Solving an inequality for a change of variables

In summary: Attempt:Can you see/guess what the answer must be? And then justify it?The answer is found by solving a system of two linear equations. The equations are as follows: y = mx + bwherem = -y_2 - y_1b = y_2 + y_1The solution is found by solving for m and b. The general method to solve this system of equations is to use the Quadratic Formula.
  • #1
Master1022
611
117
Homework Statement
If we have ## 0 < x_1 < \infty ## and ## 0 < x_2 < \infty ## and the transformations: ## y_1 = x_1 - x_2 ## and ## y_2 = x_1 + x_2 ##, find inequalities for each of ##y_1## and ##y_2##
Relevant Equations
Inequalities
Hi,

This is as part of a larger probability change of variables question, but it was this part that was giving me problems.

Question: If we have ## 0 < x_1 < \infty ## and ## 0 < x_2 < \infty ## and the transformations: ## y_1 = x_1 - x_2 ## and ## y_2 = x_1 + x_2 ##, find inequalities for each of ##y_1## and ##y_2##

Attempt:
Is there a general method to do these? The answer seems a bit arbitrary and was basically just stated and wasn't obvious to me how it was arrived at. Here is how I would attempt this:
[tex] 0 < x_2 < \infty \rightarrow 0 < \frac{y_2 - y_1}{2} < \infty [/tex]
and we can use the left inequality to get ## y_1 < y_2 ## and the right one to get ## y_2 < y_1 + \infty \rightarrow y_2 < \infty ##. Combining these give ## y_1 < y_2 < \infty ##.

Now for ## y_1##:
[tex] 0 < x_1 < \infty \rightarrow 0 < \frac{y_1 + y_2}{2} < \infty [/tex]
which we can break up into the left and right parts ## 0 < \frac{y_1 + y_2}{2} ## and ## \frac{y_1 + y_2}{2} < \infty ##. After subtracting ## y_2 ## from both sides, these yield ## -y_2 < y_1 ## and ## y_1 < \infty - y_2 \rightarrow y_1 < \infty ##. However, we can further restrict the upper bound of ## y_1 ## by using the inequality for ## y_2## to give: ## -y_2 < y_1 < y_2 ##.

How do I know whether this is sufficient for a solution (if this is even correct)? It just seems a bit arbitrary to me...

Any help would be greatly appreciated.
 
Physics news on Phys.org
  • #2
At first for each independently
[tex]?<y_1<?[/tex]
[tex]?<y_2<?[/tex]
Then compare ##y_1## and ##y_2##.
 
Last edited:
  • Like
Likes Master1022
  • #3
Master1022 said:
Homework Statement:: If we have ## 0 < x_1 < \infty ## and ## 0 < x_2 < \infty ## and the transformations: ## y_1 = x_1 - x_2 ## and ## y_2 = x_1 + x_2 ##, find inequalities for each of ##y_1## and ##y_2##
Relevant Equations:: Inequalities

Hi,

This is as part of a larger probability change of variables question, but it was this part that was giving me problems.

Question: If we have ## 0 < x_1 < \infty ## and ## 0 < x_2 < \infty ## and the transformations: ## y_1 = x_1 - x_2 ## and ## y_2 = x_1 + x_2 ##, find inequalities for each of ##y_1## and ##y_2##

Attempt:
Is there a general method to do these? The answer seems a bit arbitrary and was basically just stated and wasn't obvious to me how it was arrived at. Here is how I would attempt this:
[tex] 0 < x_2 < \infty \rightarrow 0 < \frac{y_2 - y_1}{2} < \infty [/tex]
and we can use the left inequality to get ## y_1 < y_2 ## and the right one to get ## y_2 < y_1 + \infty \rightarrow y_2 < \infty ##. Combining these give ## y_1 < y_2 < \infty ##.

Now for ## y_1##:
[tex] 0 < x_1 < \infty \rightarrow 0 < \frac{y_1 + y_2}{2} < \infty [/tex]
which we can break up into the left and right parts ## 0 < \frac{y_1 + y_2}{2} ## and ## \frac{y_1 + y_2}{2} < \infty ##. After subtracting ## y_2 ## from both sides, these yield ## -y_2 < y_1 ## and ## y_1 < \infty - y_2 \rightarrow y_1 < \infty ##. However, we can further restrict the upper bound of ## y_1 ## by using the inequality for ## y_2## to give: ## -y_2 < y_1 < y_2 ##.

How do I know whether this is sufficient for a solution (if this is even correct)? It just seems a bit arbitrary to me...

Any help would be greatly appreciated.
This is perhaps the wrong approach - and may lead you round in circles. This is a case where you are better to see/guess the answer and then demonstrate that it is the case.

Can you see/guess what the answer must be? And then justify it?
 
  • #4
PeroK said:
This is perhaps the wrong approach - and may lead you round in circles. This is a case where you are better to see/guess the answer and then demonstrate that it is the case.

Can you see/guess what the answer must be? And then justify it?
Okay thank you for the reply @PeroK . The answer reads:
"Firstly we know that:
[tex] 0 < y_1 + y_2 < \infty [/tex]
[tex] 0 < y_2 - y_1 < \infty [/tex]
This can be re-written as:
[tex] -y_2 < y_1 < y_2 [/tex]
[tex] 0 < y_2 < \infty [/tex] "

So not the most descriptive solution... Also, the wider problem was this (shown in image below), but it was only the "finding the support of y1 and y2" part I was struggling with - I am not sure if that helps to provide more context.

Screen Shot 2021-06-05 at 9.04.26 AM.png
 
  • Wow
Likes PeroK
  • #5
anuttarasammyak said:
At first for each independently
[tex]?<y_1<?[/tex]
[tex]?<y_2<?[/tex]
Then compare ##y_1## and ##y_2##.
Thanks for your reply @anuttarasammyak . I am not sure how I could obtain expressions independently in this case? Could you give an example of how I could obtain an inequality for ##y_1## different to what I had?
 
  • #6
Try ##y_2## first. Is it positive or negative ?
 
  • #7
anuttarasammyak said:
Try ##y_2## first. Is it positive or negative ?
positive. The upper bound is ## \infty ##. Then perhaps by adding two inequalities, I can get ## 0 < 2 y_2 \rightarrow 0 < y_2 ##. Then if I wanted to try to something for ## y_1 ##, I am not sure what to do (other than what I previously did), because I am not completely sure how subtracting the inequalities work
 
  • #8
OK. Do you see minimum or maximum of ##y_1## ?
 
  • #9
@Master1022 so far from being pre-calculus mathematics, this is a much more advanced question on probability distributions!

Your approach still amounts to going round in circles. I've give you the answer to your initial question, as that is trivial compared to the actual question you are to attempt:$$0 < y_2 < \infty, \ \ -\infty < y_1 < y_2$$You should try to justify that answer if you can.
 
  • Like
Likes Master1022
  • #10
PeroK said:
@Master1022 so far from being pre-calculus mathematics, this is a much more advanced question on probability distributions!

Your approach still amounts to going round in circles. I've give you the answer to your initial question, as that is trivial compared to the actual question you are to attempt:$$0 < y_2 < \infty, \ \ -\infty < y_1 < y_2$$You should try to justify that answer if you can.
Okay thank you, I will aim to justify that answer. Apologies, yes I wasn't too sure where to put this question because I wasn't going to ask about the calculus parts. However, next time I have a question from this topic, I will put it into the calc. and beyond section
 
  • Like
Likes anuttarasammyak
  • #11
anuttarasammyak said:
OK. Do you see minimum or maximum of ##y_1## ?
From the ## 0 < y_2 - y_1 ##, I can see a maximum for ## y_1 ## of ## y_2 ## (i.e. ## y_1 < y_2 ##)
 
  • #12
I said estimate them independently first. Then compare. Step by step.
 
  • #13
anuttarasammyak said:
I said estimate them independently first. Then compare. Step by step.
The upper bound for ## y_1 ## is ## \infty ##.
 
  • #14
I just had a go at graphing the four inequalities (with ##y_1## and ##y_2##) and those help me see the limits more clearly. I think that helps to double check the seemingly arbitrary expressions.
 
  • #15
OK. And the lower one ?

[EDIT]
[tex]-\infty<y_1<\infty[/tex]
[tex]0<y_2<\infty[/tex]

[tex]y_1<y_2[/tex]
also
[tex]-y_1<y_2[/tex]
So in one
[tex]|y_1|<y_2[/tex]
You were already near to it in post #4.
 
Last edited:
  • #16
Master1022 said:
How do I know whether this is sufficient for a solution (if this is even correct)? It just seems a bit arbitrary to me...

Any help would be greatly appreciated.
Your work and answer look good to me. It is systematic and thorough.

Just to be sure, I ran a little simulation using random values for y1 and y2 and the satisfaction of the y constraints match the satisfaction of the x constraints.
 
  • Informative
Likes Master1022

Related to Solving an inequality for a change of variables

1. How do I solve an inequality for a change of variables?

To solve an inequality for a change of variables, you first need to identify the variable you want to change and the desired variable you want to use. Then, you can use algebraic techniques such as substitution or rearranging terms to isolate the variable and solve for it.

2. What are the steps to solving an inequality for a change of variables?

The steps to solving an inequality for a change of variables are:

  1. Identify the variable you want to change and the desired variable you want to use.
  2. Use algebraic techniques to isolate the variable you want to change.
  3. Substitute the desired variable into the original inequality.
  4. Solve the inequality using the new variable.
  5. Check your solution by substituting it back into the original inequality.

3. Can I change variables in both sides of an inequality?

Yes, you can change variables in both sides of an inequality as long as you maintain the same inequality symbol. For example, if you change the variable x to y in the inequality x > 3, the new inequality will be y > 3.

4. What are the common mistakes to avoid when solving an inequality for a change of variables?

Some common mistakes to avoid when solving an inequality for a change of variables include:

  • Forgetting to maintain the same inequality symbol when changing variables.
  • Not checking your solution by substituting it back into the original inequality.
  • Incorrectly isolating the variable you want to change.
  • Making a mistake when substituting the new variable into the original inequality.

5. Can I solve an inequality for a change of variables with multiple variables?

Yes, you can solve an inequality for a change of variables with multiple variables. In this case, you will need to identify which variable you want to change and which variable you want to use. Then, you can use algebraic techniques to isolate the variable you want to change and solve for it. The remaining variables can be substituted back into the original inequality.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
785
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
354
  • Linear and Abstract Algebra
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
333
  • Topology and Analysis
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
771
Replies
3
Views
842
  • Precalculus Mathematics Homework Help
Replies
10
Views
2K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
Back
Top