009C Sample Final 1, Problem 5 - Revision history

Kayla Murray at 21:37, 3 December 2017

2017-12-03T21:37:11Z

Kayla Murray at 01:06, 3 December 2017

2017-12-03T01:06:05Z

Kayla Murray at 19:08, 18 March 2017

2017-03-18T19:08:09Z

Kayla Murray at 23:15, 26 February 2017

2017-02-26T23:15:58Z

Kayla Murray at 00:04, 26 February 2017

2017-02-26T00:04:33Z

Kayla Murray at 00:02, 26 February 2017

2017-02-26T00:02:03Z

Kayla Murray at 00:00, 26 February 2017

2017-02-26T00:00:52Z

Kayla Murray at 01:35, 19 April 2016

2016-04-19T01:35:17Z

Kayla Murray at 18:41, 1 March 2016

2016-03-01T18:41:59Z

Kayla Murray at 18:51, 29 February 2016

2016-02-29T18:51:37Z

@@ Line 7: / Line 7: @@
 <span class="exam">(b) Determine the interval of convergence of the power series.
-<span class="exam">(c) Obtain an explicit formula for the function <math style="vertical-align: -5px">f(x)</math>.
+<span class="exam">(c) Obtain an explicit formula for the function &nbsp;<math style="vertical-align: -5px">f(x)</math>.
 <hr>

@@ Line 9: / Line 9: @@
 <span class="exam">(c) Obtain an explicit formula for the function <math style="vertical-align: -5px">f(x)</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 72: / Line 72: @@
 !Step 2: &nbsp;
 |-
-|For &nbsp;<math style="vertical-align: -4px">x=1,</math> &nbsp; the series becomes &nbsp; <math>\sum_{n=1}^{\infty}n,</math> &nbsp; which diverges by the Divergence Test.
+|For &nbsp;<math style="vertical-align: -4px">x=1,</math> &nbsp; the series becomes &nbsp; <math>\sum_{n=1}^{\infty}n,</math> &nbsp; which diverges by the &nbsp;<math style="vertical-align: 0px">n</math>th term test.
 |}
@@ Line 78: / Line 78: @@
 !Step 3: &nbsp;
 |-
-|For &nbsp;<math style="vertical-align: -4px">x=-1,</math> &nbsp; the series becomes &nbsp;<math>\sum_{n=1}^{\infty}(-1)^n n,</math> &nbsp;which diverges by the Divergence Test.
+|For &nbsp;<math style="vertical-align: -4px">x=-1,</math> &nbsp; the series becomes &nbsp;<math>\sum_{n=1}^{\infty}(-1)^n n,</math> &nbsp;which diverges by the &nbsp;<math style="vertical-align: 0px">n</math>th term test.
 |-
 |Thus, the interval of convergence is &nbsp; <math style="vertical-align: -5px">(-1,1).</math>

@@ Line 14: / Line 14: @@
 |'''1.'''  '''Ratio Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;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> Then,
+|&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math style="vertical-align: -7px">\sum a_n</math>&nbsp; be a series and &nbsp;<math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>&nbsp; Then,
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 |-
 |'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">L=1.</math>
 |}
@@ Line 56: / Line 56: @@
 !Step 2: &nbsp;
 |-
-|Thus, we have <math style="vertical-align: -5px">|x|<1</math> and the radius of convergence of this series is <math style="vertical-align: -1px">1.</math>
+|Thus, we have &nbsp; <math style="vertical-align: -5px">|x|<1</math> &nbsp; and the radius of convergence of this series is &nbsp; <math style="vertical-align: -1px">1.</math>
 |}
@@ Line 64: / Line 64: @@
 !Step 1: &nbsp;
 |-
-|From part (a), we know the series converges inside the interval <math style="vertical-align: -5px">(-1,1).</math>
+|From part (a), we know the series converges inside the interval &nbsp;<math style="vertical-align: -5px">(-1,1).</math>
 |-
 |Now, we need to check the endpoints of the interval for convergence.
@@ Line 72: / Line 72: @@
 !Step 2: &nbsp;
 |-
-|For <math style="vertical-align: -2px">x=1,</math> the series becomes <math>\sum_{n=1}^{\infty}n,</math> which diverges by the Divergence Test.
+|For &nbsp;<math style="vertical-align: -4px">x=1,</math> &nbsp; the series becomes &nbsp; <math>\sum_{n=1}^{\infty}n,</math> &nbsp; which diverges by the Divergence Test.
 |}
@@ Line 78: / Line 78: @@
 !Step 3: &nbsp;
 |-
-|For <math style="vertical-align: -2px">x=-1,</math> the series becomes <math>\sum_{n=1}^{\infty}(-1)^n n,</math> which diverges by the Divergence Test.
+|For &nbsp;<math style="vertical-align: -4px">x=-1,</math> &nbsp; the series becomes &nbsp;<math>\sum_{n=1}^{\infty}(-1)^n n,</math> &nbsp;which diverges by the Divergence Test.
 |-
-|Thus, the interval of convergence is <math style="vertical-align: -5px">(-1,1).</math>
+|Thus, the interval of convergence is &nbsp; <math style="vertical-align: -5px">(-1,1).</math>
 |}
@@ Line 88: / Line 88: @@
 !Step 1: &nbsp;
 |-
-|Recall that we have the geometric series formula <math>\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n</math> for <math>|x|<1.</math>
+|Recall that we have the geometric series formula &nbsp; <math>\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n</math> &nbsp; for &nbsp; <math>|x|<1.</math>
 |-
 |Now, we take the derivative of both sides of the last equation to get
@@ Line 99: / Line 99: @@
 !Step 2: &nbsp;
 |-
-|Now, we multiply the last equation in Step 1 by <math style="vertical-align: 0px">x.</math>
+|Now, we multiply the last equation in Step 1 by &nbsp;<math style="vertical-align: 0px">x.</math>
 |-
 |So, we have
@@ Line 106: / Line 106: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}=f(x).</math>
 |-
-|Thus, <math>f(x)=\frac{x}{(1-x)^2}.</math>
+|Thus, &nbsp;<math>f(x)=\frac{x}{(1-x)^2}.</math>
 |}

@@ Line 42: / Line 42: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \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|\\
 &&\\
@@ Line 93: / Line 93: @@
 |-
+|
-::<math>\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}.</math>
 |}
@@ Line 104: / Line 104: @@
 |-
+|
-::<math>\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}=f(x).</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}=f(x).</math>
 |-
 |Thus, <math>f(x)=\frac{x}{(1-x)^2}.</math>

@@ Line 12: / Line 12: @@
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.'''  '''Ratio Test'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;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> Then,
 |-
+|
-:::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
 |-
+|
-:::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
 |-
+|
-:::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
 |-
 |'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval
 |-
+|
-:::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
 |}
 '''Solution:'''
@@ Line 109: / Line 108: @@
 |Thus, <math>f(x)=\frac{x}{(1-x)^2}.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' <math style="vertical-align: -3px">1</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -3px">1</math>
 |-
-|&nbsp;&nbsp; '''(b)''' <math style="vertical-align: -3px">(-1,1)</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -3px">(-1,1)</math>
 |-
-|&nbsp;&nbsp; '''(c)''' <math style="vertical-align: -18px">f(x)=\frac{x}{(1-x)^2}</math>
+|&nbsp;&nbsp; '''(c)''' &nbsp; &nbsp; <math style="vertical-align: -18px">f(x)=\frac{x}{(1-x)^2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Let
-::::::<math>f(x)=\sum_{n=1}^{\infty} nx^n</math>
+::<math>f(x)=\sum_{n=1}^{\infty} nx^n</math>
-::<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.
-::<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.
-::<span class="exam">c) Obtain an explicit formula for the function <math style="vertical-align: -5px">f(x)</math>.
+<span class="exam">(c) Obtain an explicit formula for the function <math style="vertical-align: -5px">f(x)</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 14: / Line 14: @@
 |Recall:
 |-
-|'''1. Ratio Test''' 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>. Then,
+|'''1. Ratio Test''' 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> Then,
 |-
+|
@@ Line 28: / Line 28: @@
 |-
+|
-::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1</math>.
+::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
 |}
@@ Line 66: / Line 66: @@
 |-
 |Now, we need to check the endpoints of the interval for convergence.
 |}