Results 1 to 5 of 5

Math Help - Show that cosx = lnx has a unique solution

  1. #1
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1

    Show that cosx = lnx has a unique solution

    Hi,

    I was looking at some old exams in Numerical Analysis and found this question.

    Show that cosx = lnx has a unique solution in ]0,\pi/2]. First of all, I think the sign ] is a typo, and perhaps it should be ( since ln0 is -\infty. Then again, it could perhaps be a closed interval as well...

    The first thing I think of is to look at the function f(x) = cos(x)-ln(x). Now if I assume that the interval is [0,\pi/2] then I think I could, by the intermediate value theorem show that there exists a solution. I would need to do some limit-stuff because of 0 and ln(x)...

    Another thing that comes to mind is perhaps rewriting this in the following manner;
    <br />
e^{cos(x)} = x \Rightarrow x - e^{cos(x)} = 0.

    If I let g(x) = e^{cos(x)} then,
    <br />
x-g(x) = 0 if there is some number \xi in our interval such that
    \xi - g(\xi) = 0. That is, if \xi is a fixed-point of g.

    A unique fixed point exists if g is a contraction..
    Since g'(x) = -sin(x)e^{cos(x)} we have that
    |g'(x)|<1 for all x in [0,pi/2] and so a unique solution to our problem exists.

    The problem from the exam comes with a suggested solution as well. The solution says that since the function cosx is monotonically decreasing in the given interval, and the function ln(x) is monotonically increasing in the same interval, they can only cross in one point.. I do not find this solution particularly satisfying, but what do I know.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,487
    Thanks
    1391
    We'll examine the function f(x) = \cos{x} - \ln{x}.

    If we let x = 1 we find that f(1) = \cos{1} - \ln{1} = \cos{1} > 0.

    If we let x = \frac{\pi}{2} we find that \displaystyle f\left(\frac{\pi}{2}\right) = \cos{\frac{\pi}{2}} - \ln{\frac{\pi}{2}} = -\ln{\frac{\pi}{2}} < 0.

    Since the function changes sign, we know that there's a solution.


    Now let's examine the derivative...

    f'(x) = -\sin{x} - \frac{1}{x}

    \displaystyle = -\left(\sin{x} + \frac{1}{x}\right)

     < 0 in the interval \displaystyle \left(0, \frac{\pi}{2}\right].

    Since the derivative is always negative, that means that the function is always decreasing and therefore does not turn.

    So there can only be one solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133
    I think that a Numerical Analyst would simply sketch the graphs of \cos x and \ln x, note that they intersect just once, and regard that as sufficient.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    BobP: You're probably right, but I do not see the fun in that
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    May be that is interesting to find the solutions of the equation \cos z= \ln z when z is complex. Using Newton's iterations I found...

    z= 1.30296400122... + i 0 [the only on the real axis...]

    z= -1.28661280828... + i 1.58835248621...

    z= -7.26120884757... - i 1.95753445328...

    z= -13.4349581438... -i 2.0737796597...

    z= 12.5123914472... + i 1.58339003977...

    Of course some more solution may exists and it would be interesting to valuate if their number is finite or infinite...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 23rd 2011, 03:39 AM
  2. Complex numbers: how do I show unique solution?
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 8th 2011, 07:27 PM
  3. Replies: 1
    Last Post: March 24th 2010, 12:14 AM
  4. Show that f(x)=g(x) has a unique solution
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 18th 2010, 01:44 PM
  5. Replies: 2
    Last Post: September 7th 2009, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum