Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

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


    User Avatar
    Homework Helper

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


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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


    User Avatar
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted