I’ve updated @Mistral@lemmings.world to remember the whole tree of comments when replying to you, so it can do follow-up responses. You still have to mention it on every message, though. Example in the comments.
I’ve updated @Mistral@lemmings.world to remember the whole tree of comments when replying to you, so it can do follow-up responses. You still have to mention it on every message, though. Example in the comments.
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = a, x_n = b, and x_i are points in the interval [a, b] such that x_(i-1) < x_i.
Consider the function g(x) such that g’(x) = f(x). By the Fundamental Theorem of Calculus, Part 1, we know that g(b) - g(a) is equal to the definite integral of f(x) over [a, b].
Now, let’s partition the interval [a, b] into n subintervals of length h = (b - a) / n. Let x_i be the ith point in this partition, where i = 0, 1, …, n. Then, we can approximate the definite integral as:
∫(a to b) f(x) dx ≈ [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
As n approaches infinity and the width of each subinterval approaches 0, this approximation becomes more accurate. Therefore, we have:
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] = ∫(a to b) f(x) dx
Since we know that g(b) - g(a) is equal to the definite integral of f(x) over [a, b], we have:
g(b) - g(a) = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
This completes the proof of the Fundamental Theorem of Calculus, Part 2.
Why I do this to myself lmao
Is it at least making sense or is it making stuff up?
It’s screwed up a couple times. For example, the bit about “by the properties of limits you can interchange the limits” is hilariously untrue.
Also I’ve seen a few proofs of the FTC at various levels, but I’ve never seen a proof by contradiction be used. It can be done by plugging in definitions and properties of the integral into a milquetoast delta-epsilon proof. It’s less than a page for both theorems in “baby Rudin.” The hard part IMO is building up all the results you use, but once you have them it’s an easy proof.
@Mistral@lemmings.world You said “By the properties of limits, we can interchange the order of the limits”, but you can’t in general interchange the order of limits except under strict circumstances.
You are correct. I apologize for the confusion in my previous response. I should have mentioned that the Interchange of Limits theorem, also known as Jordan’s theorem, is not applicable in general. To properly derive the Fundamental Theorem of Calculus, Part 2, we need to use a different approach.
Consider the function g(x) such that g’(x) = f(x). We want to show that:
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] = g(b) - g(a)
Let’s partition the interval [a, b] into n subintervals of length h = (b - a) / n. Let x_i be the ith point in this partition, where i = 0, 1, …, n. Then, we can approximate g(b) - g(a) as:
g(b) - g(a) ≈ [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Now, let’s define a new function h(x) as follows:
h(x) = [Σ (from i = 1 to k) f(x_i) * (x_i - x_(i-1))] for x_k < x ≤ x_(k+1)
h(x) is a step function that approximates g(x) on each subinterval. Since g(x) is continuous and has a continuous derivative on the interval [a, b], it also has a continuous second derivative. Therefore, g(x) is convex on this interval.
Now, let’s define a new function G(x) as follows:
G(x) = g(x) - h(x)
G(x) is a continuous function, and its derivative is:
G’(x) = g’(x) - h’(x) = f(x) - f(x) = 0
Since G(x) is a continuous function with a zero derivative, it must be a constant function. Let’s call this constant C. Therefore, we have:
G(
@Mistral@lemmings.world 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) - [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
G(b) = g(b) - h(b) = g(b) - [Σ (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) - [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] + [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Simplifying this expression, we get:
g(b) - g(a) = [Σ (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 -> ∞) [Σ (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).
@Mistral@lemmings.world 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) = ∫(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*}) = Σ (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*}) = ∫(a, b) f(t) dt
Now, let’s construct a sequence of partitions {P_k} such that max(P_k) -> 0 as k -> ∞. 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)) = ∫(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 ≤ x_1 ≤ … ≤ x_n-1 ≤ 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?
@Mistral@lemmings.world 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*}) = ∫(a, b) f(t) dt