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Natural logarithmic function problem

  1. Jan 12, 2006 #1
    This is the problem:

    (A) By comparing areas, show that
    [tex]\frac{1}{3}\ll\ln1.5\ll\frac{5}{12}[/tex]

    (B) Use the Midpoint Rule with [tex]n=10[/tex] to esimate [tex]\ln1.5[/tex].


    I've seen these types of "comparing areas" problems but I kind of forgot how to go about solving the problem...I was thinking perhaps using [tex]e^x[/tex] on the three parts but that was a plausible solution to this, really...the midpoint part is also somewhat strange because all three are constants so...there is really an [tex]f(x)[/tex] type of function to work with for the Riemann sum type of set-up...

    Any help/advice?
     
    Last edited: Jan 12, 2006
  2. jcsd
  3. Jan 12, 2006 #2

    benorin

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    Homework Helper

    Recall that

    [tex]\ln a=\int_{x=1}^{a} \frac{1}{x}dx[/tex]

    and since [tex] \frac{1}{x} [/tex] is an increasing function of x, one may say that the above integral is bounded above by the sum using righthand end-points and below by the sum taken with lefthand end-points. end-points.
     
    Last edited: Jan 12, 2006
  4. Jan 12, 2006 #3
    Wouldn't that be the other way around because [tex] \frac{1}{x} [/tex] is strictly decreasing for x > 0. So the left hand end point sums should be greater than those taken at the right hand.
     
  5. Jan 12, 2006 #4

    benorin

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    Yep, I was thinking of ln(x) being increasing, sorry: my bad.
     
  6. Jan 13, 2006 #5
    I'm trying to figure out how to use the Midpoint Rule for part (B) and I seem to be getting the wrong answer. The point of the rule is to use midpoints with 10 sub-intervals...so I thought I could just use:
    [tex]\frac{1}{10}[f(\frac{40.5}{120})+f(\frac{41.5}{120})+...+f(\frac{49.5}{120})][/tex]

    But that doesn't seem to be working...hmmm...


    The next problem is very similar and I am somewhat confused with it as well...the problem says:

    By comparing areas, show that
    [tex]\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\ll\ln n\ll 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}[/tex]

    I thought I could uise a riemann sum for the comparison since I fugred if I could show that the very left part is less than the very right (because the right part has a 1 added to it at the very beginning). However, I'm not sure how to show that [tex]\ln n[/tex] is between those two...
     
    Last edited: Jan 13, 2006
  7. Jan 13, 2006 #6

    krab

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    Science Advisor

    Why are you using the double inequality sign? This means MUCH less than, or MUCH greater than. Just use the symbols directly (Shift-, or Shift-. on American keyboards).
     
  8. Jan 14, 2006 #7
    Can anyone help me with the Midpoint Rule problem...and the new Riemann sum problem please?
     
  9. Jan 14, 2006 #8
    I think that for the Midpoint Rule problem...I would have sub-intervals(10) represented by [tex]\frac{1}{10}[/tex], which would be in front of the Riemann sum. I probably should not use the [tex]f(\frac{40.5}{120})[/tex]...because they seem a bit strange...I'm trying to figure out how to estimate [tex]ln 1.5[/tex] while doing a sum of multiple values....*sigh*..

    For the second "comparing areas" problem...I think I might be able to set up:
    [tex]\int_{0}^{1} \frac{1}{x}dx - 1 < \int_{0}^{1} \frac{1}{x}dx <
    \int_{0}^{1} \frac{1}{x}dx - [/tex] ?? .... I'm not sure...
     
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