Difference between revisions of "009C Sample Final 3, Problem 5"

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Line 28: Line 28:
 
|We have
 
|We have
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(\frac{-1}{3}\bigg)e^{-\frac{1}{3}x},</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f''(x)=\bigg(\frac{-1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f''(x)=\bigg(-\frac{1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math>
 
|-
 
|-
 
|and
 
|and
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(3)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<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.  
 
|If we compare these three equations, we notice a pattern.  
Line 40: Line 40:
 
|Thus,
 
|Thus,
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>  
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>  
 
|}
 
|}
  
Line 48: Line 48:
 
|Since
 
|Since
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(\frac{-1}{3}\bigg)e^{-\frac{1}{3}x},</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
 
|-
 
|-
 
|we have
 
|we have
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}.</math>  
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}.</math>  
 
|}
 
|}
  
Line 62: Line 62:
 
|Since
 
|Since
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
 
|-
 
|-
 
|we have
 
|we have
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(3)=\bigg(\frac{-1}{3}\bigg)^ne^{-1}.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(3)=\bigg(-\frac{1}{3}\bigg)^ne^{-1}.</math>
 
|-
 
|-
 
|Therefore, the coefficients of the Taylor series are
 
|Therefore, the coefficients of the Taylor series are
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>c_n=\frac{\bigg(\frac{-1}{3}\bigg)^ne^{-1}}{n!}.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>c_n=\frac{\bigg(-\frac{1}{3}\bigg)^ne^{-1}}{n!}.</math>
 
|}
 
|}
  
Line 78: Line 78:
 
|Therefore, the Taylor series for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">x_0=3</math>&nbsp; is
 
|Therefore, the Taylor series for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">x_0=3</math>&nbsp; is
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
 
|}
 
|}
  
Line 85: Line 85:
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}</math>
+
|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}</math>
 
|-
 
|-
−
|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>  
+
|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>  
 
|}
 
|}
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:56, 12 March 2017

Consider the function

(a) Find a formula for the  th derivative    of    and then find  

(b) Find the Taylor series for    at    i.e. write    in the form

Foundations:  
The Taylor polynomial of     at     is

        where


Solution:

(a)

Step 1:  
We have
       
       
and
       
If we compare these three equations, we notice a pattern.
Thus,
       
Step 2:  
Since
       
we have
       

(b)

Step 1:  
Since
       
we have
       
Therefore, the coefficients of the Taylor series are
       
Step 2:  
Therefore, the Taylor series for    at    is
       


Final Answer:  
    (a)   
    (b)   

Return to Sample Exam

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