Question

How do you find a power series converging to `f(x)=x^2e^x` and determine the radius of convergence?

Answer 1

`x^2+x^3/(1!)+x^4/(2!)+...+x^(n+2)/(n!)+..., x in (-oo, oo)`. See the graph, with local maximum `4/e^2=0.54`, nearly`, at`x = -1/2#.

Explanation:

`1+x+x^2/(2!)+...+x^n/(n!)+... to e^x, x in (-oo, oo)`

So converges (a polynomial)`xx`(the above series),

for `x in (-oo, oo)`. And so,

`x^2+x^3/(1!)+x^4/(2!)+...+x^(n+2)/(n!)+...`

converges to `x^2e^x, x in (-oo, oo)`.

graph{y(y-x^2e^x)(x+2+.001y)=0 [-10, 10, -5, 5]}