Question

What is the taylor series of `xe^x`?

Answer 1

`xe^x = x + x^2 + x^3/(2!)+x^4/(3!) + x^5/(4!) + ...`
`\ \ \ \ \ = sum_(n=0)^oo x^(n+1)/(n!)`

Explanation:

We can start with the well known Maclaurin series for `e^x`

`e^x = 1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ...`
`\ \ \ = sum_(n=0)^oo x^n/(n!)`

So then multiplying by `x` we have:

`xe^x = x{1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ... }`
`\ \ \ \ \ = x + x^2 + x^3/(2!)+x^4/(3!) + x^5/(4!) + ...`
`\ \ \ \ \ = sum_(n=0)^oo x*x^n/(n!)`
`\ \ \ \ \ = sum_(n=0)^oo x^(n+1)/(n!)`