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| | <span class="exam"> Consider the function | | <span class="exam"> Consider the function |
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| − | ::<math>f(x)=e^{-\frac{1}{3}x}</math> | + | ::<math>f(x)=e^{-\frac{1}{3}x}.</math> |
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| | <span class="exam">(a) Find a formula for the <math>n</math>th derivative <math style="vertical-align: -5px">f^{(n)}(x)</math> of <math style="vertical-align: -5px">f</math> and then find <math style="vertical-align: -5px">f'(3).</math> | | <span class="exam">(a) Find a formula for the <math>n</math>th derivative <math style="vertical-align: -5px">f^{(n)}(x)</math> of <math style="vertical-align: -5px">f</math> and then find <math style="vertical-align: -5px">f'(3).</math> |
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| | ::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math> | | ::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 3, 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 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We have
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| − | | <math>f'(x)=\bigg(\frac{-1}{3}\bigg)e^{-\frac{1}{3}x},</math>
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| − | | <math>f''(x)=\bigg(\frac{-1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math>
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| − | |and
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| − | | <math>f^{(3)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
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| − | |If we compare these three equations, we notice a pattern.
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| − | |We have
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| − | | <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x}.</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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| − | |Since
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| − | | <math>f'(x)=\bigg(\frac{-1}{3}\bigg)e^{-\frac{1}{3}x},</math>
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| − | |we have
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| − | | <math>f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}.</math>
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Since
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| − | | <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
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| − | |we have
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| − | | <math>f^{(n)}(3)=\bigg(\frac{-1}{3}\bigg)^ne^{-1}.</math>
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| − | |Therefore, the coefficients of the Taylor series are
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| − | | <math>c_n=\frac{\bigg(\frac{-1}{3}\bigg)^ne^{-1}}{n!}.</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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| − | |Therefore, the Taylor series for <math style="vertical-align: -5px">f(x)</math> at <math style="vertical-align: -3px">x_0=3</math> is
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| − | | <math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
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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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| − | | '''(a)''' <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}</math>
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| − | | '''(b)''' <math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>
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| − | |}
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| | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
