# How do you find the coefficient of x^6 in the taylor series expansion?

Jun 27, 2015

Well, if you don't know any shortcut (neither do I), it might as well be good to practice writing the full expansion out.

The Taylor series expansions are written as:

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

where ${f}^{n} \left(a\right)$ is the nth derivative of $f \left(x\right)$ at $x \to a$ (except for the $x - a$ term, in which $x$ remains as $x$). In this case, it doesn't matter what $a$ is, because we only want the coefficient. Let's just say $a = 2$.

$f \left(x\right) = {x}^{6}$
$f ' \left(x\right) = 6 {x}^{5}$
$f ' ' \left(x\right) = 30 {x}^{4}$
$f ' ' ' \left(x\right) = 120 {x}^{3}$
$f ' ' ' ' \left(x\right) = 360 {x}^{2}$
$f ' ' ' ' ' \left(x\right) = 720 x$
$f ' ' ' ' ' ' \left(x\right) = 720$

And past that is just $0$.

=> f(2)/(0!)(x-2)^(0) + (f'(2))/(1!)(x-2)^(1) + (f''(2))/(2!)(x-2)^(2) + (f'''(2))/(3!)(x-2)^(3) + (f''''(2))/(4!)(x-2)^(4) + (f'''''(2))/(5!)(x-2)^(5) + (f''''''(2))/(6!)(x-2)^(6)

$= {2}^{6} / \left(1\right) + \frac{6 \cdot {2}^{5}}{1} \left(x - 2\right) + \frac{30 \cdot {2}^{4}}{2} {\left(x - 2\right)}^{2} + \frac{120 \cdot {2}^{3}}{6} {\left(x - 2\right)}^{3} + \frac{360 \cdot {2}^{2}}{24} {\left(x - 2\right)}^{4} + \frac{720 \cdot 2}{120} {\left(x - 2\right)}^{5} + \frac{720}{720} {\left(x - 2\right)}^{6}$

Coefficients then are:

$\textcolor{b l u e}{64 , 192 , 240 , 160 , 60 , 12 , 1}$

The full expansion is:

$\textcolor{b l u e}{64 + 192 \left(x - 2\right) + 240 {\left(x - 2\right)}^{2} + 160 {\left(x - 2\right)}^{3} + 60 {\left(x - 2\right)}^{4} + 12 {\left(x - 2\right)}^{5} + {\left(x - 2\right)}^{6}}$