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Thread: Taylor series

  1. #1
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    Taylor series

    I have to expand the f(x) thanks to the taylor series, and then solve it
    How do i expand the first squareroot(senx)?
    There is no actual formula for this... I know how to expand sen^2x but how about sen^1/2 x??
    Thanks to everyone

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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Taylor series

    Quote Originally Posted by domy7997 View Post
    I have to expand the f(x) thanks to the taylor series, and then solve it
    How do i expand the first squareroot(senx)?
    There is no actual formula for this... I know how to expand sen^2x but how about sen^1/2 x??
    Thanks to everyone

    Sent from my Redmi Note 4X using Tapatalk
    Use the chain rule:
    $\displaystyle \dfrac{d}{dx} sin \left ( \sqrt{x} \right ) = cos \left ( \sqrt{x} \right ) \cdot \dfrac{1}{2 \sqrt{x}}$

    etc.

    -Dan
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  3. #3
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    Re: Taylor series

    Quote Originally Posted by topsquark View Post
    Use the chain rule:
    $\displaystyle \dfrac{d}{dx} sin \left ( \sqrt{x} \right ) = cos \left ( \sqrt{x} \right ) \cdot \dfrac{1}{2 \sqrt{x}}$

    etc.

    -Dan
    I believe you can also just write the series for $\sin(x) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{2k+1}$

    and just replace $x \to \sqrt{x}$ yielding

    $\Large \sin(\sqrt{x}) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{\frac{2k+1}{2}}$
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    Re: Taylor series

    Quote Originally Posted by romsek View Post
    I believe you can also just write the series for $\sin(x) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{2k+1}$

    and just replace $x \to \sqrt{x}$ yielding

    $\Large \sin(\sqrt{x}) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{\frac{2k+1}{2}}$
    Yeah, but what about (senx)^1/2?
    From the app i don't know how to write the math correctly sry!

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  5. #5
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    Re: Taylor series

    Quote Originally Posted by domy7997 View Post
    Yeah, but what about (senx)^1/2?
    From the app i don't know how to write the math correctly sry!

    Sent from my Redmi Note 4X using Tapatalk
    do you know in general how to find a taylor series?

    $f(x) =\sum \limits_{k=0}^\infty ~f^{(k)}(a)\dfrac{(x-a)^k}{k!}$

    $f^{(k)}(a) = \left . \dfrac{d^kf}{dx^k}(x) \right |_{x=a}$
    Last edited by romsek; Jan 10th 2019 at 07:04 PM.
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