# Natural logarithmic function problem

1. Jan 12, 2006

### Xcron

This is the problem:

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

(B) Use the Midpoint Rule with $$n=10$$ to esimate $$\ln1.5$$.

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 $$e^x$$ 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 $$f(x)$$ type of function to work with for the Riemann sum type of set-up...

Last edited: Jan 12, 2006
2. Jan 12, 2006

### benorin

Recall that

$$\ln a=\int_{x=1}^{a} \frac{1}{x}dx$$

and since $$\frac{1}{x}$$ 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
3. Jan 12, 2006

### d_leet

Wouldn't that be the other way around because $$\frac{1}{x}$$ is strictly decreasing for x > 0. So the left hand end point sums should be greater than those taken at the right hand.

4. Jan 12, 2006

### benorin

Yep, I was thinking of ln(x) being increasing, sorry: my bad.

5. Jan 13, 2006

### Xcron

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:
$$\frac{1}{10}[f(\frac{40.5}{120})+f(\frac{41.5}{120})+...+f(\frac{49.5}{120})]$$

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
$$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\ll\ln n\ll 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}$$

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 $$\ln n$$ is between those two...

Last edited: Jan 13, 2006
6. Jan 13, 2006

### krab

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).

7. Jan 14, 2006

### Xcron

Can anyone help me with the Midpoint Rule problem...and the new Riemann sum problem please?

8. Jan 14, 2006

### Xcron

I think that for the Midpoint Rule problem...I would have sub-intervals(10) represented by $$\frac{1}{10}$$, which would be in front of the Riemann sum. I probably should not use the $$f(\frac{40.5}{120})$$...because they seem a bit strange...I'm trying to figure out how to estimate $$ln 1.5$$ while doing a sum of multiple values....*sigh*..

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