Understanding Inequalities in Math: Solving Homework Problem Q7 iii

[truewt]

Hi everyone.

I am having a problem trying to understand the solutions of a homework problem that I had. Really need some help!

Basically, I am trying to establish an inequality on kT using a given set of inequalities to work with.

we have

L \leq Q \leq H
L \leq Q < Q+R_{1} \leq H
and
L \leq Q < Q+R \leq H

where R_{1} = -ln(kT+e^{-Q})and R = -ln[ \frac{kT}{1-e^{-Q}} ]

we want to establish kT = e^{-L} - e^{-H} by using some sort of sandwich from two inequalities.

However, there's this part of the proof that I could not comprehend:

We are able to reach L \leq H-L \leq H and since L \leq Q \leq H, we set Q = H-L into the inequality -ln[ \frac{kT}{1-e^{-Q}} ]+Q \leq H, and we can get kT \leq e^{-L} - e^{-H}

This is the part I do not comprehend and there appears to be such techniques used a couple of times in other math courses I had taken before.

My understanding is that this is actually incorrect; we should build some bounds and try to achieve the inequalities.

Can anybody help me out here? I really apologise for not using LaTeX to type this out, as I have no clue with LaTeX myself. Will try to edit the parts if possible.

I have attached the homework's solutions up, it's Q7 iii in concern. There's some errors to it I personally feel.

 

[truewt]

Anyone help??

 

[chiro]

Hi everyone.
However, there's this part of the proof that I could not comprehend:

We are able to reach L \leq H-L \leq H and since L \leq Q \leq H, we set Q = H-L into the inequality -ln[ \frac{kT}{1-e^{-Q}} ]+Q \leq H, and we can get kT \leq e^{-L} - e^{-H}

Hey truewt.

Upon reading the solutions text it seems what the author is doing is using the fact that since 2Q <= H and L <= Q <= H we know that L <= Q which implies 2L <= 2Q but since 2Q <= H this implies 2L <= 2Q <= H implies 2L <= H.

From this you get the inequality L <= H - L <= H.

 

[truewt]

Hey truewt.

Upon reading the solutions text it seems what the author is doing is using the fact that since 2Q <= H and L <= Q <= H we know that L <= Q which implies 2L <= 2Q but since 2Q <= H this implies 2L <= 2Q <= H implies 2L <= H.

From this you get the inequality L <= H - L <= H.

Hey chiro, thanks for replying! I do understand that part, what I am really confused is the part that follows, substituting Q by H-L...

 

[chiro]

Hey chiro, thanks for replying! I do understand that part, what I am really confused is the part that follows, substituting Q by H-L...

He is using the fact that since L <= Q <= H and L <= H - L <= H then basically we can use the substitution Q = H - L since both Q and H - L are both 'sandwiched' between L and H. I don't think they are necessarily saying that the two are equal at this point of the proof, but given the two inequalities, the substitution does fit the definition of the two inequalities.

 

[truewt]

He is using the fact that since L <= Q <= H and L <= H - L <= H then basically we can use the substitution Q = H - L since both Q and H - L are both 'sandwiched' between L and H. I don't think they are necessarily saying that the two are equal at this point of the proof, but given the two inequalities, the substitution does fit the definition of the two inequalities.

i am more inclined to think that we aren't allowed to do such substitution purely based on knowing that they are sandwiched, right? For example, 1<=2<=100 and 1<=99<=100, we can't substitute all inequalities related to 2 by 99, only very selective ones.

 

[chiro]

i am more inclined to think that we aren't allowed to do such substitution purely based on knowing that they are sandwiched, right? For example, 1<=2<=100 and 1<=99<=100, we can't substitute all inequalities related to 2 by 99, only very selective ones.

What I'm guessing that is happening is that the lower and upper bounds get closer together which proves what it needs to prove.

With regard to the inequality, there is nothing wrong with doing what the author did. You have to remember that in an inequality there are often multiple solutions to the problem as opposed to a normal equality.

The author has picked one solution that is allowed by the inequality and it probably isn't the only one that is allowed.

With regard to your 2 and 99 case you could substitute say any kind of relationship as long as the relationship is in line with the inequality.

Again it doesn't make sense if you try to think about things in the context of strictly being an equality, but when you consider that its an inequality then as long as the substitution meets the criteria for the inequality or set of inequalities, then it is one of possibly many relationships that hold under the constraint.

 

[truewt]

I still find it hard to swallow, and I am not well-trained in inequalities to refute such methodologies.. Is there anywhere I can find something to read about such methods?

I have always had trouble with techniques involving inequalities and I hope to get some help on that.

Thanks chiro!

 

[chiro]

I still find it hard to swallow, and I am not well-trained in inequalities to refute such methodologies.. Is there anywhere I can find something to read about such methods?

I have always had trouble with techniques involving inequalities and I hope to get some help on that.

Thanks chiro!

I understand your frustration since using a substitution like the one used can imply that it is 'the' solution as opposed to 'a solution fitting a set of constraints'.

I don't know any resources that go into detail about inequalities in a general context of the sort you are looking for.

What might help if you look at books that deal with solving problems that have inequalities as constraints where you get classes of solutions as opposed to just a unique solution.

 

[truewt]

Thanks, I guess such techniques really do not have a topic on its own, but rather is a concept manifested in problem solving instead. I'll try to get someone to explain it to me better, perhaps seeking my professor will help.

Thank you for your help so far!