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

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Answer 1 · 2017-03-15T13:26:45

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

with radius of convergence $R=oo$

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 $R=oo$

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