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#