# Set up expansion formula for Taylor Series f(x)=\root(3)(x) centered round a=1?

## So far I have the exponent as ${x}^{- \frac{2}{3} - n}$, and the denominator as $3$. But the rest in the numerator? What kind of factorial is $\left(1 \cdot 2 \cdot 5 \cdot 8 \cdot \ldots .\right)$?

May 10, 2018

root(3)(x)=sum_(n=0)^oo(-1)^(n+1)(x-1)^n(2*5*8*11*...*(3n-1))/(n!3^n)

#### Explanation:

$f \left(x\right) = {x}^{\frac{1}{3}}$

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

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

$f ' ' ' \left(x\right) = \frac{1 \cdot 2 \cdot 5}{3} ^ 3 {x}^{- \frac{8}{3}}$

${f}^{\left(4\right)} \left(x\right) = \frac{- 1 \cdot 2 \cdot 5 \cdot 8}{3} ^ 4 {x}^{- \frac{11}{3}}$

${f}^{\left(5\right)} \left(x\right) = \frac{1 \cdot 2 \cdot 5 \cdot 8 \cdot 11}{3} ^ 5 {x}^{- \frac{14}{3}}$

We see that the derivatives alternate signs after the first term and the $n = 1$ term is positive, so, if we ignore the $n = 0$ term, we have a clear pattern represented by an instance of ${\left(- 1\right)}^{n + 1} .$

Furthermore, the exponent can be represented by ${x}^{\frac{1}{3} - n}$.

The denominator involves ${3}^{n} .$

For the numerator, a distinct pattern emerges if we disregard $f \left(x\right)$ and consider only its derivatives. That is, we strip out $f \left(1\right) = \sqrt[3]{1} = 1$ and start at $n = 1.$

It can't be represented by a factorial (or even double factorial), but we can use the following notation to indicate the terms all have differences of $3$ and alternate between even and odd:

We have $2 \cdot 5 \cdot 8 \cdot 11 \cdot \ldots \cdot \left(3 n - 1\right)$

So,

${f}^{\left(n\right)} \left(x\right) = {\left(- 1\right)}^{n + 1} {x}^{\frac{1}{3} - n} \frac{2 \cdot 5 \cdot 8 \cdot 11 \cdot \ldots \cdot \left(3 n - 1\right)}{3} ^ n$

${f}^{\left(n\right)} \left(1\right) = {\left(- 1\right)}^{n + 1} {1}^{\frac{1}{3} - n} \frac{2 \cdot 5 \cdot 8 \cdot 11 \cdot \ldots \cdot \left(3 n - 1\right)}{3} ^ n$

${1}^{\frac{1}{3} - n} = 1$, so

${f}^{\left(n\right)} \left(1\right) = {\left(- 1\right)}^{n + 1} \frac{2 \cdot 5 \cdot 8 \cdot 11 \cdot \ldots \cdot \left(3 n - 1\right)}{3} ^ n$

Thus, using the standard form of a Taylor Series about $a$,

f(x)=sum_(n=0)^oof^((n))(a)(x-a)^n/(n!) and stripping out the first term, we get

root(3)(x)=1+sum_(n=1)^oo(-1)^(n+1)(x-1)^n(2*5*8*11*...*(3n-1))/(n!3^n)

However, if we take the above summation at $n = 0 ,$ we get -(-1)/(0!3^0)=1, so, unintentionally, the $0 t h$ term can be absorbed into the series!

root(3)(x)=sum_(n=0)^oo(-1)^(n+1)(x-1)^n(2*5*8*11*...*(3n-1))/(n!3^n)