Difference between revisions of "009C Sample Midterm 3, Problem 5"
Jump to navigation
Jump to search
m |
m |
||
Line 14: | Line 14: | ||
::<math>a_n=f(x-c)\cdot g(n)</math> | ::<math>a_n=f(x-c)\cdot g(n)</math> | ||
|- | |- | ||
− | |where <math style="vertical-align: -20%">f</math> and <math style="vertical-align: -20%">g</math> are functions of <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">n</math> respectively, and <math style="vertical-align: 0%">c</math> is a constant (frequently zero). We need to find a bound on <math style="vertical-align: -22%">|x-c|</math> such that whenever <math style="vertical-align: -22%">|x-c|< | + | |where <math style="vertical-align: -20%">f</math> and <math style="vertical-align: -20%">g</math> are functions of <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">n</math> respectively, and <math style="vertical-align: 0%">c</math> is a constant (frequently zero). We need to find a bound (radius) on <math style="vertical-align: -22%">|x-c|</math> such that whenever <math style="vertical-align: -22%">|x-c|<r</math>, the ratio test |
|- | |- | ||
| | | | ||
::<math>\left|\frac{a_{n+1}}{a_n}\right|</math> | ::<math>\left|\frac{a_{n+1}}{a_n}\right|</math> | ||
|- | |- | ||
− | |is satisfied. When we do, the interval will be <math style="vertical-align: -20%">(c- | + | |is satisfied. When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>. However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence. Most often, one will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence can be either open, half-open or closed. |
|} | |} | ||
Revision as of 14:58, 27 April 2015
Find the radius of convergence and the interval of convergence of the series.
- (a) (6 points)
- (b) (6 points)
Foundations: |
---|
When we are asked to find the radius of convergence, we are given a series where |
|
where and are functions of and respectively, and is a constant (frequently zero). We need to find a bound (radius) on such that whenever , the ratio test |
|
is satisfied. When we do, the interval will be . However, the boundary values for , and must be tested individually for convergence. Most often, one will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence can be either open, half-open or closed. |
Solution:
(a): |
---|
(b): |
---|
Final Answer: |
---|