# Find the first four terms of the Taylor Series: f(x)=xe^x given a=0?

## Why is Slader using 1st-4th differentials, to find the first four terms?

x + x^2 + x^3/(2!) + x^4/(3!)

#### Explanation:

So...

e^x = 1+x+x^2/(2!) + x^3/(3!) ... (This should be memorized)

xe^x = x(1+x+x^2/(2!) + x^3/(3!)) ...

xe^x = x + x^2 + x^3/(2!) + x^4/(3!) ...
Solved!

May 7, 2018

First four non-zero terms: $x , {x}^{2} , {x}^{3} / 2 , {x}^{4} / 6$. The $0 t h$ term is $0.$

#### Explanation:

The $n t h$ term of the Taylor series of $f \left(x\right)$ centered at $a$ is given by

f^((n))(a)(x-a)^n/(n!)

For our case, $a = 0 ,$ so the $n t h$ term is given by

f^((n))(0)x^n/(n!)

So, to find the first $4$ terms using this formula, we'll need the function itself evaluated at $a = 0$, as well as the first three derivatives (as the $0 t h$ derivative is the function itself).

$f \left(0\right) = 0 \left({e}^{0}\right) = 0$

$f ' \left(x\right) = {e}^{x} + x {e}^{x} , f ' \left(0\right) = {e}^{0} = 1$

$f ' ' \left(x\right) = 2 {e}^{x} + x {e}^{x} , f ' ' \left(0\right) = 2 {e}^{0} = 2$

$f ' ' ' \left(x\right) = 3 {e}^{x} + x {e}^{x} , f ' ' ' \left(0\right) = 3 {e}^{0} = 3$

$0 t h$ term: 0(x^0)/(0!)=0, 0! =1 is a convention we'll adopt here.

$1 s t$ term: x/(1!)=x

$2 n d$ term: 2x^2/(2!)=x^2

$3 r d$ term 3x^3/(3!)=x^3/2:

As the $0 t h$ term is $0 ,$ and we're generally interested in non-zero terms , we can take the fourth derivative and find the next term (this is, I presume, why the first four derivatives were taken, as you need four derivatives to get four non-zero terms with this function):

${f}^{\left(4\right)} \left(x\right) = 4 {e}^{x} + x {e}^{x} , {f}^{\left(4\right)} \left(0\right) = 4 {e}^{0} = 4$

$4 t h$ (non-zero) term: 4x^4/(4!)=x^4/6

Alternatively, we can use known Taylor series, which is easier with this function: (this method requires no derivatives)

e^x=sum_(n=0)^oox^n/(n!)

xe^x=xsum_(n=0)^oox^n/(n!)

xe^x=sum_(n=0)^oox^(n+1)/(n!)

Evaluate from $n = 0$ to $n = 3$:

sum_(n=0)^3x^(n+1)/(n!)=x+x^2+x^3/2+x^4/6