Holomorphic functions under certain conditions I am generally confused with problems such as the following:
Problem: Finding holomorphic functions $g$ on the open disk $C=\{z:|z|<1\}$ to $\mathbb{C}$ which satisfies the relation $g^{''}(\frac{1}{n})+g(\frac{1}{n})=0, \forall n=2,3,4,5...$
It would help greatly if someone can help me solve these kind of problems as it will help me study for my exams more easily. Much help would be appreciated. 
 A: The identity theorem implies that $g''(z) + g(z) = 0$ for all 
$z \in \Bbb D$, i.e. $g$ is a solution of the homogeneous second-order linear differential equation 
$$ \tag{*}
 w'' + w = 0 \, .
$$
It is clear that $\cos(z)$ and $\sin(z)$ are two linearly independent
solutions, and it follows from the general theory on linear differential
equations that any solution of $(*)$ is of the form
$$
 g(z) = a \cos(z) + b \sin(z)
$$
with constants $a, b \in \Bbb C$.
For holomorphic functions this can also be shown directly: Define
$$
 h(z) = g(z) - g(0) \cos(z) - g'(0) \sin(z) \, .
$$
Then
$$
h''(z) + h(z) = 0 \text{ in } \Bbb D, \quad h(0) = h'(0) = 0 \, .
$$
Now conclude that in the power series for $h$
$$
 h(z) = \sum_{n=0}^\infty b_n z^n
$$
all coefficients $b_n$ must be zero, and therefore
$$
 g(z) = g(0) \cos(z) + g'(0) \sin(z) \, .
$$
A: This problem may be viewed as consisting of essentially two sections, each of which requires a certain method of it's own:  in the first, we use a general result, the identity principle, to show that $g''(z) + g(z) = 0$ on $C$; in the second, we use more-or-less commonly known facts about power series to establish a specific form for such $g(z)$.
Since $g(z)$ is holomorphic on $C$, so are $g'(z)$ and $g''(z)$; hence
$f(z) = g''(z) + g(z) \tag 1$
is holomorphic on $C$ as well.  Since $f(z)$ is holomorphic on $C$, it is continuous on $C$, and we have
$f \left (\dfrac{1}{n} \right ) = g''\left (\dfrac{1}{n} \right ) + g \left (\dfrac{1}{n} \right ) = 0, \; n = 2, 3, 4, \ldots; \tag 2$
it follows from (2) and the continuity of $f(z)$ that
$f(0) = \displaystyle \lim_{n \to \infty} f \left ( \dfrac{1}{n} \right ) = \lim_{n \to \infty} 0 = 0; \tag 3$
since $f(z)$ is equal to the constant function $0$ on the set
$\left \{ 0, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \ldots \right \} \tag 3$
which contains the accumulation point $0$, it follows from the identity principle that 
$f(z) = 0 \; \text{on} \; C, \tag 4$
and by virtue of (1) we see this may be written as
$g''(z) + g(z) = 0 \; \text{on} \; C. \tag 5$
We may use (5) to gain information concerning the power series representation of $g(x)$ about the point $z = 0$; indeed, (5) may be written
$g''(z) = -g(z), \tag 6$
whence
$g^{(3)}(z) = g'''(z) = -g'(z), \tag 7$
$g^{(4)}(z) = -g''(z) = g(z), \tag 8$
$g^{(5)}(z) = g'(z), \tag 9$
$g^{(6)}(z) = g''(z) = -g(z), \tag{10}$
$g^{(7)}(z) = -g'(z), \tag{11}$
$g^{(8)}(z) = -g''(z) = g(z); \tag{12}$
from (6)-(12) me may hypothesize the general pattern for $n \ge 1$:
$g^{(2n)}(z) = (-1)^n g(z), \tag{13}$
$g^{(2n + 1)}(z) = (-1)^n g'(z); \tag{14}$
(13)-(14) may in fact be proved by induction: if 
$g^{(2k)}(z) = (-1)^k g(z), \tag{15}$
then
$g^{(2(k + 1))}(z) = g^{(2k + 2)}(z) = (g^{(2k)}(z))''$
$= (-1)^k g''(z) = (-1)^k(-g(z)) = (-1)^{k + 1} g(z); \tag{16}$
and if
$g^{(2k + 1)}(z) = (-1)^k g'(z), \tag{17}$
we have
$g^{(2(k + 1) + 1)}(z) = g^{(2k + 3)}(z) = (g^{2k + 1}(z))''$
$= (-1)^k g'''(z) = (-1)^k(-g'(z)) =  (-1)^{k + 1}g'(z), \tag{18}$
completing the induction in both cases and thus establishing (13) and (14), which we may evaluate at $z = 0$:
$g^{(2n)}(0) = (-1)^n g(0), \tag{19}$
$g^{(2n + 1)}(0) = (-1)^n g'(0). \tag{20}$
We next consider the power series for $g(z)$ at $z = 0$:
$g(z) = \displaystyle \sum_0^\infty \dfrac{g^{(n)}(0)}{n!}z^n, \tag{21}$
which we break into even and odd terms, thus:
$g(z) = \displaystyle \sum_0^\infty \dfrac{g^{(2n)}(0)}{(2n)!}z^{2n} + \sum_0^\infty \dfrac{g^{(2n + 1)}(0)}{(2n + 1)!}z^{2n + 1}; \tag{22}$
we substitute (19)-(20) into these series, yielding
$g(z) = \displaystyle \sum_0^\infty \dfrac{(-1)^n g(0)}{(2n)!}z^{2n} + \sum_0^\infty \dfrac{(-1)^n g'(0)}{(2n + 1)!}z^{2n + 1}$
$= \displaystyle g(0)\sum_0^\infty \dfrac{(-1)^n}{(2n)!}z^{2n} +  g'(0)\sum_0^\infty \dfrac{(-1)^n}{(2n + 1)!}z^{2n + 1}; \tag{23}$
we observe that the two series occurring on the right of (23) are those of $\cos z$ and $\sin z$, respectively, and thus write
$g(z) = g(0)\cos z + g'(0) \sin z. \tag{24}$
The above argument and derivation shows that every holomorhic $g(z)$ on $C$ satisfying (2) must be of the form (24), with $g(0), g'(0) \in \Bbb C$ specfied freely at our discretion.
In closing, I'm not sure what to say about solving such problems in general, or learning to do so, except to practice on similar problems as much as one can and try to recognize that, as one's skill and experience grows, one will likely encounter more questions which require a multiplicity of techniques to attack successfully.  Just keep at it is really the best I can offer.  And feel free to ask more questions . . .
