009C Sample Midterm 3, Problem 5 - Revision history

MathAdmin at 16:47, 28 November 2017

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MathAdmin at 23:51, 27 April 2015

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MathAdmin at 23:51, 27 April 2015

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MathAdmin at 23:50, 27 April 2015

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MathAdmin at 22:54, 27 April 2015

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MathAdmin: Created page with "Find the radius of convergence and the interval of convergence of the series. ::(a) (6 points) ${\display..."$

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Created page with "Find the radius of convergence and the interval of convergence of the series. ::(a) (6 points) <math>{\display..."

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Find the radius of convergence and the interval of convergence
of the series.

::(a) (6 points)      <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}.</math>

::(b) (6 points)      <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|When we are asked to find the radius of convergence, we are given a series where
|-
|
::<math>a_n=f(x-c)\cdot g(n)</math>
|-
|where <math style="vertical-align: -20%">f</math> and <math style="vertical-align: -20%">g</math> are functions of <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">n</math> respectively, and <math style="vertical-align: 0%">c</math> is a constant (frequently zero). We need to find a bound on <math style="vertical-align: -22%">|x-c|</math> such that whenever <math style="vertical-align: -22%">|x-c|<L</math>, the ratio test
|-
|
::<math>\left|\frac{a_{n+1}}{a_n}\right|</math>
|-
|is satisfied. When we do, the interval will be <math style="vertical-align: -20%">(c-L,c+L)</math>. However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-L</math> and <math style="vertical-align: -8%">c+L</math> must be tested individually for convergence. Most often, one will produce an alternating, convergent series while the other will produce a divergent, non-alternating series.
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a):  
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b):  
|-
|
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|The series for (a) is convergent, while the series (b) is divergent.
|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Find the radius of convergence and the interval of convergence
+<span class="exam"> Find the radius of convergence and the interval of convergence
 of the series.
-::<span class="exam">(a) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}.</math>
+<span class="exam">(a) &nbsp;<math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}</math>
-::<span class="exam">(b) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}.</math>
+<span class="exam">(b) &nbsp;<math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}</math>
-&nbsp;'''Solution:'''
+[[009C Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 74: / Line 74: @@
 ::<math>\left|a_{n}\right|\,=\,\left|\frac{(x+1)^{n}}{n^{2}}\right|\,=\,\frac{1}{n^{2}},</math>
 |-
-|which defines a p-series with <math style="vertical-align: -20%">p=2</math>. Thus, the series defined by each boundary point is absolutely convergent (and therefore convergent), and the interval of convergence is <math style="vertical-align: -20%">[-2,0]</math>.
+|which defines a p-series with <math style="vertical-align: -15%">p=2</math>. Thus, the series defined by each boundary point is absolutely convergent (and therefore convergent), and the interval of convergence is <math style="vertical-align: -20%">[-2,0]</math>.
 |}

@@ Line 69: / Line 69: @@
 \end{array}</math>
 |-
-|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: -2%">x=-1</math>, or when <math style="vertical-align: -10%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
+|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: -2%">x=-1</math>, or when <math style="vertical-align: -5%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
 |-
+|

@@ Line 69: / Line 69: @@
 \end{array}</math>
 |-
-|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: 0%">x=-1</math>, or when <math style="vertical-align: -10%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
+|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: -2%">x=-1</math>, or when <math style="vertical-align: -10%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
 |-
+|

@@ Line 69: / Line 69: @@
 \end{array}</math>
 |-
-|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: -5%">x=-1</math>, or when <math style="vertical-align: -10%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
+|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: 0%">x=-1</math>, or when <math style="vertical-align: -10%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
 |-
+|

@@ Line 19: / Line 19: @@
 ::<math>\left|\frac{a_{n+1}}{a_n}\right|<1.</math>
 |-
-When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence.  Many times, one boundary value will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence may not be strictly open.
+When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence.  Many times, one boundary value will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence may open, half-open or closed.
 |}
@@ Line 26: / Line 26: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !(a): &nbsp;
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !(b): &nbsp;
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
+|For (a), the radius is 1 and the interval of convergence is <math style="vertical-align: -20%">(-1,1]</math>.
 |}
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 17: / Line 17: @@
 |-
+|
-::<math>\left|\frac{a_{n+1}}{a_n}\right|</math>
+::<math>\left|\frac{a_{n+1}}{a_n}\right|<1.</math>
 |-
-|is satisfied.  When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence.  Most often, one will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence can be either open, half-open or closed.
+When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence.  Many times, one boundary value will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence may not be strictly open.
 |}

@@ Line 14: / Line 14: @@
 ::<math>a_n=f(x-c)\cdot g(n)</math>
 |-
-|where <math style="vertical-align: -20%">f</math> and <math style="vertical-align: -20%">g</math> are functions of <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">n</math> respectively, and <math style="vertical-align: 0%">c</math> is a constant (frequently zero).  We need to find a bound on <math style="vertical-align: -22%">|x-c|</math> such that whenever <math style="vertical-align: -22%">|x-c|<L</math>, the ratio test
+|where <math style="vertical-align: -20%">f</math> and <math style="vertical-align: -20%">g</math> are functions of <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">n</math> respectively, and <math style="vertical-align: 0%">c</math> is a constant (frequently zero).  We need to find a bound (radius) on <math style="vertical-align: -22%">|x-c|</math> such that whenever <math style="vertical-align: -22%">|x-c|<r</math>, the ratio test
 |-
+|
 ::<math>\left|\frac{a_{n+1}}{a_n}\right|</math>
 |-
-|is satisfied.  When we do, the interval will be <math style="vertical-align: -20%">(c-L,c+L)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-L</math> and <math style="vertical-align: -8%">c+L</math> must be tested individually for convergence.  Most often, one will produce an alternating, convergent series while the other will produce a divergent, non-alternating series.
+|is satisfied.  When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence.  Most often, one will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence can be either open, half-open or closed.
 |}

009C Sample Midterm 3, Problem 5 - Revision history

MathAdmin at 16:47, 28 November 2017

MathAdmin at 23:51, 27 April 2015

MathAdmin at 23:51, 27 April 2015

MathAdmin at 23:50, 27 April 2015

MathAdmin at 23:50, 27 April 2015

MathAdmin at 23:50, 27 April 2015

MathAdmin at 23:15, 27 April 2015

MathAdmin at 22:58, 27 April 2015

MathAdmin at 22:54, 27 April 2015

MathAdmin: Created page with "Find the radius of convergence and the interval of convergence of the series. ::(a) (6 points) {\display..."

MathAdmin: Created page with "Find the radius of convergence and the interval of convergence of the series. ::(a) (6 points) ${\display..."$