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Logaritmetic problem

  1. May 8, 2007 #1
    show that (for all posetive x)
    x-1/2 x^2<ln(1+x)<x

    i understand it in graf please prove it for my in methemathic
     
  2. jcsd
  3. May 8, 2007 #2
    does the 1st inequality mean [tex] x-\frac{x^2}{2} < \ln (1+x) [/tex] ?
    i suppose this will be true for only |x|<1
     
    Last edited: May 8, 2007
  4. May 8, 2007 #3
    yes you right x - (x^2)/2
    and in the book (calculus , rechard a.silverman) has writen for x>0
    Do you need the adress of problem?
     
  5. May 9, 2007 #4
    ok sorry about that...the problem's fine , i mistook the signs.
    The identity may be proved by using the expansion series for ln(1+x) but this holds for |x|<1.
     
  6. May 9, 2007 #5

    VietDao29

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    Homework Helper

    Or, you can try to take the derivative and prove that it's either an increasing function, or a decreasing function (Have you covered how to differentiate a function?). It goes like this:

    I am going to prove the ln(1 + x) < x for all x > 0 part.
    Re-arrange the inequality gives:
    ln(1 + x) - x < 0

    Let f(x) = ln(1 + x) - x
    Now, we have that f(0) = ln(1 + 0) - 0 = 0
    [tex]f'(x) = \frac{1}{1 + x} - 1[/tex]
    For x > 0, we have 1 + x > 1, hence 1 / (x + 1) < 1, hence 1 / (x + 1) - 1 < 0, so f'(x) < 0 for all x > 0. f(x) is a decreasing function for x > 0.

    So, for x > 0, we have:
    f(x) < f(0) <=> f(x) < 0 <=> ln(1 + x) - x < 0 <=> ln(1 + x) < x
     
  7. May 12, 2007 #6
    thank you very much but how we can prove the other side
     
  8. May 12, 2007 #7

    VietDao29

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    Homework Helper

    Have you tried anything? It's pretty much the same.
     
  9. May 14, 2007 #8
    f(x)=ln(x+1)-x+x^2/2
    f'(x)=1/(x+1)-1+x=x^2/(x+1)>0
    f'(x)>0 then it is increasing
    So, for x > 0, we have:
    f(x) > f(0) <=> f(x) > 0 <=> ln (x+1)-x+x^2/2>0 <=> ln (x+1)>x-x^2/2
    is this proof Ok?
     
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