Difference between revisions of "009C Sample Final 1, Problem 5"

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|'''1.'''  '''Ratio Test'''  
 
|'''1.'''  '''Ratio Test'''  
 
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|-
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|&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,
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|&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,
 
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|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.  
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&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.  
 
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|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
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&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 
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|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
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&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 
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|-
 
|'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval  
 
|'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval  
 
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|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
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&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">L=1.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|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>
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|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>
 
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|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|From part (a), we know the series converges inside the interval <math style="vertical-align: -5px">(-1,1).</math>
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|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.
 
|Now, we need to check the endpoints of the interval for convergence.
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|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.
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|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.
 
|}
 
|}
  
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
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|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.
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|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.
 
|-
 
|-
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|Thus, the interval of convergence is <math style="vertical-align: -5px">(-1,1).</math>
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|Thus, the interval of convergence is &nbsp; <math style="vertical-align: -5px">(-1,1).</math>
 
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|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|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>
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|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
 
|Now, we take the derivative of both sides of the last equation to get
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, we multiply the last equation in Step 1 by <math style="vertical-align: 0px">x.</math>
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|Now, we multiply the last equation in Step 1 by &nbsp;<math style="vertical-align: 0px">x.</math>
 
|-
 
|-
 
|So, we have  
 
|So, we have  
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&nbsp; &nbsp; &nbsp; &nbsp;<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>
 
|-
 
|-
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|Thus, <math>f(x)=\frac{x}{(1-x)^2}.</math>
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|Thus, &nbsp;<math>f(x)=\frac{x}{(1-x)^2}.</math>
 
|}
 
|}
  

Revision as of 15:15, 26 February 2017

Let

(a) Find the radius of convergence of the power series.

(b) Determine the interval of convergence of the power series.

(c) Obtain an explicit formula for the function .

Foundations:  
1. Ratio Test
       Let    be a series and    Then,

       If    the series is absolutely convergent.

       If    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.

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  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.}


Solution:

(a)

Step 1:  
To find the radius of convergence, we use 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{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)x^{n+1}}{nx^n}}\bigg|\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{n+1}{n}x\bigg|}\\ &&\\ & = & \displaystyle{|x|\lim_{n \rightarrow \infty}\frac{n+1}{n}}\\ &&\\ & = & \displaystyle{|x|.}\\ \end{array}}

Step 2:  
Thus, 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 |x|<1}   and the radius of convergence of this 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 1.}

(b)

Step 1:  
From part (a), we know the series converges inside the interval  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).}
Now, we need to check the endpoints of the interval for convergence.
Step 2:  
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=1,}   the series becomes   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=1}^{\infty}n,}   which diverges by the Divergence Test.
Step 3:  
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=-1,}   the series becomes  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=1}^{\infty}(-1)^n n,}  which diverges by the Divergence Test.
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).}

(c)

Step 1:  
Recall that we have the geometric series formula   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}{1-x}=\sum_{n=0}^{\infty} x^n}   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|<1.}
Now, we take the derivative of both sides of the last equation to get

       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}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}.}

Step 2:  
Now, we multiply the last equation in Step 1 by  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.}
So, 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 \frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}=f(x).}

Thus,  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 f(x)=\frac{x}{(1-x)^2}.}


Final Answer:  
   (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 1}
   (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 (-1,1)}
   (c)     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 f(x)=\frac{x}{(1-x)^2}}

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