Convergence of Analysis: Troubleshooting and Simplification Techniques

Click For Summary
SUMMARY

The discussion focuses on troubleshooting mathematical inequalities, specifically the expression involving the terms \(\frac{1-\frac{1}{2r}}{1+\frac{1}{r}}\) and \(\frac{1}{(1+\frac{1}{r})^{3/2}}\). Participants suggest simplifying the problem by squaring both sides and multiplying by an appropriate power of \(r\). This approach is aimed at clarifying the relationship between the two expressions and facilitating the solution of parts (iv) and (v) of the problem. The conversation emphasizes the importance of manipulation techniques in mathematical analysis.

PREREQUISITES
  • Understanding of algebraic manipulation techniques
  • Familiarity with inequalities and their properties
  • Knowledge of mathematical expressions involving rational functions
  • Basic skills in calculus, particularly in handling limits and continuity
NEXT STEPS
  • Explore techniques for manipulating inequalities in algebra
  • Learn about the properties of rational functions and their graphs
  • Study the application of squaring in solving inequalities
  • Investigate advanced algebraic techniques such as the AM-GM inequality
USEFUL FOR

Students and educators in mathematics, particularly those focusing on algebra and inequality analysis, as well as anyone seeking to enhance their problem-solving skills in mathematical expressions.

gomes.
Messages
58
Reaction score
0
Stuck on part iii, I am not sure what I've done wrong as my answer seems correct! any help wil be appreciated.

And err, I can't seem to do part (iv) and (v), any help will be appreciated, thanks!
 

Attachments

  • analysis.jpg
    analysis.jpg
    24.1 KB · Views: 372
Physics news on Phys.org
It seems that you want to show

\frac{1-\frac{1}{2r}}{1+\frac{1}{r}}\leq \frac{1}{(1+\frac{1}{r})^{3/2}}

This is equivalent to

(1-\frac{1}{2r})(1+\frac{1}{r})^{3/2}\leq 1+\frac{1}{r}

Try to work this out. Things you could do to simplify this is squaring both terms and multiplying both sides by a suitable power of r.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
913
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K