Ch5: Series part 1 Flashcards
limit of sequence
The sequence $(z_n)$ converges to a limit $z \in C$ if
$\forall \ \epsilon > 0 \ \exists \ N \in \mathbb{N} $
st if
$n \geq N \Rightarrow |z-z_n| < \epsilon$
Bounded sequence
seq $z_n$ is bounded if $\exists \ M$ (M is constant) st that $|z_n| < M \ \forall \ n \in N$
Thm connect bounds of sequences
(1) If the limit of a complex sequence exists, then it is unique.
(2) A convergent sequence is bounded.
thm page 180 that you need to prove
Suppose that $z_n = x_n + iy_n (n \in N)$ and $z = x+iy$.
Then $z_n$ → z if and only if $x_n$ → x and
$y_n$ → y.
Complex power series def
A series of the form $\sum_{n=0}^\infty a_n (z - z_0)^n$
where $z_0 \in C$ is a constant $a_n \in C$ are constants (the coefficients)
and $z \in C$ is a variable
NB power series formula
$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ whenever $|z| <1$
if $z \not = 1$ and $N \in N$ then $$\frac{1}{1-z} = \sum_{n=0}^{N-1} z^N +\frac{z^N}{1-z}$$
Corallary on power seires
if $z,s \in C, s \not = 0, s\not = z, N\in N$ then
$$\frac{1}{s-z} = \sum_{n=0}^{N-1} \frac{z^n}{s^{n
+1}} +\frac{z^N}{(s-z)s^N}$$
Doubly infinite series convergence
Suppose that s ∈ C and $\forall\ \epsilon > 0\ \exists\ N ∈ N$ st if $m, n ≥ N$, then
|$(z_{−m}+z_{−m+1}+· · ·+z_{−1}+z_0+z_1+· · ·+z_n)−s| < \epsilon$.
Then the doubly infinite series $\sum_{n=-\infty}^{\infty} z_n$
is
convergent with sum s
Note about doubly infinite seires
Note that such a doubly infinite series converges iff both the negative bits and poositvie bits converge