# Function analysis

1. Apr 5, 2007

### standardflop

Hello,

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

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

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

Thank you.

2. Apr 5, 2007

### HallsofIvy

Staff Emeritus
Since $0< y_s< 1[/tex], then [itex]0< z< 1$
Adding the first two terms, what you have is
$$\frac{-1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}$$
Now what is the smallest that first term could be? What is the largest the second term could be? (Remember the condition that
$$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$).

3. Apr 5, 2007

### standardflop

The second term is largest when $z \rightarrow 0$, where it takes the values $$\frac{\gamma \beta}{\theta^2}$$. But as i see it, the first term is a problem since it goes toward $\pm \infty$ (or undef.?) when $z \rightarrow 0 \ \wedge \ z \rightarrow 1$ respectively.? So can this term be bounded in the interval $0<z<1$ ?

Also i believe your first term has the wrong sign.

Last edited: Apr 6, 2007
4. Apr 6, 2007

### HallsofIvy

Staff Emeritus
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
$$\frac{1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}$$
or, what I really intended,
$$\frac{-1}{z(1-z)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}$$
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?

5. Apr 6, 2007

### standardflop

The largest possible value of the second term, minus the (abs.) smallest possible value of the first is
$$\frac{\gamma \beta}{\theta^2} - 4 < 0$$ because $$\vline \ \max_{0<z<1} \ \frac{1}{z(z-1)} \ \vline = 4$$

which leads to $$\frac{\gamma \beta}{\theta^2} < 4$$ and not the expected $$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$

Last edited: Apr 6, 2007
6. Apr 7, 2007

### standardflop

=> or is this to be understood as a less strict requirement, because $$\frac{\gamma \beta}{\theta (\beta + \theta)} < \frac{\gamma \beta}{\theta^2}$$ ? Therefore if $$\frac{\gamma \beta}{\theta^2} < 4$$ then also $$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$ ?