I’ve updated @[email protected] 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.worldB
      1·
      9 months ago

      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.worldB
          1·
          9 months ago

          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.worldB
              1·
              9 months ago

              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(

              • PM_ME_VINTAGE_30S [he/him]@lemmy.sdf.orgEnglish
                1·
                9 months ago

                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?