009C Sample Final 3, Problem 5 - Revision history

MathAdmin: Created page with " Consider the function :: $f(x)=e^{-\frac{1}{3}x}$ (a) Find a formula for the $n$ th derivative
2017-03-19T18:48:40Z

Created page with "<span class="exam"> Consider the function ::<math>f(x)=e^{-\frac{1}{3}x}</math> <span class="exam">(a) Find a formula for the <math>n</math>th derivative <math s..."

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<span class="exam"> Consider the function

::<math>f(x)=e^{-\frac{1}{3}x}</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>

<span class="exam">(b) Find the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -5px">x_0=3,</math>  i.e. write  <math style="vertical-align: -5px">f(x)</math>  in the form

::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The Taylor polynomial of <math style="vertical-align: -5px">f(x)</math>  at <math style="vertical-align:0px">a</math> is
|-
|
       <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>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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>
|-
|and
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|       <math>f^{(3)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
|-
|If we compare these three equations, we notice a pattern.
|-
|Thus,
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|       <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|       <math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
|-
|we have
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|       <math>f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since
|-
|       <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
|-
|we have
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|       <math>f^{(n)}(3)=\bigg(-\frac{1}{3}\bigg)^ne^{-1}.</math>
|-
|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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Therefore, the Taylor series for  <math style="vertical-align: -5px">f(x)</math>  at  <math style="vertical-align: -3px">x_0=3</math>  is
|-
|       <math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(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>
|-
|    '''(b)'''    <math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>
|}
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