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

Question

Paisley Johnson · Accepted Answer

Write out a power series that when multiplied by #1+x^2# gives #x#.

Find #sum_(n=0)^oo (-1)^n x^(2n+1)# works and has radius of convergence #1#.

Consider #sum_(n=0)^oo (-1)^n x^(2n+1) = x - x^3 + x^5 - x^7 +...# #(1+x^2)sum_(n=0)^oo (-1)^n x^(2n+1)# #=sum_(n=0)^oo (-1)^n x^(2n+1) + x^2 sum_(n=0)^oo (-1)^n x^(2n+1)# #=sum_(n=0)^oo (-1)^n x^(2n+1) - sum_(n=1)^oo (-1)^n x^(2n+1)# #=(-1)^0x^1=x# So: #sum_(n=0)^oo (-1)^n x^(2n+1) = x / (1+x^2) = f(x)# ...if the sums converge. The sum #sum_(n=0)^oo (-1)^n x^(2n+1)# is a geometric series with common ratio #-x^2#. To converge, the absolute value of the common ratio must be less than #1#. That is #abs(-x^2) < 1#, so #abs(x) < 1# That is: the radius of convergence is #1#.

Cameron Alexander · Answer

undefined To find a power series representation for $ f(x) = \frac{x}{1+x^2} $, we can use the geometric series formula:

$$ \frac{1}{1 - u} = 1 + u + u^2 + u^3 + \dots $$

First, rewrite $ f(x) $ as:

$$ f(x) = x \cdot \frac{1}{1+x^2} $$

We can represent $ \frac{1}{1+x^2} $ as a geometric series by letting $ u = -x^2 $. Thus:

$$ \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \dots $$

Multiplying by $ x $:

$$ f(x) = x(1 - x^2 + x^4 - x^6 + \dots) $$

$$ f(x) = x - x^3 + x^5 - x^7 + \dots $$

This is the power series representation for $ f(x) = \frac{x}{1+x^2} $.

The radius of convergence $ R $ of a power series can be found using the ratio test formula:

$$ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| $$

In this case, the terms of the series alternate in sign and decrease in magnitude, suggesting that the series converges for all $ x $. Thus, the radius of convergence $ R $ is $ \infty $.