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?

2 Answers

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 0th term is 0.

Explanation:

The nth term of the Taylor series of f(x) centered at a is given by

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

For our case, a=0, so the nth 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 0th derivative is the function itself).

f(0)=0(e^0)=0

f'(x)=e^x+xe^x, f'(0)=e^0=1

f''(x)=2e^x+xe^x, f''(0)=2e^0=2

f'''(x)=3e^x+xe^x, f'''(0)=3e^0=3

0th term: 0(x^0)/(0!)=0, 0! =1 is a convention we'll adopt here.

1st term: x/(1!)=x

2nd term: 2x^2/(2!)=x^2

3rd term 3x^3/(3!)=x^3/2:

As the 0th 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^((4))(x)=4e^x+xe^x, f^((4))(0)=4e^0=4

4th (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