Proving Log 2 with Riemann Sums

In summary, the author is having trouble proving that log2 is equal to a transcendental function. He has been working on the problem for a while and has come up with a way to do it using Riemann sums, but he is unsure about how to do it. However, somebody with more experience in this area may be able to help him out.
  • #1
tonix
18
0
I have been working on this problem for a while.
I am supposed to prove that [tex]
log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2^n}[/tex].

The problem is that I have a hard time figuring out how I am supposed to prove that something is equal to a transcendental function without assuming its existence.

First, I am supposed to let
[tex]
\lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2^n} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}[/tex]

So far so good... but then I should use Riemann sums to prove that this is equal to log 2. How can I do that?
 
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  • #2
your bracketing seems off, you mean the terms are like 1/(n+1), right.
 
  • #3
matt grime said:
your bracketing seems off, you mean the terms are like 1/(n+1), right.

I changed the equations to latex, hope it is clear now.
 
  • #4
I think you mean:

[tex]
\log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}[/tex]

Can you write log(2) in terms of an integral? How did you define the function log(x)?
 
  • #5
shmoe said:
I think you mean:

[tex]
\log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}[/tex]

Can you write log(2) in terms of an integral? How did you define the function log(x)?

No, I (or better the author of the book) mean [tex]2^n[/tex]. The problem is given just like that. log(x) is just defined the usual way.
 
  • #6
The reason shmoe said 2n and not 2 to the n is that you rewriting of the series in that second form does not agree with the original one.
 
  • #7
matt grime said:
The reason shmoe said 2n and not 2 to the n is that you rewriting of the series in that second form does not agree with the original one.

argh...
yes. I just checked the errata and it indeed is 2n and not [tex]2^n[/tex].
Well, given that it is 2n, does anybody have any ideas how to prove it using Riemann sums.
 
  • #8
What is

[tex]\int_1^2 \frac{1}{x} dx[/tex]

?

What is the Riemann sum of the above integral?
 
  • #9
Data said:
What is

[tex]\int_1^2 \frac{1}{x} dx[/tex]

?

What is the Riemann sum of the above integral?
um, what is n?

or do you just want to know the answer without using Riemann?
 
  • #10
for arbitrary n of course.
 
  • #11
ah... I get it now. Thanks.
 

1. What is a Riemann Sum?

A Riemann Sum is a mathematical method used to approximate the area under a curve by dividing the area into smaller, simpler shapes and calculating the sum of their areas.

2. How does a Riemann Sum relate to logarithms?

A Riemann Sum can be used to approximate the value of a logarithm by dividing the interval between 1 and the desired value into smaller subintervals and calculating the sum of their widths.

3. What is the relationship between Riemann Sums and limits?

Riemann Sums are used to approximate the area under a curve, and as the number of subintervals approaches infinity, the approximation gets closer to the actual area. This is similar to how limits are used to find the exact value of a function at a specific point.

4. How can Riemann Sums be used to prove Log 2?

By using a Riemann Sum with a specific choice of subintervals and widths, it is possible to approximate the area under the curve of the logarithmic function y=log(x) between x=1 and x=2. This approximation can then be used to prove that log(2) is equal to a specific value.

5. What are the limitations of using Riemann Sums to prove logarithms?

Riemann Sums can only provide an approximation of the value of a logarithm, and the accuracy of the approximation depends on the number of subintervals and their widths. Also, some logarithmic values cannot be expressed exactly as a sum of subintervals, making it impossible to prove their exact value using Riemann Sums.

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