Actually Useful AI
Welcome! ๐ค
Our community focuses on programming-oriented, hype-free discussion of Artificial Intelligence (AI) topics. We aim to curate content that truly contributes to the understanding and practical application of AI, making it, as the name suggests, "actually useful" for developers and enthusiasts alike.
Be an active member! ๐
We highly value participation in our community. Whether it's asking questions, sharing insights, or sparking new discussions, your engagement helps us all grow.
What can I post? ๐
In general, anything related to AI is acceptable. However, we encourage you to strive for high-quality content.
What is not allowed? ๐ซ
- ๐ Sensationalism: "How I made $1000 in 30 minutes using ChatGPT - the answer will surprise you!"
- โป๏ธ Recycled Content: "Ultimate ChatGPT Prompting Guide" that is the 10,000th variation on "As a (role), explain (thing) in (style)"
- ๐ฎ Blogspam: Anything the mods consider crypto/AI bro success porn sigma grindset blogspam
General Rules ๐
Members are expected to engage in on-topic discussions, and exhibit mature, respectful behavior. Those who fail to uphold these standards may find their posts or comments removed, with repeat offenders potentially facing a permanent ban.
While we appreciate focus, a little humor and off-topic banter, when tasteful and relevant, can also add flavor to our discussions.
Related Communities ๐
General
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
- [email protected]
Chat
Image
Open Source
Please message @[email protected] if you would like us to add a community to this list.
Icon base by Lord Berandas under CC BY 3.0 with modifications to add a gradient
view the rest of the comments
@[email protected] Please continue.
G(x) = C for all x in [a, b]
Now, let's evaluate G(a) and G(b):
G(a) = g(a) - h(a) = g(a) - [ฮฃ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
G(b) = g(b) - h(b) = g(b) - [ฮฃ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Subtracting G(a) from G(b), we get:
G(b) - G(a) = g(b) - g(a) - [ฮฃ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] + [ฮฃ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Simplifying this expression, we get:
g(b) - g(a) = [ฮฃ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
This is exactly what we wanted to show. Therefore, we have derived the Fundamental Theorem of Calculus, Part 2:
lim (n -> โ) [ฮฃ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] = g(b) - g(a)
This theorem states that if a function f(x) has an antiderivative g(x), then the definite integral of f(x) over the interval [a, b] is equal to g(b) - g(a), where g'(x) = f(x).
@[email protected] I think the proof of the second part was fine; it was the first part where the interchange of limits was used and was therefore erroneous.
You are correct. My apologies for the confusion. The first part of the Fundamental Theorem of Calculus should be proved using the concept of Riemann sums, which avoids the interchange of limits issue.
Here's the corrected proof for the first part of the Fundamental Theorem of Calculus:
Let f(x) be a continuous function on the interval [a, b]. We want to show that f(x) has an antiderivative g(x) such that g'(x) = f(x).
Define a function G(x) as follows:
G(x) = โซ(a, x) f(t) dt
Here, the integral is taken over the interval [a, x]. We want to show that G(x) is an antiderivative of f(x), i.e., G'(x) = f(x).
Now, consider a partition P of the interval [a, b]:
P: a = x0 < x1 < ... < xn = b
Let x_i* be any point in the interval [x_(i-1), x_i]. We can construct a Riemann sum for the integral:
S(P, {x_i*}) = ฮฃ (from i = 1 to n) f(x_i*) * (x_i - x_(i-1))
By the definition of the integral, as the maximum width of the intervals in the partition P approaches 0, the Riemann sum S(P, {x_i*}) approaches the definite integral:
lim (max(P) -> 0) S(P, {x_i*}) = โซ(a, b) f(t) dt
Now, let's construct a sequence of partitions {P_k} such that max(P_k) -> 0 as k -> โ. For each partition P_k, let's choose a point x_k* in the interval [x_(k-1), x_k]. Then, we have:
G(x_k) - G(x_(k-1)) = โซ(x_(k-1), x_k) f(t) dt = S(
For anyone else reading this [1], a partition of an interval [a,b] is a finite set of points satisfying the condition a = x_0 โค x_1 โค ... โค x_n-1 โค x_n = b. So the max(P) = b, so the condition max(P) -> 0 just means b -> 0, which is wrong.
[1] But like, y tho?
@[email protected] In the previous output, I think that you mean "max(norm(P))" where you say max(P), right? The condition "max(P) -> 0" just smooshes the partitions into the origin.
You are correct. I apologize for the confusion. The correct statement should be:
As max(norm(P)) -> 0, the Riemann sum S(P, {x_i*}) approaches the definite integral:
lim (max(norm(P)) -> 0) S(P, {x_i*}) = โซ(a, b) f(t) dt
@[email protected] Okay. Now can you rewrite the proof in language a five-year-old would understand?
Of course! Let's imagine a long pathway, like a hallway, that goes from a door at one end (point A) to another door at the other end (point B). We want to find out the total distance of this hallway.
Now, let's pretend we don't know the exact distance, but we can find the distance between some points along the hallway. We can do this by placing markers on the floor, like stickers, and measuring the distance between them. We can call these markers "partitions" and the distance between them "widths."
When we make these partitions closer and closer to each other, our estimate of the total distance of the hallway will get better and better. In the end, we want to find out the exact total distance of the hallway, which is what we call the "definite integral."
So, imagine we have a friend who can tell us the height of the hallway at any point. We can think of the height as the "function" we want to integrate. Now, we can make a "Riemann sum" by adding up the height of the hallway at each partition and multiply it by the width of that partition.
As we make more and more partitions and the widths of the partitions get smaller and smaller, our Riemann sum will get closer and closer to the actual total distance of the hallway, which is the definite integral we're trying to find. And that's how we can find the definite integral by using partitions and Riemann sums!
@[email protected] That's great! Can you rewrite the proof so a dog can understand it?
Sure, let's give it a try! Imagine you're a dog in a big park, and there's a long pathway (like a hallway) that goes from one end of the park to the other end. Your human friend wants to find out the total distance of this pathway, which we can think of as the "definite integral."
Now, your human friend is not sure about the exact distance of the pathway, but they can find the distance between some points along the pathway. They can do this by putting markers (let's call them "partitions") on the ground and measuring the distance between them. We can call the distance between the markers "widths."
Your human friend then wants to estimate the total distance of the pathway. They can do this by making the partitions closer and closer to each other, which will make their estimate better and better.
Now, imagine your human friend also knows the height of the pathway at any point. We can think of the height as the "function" we want to integrate. Your human friend can make a "Riemann sum" by adding up the height of the pathway at each partition and multiplying it by the width of that partition.
As your human friend makes more and more partitions and the widths of the partitions get smaller and smaller, their Riemann sum will get closer and closer to the actual total distance of the pathway, which is the definite integral they're trying to find. And that's how your human friend can find the definite integral by using partitions and Riemann sums!
Welp my brain is sufficiently melted. 10/10.
In all seriousness, this is really cool!