# How do you find the taylor series for f(x)=1/(1+x^3)?

Jun 19, 2015

The general process I would use is to take all the derivatives you need to do, first.

That way you can simplify them first out of context, when things aren't yet complicated (you'll see). Then, plug them into the Taylor series formula, and use a calculator to numerically simplify the result.

The result would be this.

The formula for the Taylor series is:

sum_(n=0)^(oo)f^n(a)/(n!)(x-a)^n

Some things to remember are:

• $x \to a$ for $f \left(x\right) \to f \left(a\right)$ but not $\left(x - a\right)$.
• $n$ increases but $a$ stays the same.
• Plug in $a$ last.
• Remember to do the Chain Rule and Product Rule!

So first, I would take the $n$th derivative of this. I want to stop at $n = 3$, though (you'll see why). Here goes:

${f}^{0} \left(x\right) = \textcolor{b l u e}{f \left(x\right) = \frac{1}{1 + {x}^{3}}}$

$\textcolor{b l u e}{f ' \left(x\right)} = - 1 {\left(1 + {x}^{3}\right)}^{- 2} \left(3 {x}^{2}\right) = \left(- 3 {x}^{2}\right) {\left(1 + {x}^{3}\right)}^{-} 2$

$\textcolor{b l u e}{= - \frac{3 {x}^{2}}{1 + {x}^{3}} ^ 2}$

$\textcolor{b l u e}{f ' ' \left(x\right)} = - \left[3 {x}^{2} \left(- 2 {\left(1 + {x}^{3}\right)}^{-} 3 \left(3 {x}^{2}\right)\right) + {\left(1 + {x}^{3}\right)}^{-} 2 \left(6 x\right)\right]$

$= 18 {x}^{4} {\left(1 + {x}^{3}\right)}^{-} 3 - 6 x {\left(1 + {x}^{3}\right)}^{-} 2$

$\textcolor{b l u e}{= \frac{18 {x}^{4}}{1 + {x}^{3}} ^ 3 - \frac{6 x}{1 + {x}^{3}} ^ 2}$

$\textcolor{b l u e}{f ' ' ' \left(x\right)} = \left[18 {x}^{4} \left(- 3 {\left(1 + {x}^{3}\right)}^{-} 4 \left(3 {x}^{2}\right)\right) + \left({\left(1 + {x}^{3}\right)}^{-} 3\right) 72 {x}^{3}\right] - \left[6 x \left(- 2 {\left(1 + {x}^{3}\right)}^{-} 3 \left(3 {x}^{2}\right)\right) + \left({\left(1 + {x}^{3}\right)}^{-} 2\right) 6\right]$

$= \left[- 162 {x}^{6} {\left(1 + {x}^{3}\right)}^{-} 4 + 72 {x}^{3} {\left(1 + {x}^{3}\right)}^{-} 3\right] - \left[- 36 {x}^{3} {\left(1 + {x}^{3}\right)}^{-} 3 + 6 {\left(1 + {x}^{3}\right)}^{-} 2\right]$

$= - 162 {x}^{6} {\left(1 + {x}^{3}\right)}^{-} 4 + 72 {x}^{3} {\left(1 + {x}^{3}\right)}^{-} 3 + 36 {x}^{3} {\left(1 + {x}^{3}\right)}^{-} 3 - 6 {\left(1 + {x}^{3}\right)}^{-} 2$

$= - 162 {x}^{6} {\left(1 + {x}^{3}\right)}^{-} 4 + 108 {x}^{3} {\left(1 + {x}^{3}\right)}^{-} 3 - 6 {\left(1 + {x}^{3}\right)}^{-} 2$

$\textcolor{b l u e}{= - \frac{162 {x}^{6}}{1 + {x}^{3}} ^ 4 + \frac{108 {x}^{3}}{1 + {x}^{3}} ^ 3 - \frac{6}{1 + {x}^{3}} ^ 2}$

I'm not going to simplify these anymore though :-P Alright, let's use them. Plug in $a$ and whip out your calculator!

sum_(n=0)^(oo)f^n(a)/(n!)(x-a)^n = sum_(n=0)^(oo)f^n(a)*(1/(n!))(x-a)^n

= (1/(1+a^3))(1/(0!))(x-a)^0 +
(-(3a^2)/(1+a^3)^2)(1/(1!))(x-a)^1 +
((18a^4)/(1+a^3)^3 - (6a)/(1+a^3)^2)(1/(2!))(x-a)^2 +
(-(162a^6)/(1+a^3)^4 + (108a^3)/(1+a^3)^3 - 6/(1+a^3)^2)(1/(3!))(x-a)^3
$+ \ldots$

At $a = 1$ for simplicity (you see why!):

$= \frac{1}{2} - \frac{3}{4} \left(x - 1\right) + \frac{\frac{18}{8} - \frac{12}{8}}{2} {\left(x - 1\right)}^{2} + \frac{- \frac{162}{16} + \frac{216}{16} - \frac{24}{16}}{6} {\left(x - 1\right)}^{3}$

$= \textcolor{b l u e}{\frac{1}{2} - \frac{3}{4} \left(x - 1\right) + \frac{3}{8} {\left(x - 1\right)}^{2} + \frac{5}{16} {\left(x - 1\right)}^{3} + \ldots}$

If you look here, Wolfram Alpha gives the same result! YEAH!

$= \textcolor{b l u e}{\frac{1}{2} + \left(- \frac{3}{4}\right) \left(x - 1\right) + \frac{3}{8} {\left(x - 1\right)}^{2} + \frac{5}{16} {\left(x - 1\right)}^{3} + \left(- \frac{21}{32}\right) {\left(x - 1\right)}^{4} + \frac{21}{64} {\left(x - 1\right)}^{5} + \ldots}$