Please explain taylor-maclaurin factorial when the problem is to obtain the Taylor-Maclaurin series for e4x as far as the term in x5.

Obtain the Taylor-Maclaurin series for e4x as far as the term in x5.

**Ok, I know that:**

e4x = 1 + (4x) + ((4x)^2)/2! + ((4x)^3)/3! + ((4x)^4)/4! + ((4x)^5)/5!

**What I don't understand** is the next part, apparently this is the answer:

= 1 + 4x + 8x^2 + (32x^3)/3 + (32x^4)/3 + (128x^5)/15

**What I mainly don't understand** is how the factorials are working here. I would really apprieciate if someone could explain.

### 1 Answer | Add Yours

`(4x)^2/(2!)=(16x^2)/(2*1)=8x^2`

`(4x)^3/(3!)=[(4*4*4)*x^3]/(3*2*1)=[2*4*4*x^3]/3=(32x^3)/3`

What you need to do in every one of these terms is expand the exponent as well as the factorial, simplify, then multiply.

Next term: `[(4x)^4]/[4!]=[4*4*4*4*x^4]/[4*3*2*1]=[1*2*4*4*x^4]/3=[32x^4]/3`

The final term: `(4x)^5/(5!)=[4*4*4*4*4*x^5]/[5*4*3*2*1]=`

`[1*2*4*4*4*x^5]/[5*3]=[128x^5]/15`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes