Results 1 to 13 of 13

Math Help - Show that the function is monotonic

  1. #1
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318

    Show that the function is monotonic

    Show that the function is monotonic on the interval ]0,\frac{\pi}{4}[.

    sinx + cosx + tanx + cotx

    I created another topic in Trigonometry about solving the derivative for this function, and it seems the derivative is a bit complicated. Are there any other ways to solve this that doesn't require finding the roots of the derivative (I can't find all the roots)?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    What does monotonic mean? It means that the function is never decreasing or never increasing. Therefore, a monotonic function must satisfy f'(x)\le 0 OR f'(x)\ge 0
    Last edited by colby2152; January 14th 2008 at 07:55 AM. Reason: Plato pointed out something
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by colby2152 View Post
    What does monotonic mean? It means that the function is never decreasing or never increasing. Therefore, it never has a solution to f'(x)=0
    Is that true of f(x) = x^3?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Plato View Post
    Is that true of f(x) = x^3?
    Good point Plato. I wasn't totally comfortable with my statement, and you find the flaw. I should have said that values of the derivative must be greater than OR equal to zero.

    The following properties are true for a monotonic function f : R → R:

    * f has limits from the right and from the left at every point of its domain;
    * f has a limit at infinity (either ∞ or −∞) of either a real number, ∞, or −∞.
    * f can only have jump discontinuities;
    * f can only have countably many discontinuities in its domain.

    These properties are the reason why monotonic functions are useful in technical work in analysis. Two facts about these functions are:

    * if f is a monotonic function defined on an interval I, then f is differentiable almost everywhere on I, i.e. the set of numbers x in I such that f is not differentiable in x has Lebesgue measure zero.
    * if f is a monotonic function defined on an interval [a, b], then f is Riemann integrable.

    An important application of monotonic functions is in probability theory. If X is a random variable, its cumulative distribution function

    FX(x) = Prob(X ≤ x)

    is a monotonically increasing function.

    A function is unimodal if it is monotonically increasing up to some point (the mode) and then monotonically decreasing.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    I know what a monotonic function is. I just need to prove that this particular function is monotonic, and I don't know how to do that.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,842
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Spec View Post
    I know what a monotonic function is. I just need to prove that this particular function is monotonic, and I don't know how to do that.
    \frac{d}{dx} ( sinx + cosx + tanx + cotx ) = cos(x) - sin(x) + sec^2(x) +- csc^2(x)

    Is this always positive or negative on your interval? (I'd say the easiest way to show this is to graph the thing.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    We're not supposed to use any tools like calculators, so graphing it is not an acceptable solution.

    cosx is monotonically decreasing on the interval.
    -sinx is monotonically decreasing on the interval.
    \frac{1}{cos^2x} is monotonically increasing on the interval.
    -\frac{1}{sin^2x} is monotonically increasing on the interval.

    I don't really know what to do with that.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Spec View Post
    We're not supposed to use any tools like calculators, so graphing it is not an acceptable solution.

    cosx is monotonically decreasing on the interval.
    -sinx is monotonically decreasing on the interval.
    \frac{1}{cos^2x} is monotonically increasing on the interval.
    -\frac{1}{sin^2x} is monotonically increasing on the interval.

    I don't really know what to do with that.
    You don't need a calculator to graph a function. Calculate the values on your own and plot them. I would plot f(0), f(\frac{\pi}{6}), f(\frac{\pi}{4}).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    As Dan was showing you:

    \frac{d}{dx} ( sinx + cosx + tanx + cotx ) = cos(x) - sin(x) + sec^2(x) +- csc^2(x)

    Now, try to solve this algebraically:

    cos(x) - sin(x) + sec^2(x) - csc^2(x) \le 0

    OR

    cos(x) - sin(x) + sec^2(x) - csc^2(x) \ge 0

    ------------------------

    cos(x) - sin(x) + sec^2(x) - csc^2(x) \le 0

    cos(x) - sin(x) + \frac{1}{cos^2(x)} - \frac{1}{sin^2(x)} \le 0

    cos^3(x)sin^2(x)-cos^2(x)sin^3(x)+sin^2(x)-cos^2(x) \le 0

    sin^2(x)(cos^3(x)+1) \le cos^2(x)(sin^3(x)+1), looks true to me!

    OR

    sin^2(x)(cos^3(x)+1) \ge cos^2(x)(sin^3(x)+1), FALSE: Try x = 0

    In that interval [0, \frac{\pi}{4}] that you provided, both the cosine and sine function are positive. Therefore, both of those inequalities are true. Sine minus cosine is never more than one when the signs are the same. The function is monotonically decreasing.
    Last edited by colby2152; January 14th 2008 at 09:35 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    Quote Originally Posted by colby2152 View Post
    sin^2(x)(cos^3(x)+1) \le cos^2(x)(sin^3(x)+1), looks true to me!
    I don't really know if "looks true to me" constitues a proof.

    I still consider this to be unsolved. Solving this inequality is the hard part:

    cosx - sinx + \frac{sin^2x - cos^2x}{cos^2x \cdot sin^2x} < 0, \forall x \in \left]0, \frac{\pi}{4}\right[
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    <br />
\begin{gathered}<br />
  f'(x) = \cos x - \sin x + \frac{{\sin ^2 x - \cos ^2 x}}<br />
{{sin^2 x\cos ^2 x}} \hfill \\<br />
   \hfill \\<br />
   \leftrightarrow \left( {\cos x - \sin x} \right)\left( {1 - \frac{{\sin x + \cos x}}<br />
{{sin^2 x\cos ^2 x}}} \right) \hfill \\ <br />
\end{gathered}

    now lets look at the following function:


    g(x) = 1 - \frac{{\sin x + \cos x}}<br />
{{sin^2 x\cos ^2 x}}

    notice that:


    \begin{gathered}<br />
  \mathop {\lim }\limits_{x \to 0} g(x) =  - \infty  \hfill \\<br />
  g\left( {\frac{\pi }<br />
{4}} \right) < 0 \hfill \\ <br />
\end{gathered}

    you can easily verify that g(x) is continues at the half open interval (0,pi/4]
    thus if we show that g(x) has no extrema points (maximum) in this interval, it must be negative throughout this interval.


    <br /> <br />
g'(x) =  - \frac{{\left( {\cos x - \sin x} \right)sin^2 x\cos ^2 x - 2\left( {\sin x\cos ^3 x - \cos x\sin ^3 x} \right)\left( {\sin x + \cos x} \right)}}<br />
{{(...)^2 }} = <br /> <br />

    <br />
 =  - \frac{{\left( {\cos x - \sin x} \right)sin^2 x\cos ^2 x - 2\sin x\cos x\left( {\cos ^2 x - \sin ^2 x} \right)\left( {\sin x + \cos x} \right)}}<br />
{{(...)^2 }} = <br />


    \begin{gathered}<br />
   =  - \frac{{\left( {\cos x - \sin x} \right)sin^2 x\cos ^2 x - 2\sin x\cos x\left( {\cos x - \sin x} \right)\left( {\sin x + \cos x} \right)^2 }}<br />
{{(...)^2 }} =  \hfill \\<br />
   \hfill \\<br />
   =  - \frac{{\left( {\cos x - \sin x} \right)\sin x\cos x\left[ {\sin x\cos x - 2\left( {\sin x + \cos x} \right)^2 } \right]}}<br />
{{(...)^2 }} =  \hfill \\ <br />
\end{gathered}

    <br />
\begin{gathered}<br />
   =  - \frac{{\left( {\cos x - \sin x} \right)\sin x\cos x\left( { - 3\sin x\cos x - 2} \right)}}<br />
{{(...)^2 }} =  \hfill \\<br />
   \hfill \\<br />
   = \frac{{\left( {\cos x - \sin x} \right)\sin x\cos x\left( {3\sin x\cos x + 2} \right)}}<br />
{{(...)^2 }} \hfill \\ <br />
\end{gathered} <br /> <br />

    now let's look at the constituting parts of this expression:

    1. <br />
\left( {\cos x - \sin x} \right)<br /> <br />
    is positive since cosx is bigger than sinx for all x in the given interval.

    2. <br />
{\sin x\cos x}<br /> <br />

    both sinx and cos x are positive in this interval.

    3.
    {\left( {3\sin x\cos x + 2} \right)}

    trivial...
    4. the denominator is squared therefore it's also positive.

    therefore g(x) is strictly negative in this interval and therefore f'(x) is strictly negative in this interval since:

    <br />
f'(x) = g(x)\left( {\cos x - \sin x} \right)<br /> <br />
    Last edited by Peritus; January 25th 2008 at 08:23 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Spec View Post
    Solving this inequality is the hard part:

    \cos x - \sin x + \frac{\sin^2x - \cos^2x}{\cos^2x \cdot \sin^2x} < 0, \forall x \in \left]0, \frac{\pi}{4}\right[
    Notice that this expression is zero when x = π/4. Write it as f'(x) = \cos x - \sin x + \frac1{\cos^2x} - \frac1{\sin^2x}, and differentiate it once more, to get

    . . . . . f''(x) = -\sin x - \cos x +\frac{2\sin x}{\cos^3x} + \frac{2\cos x}{\sin^3x} = \frac{(2-\cos^3x)\sin^4x + (2-\sin^3x)\cos^4x}{\sin^3x\cdot\cos^3x}.

    This is clearly positive, so f'(x) is increasing in the interval (0,π/4). But since it is zero at x=π/4, it must be negative when 0<x<π/4.

    So f'(x) < 0 and therefore f is monotonically decreasing in that interval.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member TriKri's Avatar
    Joined
    Nov 2006
    Posts
    357
    Thanks
    1
    I haven't read all answers, but you don't have to find the roots of f'(x). You just have to show that f'(x) never crosses the x-axis at that specific interval. For example, if you can't show directly that f'(x) is all negative or all positive in that interval, you could for example look at f'(0). If you see that f'(0) is negative, and you somehow can show that f''(0) is all negative in the interval, then f'(0) will only decrease and hence not cross the x-axis in that interval. Hence f(x) will be monotonic.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 29th 2011, 02:08 PM
  2. Integrability of monotonic decreasing function
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 11th 2009, 06:55 AM
  3. monotonic function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 15th 2009, 10:21 AM
  4. Replies: 3
    Last Post: December 10th 2008, 11:32 AM
  5. Monotonic Convergence of a function
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 3rd 2008, 04:28 PM

Search Tags


/mathhelpforum @mathhelpforum