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2017-03-12T16:55:15Z

Created page with "<span class="exam">Find <math>n</math> such that the Maclaurin polynomial of degree <math>n</math> of <math style="vertical-align: -5px">f(x)=\co..."

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<span class="exam">Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''Taylor's Theorem'''
|-
|        Let  <math style="vertical-align: -5px">f</math>  be a function whose  <math style="vertical-align: -4px">(n+1)^{\mathrm{th}}</math> derivative exists on an interval  <math style="vertical-align: 0px">I</math>,  and let  <math style="vertical-align: 0px">c</math>  be in  <math style="vertical-align: 0px">I.</math>
|-
|        Then, for each  <math style="vertical-align: 0px">x</math>  in  <math style="vertical-align: -4px">I,</math>  there exists  <math style="vertical-align: -3px">z_x</math>  between  <math style="vertical-align: 0px">x</math>  and  <math style="vertical-align: 0px">c</math>  such that
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|        <math>f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+\cdots+\frac{f^{(n)}(c)}{n!}(x-c)^n+R_n(x),</math>
|-
|        where  <math style="vertical-align: -18px">R_n(x)=\frac{f^{n+1}(z_x)}{(n+1)!}(x-c)^{n+1}.</math>
|-
|        Also,  <math style="vertical-align: -17px">|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using Taylor's Theorem, we have that the error in approximating  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  with
|-
|the Maclaurin polynomial of degree  <math style="vertical-align: 0px">n</math>  is  <math style="vertical-align: -16px">R_n\bigg(\frac{\pi}{3}\bigg)</math>  where
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|        <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}\bigg|\bigg(\frac{\pi}{3}-0\bigg)^{n+1}\bigg|.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We note that
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|       <math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1</math>  or  <math style="vertical-align: -5px">|f^{n+1}(z)|=|\sin(z)|\le 1.</math>
|-
|Therefore, we have
|-
|        <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}.</math>
|-
|Now, we have the following table.
|-
|<table border="1" cellspacing="0" cellpadding="6" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> \approx\frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> 0.548311 </math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> 0.191396</math></td>
</tr>
<tr>
<td align = "center"><math>3</math></td>
<td align = "center"><math> 0.050107 </math></td>
</tr>
<tr>
<td align = "center"><math>4</math></td>
<td align = "center"><math> 0.01049 </math></td>
</tr>
<tr>
<td align = "center"><math>5</math></td>
<td align = "center"><math> 0.00183 </math></td>
</tr>
<tr>
<td align = "center"><math>6</math></td>
<td align = "center"><math> 0.000274 </math></td>
</tr>
<tr>
<td align = "center"><math>7</math></td>
<td align = "center"><math> 0.0000358 </math></td>
</tr>
</table>
|-
|So,  <math style="vertical-align: 0px">n=7</math>  is the smallest value of  <math style="vertical-align: 0px">n</math>  where the error is less than or equal to 0.0001.
|-
|Therefore, for  <math style="vertical-align: 0px">n=7</math>  the Maclaurin polynomial approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>n=7</math>
|}
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

009C Sample Final 2, Problem 8 - Revision history

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