Integrability question: intro to analysis

[icantadd]

Homework Statement

find the integral of f(x) = x by finding a number A such that L(p,f) <= A <= U(p,f) for all partitions p of [0,1].

where a partition p of an interval [a,b] is of the form {x0,x1, ... , xn}

Homework Equations

L(p,f) is the lower sum of f with respect to the partition p. In this case f is increasing so
L(p,f) = $$\sum_{i=1}^{n} x_{i-1}*(x_{i}-x_{i-1})$$
and
U(p,f) = $$\sum_{i=1}^{n} x_{i}*(x_{i}-x_{i-1})$$ is the upper sum of f with respect to partition p.

The Attempt at a Solution

What I thought would work is to show inductively that L(p,f) is less than 1/2 and the U(p,f) is greater than 1/2. But then I get stuck at how to use the recursive step.

StepI: L(p,f) = 0*(1-0) = 0 < 1/2
StepII: Assume $$\sum_{i=1}^{n} x_{i-1}*(x_{i}-x_{i-1})$$ <= 1/2. To prove $$\sum_{i=1}^{n+1} x_{i-1}*(x_{i}-x_{i-1})$$

Then here is where I am stuck. How do I inject an extra value into this partition. I guess that the sum from 1 to n+1 above should be less than something which equals the inductive hypothesis which is less than 1/2, but I don't know what trick to use to get there.

Can anyone maybe clear this up, or if there is a perhaps better way to do this, hint me in a different direction. Thank you greatly in advance!

 

[Mark44]

One comment.
The first term in your lower sum has $$x_{i-1}$$ in it, when i = 1, which means that it is referring to $$x_{0}$$. There is no such term in your partition.

 

[icantadd]

You are right. I meant P = {x0,x1,...,xn}. I fixed it above. Note also that x0 should equal 0 and that xn (or x_(n+1)) should equal 1. Thank you.

 

[Mark44]

After thinking about this some more, I've come up with something that might be of help. Use a partition that is equally spaced, with each subinterval of length 1/n. If someone has a quibble that you need to show this for any partition P, just specify n large enough so that all your subintervals are smaller than the smallest subinterval in the other partition.

Let P = {0, 1/n, 2/n, 3/n, ..., (n-1)/n, n/n = 1}
A lower bound for the area beneath y = x is given by:

L_n = 0*1/n + 1/n * 1/n + 2/n * 1/n + ... + (n - 1)/n * 1/n

Here, I'm using the y-values at the left end of each subinterval, since we're dealing with a monotonically increasing function.

The sum above has n terms, where 1/n is present twice in each term, so it can be factored out.

$$L_n = \frac{1}{n^2} [0 + 1 + 2 + 3 + ... + (n - 1)]$$
So $$L_n = \frac{1}{n^2} \frac{n(n - 1)}{2}$$, using the well-known fact that the sum of the first (n - 1) integers is n(n - 1)/2. [A perhaps more well-known fact is that the sum of the first n integers is n(n+1)/2.]

From the above, we can see that $$L_n = \frac{n^2(1 -1/n)}{2n^2}$$. It's pretty easy to see that this is less than 1/2 for all n > 0.

 

[icantadd]

Yes, I agree. To satisfy any partition it is obvious that 1/n can be as fine as It like. I have trouble saying things are obvious, just because I don't know where to draw the line yet. But, it is obvious that taking the value of the function at the leftmost point (as you have done), will always underestimate the area under that interval, and so if you make those leftmost points arbitrarily close, then you have that some thing that is a bigger underestimate. The solution you provided does this, it says, take as big and underestimate as you like, it will still be less than 1/2.

Thank you Mark. Here is a question I have for you: Any time I see something indexed by natural numbers, and I want to show something that relies on those indexes, such as a partition, I immediately think induction. Yet, this problem works out very pretty without induction. Is there a way you look at a problem and say, "oh I should(not) use induction here?"

 

[boombaby]

hey, but why not using this one:
$$x_{i-1} (x_{i}-x_{i-1}) < \frac{x_{i}+x_{i-1}}{2} (x_{i}-x_{i-1})=\frac{x^{2}_{i}-x^{2}_{i-1}}{2}$$
you can sum them up easily. and draw a graph you will see why this inequality works pretty well

Mark44,
what if I let P={0,sqrt(2),1}? so that P*={0,1/n,2/n,...,1} would not be a refinement of P. I'm thinking that L(P*,f)>=L(P,f) is not guaranteed if P* is not a refinement of P.

 

[icantadd]

hey, but why not using this one:

what if I let P={0,sqrt(2),1}? so that P*={0,1/n,2/n,...,1} would not be a refinement of P. I'm thinking that L(P*,f)>=L(P,f) is not guaranteed if P* is not a refinement of P.

No you have a syntax error. sqrt(2) > 1 and P is not a valid partition of [0,1]. What you wish to get at is that Q is a refinement of P in case P must be a subset of Q. So sure, you could make a partition of of irrational numbers, but to the best of my knowledge a partition must be countable, and the partition Mark gave will be as refined as one needs i.e. you can identify a max partition size denoted u(P) such that u(P) is as small as one needs to make a sequence of Lower (or upper) sums converge to the Lower (or upper) integral.

 

[boombaby]

..that sqrt(2) was a mistake...sqrt(2)/2 would be ok...
but yes, Q is a refinement of P iff P is a subset of Q.

You can use arbitrary partition only if the existence of the integral is guaranteed first..I cannot give a counter example right now, since finding one seems somewhat difficult for me...

 

[icantadd]

..that sqrt(2) was a mistake...sqrt(2)/2 would be ok...
but yes, Q is a refinement of P iff P is a subset of Q.

You can use arbitrary partition only if the existence of the integral is guaranteed first..I cannot give a counter example right now, since finding one seems somewhat difficult for me...

Hmm, I need to think about that. I only know Riemann integrability, so I may be missing some points in real Real analysis, but I have a theorem that states that a function f is Riemann integrable iff for any e there exists a partition such that the upper sum with respect to that partition minus the lower sum is less than or equal to e. So, I only need to show how small the size of the biggest interval created by a partition, is with respect to e. I don't know. I should be able to make a sequence of P's that converge to the integral; I'm not sure why an infinitely refined partition would not take care of such needs. Any ideas??

 

[boombaby]

ah..you remind me. There exists a such partition is okay...I'm sorry to bring lots of troubles. My head was dumb...:)

 

[Mark44]

hey, but why not using this one:
$$x_{i-1} (x_{i}-x_{i-1}) < \frac{x_{i}+x_{i-1}}{2} (x_{i}-x_{i-1})=\frac{x^{2}_{i}-x^{2}_{i-1}}{2}$$
you can sum them up easily. and draw a graph you will see why this inequality works pretty well

Mark44,
what if I let P={0,sqrt(2)/2,1}? so that P*={0,1/n,2/n,...,1} would not be a refinement of P. I'm thinking that L(P*,f)>=L(P,f) is not guaranteed if P* is not a refinement of P.

(I edited your partition to fix the error.)
I didn't say anything about a refinement of a given partition, just that the new partition's subinterval size be less than the smallest subinterval of the given partition. With your partition P as described above, a new partition P* = {0, 1/4, 1/2, 3/4, 1} will give an estimate that is closer to 1/2 than will the partition P.