$n$th derivative of $\frac{1}{\sqrt{1-2x}}$

Question

I was asked to find the nth derivative of the given function: $$\frac{1}{\sqrt{1-2x}}$$

I tried to use binomial theorem, finally with some effort I could find a way to express the function in the following form:

$$\frac{1}{\sqrt{1-2x}}=\sum_{k=0}^{∞}{{2k}\choose{k}}\left(\frac{1}{2}\right)^{k}x^{k}$$

then I found a formula for the nth derivative of the given function:

$$\frac{d^{n}}{dx^{n}}\left(\frac{1}{\sqrt{1-2x}}\right)=\sum_{k=0}^{∞}{{2k}\choose{k}}\left(\frac{1}{2}\right)^{k}\left(\prod_{m=0}^{n-1}\left(k-m\right)\right)x^{\left(k-n\right)}$$ which is non-zero for $k\le n$

Also I wanted to know what is the radius of convergence of this expression then I used ratio test such that:

$$\lim\limits_{k \to ∞}\left|\frac{a_{k+1}}{a_{k}}\right|=\lim\limits_{k \to ∞}\left|\frac{\ x^{\left(k+1\right)}\left(2k+2\right)!\left(\frac{1}{2}\right)^{\left(k+1\right)}\ }{\left(\left(k+1\right)!\right)^{2}}\cdot\frac{\left(\left(k\right)!\right)^{2}}{x^{k}\left(2k\right)!\left(\frac{1}{2}\right)^{\left(k\right)}}\right|=$$$$\lim\limits_{k \to ∞}\left|\frac{x\left(2k+2\right)\left(2k+1\right)\left(\frac{1}{2}\right)}{\left(k+1\right)^{2}}\right|=$$$$\lim\limits_{k \to ∞}\frac{1}{2}\left|\frac{4k^{2}+6k+2}{k^{2}+2k+1}\cdot x\right|=$$$$\lim\limits_{k \to ∞}\frac{1}{2}\left|4x\right|$$

hence the expression is valid for $\left|x\right|<\frac{1}{2}$, the same radius of convergence is for the derivative of the given function.

my question is that: is there any way which calculate the nth derivative of the function for all the domain of the function?

Answer 1

Suppose we have $$f(x)=\frac{1}{\sqrt{1-2x}}=(1-2x)^{-1/2}.$$

The first, second, and third derivatives would be

\begin{align} f'(x)&=-\frac12(1-2x)^{-3/2}\cdot(-2)=(1)(1-2x)^{-3/2}\\ f''(x)&=-\frac32(1-2x)^{-5/2}\cdot(-2)=(1)(3)(1-2x)^{-5/2}\\ f'''(x)&=-\frac52(3)(1-2x)^{-7/2}\cdot(-2)=(1)(3)(5)(1-2x)^{-7/2} \end{align}

By inspection, the pattern appears to be $$f^{(n)}(x)=(2n-1)!!(1-2x)^{-(2n+1)/2}=\boxed{\frac{(2n-1)!!}{\sqrt{(1-2x)^{2n+1}}}}$$ where $n!!$ is the double factorial function, defined for odd numbers as $(2n-1)!!=(2n-1)(2n-3)\cdots(5)(3)(1)$.

Answer 2

Hint:

Computing the first few derivatives, using the fractional notation for exponents, you should check the formula should be $$f^{(n)}(x)=\frac{(2n-1)!!}{\bigl(1-2x\bigr)^{\tfrac{2n+1}2}}$$ then prove it by an easy induction.