Question

How do you find the Maclaurin series for `f(x)=(9x^2)e^(−7x)`?

Answer 1

`f(x) = 9x^2 - 63x^3 + (441x^4)/(2)-(1029x^5)/(2) + ...`

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 replacing `x` with `-7x` we have:

`e^(-7x) = 1 + (-7x) + (-7x)^2/(2!)+(-7x)^3/(3!) + (-7x)^4/(4!) + ...`
`" " = 1 - 7x + (7^2x^2)/(2!)-(7^3x^3)/(3!) + (7^4x^4)/(4!) + ...`
`" " = 1 - 7x + (49x^2)/(2)-(343x^3)/(6) + ...`
`" " = sum_(n=0)^oo (-7x)^n/(n!)`

So the the series for `f(x)` is

`f(x) = 9x^2 \ e^(-7x)`
`" " = 9x^2 {1 - 7x + (49x^2)/(2)-(343x^3)/(6) + ... }`
`" " = 9x^2 - 63x^3 + (441x^4)/(2)-(3087x^5)/(6) + ...`
`" " = 9x^2 - 63x^3 + (441x^4)/(2)-(1029x^5)/(2) + ...`
`" " = sum_(n=0)^oo (9x^2)(-7x)^n/(n!)`

The initial series converges `AA x in RR` and so therefore the modified series does also.