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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. | + | <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 \bigg(\frac{\pi}{3}\bigg)</math> within 0.0001 of the actual value. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 2, Problem 8 Solution|'''<u>Solution</u>''']] |
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| − | |'''Taylor's Theorem''' | |
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| − | | Let <math>f</math> be a function whose <math>n+1</math>th derivative exists on an interval <math>I</math> and let <math>c</math> be in <math>I.</math>
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| − | | Then, for each <math>x</math> in <math>I,</math> there exists <math>z_x</math> between <math>x</math> and <math>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>
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| − | | where <math>R_n(x)=\frac{f^{n+1}(z_x)}{(n+1)!}(x-c)^{n+1}.</math>
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| − | | Also, <math>|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
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| − | |}
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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Using Taylor's Theorem, we have that the error in approximating <math>\cos \frac{\pi}{3}</math> with
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| − | |the Maclaurin polynomial of degree <math>n</math> is <math>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>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |We note that <math>|f^{n+1}(z)|=|\cos(z)|\le 1</math> or <math>|f^{n+1}(z)|=|\cos(z)|\le 1.</math>
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| − | |Therefore, we have
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| − | | <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}.</math>
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| − | |Now, we have the following table.
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| − | |<table border="1" cellspacing="0" cellpadding="6" align = "center">
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| − | <tr>
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| − | <td align = "center"><math> n</math></td>
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| − | <td align = "center"><math> \approx\frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}</math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>1</math></td>
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| − | <td align = "center"><math> 0.548311 </math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>2</math></td>
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| − | <td align = "center"><math> 0.191396</math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>3</math></td>
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| − | <td align = "center"><math> 0.050107 </math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>4</math></td>
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| − | <td align = "center"><math> 0.01049 </math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>5</math></td>
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| − | <td align = "center"><math> 0.00183 </math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>6</math></td>
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| − | <td align = "center"><math> 0.000274 </math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>7</math></td>
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| − | <td align = "center"><math> 0.0000358 </math></td>
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| − | </tr>
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| − | </table>
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| − | |So, <math>n=7</math> is the smallest value of <math>n</math> where the error is less than or equal to 0.0001.
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| − | |Therefore, for <math>n=7</math> the Maclaurin polynomial approximates <math>\cos \frac{\pi}{3}</math> within 0.0001 of the actual value.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>n=7.</math>
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| − | |}
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| | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
