MHB Finding Maclaurin series of a function

[tmt1]

I need to find the Maclaurin series for this function:

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

And I need to find $f^n(a)$

First, I need the first few derivatives:

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

$$f''(x) ={ \frac{3}{4}} (1 - x)^{- \frac{5}{2}}$$

$$f'''(x) ={- \frac{15}{8}} (1 - x)^{- \frac{7}{2}}$$

So, I get something like $(1 - x)^{-(\frac{1}{2} + n)}$

but I don't know how to get an expression for the left coefficient.

 

[I like Serena]

I need to find the Maclaurin series for this function:

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

And I need to find $f^n(a)$

First, I need the first few derivatives:

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

$$f''(x) ={ \frac{3}{4}} (1 - x)^{- \frac{5}{2}}$$

$$f'''(x) ={- \frac{15}{8}} (1 - x)^{- \frac{7}{2}}$$

So, I get something like $(1 - x)^{-(\frac{1}{2} + n)}$

but I don't know how to get an expression for the left coefficient.

Hi tmt! ;)

Those coeffiecients are:
$$-\frac 1 2, \quad -\frac 12 \cdot -\frac 32, \quad -\frac 12 \cdot -\frac 32 \cdot -\frac 52, \quad ..., \quad
\underbrace{-\frac 12 \cdot -\frac 32 \cdot -\frac 52 \cdot ... \cdot -\frac n2}_{n\text{ factors}}
$$
Btw, it may be sufficient to simply specify them as:
$$-\frac 12, \frac 34, -\frac {15}8, \frac {105}{16}, ...$$

Taking it a step further, the generalization of the binomial coefficient for real numbers defines:
$$\binom{\alpha}{n} = \frac{\alpha \cdot (\alpha - 1) \cdot (\alpha - 2) \cdot ... \cdot (\alpha - n + 1)}{n!}
$$
So we can write them as:
$$\binom{-\frac 12}{n} \cdot n! = \underbrace{-\frac 12 \cdot -\frac 32 \cdot -\frac 52 \cdot ... \cdot -\frac n2}_{n\text{ factors}}
$$