Question

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

Answer 1

`e^(x/2) = sum_(n=0)^oo x^n/(2^n(n!))`

with radius of convergence `R=oo`

Explanation:

Start from the MacLaurin expansion of `e^t`.

Since:

`d^n/(dt^n) e^t = e^t`

`[d^n/(dt^n) e^t]_(t=0) = e^0 = 1`

we have that:

`e^t = sum_(n=0)^oo t^n/(n!)`

Using the ratio test:

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( t^(n+1)/((n+1)!))/(t^n/(n!))) = lim_(n->oo) abs t /(n+1) = 0`

so the series is convergent for any `t in RR`, that is it has radius of convergence `R=oo`

So, substituting `t=x/2`:

`e^(x/2) = sum_(n=0)^oo (x/2)^n/(n!) = sum_(n=0)^oo x^n/(2^n(n!))`

still with radius of convergence `R=oo`