as the title states: Find all analytic functions $f: \mathbb{C} \longrightarrow \mathbb{C}$ such that $f(3z)-f(2z)=5f(z)$ where $z \in \mathbb{C}$ and $f(1)=3$
Starting with the taylor series of expansion of $f(z)$ we have $$f(z) = \sum_{n=0}^{\infty}a_n z^n$$ then substituting the above we have
$$f(3z)-f(2z)=5f(z) \implies \sum_{n=0}^{\infty}a_n3^nz^n – \sum_{n=0}^{\infty}a_n2^nz^n = 5\sum_{n=0}^{\infty}a_nz^n$$
then via the uniqueness of power series coeffecients we have
$$a_n3^n – a_n2^n = 5a_n$$
which has a single solution at $n=2$ then $a_n = 0~\forall n \in \mathbb{N}\setminus \{2\}$ otherwise.
then we have
$$f(z)=\sum_{n=0}^{\infty}a_nz^n = a_2z^2$$
from here i'm tempted to use the formula for $a_2$ ie
$$a_n = \frac{1}{2 \pi i}\int_{S_{r}^{+}(0)}\frac{f(w)}{w}dw $$ but since it's been given to me that $f(1)=3$ i'm tempted to just state that $a_2 = 3$. as i can't really think of any other way of determining f.
Thoughts? is this correct?
Best Answer
It is not too hard to see (differentiate and set $z=0$) that $(3^k-2^k-5)f^{(k)}(0) = 0$. Hence $f^{(k)}(0) = 0$ for all $k \neq 2$.