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

Question

Isabella Ackerman · Accepted Answer

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

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#.

Axel Alvarez · Answer

undefined To find a power series representation for $ f(x) = \frac{1}{1 + 4x^2} $, we can use the geometric series expansion $ \frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n $ for $ |u| < 1 $.

Substitute $ u = -4x^2 $ into the geometric series expansion, yielding $ \frac{1}{1 + 4x^2} = \sum_{n=0}^{\infty} (-4x^2)^n $.

This simplifies to $ \sum_{n=0}^{\infty} (-1)^n \cdot 4^n \cdot x^{2n} $.

The radius of convergence $ R $ for this power series can be determined using the ratio test, which states that $ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| $, where $ a_{n} $ is the coefficient of $ x^n $ in the power series.

For our series, $ a_{n} = (-1)^n \cdot 4^n $, so $ \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{(-1)^n \cdot 4^n}{(-1)^{n+1} \cdot 4^{n+1}} \right| = \lim_{n \to \infty} \left| \frac{1}{4} \right| = \frac{1}{4} $.

Thus, the radius of convergence $ R $ is $ \frac{1}{4} $.