Going from a sum to an integral

[MrGandalf]

Homework Statement

This isn't a problem, it is just a small verification I need in a much larger proof.

Over the interval $$[0,t]$$ we define a partition:
$$0 = s_0 < s_1 < \ldots < s_{n-1} < s_n = t$$

I have:
$$\sum_{i<j}(s_{j+1} - s_j)(s_{i+1} - s_i)$$

Homework Equations

What I need is for this to be equal to
$$\frac{1}{2}t^2$$

The Attempt at a Solution

When we pass to the limit, $$n\rightarrow\infty$$, I think we get something like
$$\int_0^t\int_0^sduds = \int_0^tsds = \frac{1}{2}t^2$$
but I am unable to show the connection.

This seems reasonable to me since we have the $$i<j$$ in the sum.

Any hints in the right direction will be appreciated. If you can just verify that I can do this, that will also be okay. :)

 

[stevenb]

Homework Statement

This isn't a problem, it is just a small verification I need in a much larger proof.

Over the interval $$[0,t]$$ we define a partition:
$$0 = s_0 < s_1 < \ldots < s_{n-1} < s_n = t$$

I have:
$$\sum_{i<j}(s_{j+1} - s_j)(s_{i+1} - s_i)$$

Homework Equations

What I need is for this to be equal to
$$\frac{1}{2}t^2$$

The Attempt at a Solution

When we pass to the limit, $$n\rightarrow\infty$$, I think we get something like
$$\int_0^t\int_0^sduds = \int_0^tsds = \frac{1}{2}t^2$$
but I am unable to show the connection.

This seems reasonable to me since we have the $$i<j$$ in the sum.

Any hints in the right direction will be appreciated. If you can just verify that I can do this, that will also be okay. :)

One thing you have to be careful with in limits like this is to make sure that all intervals "lengths" (i.e. $$s_{i+1}-s_i$$) are going to zero as $$n$$ goes to infinity. In other words it's not just the fact that $$n$$ goes to infinity, but also the fact that the largest interval goes to zero which is important.

I'm still thinking about your full question.

 

[MrGandalf]

Hi, stevenb, thanks for answering. But I think I got it.

I will edit this post with my full answer, so you don't waste your precious time verifying this for me.

Using the rule of a squared sum, which is what I started with:

$$\left(\sum_i a_i\right)^2 = \sum_i a^2 + 2\sum_{i<j}a_ia_j$$

In my case, every [ltex]a[/ltex] corresponds with a partition of the interval. So for the first term on the right hand side:
$$\sum_i (s_{i+1} - s_i)^2 \leq \max_i\Delta s_i\sum_i (s_{i+1} - s_i) = \max_i\Delta s_i(t) \rightarrow 0$$
as $$n\rightarrow\infty$$ since $$\Delta s_i\rightarrow 0$$. Also used that I have a telescoping sum.

Passing to the limit in the rule of sums I have over
$$\left(\int_0^t ds\right)^2 = 0 + 2\int_0^t\int_0^sduds$$
which works as a verification for me, because this is something I have used before. (It works for reasonable functions f as well).

On the left side, when we integrate:
$$\left(\int_0^tds\right)^2 = t^2$$

On the right side:
$$2\int_0^t\int_0^sduds = 2\int_0^tsds = 2(\frac{1}{2}t^2) = t^2.$$
Greeeat success! :)

This doesn't count as a proof or anything, but this sure is a strong indication that I can do this.

 

[stevenb]

It's not clear to me why you are saying $$\sum_i a_i^2$$ is equal to zero in the limit.

 

[MrGandalf]

I am summing the square of of many small intervals $$(s_{j+1} - s_j) := \Delta s_j$$.

I take the biggest partition and multiply ever term in the sum with that instead, and that is where I get the inequality:
$$(s_{j+1} - s_j)^2 = (\Delta s_j)^2 \leq \max_k(\Delta s_k)\Delta s_j$$

When I pass to the limit, the maximum partition becomes arbitrarily small, so the entire sum vanishes.

I know I can do this since the professor has used this exact argument a number of times in similar derivations. :)

 

[stevenb]

I am summing the square of of many small intervals $$(s_{j+1} - s_j) := \Delta s_j$$.

I take the biggest partition and multiply ever term in the sum with that instead, and that is where I get the inequality:
$$(s_{j+1} - s_j)^2 = (\Delta s_j)^2 \leq \max_k(\Delta s_k)\Delta s_j$$

When I pass to the limit, the maximum partition becomes arbitrarily small, so the entire sum vanishes.

I know I can do this since the professor has used this exact argument a number of times in similar derivations. :)

I agree that each individual term is zero, but aren't you then doing an infinite sum of terms? Wouldn't this then potentially become a finite-valued sum in itself?

 

[MrGandalf]

Hmmm, not sure. But do we have to worry about that when we have showed that it has an upper bound of 0? (Which also works for the absolute value I think, so it can't be negative).

Also, the "result" I reached over isn't really very solid, as it didn't work when I tried inserting $$x^2$$, but I think it holds for my case so I'm golden!

 

[stevenb]

Hmmm, not sure. But do we have to worry about that when we have showed that it has an upper bound of 0? (Which also works for the absolute value I think, so it can't be negative).

Also, the "result" I reached over isn't really very solid, as it didn't work when I tried inserting $$x^2$$, but I think it holds for my case so I'm golden!

I think you are on the right track and I think I follow what you are trying to do. It seems to work out, but I'm not convinced that someone else will follow your notation and logic.

I can see the first term going to zero in the limit, but you should show this clearly as a limit.

It's easier if we consider equally divided intervals as follows.

$$\lim_{n \to \infty}\sum_i a_i^2 = \lim_{n \to \infty}\sum_i(t/n)^2 = \lim_{n \to \infty}(t/n)^2\sum_i 1= \lim_{n \to \infty}(t/n)^2 n =\lim_{n \to \infty}t^2/n =0$$

Doing it for unequal intervals as a general case requires greater care and again the notation should show that not only is the number of intervals going to infinity, but the largest interval is going to zero. (I think you are doing that, but it just is a little unclear in the notation.)

 

[MrGandalf]

Yes, I agree, and your reasoning seems a bit nicer.

However, I'm not free to choose the intervals as I please, since I have derived them from the expectation of an elementary approximation of a Brownian motion. In addition this is just a small sub problem I ran into during a 7 page proof in stochastic analysis, so the real issue here isn't the various analysis details. :-)

But thanks for your answers! They were helpful.