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| | <span class="exam"> Find the Taylor Polynomials of order 0, 1, 2, 3 generated by <math style="vertical-align: -5px">f(x)=\cos(x)</math> at <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math> | | <span class="exam"> Find the Taylor Polynomials of order 0, 1, 2, 3 generated by <math style="vertical-align: -5px">f(x)=\cos(x)</math> at <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009C Sample Final 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |The Taylor polynomial of <math style="vertical-align: -5px">f(x)</math> at <math style="vertical-align: -1px">a</math> is
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| − | <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>
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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 5 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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| − | |Let <math style="vertical-align: -14px">a=\frac{\pi}{4}.</math>
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| − | |First, we make a table to find the coefficients of the Taylor polynomial.
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| − | <table border="1" cellspacing="0" cellpadding="11" 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> f^{(n)}(x) </math></td>
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| − | <td align = "center"><math> f^{(n)}(a) </math></td>
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| − | <td align = "center"><math> \frac{f^{(n)}(a)}{n!} </math></td>
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| − | </tr>
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| − | <tr>
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| − | <td align = "center"><math>0</math></td>
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| − | <td align = "center"><math> \cos x </math></td>
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| − | <td align = "center"><math> \frac{\sqrt{2}}{2}</math></td>
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| − | <td align = "center"><math> \frac{\sqrt{2}}{2}</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> -\sin x</math></td>
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| − | <td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
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| − | <td align = "center"><math> -\frac{\sqrt{2}}{2} </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> -\cos x </math></td>
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| − | <td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
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| − | <td align = "center"><math> -\frac{\sqrt{2}}{4} </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> \sin x </math></td>
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| − | <td align = "center"><math> \frac{\sqrt{2}}{2} </math></td>
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| − | <td align = "center"><math> \frac{\sqrt{2}}{12}</math></td>
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| − | </tr>
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| − | </table>
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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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| − | |Let <math style="vertical-align: -4px">T_n</math> be the Taylor polynomial of order <math>n.</math>
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| − | |Since <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!},</math> we have
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| − | | <math>T_0=\frac{\sqrt{2}}{2}</math>
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| − | | <math>T_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)</math>
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| − | | <math>T_2=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2</math>
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| − | | <math>T_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2+\frac{\sqrt{2}}{12}\bigg(x-\frac{\pi}{4}\bigg)^3.</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | Let <math style="vertical-align: -4px">T_n</math> be the Taylor polynomial of order <math>n.</math>
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| − | | <math>T_0=\frac{\sqrt{2}}{2}</math>
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| − | | <math>T_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)</math>
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| − | | <math>T_2=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2</math>
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| − | | <math>T_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2+\frac{\sqrt{2}}{12}\bigg(x-\frac{\pi}{4}\bigg)^3</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>''']] |
