Difference between revisions of "009C Sample Final 2, Problem 8"

From Grad Wiki
Jump to navigation Jump to search
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
<span class="exam">Find &nbsp;<math>n</math>&nbsp; such that the Maclaurin polynomial of degree &nbsp;<math>n</math>&nbsp; of &nbsp;<math style="vertical-align: -5px">f(x)=\cos(x)</math>&nbsp; approximates &nbsp;<math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>&nbsp; within 0.0001 of the actual value.
+
<span class="exam">Find &nbsp;<math>n</math>&nbsp; such that the Maclaurin polynomial of degree &nbsp;<math>n</math>&nbsp; of &nbsp;<math style="vertical-align: -5px">f(x)=\cos(x)</math>&nbsp; approximates &nbsp;<math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>&nbsp; within 0.0001 of the actual value.
  
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
<hr>
−
!Foundations: &nbsp;
+
[[009C Sample Final 2, Problem 8 Solution|'''<u>Solution</u>''']]
−
|-
 
−
|'''Taylor's Theorem'''
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; be a function whose &nbsp;<math style="vertical-align: -2px">n+1</math>th derivative exists on an interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; and let &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; be in &nbsp;<math style="vertical-align: 0px">I.</math>
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; Then, for each &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in &nbsp;<math style="vertical-align: -4px">I,</math>&nbsp; there exists &nbsp;<math style="vertical-align: -3px">z_x</math>&nbsp; between &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; such that
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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>
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; where &nbsp;<math style="vertical-align: -18px">R_n(x)=\frac{f^{n+1}(z_x)}{(n+1)!}(x-c)^{n+1}.</math>
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; Also, &nbsp;<math style="vertical-align: -17px">|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
 
−
|}
 
  
  
−
'''Solution:'''
+
[[009C Sample Final 2, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
|Using Taylor's Theorem, we have that the error in approximating &nbsp;<math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>&nbsp; with
 
−
|-
 
−
|the Maclaurin polynomial of degree &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; is &nbsp;<math style="vertical-align: -16px">R_n\bigg(\frac{\pi}{3}\bigg)</math>&nbsp; where
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
 
−
|-
 
−
|We note that
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1</math>&nbsp; or &nbsp;<math style="vertical-align: -5px">|f^{n+1}(z)|=|\sin(z)|\le 1.</math>
 
−
|-
 
−
|Therefore, we have
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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, &nbsp;<math style="vertical-align: 0px">n=7</math>&nbsp; is the smallest value of &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; where the error is less than or equal to 0.0001.
 
−
|-
 
−
|Therefore, for &nbsp;<math style="vertical-align: 0px">n=7</math>&nbsp; the Maclaurin polynomial approximates &nbsp;<math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>&nbsp; within 0.0001 of the actual value.
 
−
|}
 
−
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Final Answer: &nbsp;
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>n=7</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 15:57, 3 December 2017

Find    such that the Maclaurin polynomial of degree    of    approximates    within 0.0001 of the actual value.

Solution


Detailed Solution


Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009C_Sample_Final_2,_Problem_8&oldid=4646"