How to formally use Taylor expansions for $n$th derivatives and generating functions? When deriving Catalan numbers, the generating function takes on this form:
$$C(x) = \frac{1}{2} (1 - \sqrt{1-4x}) = \frac{1}{2} (1 - f(x))$$
where $f(x) = \sqrt{1-4x}$
How does one formally show what it evaluates to? I can do it somewhat "informally" like so:
$$f(x) = \sqrt{1-4x} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
$$f(x) = \frac{f^{(0)}(0)}{0!} x^0 + \frac{f^{(1)}(0)}{1!} x^1 + \frac{f^{(2)}(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + ...$$
Now at this point I can literally start computing the successive derivatives of $f(x)$ and then plug in $0$ for each $x$ and look for patterns and "eyeball" a closed-form for it.
But is there a way at this point to formally show what the $n$th derivative is? Or is this more of an art than a science that changes depending on the function?
I could eyeball the $n$th derivative formula as well:
$f^{0}(x) = (1 - 4x)^{\frac{1}{2}}$
$f^{1}(x) = (\frac{1}{2})(-4)(1 - 4x)^{-\frac{1}{2}}$
$f^{2}(x) = (-\frac{1}{2})(\frac{1}{2})(-4)(-4)(1 - 4x)^{-\frac{3}{2}}$
$f^{3}(x) = (-\frac{3}{2})(-\frac{1}{2})(\frac{1}{2})(-4)(-4)(-4)(1 - 4x)^{-\frac{5}{2}}$
$f^{4}(x) = (-\frac{5}{2})(-\frac{3}{2})(-\frac{1}{2})(\frac{1}{2})(-4)(-4)(-4)(-4)(1 - 4x)^{-\frac{7}{2}}$
At this point I can eyeball the pattern and solve it manually. But again, what if I couldn't do that? Is there a more direct and methodical way to this?

May be, you could use the generalized binomial expansion $$(1+x)^a=\sum_{n=0}^\infty \binom{a}{n} x^n$$ which makes $$f(x)=\sqrt{1-4x}=\sum_{n=0}^\infty (-1)^n\binom{\frac 12}{n} (4x)^n=1+\sum_{n=1}^\infty (-1)^n\binom{\frac 12}{n} (4x)^n$$ and then $$C(x) = \frac{1}{2} (1 - \sqrt{1-4x})=\frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\binom{\frac 12}{n} (4x)^n$$
