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Thread: Approximation

  1. #1
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    Approximation

    Show that for small $\displaystyle x$, $\displaystyle \frac{sinx-1}{cos(x-\frac{\pi}{4})} \approx \sqrt2(-1+2x-\frac{5}{2}x^2)$



    $\displaystyle \frac{sinx-1}{cos(x-\frac{\pi}{4})}\approx\frac{x-1}{1-\frac{(x-\frac{\pi}{4})^2}{2}}$
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  2. #2
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    Re: Approximation

    Maybe you can do something with limits ($\displaystyle x \to 0$), or isn't that the intention of this exercice? ...
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    Re: Approximation

    But if I do that, then x will not be shown in the approximated answer
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  4. #4
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    Re: Approximation

    Quote Originally Posted by Punch View Post
    Show that for small $\displaystyle x$, $\displaystyle \frac{sinx-1}{cos(x-\frac{\pi}{4})} \approx \sqrt2(-1+2x-\frac{5}{2}x^2)$



    $\displaystyle \frac{sinx-1}{cos(x-\frac{\pi}{4})}\approx\frac{x-1}{1-\frac{(x-\frac{\pi}{4})^2}{2}}$
    For starters, $\displaystyle \displaystyle \begin{align*} \cos{\left(x - \frac{\pi}{4}\right)} &= \cos{(x)}\cos{\left(\frac{\pi}{4}\right)} + \sin{(x)}\sin{\left(\frac{\pi}{4}\right)} \\ &= \frac{1}{\sqrt{2}}\cos{(x)} + \frac{1}{\sqrt{2}}\sin{(x)} \\ &= \frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}} \end{align*}$

    So

    $\displaystyle \displaystyle \begin{align*} \frac{\sin{(x)} - 1}{\cos{\left(x - \frac{\pi}{4}\right)}} &= \frac{\sin{(x)} - 1}{\frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}}} \\ &= \sqrt{2}\left[\frac{\sin{(x)} - 1}{\cos{(x)} + \sin{(x)}}\right] \end{align*}$

    Now I would advise substituting the MacLaurin Series for $\displaystyle \displaystyle \sin{(x)}$ and $\displaystyle \displaystyle \cos{(x)}$ and seeing what you can do...
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  5. #5
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    Re: Approximation

    Quote Originally Posted by Punch View Post
    Show that for small $\displaystyle x$, $\displaystyle \frac{sinx-1}{cos(x-\frac{\pi}{4})} \approx \sqrt2(-1+2x-\frac{5}{2}x^2)$



    $\displaystyle \frac{sinx-1}{cos(x-\frac{\pi}{4})}\approx\frac{x-1}{1-\frac{(x-\frac{\pi}{4})^2}{2}}$


    Expand $\displaystyle \sin(x)-1$ and $\displaystyle 1/\cos(x-\pi/4)$ about $\displaystyle x=0$, as far as the term in $\displaystyle x^2$ (if there is one) and multiply out and truncate at the $\displaystyle x^2$ term.

    CB
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  6. #6
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    Re: Approximation

    Quote Originally Posted by Prove It View Post
    For starters, $\displaystyle \displaystyle \begin{align*} \cos{\left(x - \frac{\pi}{4}\right)} &= \cos{(x)}\cos{\left(\frac{\pi}{4}\right)} + \sin{(x)}\sin{\left(\frac{\pi}{4}\right)} \\ &= \frac{1}{\sqrt{2}}\cos{(x)} + \frac{1}{\sqrt{2}}\sin{(x)} \\ &= \frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}} \end{align*}$

    So

    $\displaystyle \displaystyle \begin{align*} \frac{\sin{(x)} - 1}{\cos{\left(x - \frac{\pi}{4}\right)}} &= \frac{\sin{(x)} - 1}{\frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}}} \\ &= \sqrt{2}\left[\frac{\sin{(x)} - 1}{\cos{(x)} + \sin{(x)}}\right] \end{align*}$

    Now I would advise substituting the MacLaurin Series for $\displaystyle \displaystyle \sin{(x)}$ and $\displaystyle \displaystyle \cos{(x)}$ and seeing what you can do...
    Following up, $\displaystyle \sqrt2[\frac{sinx-1}{cosx+sinx}]=\sqrt2[\frac{x-1}{1-\frac{x^2}{2}+x}]$

    but I cant convert it into the given form
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  7. #7
    Grand Panjandrum
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    Re: Approximation

    Quote Originally Posted by Punch View Post
    Following up, $\displaystyle \sqrt2[\frac{sinx-1}{cosx+sinx}]=\sqrt2[\frac{x-1}{1-\frac{x^2}{2}+x}]$

    but I cant convert it into the given form
    $\displaystyle \frac{1}{1-\frac{x^2}{2}+x}=\frac{1}{1-\left(\frac{x^2}{2}-x\right)}=1+\left(\frac{x^2}{2}-x\right)+\left(\frac{x^2}{2}-x\right)^2+..\\ \\ \\ \phantom{xxxxxxxxxxx}=1-x+\frac{3x^2}{2}+..$

    CB
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