Section10.6-Power Series

# X 3 1 x 3 2 x 3 3 x 3 n 2 3 n 0

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ! x 3 n 2 3 n 0 Find the radius and interval of convergence: an 1 n 1 ! x 3 n an n! x 3 n 1 n 1 x 3 lim n an 1 an unless | x – 3 | = 0 Interval of convergence: x = 3 Radius of convergence, R = 0 Example 3 x 2 n 1 n 0 n Find the interval and radius of convergence: an 1 an 3 x2 n n2 3 x2 n 1 3 x2 x2 lim n 1 n n 1 3 x2 n 1 n2 n 1 n2 an 1 x2 an n Example ‐ continued lim n an 1 x2 an Convergent for | x – 2 | &lt; 1 Divergent for | x – 2 | &gt; 1 What about | x – 2 | = 1 ? Ratio Test is inconclusive. Use another test. Example – continued 3 x 2 n 1 n 0 n x 2 1 3 1 n 1 n 0 n Convergent Checking | x – 2 | = 1 x 2 1 3 n 1 n0 Divergent Series is convergent for ‐1 ≤ x – 2 &lt; 1 ; 1 ≤ x &lt; 3 Interval of convergence: [ 1, 3 ) Radius of convergence, R = 1 Theorem bn x c n For a given power series, , there are n 0 only three possibilities: i) The...
View Full Document

## This note was uploaded on 02/24/2014 for the course APMA 1110 taught by Professor Morris during the Fall '11 term at UVA.

Ask a homework question - tutors are online