Need help in manipulating rational absolute value inequalities

[bamajon1974]

Homework Statement:

For context, this question is not for homework but in reference to bounding terms for epsilon-delta limit proofs. I know you can bound the numerator and denominator separately but I really would like to know how manipulate the quotient together as I have forgotten some elementary algebra.

Relevant Equations:

Please see below.

How does one manipulate rational absolute inequalities?

For example, I want to transform the absolute value inequality $|x-3|<1$ to $\frac{|x+3|}{5x^2}<A \$, for some number $\text{A}$, to find an upper and lower bound on the latter term using the constraint in the first term, and not sure what to do with the denominator and changing inequality direction.

I can expand the absolute value inequality as follows: $|x-3|<1 \implies -1<x-3<1 \implies 2<x<4 \implies 5<x+3<7.$ How do I introduce the $5x^2$ term in the denominator? Dividing all three sides by $5x^2$ would add a variable to the numbers and possibly change signs. Not sure where to go from here.

For context, this question is not for homework but in reference to bounding terms for epsilon-delta limit proofs. I know you can bound the numerator and denominator separately but I really would like to know how manipulate the quotient together as I have forgotten some elementary algebra.

Thanks!

 

[PeroK]

Your lower bound for $x$ implies an upper bound for $\frac 1{5x^2}$

 

[BvU]

Dividing all three sides by 5x²

Sounds like a bad idea. since $5x^2$ is positive (*), multiplying seems more logical: $|x+3|<5x^2$ and you have to study two cases:$$x\le-3 \ \& \ 5x^2 +x+3>0 \quad \Rightarrow x\le -3$$and$$
x>-3 \ \&\ 5x^2-x-3 >0 $$which requires solving a quadratic equation

(*) Edit: Oops, have to exclude $x=0$

$\$

 

[bamajon1974]

Your lower bound for $x$ implies an upper bound for $\frac 1{5x^2}$

Your lower bound for $x$ implies an upper bound for $\frac 1{5x^2}$

I made a mistake and $\frac{|x+3|}{5x^2}<A$ for some number A to be determined, not 1.

I am guessing you would have to consider the numerator and denominator separately. So for the numerator, $5<x+3<7$ and for the denominator, $\frac{1}{80}<\frac{1}{5x^2}<\frac{1}{20}.$

Then, can I combine the two expressions, i.e., $\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20}$ and, to find an upper bound? That is, to produce $-\frac{7}{20}<\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies \frac{|x+x|}{5x^2}<\frac{7}{20}?$ It seems arbitrary to separate the numerator and denominator.

Thanks.

 

[bamajon1974]

Sounds like a bad idea. since $5x^2$ is positive (*), multiplying seems more logical: $|x+3|<5x^2$ and you have to study two cases:$$x\le-3 \ \& \ 5x^2 +x+3>0 \quad \Rightarrow x\le -3$$and$$
x>-3 \ \&\ 5x^2-x-3 >0 $$which requires solving a quadratic equation

(*) Edit: Oops, have to exclude $x=0$

$\$

I made a mistake in the original post and corrected. What I want to do is use $$|x-3|<1$$ to bound $\frac{|x+3|}{5x^2}$ and equal to some number.

 

[bamajon1974]

I made a mistake in the original post and corrected. What I want to do is use $$|x-3|<1$$ to bound $\frac{|x+3|}{5x^2}$ and equal to some number. Does that clarify?

 

[PeroK]

It seems arbitrary to separate the numerator and denominator.

What's not clear is why this is an issue. Bounding both is precisely what you want to do.

 

[bamajon1974]

What's not clear is why this is an issue. Bounding both is precisely what you want to do.

Ok then I should clarify...I understand the process of bounding. The question I have is the algebra involved in bounding, i.e., is it legal to find the bounds f the numerator and denominator separately or as a single quotient. In the end, I found my bounds but the issue was just a side question I asked,

 

[SammyS]

Ok then I should clarify...I understand the process of bounding. The question I have is the algebra involved in bounding, i.e., is it legal to find the bounds f the numerator and denominator separately or as a single quotient. In the end, I found my bounds but the issue was just a side question I asked,

Remember that $|a| |b|=|ab|$.

Therefore, $\displaystyle \left| \frac{1}{5x^2} \right| |x+3| = \left| \frac{x+3}{5x^2} \right|$.

 

[PeroK]

is it legal to find the bounds of the numerator and denominator separately or as a single quotient.

Whatever works best. As long as you follow the rules of algebra there's no issue.

By the way, it's often possible to simplify these calculations by taking a round number as the bound. E.g. If we have $2 < x < 4$, then $\frac 1 {5x^2} < \frac 1 {20} < 1$.

And, as $| x + 3| < 7$, we have:$$|x - 3| < 1 \ \Rightarrow \ \frac{|x + 3|}{5x^2} < 7$$Which is just as good as $\frac 7 {20}$.

 

[bamajon1974]

Remember that $|a| |b|=|ab|$.

Therefore, $\displaystyle \left| \frac{1}{5x^2} \right| |x+3| = \left| \frac{x+3}{5x^2} \right|$.

I think I see now. Equivalently, $|a/b| = |a|/|b|.$ In the context of my question, it is $|a|/b$, since $|a|=|x+3|$ and $|b|$ is $|5x^2|=5x^2$ which always positive. Given that $|x-3|<1$ and $|a/b| = |a|/|b|.$, I can find upper and lower bounds of the numerator and denominator separately and combine them back into one single absolute value inequality, right?

Maybe something like this?

If $$|x-3|<1, $$ then $$|x-3|<1 \implies -1<x-3<1 \implies 2<x<4 \implies 5<x+3<7$$ for the numerator and $$\frac{1}{4}<\frac{1}{x}<\frac{1}{2} \implies \frac{1}{16}<\frac{1}{x^2}<\frac{1}{4} \implies \frac{1}{80}<\frac{1}{5x^2}<\frac{1}{20}$$ for the denominator.

Dividing produces $$\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies -\frac{7}{20}<\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies \frac{|x+3|}{5x^2}<\frac{7}{20}.$$

 

[PeroK]

Yes, that's the idea.