1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of a simple inequality

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove [tex] 1 - x \leq e^{-x}[/tex] for [tex]0 \leq x \leq 1[/tex] by calculus.

    3. The attempt at a solution
    Sketching shows that the statement seems to be true.

    Let's assume that
    [tex]f(x) = x + e^{-x} -1[/tex], and
    [tex]f(x) \geq 0[/tex].

    If the function is continuous and increasing, then the function gets all the
    values between the interval [0, [tex]e^{-1}[/tex]], by Boltzman's Min-Max theorem.
    Thus, the initial assumptions are true.

    Please, point out any mistakes.
     
  2. jcsd
  3. Feb 25, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Here's your basic mistake- you are assuming what you want to prove!
    [tex]f(x)\ge 0[/tex] is the same as [tex]x+ e^{-x}- 1\ge 0[/itex] which is the same as [tex]x- 1\ge -e^{-x}[/tex] and, multiplying both sides by -1, the same as [tex]1- x\le e^{-x}[/tex]

    You need to learn that your whole concept here is WRONG! You cannot assume the conclusion and then prove that the assumptions are true.
     
  4. Feb 25, 2009 #3
    I changed the proof.

    3. The attempt at a solution
    We can write that
    [tex]f(x) = x + e^{-x} -1[/tex].

    We want to prove that
    [tex]f(x) \geq 0[/tex].

    If the function is continuous and increasing, then the function gets all the
    values between the interval [0, [tex]e^{-1}[/tex]], by Boltzman's Min-Max theorem.

    Thus, the statement is true if we can prove the continuity and the fact the
    function increases.

    The function is continuous.

    Proof:
    At any point [tex] c \in \Re, f(c)[/tex] is defined.
    The limit of [tex]f(x)[/tex] exists as [tex]x[/tex] approaches [tex]c[/tex]
    either from the left or from the right, and it equals [tex]f(c)[/tex].

    The function is clearly increasing, since [tex]f(0) < f(1)[/tex], and any point [tex]k[/tex] in the
    interval [0, 1] is less or equal to f(1) that is [tex]f(k) \leq f(1)[/tex].

    Thus, the statement must be true.

    Please, point out any mistakes.
     
  5. Feb 25, 2009 #4

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    You have not proved that as far as I can see.

    You were asked to prove it by calculus.

    You have not used any, but doing so would make it easy.
     
  6. Feb 25, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Have you proved that limit? It might be that you are allowed to assert that ex is continuous and so f(x) is continuous. If not, you wil have to give a more detailed proof than simply stating that the limit is the value of the function. What, exactly, are you allowed to assume about ex for this problem?

    Again, you cannot simply assert that "any point [tex]k[/tex] in the interval [0, 1] is less or equal to f(1) that is [tex]f(k) \leq f(1)[/tex]" Nor would proving that statement tell you anything useful; f might start out at f(0)= 0 drop below 0 before going back up to [itex]e^{-1}[/itex]. Simply showing that "f(0)< f(1) and f(x)< f(1) for all x in [0, 1]" does NOT show that the function is increasing.

    What is f'(x)? What is true about f'(x) for x between 0 and 1? What does that tell you?

     
  7. Feb 25, 2009 #6
    I get
    [tex]f'(x) = 1 - e^{-x}[/tex]

    I see that
    [tex]f'(0) = 0[/tex] and

    [tex]f'(1) = 1 - e^{-1}[/tex],
    which is greater than 0.

    I note that f'(x) is the interval (0, 1] greater than zero.
    This suggests me that the function is increasing.
    (Is it this enough to prove that the function increases?)

    It seems to be hard to prove for the general event.
    I will try nevertheless.

    We have a problem
    [tex] lim_{x -> c} f(x) [/tex],
    where [tex]c[/tex] is a real number and belongs to the interval [0, 1].

    Proof:

    Let [tex]\epsilon > 0[/tex].
    We need to find [tex]\delta > 0[/tex] such that
    [tex] 0 < |x - c| < \delta => |(x + e^{-x} -1) - C| < \epsilon[/tex],
    where [tex]f(c) = C[/tex].

    The choice is of [tex]\delta[/tex] is hard because we have a general proof.
    Perhaps, we should prove first that the minimum point has a limit.
    Then, we could similarly say that the maximum point has a limit.
    Thus, it may be possible to say that each point in the interval has a limit.

    Please, pinpoint my mistakes.
     
  8. Feb 26, 2009 #7

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    In a word, yes.
    At least as long as the function exists and is continuous in the interval of interest.

    Something starts as 0 and increases ever after, it must be > 0 ever after surely QED? - I cannot see your problems unless it is that you have to express that simple idea in some difficult language, in which case I have butted in, but I hope the difficult language does not stop you seeing the simple ideas. :smile:
     
  9. Feb 27, 2009 #8
    Please, let me know how I should do the proof correctly.
    Do I need to use epsilon and delta in the proof of the limit?
     
  10. Feb 28, 2009 #9

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    I don't see you need epsilon, delta, or limit!
    You are only trying to prove something is bigger than 0. If something increases it gets bigger.

    Say your proof in your own words. If those are not the words you are supposed to say let your Prof. or someone here correct them.
     
    Last edited: Mar 1, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof of a simple inequality
Loading...