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Thread: Find the regions where sin(x) is greater than or equal to X

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    Find the regions where sin(x) is greater than or equal to X

    I know the graphs intersect three times, one at (0,0) and the other 2 points are reflected about the y-axis. Is there a solution for this exercise? I can't think of any of these points where a value of X is less than its sin value. And also, is there any way I can solve for the other 2 points algebraically? Because I've been approximating and it seems my approximates aren't that good...

    Thanks in advance guys, I appreciate it.
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    Have you tried graphing the functions? The answer
    Spoiler:
    $\displaystyle \displaystyle x < 0$
    should then be obvious.
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    Quote Originally Posted by Pupil View Post
    I know the graphs intersect three times, one at (0,0) and the other 2 points are reflected about the y-axis. Is there a solution for this exercise? I can't think of any of these points where a value of X is less than its sin value. And also, is there any way I can solve for the other 2 points algebraically? Because I've been approximating and it seems my approximates aren't that good...

    Thanks in advance guys, I appreciate it.
    $\displaystyle \frac{d}{dx}Sinx=Cosx$

    The maximum value of the derivative of $\displaystyle Sinx$ is 1.

    The derivative of $\displaystyle f(x)=x$ is always 1.

    Hence $\displaystyle Sinx=x$ only when $\displaystyle x=0$

    Therefore $\displaystyle Sinx>x$ if $\displaystyle x<0$
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    modem error....
    please delete
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    Alright guys, thanks for the help. I understand the inequality now.

    Just one more question: Is there any way I can get the exact values by algebraic methods? If I try to do it graphically, my approximations are not as nearly as accurate. I know the graphs intersect 3 times and one is (0,0), but how would I determine the other points without approximating? Because frankly, my graphing software sucks. ><
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    Quote Originally Posted by Pupil View Post
    Alright guys, thanks for the help. I understand the inequality now.

    Just one more question: Is there any way I can get the exact values by algebraic methods? If I try to do it graphically, my approximations are not as nearly as accurate. I know the graphs intersect 3 times and one is (0,0), but how would I determine the other points without approximating? Because frankly, my graphing software sucks. ><
    If the graph of the functions intersect to the right of the y-axis,
    then Sinx would have had to have a gradient >1 at some stage but it doesn't!

    Why are you convinced that there is more than one point of intersection?
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    Quote Originally Posted by Archie Meade View Post
    If the graph of the functions intersect to the right of the y-axis,
    then Sinx would have had to have a gradient >1 at some stage but it doesn't!

    Why are you convinced that there is more than one point of intersection?
    Ah, it was just a mistake with my software then. But I have other questions where there is more than 1 intersection, and my approximation with the software is not that good.

    So how would I be able to solve for the exact points of intersection? Ex: finding the 4 points where $\displaystyle cos(x) = 1/2^x$ Thanks for all the help, by the way.
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    Quote Originally Posted by Pupil View Post
    Ah, it was just a mistake with my software then. But I have other questions where there is more than 1 intersection, and my approximation with the software is not that good.

    So how would I be able to solve for the exact points of intersection? Ex: finding the 4 points where $\displaystyle cos(x) = 1/2^x$ Thanks for all the help, by the way.
    The first point of intersection is quite straightforward, x=0.

    You could use Newton's method to find the other two from $\displaystyle 0$ to $\displaystyle 2{\pi}$, by solving

    $\displaystyle cosx-\frac{1}{2^x}=0$

    However, that will give "very close approximations" !
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    Quote Originally Posted by Archie Meade View Post
    The first point of intersection is quite straightforward, x=0.

    You could use Newton's method to find the other two from $\displaystyle 0$ to $\displaystyle 2{\pi}$, by solving

    $\displaystyle cosx-\frac{1}{2^x}=0$

    However, that will give "very close approximations" !
    Okay, it seems similar to solving trig equations. However, since we can't use the unit circle, so how would I go about approximating the points? Oh and its $\displaystyle 0.5^x$

    Thanks in advance.
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    Quote Originally Posted by Pupil View Post
    Okay, it seems similar to solving trig equations. However, since we can't use the unit circle, so how would I go about approximating the points? Oh and its $\displaystyle 0.5^x$

    Thanks in advance.
    The Newton-Raphson formula is

    $\displaystyle \displaystyle\ x_{n+1}=x_n-\frac{f\left(x_n\right)}{f'\left(x_n\right)}$

    $\displaystyle x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$

    $\displaystyle x_2=x_1-\frac{f\left(x_1\right)}{f'\left(x_1\right)}$

    $\displaystyle x_3=x_2-\frac{f\left(x_2\right)}{f'\left(x_2\right)}$

    It's an iterative process. $\displaystyle x_0$ is a first guess at a solution.

    Then $\displaystyle x_1$ is the calculated first approximation. $\displaystyle x_2$ will be much closer, as will $\displaystyle x_3$

    $\displaystyle \frac{1}{2^x}=2^{-x}=0.5^x$

    $\displaystyle f(x)=cosx-0.5^x$

    $\displaystyle f'(x)=-sinx+\left(0.5^x\right)ln(0.5)$

    Take an initial guess at the next root of $\displaystyle cosx-0.5^x=0,$ say $\displaystyle x=\frac{\pi}{3}$.

    Then

    $\displaystyle \displaystyle\ x_1=\frac{\pi}{3}-\frac{cos\left(\frac{\pi}{3}\right)-0.5^{\frac{\pi}{3}}}{-sin\left(\frac{\pi}{3}\right)+\left(0.5^{\frac{\pi }{3}}\right)ln(0.5)}$

    Use the result as $\displaystyle x_1$ in the 2nd iteration, which should be enough.

    Use $\displaystyle \frac{3{\pi}}{2}$ as an initial guess for the 3rd root of $\displaystyle cosx-0.5^x=0$
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