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\title{Mathematics 311:  Complex Analysis --- Assignment 5}
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{\bf Reading}:  Marsden, sections 3.1, 3.2.

{\bf Problems}:
1.  The Bernoulli numbers~$B_n$ are defined to be the coefficients 
of the power series expansion
$$
\frac{z}{e^z-1}=\sum_{n=0}^\infty\frac{B_n}{n!}z^n.
$$

(a) Show that $B_0=1$, $B_1=-1/2$, $B_2=1/6$.
(Hint: Expand the double power series $z=(e^z-1)\sum(B_n/n!)z^n$.)

(b) Show that in general these numbers satisfy the recurrence formula
$$
\binom{n}{0}B_0+\binom{n}{1}B_1+\binom{n}{2}B_2+\cdots 
+\binom{n}{n-1}B_{n-1}=0,\quad (n>1),
$$
so they are all rational numbers.  Note that this formula can be written 
symbolically $(B+1)^n=B^n$, for expanding the left--hand side by the binomial
formula then replacing $B^k$ by $B_k$ yields the recurrence.
(Hint: Rearrange the double sum in (a).)

(c) Show that
$$
\frac{z}{e^z-1}+\frac{z}2 = \frac{z}2\frac{e^z+1}{e^z-1},
$$
note that this is an even function and conclude that $B_n=0$ whenever $n$ is 
odd and greater than~$1$.

(d) Show by using the preceding formula that
$$
z\cot z = \sum_{n=0}^\infty(-1)^n\frac{2^{2n}B_{2n}}{(2n)!}z^{2n}.
$$

(e) Show by using the preceding formula and the formula
$\tan z=\cot z-2\cot2z$ that
$$
\tan z 
=\sum_{n=1}^\infty(-1)^{n-1}\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}.
$$

The notation for the Bernoulli numbers differs somewhat, so be careful when
comparing formulas from various sources.  The Bernoulli numbers vary rather 
mysteriously and erratically; for your information,
\begin{align*}
&B_4=-1/30,\quad B_6=1/42,\quad B_8=-1/30,\quad B_{10}=5/66,\\
&B_{12}=-691/2730,\quad B_{14}=7/6,\quad B_{20}=-174611/330,\\
&B_{30}=8615841276005/14322. 
\end{align*}

2.  Derive the power series expansions
\begin{align*}
&e^z=\sum_{n=0}^\infty\frac{z^n}{n!},\qquad
\log(1+z)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}z^n\\
&\sin z=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}z^{2n+1},\qquad
\cos z=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}
\end{align*}
and find the radius of convergence of each.

3.  Let $f(z)$ be the analytic function in the disk $D=\{z:|z|<1\}$ 
for which $f'(z)=1/(1+z^2)$ and $f(0)=0$.  
From the power series expansion
$$
\frac1{1+z^2} = \sum_{n=0}^\infty(-1)^nz^{2n}
$$
deduce the power series expansion
$$
f(z) = \sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{2n+1}
$$
and indicate why your deduction is justified.  What is the radius of 
convergence of this series? 
Show that
$$
\frac{d}{dz}(f(\tan z)) = 1\text{ and }f(\tan z)=z
$$
in a disk about the origin, so that $f(z)$ may be viewed as the inverse
tangent function there.

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