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\begin{document}

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\begin{center}
History and Applications of Calculus 
Homework 5\\
Sir Isaac Newton and 
Gottfried Wilhelm Leibniz\\
\end{center}

\bigskip

\noindent

\vspace{2mm}
\noindent
1.
\begin{itemize}
\item[(a)] 
Use the substitution $x = \tanh u$ to calculate $\int
\frac{dx}{1-x^2}$.
\item[(b)] 
Instead, calculate the integral in (b) by 
factoring $1-x^2=(1+x)(1-x)$ then using the method of
partial fractions. By comparing your answer to (a), deduce that
$$
\arctanh x = \ln\sqrt{\frac{1+x}{1-x}}.
$$
\item[(c)] 
Let $x = \tanh y = \frac{e^{2y}-1}{e^{2y}+1}$ (so $y = \arctanh x$).
By solving the equation
$$
x = \frac{e^{2y}-1}{e^{2y}+1}
$$
for $y$, give another proof that $\arctanh x = \ln \sqrt{\frac{1+x}{1-x}}$.
\item[(d)] 
Find a formula for $\arcsinh x$ involving $\ln$.
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
For (a), you'll need the identity $\cosh^2 u - \sinh^2 u = 1$ from
last week's homework; divide
both sides by $\cosh^2 u$ to get an identity involving $\tanh^2 u$.
For (b) make sure you think about the ``$+c$'' you get when
calculating indefinite integrals like these.
For (d) mimic what you did in (c). You should end up with an equation
involving $e^{2y} = (e^y)^2$ and $e^y$ which you can solve using the
quadratic formula before taking logs...

\vspace{2mm}
\noindent
2. It may have been a while since you had to use ``integration by
parts,'' but you'll need it in this question!
For $n=1,2,3,\dots$, let 
$$
I_n := \int_0^{\pi/2} \sin^n x \, dx.
$$
\begin{itemize}\item[(a)] 
Use integration by parts to prove the following recurrence
relation for the number $I_n$:
$$
I_n = \frac{n-1}{n} I_{n-2}.
$$
\item[(b)] 
Show that 
\begin{align*}
\int_0^{\pi/2} \sin^{2n+1} \,dx &= \frac{2}{3} \cdot \frac{4}{5}
                                  \cdot \frac{6}{7}\cdot \cdots \cdot
                                  \frac{2n}{2n+1},\\
\int_0^{\pi/2} \sin^{2n} x \, dx &= \frac{\pi}{2} \cdot \frac{1}{2}
                                   \cdot \frac{3}{4}\cdot
                                   \frac{5}{6}\cdot \cdots \cdot
                                   \frac{2n-1}{2n}.
\end{align*}
Conclude that
$$
\frac{\pi}{2} = \frac{2}{1}\cdot \frac{2}{3} \cdot \frac{4}{3} \cdot
\frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots \cdot
\frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \frac{\int_0^{\pi/2} \sin^{2n}
  x\,dx}{\int_0^{\pi/2} \sin^{2n+1}x \, dx}.
$$
\item[(c)] 
Show that the quotient of the two integrals in the expression just
obtained is between 1 and $1+\frac{1}{2n}$.
This shows that the products
$$
\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot
\frac{4}{3}\cdot\frac{4}{5}
\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\cdots\cdot
\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}
$$
can be made as close to $\pi/2$ as desired by taking $n$ sufficiently
large. It is usually written as an infinite product known as Wallis'
product:
$$
\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot
\frac{4}{3}\cdot\frac{4}{5} \cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\cdots.
$$
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
For (a), you should do the ``obvious'' integration by parts, replace
the $\cos^2$ by $\sin^2$ using the usual trig identity, then
rearrange. 
For (c), start with the
inequalities
$$
0 < \sin^{2n+1} x < \sin^{2n} x \leq \sin^{2n-1} x
$$
for $0 < x < \pi/2$.


\vspace{2mm}
\noindent
3. Now you'll need to remember how to calculate the Taylor series of a
smooth ($=$ infinitely differentiable) function $f(x)$ at $x=0$:
$$
f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3
+ \cdots.
$$
The series may or may not converge to the function $f(x)$, or it may only inverge inside some
``radius of convergence,'' that is the point of Taylor's theorem. 
When it converges, it is perfectly legitimate to integrate or
differentiate the function by integrating or differentiating the
Taylor series ``term by term'' (it works for $x$ inside the radius of convergence).
\begin{itemize}
\item[(a)] 
The generalized binomial coefficient $\binom{\alpha}{n}$ is
$$
\frac{\alpha (\alpha-1)\cdots (\alpha-n+1)}{n!}
$$
which makes sense for any $\alpha \in \R$. Show that the Taylor series
of
the function $(1+x)^\alpha$ is 
$$
(1+x)^\alpha = \sum_{n \geq 0} \binom{\alpha}{n} x^n.
$$
The radius of convergence of this series is $1$ (this is a bit harder
to prove and we won't do it here!).
\item[(b)]
The Taylor series in (a) is called the {\em binomial series}.
If $\alpha$ is actually a positive integer, show that
$\binom{\alpha}{n} = 0$ for $n > \alpha$. In this case, the infinite
sum in the binomial
series collapses to just a finite sum, and the formula is called the
{\em binomial theorem}, which should be familiar to you!
\item[(c)] 
If you replace $x$ by $-x$ and take $\alpha = -1$ in the binomial
series from (a), you get the {\em geometric series} for the function
$$
\frac{1}{1-x}.
$$
Work this series out explicitly using the formula from (a)!
\item[(d)] 
Now replace $x$ by $x^2$ and take $\alpha = -1$
to get the Taylor series of the
function $\frac{1}{1+x^2}$. Then integrate both sides to prove that
$$
\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.
$$
(The radius of convergence is 1 again.)
\item[(e)] 
Similarly derive the Taylor series of the function $\arcsin x$.
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
For (e), remember the derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$.

\vspace{2mm}
\noindent
4. Here is another much more efficient way to calculate the number $\pi$.
\begin{itemize}
\item[(a)] Show that
$$
\arctan X+\arctan Y = \arctan\left(\frac{X + Y}{ 1 - X Y}\right).
$$
\item[(b)] Use the identity from (a) to show that 
\begin{align*}
\frac{\pi}{4} &= \arctan \frac{1}{2} + \arctan \frac{1}{3},\\
\frac{\pi}{4} &= 4 \arctan \frac{1}{5} - \arctan\frac{1}{239}.
\end{align*}
Using the second of these identities and some of the first few terms
of the Taylor series for
$\arctan x$, show that $\pi = 3.14159\cdots$. If you use a
few more terms in the Taylor series, you can get lots of digits of
$\pi$ very quickly this way!
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
(a) Remember the multiple angle formula for tangent: $\tan(x+y) =
\frac{tan x + \tan y}{1-\tan x \tan y}$. Try letting $X = \tan x$ and
$Y = \tan y$ so $x = \arctan X$ and $y = \arctan Y$.
(b) Don't forget $\pi/4 = \arctan 1$.

\vspace{2mm}
\noindent
5. Calculate the Taylor series of $\tan x$ up to and including the
$x^7$ term and the Taylor series of $\sec x$ up to and including the
$x^6$ term. 


\vspace{2mm}
 \noindent
{\em Hints.}
The pattern behind the coefficients of these two Taylor series is not at
all obvious and quite interesting, you will see more about it next week!

You should start off just by doing the calculation from the definition
of Taylor series: its $f(0) + f'(0) x + f''(0) x^2 / 2! + f'''(0) x^3
/ 3!+\cdots$. You can definitely get the first couple of terms this
way, but you'll probably discover that it is hard to calculate the
derivatives $f^{(n)}(x)$ for these two functions after $n=2$...

There is a better way! I'll explain for $\tan x$. Note this function
is odd, so the series you are after is going to look like $a_1 x + a_3
x^3 / 3! + a_5 x^5 / 5!  + a_7 x^7 / 7! + \cdots$ for some unknown $a_n$'s.
Since $\tan x = \sin x / \cos x$, you know that $\tan x \cos x = \sin
x$. In terms of series:
\begin{multline*}
(a_1 x + a_3 x^3 / 3! + a_5 x^5 / 5! + a_7 x^7 / 7! +
\cdots)(1-x^2/2+x^4/24-x^6/720+\cdots) \\= x - x^3/6+x^5/120 -
x^7/7!+\cdots.
\end{multline*}
Now FOIL out the left hand side and equate coefficients of powers of
$x$ to calculate $a_1,a_3,a_5,a_7,...$

I'll leave $\sec x$ to you, there is a similar technique to do it!

There are other approaches too... for example
you could use that $$\tan(\arctan x) = x,$$ 
take $\tan y = a_1y  + a_3y^3 / 3!
+ a_5 y^5 / 5!+a_7 y^7 / 7!+ \cdots$ then replace $y$ by
the Taylor series of $\arctan x$ which you know and ``open
parenthesis.'' Then you can equate coefficients to work out
$a_1,a_3,a_5$,...
Or you could try using that $\frac{d}{dx} (\tan x) = \sec^2 x$ and
$\frac{d}{dx}(\sec x) = \sec x \tan x$.

\end{document}
