j-th order derivative of function $f(x) = 1 - \sqrt{1 - x}$

Question

Can you please help me with this part of book (Stochastic Calculus for Finance I: The Binomial Asset Pricing Model; Shreve; p.140):

Define $f(x) = 1 - \sqrt{1 - x}$ so that

$$f^{’}(x)=\frac{1}{2}(1-x)^{-\frac{1}{2}}$$ $$f^{’’}(x)=\frac{1}{4}(1-x)^{-\frac{3}{2}}$$ $$f^{’’’}(x)=\frac{3}{8}(1-x)^{-\frac{5}{2}}$$

and, in general, the jth-order derivative of $f$ is

$$f^{(j)}(x)=\frac{1\cdot 3\cdots(2j-3)}{2^j}(1-x)^{-\frac{2j-1}2},\qquad j=1,2,3,\ldots$$

It's probably simple problem but i don't see how we got $${1\cdot 3\cdots(2j-3)}$$ or maybe I don't understand what does it exactly mean.

Answer 1

I'm assumming you want to derive the $j$-th derivative of $f$, not just prove what is written. If so, then that could be done by finding the pattern and then proving it.

Finding the pattern

We'll start by looking at first $6$ derivatives of function $f$ (you can check how ever many until you're comfortable in your formula for $j$-th derivative).
$$f'(x)=\frac{1}{2(1-x)^{\frac{1}{2}}}=\frac{1}{2^1}(1-x)^{-\frac{1}{2}}$$ $$f''(x)=\frac{1}{4(1-x)^{\frac{3}{2}}}=\frac{1}{2^2}(1-x)^{-\frac{3}{2}}$$ $$f^{(3)}(x)=\frac{3}{8(1-x)^{\frac{5}{2}}}=\frac{1\cdot3}{2^3}(1-x)^{-\frac{5}{2}}$$ $$f^{(4)}(x)=\frac{15}{16(1-x)^{\frac{7}{2}}}=\frac{1\cdot3\cdot5}{2^4}(1-x)^{-\frac{7}{2}}$$ $$f^{(5)}(x)=\frac{105}{32(1-x)^{\frac{9}{2}}}=\frac{1\cdot3\cdot5\cdot7}{2^5}(1-x)^{-\frac{9}{2}}$$ $$f^{(6)}(x)=\frac{945}{64(1-x)^{\frac{11}{2}}}=\frac{1\cdot3\cdot5\cdot7\cdot9}{2^6}(1-x)^{-\frac{11}{2}}$$ Here it seems like we can see a pattern in the denominator (number part) it looks like $2^j$. Also term $(1-x)$ is in every one of these derivatives, and we can see it is to power of a negative odd number divided by two. It is obvious that we have $(1-x)^{-\frac{2j-1}{2}}$ as a factor of $f^{(j)}(x)$.
The last pattern we can see is that all the number factors in the numerator (e.g. $1\cdot3\cdot5$) are odd numbers, but we need to find the last factor of this product (dependent on j). Since this isn't as obvious as was the case with $(1-x)^{-\frac{2j-1}{2}}$.
So we can make a system of two linear equations. So let's take $j=5$ and $j=6$, although you can take any other $j$, it should yield the correct result. I'll take $y$ and $z$ as values we're trying to find. Looking at the fifth and sixth derivatives their last factors are $7$ and $9$, respectively. We know that this last factor can be written in the form $yj+z$, so we have two equations:
$5y+z=7$
$6y+z=9$
Solving this we get $y=2$ and $z=-3$. So the last factor is $2j-3$.
Combining all of this together we get $$f^{(j)}(x)=\frac{1\cdot 3\cdots(2j-3)}{2^j}(1-x)^{-\frac{2j-1}2},\qquad j=2,3,4,\ldots$$ But you can also see that the formula won't be correct for $j=1$.

Proving the pattern

Simplest way to prove this is using induction.
We already showed that it is true for up to sixth derivative, excluding first derivative.
We'll use induction on $j$, suppose that $j$-th derivative is given as stated. Let's look at $$(f^{j}(x))'=\frac{1\cdot 3\cdots(2j-3)\cdot(2j-1)}{2^j\cdot2}(1-x)^{-\frac{2j+1}2}=\frac{1\cdot 3\cdots(2j-3)\cdot(2j-1)}{2^{j+1}}(1-x)^{-\frac{2j+1}2}=f^{(j+1)}(x)$$

Using induction we showed that $f^{(j)}(x)$ that we derived is in fact correct.

Answer 2

You have that $$\frac{\mathrm d }{\mathrm d x}(1-x)^{-\frac{n}{2}}{}=\frac{n}{2}(1-x)^{-\frac{n+2}{2}}.$$

Therefore $$\frac{\mathrm d^m }{\mathrm d x^m}(1-x)^{-\frac{n}{2}}=\frac{n(n+2)...(n+2(m-1))}{2^m}(1-x)^{-\frac{n+2m}{2}}.$$