MathAdmin: Created page with "(a) Consider the function $f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.$ Find the first three terms of its Bin..."

2017-03-12T16:54:41Z

Created page 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 Bin..."

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<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 Binomial Series.

<span class="exam">(b) Find its radius of convergence.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The Taylor polynomial of   <math style="vertical-align: -5px">f(x)</math>   at   <math style="vertical-align: -1px">a</math>   is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math> where <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|-
| '''2.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by finding the coefficients of the Maclaurin series for  <math>f(x)=\frac{1}{(1-\frac{1}{2}x)^2}.</math>
|-
|We make a table to find the coefficients of the Maclaurin series.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(0) </math></td>
<td align = "center"><math> \frac{f^{(n)}(0)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^2} </math></td>
<td align = "center"><math> 1</math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^3}</math></td>
<td align = "center"><math> 1 </math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> \frac{\frac{3}{2}}{(1-\frac{1}{2}x)^4} </math></td>
<td align = "center"><math> \frac{3}{2} </math></td>
<td align = "center"><math> \frac{3}{4} </math></td>
</tr>
</table>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the first three terms of the Binomial Series is
|-
|        <math>1+x+\frac{3}{4}x^2.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By taking the derivative of the known series
|-
|   <math>\frac{1}{1-x}\,=\,1+x+x^2+\cdots,</math>
|-
|we find that the Maclaurin series of  <math>\frac{1}{(1-x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)x^n.</math>
|-
|Letting <math style="vertical-align: -5px"> x/2 </math>   play the role of <math style="vertical-align: -4px">x,</math> the Maclaurin series of  <math>\frac{1}{(1-\frac{1}{2}x)^2}</math>  is
|-
|       <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:  
|-
|Now, we use the Ratio Test to determine the radius of convergence of this power series.
|-
|We have
|-
|        <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  <math style="vertical-align: -14px">\frac{|x|}{2}<1.</math>  So,  <math style="vertical-align: -6px">|x|<2.</math>
|-
|Hence, the radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>1+x+\frac{3}{4}x^2</math>
|-
|   '''(b)'''    The radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

009C Sample Final 2, Problem 7 - Revision history

MathAdmin: Created page with "(a) Consider the function f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}. Find the first three terms of its Bin..."

MathAdmin: Created page with "(a) Consider the function $f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.$ Find the first three terms of its Bin..."