How do you do the taylor series for expanding #1/(2+x^2)#, about a = 0 up to order n = 2?

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Answer 1 · 2015-05-24T22:15:29

Firstly we would find the Taylor series for the function centered at a=0, thus moreso finding is Maclaurin series.

we can make use of the common Maclaurin series of:

#1/(1-x) = sum_(n=0)^oo x^n#

Now lets try and write our function to look similar to the common Maclaurin Series:

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

now we can just substitute #(-x^2/2)# in place of #x# in our common Maclaurin series, thus we get:

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

which can simplify a bit:

#= sum_(n=0)^oo (-1)^n (x^2/2)^n#

Now to get the expansion up to order n=2, we just follow the sum and get:

#(-1)^0 (x^2/2)^0 + (-1)^1 (x^2/2)^1 + (-1)^2 (x^2/2)^2#

which can simplify to:

#1 - x^2/2 + x^4/4#

note that this series is only convergent at #abs(x^2/2) < 1#

thus:

#-1< x^2/2 < 1#

#-2< x^2 < 2#

#-sqrt(2) < x < sqrt(2)#