General formula for the Taylor expansion of $f(x)=(1+x)^\frac{1}{k}$ at $a = 0$

Question

Let $f(x)=(1+x)^\frac{1}{k}$ where $k \in \mathbb{N}$.

I'm seeking a general formula for the Taylor expansion of $f(x)$ at the point $a = 0$ up to $n \in \mathbb{N}$ but I struggle with the generality. Note: $f^m(x)$ is the $m$-th derivative of $f(x)$.

Let's compute the first few terms:

$$f^0(x)=(1+x)^\frac{1}{k},$$ $$f^0(0)=1,$$ $$f^1(x)=\frac{d}{dx}(1+x)^\frac{1}{k}=\frac{1}{k}(1+x)^\frac{1-2k}{k},$$ $$f^1(0)=\frac{1}{k},$$ $$f^2(x)=\frac{d}{dx}\frac{1}{k}(1+x)^\frac{1-2k}{k}=\frac{1}{k}\cdot\frac{1-k}{k}\cdot(1+x)^\frac{1-2k}{k},$$ $$f^2(0) = \frac{1}{k} \cdot \frac{1-k}{k},$$ $$f^3(x)=\frac{d}{dx}\frac{1}{k}\cdot\frac{1-k}{k} \cdot (1+x)^\frac{1-2k}{k}=\frac{1}{k} \cdot \frac{1-k}{k} \cdot \frac{1-2k}{k}\cdot(1+x)^\frac{1-3k}{k},$$ $$f^3(0)=\frac{1}{k} \cdot \frac{1-k}{k} \cdot \frac{1-2k}{k}.$$ So the first few terms in Taylor series are $$T^{f,0}_{n} = \frac{1}{0!} + \frac{1}{1!\cdot k}x + \frac{1-k}{2! \cdot k^2}x^2 + \frac{(1-k)(1-2k)}{3! \cdot k^3}x^3 + \frac{(1-k)(1-2k)(1-3k)}{4! \cdot k^4}x^4 \cdot \cdot \cdot \ $$ I imagine it to be something like $\sum_{n=0}^{\infty} \frac{\prod_{i=1}^{n} ?}{?! \cdot k^?}$.

Answer 1

If $ \alpha\in\mathbb{R} $, then the dérivatives of $ x\overset{f}{\mapsto}\left(1+x\right)^{\alpha} $ are not difficult to find (In your case we se $ a_{k}=\frac{1}{k},\ k\in\mathbb{N}^{*} $) :

\begin{aligned} \left(\forall x\in\left]-1,1\right[\right),\ \ \ f'\left(x\right)&=\alpha\left(1+x\right)^{\alpha -1}\\ \left(\forall x\in\left]-1,1\right[\right),\ \ f''\left(x\right)&=\alpha\left(\alpha -1\right)\left(1+x\right)^{\alpha -2}\\ \left(\forall x\in\left]-1,1\right[\right),\ f'''\left(x\right)&=\alpha\left(\alpha -1\right)\left(\alpha -2\right)\left(1+x\right)^{\alpha -3} \end{aligned}

In general, we can prove by induction that : $$ \left(\forall p\in\mathbb{N}\right)\left(\forall x\in\left]-1,1\right[\right),\ f^{\left(p\right)}\left(x\right)=\alpha\left(\alpha -1\right)\cdots\left(\alpha -p+1\right)\left(1+x\right)^{\alpha -p} $$

Let $ n,k $ be positive integers. Setting, as you wanted, $ \alpha =\alpha_{k}=\frac{1}{k} $, and denoting, for $ p\in\mathbb{N} $, $ \frac{\alpha_{k}\left(\alpha_{k} -1\right)\cdots\left(\alpha_{k} -p+1\right)}{p!} $ as $ \binom{\alpha_{k}}{p} $, we get can the Taylor expansion :

$$ \left(1+x\right)^{\frac{1}{k}}=\sum_{p=0}^{n}{\binom{\frac{1}{k}}{p}x^{p}}+\underset{\overset{x\to 0}{}}{\mathcal{O}}\left(x^{n+1}\right) $$

Answer 2

Hint:

The general formula for the binomial expansion of $(1+x)^\alpha$ simply uses the generalised binomial coefficients: $$\binom{\alpha}{n}=\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+1)}{n!}.$$ The main difference with the case when $\alpha$ is a positive integer is that these generalised binomial coefficients are never $0$, so that we indeed obtain an infinite series, not a polynomial.