Problem 83 - Calculus Single Variable

Question

Suppose that \(f\) satisfies the hypotheses of Rolle's Theorem on \([a, b]\). In this exercise we will construct a sequence \(\left\\{\left[a_{n}, b_{n}\right]\right\\}_{n=0}^{\infty}\) of intervals with the properties (1) \(\left[a_{n}, b_{n}\right] \subset\left[a_{n-1}, b_{n-1}\right] \quad\) for \(n \geq 1\); (2) \(b_{n}-a_{n}=\left(b_{n-1}-a_{n-1}\right) / 2\) for \(n \geq 1\) (3) \(f\left(a_{n}\right)=f\left(b_{n}\right)\) for \(n \geq 0 ;\) and (4) There exists a point \(c \in(a, b)\) for which $$ \begin{aligned} c=\lim _{n \rightarrow \infty} a_{n} &=\lim _{n \rightarrow \infty} b_{n} \text { and } \\ & f^{\prime}(c)=\lim _{n \rightarrow \infty} \frac{f\left(b_{n}\right)-f\left(a_{n}\right)}{b_{n}-a_{n}}=0 . \end{aligned} $$ If \(f\left((a+b / 2)=f(a),\right.\) then set \(a_{0}=(a+b) / 2\) and \(b_{0}=b\). Otherwise, set \(a_{0}=a\) and \(b_{0}=b\). Let \(g(x)=\) \(f\left(x+\left(b_{0}-a_{0}\right) / 2\right)-f(x) .\) Use the Intermediate Value Theorem to find an \(a_{1} \in\left[a_{0},\left(a_{0}+b_{0}\right) / 2\right)\) such that \(g\left(a_{1}\right)=0 . \quad\) Let \(\quad b_{1}=a_{1}+(b-a) / 2 .\) Show that \(f\left(a_{1}\right)=f\left(b_{1}\right) .\) Replacing \(a_{0}\) and \(b_{0}\) with \(a_{1}\) and \(b_{1}\) in the definition of \(g,\) find \(a_{2}\) and \(b_{2}\) for which properties (1),(2) and (3) hold for \(n=2\). Repeat this construction indefinitely. Prove property (4), thereby establishing Rolle's Theorem

Step-by-Step Solution

Answer

Rolle's Theorem holds as \(f'(c) = 0\) for \(c\) such that \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = c\) with \(f(a_n) = f(b_n)\).

Step 1: Initial Setup

Determine the initial interval \([a_0, b_0]\). If \( f\left(\frac{a+b}{2}\right) = f(a) \), set \(a_0 = \frac{a+b}{2}\) and \(b_0 = b\). Otherwise, set \(a_0 = a\) and \(b_0 = b\). Let's determine the function \(g(x) = f\left(x + \frac{b_0 - a_0}{2}\right) - f(x)\).

Step 2: Finding a1

Use the Intermediate Value Theorem (IVT) on \([a_0, \frac{a_0 + b_0}{2}]\) to find \(a_1\) such that \(g(a_1) = 0\). This is possible because \(g(a_0)\) and \(g\left(\frac{a_0 + b_0}{2}\right)\) have opposite signs. Set \(b_1 = a_1 + \frac{b-a}{2}\).

Step 3: Verify g at a1

By choosing \(a_1\) such that \(g(a_1) = 0\), we have \(f\left(a_1 + \frac{b_0 - a_0}{2}\right) = f(a_1)\). This implies \(f(b_1) = f(a_1)\), where \(b_1 = a_1 + \frac{b-a}{2}\) matches the requirement for \(f(a_1) = f(b_1)\).

Step 4: Constructing a2 and b2

Set up \([a_2, b_2]\) similar to the preceding step with the interval \([a_1, b_1]\). Define \(g(x) = f\left(x + \frac{b_1 - a_1}{2}\right) - f(x)\) on \([a_1, \frac{a_1 + b_1}{2}]\) and find \(a_2\) using IVT as before. Define \(b_2 = a_2 + \frac{b-a}{4}\) to maintain \(b_2 - a_2 = \frac{b_1 - a_1}{2}\).

Step 5: Repeat Construction to Infinity

Continue constructing \([a_n, b_n]\) where each interval halves the width of the previous one and ensuring \(f(a_n) = f(b_n)\). This ensures the sequence \(\{a_n\}\) and \(\{b_n\}\) converge to the same limit \(c\).

Step 6: Prove Property 4

Since \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = c\) and \(f(a_n) = f(b_n)\), compute \(\lim_{n \to \infty} \frac{f(b_n) - f(a_n)}{b_n - a_n} = 0\). By continuity and differentiability (hypotheses of Rolle's Theorem), \(f'(c) = 0\).

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that guarantees the existence of a value within a certain interval for continuous functions. Suppose we have a continuous function, say \( f \), on a closed interval \([a, b]\). If \( f(a) \) and \( f(b) \) are of opposite signs, the IVT asserts that there is at least one point \( c \) in the interval \((a, b)\) where \( f(c) = 0 \).
This theorem is immensely useful for proving that solutions to equations exist within certain bounds.

In our context, it helps in constructing subsequences of intervals \([a_n, b_n]\), ensuring that within these shrinking intervals, specific values meet certain conditions consistently. Thanks to this theorem, we were able to find \( a_1 \), ensuring that \( g(a_1) = 0 \), thus enabling the sequence process to move forward for the subsequent steps.

Function Sequences

Function sequences, particularly those converging within intervals, play a crucial role in calculus and analysis. A function sequence, in essence, is a sequence of functions \( \{f_n(x)\} \) which converges to a specific function \( f(x) \) as \( n \rightarrow \infty \).

In our exercise, we construct a sequence of nested intervals \([a_n, b_n]\) such that each interval is contained within the preceding one and has halved its width. This ensures the convergence of both endpoints \( a_n \) and \( b_n \) of the interval to the same limit \( c \).

• The shrinking intervals are crucial because they provide a way to zoom in closer to this point \( c \), where special properties of the function can be observed or proven.
• The beauty of function sequences lies in this ability to make sweeping generalizations not just at a point but across an entire interval.

Such sequences also highlight the rigorous nature of calculus proofs, ensuring that no steps are skipped in determining convergence.

Calculus Proofs

Proofs are the backbone of mathematics, providing certainty and establishing truth. Calibrating with calculus, proofs involve step-by-step logical reasoning to demonstrate specific properties or statements. In this exercise, we're interested in proving the conclusion drawn by Rolle's Theorem.

Our step-by-step approach:

• We start by constructing sequences of intervals that are progressively narrower.
• Using properties like continuity and differentiability, we ensure these sequences adhere to specified requirements.
• The ultimate goal is to arrive at the point \( c \) where the derivative \( f'(c) = 0 \), validating Rolle's Theorem.

Proofs in calculus not only involve algebraic manipulations but also a deep understanding of the underlying concepts involved, like sequences, limits, and continuity, which culminate in establishing rigorous results like this one.

Differentiable Functions

Differentiable functions are a cornerstone of calculus, defined by their ability to have a derivative at every point in their domain. A function is differentiable on an interval when it has a derivative at every point within that interval, meaning it changes at a consistent rate.

For Rolle's Theorem, differentiability is one of the crucial hypotheses. It ensures that not only the function is smooth (without any jumps or sharp corners) but also that we can find a horizontal tangent (i.e., derivative equals zero) somewhere on the curve.

• This characteristic is critical because without differentiability, we can't guarantee the existence of such a tangent, thereby failing the theorem.
• From a learning standpoint, understanding differentiability helps in appreciating the beauty of how smoothly functions behave and can be analyzed.

In our exercise, the differentiability of \( f \) allowed us to apply the theorem confidently, knowing that the conditions required are firmly satisfied.