Prove that area under curve by rectangle is less than integration

Click For Summary

Discussion Overview

The discussion revolves around comparing the area under the curve of the function y=1/x with the areas of rectangles formed beneath that curve. Participants aim to demonstrate that the sum of the series 1/2 + 1/3 + ... + 1/n is less than log(n+1), exploring concepts related to harmonic series and Riemann sums.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant initially states the problem involves the curve y=x^2, suggesting a comparison with rectangles to show a relationship with the harmonic series.
  • Another participant corrects this by asserting the curve should be y=1/x, which is crucial for the area comparison.
  • A third participant proposes using Riemann sums to approximate the area under the curve, indicating that the sum of the rectangles provides a downward approximation of the area.
  • Further, a participant mentions that the sum 1/2 + 1/3 + ... + 1/n is less than or equal to log(n+1) by the definition of lower sums, suggesting that to prove strict inequality, one must subdivide further.
  • One participant acknowledges the correction regarding the function and expresses appreciation for the explanations provided.

Areas of Agreement / Disagreement

There is disagreement regarding the correct function to analyze initially, with some participants asserting y=1/x while another suggested y=x^2. The discussion remains unresolved as participants explore different approaches without reaching a consensus on the best method to prove the inequality.

Contextual Notes

Participants have not fully resolved the assumptions regarding the definitions of the functions involved or the implications of the Riemann sums. The discussion also reflects varying levels of understanding of the harmonic series and its relationship to logarithmic functions.

basil
Messages
8
Reaction score
0
hi all,

I am suppose to compare the area of curve y=x2 with rectangles beneath that curve to show that,

1/2 + 1/3 + ...+ 1/n < log(n+1)

i believe this some sort of harmonic series. Is there a way around this problem?

Regards
 
Physics news on Phys.org
basil said:
I am suppose to compare the area of curve y=x2 [...]

Don't you mean y=1/x?
 
I think it is y=1/x. So, take an arbitrary interval [0,l] and employ riemann sums: divide it in n subintervals, each delta x= l/n wide. The area of all rectangles of basis delta x and height 1/i(delta x), where i runs from 1 to n, is an approximation downwards of the area under the curve. In particular, it will be less than the area from x=delta x to x=l, which is just log n, plus the area of the first rectangle, that is 1. So, sum(from i=1 to n) 1/i <log n +1 and we are done.
 
basil said:
hi all,

I am suppose to compare the area of curve y=x2 with rectangles beneath that curve to show that,

1/2 + 1/3 + ...+ 1/n < log(n+1)

i believe this some sort of harmonic series. Is there a way around this problem?

Regards

1/2 + 1/3 + ...+ 1/n < or = log(n+1) is by definition of lower sum. To prove strict inequality subdivide more and show that the sum is bigger.
 
Oh yes, its

[itex]\frac{1}{x}[/itex]

Nevertheless, the explanation was beneficial. Thanks guys.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K