it with the series
obtained in class by long division. {\sl Hint: It's probably more
  convenient for successive derivatives if you write the derivative 
  of $\tan x$ as $1+\tan^2 x$ rather  than $1/\cos^2 x$.}

\item
The nice features of term-by-term differentiation being allowed for
power series do not carry over to other series. 
For instance $f(x):= \sum_{n=1}^\infty \frac{\cos n x}{n^2}$ is a
useful series that is {\em not\/} a power series.

(a) Show by a direct comparison test, that this series converges
absolutely for every real~$x$.

(b) Attempt to take a second derivative via term-by-term
differentiation (twice). What series do you get as a result? Does it
converge for any~$x$~? (A somewhat heuristic answer is acceptable
here, even if it is not logically watertight.)



{\sf Note: You have no tools at this level to guess the value of this
  sum $f(x)$. Senior level mathematics would tell you that actually 
$f(x) = \frac{\pi^2}{6}-\frac\pi2\abs{x} + \frac14x^2$ 
for $\abs{x}\le\pi$. --- The only reason why you see this
  example in Calc 2 is to convince you that some of the `nice'
  features of calculation with power series are not `automatic'
  carry-overs from the rules for finite sums. }
\end{enumerate}
\end{document}
