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Math Help - [SOLVED] Series

  1. #1
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    Question [SOLVED] Series

    I am given:

    And it says to find the first few coefficients, c0,c1...,to c4, which I can't find. Then it says to find the radius of convergence R of the power series. I'm not sure why they have the fraction equaling the series. I know I can use the ratio or root test to find the radius of convergence, but what exactly is the above problem stating?
    Thanks,
    Matt
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  2. #2
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    Okay, I got help and figured out that I should use the form:
    \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n
    So then I got to:
    \frac{6x}{10}\sum_{n=0}^{\infty}\left(\frac{-x}{10}\right)^n
    and (-1)^n\left(\frac{1}{10}\right)^n x^n
    Now, I figured out that the radius of convergence is 10, but I'm still unsure for the coefficients.
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  3. #3
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    Hello, matt3D!

    \frac{6x}{10+x} \:=\:\sum^{\infty}_{n=0} c_nx^n

    (a) Find the first few coefficients, to c_4
    (b) Find the radius of convergence R of the power series.
    Use long division . . .

    \begin{array}{ccccccc}<br />
& & & \frac{6}{10}x & -\frac{6}{10^2}x^2 & +\frac{6}{10^3}x^3 & - \quad\cdots \\<br />
& & ---&---&---&--- \\<br />
10+x & ) & 6x \\<br />
& & 6x & +\frac{6}{10}x^2 \\<br />
& & --- & --- \\<br />
& & & -\frac{6}{10}x^2 \\<br />
& & & -\frac{6}{10}x^2 & -\frac{6}{10^2}x^2 \\<br />
& & & --- & --- \\<br />
& & & & \frac{6}{10^2}x^3 \\<br />
& & & & \frac{6}{10^2}x^3 & + \frac{6}{10^3}x^4 \\<br />
& & & & ---&---\end{array}



    We see that: . \frac{6x}{10+x} \:=\:\frac{6}{10}x - \frac{6}{10^2}x^2 + \frac{6}{10^3}x^3 - \frac{6}{10^4}x^4 + \cdots


    (a) The first five coefficients are: . \begin{Bmatrix}c_o &=& 0 \\ \\[-4mm]c_1 &=&\frac{6}{10} \\ \\[-4mm]c_2 &=& \text{-}\frac{6}{10^2} \\ \\[-4mm]c_3 &=& \frac{6}{10^3} \\ \\[-4mm]c_4 &=& \text{-}\frac{6}{10^4}\end{Bmatrix}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The general term is: . a_n \:=\:(\text{-}1)^{n+1}\frac{6}{10^n}x^n . . . for n \geq 1


    Ratio Test: . \frac{a_{n+1}}{a_n} \;=\; \left|\frac{6x^{n+1}}{10^{n+1}}  \cdot \frac{10^n}{6x^n}\right| \;=\;\left|\frac{x}{10}\right|

    And we have: . \left|\frac{x}{10}\right| \:<\:1 \quad\Rightarrow\quad |x| \:< \:10


    (b) Therefore, the radius of convergence is: . R \:=\:10

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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by matt3D View Post
    I am given:

    And it says to find the first few coefficients, c0,c1...,to c4, which I can't find. Then it says to find the radius of convergence R of the power series. I'm not sure why they have the fraction equaling the series. I know I can use the ratio or root test to find the radius of convergence, but what exactly is the above problem stating?
    Thanks,
    Matt
    Let's find a more general power series. Consider

    \frac{ax^m}{b+cx^p}

    Simple algebra shows that this is equivalent to

    \frac{ax^m}{b}\cdot\frac{1}{1+\frac{c}{b}x^p}

    =\frac{ax^m}{b}\cdot\frac{1}{1+\left(\frac{c^{\tfr  ac{1}{p}}x}{b^{\tfrac{1}{p}}}\right)^p}

    Now we know that

    \frac{1}{1+x^p}=\sum_{n=0}^{\infty}(-1)^nx^{np}

    \therefore\quad\frac{1}{1+\left(\frac{c^{\tfrac{1}  {p}}x}{b^{\tfrac{1}{p}}}\right)^p}=\sum_{n=0}^{\in  fty}(-1)^n\left(\frac{c^{\frac{1}{p}}x}{b^{\frac{1}{p}}}  \right)^{np}

    =\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np}}{b^n}

    \therefore\quad\frac{ax^n}{b+cx^p}=\frac{ax^m}{b}\  cdot\frac{1}{1+\left(\frac{c^{\tfrac{1}{p}}x}{b^{\  tfrac{1}{p}}}\right)^p}

    =\frac{ax^m}{b}\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np}}{b^n}

    =a\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np+m}}{b^{n+1}}

    Using the root test we check for the ROC...to do this we must find all x such that

    \lim_{n\to\infty}\sqrt[n]{\left|\frac{(-1)^nc^nx^{np+m}}{b^{n+1}}\right|}
    =\frac{c|x|^p}{b}<1

    Solving for x gives

    |x|<\sqrt[p]{\frac{b}{c}}

    I'll leave you the endpoints...they are fairly easy.
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  5. #5
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    Wow, thanks Soroban, I didn't know you could use long division. Mathstud28 I'm still trying to figure your post out, but thanks for the info!
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