math help? *solved*

[eLiu]

granted this IS homework...but i just want someone to point me in the right direction, as I don't know how to start this one...

I need to analyze this function:
f(x) = sum ( [n*x]/n^2 , n=1..infinity)

Where [y] denotes the fractional part of y. For example, [50.512] = 0.512.

So that's the sum from 1 to infinity of [n*x]/n^2...if the notation was at all unclear.

So the assignment is: find all discontinuities of f(x) and show that they form a countable dense set. Show that f is nevertheless Riemann-integrable on every bounded interval.

Yeah...uhm...lol. I'm having a lot of trouble tryin to visualize what's going on here. I think there will be discontinuities at all integral values of x...but other than that, I'm not entirely sure. The fact that like, for x = 0.2, [n*x] generates 0.2, 0.4, 0.6, 0.8, 0, 0.2...etc and that x = 0.3, [n*x] = 0.3, 0.6, 0.9, 0.2, 0.5, 0.8, 0.1, 0.4, 0.7, 0, 0.3...etc boggles my mind...

And I don't even know what to do with the irrationals.

I do suspect that the function is discontinuous at EVERY rational number...but I don't know how to prove this at all....nor how to really visualize it.

-Eric

 

[TuxDave]

Well, I have no clue what the final answer is but if I was stuck with this hellish problem, the first thing I would do is run a matlab script that'll generate a large series of points very close to each other just to see how this function behaves. That way I can see if it is indeed discontinuous at all point.

 

[DrPizza]

just at a glance, it does seem like there might be discontinuities at integral values, since f(integer) = 0
But, what's f(.00001)? Is it close to 0? If I didn't have a paper to write, this looks like a neat problem to spend a bit of time on...

 

[eLiu]

Well I tried to graph the function in maple to no avail...asking maple to evaluate some values isn't really helping much either.

Like apparently f(0.1) = infinity, f(0.01)=infinity, f(0.001) = infinity, f(0.21) = infinity...f(0.2) returns no value...f(pi), f(exp(1)) return no value...I'm confused, lol.

Well the reason I suspected all the rationals is because it wants me to prove the discontinuities form a countable, dense set...which would basically imply the rationals, i think (or some subset of the rationals with a "very" limited number of removals).

Still... I don't know >.<

 

[TuxDave]

Originally posted by: eLiu
Well I tried to graph the function in maple to no avail...asking maple to evaluate some values isn't really helping much either.

Like apparently f(0.1) = infinity, f(0.01)=infinity, f(0.001) = infinity, f(0.21) = infinity...f(0.2) returns no value...f(pi), f(exp(1)) return no value...I'm confused, lol.

Well the reason I suspected all the rationals is because it wants me to prove the discontinuities form a countable, dense set...which would basically imply the rationals, i think (or some subset of the rationals with a "very" limited number of removals).

Still... I don't know >.<

Are you including the divide by n^2 term in your infinite summation? There's no way that the sequence should not converge is [nx] is bounded to at most 1.

 

[eLiu]

Yup the n^2 term is in there...which is exactly why maple is baffling me right now, lol

edit: btw the sum is written like this in maple...maybe i doing something wrong...?

sum( (n*x - floor(n*x))/n^2 , n=1..infinity)

where floor() is the floor function...i.e. floor(51.526) = 51.

 

[TuxDave]

How about you go to like... 100 or so... your evaluation will be bounded by an error of 0.01% which is good enough.

 

[eLiu]

hm good call...

seems like the discontinuities occur at m/(2^n), where m,n are integers. Like... f(0.75)=0.999679..., g(0.749999) = 1.1022305...

This is summing 1 to 1000.

Interesting...

I tried some other rationals like 1/3 and some irrationals like pi...those didn't seem to have discontinuities. From what I can tell, the graph of the sum (1 to 1000) reflects the m/(2^n) idea as well...

edit: now if only i oculd find a way to verify this...booo

 

[TuxDave]

geez... I still have no idea how to prove it. Do ya think proof by 'maple' and '1000 samples is good enough' is sufficient as proof?

 

[eLiu]

Originally posted by: TuxDave
geez... I still have no idea how to prove it. Do ya think proof by 'maple' and '1000 samples is good enough' is sufficient as proof?

haha...nope. It's a course in real analysis...so even worse, it has to be a rigorous proof. Booo...

 

[jman19]

Originally posted by: eLiu

Originally posted by: TuxDave
geez... I still have no idea how to prove it. Do ya think proof by 'maple' and '1000 samples is good enough' is sufficient as proof?

haha...nope. It's a course in real analysis...so even worse, it has to be a rigorous proof. Booo...

Ugh, real analysis made me a sad panda when I had to take it

If I can think of anything, though, I'll let you know.

 

[silverpig]

Proof by induction.

1. Show for the example given here.
2. Assume true for some number.
3. Uh... so yeah, it's true for the next one too. (Let me assume that I pick a good number for 2. such that this is satisfied).
4. QED

 

[eLiu]

Oops crap...I was definitely in error when I said m/2^n...turns out I was doing it kinda wrong for some of the other rationals. Seems like this pattern will occur for all the rationals after all...I see the discontinuities at say 1/7, 1/3, 1/11, 1/13, and so on...

 

[CSMR]

Originally posted by: eLiu
I need to analyze this function:
f(x) = sum ( [n*x]/n^2 , n=1..infinity)

Where [y] denotes the fractional part of y. For example, [50.512] = 0.512.

So that's the sum from 1 to infinity of [n*x]/n^2...if the notation was at all unclear.

So the assignment is: find all discontinuities of f(x) and show that they form a countable dense set. Show that f is nevertheless Riemann-integrable on every bounded interval.
-Eric

Have you found all discontinuities? I'll give you a hint:
If f(x)=g1(x)+g2(x)+g3(x)+....
sum sup(|gi(x)|)<infinity
gi(x) is continuous at a for every i
Show that f(x) is continuous at a.

From this you will be able to find a large set of points on which your f is continuous. On the remainder you have to show that it is discontinuous, which is not hard.

(Your guess that the function is discontinuous at every rational number is correct. But it is discontinuous at other points too, so this is the wrong track.)

Don't try to visualise. This function is discontinuous on a countable dense set. It is not visualisable.

 

[CSMR]

Originally posted by: eLiuShow that f is nevertheless Riemann-integrable on every bounded interval.

To do this:
Show that you can measure the total extent of the discontinuities in your bounded interval. (Edit - this is untrue, ignore this!)

 

[eLiu]

when you're summing the sup's of |gi(x)|--are you taking the sup over all x? Can you elaborate on that step a little? I'm not sure I'm following it completely...

Also, on the integration note, what do you mean by "measure the total extent of the discontinuities?" Also, when you say a small grid...do you mean like partitions Pi (i as in subscript i) chosen in such a way as to exclude the discontinuities...?

Sorry if I'm sounding stupid...I guess I've been thinking about this thing wrong all along...lol. If it were just rationals, I think maybe i got something going for integrability there...though somehow I think it's wrong.

^^I did that by looking at just the (nx)/n^2 (not the sum); call it fn. So 0<=fn<1/n^2, => fn is uniformly convergent. fn is continuous on intervals [m/n, (m+1)/n) &amp; it is discontinuous at each m/n. So in some interval [a,b] there are finitely many m/n, so (nx)/n^2 is Riemann-integrable by a theorem proved in Rudin. Then by another theorem, the sum ((nx)/n^2) is also Riemann-integrable.

But if there are more discontinuities than just the rationals...well...darn. I'll think about this more tomorrow...

thanks for the help so far,
-Eric

 

[CSMR]

I'm sorry, I misread the formula as n^x rather than n*x. The discontinuities are at the rational numbers as you thought. Everything else I said still works.

Originally posted by: eLiu
when you're summing the sup's of |gi(x)|--are you taking the sup over all x? Can you elaborate on that step a little? I'm not sure I'm following it completely...

Over all x, or over a region around a - it doesn't matter. That condition ensures uniform convergence.

Also, on the integration note, what do you mean by "measure the total extent of the discontinuities?"

I thought you could measure the discontinuity at each point and then add up all the discontinuities. But the total turns out to be infinite, so this is a bad approach - sorry if I put you off track!

^^I did that by looking at just the (nx)/n^2 (not the sum); call it fn. So 0<=fn<1/n^2, => fn is uniformly convergent. fn is continuous on intervals [m/n, (m+1)/n) &amp; it is discontinuous at each m/n. So in some interval [a,b] there are finitely many m/n, so (nx)/n^2 is Riemann-integrable by a theorem proved in Rudin. Then by another theorem, the sum ((nx)/n^2) is also Riemann-integrable.

That's right, well argued.

 

[eLiu]

So for the discontinuities part...

Can I just say that (n*x)/n^2 is continuous at all irrational x? Because I've shown in another problem that (x) has first order discontinuities at every integer. So (n*x)/n^2 would only change the height/width of the graph...

Then the sum of continuous functions is continuous, so f(x) is continuous where fn (fn being (nx)/n^2) is continuous for all n...those x's being all irrationals.

So this means that the only points where f(x) could be discontinuous are the rationals. And here I'm kind of stuck again...b/c I don't know how to show that the discontinuities must occur at every rational.

Another thing I've noticed in Maple (but can't prove) is that say A = f(a), where a is some rational number. Now let B = left side limit of f(a). Now let C = A-B. A/C - B/C = 1. I've tried a few points like that and it's held for all of them...so I assume it's true by fake-induction, lol.

Still I'm not sure how to show it...could you give me another hint or tell me if I'm even headed in the right direction?

-Eric

 

[CSMR]

Originally posted by: eLiuCan I just say that (n*x)/n^2 is continuous at all irrational x?

Definitely a useful thing to note.

Then the sum of continuous functions is continuous, so f(x) is continuous where fn (fn being (nx)/n^2) is continuous for all n...those x's being all irrationals.

You have the right conclusion, but the wrong argument. It is true that a finite sum of continuous functions is continuous, and that a uniformly convergent sum of continuous functions is continuous - but the functions you are dealing with are not continuous.

Can you prove that the limit of a uniformly convergent sequence of continuous functions is continuous?
It's just the same proof to show that the limit of a uniformly convergent sequence of functions which are continuous at a particular point is continuous at that point.

Assume that the limit is discontinuous at a, write this out, then show that some partial sum has to be discontinuous at a.

So this means that the only points where f(x) could be discontinuous are the rationals. And here I'm kind of stuck again...b/c I don't know how to show that the discontinuities must occur at every rational.

You can say something about the nature of the discontinuities. If you have lots of functions which are discontinuous at a the sum may be continuous at a. There is a simple property of the discontinuities in the functions you are considering that makes this impossible.

Another thing I've noticed in Maple (but can't prove) is that say A = f(a), where a is some rational number. Now let B = left side limit of f(a). Now let C = A-B.

Anything you notice about C?

A/C - B/C = 1

:thumbsdown:
This is true by the definition of C!

 

[eLiu]

Originally posted by: CSMR

Originally posted by: eLiuCan I just say that (n*x)/n^2 is continuous at all irrational x?

Definitely a useful thing to note.

Then the sum of continuous functions is continuous, so f(x) is continuous where fn (fn being (nx)/n^2) is continuous for all n...those x's being all irrationals.

You have the right conclusion, but the wrong argument. It is true that a finite sum of continuous functions is continuous, and that a uniformly convergent sum of continuous functions is continuous - but the functions you are dealing with are not continuous.

Can you prove that the limit of a uniformly convergent sequence of continuous functions is continuous?
It's just the same proof to show that the limit of a uniformly convergent sequence of functions which are continuous at a particular point is continuous at that point.

Assume that the limit is discontinuous at a, write this out, then show that some partial sum has to be discontinuous at a.

So this means that the only points where f(x) could be discontinuous are the rationals. And here I'm kind of stuck again...b/c I don't know how to show that the discontinuities must occur at every rational.

You can say something about the nature of the discontinuities. If you have lots of functions which are discontinuous at a the sum may be continuous at a. There is a simple property of the discontinuities in the functions you are considering that makes this impossible.

Another thing I've noticed in Maple (but can't prove) is that say A = f(a), where a is some rational number. Now let B = left side limit of f(a). Now let C = A-B.

Anything you notice about C?

A/C - B/C = 1

:thumbsdown:
This is true by the definition of C!

hahaha...wow i must be more tired than i thought...lol. I guess sometimes you get too eager to solve a problem and anything becomes "amazing." D'oh! Man I'm feeling stupid now, heh.

And yeah, I've done the proof for the limit of a uniformly continuous sequence of functions is continuous and vice versa.

Also...for irrationals, the sum is uniformly convergent isn't it? B/c it's always less than sum(1/n^2) which converges... So why wouldn't the sum of functions fn which are continuous at irrational #s be continuous?

But yeah...I've noted that the discontinuities are "jump" discontinuities, and that the size of the "jump" increases as the denominator of the rational # increases--e.g. the largest jump occurs between f-(1/2) and f(1/2). f-(1/2) being the left-hand limit.

Intuition tells me that there should be some way to compute the size of the jump...I think it might go like this:

If I took point where fn(x) = 0 and replaced it by 1/n^2...and summed up all those 1/n^2's...I would definitely get something that converges for every rational #...would that sum equate to the difference between my "A" and "B"? That is, f1(1/2) = 1/2...f2(1/2) = 0, f3 = 1/2, f4 = 0...etc. So I would take 1/2^2, 1/4^2, 1/6^2, etc and sum them. I think that these values are exactly those that cause the left-hand limit of 1/2 (or whatever rational#) to be larger than the value at the number itself.

So hte size of the gap should be sum(1/(q*n)^2, n=1..infinity). Where q is the denominator of my rational number p/q, with p, q being relatively prime integers. Relative prime-ness guarantees that I get the same behavior for say, p/7, regardless of whether it's 1/7, 2/7, 3/7, or whatever.


If that is correct, then does that mean that I can show discontinuity at the rationals by just writing down my "derivation" of the gap-size? Seems like that would be sufficient.

 

[CSMR]

Originally posted by: eLiu
And yeah, I've done the proof for the limit of a uniformly continuous sequence of functions is continuous and vice versa.

Also...for irrationals, the sum is uniformly convergent isn't it? B/c it's always less than sum(1/n^2) which converges... So why wouldn't the sum of functions fn which are continuous at irrational #s be continuous?

Well it is true that a uniformly convergent sum of functions which are continuous at a point is continuous at that point... you just need to show that. (If you take uniformly away that becomes false of course.)

I've noted that the discontinuities are "jump" discontinuities

good

the left-hand limit of 1/2 (or whatever rational#) to be larger than the value at the number itself.

good

So hte size of the gap should be sum(1/(q*n)^2, n=1..infinity). Where q is the denominator of my rational number p/q, with p, q being relatively prime integers.

That's right.

If that is correct, then does that mean that I can show discontinuity at the rationals by just writing down my "derivation" of the gap-size? Seems like that would be sufficient.

Yes, you can do that. A good way to formalize your argument is given a point p/q to construct new functions f'i which have the discontinuity removed, and note that sum f'i converges uniformly to f', with f'-f a step function, and f' continuous at p/q.
Alternatively prove that for any sequence ai tending up to p/q and bi tending down to p/q f(ai)-f(bi) cannot converge to zero.

 

[eLiu]

cool...thanks for all your help I figured it out methinks...

sorry I didn't respond earlier--lots of work from other classes...finals coming up >.<