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| | + | [[009C Sample Midterm 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''Ratio Test'''
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| − | |-
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| − | | Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
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| − | | Then,
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| − | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
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| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
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| − | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
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| − | |}
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| | + | [[009C Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Assume that the power series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math> converges.
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| − | |Let <math style="vertical-align: 0px">R</math> be the radius of convergence of this power series.
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| − | |We can use the Ratio Test to find <math style="vertical-align: 0px">R.</math>
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| − | |-
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| − | |Using the Ratio Test, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
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| − | \end{array}</math>
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| − | |Since the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math> is <math style="vertical-align: -5px">R,</math> we have
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| − | | <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</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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| − | |Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n.</math>
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| − | |Using the Ratio Test, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}2^nx^{n+1}}{c_n2^{n+1}x^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{2c_n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
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| − | \end{array}</math>
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| − | |Hence, the radius of convergence of this power series is
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| − | | <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=2R.</math>
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| − | |Therefore, this power series converges.
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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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| − | |Assume that the power series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math> converges.
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| − | |-
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| − | |Let <math style="vertical-align: 0px">R</math> be the radius of convergence of this power series.
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| − | |We can use the Ratio Test to find <math style="vertical-align: 0px">R.</math>
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| − | |-
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| − | |Using the Ratio Test, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
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| − | \end{array}</math>
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| − | |-
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| − | |Since the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math> is <math style="vertical-align: -5px">R,</math> we have
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| − | | <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</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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| − | |Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n(-x)^n .</math>
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| − | |Using the Ratio Test, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}(-x)^{n+1}}{c_n(-x)^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}(-x)}{c_n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
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| − | \end{array}</math>
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| − | |Hence, the radius of convergence of this power series is
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| − | | <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.</math>
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| − | |Therefore, this power series converges.
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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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| − | | '''(a)''' converges
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| − | | '''(b)''' converges
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| − | |}
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| | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
