Subtracting LUB and GLB to approach function

In summary, the problem states that there is a sum of Mv(x-x-1) - mv(x-x-1) which must be less than ε, but it is proving difficult to solve. The problem attachments include a simpler problem where the function being squeezed can be increased anywhere and upper and lower bounds are not used, as well as a solution using right and left end points. The problem is stated to be more difficult than it seems, and some of the information in the attachments is not understood.
  • #1
ArcanaNoir
779
4

Homework Statement



I am trying to understand the equation U-L<ε as part of a proof. I have attached the original problem, [102], as well as the hints page.

Homework Equations



[tex] \sum\limits_{v=1}^n M_v(x_v-x_{v-1})-\sum\limits_{v=1}^n m_v(x_v-x_{v-1})<\epsilon [/tex]

The Attempt at a Solution


[tex] \sum\limits_{v=1}^n M_v(x_v-x_{v-1})-\sum\limits_{v=1}^n m_v(x_v-x_{v-1})=\sum\limits_{v=1}^n (M_v-m_v)(x_v-x_{v-1}) [/tex]

So we have the length of the interval times the distance between the upper and lower bounds all added together. I'm not really strong on upper and lower bounds. I don't see how exactly we ensure that this will be less than epsilon.

I have solved a simpler problem where I let the function we were squeezing be everywhere increasing, and then instead of upper and lower bounds I used right and left end points. Then for a given epsilon I was able to choose k intervals such that [tex] \frac{b-a}{k}(f(b)-f(a))<\epsilon [/tex]
 

Attachments

  • PS 102 Hints.pdf
    667.9 KB · Views: 259
  • PS 102.pdf
    354.2 KB · Views: 237
  • 102 Increasing everywhere.pdf
    75.4 KB · Views: 182
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  • #2
Hey Arcana! :smile:

It's been a long time that you posted questions.

You will need some more information on ##M_v, m_v, x_v## to solve that.
Do you have any?
 
  • #3
What kind of information? Is that a "I should think about those things" kind of question, or a "you actually do need more information" kind of question?
 
  • #4
It's: "I actually do need more information".

Although I have to admit that I did not look at your attachments (too much work).

With the information you provided in your opening post, the problem cannot be solved.
 
  • #5
[edit]Oh fiddlesticks that isn't right...
 
  • #6
That looks much better (except for the typos).
Can it be that x is greater or equal to m, instead of less or equal?

But... what is the question?
 
  • #7
I like Serena said:
It's: "I actually do need more information".

Although I have to admit that I did not look at your attachments (too much work).

With the information you provided in your opening post, the problem cannot be solved.

Sorry but the information you need is in the attachments. It's an awful lot to type. It will make the situation oh so much clearer. You can ignore the "[102] increasing everywhere" attachment. pweeeeease?
 
  • #8
What... you don't like typing latex?

Edit: Perhaps you can summarize in one sentence?
 
  • #9
Not on my iPad...
It's already a beautiful PDF :)
 
  • #10
Ah well, that will have to wait.
I'm off to bed. :zzz:
 
  • #11
M is th least upper bound and m is the greatest lower bound f a function f(x). The sum with M is the upper sum and the m sum is the lower sum.

What you got against PDFs anyway? The whole problem and solution are basically given in the PDF, I'm just trying to understand the solution.
 
  • #12
No problem, it's not urgent :)
 
  • #13
ArcanaNoir said:
Sorry but the information you need is in the attachments. It's an awful lot to type. It will make the situation oh so much clearer. You can ignore the "[102] increasing everywhere" attachment. pweeeeease?

You might note you have had 189 people look at this thread and at most 3 views of any of your attachments. Some of us come by with a little time to give assistance and it's a lot to expect to think we will download and read through 3 attachments just to figure out what the problem is. If you can't take the time to at least summarize the problem here, you will be lucky if anyone bothers.
 
  • #14
LCKurtz said:
You might note you have had 189 people look at this thread and at most 3 views of any of your attachments. Some of us come by with a little time to give assistance and it's a lot to expect to think we will download and read through 3 attachments just to figure out what the problem is. If you can't take the time to at least summarize the problem here, you will be lucky if anyone bothers.

Yeah I know, but this isn't a quick little problem. Its okay if no one has time to read the attachments and help, I'm not obligated to complete this problem for any class. I won't be all pissy if I don't get any help. I know how it works.
 
  • #15
Hmm, I took a look... but these problems and their solutions don't look very sharp... at least not to me.

Apparently you're supposed to prove the existence of ##\psi(x), \Psi(x)## given f(x) and any ε>0 with ##\psi(x) \le f(x) \le \Psi(x)##.

But... I would just pick ##\psi(x) \overset{def}{=} \Psi(x) \overset{def}{=} f(x)##.
Those would do the job.

I don't really understand why they have to introduce ##M_v## and ##m_v##.
But perhaps that's just because I'm missing some context here.

And if you really want to use step functions with some partition...
then each of those summations are defined to approach the integral of f(x) due to the definition of the Riemann integral.
So you can always find each one with its summation less than say ε/2, which would do the job.

As I said, these problems with their solutions are really fuzzy as I see it.
I prefer texts that are a bit sharper and more to the point.
 
  • #16
Thanks for taking a look, ILS. What do you mean "find each one with its summation less than ε/2"? For example, the upper sum will be larger than f(x), so how can it be less than ε?
 
  • #17
What you have, is that:
$$\lim_{n \to \infty} \sum^n M_v (x_v - x_{v-1}) = \int f(x) dx$$

This means that for any ##\epsilon > 0## there exist equidistant partitions and N with:
$$\forall n > N: |\int f(x) dx - \sum^n M_v (x_v - x_{v-1})| < \epsilon / 2$$
assuming that ##M_v \in f([x_{v-1}, x_v))## for each v.
 
Last edited:
  • #18
How do we know that the limit as n goes to infinity is 0?
Your tex isn't working either.
 
  • #19
ArcanaNoir said:
How do we know that the limit as n goes to infinity is 0?

Do you have a definition of an integral handy?


Your tex isn't working either.

What?
My tex is beautiful!
What's wrong with it?
 
  • #20
Yeah sorry.. My tex isn't working either. Which is weird because I'm on my iPad so it's not like I touched any settings.


oh lovely on the definition of an integral. I see it is all there. Thanks, I will spend some time staring at that.

Right after I figure out what happened to my tex...

Thanks!
 
  • #21
Perhaps I should add that those ##m_v## would define a set of narrow rectangles that are below the graph of f(x).
That is, they would be a set of inscribed rectangles with height ##m_v##.
The summed area of those rectangles approach the integral as the rectangles become narrower.

This is the way the Riemann integral is defined.
 
  • #22
Yeah I'm okay with how the sums are like integrals, I just wasn't sure how to get the difference between the upper and lower sums to be less than epsilon.
 

Related to Subtracting LUB and GLB to approach function

1. What is the purpose of subtracting LUB and GLB when approaching a function?

The purpose of subtracting the least upper bound (LUB) and greatest lower bound (GLB) when approaching a function is to determine the limit of the function. By subtracting LUB and GLB, we can find the smallest and largest possible outputs of the function, which can help us determine the behavior of the function at a specific point.

2. How do you subtract LUB and GLB when approaching a function?

To subtract LUB and GLB when approaching a function, you first need to identify the point at which you want to find the limit. Then, you take the limit of the function as x approaches that point from both the left and right sides. Finally, you subtract the two limits to find the LUB and GLB.

3. What is the difference between LUB and GLB when approaching a function?

LUB and GLB are two different types of limits that can be used to approach a function. LUB is the limit as x approaches a point from the left side, while GLB is the limit as x approaches a point from the right side. LUB and GLB can have different values, which is why we subtract them to find the limit of the function.

4. When is it necessary to use LUB and GLB when approaching a function?

LUB and GLB are necessary when approaching a function at a point where the function is not continuous. In other words, when there is a jump or a discontinuity at that point, LUB and GLB are needed to determine the limit of the function.

5. Can LUB and GLB be used with all types of functions?

Yes, LUB and GLB can be used with all types of functions, including polynomial, rational, exponential, and trigonometric functions. However, they are most commonly used when approaching piecewise functions or functions with discontinuities.

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