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I Proving an inequality from an equation

  1. Jun 23, 2016 #1
    Given that ##v_0## and ##v_f## are positive variables related by the equation
    Screen Shot 2016-06-24 at 4.52.38 am.png
    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##.
     
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  3. Jun 23, 2016 #2

    mfb

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    Yes. Hint: -ln(1-x) > ln(1+x) for positive x.
     
  4. Jun 23, 2016 #3
    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}##.
     
  5. Jun 24, 2016 #4

    mfb

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    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).
     
  6. Jun 24, 2016 #5
    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.
     
  7. Jun 24, 2016 #6

    mfb

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    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##.
     
  8. Jun 24, 2016 #7
    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
  9. Jun 24, 2016 #8

    mfb

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    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
     
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