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

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|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>
 
|}
 
|}
  
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|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>
 
|}
 
|}
  
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!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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=3,}   i.e. write  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   in the form

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sum_{n=0}^\infty a_n(x-3)^n.}
Foundations:  
The Taylor polynomial of   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   at   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a}   is

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty}c_n(x-a)^n} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_n=\frac{f^{(n)}(a)}{n!}.}


Solution:

(a)

Step 1:  
We have
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},}
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f''(x)=\bigg(-\frac{1}{3}\bigg)^2 e^{-\frac{1}{3}x},}
and
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(3)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x}.}
If we compare these three equations, we notice a pattern.
Thus,
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.}
Step 2:  
Since
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},}
we have
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}.}

(b)

Step 1:  
Since
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x},}
we have
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(n)}(3)=\bigg(-\frac{1}{3}\bigg)^ne^{-1}.}
Therefore, the coefficients of the Taylor series are
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_n=\frac{\bigg(-\frac{1}{3}\bigg)^ne^{-1}}{n!}.}
Step 2:  
Therefore, the Taylor series for  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=3}   is
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.}


Final Answer:  
    (a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}}
    (b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n}

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