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Math Help - Maclaurin Series

  1. #1
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    Maclaurin Series

    Use an appropriate local quadratic approximation to approximate tan61

    http://img212.imageshack.us/img212/1256/52475955jo9.png

    this is the answer from the solution mannual

    Can anybody explain this more clearly?
    Thank you
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by soleilion View Post
    Use an appropriate local quadratic approximation to approximate tan61

    http://img212.imageshack.us/img212/1256/52475955jo9.png

    this is the answer from the solution mannual

    Can anybody explain this more clearly?
    Thank you
    Work in radians and expand as a Taylor series about 60^{\circ}=\pi/3 {\rm{rad}}

     <br />
f(x+\varepsilon)=f(x)+\varepsilon~\left. \frac{df}{dx}\right|_{x}+\frac{\varepsilon^2}{2}~\  left. \frac{d^2f}{dx^2}\right|_{x}+..<br />

    Now:

    \frac{d}{dx}\tan(x)=(\sec(x))^2=1+(\tan(x))^2,

    and

    \frac{d^2}{dx^2}\tan(x)=2 (\sec(x))^2 \tan(x)=2\left[\frac{d}{dx}\tan(x)\right] \tan(x).

    Now all you need is that \tan(\pi/3)=\sqrt{3}, and the \varepsilon corresponding to 1^{\circ} is \varepsilon=\pi/180

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Work in radians and expand as a Taylor series about 60^{\circ}=\pi/3 {\rm{rad}}

     <br />
f(x+\varepsilon)=f(x)+\varepsilon~\left. \frac{df}{dx}\right|_{x}+\frac{\varepsilon^2}{2}~\  left. \frac{d^2f}{dx^2}\right|_{x}+..<br />

    Now:

    \frac{d}{dx}\tan(x)=(\sec(x))^2=1+(\tan(x))^2,

    and

    \frac{d^2}{dx^2}\tan(x)=2 (\sec(x))^2 \tan(x)=2\left[\frac{d}{dx}\tan(x)\right] \tan(x).

    Now all you need is that \tan(\pi/3)=\sqrt{3}, and the \varepsilon corresponding to 1^{\circ} is \varepsilon=\pi/180

    RonL
    As the answer from my website link, why it just shows f'', second prime? why not third or more or less than second?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by soleilion View Post
    As the answer from my website link, why it just shows f'', second prime? why not third or more or less than second?
    Because it asks for a locally quadratic approximation which means that the highest order term in \varepsilon should be \varepsilon^2 which means we only go as far as the second derivative.

    RonL
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