Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Real Root Help!

  1. May 26, 2005 #1
    Show that the equation:
    [tex]2x - 1 - \sin x = 0[/tex]
    has exactly one real root.


    [tex]\frac{d}{dx} (2x - 1 - \sin x) = 2 - \cos x = 0[/tex]
    [tex]2 = \cos x[/tex]
    [tex]x = \cos^{-1} 2[/tex]

    Is there a better way to approach the root?
    any suggestions?
     
  2. jcsd
  3. May 26, 2005 #2
    Do you mean root or critical point?
     
  4. May 26, 2005 #3

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Suppose f(x) = 2x - 1 - sin x

    Then f'(x) = 2 - cos x

    For any value of x, f'(x) is greater than 0, so f(x) is a constantly increasing function.

    Note that

    [tex]f(-\pi) = 2(-\pi) - 1 - \sin(-\pi) = -2\pi[/tex]
    [tex]f(\pi) = 2\pi - 1 - \sin(\pi) = 2\pi[/tex]

    Since f continually increases between x=-pi and x=pi, at exactly one value of x between -pi and pi, f(x) must be equal to zero.

    Thus, the equation has exactly one solution (root), and it lies between -pi and pi.
     
  5. May 26, 2005 #4
    Critical Criterion...

    The question is exactly as stated: 'exactly one real root.', and does not give any parameters for a 'critical point'.


    OK, however, is there a real equation solution for [tex]x[/tex]?
     
    Last edited: May 26, 2005
  6. May 26, 2005 #5

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There is no analytic solution for x, since the equation is transcendental.

    The numerical solution is approximately x = 0.887862
     
  7. May 26, 2005 #6

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    This has no solutions (cos x is always between -1 and 1).

    So the graph has no 'turning point'. For large values we have f(x)>0 and for small values f(x)<0. Since f is continuous there's a point in between where it is zero (Bolzano, or the intermediate value theorem). Also, there can be only one such point by using the mean value theorem.
     
  8. May 26, 2005 #7

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    If it helps to visualize the root is approximately:

    0.8878622115708660240357015114947117349741536783585225169984587779581531172428676120318808231042402676

    It's quite simple, function is constantly decreasing so it's not going to cross the axis again. It's simple to show that it crosses the axis as the function is continuous and is positive for a given value on one side and negative for a given value on the other side.
     
  9. May 26, 2005 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The equation [itex]\cos x =2 [/itex] can be solved easily.Make the sub [itex]e^{-ix} =t[/itex] and solve for "t".

    Daniel.
     
  10. May 26, 2005 #9

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    isn't it kind of obvious that the line like 2x-1 with slope 2, meets the graph of sin(x) at one poiint?
     
  11. May 28, 2005 #10

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not really. You can draw lots of straight lines which meet the graph of sin(x) at more than one point.
     
  12. May 28, 2005 #11

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    not too many with slope greater than one.
     
  13. May 28, 2005 #12
    Calculus Nexus...



    Uncertain if this is the correct identity approach:
    [tex]t = e^{-ix} = \cos (-x) + i \sin (-x) = \cos x - i \sin x[/tex]
    [tex]\cos x = 2[/tex]
    [tex]t = 2 - i \sin x[/tex]

    Note that it is relatively easy to write a subroutine to 'scan' the equation for solutions for 'x', which is similar to a 'trace' routine on a graphing calculator, however, the key here is to achieve an exact solution for 'x' without such methods.

    An algebraic-trigonometric solution does not appear to exist, however, a calculus solution may exist.

     
    Last edited: May 28, 2005
  14. May 28, 2005 #13
    Orion1,

    "An algebraic-trigonometric solution does not appear to exist, however, a calculus solution may exist."

    2x-1 - sin(x) = 0 is a transcendental equation, and it can't be solved for x.

    But your problem didn't ask for the solution, it just asked you to show that there is exactly one real solution. Mathwonk gave you a big hint.

    And BTW you can't take the derivative of an equation in one variable and assume that what you get has the same roots as the the original equation.
     
    Last edited: May 28, 2005
  15. Jun 1, 2005 #14
    The standard approach to these problems is to use the intermediate value problem to show that roots exists, then, argue by contradiction, and use the mean value theorem to show that there cannot be two or more roots.
     
  16. Jun 1, 2005 #15

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Orion,

    [tex] \cos x=\frac{1}{2}\left(e^{ix}+e^{-ix}\right) [/tex]

    If you make the sub i hinted,u'll get a quadratic in "t".

    Daniel.
     
  17. Jun 1, 2005 #16
    But is any of that even necessary? Since cos x := 2 for any x, the function has no critical points, so it can cross the x-axis at most one time.
     
  18. Jun 1, 2005 #17

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I was addressing other problem,namely solving the equation [itex] \cos x=2 [/itex],which means finding all possible "x" for which [itex]\cos x [/itex] is equal to [itex]2[/itex].

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Real Root Help!
  1. Real roots criterion (Replies: 1)

Loading...