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I Prove ln(x) <= x-1 for positive x

  1. Jan 15, 2017 #1
    proof ##\ln x##<=x-1 for positive x.
    if i show that at the start of the segment, the value in the start of the segment (0, +inf), of inx is smaller than x-1, and the derivative of inx is always smaller than x-1, is that suffice to proof that inx is always smaller than x-1? if not why?
    Last edited: Jan 15, 2017
  2. jcsd
  3. Jan 15, 2017 #2


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    What is inx?
  4. Jan 15, 2017 #3
    Natural logarithm ;)
  5. Jan 15, 2017 #4


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  6. Jan 15, 2017 #5
  7. Jan 15, 2017 #6
  8. Jan 15, 2017 #7
    solved. thanks
  9. Jan 15, 2017 #8


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    But the derivative of ln(x) is 1/x, which gets huge near 0.
  10. Jan 15, 2017 #9
    that's right, so:
    f(x) = x-1 - Inx . f''(x) = 1 - 1/x = 0 ==> x=1. that should be either maximus or minimum
    f(1) = 0.
    now if we take the second derivative:
    f''(x) = 1/x^2
    which means the derivative is increasing constantly and f(1) is the minimum point inthe graph, and hence x-1>=Inx
  11. Jan 15, 2017 #10


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    It looks like you are ok here.

    In my view, the slickest and most intuitive approach to this by far is to use Jensen's Inequality-- as the logarithm is (strictly) negative convex. If you need to, you can derive Jensen's Inequality, and the above relation falls out of it.... that is, construct a Taylor Polynomial for the logarithm at x = 1, with the remainder term being O(n^2). I'd suggest using the Lagrange form for the remainder term. Then use the fact that the logarithm has a continuously negative second derivative. Graphically, this means ln(x) will always be below the tangent line for said logarithm evaluated at x = 1, with equality only where x = 1.
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