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Math Help - Derivative

  1. #1
    Member vernal's Avatar
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    Derivative

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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Derivative

    Quote Originally Posted by vernal View Post
    You can apply the general formula...

    D^{\alpha} f(x) = \frac{1}{\Gamma(1-\alpha)} \frac{d}{d x} \int_{0}^{x} \frac{f(t)}{(x-t)^{\alpha}}\ dt

    ... setting \alpha=\frac{1}{2} and f(x)=\sin x

    Kind regards

    \chi \sigma
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  3. #3
    Member vernal's Avatar
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    Re: Derivative

    Quote Originally Posted by chisigma View Post
    You can apply the general formula...

    D^{\alpha} f(x) = \frac{1}{\Gamma(1-\alpha)} \frac{d}{d x} \int_{0}^{x} \frac{f(t)}{(x-t)^{\alpha}}\ dt

    ... setting \alpha=\frac{1}{2} and f(x)=\sin x

    Kind regards

    \chi \sigma
    thanks, can you solve??
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Derivative

    A 'brute force approach' probably fails, so that we try a 'step by step' solution. First step is to perform an integration be part obtaining...

    \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = -2\ |\sin t\ \sqrt{x-t}|_{0}^{x} + 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt (1)
    From (1) we derive...

    \frac{d}{dx} \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = 2 \frac{d}{dx}  \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= \int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt (2)

    Now with a little of patience or using Wolfram if You don't have patience [...] You find that...

     \int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt = \sqrt{2 \pi}\ \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\} (3)

    … where C and S are the Fresnel Cosine and Fresnel Sine functions, so that, remembering that is \Gamma(\frac{1}{2})= \sqrt{\pi} You obtain finally…

    D^{\frac{1}{2}} \sin x =  \sqrt{2}\   \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\} (4)

    Kind regards

    \chi \sigma
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  5. #5
    Member vernal's Avatar
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    Re: Derivative

    Quote Originally Posted by chisigma View Post
    A 'brute force approach' probably fails, so that we try a 'step by step' solution. First step is to perform an integration be part obtaining...

    \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = -2\ |\sin t\ \sqrt{x-t}|_{0}^{x} + 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt (1)
    From (1) we derive...

    \frac{d}{dx} \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = 2 \frac{d}{dx}  \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= \int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt (2)

    Now with a little of patience or using Wolfram if You don't have patience [...] You find that...

     \int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt = \sqrt{2 \pi}\ \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\} (3)

    … where C and S are the Fresnel Cosine and Fresnel Sine functions, so that, remembering that is \Gamma(\frac{1}{2})= \sqrt{\pi} You obtain finally…

    D^{\frac{1}{2}} \sin x =  \sqrt{2}\   \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\} (4)

    Kind regards

    \chi \sigma
    thanks for your help
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  6. #6
    Member vernal's Avatar
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    Re: Derivative

    Quote Originally Posted by chisigma View Post
    A 'brute force approach' probably fails, so that we try a 'step by step' solution. First step is to perform an integration be part obtaining...

    \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = -2\ |\sin t\ \sqrt{x-t}|_{0}^{x} + 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt (1)
    From (1) we derive...



    \chi \sigma
    please say what way did use?
    Last edited by vernal; December 29th 2011 at 12:38 PM.
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  7. #7
    Member vernal's Avatar
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    Re: Derivative

    oh sorry:


    then :


    and


    Is that true?
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: Derivative

    Quote Originally Posted by vernal View Post
    oh sorry:


    then :


    and


    Is that true?
    Yes, it is!... To understand why that 'little excamotage' is necessary I suggest You to try to apply the 'normal procedure' that consists, given a function...

    F(x)=\int_{\alpha(x)}^{\beta(x)} f(x,t)\ d t (1)

    ... in the computation of...

    \frac{d}{dx} F(x)= \int_{\alpha(x)}^{\beta(x)} f_{x} (x,t)\ dt - \frac{d \alpha}{d x}\ f(x,\alpha) + \frac{d \beta}{d x}\ f(x,\beta) (2)

    ... when f(x,t)= \frac{\sin t}{\sqrt{x-t}}, \alpha(x)=0 and \beta(x)=x...

    Kind regards

    \chi \sigma
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