2.1 Avoiding
Infinite Series
- Using the Archimedean understanding, the value of
 is
the target value, T, of
the sequence of finite sums, 
What is this target value? Show that if we take any x > T,
then all of the finite sums are less than x.
Show that if we take any x < T
, then from some point on (and this point will depend on the size of
x), all of the finite sums are larger than x.
- A topic
that often sets off lively discussion in the mathematics classroom
is whether or not 0.9999... is
equal to 1.
What would be the Archimedean understanding of the meaning of
this decimal of infinitely repeating 9's?
Show that with this understanding, 0.9999... does
equal 1.
- Use the Archimedean understanding to prove that a series cannot
have more than one value.
- What is the Archimedean understanding of the infinite series
 ?
Show that if this series has a value, then it must be at least 1.
Show that if this series has a value, then it must be at most 0.
From this you can conclude that it has no value.
- Using the Rearranger applet,
calculate the partial sums of a rearranged alternating harmonic series
in which you take
the first r positive
summands followed by the first s negative
summands and
then alternate: the next r positive
summands followed by the next s negative
summands. For example, if r =
1 and s =
2, this would return the partial sums of the series 
This program will work up to about
 iterations,
at which point you can trust the first seven digits to the
right of the decimal.
The Inverse
Symbolic Calculator (ISC) takes a decimal
and returns the exact numbers (such as ln
2)
that agree with that decimal to the number of places given. Use the ISC
to find the probable limit of this rearrangement of the alternating harmonic
series.
Given any pair of positive integers, (r,s),
we can define the function R(r,s) to
be the target value of the rearranged alternating harmonic series in
which the next r positive
summands alternate with the the next s negative
summands. Using applet and the ISC, guess this formula.
Keep track of your experiments and then write a description of what you
tried, what you discovered, what you believe to be the case, what you
tried to prove and any cases that you were able to prove.
- Prove that the value of R(1,2) is (1/2)
ln 2.You may assume that the associative law holds so that
you can group the first and second, fourth and fifth, seventh and
eighth, etc. terms.
- The following is a sketch of an argument that could be used to
convince someone that R(2,3) =
(1/2) ln (8/3). Define f(x) to be
the following power series which can be rewritten in terms of logarithms
and simplified.
What happens when we take the limit as x approaches 1
from below?
- Using the idea from the previous exercise, find the general formula
for the value of R(r,s).
In later chapters, we will see how to justify the individal steps
in
what now is simply a convincing argument. Hint:
You want to assign powers of x to
the summands so that:
- The r positive
summands and the s negative
negative summands in the kth
set all have exponents that are larger than the exponents
of
the terms in the k–1st
set and that are smaller than the exponents of the terms
in the k+1st
set.
- If you separate the positive terms from the negative
terms, the exponents for the positive terms form
an arithmetic sequence, and the exponents for the negative
terms form
an arithmetic sequence.
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\subhead{Exercises}\begin{enumerate}
\item Assume that $u = r^{\lambda}(1 + a_1r + a_2r^2 + \cdots)$
is a solution to equation~(\ref{eqn:2.7.1}). Prove equations~(\ref{eqn:2.7.1a}--\ref{eqn:2.7.1c})
and then show that they imply equation~(\ref{eqn:2.7.2}).
\item \label{ex:2.7.2} Find the first five nonzero terms in the
power series for each of Gregory's functions:
\begin{enumerate}
\item $\tan x$,
\item $\arctan x$,
\item $\sec x$,
\item $\ln (\sec x)$,
\item $\sec^{-1}(\sqrt{2}\:e^x)$, ($\sec^{-1}$ is the arc secant),
\item $\ln [ \tan(x/2 + \pi/4)]$,
\item $2 \arctan (\tanh x/2)$, [$\tanh x = (e^x - e^{-x})/(e^x +
e^{-x})$ is the hyperbolic tangent].
\end{enumerate}
\Math hint: use the command {\tt Series[ f[x], \{x,0,4\} ]} to find
the expansion of $f(x)$ as far as the coefficient of $x^4$. If some
of these coefficients are zero, it will be necessary to expand the
function to a higher power of $x$. Leave the coefficients as rational
numbers. The \Math functions used in this exercise are {\tt Tan[x],
ArcTan[x], Sec[x], Log[x], ArcSec[x], Sqrt[x], Exp[x]} (which is
the exponential function $e^x$), and {\tt Tanh[x]}.
\item For each problem in exercise~\ref{ex:2.7.2}, graph the given
function and compare it to the graph of the first five terms of
its power series. For what values of $x$ do you have a good approximation?
\item Prove Bernoulli's identity, equation~(\ref{eqn:2.7.13}), by
using repeated integration by parts:
\begin{eqnarray*}
\int_0^x f(t)\:dt & = & x\,f(x) - \int_0^x tf'(t)\:dt \\
& = & x\,f(x) - \frac{x^2}{2}\,f'(x) + \int_0^x \frac{t^2}{2}\,f''(t)\:dt
\\
& = & \ldots
\end{eqnarray*}\item Show that Taylor's series implies Bernoulli's
identity by first using Taylor's series to prove that
\begin{eqnarray*}
f(x) - f(0) & = & f'(0)\,x + \frac{f''(0)}{2!}\,x^2 + \frac{f'''(0)}{3!}\,x^3
+ \cdots, \\[3pt]
f'(x) - f'(0) & = & f''(0)\,x + \frac{f'''(0)}{2!}\,x^2
+ \frac{f^{(4)}(0)}{3!}\,x^3 + \cdots, \\[3pt]
f''(x) - f''(0) & = & f'''(0)\,x + \frac{f^{(4)}(0)}{2!}\,x^2
+ \frac{f^{(5)}(0)}{3!}\,x^3 + \cdots, \\[3pt]
f'''(x) - f'''(0) & = & f^{(4)}(0)\,x + \frac{f^{(5)}(0)}{2!}\,x^2
+ \frac{f^{(6)}(0)}{3!}\,x^3 + \cdots, \\
& \vdots &
\end{eqnarray*}
Now eliminate $f'(0)$, $f''(0)$, $f'''(0)$, \ldots\ to obtain
\[ f(x) = f(0) + f'(x)\,x - \frac{f''(x)}{2!}\,x^2 + \frac{f'''(x)}{3!}\,x^3
- \frac{f^{(4)}(x)}{4!}\,x^4 + \cdots. \]
\item Use Taylor series to find the power series for $(1+x)^{\alpha}$.
What happens when $\alpha$ is a positive integer? What happens when
$\alpha = 0$?
\item Use the \Math command {\tt binom[.5,n,x]} defined in exercise~\ref{ex:2.5.2}
in section~2.2 to calculate partial sums of the binomial series
for the square root:
\begin{equation} \sqrt{1+x} = 1 + (1/2)\,x + \frac{(1/2)(1/2-1)}{2!}\,x^2
+ \frac{(1/2)(1/2-1)(1/2-2)}{3!}\,x^3 + \cdots. \label{eqn:5.1.9}
\end{equation}
Calculate the partial sums as $n$ goes from 100 to 1000 in steps
of 100 when $x = 200/199$. Describe what you see.
\item Calculate the partial sum of the series in equation~(\ref{eqn:5.1.9})
as $n$ goes from 100 to 1000 in steps of 100 when $x=1$ and when
$x=-1$. Describe what you see. Make a guess of whether or not this
series converges for these values of $x$. Explain your reasoning.
\item Prove that
\[ \left( 1 - \frac{3}{2n}\right)\frac{200}{199} > 1\qquad {\rm
if\ and\ only\ if}\qquad n > 300. \]
In general, prove that when $n \geq 1 + \alpha > 0$, then
\[ \left| \frac{(\alpha - n +1)x}{n}\right| > 1\]
if and only if $|x| > 1$ and
\[ n > \frac{1+\alpha}{1-|x|^{-1}}. \]
What happens if $1 + \alpha \leq 0$?
\item Calculate the partial sum for $\sqrt{1+(200/199)}$:
\[ 1 + (1/2)\,\left(\frac{200}{199}\right) + \cdots + \frac{(1/2)(-1/2)\ldots(-297.5)}{299!}\,\left(\frac{200}{199}\right)^{299}
\]
and compare your result to the calculator value of $\sqrt{399/199}$.
Is the error within the predicted bounds? How close are you to the
outer bound?
\item \label{ex:2.5.33} Find the Lagrange remainder for an approximation
to an arbitrary binomial series:
\[ (1+x)^{\alpha} \approx 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}\,x^2
+ \cdots + \frac{\alpha(\alpha - 1) \ldots (\alpha - n + 2)}{(n-1)!}\,x^{n-1}.
\]
\item Simplify the Lagrange remainder of the previous exercise when
$\alpha = -1$. What happens to this remainder when $x=1$ and $n$
increases? Does it approach 0?
\item What is wrong with the following argument? Using the Lagrange
remainder, we know that
\[ (1+x)^{-1} = 1 - x + x^2 - x^3 + \cdots + (-1)^{n-1}x^{n-1} +
D_n(0,x), \] where $D_n(0,x) = (-1)^nx^n/(1+c)^{n+1}$ for some $c$
between 0 and $x$. If $x=1$, then $c$ is positive and the absolute
value of the Lagrange remainder is $1/(1+c)^{n+1}$ which approaches
0 as $n$ increases.
\item Experiment with different values of $\alpha$ between $-1$
and 1/2 in the Lagrange remainder for the binomial series (exercise~\ref{ex:2.5.33}).
Does the remainder approach zero when $x=1$ and $n$ increases? Describe
and discuss the results of your experiments.
\item \label{ex:2.5.36} Find the Lagrange form of the remainder
for the partial sum approximation to $\ln(1+x)$:
\[ \ln(1+x)\approx x - \frac{x^2}{2} + \cdots + (-1)^{n-2}\frac{x^{n-1}}{n-1}.
\]
\item Continuing exercise~\ref{ex:2.5.36}, how many terms of the
partial series should you take if you want to minimize the error
in approximating $\ln(1 + 200/199)$?
\item Consider Lagrange's form of the remainder for $D_6(0,x)$ when
the function is $f(x) = \sqrt[3]{1+x} = (1+x)^{1/3}$. Find an upper
bound for the absolute value of Lagrange's remainder that does not
depend on $c$. Graph $|D_6(0,x)|$ and the bounding function for
$x \geq 0$. How large can $x$ be before $|D_6(0,x)|$ exceeds 0.5?
\item For the function $f(x) = \ln(1+x)$, graph the two functions
that bound the remainder when $n=6$:
\[ y = \frac{f^{(7)}(0)}{7!}\,x^7 \quad {\rm and} \quad y = \frac{f^{(7)}(x)}{7!}\,x^7.
\]
Now graph the actual remainder and see how it compares with these
boundaries:
\[ y = \ln(1+x) - \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}
+ \frac{x^5}{5} - \frac{x^6}{6}\right) .\]
\end{enumerate}
\newpage\subhead{Exercises}\begin{enumerate}
\item Rewrite the series
\[ 1 - x^2 + x^5 - x^7 + x^{10} - x^{12} + \cdots \]
as a rational function of $x$. Set $x=1$. What value does this give
for the series $1 - 1 + 1 - 1 + \cdots$\,?
\item Find a power series in $x$ that would imply that
\[ 1 - 1 + 1 - 1 + \cdots = \frac{4}{7} \]
when $x$ is set equal to 1.
\item Given {\it any\/} nonzero integers $m$ and $n$, find a power
series in $x$ that would imply that
\[ 1 - 1 + 1 - 1 + \cdots = \frac{m}{n} \]
when $x$ is set equal to 1.
\end{enumerate}
\subhead{Exercises}\begin{enumerate}
\item Consider Wallis's infinite product for $\pi/2$ given in equation~(\ref{eqn:2.3.16}).
\begin{enumerate}
\item Show that the product of the $k$th {\it pair\/} of fractions
is $4k^2/(4k^2 - 1)$.
\item The following {\it Mathematica\/} command will calculate the
twice the product of the first {\tt n} pairs of fractions:
\begin{verbatim}wallis[n_] := 2 NProduct[ 4k^2/(4k^2-1),{k,1,n}]
\end{verbatim}
\item How large a value of {\tt n} is needed in order to get 3 digit
accuracy using this product?
\item We can improve our accuracy if we average the upper and lower
bounds. Prove that this average is
\[ \frac{2^2 \cdot 4^2 \cdots (2n) \cdot (2n + 1/2)}{1 \cdot 3^2
\cdots (2n-1)^2\cdot (2n+1)}. \]
\item Modify the {\it Mathematica\/} command so that it computes
this average. How large a value of {\tt n} is needed in order to
get 3-digit accuracy using the modified program?
\end{enumerate}
\item Prove equation~(\ref{eqn:2.3.15}) by showing that
\begin{equation}
\int_0^1 \left(1 - x^{1/p}\right)^q\:dx = \frac{q}{p+q}\:\int_0^1
\left(1 - x^{1/p}\right)^{q-1}\:dx. \label{eqn:2.3.20}
\end{equation}
Hint: begin by rewriting the left side as
\begin{eqnarray*}
&& \hspace{-15mm} \int_0^1 \left(1 - x^{1/p}\right)^{q-1}\:dx
- \int_0^1 \left(1 - x^{1/p}\right)^{q-1}x^{1/p}\:dx \\[2pt]
& = & \int_0^1 \left(1 - x^{1/p}\right)^{q-1}\:dx + p\int_0^1
\left(1 - x^{1/p}\right)^{q-1} \left(\frac{-x^{(1-p)/p}}{p}\right)x\:dx,
\end{eqnarray*}
and then use integration by parts on the second piece.
\item Using equation~(\ref{eqn:2.3.15}) and induction, prove that
if $q$ is a positive integer then
\begin{equation}
\int_0^1 \left(1 - x^{1/p}\right)^q\:dx = \frac{q!}{(p+1)(p+2)\cdots(p+q)}.
\end{equation}
\item Prove that
\begin{equation}
\int_0^1 \left(1 - x^{1/p}\right)^q\:dx = \int_0^1 \left(1 - x^{1/q}\right)^p\:dx,
\end{equation}
and therefore $f(p,q) = f(q,p)$. Hint: use the substitution $u =
(1-x^{1/p})^q$.
\item Prove that if $p$ is positive and $q_1 > q_2$, then
\[ \left( 1 - x^{1/p} \right)^{q_1} < \left( 1 - x^{1/p} \right)^{q_2},
\]
for all $x$ between 0 and 1, and therefore $f(p,q_1) > f(p,q_2)$.
Prove that if $p$ is negative and $q_1 > q_2$, then $f(p,q_1)
< f(p,q_2)$.
\item If we restrict our attention to the rows and columns of $f(p,q)$
for which $p$ and $q$ are integers, we see a pattern that should
be familiar:
\begin{center}
\begin{tabular}{|l||c|c|c|c|c|c|c|c|c|c|} \hline
\multicolumn{11}{|c|}{$f(p,q)$}\\ \hline
p$\downarrow$ q $\rightarrow$ &{\scriptsize $-1/2$}& 0 &
{\scriptsize 1/2}& 1 &{\scriptsize 3/2}& 2 &{\scriptsize
5/2}& 3 &{\scriptsize 7/2}& 4 \\ \hline \hline
{\scriptsize $-1/2$}& & & & & & & &
& & \\ \hline
0 & & 1 & & 1 & &1 & &1 & &1
\\ \hline
{\scriptsize 1/2}& & & & & & & &
& &\\ \hline
1 & & 1 & &2 & &3 & &4 & &5
\\ \hline
{\scriptsize 3/2}& & & & & & & &
& & \\ \hline
2 & & 1 & &3 & &6 & &10 & &15
\\ \hline
{\scriptsize 5/2}& & & & & & & &
& & \\ \hline
3 & & 1 & &4 & &10 & &20 & &35
\\ \hline
{\scriptsize 7/2}& & & & & & & &
& & \\ \hline
4 & & 1 & &5 & &15 & &35 & &70\\
\hline
\end{tabular}\end{center}
Where have you seen this pattern before? What does it suggest about
the meaning of $f(p,q)$?
\item When $p$ and $q$ are integers, we have the relationship
\begin{equation} f(p,q) = f(p,q-1) + f(p-1,q). \end{equation}
Does this continue to hold true when $p$ and $q$ are not both integers?
Either give an example where it does not work or prove that it is
always true.
\item Show that if we use the row $p = -1/2$ to approximate $\pi/2$:
\[ 1 > \frac{4}{2\pi} > \frac{1}{2} > \frac{4}{3\pi} >
\frac{3}{8} > \frac{16}{15\pi} > \frac{5}{16} > \frac{32}{35\pi}
> \frac{35}{128} > \cdots, \]
then we get the same bounds for $\pi/2$.
\item What bounds do we get for $\pi/2$ if we use the row $p = 3/2$?
\item What bounds do we get for $\pi/2$ if we use the diagonal:
\[ 1 < \frac{4}{\pi} < 2 < \frac{32}{3\pi} < 6 <
\frac{512}{15\pi} < 20 < \frac{4094}{35\pi} < 70 < \cdots?
\]
\item Fill in the missing entries in the table of ratios:
\begin{center}
\begin{tabular}{|l||c|c|c|c|c|c|c|c|c|c|} \hline
\multicolumn{11}{|c|}{$f(p,q)/f(p,q-1)$}\\ \hline
p$\downarrow$ q $\rightarrow$ & {\scriptsize $-1/2$} & 0
& {\scriptsize 1/2} & 1 &{\scriptsize 3/2}& 2 &{\scriptsize
5/2}& 3 &{\scriptsize 7/2}& 4 \\ \hline \hline
{\scriptsize $-1/2$} && && {\scriptsize 1/2} &&
{\scriptsize 3/4} && {\scriptsize 5/6} && {\scriptsize
7/8} \\ \hline
0 && & 1 & 1 & 1&1 &1 &1 &1
&1 \\ \hline
{\scriptsize 1/2}& && &{\scriptsize 3/2} & &{\scriptsize
5/4} & &{\scriptsize 7/6} & &{\scriptsize 9/8}\\
\hline
1 && & 3 & 2 &{\scriptsize 5/3} &{\scriptsize
3/2} & {\scriptsize 7/5 }& {\scriptsize 4/3} &{\scriptsize
9/7} &{\scriptsize 5/4} \\ \hline
{\scriptsize 3/2}&& && {\scriptsize 5/2} & &{\scriptsize
7/4} & &{\scriptsize 3/2} & &{\scriptsize 11/8}
\\ \hline
2 & && 5 & 3 &{\scriptsize 7/3} &2 &
{\scriptsize 9/5}& {\scriptsize 5/3} &{\scriptsize 11/7}
& {\scriptsize 3/2} \\ \hline
{\scriptsize 5/2}&& && {\scriptsize 7/2} & &{\scriptsize
9/4} & &{\scriptsize 11/6} & & {\scriptsize 13/8}\\
\hline
3 & && 7 &4 & 3 &{\scriptsize 5/2} &{\scriptsize
11/5} & 2 &{\scriptsize 13/7} & {\scriptsize 7/4} \\
\hline
{\scriptsize 7/2}&& && {\scriptsize 9/2} & &{\scriptsize
11/4} & &{\scriptsize 13/6} & &{\scriptsize 15/8}
\\ \hline
4 & && 9 & 5 &{\scriptsize 11/3} &3 &{\scriptsize
13/5}&{\scriptsize 7/3} &{\scriptsize 15/7} & 2\\ \hline
\end{tabular}\end{center}
\item As far as you can, extend the table of values for $f(p,q)$
into negative values of $p$ and $q$.
\item Use the method of this section to find an infinite product
that approaches the value of
\[ \int_0^1 \left(1 - x^3\right)^{1/3}\:dx. \]
\end{enumerate}
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