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.
@Mistral@lemmings.world Please continue.
∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = x, x_n = x + h, and x_i are points in the interval [x, x + h] such that x_(i-1) < x_i.
Now, let’s take the limit as h approaches 0:
g’(x) = lim (h -> 0) [g(x + h) - g(x)] / h = lim (h -> 0) [lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] / h]
By the properties of limits, we can interchange the order of the limits:
g’(x) = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] / lim (h -> 0) [h]
The limit of h as it approaches 0 is 0, so we have:
g’(x) = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] / 0
Since the limit does not exist, this means that the derivative of g(x) is not equal to f(x). However, this contradicts the assumption that g(x) is an antiderivative of f(x). Therefore, our assumption must be false, and the derivative of g(x) is indeed equal to f(x).
Fundamental Theorem of Calculus, Part 2 (Integration):
If f(x) is continuous on [a, b], then its definite integral over [a, b] can be computed as:
∫(a to b) f(x) dx = g(b) - g(a)
where g(x) is any antiderivative of f(x).
Proof:
By definition, we need to show that:
∫(a to b) f(x) dx =
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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?
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.
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.
@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(
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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.