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Math Help - Taylor Series #2

  1. #1
    Senior Member mollymcf2009's Avatar
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    Taylor Series #2

    Find the Taylor Series for f(x) centered at a=25. (assume f has a power series expansion)

    f(x) = \frac{1}{\sqrt{x}}

    I'm not going to type all these derivatives, but here are the 4th and nth:

    f^4(x) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) x^{-\frac{9}{2}}

    So, when I calculated my nth derivative, I got:

    f^n(25) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) (-\frac{9}{2})...-\frac{(2n+1)}{2}(25)^{2n+1}


    No idea....

    EDIT: Do I need to use the binomial series here?
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  2. #2
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    Quote Originally Posted by mollymcf2009 View Post
    Find the Taylor Series for f(x) centered at a=25. (assume f has a power series expansion)

    f(x) = \frac{1}{\sqrt{x}}

    I'm not going to type all these derivatives, but here are the 4th and nth:

    f^4(x) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) x^{-\frac{9}{2}}

    So, when I calculated my nth derivative, I got:

    f^n(25) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) (-\frac{9}{2})...-\frac{(2n+1)}{2}(25)^{2n+1}


    No idea....

    EDIT: Do I need to use the binomial series here?
    f^{(n)}(x)=(-1)^n\frac{3 \times 5 \times \cdots \times (2n-1)}{2^n}x^{-\frac{2n+1}{2}}=(-1)^n\frac{(2n)!}{4^n n!}x^{-\frac{2n+1}{2}}. thus f^{(n)}(25)=(-1)^n \frac{(2n)!}{4^n5^{2n+1}n!}. hence \frac{1}{\sqrt{x}} =\sum_{n=0}^{\infty} (-1)^n \frac{\binom{2n}{n}}{4^n5^{2n+1}}(x - 25)^n.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    f^{(n)}(x)=(-1)^n\frac{3 \times 5 \times \cdots \times (2n-1)}{2^n}x^{-\frac{2n+1}{2}}=(-1)^n\frac{(2n)!}{4^n n!}x^{-\frac{2n+1}{2}}. thus f^{(n)}(25)=(-1)^n \frac{(2n)!}{4^n5^{2n+1}n!}. hence \frac{1}{\sqrt{x}} =\sum_{n=0}^{\infty} (-1)^n \frac{\binom{2n}{n}}{4^n5^{2n+1}}(x - 25)^n.
    Thanks! I am really struggling with this entire concept. Where did this x^{-\frac{2n+1}{2}} come from? And where did this 4^n5^{2n+1} come from?
    Sorry, I have been trying to understand this stuff for over a week and I can not get it...I desperately need someone to show it to me like I'm a 5 year old. Thanks so much for any help!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Find the Taylor Series for f(x) centered at a=25. (assume f has a power series expansion)

    f(x) = \frac{1}{\sqrt{x}}

    I'm not going to type all these derivatives, but here are the 4th and nth:

    f^4(x) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) x^{-\frac{9}{2}}

    So, when I calculated my nth derivative, I got:

    f^n(25) = (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2}) (-\frac{9}{2})...-\frac{(2n+1)}{2}(25)^{2n+1}


    No idea....

    EDIT: Do I need to use the binomial series here?
    Try to spot the pattern:

    f\left(x\right)=\frac{1}{\sqrt{x}}\implies f\left(25\right)=\frac{1}{5}=\left(-\tfrac{1}{2}\right)^0\frac{1}{5^1}
    f^{\prime}\left(x\right)=-\tfrac{1}{2}x^{-\frac{3}{2}}\implies f^{\prime}\left(25\right)=-\frac{1}{2}\frac{1}{5^3}=\left(-\tfrac{1}{2}\right)^1\cdot1\cdot\frac{1}{5^3}
    f^{\prime\prime}\left(x\right)=-\tfrac{1}{2}\left(-\tfrac{1}{2}\right)\cdot1\cdot3\cdot x^{-\frac{5}{2}}\implies f^{\prime\prime}\left(25\right)=\left(-\tfrac{1}{2}\right)^2\cdot1\cdot3\cdot\frac{1}{5^5  }

    Thus, the nth derivative should take on the form:

    f^{(n)}\left(x\right)=\left(-\frac{1}{2}\right)^n\prod_{k=1}^{n}(2k-1)x^{-\frac{(2n+1)}{2}}\implies f^{(n)}\left(25\right)=\left(-\frac{1}{2}\right)^n\prod_{k=1}^{n}(2k-1)5^{-(2n+1)}

    Now, the taylor series should be f\!\left(x\right)=\sum_{k=0}^{\infty}\frac{1}{k!}\  left(-\tfrac12\right)^k\left(\displaystyle\prod_{j=1}^k(  2j-1)\right)5^{-(2k+1)}\left(x-25\right)^k

    I plugged this into Maple, and it gave me this simplification (which really isn't much...)

    f(x)=\tfrac{1}{2}\sum_{k=0}^{\infty}\frac{1}{\sqrt  {\pi}k!}\left(-\tfrac{1}{2}\right)^k2^{k+1}\Gamma\left(k+\tfrac{1  }{2}\right)5^{-(2k+1)}(x-25)^k

    But I see NonCommAlg beat me to this...and his way looks much, much neater...
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