Talk:Calculus/Building up to the Riemann-Darboux Definition

Go ahead and ask any questions you may be having here, or if you want me to speed up, slow down, or skip sections of the course. Feedback is encouraged. Fephisto 00:11, 31 July 2006 (UTC)

Basically, tell me when you want the next part of the lecture installed. Fephisto 17:40, 2 August 2006 (UTC)

You can start writing it, as I think I have undestrood the previous part. --vitalij 07:56, 4 August 2006 (UTC)

Sorry it's taken me a little while, I'll try to see if I can write more to keep ahead of you. I may be gone for this weekend, if I am, there will be at least these few exercises to keep you company, if not, I'll be sure to catch up with how you're pacing along. If you want to work during the weekend go ahead, either way, have a nice weekend.Fephisto 01:19, 5 August 2006 (UTC)

It takes time to write something good, so don't worry about how long you take to write something. So don't worry about the time you take to write something. --vitalij 06:52, 8 August 2006 (UTC)

What is general an what is about numbers?
I like very much the style of this text, but one thing is not really clear: which definitions can in general be applied to any field and which can be applied only to real numbers? There is another question: how many upper/lower bounds does a set have? Why we cannot just take the maximum/minimum? - First of all, keep in mind that real numbers are an ordered field. The supremum can work for any ordered field (thus it works for the real numbers). And, whenever I say for a field F, it means that it can work for the real numbers.

Technically a set can have an infinite amout of upper and lower bounds. A set that includes the numbers from 1 to 5 can have the upper bounds 6, 7, 8, 9, 10, 9.872347823..., etc.. The only thing that makes the least upper bound special is that it's lower than any other upper bound.

The reason we can't take the maximum of a set is because of a very slight technicality. Imagine we had a set that went from 1-5 but did not include 5, it approaches 5 ever so slightly, 5 is not the maximum number of the set (because it isn't included in it), but it is the least upper bound /of/ the set.

In case you were asking how many least upper/greatest lower bounds a set has, there is a proof (that I can show, or guide you along), that shows that the least upper/greatest lower bound is unique, i.e., for each specific set, there is one least upper and greatest lower bound.Fephisto 17:37, 2 August 2006 (UTC)

A Small Exercise
Exercise Now, prove that if $$F \supe \mathbf{R}$$ and $$E \supe F$$ then sup E &leq; sup F. Execution I will try to prove that $$sup E > sup F$$ implies absurd. If $$sup E > sup F$$ then there exist an element $$e$$ in $$E$$ that is bigger than any other element in $$F$$. But this is impossible because any element of $$E$$ is also an element of $$F$$. Do you think it is a good proof? --vitalij 07:08, 3 August 2006 (UTC)

Feedback
I really apologize for all the typos that were made in that script. I usually make things in LaTeX, but I guess it didn't transfer over well, either way it looks like you got my meaning.

In Mathematics we tend to be very critical of absolute proofs, what's the reason for this? What's the reason for that? What definition is being used here? What theorem is being used here? But you don't have access to a large number of theorems or definitions yet, and the general idea is fine.

This is being really picky, and showing each step and why, in case you wanted to know, all I'm looking for is that you understand the gneral areas of the concept of supremum, and it looks like you do:

Let's see, every element x in E is less than sup E, and every element y in F is less than sup F:

sup E >= x by def of supremum

sup F >= y by def of supremum

sup E > sup F by assumption

By another theorem (an Archimedean Theorem, which if you want I can dig up) implies that there is an element b such that sup E >= b > sup F.

An Archimedean Theorem Yes it would be useful to see something about that.

b is not in F by the definition of supremum

b is an element in E and not F, which contradicts with our other assumption, reductio ad absurdum.

Summing Exercise
Exercise Let f: [0,1] $$ \rightarrow \mathbf{R}$$ be f(x) = x, and let $$n \in \mathbf{N}$$ to form a uniform partition P = {i/n | 0,...,n} for i from 1 to n. Find U(P,f) and L(P,f) explicitly. Execution(Not complete) The intervals are [i-1/n, i/n] so for each interval $$i$$ I have sup f=i/n and inf f=i-1/n. U(P,f) = $$\sum_{i=1}^n M_i (x_i - x_{i-1})$$= $$\sum_{i=1}^n (i/n)(1/n)$$= $$1/n^2 \sum_{i=1}^n i$$

L(P,f) = $$\sum_{i=1}^n m_i (x_i - x_{i-1})$$= $$\sum_{i=1}^n (i+1/n)(1/n)$$= $$1/n^2 \sum_{i=1}^n i+1$$ Maybe I did something wrong. :-) What I haven't understood about that? --vitalij 07:34, 3 August 2006 (UTC)

Feedback
The main part was evaluating Mi and mi, good! That and this will be used extensively later on.

$$\sum_{i=1}^n m_i (x_i - x_{i-1})= \sum_{i=1}^n (i-1/n)(1/n)= 1/n^2 \sum_{i=1}^n i-1$$

Because mi=inf f=i-1/n right?

Also, $$\sum_{i=1}^n i = (n/2)(n+1)$$ So you can simplify a little more, the two sums look really similar, and will get more similar to come!

An exercise on partitions and refinements
Could you please add an exercise on partitions and refinements? Because I think I should work more on this topic. --vitalij 07:45, 3 August 2006 (UTC)

Feedback
This is, ironically, the next part! I'll add the next part including some more stuff on partitions and refinements (mostly refinements), and see if I can find more exercises. (After this the definition of an integrable function can easily be slid in.) Fephisto 06:11, 4 August 2006 (UTC)

More Feedback
If you want to know more about partitions we can go ahead and do that. I mainly focused on refinements, which are essential; but I'm not sure if that's what you wanted to cover. Fephisto 01:19, 5 August 2006 (UTC)

It is exactly what I wanted. Thank you. --vitalij 08:34, 7 August 2006 (UTC)

The point refinement exercise
Exercise We have a partition P of [a,b] and a refined partition P with ONE extra point P'=P &cup; {z}, and if $$f: [a,b] \rightarrow \mathbf{R}$$ is any bounded function, then by our intuition we stated that $$L(P,f) \leq L(P',f) \leq U(P',f) \leq U(P,f)$$. Prove it. Execution I define $$A={a \in P | az}$$. Let $$a'=min A$$ and $$b'=min B$$. This way $$[a',b']$$ is the minimal interval that contains z. As $$a \neq z$$ and $$b \neq z$$, I know that both elements are in P. So it is an interval of the partition P, suppose it is the k-th interval. I define: $$S=\sum _{i=0}^{k-1}M_i(x_i-x_{i-1})+\sum _{i=k+1}^{n}M_i(x_i-x_{i-1})$$ So I can say: $$U(P,f)=S+(sup_{[a',b']} f)(b'-a')=S+(sup_{[a',b']} f)[(z-a')+(b'-z)]$$ $$S+(sup_{[a',b']} f)[(z-a')+(b'-z)]= S+ (sup_{[a',b']} f) (z-a')+(sup_{[a',b']} f)(b'-z)$$ While: $$U(P',f)=S+(sup_{[a',b']} f)(b'-a')= S+ (sup_{[a',z]} f) (z-a')+(sup_{[z,b']} f)(b'-z)$$ Now, seeing that [a',z] and [z,b'] are subintervals of [a',b']. Using the Small Exercise I can say: $$sup_{[a',z]} f\leq sup_{[a',b']}f$$ and $$sup_{[z,b']} f\leq sup_{[a',b']}f$$ From this follows the statement I wanted to prove. For the lower sums the proof is similar. Do you think I missed something? --vitalij 07:30, 8 August 2006 (UTC)

Feedback
Good! However, I think a'=minA is a typo, was a'=maxA what you were going for? Also, you can go straight from U(P',f) to $$=S+ (sup_{[a',z]} f) (z-a')+(sup_{[z,b']} f)(b'-z)$$ since that's technically how it works from the definition, either way, both ways work. But yeah, good. Fephisto 04:36, 10 August 2006 (UTC)