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Subtracting LUB and GLB to approach function

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

$$\sum\limits_{v=1}^n M_v(x_v-x_{v-1})-\sum\limits_{v=1}^n m_v(x_v-x_{v-1})<\epsilon$$

3. The attempt at a solution
$$\sum\limits_{v=1}^n M_v(x_v-x_{v-1})-\sum\limits_{v=1}^n m_v(x_v-x_{v-1})=\sum\limits_{v=1}^n (M_v-m_v)(x_v-x_{v-1})$$

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 $$\frac{b-a}{k}(f(b)-f(a))<\epsilon$$
Attached Files
 PS 102 Hints.pdf (667.9 KB, 5 views) PS 102.pdf (354.2 KB, 5 views) 102 Increasing everywhere.pdf (75.4 KB, 2 views)
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 Recognitions: Homework Help Hey Arcana! 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?
 Recognitions: Gold Member 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?

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Subtracting LUB and GLB to approach function

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.
 Recognitions: Gold Member Oh fiddlesticks that isn't right...
 Recognitions: Homework Help 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?

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 Quote by I like Serena 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?
 Recognitions: Homework Help What... you don't like typing latex? Edit: Perhaps you can summarize in one sentence?
 Recognitions: Gold Member Not on my iPad... It's already a beautiful PDF :)
 Recognitions: Homework Help Ah well, that will have to wait. I'm off to bed.
 Recognitions: Gold Member 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.
 Recognitions: Gold Member No problem, it's not urgent :)

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 Quote by ArcanaNoir 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.

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 Quote by LCKurtz 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.
 Recognitions: Homework Help 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.
 Recognitions: Gold Member 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?
 Recognitions: Homework Help 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.