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

Question

Isabella Ackerman · Accepted Answer

Use the Maclaurin series for #1/(1-t)# and substitution to find:

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

with radius of convergence #1#.

The Maclaurin series for #1/(1-t)# is #sum_(n=0)^oo t^n# since #(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1# Substitute #t = x^2# to get: #1/(1-x^2) = sum_(n=0)^oo (x^2)^n = sum_(n=0)^oo x^(2n)# Multiply by #x# to get: #x/(1-x^2) = x sum_(n=0)^oo x^(2n) = sum_(n=0)^oo x^(2n+1)# This is a geometric series with common ratio #x^2#, so it will converge if #abs(x^2) < 1#, which is when #abs(x) < 1#. That is, the radius of convergence is #1#.

Luke Alvarez · Answer

undefined To find a power series representation for $ \frac{x}{1-x^2} $, you can use the method of partial fractions and geometric series.

First, rewrite $ \frac{x}{1-x^2} $ as $ \frac{x}{(1-x)(1+x)} $. Then, express $ \frac{x}{(1-x)(1+x)} $ as the sum of two fractions using partial fractions:

$$ \frac{x}{(1-x)(1+x)} = \frac{A}{1-x} + \frac{B}{1+x} $$

Find the values of $ A $ and $ B $. After finding them, you can express each fraction as a power series. The power series representation for $ \frac{1}{1-x} $ is $ \sum_{n=0}^{\infty} x^n $ (geometric series), and for $ \frac{1}{1+x} $ is $ \sum_{n=0}^{\infty} (-1)^n x^n $.

Thus, the power series representation for $ \frac{x}{1-x^2} $ will be:

$$ \frac{x}{1-x^2} = A \sum_{n=0}^{\infty} x^n + B \sum_{n=0}^{\infty} (-1)^n x^n $$

$$ = (A + B) \sum_{n=0}^{\infty} x^n + (A - B) \sum_{n=0}^{\infty} (-1)^{n+1} x^n $$

$$ = (A + B) \sum_{n=0}^{\infty} x^n + (A - B) \sum_{n=0}^{\infty} (-x)^n $$

$$ = (A + B) \sum_{n=0}^{\infty} x^n + (A - B) \sum_{n=0}^{\infty} (-1)^n x^n $$

$$ = (A + B + (A - B)) \sum_{n=0}^{\infty} x^n $$

$$ = (2A) \sum_{n=0}^{\infty} x^n $$

$$ = 2A \sum_{n=0}^{\infty} x^n $$

$$ = \sum_{n=0}^{\infty} 2A \cdot x^n $$

$$ = \sum_{n=0}^{\infty} C_n \cdot x^n $$

Where $ C_n = 2A $ for all $ n $.

The radius of convergence of the power series will be the same as the radius of convergence of the original function, which is the distance from the center of the series to the nearest singularity. In this case, the singularities are at $ x = -1 $ and $ x = 1 $, so the radius of convergence is $ R = 1 $.

Charles Arocha · Answer

undefined To find a power series representation for $ \frac{x}{1-x^2} $, we can use the method of partial fractions. First, we can write:

$$ \frac{x}{1-x^2} = \frac{A}{1+x} + \frac{B}{1-x} $$

Multiplying both sides by $ (1-x^2) $, we get:

$$ x = A(1-x) + B(1+x) $$

Expanding and simplifying, we get:

$$ x = (A + B) + (B - A)x $$

This implies that $ A + B = 0 $ and $ B - A = 1 $. Solving these equations simultaneously, we find that $ A = -\frac{1}{2} $ and $ B = \frac{1}{2} $.

Therefore, we can write:

$$ \frac{x}{1-x^2} = -\frac{1}{2} \cdot \frac{1}{1+x} + \frac{1}{2} \cdot \frac{1}{1-x} $$

Using the geometric series formula, $ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n $ for $ |x| < 1 $, we have:

$$ \frac{x}{1-x^2} = -\frac{1}{2} \sum_{n=0}^{\infty} (-x)^n + \frac{1}{2} \sum_{n=0}^{\infty} x^n $$

Simplifying further, we get:

$$ \frac{x}{1-x^2} = -\frac{1}{2} \sum_{n=0}^{\infty} (-1)^n x^n + \frac{1}{2} \sum_{n=0}^{\infty} x^n $$

$$ = -\frac{1}{2} \sum_{n=0}^{\infty} (-1)^n x^n + \frac{1}{2} \sum_{n=0}^{\infty} x^n $$

$$ = \sum_{n=0}^{\infty} \frac{(-1)^n - 1}{2} x^n $$

$$ = \sum_{n=0}^{\infty} (-1)^n \frac{1 - (-1)^n}{2} x^n $$

$$ = \sum_{n=0}^{\infty} (-1)^n \frac{1 + (-1)^{n+1}}{2} x^n $$

$$ = \sum_{n=0}^{\infty} (-1)^n \frac{1 + (-1)^n}{2} x^n $$

The radius of convergence for this series is 1, which is the distance from the center (x=0) to the nearest singularity (x=±1).