# I Proving an inequality from an equation

1. Jun 23, 2016

### Happiness

Given that $v_0$ and $v_f$ are positive variables related by the equation

where $g$ and $\alpha$ are positive constants.

Can you show that $v_f<v_0$ for all positive values of $v_0$ using a non-graphical method?

Physically, $v_0$ and $v_f$ are the initial and final speeds (at the same height) of a ball thrown upwards in a medium with a drag force $F=-m\alpha v$.

2. Jun 23, 2016

### Staff: Mentor

Yes. Hint: -ln(1-x) > ln(1+x) for positive x.

3. Jun 23, 2016

### Happiness

But this is only true for positive $x<1$. How then can we show $v_f<v_0$ for all positive values of $v_0$?

Using the hint, I can only get to $y-\ln(1+y)>\ln(1+x)-x$, where $x=\frac{\alpha v_0}{g}$ and $y=\frac{\alpha v_f}{g}$.

4. Jun 24, 2016

### Staff: Mentor

Your logarithm is not defined otherwise, a vf where the logarithm is not defined cannot be a solution.

But does that have a solution at all?
Let $c=\frac{g}{\alpha}$, and using the inequality for the denominator, I get

$$v_0 + v_f \leq c \ln(1+v_0/c) + c \ln(1+v_f/c)$$
That is wrong, as x > ln(1+x).

5. Jun 24, 2016

### Happiness

Using the hint

$c\ln(\frac{1+v_0/c}{1-v_f/c})>c\ln(1+v_0/c)+c\ln(1+v_f/c)$

So

$v_0+v_f>c\ln(1+v_0/c)+c\ln(1+v_f/c)$

Indeed, $v_f<\frac{g}{\alpha}$. It can be easily shown that $\frac{g}{\alpha}$ is the terminal velocity. So, indeed, only positive $x<1$ needs to be considered.

6. Jun 24, 2016

### Staff: Mentor

Oh right, wrong direction. Hmm, making it symmetric does not help. Okay, more mathematics.

Introduce V as $V=\frac{v_0 \alpha}{g}$ and $a>0$ as $v_f=a v_0$. Then we get
$$V+aV = ln(1+V) - ln(1-aV)$$
Here is a plot

Consider f(x)=x-ln(1+x) where x>0.
$f'(x)=\frac{x}{x+1} > 0$
Rewrite the equation from above:
$$V-ln(1+V)- aV + ln(1-aV) = 2aV$$
$$f(V) - f(aV) = 2 aV$$As a>0, V>0, we get
$$f(V) > f(aV)$$
We established that f is increasing monotonically, therefore a<1 which corresponds to $v_f < v_0$.

7. Jun 24, 2016

### Happiness

This doesn't seem right.

$f(V) - f(aV)=V-\ln(1+V)-\bigg(aV - \ln(1+aV)\bigg)=V-\ln(1+V)- aV + \ln(1+aV)$

The last term is different from the LHS in the previous step.

And the RHS should be $-2aV$.

Last edited: Jun 24, 2016
8. Jun 24, 2016

### Staff: Mentor

That should work now:
Consider f(x)=x-ln(1+x) where x>-1.
$$f'(x)=\frac{x}{x+1}$$
Note that |f'(-x)| > |f'(x)| for 0<x<1.
Via integration from 0 to x, we can conclude that f(-x)>f(x) for positive x.
$$V+aV = ln(1+V) - ln(1-aV)$$
$$V- ln(1+V) = -aV - ln(1+(-aV))$$
$$f(V) = f(-aV) > f(aV)$$
$$f(V) > f(aV)$$
The function f has a minimum at x=0 and is monotonically increasing for positive x.
=> a < 1