Difference between revisions of "009C Sample Midterm 2, Problem 5"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
|If a power series converges, then it has a nonempty interval of convergence.
+
|'''Ratio Test'''
 +
|-
 +
|&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; &nbsp; &nbsp; &nbsp; Then,
 +
|-
 +
|
 +
&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 &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 
|}
 
|}
  
Line 24: Line 37:
 
|Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 
|Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 
|-
 
|-
−
|So, the power series
+
|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp;
+
|Using the Ratio Test, we have
 
|-
 
|-
−
|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\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|}\\
 +
&&\\
 +
& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
 +
\end{array}</math>
 +
|-
 +
|Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
 
|}
 
|}
  
Line 34: Line 56:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Let &nbsp;<math style="vertical-align: -5px">a\in (-2R,2R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -13px">\frac{a}{2} \in (-R,R).</math>&nbsp;
+
|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>
 
|-
 
|-
−
|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+
|Using the Ratio Test, we have
 
|-
 
|-
−
|&nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{a}{2}\bigg)^n</math>&nbsp; converges.
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\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|}\\
 +
&&\\
 +
& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
 +
\end{array}</math>
 
|-
 
|-
−
|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R),</math>&nbsp;
+
|Hence, the radius of convergence of this power series is
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=2R.</math>
 
|-
 
|-
−
|converges in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R).</math>&nbsp;
+
|Therefore, this power series converges.  
 
|}
 
|}
  
Line 55: Line 81:
 
|Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 
|Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 
|-
 
|-
−
|So, the power series
+
|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 +
|-
 +
|Using the Ratio Test, we have
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\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|}\\
 +
&&\\
 +
& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
 +
\end{array}</math>
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp;  
+
|Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
 
|-
 
|-
−
|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
 
|}
 
|}
  
Line 65: Line 100:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Let &nbsp;<math style="vertical-align: -5px">a\in (-R,R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">-a \in (-R,R).</math>&nbsp;
+
|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>
 
|-
 
|-
−
|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+
|Using the Ratio Test, we have
 
|-
 
|-
−
|&nbsp;<math>\sum_{n=0}^\infty c_n(-a)^n</math>&nbsp; converges.
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\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|}\\
 +
&&\\
 +
& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
 +
\end{array}</math>
 
|-
 
|-
−
|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+
|Hence, the radius of convergence of this power series is
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n(-x)^n</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.</math>
 
|-
 
|-
−
|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+
|Therefore, this power series converges.
 
|}
 
|}
  

Revision as of 11:55, 24 April 2017

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 \sum_{n=0}^\infty c_nx^n}   converges, does it follow that the following series converges?

(a)  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 \sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n}

(b)  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 \sum_{n=0}^\infty c_n(-x)^n }


Foundations:  
Ratio Test
        Let  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 \sum a_n}   be a series 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.}
        Then,

        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 L<1,}   the series is absolutely convergent.

        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 L>1,}   the series is divergent.

        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 L=1,}   the test is inconclusive.


Solution:

(a)

Step 1:  
Assume that the power series  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 \sum_{n=0}^\infty c_nx^n}   converges.
Let  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}   be the radius of convergence of this power series.
We can use the Ratio Test to find  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.}  
Using the Ratio Test, we have

        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} \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|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.} \end{array}}

Since the radius of convergence of the series  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 \sum_{n=0}^\infty c_nx^n}   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 R,}   we have
        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=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.}
Step 2:  
Now, we use the Ratio Test to find the radius of convergence of the series 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 \sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n.}
Using the Ratio Test, we have
        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} \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|}\\ &&\\ & = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.} \end{array}}
Hence, the radius of convergence of this power series 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 \frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=2R.}
Therefore, this power series converges.

(b)

Step 1:  
Assume that the power series  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 \sum_{n=0}^\infty c_nx^n}   converges.
Let  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}   be the radius of convergence of this power series.
We can use the Ratio Test to find  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.}  
Using the Ratio Test, we have

        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} \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|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.} \end{array}}

Since the radius of convergence of the series  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 \sum_{n=0}^\infty c_nx^n}   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 R,}   we have
        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=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.}
Step 2:  
Now, we use the Ratio Test to find the radius of convergence of the series 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 \sum_{n=0}^\infty c_n(-x)^n .}
Using the Ratio Test, we have
        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} \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|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.} \end{array}}
Hence, the radius of convergence of this power series 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 \frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.}
Therefore, this power series converges.


Final Answer:  
    (a)     converges
    (b)     converges

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