Results 1 to 10 of 10

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

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    100

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1010
    Have you tried graphing the functions? The answer
    Spoiler:
    \displaystyle x < 0
    should then be obvious.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    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.
    \frac{d}{dx}Sinx=Cosx

    The maximum value of the derivative of Sinx is 1.

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

    Hence Sinx=x only when x=0

    Therefore Sinx>x if x<0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    modem error....
    please delete
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2010
    Posts
    100
    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. ><
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2010
    Posts
    100
    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 cos(x) = 1/2^x Thanks for all the help, by the way.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    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 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 0 to 2{\pi}, by solving

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

    However, that will give "very close approximations" !
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2010
    Posts
    100
    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 0 to 2{\pi}, by solving

    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 0.5^x

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    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 0.5^x

    Thanks in advance.
    The Newton-Raphson formula is

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

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

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

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

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

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

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

    f(x)=cosx-0.5^x

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

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

    Then

    \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 x_1 in the 2nd iteration, which should be enough.

    Use \frac{3{\pi}}{2} as an initial guess for the 3rd root of cosx-0.5^x=0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. diving an area into two equal regions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 12th 2010, 05:07 PM
  2. Prove that g(x) greater than/equal to 0?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2009, 02:23 AM
  3. Replies: 10
    Last Post: March 24th 2009, 11:11 PM
  4. Replies: 2
    Last Post: March 23rd 2009, 07:11 AM
  5. greater than, equal to or less than zero
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 23rd 2008, 04:09 AM

Search Tags


/mathhelpforum @mathhelpforum