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Homework Help: Function analysis

  1. Apr 5, 2007 #1
    Hello,

    given is the function [tex] h(y_s) = \ln (1-y_s) - \ln y_s - \gamma + \frac{\gamma}{\theta + \beta (1-y_s)} [/tex]
    my job is now to show that [itex] h'(y_s) < 0, \forall y_s \in ]0,1[ [/itex] when
    [tex] \frac{\gamma \beta}{\theta (\beta + \theta)} < 4 [/tex]

    I guess that all constants can be assumed to be real and positive.
    My first tought was to introduce the new variable [itex] z=1-y_s [/itex] so that after differentiation i would get

    [tex] - 1/z + 1/(z-1) + \frac{\gamma \beta}{(\theta + \beta z)^2} < 0[/tex]
    but i cant derive the above expression from this inequality. Any help would be greatly appreciated.

    Thank you.
     
  2. jcsd
  3. Apr 5, 2007 #2

    HallsofIvy

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    Since [itex]0< y_s< 1[/tex], then [itex]0< z< 1[/itex]
    Adding the first two terms, what you have is
    [tex]\frac{-1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}[/tex]
    Now what is the smallest that first term could be? What is the largest the second term could be? (Remember the condition that
    [tex] \frac{\gamma \beta}{\theta (\beta + \theta)} < 4 [/tex]).
     
  4. Apr 5, 2007 #3
    The second term is largest when [itex]z \rightarrow 0[/itex], where it takes the values [tex]\frac{\gamma \beta}{\theta^2}[/tex]. But as i see it, the first term is a problem since it goes toward [itex] \pm \infty[/itex] (or undef.?) when [itex]z \rightarrow 0 \ \wedge \ z \rightarrow 1[/itex] respectively.? So can this term be bounded in the interval [itex] 0<z<1[/itex] ?

    Also i believe your first term has the wrong sign.
     
    Last edited: Apr 6, 2007
  5. Apr 6, 2007 #4

    HallsofIvy

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    Sorry about the sign. When I did it I was thinking 1/(1-z)- 1/z but then used your "reversed" z-1. Yes, it is
    [tex]\frac{1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}[/tex]
    or, what I really intended,
    [tex]\frac{-1}{z(1-z)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}[/tex]
    That first term, whether 1/(z(z-1)) or -1/(z(1-z)) is always negative. And when I said "smallest value", I should have made it clear that I was thinking of the absolute value because I was focusing on the subtraction- the largest possible value of the second term, minus the smallest possible value of the first is what?
     
  6. Apr 6, 2007 #5
    The largest possible value of the second term, minus the (abs.) smallest possible value of the first is
    [tex] \frac{\gamma \beta}{\theta^2} - 4 < 0 [/tex] because [tex] \vline \ \max_{0<z<1} \ \frac{1}{z(z-1)} \ \vline = 4 [/tex]

    which leads to [tex] \frac{\gamma \beta}{\theta^2} < 4 [/tex] and not the expected [tex] \frac{\gamma \beta}{\theta (\beta + \theta)} < 4 [/tex]
     
    Last edited: Apr 6, 2007
  7. Apr 7, 2007 #6
    => or is this to be understood as a less strict requirement, because [tex] \frac{\gamma \beta}{\theta (\beta + \theta)} < \frac{\gamma \beta}{\theta^2} [/tex] ? Therefore if [tex] \frac{\gamma \beta}{\theta^2} < 4 [/tex] then also [tex] \frac{\gamma \beta}{\theta (\beta + \theta)} < 4 [/tex] ?
     
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