Difference between revisions of "009C Sample Final 2, Problem 7"

From Grad Wiki
Jump to navigation Jump to search
(Replaced content with "<span class="exam">(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its...")
 
(3 intermediate revisions by 2 users not shown)
Line 3: Line 3:
 
<span class="exam">(b) Find its radius of convergence.
 
<span class="exam">(b) Find its radius of convergence.
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
<hr>
!Foundations: &nbsp;
+
[[009C Sample Final 2, Problem 7 Solution|'''<u>Solution</u>''']]
|-
 
|
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|}
 
  
  
'''Solution:'''
+
[[009C Sample Final 2, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
'''(a)'''
 
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|
 
|-
 
|
 
|}
 
 
'''(b)'''
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|The Maclaurin series of &nbsp;<math>\frac{1}{(1-x)^2}</math>&nbsp; is
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty (n+1)x^n.</math>
 
|-
 
|So, the Maclaurin series of &nbsp;<math>\frac{1}{(1-\frac{1}{2}x)^2}</math>&nbsp; is
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty (n+1)\bigg(\frac{1}{2}x\bigg)^n=\sum_{n=0}^\infty \frac{(n+1)x^n}{2^n}.</math>
 
|-
 
|
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|Now, we use the Ratio Test to determine the radius of convergence of this power series.
 
|-
 
|We have
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+2)x^{n+1}}{2^{n+1}} \frac{2^n}{(n+1)x^n}\bigg|}\\
 
&&\\
 
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{|x|}{2} \frac{n+2}{n+1}}\\
 
&&\\
 
& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{n+2}{n+1}}\\
 
&&\\
 
& = & \displaystyle{\frac{|x|}{2}.}
 
\end{array}</math>
 
|-
 
|Now, the Ratio Test says this series converges if &nbsp;<math style="vertical-align: -14px">\frac{|x|}{2}<1.</math>&nbsp; So, &nbsp;<math style="vertical-align: -6px">|x|<2.</math>
 
|-
 
|Hence, the radius of convergence is &nbsp;<math style="vertical-align: 0px">R=2.</math>
 
|}
 
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
|-
 
|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;
 
|-
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: 0px">R=2.</math>
 
|}
 
 
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 18:33, 2 December 2017

(a) Consider the function    Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.


Solution


Detailed Solution


Return to Sample Exam