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#