1. ## General term

I'm having trouble finding the coefficients for the maclaurin series for $f(x)=\sqrt{1-x}$. i have

$\frac{\Pi_{k=2}^n(2k-3)}{2^nn!}$. Is this right?

2. By differentiation ,

$f(x) = \sqrt{1-x} = (1-x)^{ \frac{1}{2} }$

$f'(x) = -\frac{1}{2} \cdot (1-x)^{- \frac{1}{2} }$

$f''(x) = - \frac{1}{2} \cdot \frac{1}{2} \cdot (1-x)^{- \frac{3}{2} }$

$f'''(x) = - \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot (1-x)^{- \frac{5}{2} }$

Observation is a good tool to conclude that

$f^{(n)}(x) = - \frac{1}{2} \cdot \frac{1 \cdot 3 ... \cdot (2n-3) }{ 2^{n-1} } \cdot (1-x)^{- \frac{2n-1}{2} }$ ( Isolate the first $\frac{1}{2}$ ) $n \geq 2$

Therefore ,

$f^{(n}(0) = - \frac{1}{2} \cdot \frac{1 \cdot 3 ... \cdot (2n-3) }{ 2^{n-1} }$

But if we " fill up " the spaces between the odd numbers ,

$1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-3)$

$= 1 \cdot 2 \cdot 3 \cdot 4 ... \cdot (2n-3) \cdot (2n-2)$ $\cdot \frac{1}{2 \cdot 4 \cdot ... \cdot (2n-2)}$

$= \frac{ (2n-2)!}{ 2^{n-1} \cdot (n-1)!}$

we will find that

$f^{(n)}(0)$ is also equal to :

$- \frac{ (2n-2)! }{ 2^{2n-1} (n-1)! }$

When accompanying with $\frac{1}{n!}$ to form the coefficient for the Maclaurin series , we can obtain an elegant expression .

$- \frac{ (2n-2)! }{ 2^{2n-1} (n-1)! n! }$

$= - \frac{ (2n-2)! }{ 2^{2n-1} (n-1)! n! } \cdot \frac{ (2n-1)(2n) }{ n } \cdot \frac{n}{2n(2n-1)}$

$= - \frac{ \binom{2n}{n} }{ 2^{2n} (2n-1) }$

$= - \frac{1}{4^n} \frac{ \binom{2n}{n} }{ 2n-1}$