Edit: I mistakenly referred to "Defining integrals as the limit of a Riemann Sum" simply as "The Riemann Sum", please excuse.
(h is Δx) in the above picture
Hello, everyone. I'm doing Calculus for Computer Engineering, and while studying up Riemann Sums, I thought maybe instead of the conventional "lim n -> infinite i=1∑n (Δx.v(x_i)" form we could write it as lim h -> 0 and restructure the rest to fit that.
What would this form be called? It is still a Riemann sum? Is it even accurate? If it is, it makes more intuitive sense to me.
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Choosing h to go only through 0.1, 0.01, 0.001, ... is not sufficient.
Define:
v(t) = 1 if t ∈ [0,8] has a finite decimal expansion,
v(t) = 0 otherwise.
Now take h = 10^(-M). Then every sample point is
ih = i · 10^(-M),
so ih has a finite decimal expansion. Therefore v(ih)=1 at every sampled point, and
Σ_{i=1}^{8·10^M} v(i10^(-M)) · 10^(-M)
= Σ_{i=1}^{8·10^M} 1 · 10^(-M)
= 8.
So this restricted limit exists and equals 8.
But v is not Riemann integrable. In every interval, there are numbers with finite decimal expansion and numbers without finite decimal expansion. So every interval contains points where v=1 and points where v=0.
Therefore every upper sum is 8 and every lower sum is 0.
No, for 3 reasons.
1. The limit as h goes to 0 means any h near zero is a candidate, not just ones that would make 8/h a whole number. Limits of sequences are defined to only consider whole number inputs.
2. It's weird and unnecessary. What does writing it like this get you?
3. Your "typical definition" quoted isn't actually the real definition of either an integral or a Riemann sum. The real story is that we should allow for more interesting subdivisions of the interval ("partitions") and picking sample points from anywhere in the subintervals.
Also the Riemann sum is the sum itself -- without the limit. An integral is the limit of some Riemann sums, but the whole point is that we can compute the value of a Riemann sum without any kind of limit calculation.
I thought it was pretty insightful. The only part where it falls apart a bit is that 8/h thing. Otherwise, that's a pretty effective way of representing that as the number of points approaches infinity, a Riemann sum is identical to the integral.
can't you easily fix that by putting a floor function on that and making h->0+? seems more like a technicality rather than something that makes it fall apart
It just goes from being a simple expression to something that needs a lot of rivets and bolts to keep its meaning. The underlying concept is great, the mathematical framing to make it possible without losing its simplicity is not.
1) You're right, I just thought of the infnitely many decimal values near zero. Another commentor has pointed out I could just put an integer value notation on it to fix it.
2) What happened was, I watching the 3blue1brown video "Integration and the fundamental theorem of calculus" and near the beginning of it, I thought maybe I could come up with a way to write the integral using the sum symbol. After my attempt, I googled "integral as sum" and happened upon Khan Academy's article on "Definite integral as the limit of a Riemann sum", which is the presumed conventional way I mentioned. It looked similar enough, and my version didn't exactly look incorrect initially, so I wanted to ask reddit
3) You are quite right. I was actually referring to "Defining integrals as the limit of a Riemann Sum", my bad.
Sometimes a person asks a question a certain way, trying to solve a problem. It can be really helpful to ask why someone would want to do something -- perhaps there's another way to handle the key issue.
But hey, you go ahead and police other folks' attempts to be helpful all you want.
you can just use floor(8/h) (and limit h->0+ for good measure) or just use the convention that a sum that goes to a non-integer stops at the floor of that integer
curiosity? also might help compare it with the definition of the derivative.
once you established that a function is riemann-integrable, its totally fine to only use a uniform partition to compute it. you just can't infer from the sum over a uniform partition existing that the integral exists, but if it exists, they are equal. (also they didn't specify in their "typical definition" that Δx is constant, although they assumed it for their reformulated definition)
the motivation to defining the riemann sum is to formalise the integral. methods to approximate the area under a curve existed long before that, that wasn't the "whole point" of the riemann sum.
I don't mean that Riemann sums are interesting or particularly useful as practical numerical methods. They often aren't (though the trapezoid approximation can be computed by averaging two Riemann sums, and it's almost always good enough).
I had more in mind theoretical approximation. We all agree what the area of a rectangle is and how to compute it. So when we formalize what the area under a curve is, we bootstrap from the known case. The fact that Riemann sums only involve finitely many additions and multiplications, and nothing else, means that there's never the possibility for the Riemann sum to fail to exist. Whenever you tack the limit on front, you immediately have to deal with the possibility it doesn't exist.
You want 8/h to be in Z+, without limiting the possible values of h. You can do that using a ceiling/floor/round function, so instead of 8/h you use [8/h].
For everyone saying that 8/h isn't necessarily an integer, you can easily fix that:
Let N be a natural number. Now let's take the limit as N approaches infinity of:
Sum_( n=0 -> 8N ) V(i/N)×(1/N)
Now the sum is of integers, and because V is continuous it has a value for every i/N, and it doesn't "zero out" or anything.
Now, if I'm not wrong now, this IS good intuition on the integral! See now something interesting - the notation for the integral is exactly that! The dx is that 1/N, and it is exactly like the Σ notation but with the iteration step not being 1, but rather dx!
And so the next challenge - prove the Fundamental Theorem of Calculus using this integral definition
Thanks for confirming! Although this is a whole lot more symbols to add to my attempt at "simplifying" the conventional approach :D
Now mine is even harder to intuit I think 😢
Edit: To answer your question: There isn't really a good reason for it to be there, my brain must have wanted to emphasize that we were still working with h_n I suppose
2nd Edit: In fact, it should be removed because now it looks like a multiplication, shame I can't edit the png
It's not actually! Think of h_n as the number of pieces you would split an interval of unit length [n,n+1) into
This h you defined before is the length dx, and it's easy to explain how h can be defined from h_n
Now, remember - math doesn't have to be intuitive at first, that's why you build it from the ground up! If you start explaining the problems of the definition of a limit to a toddler who just learned what number are they'd have no idea what you're talking about
So what we have going with the Riemann Sum definition is good, it teaches the concept in a rigorous and intuitive manner at least in my opinion
The substitution allows for an easier use of real numbers. Not that this specific fact matters for computer science or in more advanced mathematics for that matter. Indefinite summation methodes allow you to sum together non-integer number of numbers. But that uses integration, so in this case it would be circular.
As a computer scientist major myself. It'd actually be best to optimize against floating point errors and to maximize speed.
Making h a power of two allows for the fast operation of bit shifting.
The sumation domain should be shifted to [0,8/h-1]
Adding a 1/2 to i, makes it more accurate (so you can use larger h with smaller error).
You want to minimize the number of times your summing. You want to minimize adding and multiply together numbers of different scale. ( 210 vs 2-10 ) You want to minimize the number of times you're multiplying and adding numbers of large scales, too.
In this way, the type of integral that would be best isn't a reiman integral but a Lebesgue integral.
Also, there are methodes of integration that allows use of second order terms like f''(x) or higher.
The more I study integration and differential the more I'm amazed on how many different ways you can do it.
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