# How do you find the Taylor's formula for f(x)=x^p for x-1?

Feb 20, 2017

${x}^{p} = {\sum}_{k = 0}^{\infty} \left(\begin{matrix}p \\ k\end{matrix}\right) {\left(x - 1\right)}^{k}$

where $\left(\begin{matrix}p \\ k\end{matrix}\right)$ is the generalized binomial coefficient.

#### Explanation:

We have to distinguish two cases:

(1) If $p \in \mathbb{N}$ then the derivatives of order grater than $p$ of $f \left(x\right)$ are identically null. In such case:

$\frac{\mathrm{df}}{\mathrm{dx}} = p {x}^{p - 1}$

...

(df^k)/dx^k = p(p-1)...(p-k+1)x^(p-k) = (p!)/(p-k!)x^(p-k)

and for $x = 1$

[(df^k)/dx^k ]_(x=1) = (p!)/(p-k!)

so the Taylor series has a finite number of terms:

x^p = sum_(k=0)^p (p!)/(p-k!)(x-1)^k/(k!) = sum_(k=0)^p ((p),(k)) (x-1)^k

where ((p),(k)) = (p!)/((k!)(p-k)!) is a binomial coefficient.

(2) If $p \notin \mathbb{N}$ then we can use the definition of generalized binomial coefficient:

((p),(k)) = (p(p-1)(p-2)...(p-k+1))/(k!)

where $p \in \mathbb{R}$ and $k \in \mathbb{N}$

In this case the derivatives of all orders are:

(df^k)/dx^k = p(p-1)...(p-k+1)x^(p-k) = ((p),(k)) k! x^(p-k)

and for $x = 1$:

[(df^k)/dx^k ]_(x=1) = ((p),(k)) (k!)

so the Taylor series is:

x^p = sum_(k=0)^oo ((p),(k)) (k!)(x-1)^k/(k!) = sum_(k=0)^oo ((p),(k)) (x-1)^k

Note that if we use the definition of generalized binomial coefficient, we have:

((p),(k)) = (p(p-1)(p-2)...color(blue)((p-p))..(p-k))/(k!) = 0

for $p , k \in \mathbb{N}$ and $k > p$

then we can unify the notation as:

${x}^{p} = {\sum}_{k = 0}^{\infty} \left(\begin{matrix}p \\ k\end{matrix}\right) {\left(x - 1\right)}^{k}$

for any $p$.