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| | <span class="exam"> Let | | <span class="exam"> Let |
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| − | ::::::<math>f(x)=\sum_{n=1}^{\infty} nx^n</math>
| + | ::<math>f(x)=\sum_{n=1}^{\infty} nx^n</math> |
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| − | <span class="exam">a) Find the radius of convergence of the power series. | + | <span class="exam">(a) Find the radius of convergence of the power series. |
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| − | <span class="exam">b) Determine the interval of convergence of the power series. | + | <span class="exam">(b) Determine the interval of convergence of the power series. |
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| − | <span class="exam">c) Obtain an explicit formula for the function <math>f(x)</math>. | + | <span class="exam">(c) Obtain an explicit formula for the function <math style="vertical-align: -5px">f(x)</math>. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |Review ratio test.
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| − | |} | |
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009C Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |To find the radius of convergence, we use the ratio test. We have
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)x^{n+1}}{nx^n}}\bigg|\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{n+1}{n}x\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n \rightarrow \infty}\frac{n+1}{n}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|}\\
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| − | \end{array}</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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| − | |-
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| − | |Thus, we have <math>|x|<1</math> and the radius of convergence of this series is <math>1</math>.
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| − | |}
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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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| − | |-
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| − | |From part (a), we know the series converges inside the interval <math>(-1,1)</math>.
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| − | |-
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| − | |Now, we need to check the endpoints of the interval for convergence.
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| − | |-
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| − | |}
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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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| − | |-
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| − | |For <math>x=1</math>, the series becomes <math>\sum_{n=1}^{\infty}n</math>, which diverges by the Divergence Test.
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |For <math>x=-1</math>, the series becomes <math>\sum_{n=1}^{\infty}(-1)^n n</math>, which diverges by the Divergence Test.
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| − | |-
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| − | |Thus, the interval of convergence is <math>(-1,1)</math>.
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |Recall we have the geometric series formula <math>\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n</math> for <math>|x|<1</math>.
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| − | |-
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| − | |Now, we take the derivative of both sides of the last equation to get
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| − | |-
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| − | |<math>\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}</math>.
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| − | |}
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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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| − | |-
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| − | |Now, we multiply the last equation in Step 1 by <math>x</math>.
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| − | |-
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| − | |So, we have <math>\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}=f(x)</math>.
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| − | |-
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| − | |Thus, <math>f(x)=\frac{x}{(1-x)^2}</math>.
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| − | |}
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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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| − | |-
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| − | |'''(a)''' <math>1</math>
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| − | |-
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| − | |'''(b)''' <math>(-1,1)</math>
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| − | |-
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| − | |'''(c)''' <math>f(x)=\frac{x}{(1-x)^2}</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
