Difference between revisions of "009C Sample Final 3, Problem 5"
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|We have | |We have | ||
|- | |- | ||
− | | <math>f'(x)=\bigg(\frac{ | + | | <math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math> |
|- | |- | ||
− | | <math>f''(x)=\bigg(\frac{ | + | | <math>f''(x)=\bigg(-\frac{1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math> |
|- | |- | ||
|and | |and | ||
|- | |- | ||
− | | <math>f^{(3)}(x)=\bigg(\frac{ | + | | <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. | ||
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|Thus, | |Thus, | ||
|- | |- | ||
− | | <math>f^{(n)}(x)=\bigg(\frac{ | + | | <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math> |
|} | |} | ||
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|Since | |Since | ||
|- | |- | ||
− | | <math>f'(x)=\bigg(\frac{ | + | | <math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math> |
|- | |- | ||
|we have | |we have | ||
|- | |- | ||
− | | <math>f'(3)=\bigg(\frac{ | + | | <math>f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}.</math> |
|} | |} | ||
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|Since | |Since | ||
|- | |- | ||
− | | <math>f^{(n)}(x)=\bigg(\frac{ | + | | <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x},</math> |
|- | |- | ||
|we have | |we have | ||
|- | |- | ||
− | | <math>f^{(n)}(3)=\bigg(\frac{ | + | | <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 | ||
|- | |- | ||
− | | <math>c_n=\frac{\bigg(\frac{ | + | | <math>c_n=\frac{\bigg(-\frac{1}{3}\bigg)^ne^{-1}}{n!}.</math> |
|} | |} | ||
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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 | |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{ | + | | <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: | !Final Answer: | ||
|- | |- | ||
− | | '''(a)''' <math>f^{(n)}(x)=\bigg(\frac{ | + | | '''(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{ | + | | '''(b)''' <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: |
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The Taylor polynomial of at is |
where |
Solution:
(a)
Step 1: |
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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) |