Difference between revisions of "009C Sample Midterm 3, Problem 5"

From Math Wiki
Jump to navigation Jump to search
Line 19: Line 19:
 
::<math>\left|\frac{a_{n+1}}{a_n}\right|<1.</math>
 
::<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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!(a): &nbsp;
 
!(a): &nbsp;
 +
|-
 +
|We need to choose a radius <math style="vertical-align: 0%">r</math> such that whenever <math style="vertical-align: -20%">|x|<r</math>,
 +
|-
 +
|
 +
::<math>\begin{array}{rcl}
 +
\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = & \lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+1}x^{n+1}}{n+2}}{\frac{(-1)^{n}x^{n}}{n+1}}\right|\\
 +
\\
 +
& = & \lim_{n\rightarrow\infty}\left|x\cdot\frac{n+1}{n+2}\right|\\
 +
\\
 +
& = & |x|\cdot\lim_{n\rightarrow\infty}\left|\frac{n+1}{n+2}\right|\\
 +
\\
 +
& = & |x|<1.
 +
\end{array}</math>
 +
|-
 +
|In this case, the radius is 1, and the interval will be centered at 0. We then need to take a look at the boundary points. If <math style="vertical-align: -20%">x=1,</math> then
 +
|-
 +
|
 +
::<math>a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}}{n+1},</math>
 +
|-
 +
|so the series is an alternating harmonic series which converges. On the other hand, if <math style="vertical-align: -20%">x=-1,</math> then
 
|-
 
|-
 
|
 
|
 +
::<math>a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}(-1)^{n}}{n+1}\,=\,\frac{1}{n+1},</math>
 +
|-
 +
|a standard harmonic series which does not converge. Thus, the interval of convergence is <math style="vertical-align: -20%">(-1,1]</math>.
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!(b): &nbsp;
 
!(b): &nbsp;
 +
|-
 +
|We need to choose a radius <math style="vertical-align: 0%">r</math> such that whenever <math style="vertical-align: -20%">|x|<r</math>,
 +
|-
 +
|
 +
::<math>\begin{array}{rcl}
 +
\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = & \lim_{n\rightarrow\infty}\left|\frac{\frac{(x+1)^{n+1}}{(n+1)^{2}}}{\frac{(x+1)^{n}}{n^{2}}}\right|\\
 +
\\
 +
& = & \lim_{n\rightarrow\infty}\left|(x+1)\cdot\left(\frac{n}{n+1}\right)^{2}\right|\\
 +
\\
 +
& = & |x+1|\cdot\lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{2}\\
 +
\\
 +
& = & |x+1|<1.
 +
\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
 
|-
 
|-
 
|
 
|
 +
::<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>.
 +
 
|}
 
|}
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|
+
|For (a), the radius is 1 and the interval of convergence is <math style="vertical-align: -20%">(-1,1]</math>.
 +
|-
 +
|For (b), the radius is also 1, but the interval of convergence is <math style="vertical-align: -20%">[-2,0]</math>.
 
|}
 
|}
 
[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:50, 27 April 2015

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

(a) (6 points)     
(b) (6 points)     
When we do, the interval will be Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (c-r,c+r)} . However, the boundary values for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c-r} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c+r} 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.
Foundations:  
When we are asked to find the radius of convergence, we are given a series where
where and are functions of and respectively, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is a constant (frequently zero). We need to find a bound (radius) on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|} such that whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|<r} , the ratio test
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\frac{a_{n+1}}{a_n}\right|<1.}

 Solution:

(a):  
We need to choose a radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} such that whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x|<r} ,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = & \lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+1}x^{n+1}}{n+2}}{\frac{(-1)^{n}x^{n}}{n+1}}\right|\\ \\ & = & \lim_{n\rightarrow\infty}\left|x\cdot\frac{n+1}{n+2}\right|\\ \\ & = & |x|\cdot\lim_{n\rightarrow\infty}\left|\frac{n+1}{n+2}\right|\\ \\ & = & |x|<1. \end{array}}
In this case, the radius is 1, and the interval will be centered at 0. We then need to take a look at the boundary points. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1,} then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}}{n+1},}
so the series is an alternating harmonic series which converges. On the other hand, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-1,} then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}(-1)^{n}}{n+1}\,=\,\frac{1}{n+1},}
a standard harmonic series which does not converge. Thus, the interval of convergence is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1,1]} .
(b):  
We need to choose a radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} such that whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x|<r} ,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = & \lim_{n\rightarrow\infty}\left|\frac{\frac{(x+1)^{n+1}}{(n+1)^{2}}}{\frac{(x+1)^{n}}{n^{2}}}\right|\\ \\ & = & \lim_{n\rightarrow\infty}\left|(x+1)\cdot\left(\frac{n}{n+1}\right)^{2}\right|\\ \\ & = & |x+1|\cdot\lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{2}\\ \\ & = & |x+1|<1. \end{array}}
In this case, the radius is 1, and the interval will be centered at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-1} , or when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+1=0} . We then need to take a look at the boundary points. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-2} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} , then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|a_{n}\right|\,=\,\left|\frac{(x+1)^{n}}{n^{2}}\right|\,=\,\frac{1}{n^{2}},}
which defines a p-series with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p=2} . Thus, the series defined by each boundary point is absolutely convergent (and therefore convergent), and the interval of convergence is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-2,0]} .
Final Answer:  
For (a), the radius is 1 and the interval of convergence is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1,1]} .
For (b), the radius is also 1, but the interval of convergence is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-2,0]} .

Return to Sample Exam

Retrieved from "https://wiki.math.ucr.edu/index.php?title=009C_Sample_Midterm_3,_Problem_5&oldid=420"