Use the binomial series to expand the function as a power series. Find the radius of convergence. x/ â 9+ x^2

Question

Use the binomial series to expand the function as a power series. Find the radius of convergence.

x/ â 9+ x^2

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MichaelMet 5 years 2021-07-30T23:50:46+00:00 1 Answers 16 views 0

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  1. haidang
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    2021-07-30T23:52:14+00:00
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    Answer:

    Step-by-step explanation:

    The given function is:

    f(x) = \dfrac{x}{\sqrt{9+x^2}}

    Using the binomial series:

    = x(9+x^2)^{-1/2} \\ \\ = x *9^{-1/2}(1+\dfrac{x^2}{9})^{-1/2} \\ \\ = \dfrac{x}{3}(1+ \dfrac{x^2}{9})^{-1/2}

    = \dfrac{x}{3} \sum \limits ^{\alpha }_{n=0}(^{-\frac{1}{2}}_n)(\dfrac{x^2}{9})^n

    \implies \dfrac{x}{3}\Bigg [ 1 + (-\dfrac{1}{2})*(\dfrac{x^2}{9})+ \dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)}{2!}(\dfrac{x^2}{9}) ^2 + \dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1) (-\dfrac{1}{2}-2)}{3!} ) (\dfrac{x^2}{9}) ^3+ ... \Bigg ]

    = \dfrac{x}{3}\Bigg [ 1 - \dfrac{x^2}{18}+ \dfrac{3}{5832}*\dfrac{x^4}{1}-\dfrac{15}{34992}x^6+... \Bigg ]

    \mathbf{= \dfrac{x}{3}- \dfrac{x^3}{54}+ \dfrac{1}{5832}x^4 - \dfrac{5}{34992}x^7 + ...}

    To compute the radius of convergence:

    f(x) = \dfrac{\lambda }{3} \sum \limits ^{\alpha }_{n=0} (1+\dfrac{x^2}{9})^{-1/2}

    f(x) = \dfrac{\lambda }{3} \sum \limits ^{\alpha }_{n=0} (^{-1/2} _n ) (\dfrac{x^2}{9})^n \\ \\ \implies \dfrac{\lambda }{3} \sum \limits ^{\alpha }_{n=0} (^{-1/2} _n ) (\dfrac{x^2}{9})^n \\ \\ \implies \sum \limits ^{\alpha}_{n=0} (^{-1/2} _n ) \dfrac{1}{3*9^n}*x^{2n} \\ \\ \implies \sum \limits ^{\alpha}_{n=0} (^{-1/2} _n ) \dfrac{1}{3^{2n+1}}*x^{2n}

    Suppose a_n = (^{-1/2}_{n})*\dfrac{1}{3^{2n+1}}*x^{2n}

    Then, rewriting the equation above as:

    a_{n+1} = (^{-1/2}_{n+1})*\dfrac{1}{3^{2n+3}}*x^{2n+2}

    As such;

    \lim_{n \to x} \Big| \dfrac{a_n+1}{a_n} \Big| = \lim_{n \to x} \Bigg | \dfrac{ (^{-1/2}_{n+1}) \dfrac{x^{2n+2}}{3^{2n+3}} }{(^{-1/2}_{n} )\dfrac{x^2}{3^{2n+1}}}} \Bigg |

    \implies \lim_{n \to \alpha} \Bigg | \dfrac{ (^{-1/2}_{n+1}) \dfrac{x^{2}}{3^{2}} }{(^{-1/2}_{_n} )} \Bigg |

    \implies \lim_{n \to \alpha} \Bigg | \dfrac{\dfrac{(-1/2)!}{(-1/2-n -1)!(n+1)!}*\dfrac{x^2}{9} }{ \dfrac{(-1/2!)}{(-1/2-n)!(n!)} } \Bigg| \\ \\ \\ \implies \lim_{n \to \alpha} \Bigg | \dfrac{(-1/2 -n)! (n!) }{(-1/2 -n-1)! (n+1)! } *\dfrac{x^2}{9} \Bigg| \\ \\ \\ \implies \lim_{n \to \alpha} \Bigg | \dfrac{(-1/2 -n) (-1/2 -n-1)! \ n! }{(-1/2 -n-1)! (n+1) n! } *\dfrac{x^2}{9} \Bigg|

    \implies \lim_{n \to \alpha} \Bigg | \dfrac{(-1/2 -n)}{n+1 } *\dfrac{x^2}{9} \Bigg|

    \implies \lim_{n \to \alpha} \Bigg | \dfrac{n( -\dfrac{1}{2n -1 } ) }{n(1+\dfrac{1}{n}) } *\dfrac{x^2}{9} \Bigg| \\ \\ \\ \implies \Big| \dfrac{x^2}{9} \Big| \lim_{n \to \alpha} \Big | \dfrac{-\dfrac{1}{2n} -1}{1+ \dfrac{1}{n}} \Big| \\ \\ \implies | \dfrac{x^2}{9}| |\dfrac{0-1}{1}|

    \implies | \dfrac{x^2}{9}|

    However, the series converges if and only if:

    | \dfrac{x^2}{9}| < 1

    ∴

    \dfrac{|x^2|}{9} < 1

    ={|x^2|}< 9 \\ \\ ={|x|} < \sqrt{9} \\ \\ = \mathbf{{|x|} < 3}

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