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Math Help - Calculus problem...

  1. #1
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    Calculus problem...

    How would you show that the functions

    <br />
F(x)=\frac{2}{\pi}\arctan\Big(\frac{x}{a}\Big), x \geq 0<br />

    with a>0

    and

    G(x)=1-\exp(-\lambda x),x\geq0

    with \lambda>0

    meet (only)once at some x>0?
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  2. #2
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    Quote Originally Posted by willy0625 View Post
    How would you show that the functions

    <br />
F(x)=\frac{2}{\pi}\arctan\Big(\frac{x}{a}\Big), x \geq 0<br />

    with a>0

    and

    G(x)=1-\exp(-\lambda x),x\geq0

    with \lambda>0

    meet (only)once at some x>0?
    The will intersect where they are equal.

    So \frac{2}{\pi}\arctan{\frac{x}{a}} = 1 - e^{-\lambda x}

    or \frac{2}{\pi}\arctan{\frac{x}{a}} - 1 - e^{-\lambda x} = 0.


    If there was more than one intersection, then the graph must turn around on itself (by Rolle's Theorem). So see if there are any turning points.
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  3. #3
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    I thought about this for a while and my lecturer gave me the following result to prove the uniqueness of the point of the intersection.


     \noindent{\bf Claim} Let F and G be function defined as

     F(t):= \frac{1}{\pi}\arctan\Big(\dfrac{t}{a}\Big)

     G(t):= 1-\exp(-\lambda t)

    for t\in(0,\infty).

    Suppose there exist x,y \in(0,\infty) with 0<x<y such that


     F(y)  >  G(y)  \quad\cdots (1)
     F(x)  <  G(x)  \quad\cdots (2)
     F'(y) >  G'(x)  \quad\cdots (3)


    Then there exists a unique t^{*}\in[x,y] such that
    F(t^{*})=G(t^{*}).

    {\bf Proof}
    Let 0,\infty) \rightarrow [0,1]" alt=" H0,\infty) \rightarrow [0,1]" /> be a function defined by
      H(t):= F(t)-G(t)
    and let  x,y\in(0,\infty) WLOG  0<x<y.
    It follows that H is continuous since it is the difference of continuous functions F and G.

    Then by (1) and (2), we have
      H(y)=F(y)-G(y) > 0
      H(x)=F(x)-G(x) < 0

    Then it follows from the Bolzano's Intermediate Value Theorem that there exists a  t^{*}\in[x,y] such that
      H(t^{*})=0
    which is equivalently,
      F(t^{*})=G(t^{*})

    \hspace{\stretch{1}} \blacksquare

    I think this point is unique in the interval  (0,\infty).

    I'm not so sure about this because although I have proved that there is a point of intersection, yet the rate of increase for each function changes over different values for x^{*}<x. I do not think the assumption (3) guarantees that G never over takes F again since we can have for some value z that is greater than y such that F(z) < G(z).
    Last edited by willy0625; March 15th 2010 at 12:38 AM.
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