How do you find the Taylor series of #f(x)=e^(x/2)# centered around #a=-1# ?

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Answer 1 · 2018-02-19T14:47:16

#e^(x/2) = 1/sqrte sum_(n=0)^oo (x+1)^n/(2^n(n!))#

The general formula of the Taylor series is:

#f(x) = sum_(n=0)^oo f^((n))(a)/(n!) (x-a)^n#

Now:

#d/dx (e^(x/2)) = 1/2 e^(x/2)#

and by induction we can easily prove that:

#(d^n)/dx^n (e^(x/2)) = 1/2^n e^(x/2)#

so that:

#f^((n)) (-1) = 1/2^n e^(-1/2) = 1/(2^nsqrte)#

Then:

#e^(x/2) = 1/sqrte sum_(n=0)^oo (x+1)^n/(2^n(n!))#

We can also proceed starting from the MacLaurin series of #e^t#:

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

substitute #t=(x+1)/2# and note that:

#e^((x+1)/2) = e^(x/2+1/2) = e^(1/2)e^(x/2)#

so:

#e^(1/2)e^(x/2) = sum_(n=0)^oo ((x+1)/2)^n/(n!)#

that is:

#e^(x/2) = 1/sqrte sum_(n=0)^oo (x+1)^n/(2^n(n!))#