power series expansion at 0

Question

How can I find the power series expansion of $f(z) = \sqrt{1-4z}$ at $0$?

I need to get the form: $\sum\nolimits_{n=0}^\infty a_nz^n$. I tried to use the fact that the $a_n$ are of the form $\frac{f^{(n)}(0)}{n!}$, but this doesn't help me much. Thanks for any hint.

Answer 1

Hint:

$$f'(x)=-2(1-4x)^{\frac12}=-2\times(f(x))^{-1}$$ $$f''(x)=-4(1-4x)^{\frac32}=-2\times2\times(f(x))^{-3}$$ $$f^{(3)}(x)=-24(1-4x)^{\frac52}=-4\times(2+4)\times(f(x))^{-5}$$ $$f^{(4)}(x)=-240(1-4x)^{\frac72}=-24\times(6+4)\times(f(x))^{-7}$$ $$f^{(5)}(x)=-3360(1-4x)^{\frac92}=-240\times(10+4)\times(f(x))^{-9}$$

Answer 2

Recall the binomial series expansion for $|z|<1$ and $\alpha\in\mathbb{C}$ is \begin{align*} (1+z)^\alpha=\sum_{n=0}^\infty \binom{\alpha}{n}z^n \end{align*}

Setting $\alpha=\frac{1}{2}$ we obtain for $|z|<\frac {1}{4}$ \begin{align*} \sqrt{1-4z}&=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n}(-4)^nz^n\\ &=1-2z-2z^2-4z^3-10z^4-\cdots \end{align*}