How do you find a power series representation for #f(x)= 1/(3-x)# and what is the radius of convergence?

Question

Isaiah Allen · Accepted Answer

The radius of convergence is -3< x <3

#f(x)= 1/(3-x) = 1/(3(1-x/3))# Now compare this epression with the sum of an infinite geometric series 1 +x+x^2 +x^3 +...... . The sum is #1/(1-x)# with -1< x <1. Now substitute x with #x/3#, then #1/(1-x/3) = 1 +x/3 +(x/3)^2 +(x/3)^3 +........# where #-1< x/3 <1 -> -3< x < 3# Hence #f(x)= 1/(3-x)= 1/3 (1 +(x/3) + (x/3)^2 +........)# The radius of convergence is therefore -3< x <3

Christopher Agresta · Answer

undefined To find a power series representation for $ f(x) = \frac{1}{3-x} $, we can use the geometric series formula $ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n $ for $ |x| < 1 $. Let's rewrite $ f(x) $ in a form suitable for the geometric series:

$$ f(x) = \frac{1}{3-x} = \frac{1}{3(1-\frac{x}{3})} $$

Now, replace $ \frac{x}{3} $ with $ u $, which gives $ x = 3u $:

$$ f(x) = \frac{1}{3(1-u)} = \frac{1}{3} \cdot \frac{1}{1-u} $$

This is in the form $ \frac{1}{1-x} $, so we can apply the geometric series formula:

$$ \frac{1}{3} \cdot \sum_{n=0}^{\infty} u^n = \sum_{n=0}^{\infty} \frac{u^n}{3} $$

Substitute $ u = \frac{x}{3} $ back in:

$$ \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}} $$

The radius of convergence can be found using the ratio test. Let's apply the ratio test to the series:

$$ a_n = \frac{1}{3^{n+1}} $$

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{1}{3^{n+2}} \cdot \frac{3^n}{1} \right| = \frac{1}{3} $$

Since the limit is less than 1, the series converges for all $ x $ within a radius of $ R = 3 $. Thus, the radius of convergence is 3.