Problem 15
Question
Suppose that \(f\) is continuous on \([a, b]\) and that \(f(a) f(b)<0\). Prove
Theorem 14 by filling in the details of the following argument.
(a) Apply the process of repeated bisection to construct two sequences
\(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) such that \(a \leq
a_{n}
Step-by-Step Solution
Verified Answer
The sequences \( \{a_{n}\} \) and \( \{b_{n}\} \) are constructed through the process of repeated bisection, which guarantees that \( a_{n} ≤ b_{n} \) and that \( f(a_{n})f(b_{n}) ≤ 0 \). The limit of the sequences under the conditions \( f(c)=0 \), \( a < c < b \), and \( f(a) \neq f(b) \) is \( c \).
1Step 1: Construction of Sequences
The process of repeated bisection starts by defining the first terms of each sequence as \( a_{1} = a \) and \( b_{1} = b \). For each next term, we define the midpoint \( m = \frac{a_{n} + b_{n}}{2} \). If \( f(m) = 0 \), process can stop (as we've found a root). If not, we check if \( f(a_{n}) f(m) < 0 \), if true then our next \( a \) remains the same and \( b_{n+1}=m \); else, \( a_{n+1}=m \) and \( b \) remains the same. This ensures that \( f(a_{n})f(b_{n}) ≤ 0 \) and \( b_{n} − a_{n} → 0 \) as \( n → ∞ \).
2Step 2: Proving the Limit
Considering \( \{a_{n}\} \) is increasing (as we are always forcing \( a_{n} \) towards \( b_{n} \)), bounded above (by \( b \)) and \( \{b_{n}\} \) is decreasing (we are pulling it towards \( a_{n} \)) and bounded below (by \( a \)), we can apply the Monotone Sequence Theorem which asserts that both sequences will converge. Assume that these two limits are the same and call this common limit \( c \). Thus, \( \lim_{n → ∞} a_{n} = \lim_{n → ∞} b_{n} = c \).
3Step 3: f(c) = 0
To show that \( f(c) = 0 \), we know that \( f \) is continuous on \( [a, b] \) and that the limit of a product is the product of the limits. So, if \( \lim_{n → ∞} f(a_{n})f(b_{n}) ≤ 0 \) then \( \lim_{n → ∞} f(b_{n}) = \lim_{n → ∞} f(a_{n}) = f(c) \). We also know that \( \lim_{n → ∞} f(a_{n}) ≤ 0 \) and \( \lim_{n → ∞} f(b_{n}) ≥ 0 \), therefore the only value that \( f(c) \) can take and respect these conditions is 0. Thus, f(c) = 0.
Key Concepts
Bisection MethodContinuous FunctionsConvergence of SequencesLimits and Continuity
Bisection Method
The bisection method is a root-finding algorithm that repeatedly narrows down an interval containing a root of a continuous function. You start with an interval \[a, b\] such that the function \(f\) has opposite signs at the endpoints, which means \(f(a)f(b)<0\). This opposite sign condition indicates there is at least one root within the interval due to the Intermediate Value Theorem.
Here's how the bisection method works in a nutshell:
Here's how the bisection method works in a nutshell:
- Find the midpoint \(m\) of the interval: \(m = \frac{a + b}{2}\).
- Check the value of the function at the midpoint: \(f(m)\).
- If \(f(m) = 0\), you've found a root and can stop.
- If \(f(m) \eq 0\), determine the half-interval in which the root lies by evaluating the sign of \(f(a)f(m)\).
- Update the interval to the half that contains the root and repeat the process.
Continuous Functions
A continuous function is one where small changes in the input lead to small changes in the output. In mathematical language, a function \(f\) is continuous at a point \(x = a\) if \(\lim_{x \to a} f(x) = f(a)\). In simpler terms, this means as \(x\) gets very close to \(a\), the function's value \(f(x)\) gets very close to \(f(a)\).
For intervals, a function is said to be continuous on an interval [a, b] if it is continuous at every point in that interval. The importance of continuity in the context of the Intermediate Value Theorem is that it guarantees the presence of a root within an interval where the function changes sign. This behavior is crucial for the bisection method because it ensures that the interval \( [a, b]\) indeed contains a value \(c\) for which \(f(c) = 0\), given that \(f(a)\) and \(f(b)\) have opposite signs.
For intervals, a function is said to be continuous on an interval [a, b] if it is continuous at every point in that interval. The importance of continuity in the context of the Intermediate Value Theorem is that it guarantees the presence of a root within an interval where the function changes sign. This behavior is crucial for the bisection method because it ensures that the interval \( [a, b]\) indeed contains a value \(c\) for which \(f(c) = 0\), given that \(f(a)\) and \(f(b)\) have opposite signs.
Convergence of Sequences
Convergence is a fundamental concept in calculus. A sequence \(\{x_n\}\) converges to a limit \(L\) if the terms in the sequence can be made arbitrarily close to \(L\) by making \(n\) large enough. Formally, \(\lim_{n \to \infty}x_n = L\) means that for any \(\varepsilon > 0\), however small, there exists a natural number \(N\) so that for all \(n > N\), the distance between \(x_n\) and \(L\) is less than \(\varepsilon\).
In the context of the bisection method, you create two sequences \(\{a_n\}\) and \(\{b_n\}\) that converge to the same limit \(c\), the root of the function. This is because the difference between \(b_n\) and \(a_n\), \(b_n - a_n\), becomes very small as \(n\) increases, ensuring that the two sequences trap the root \(c\) within an ever-shrinking interval. The bisection method harnesses this principle to ensure that the sequence of midpoints converges to the actual root.
In the context of the bisection method, you create two sequences \(\{a_n\}\) and \(\{b_n\}\) that converge to the same limit \(c\), the root of the function. This is because the difference between \(b_n\) and \(a_n\), \(b_n - a_n\), becomes very small as \(n\) increases, ensuring that the two sequences trap the root \(c\) within an ever-shrinking interval. The bisection method harnesses this principle to ensure that the sequence of midpoints converges to the actual root.
Limits and Continuity
The concepts of limits and continuity are intimately connected. A limit describes the behavior of a function as the input approaches a particular value. Continuity, on the other hand, relies on the concept of a limit and ensures that the function doesn’t have any 'jumps' or 'breaks'.
An important aspect of both limits and continuity is their role in the Intermediate Value Theorem. This theorem states that if a function \(f\) is continuous on a closed interval \[a, b\] and \(d\) is any number between \(f(a)\) and \(f(b)\), then there is at least one number \(c\) in the interval \[a, b\] such that \(f(c) = d\).
When applying this theorem in the bisection method, both sequences \(\{a_n\}\) and \(\{b_n\}\) have limits. The trapped root \(c\) is the limit towards which both sequences converge. Thanks to the property of continuity, we can confirm that \(f(c)\) is the desired output - typically zero for root-finding scenarios - thereby proving that the limit as \(n\) approaches infinity for the values of \(f(a_n)\) and \(f(b_n)\) is indeed \(f(c)\).
An important aspect of both limits and continuity is their role in the Intermediate Value Theorem. This theorem states that if a function \(f\) is continuous on a closed interval \[a, b\] and \(d\) is any number between \(f(a)\) and \(f(b)\), then there is at least one number \(c\) in the interval \[a, b\] such that \(f(c) = d\).
When applying this theorem in the bisection method, both sequences \(\{a_n\}\) and \(\{b_n\}\) have limits. The trapped root \(c\) is the limit towards which both sequences converge. Thanks to the property of continuity, we can confirm that \(f(c)\) is the desired output - typically zero for root-finding scenarios - thereby proving that the limit as \(n\) approaches infinity for the values of \(f(a_n)\) and \(f(b_n)\) is indeed \(f(c)\).
Other exercises in this chapter
Problem 14
(a) Show that any heated tetrahedron must have three points located on its edges or vertices that have the same temperature. (b) Can you prove there must actual
View solution Problem 15
Formulate the definition of continuity for a complex-valued function \(f\). Show that \(f\) is continuous if and only if its real and imaginary parts are contin
View solution Problem 16
(a) Setting \(z=x+i y=(x, y)\), consider the function \(f\) defined from complex numbers to complex numbers \(\left(\mathbf{R}^{2}\right.\) to \(\left.\mathbf{R
View solution Problem 16
Let \(F(x, y)=(x-y)^{2}\). Then show that \(\max _{0 \leq x \leq 1} \min _{0 \leq y \leq 1} F(x, y)=0\) and \(\min _{0 \leq y \leq 1} \max _{0 \leq x \leq 1} F(
View solution