This one is fun. Find what values of make true. Next find all values that make true.

Note: No calculus.

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- December 20th 2008, 01:13 PMMathstud28Trigonometric inequality
This one is fun. Find what values of make true. Next find all values that make true.

Note: No calculus. - December 20th 2008, 01:59 PMNonCommAlg
- December 20th 2008, 02:05 PMMathstud28
- December 20th 2008, 02:21 PMNonCommAlg
try your inequality's cousin: (Wink) prove that: for all

i'd like to see different approaches. of course, using calculus probably would be the most convenient one and you may use it. but are there other ways to do it? (Wondering) - December 20th 2008, 03:23 PMMathstud28
Besides the obvious way of doing it with maxs and stuff how about this method. It uses some shady postulates (that may be untrue), but this is for fun so why not just suggest it?

Suppose the inequality presented is true, then so must be the inequality , or alternatively . It can be verified that both these functions are monotonically increasing, positive, and continuous on the specified interval.

Now here is where the geometrically logical but probably incorrect "lemma" I am using comes into play. It goes someting to the tune that if and are monotonic as well as continuous on the interval , then . Now suppose that this is true, it can be easily verified that

Now supposing that my "made up" lemma is correct, this proves the inequality.

I have learned to not always trust my geometric intuition...so Im not too sure about this (Worried) - December 20th 2008, 04:00 PMNonCommAlg
unfortunately your made up lemma is not always true, e.g. then but neither nor on the unit interval.

the trouble here comes from this fact that and intersect! so you need to assume that do not intersect in the interval. also i don't think we need to be positive. - December 20th 2008, 04:04 PMMathstud28
Dangit! I forgot to say that. I knew that they cannot intersect...and no we do not need them positive we need . I have come up with a "proof" of my lemma if anyone wants to see it.

EDIT: Wait, they cannot intersect except possibly at the interval because then after that point.

For example but it is reversed afterwards...but you applied my lemma to the interval , and

EDIT EDIT: We dont even need we must just have

EDIT EDIT EDIT: I am busy now but I will come back later and write out this lemma in a clear manner...I will then attempt to prove it - December 20th 2008, 04:15 PMNonCommAlg
- December 20th 2008, 04:30 PMbkarpuz
I have the following solution using calculus.

Let for .

Then, we see that is periodic, hence it suffices to prove on .

On the other hand, we have for all , which indicates that it suffices to prove for all .

Clearly, is positive (and decreasing) on , for all , and is negative (and increasing) on , .

Therefore, on .

To complete the proof we have to prove on .

Similar reasoning to the discussion above about increasing and decreasing natures of the functions and together with the fact for all , we get for all , and the proof is hence completed.

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.................Graph of .....................................Graph of .....................................Graph of

- December 20th 2008, 10:17 PMMathstud28
So the statement is this: Suppose that posses the following charcteristics on : they are positive, they are continuous, they do not intersect, and they are monotonic. Then on it is true that

First let us prove that . Consider any partition of consisting of the set of points . Now as in the usual way define , , and . Now it is clear that since that and since this implies that . Finally we can conclude that . And since are continuous, thus Riemann integrable,

Now let us prove that . Define as before.

1. Now it is clear that either or for all . To see this first define it is clear that is continuous. Then suppose that there were two values such that and then there exists a such that and there exists a such that . Now because is continuous and connected this implies there exists a such that which contradicts that the functions do not intersect.

2. So from the fact that we can see that

3. So all that is left to do is prove that . To do this once again define . Let be the point such that , and let be defined similarly. Now since is compact it follows that . Now consider when are monotonically increasing, it is clear now that . So . So now suppose there was a point such that , then at that point and by the connectedness of and the continuity of there must be a point in such that , but this contradicts the two functions not intersecting. The proof is done similarly for being monotonically decreasing.

4. Now since the interval was arbitrary in 3. this completes the proof - December 21st 2008, 12:15 AMNonCommAlg
ok, i didn't read your proof but i'm sure it's a good practice for you since you're studying Rudin! (Nod) first of all, you don't need to assume are positive or monotonic. "continuous" and

"not intersecting" are only conditions we need: let suppose first that on the interval. it's not hard to see that the integral of a positive continuous function is positive.*****

thus conversely, suppose since do not intersect, we have everywhere on [a,b]. so by the intermediate value theorem, either or everywhere

on [a,b]. but if then and hence by*****we'll have and hence which is a contradiction. Q.E.D.

*****in general, if is continuous, non-negative and not identically 0 on [a,b], then__Hint__: since is not identically 0 over [a,b], there exists a subinterval of [a,b] over which: - December 21st 2008, 12:23 AMMathstud28
Yeah, I am not trying to be easy, I am trying to be as rigorous as possible...now this may not always be the best way...but it helps me learn all the material since I end up using three fifths of it in one proof.

And I understand your proof, but the reason it is so short is that a lot of the stuff you just stated I proved...now of course for a mathematician such as yourself this is obvious...but I thought for us other folks it would be best to show it.

Thanks for your time NonCommAlg (Nod)