Absolute quadratic inequalities.

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Discussion Overview

The discussion revolves around solving absolute quadratic inequalities, specifically focusing on the inequality |3n-4| < 9εn² + 3ε, where ε is a small positive number. Participants explore methods for breaking down the inequality and the implications for determining bounds on n, particularly in the context of delta-epsilon proofs related to limits at infinity.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses confusion about how to split the inequality into four inequalities and questions the logic behind it.
  • Another suggests factoring the quadratic and analyzing the sign of the factors as a method for solving the inequality.
  • A different participant proposes solving the corresponding equations to find boundary points that define intervals where the inequality holds true.
  • Concerns are raised about which of the solutions from the quadratic equations could serve as bounds for n in the context of a delta-epsilon proof.
  • One participant indicates that it may not be necessary to solve the inequality completely, suggesting that demonstrating the inequality holds for sufficiently large n could suffice.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to take for solving the inequality. Multiple competing views on how to handle the absolute value and the implications for n remain present throughout the discussion.

Contextual Notes

Participants express uncertainty regarding the implications of the quadratic inequalities and the conditions under which certain solutions may be valid. There is also a lack of clarity on how to effectively apply the results to the delta-epsilon context.

Kelvie
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A bit of a newbie question, but I was wondering how does one go about solving these?

For example: (I was working on a problem posted on another thread on Homework Help)
[tex] |3n-4| < 9\epsilon n^2 + 3 \epsilon[/tex]
Epsilon is a small positive number of course :P

The tricky part is when I split it up..
[tex] \begin{align*}<br /> -9\epsilon n^2 - 3n - 3\epsilon + 4 < 0 \\<br /> 9\epsilon n^2 -3n + 3\epsilon + 4 > 0<br /> \end{align*}[/tex]

Wouldn't the solution for n then be 4 inequalities? That doesn't make sense, does it?
 
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try factoring the quadratic and analyze according to the sign of the factors
 
A general method for solving inequalities is to solve the equation first. The points mark the boundaries of the intervals on which the inequality is true.

In this case, solve the equation |3n-4|= 9&epsilon;n2+ 3&epsilon;

If 3n-4> 0 that is equivalent to 3n-4= 9&epsilon;n2+ 3&epsilon; which is the quadratic equation 9&epsilon;n2- 3n+ (4+3&epsilon)= 0.

If 3n-4< 0 that is equivalent 4- 3n= 9&epsilon;n2+ 3&epsilon; which is the quadratic equation 9&epsilon;n2+ 3n- (4+3&epsilon)= 0.

Those two equations have 4 solutions which divide all real numbers into 5 intervals. You can choose one point in each interval to determine whether you get ">" or "<".
 
Hmm.. I was afraid it would come to this.

I was trying to solve this for a delta-epsilon proof of a limit at infinity (finding what N of epsilon could be that is < |n|.

I got my two quadratic equations, so technically, the smallest one could be N? or the largest? Or do all of them work?

Because there are two quadratic inequalities to solve, this puts a bound on what n could be, so it would be redundant to say that |n| is greater than all N of epsilon.

Or am I missing something very painfully obvious? :P
 
In that case, you don't really need to solve the inequality, just show that it is true for "sufficiently large" n. Certainly as soon as n> 4/3, 3n- 4 will be positive so you don't need the second of the ways you are "splitting up" the absolute value.
From [tex]9\epsilon n^2 + 3n + 3\epsilon - 4 0 0[/tex]
you can use the quadratic formula to determine where left side is larger than 0. If that is true for all n larger than some number, you are done.
 

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