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Question involving Logs

  1. Feb 6, 2005 #1
    I was trying to do some homework problems and one of the questions i didn't get was this: Show that x-1/x </= ln x </= x-1 for all x>0. Using mean value thm.

    Now in class teacher gave us a lemma that would help solve this problem, which is:
    Suppose that f(a)=g(a), and that f & g are continuous on [a, infinity). Suppose also that f'(x)<g'(x) for all x in (a, infinity). Then f(x) < g(x) for all x in (a, infinity).

    The only thing i understand for now is that, by graphing the two functions, there seems to be a relation between the gaps of the two functions crossing at a single point, say a. So on the right side of the cross section, one function increases faster while other decreases, or so, etc. but how would i prove this using the lemma, i dont even have proof of the lemma.
  2. jcsd
  3. Feb 6, 2005 #2


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    I take it that for the first function you mean (x-1)/x, because x-(1/x) doesn't satisfy that inequality. If so, note that the functions are all tangent at x=1. As you move right from a tangent point, what has to be true of the slopes of two functions to ensure one is greater than the other? (this is the lemma your teacher was talking about) What about as you move left?
  4. Feb 6, 2005 #3
    as u move left, the slopes of the two functions would be increasing and/or decreasing really fast relative to one another....
  5. Feb 6, 2005 #4


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    I don't know what you mean by that, but just look at the graphs. For f(x) to be less than g(x) away from their tangent point, it must increase slower than g as you move right, and decrease faster as you move left. The rate of increase is just the derivative. If you're looking for a proof of this fact, you can write f(x) as the integral of f'(x) and g(x) as the integral of g'(x), and then use the fact that if one integrand is greater than another at all points on an interval, its integral over that interval must be greater as well.
  6. Feb 6, 2005 #5
    can u tell me if there is any specific definitions involved in solving or proving this, i mean i understand what your saying, but it would be better if i kno what definitions or facts i can use to solve it too u kno
  7. Feb 6, 2005 #6


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    I've given you all I think you need. You mentioned the mean value theorem, but I can't see how that would fit in. The only equation I can give you that might help is:

    [tex] \mbox{if} \ f(x)>0 \ \mbox{for} \ a<x<b, \ \mbox{then} \ \int_a^b f(x) dx > 0 [/tex]

    Substitute in for f(x) the difference between the derivatives of two functions, and if that difference is always positive, the difference between the functions is positive as well.
  8. Feb 7, 2005 #7


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    A previous msg suggested using the Fundamental Theorem of Calculus for your proof. However, the Mean Value Theorem is more direct.

    If functions f(x) and g(x) are differentiable on r<x<s and there exist points "a" and "b", where r<a<s and r<b< s, such that f(a)=g(a), then we can define the function d(x)={f(x) - g(x)} which is differentiable on r<x<s. The Mean Value Theorem guarantees the existence of another point "c" where a<c<b such that:

    [tex] d'(c) = \frac {d(b) - d(a)} {b - a} [/tex]

    Substituting for the function d(x), collecting terms, and using f(a)=g(a), we get:

    [tex] f'(c) - g'(c) = \frac {f(b) - g(b)} {b - a} [/tex]

    Since point b was arbitrary, we get:
    a) If b>a, then f'(x)>g'(x) for all r<x<s IMPLIES f(x)>g(x) for all r<x<s
    b) If b<a, then f'(x)>g'(x) for all r<x<s IMPLIES f(x)<g(x) for all r<x<s

    The above proves the stated Lemma. The proof for your problem statement involving logs would then proceed like this:

    We define the following:
    f(x) = (x - 1)/x
    -----> f'(x) = 1/(x^2)
    g(x) = ln(x)
    -----> g'(x) = 1/x
    h(x) = x - 1
    -----> h'(x) = 1

    Note that:
    At x = 1:
    f(1) = (1 - 1)/1 = 0
    g(1) = ln(1) = 0
    h(1) = 1 - 1 = 0

    Thus, all functions are equal at x=1. Next, compare derivatives:
    For x > 1:
    f'(x) = 1/(x^2) < 1/x = g'(x) for all x>1
    -----> f'(x) < g'(x) for all x>1
    g'(x) = 1/x < 1 = h'(x) for all x>1
    -----> g'(x) < h'(x) for all x>1

    Hence, by the Lemma:
    f(x) < g(x) < h(x) for all x>1
    (x - 1)/x < ln(x) < (x - 1) for all x>1

    What about the interval {0 < x < 1}?
    We already know the functions are EQUAL at x=(1), so we need to compare functions for 0<x<1, and begin by comparing their derivatives:
    For 0 < x < 1:
    f'(x) = 1/(x^2) > 1/x = g'(x) for all 0<x<1
    -----> f'(x) > g'(x) for all 0<x<1
    g'(x) = 1/x > 1 = h'(x) for all 0<x<1
    -----> g'(x) > h'(x) for all 0<x<1

    Hence, by the Lemma modified for points LESS than 1:
    f(x) < g(x) < h(x) for all 0 < x < 1
    (x - 1)/x < ln(x) < (x - 1) for all 0 < x < 1

    Proof is complete for 0 < x < Infinity
    [tex] Q.E.D. [/tex]
    Last edited: Feb 7, 2005
  9. Feb 7, 2005 #8
    What do you know about Integration?

    Do you know that [tex] inf[L_{f}(P)] \leq \int_{1}^{x} \frac{1}{t} dt \leq sup[U_{f}(P)] [/tex]
    where inf means greatest lower bound and
    sup means least upper bound and
    Lf (P) = least value of the function on that interval times the width of the interval while
    Uf(P) means greatest values of the function over that interval times the width of the interval

    and if you chose a partition P[0,b) for b>0 i'm sure you could prove it using much less paper :smile: and much more efficiently too.
    Last edited: Feb 7, 2005
  10. Feb 7, 2005 #9


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    Your comments are acknowledged.
    However, in this situation, the Mean Value Theorem is the best and most direct path to the required proof. The essential proof is completed in 9 lines. The remainder provides details required for the specific homework problem. Don't try to cross a stream with the Titanic when a canoe will do!! :wink:
    Last edited: Feb 7, 2005
  11. Feb 7, 2005 #10
    i was able to finish the proof in around 15 lines, cuz teacher says reasons are most important, but xanthym's detailed version involves everything or more than needed for the solution, thnx alot

    also i tried using partition and lower/upper sums but it seems to require more work,
    Last edited: Feb 7, 2005
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