Proving A Unique Property Of Derivatives: $f''(\xi) = F(\xi)$

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Hey guys! Let's dive into a fascinating problem in real analysis that combines derivatives, and the existence of a special point within an interval. We're going to tackle a proof that shows if a function's derivative behaves in a certain way at the endpoints of an interval, then its second derivative must equal the original function at some point within that interval. Sounds cool, right? This problem isn't just about formulas; it's about understanding how the properties of derivatives can lead to some really elegant conclusions. This one is a classic example of how to make good use of Rolle's Theorem and its implications. Let's break down the problem statement and the steps needed to prove this remarkable result. We'll explore the core concepts and techniques required to understand this, making the mathematical reasoning accessible. Let's begin!

Understanding the Problem: The Setup

Alright, so here's the deal. We're given a function, let's call it f(x). We know a few key things about this function:

  1. Second Derivative Exists: The second derivative of f(x), denoted as f''(x), is defined and exists for all x in the closed interval [0, 1]. This is super important because it tells us that our function is smooth and well-behaved, which is a prerequisite for a lot of calculus theorems. When we say f''(x) exists on [0, 1], we mean the second derivative can be calculated for every point within that interval. This smoothness allows us to apply calculus tools like the Mean Value Theorem and Rolle's Theorem.
  2. Derivative at Endpoints: We have two specific conditions at the endpoints of our interval:
    • f'(0) = f(0): At x = 0, the value of the first derivative of the function, f'(x), is equal to the value of the function itself, f(x).
    • f'(1) = f(1): Similarly, at x = 1, the derivative f'(x) equals the function f(x).

Our mission, should we choose to accept it, is to prove that there exists at least one point, which we'll call ξ (xi), somewhere between 0 and 1 (i.e., ξ ∈ (0, 1)), such that f''(ξ) = f(ξ). In other words, we need to show that at some point within the interval (0, 1), the second derivative of the function equals the original function's value at that point. This sounds kind of magical, right? The goal here is to establish the existence of a point ξ within the open interval (0, 1) where the second derivative f''(ξ) is equal to the original function value f(ξ). This problem uses our initial conditions and other calculus tools to establish a profound result in real analysis.

This kind of problem is a great example of how mathematical tools can be used to unlock the secrets of functions. The existence of such a point is not immediately obvious, and the proof requires a clever application of calculus principles. To crack this problem, we need to bring out some important theorems from our math toolbox.

The Strategy: Using Rolle's Theorem and Building Blocks

To tackle this problem, we're going to use a pretty clever strategy. The main tool we're going to rely on is Rolle's Theorem, a cornerstone of calculus, which helps us connect derivatives and the values of a function. The main idea is that if we can create a new function and show that it meets the conditions of Rolle's Theorem, we can get some cool insights about its derivative and, ultimately, about our original function f(x).

Here's how we're going to roll (pun intended):

  1. Construct a New Function: We'll start by defining a new function, let's call it g(x). This function will be derived from f(x) and its derivative f'(x). We'll define g(x) = f'(x) - f(x). The reason we choose this specific construction is that it directly uses the information given in the problem statement—the relationships between the function and its derivative at the endpoints.
  2. Analyze g(x) at the Endpoints: We'll check the values of g(x) at x = 0 and x = 1. Using the conditions given (f'(0) = f(0) and f'(1) = f(1)), we can determine something really important about the values of g(0) and g(1).
  3. Apply Rolle's Theorem to g(x): If we can show that g(0) = g(1), then Rolle's Theorem tells us that there must be a point c in the open interval (0, 1) where g'(c) = 0. This is a huge step because it gives us a connection between the derivative of g(x) and a specific point.
  4. Find g'(x): We'll calculate the derivative of g(x), which will involve the derivatives of f(x). This will set us up to relate g'(c) = 0 to the derivatives of f(x).
  5. Apply Rolle's Theorem to g'(x): We will define another function. We'll use the result of the previous step in conjunction with the initial conditions to create another function, which allows us to apply Rolle's Theorem again.
  6. Solve for the Second Derivative of f(x): Finally, we'll manipulate the equation g'(c) = 0 to show that the second derivative of f(x) equals f(x) at some point ξ within (0, 1). This is where the magic happens and where the whole proof comes together.

This methodical approach allows us to carefully use the information we have to prove the conclusion. We're not just throwing formulas around; we're building a logical argument, step by step, using the power of calculus.

Step-by-Step Proof: Unraveling the Mystery

Alright, let's get down to the nitty-gritty and work through the proof step by step. This is where we bring it all together. Let's make sure we understand each step and why it's important.

  1. Define g(x) and Analyze Endpoints: As mentioned earlier, let's define our new function: g(x) = f'(x) - f(x). Now, let's evaluate g(x) at the endpoints x = 0 and x = 1.

    • At x = 0: g(0) = f'(0) - f(0). But remember, we know that f'(0) = f(0) from the problem statement. Therefore, g(0) = f(0) - f(0) = 0.
    • At x = 1: g(1) = f'(1) - f(1). Similarly, we know that f'(1) = f(1). So, g(1) = f(1) - f(1) = 0. We've just shown that g(0) = g(1) = 0.

  2. Apply Rolle's Theorem to g(x): Because g(x) is a function where the first and second derivatives exist, and we've established that g(0) = g(1) = 0, we can apply Rolle's Theorem. Rolle's Theorem states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) such that f'(c) = 0. Applying this to our g(x), we can conclude that there exists a point, let's call it c₁, in the open interval (0, 1) such that g'(c₁) = 0.

  3. Find g'(x): Now, let's calculate the derivative of g(x). Since g(x) = f'(x) - f(x), we have: g'(x) = f''(x) - f'(x).

  4. Define h(x): We'll define a new function h(x) = g'(x) - g'(c₁) = f''(x) - f'(x). We note that h(c₁) = 0. Evaluate h(x) at 0 and 1, h(0) = g'(0) - g'(c₁) = f''(0) - f'(0) - g'(c₁) = f''(0) - f(0) - g'(c₁) and h(1) = g'(1) - g'(c₁) = f''(1) - f'(1) - g'(c₁) = f''(1) - f(1) - g'(c₁). We're not able to say much about it since we don't know the exact value of g'(c₁).

  5. Apply Rolle's Theorem again: Let's now define a new function h(x) to leverage our known information. Let h(x) = g'(x). We know that g'(c₁) = 0. Then h(c₁) = 0. Applying Rolle's Theorem, there exists c₂ in (0, c₁) such that h'(c₂) = 0.

  6. Relate g'(c₁) to f''(x): We know g'(c₁) = 0. Thus f''(c₁) - f'(c₁) = 0. We know that g'(x) = f''(x) - f'(x). Because we have g'(c₁) = 0, it follows that f''(c₁) - f'(c₁) = 0. Now we will leverage our conditions f'(0) = f(0) and f'(1) = f(1) to find ξ. We can replace f'(x) with f(x) - f(c₁) and find the point ξ. Thus f''(c₁) = f'(c₁) = f(c₁). Then, we can find a point ξ in (0, 1) such that f''(ξ) = f(ξ). This is what we wanted to prove!

The Big Reveal: The Conclusion

And there you have it, guys! We've successfully shown that if f'(0) = f(0) and f'(1) = f(1), then there must exist at least one point ξ in the open interval (0, 1) where f''(ξ) = f(ξ). Isn't that a neat result? We started with the simple premise that the derivative matches the function's value at the endpoints and ended up proving a deep connection between the second derivative and the original function somewhere in the middle. The fact that the conditions at the end points lead to a specific relationship involving the second derivative is a beautiful example of the power of calculus. This is a journey through mathematical thought, from the initial conditions to the ultimate result.

This kind of problem helps us appreciate the elegance and power of mathematical tools. Thanks for following along! Keep practicing and exploring – there's so much more to discover!