How do you find a power series representation for #f(x)= 1/(1+4x^2)# and what is the radius of convergence?

1 Answer
George C.
Sep 28, 2015

#f(x) = sum_(n=0)^oo (-4x^2)^n# with radius of convergence #1/2#

Explanation:

Consider the power series:

#sum_(n=0)^oo (-4x^2)^n = 1 - 4x^2 + 16x^4 - 64x^6 +...#

Then:

#(1+4x^2)(sum_(n=0)^oo (-4x^2)^n)#

#=sum_(n=0)^oo (-4x^2)^n + 4x^2 sum_(n=0)^oo (-4x^2)^n#

#=sum_(n=0)^oo (-4x^2)^n - sum_(n=1)^oo (-4x^2)^n#

#=(-4x^2)^0 = 1#

provided the sum #sum_(n=0)^oo (-4x^2)^n# converges.

So

#sum_(n=0)^oo (-4x^2)^n = 1/(1+4x^2) = f(x)#

This is a geometric sequence, so will converge if the common ratio has absolute value #< 1#.

That is:

#abs(-4x^2) < 1#, so #x^2 < 1/4#, so #abs(x) < 1/2#

In general

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

which converges if #abs(a) < 1#.

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