### Theorem Let \( D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\} \) be the closed unit disk, visualized below:  ### Explanation of Diagram: The above diagram shows the closed unit disk \( D \) centered at the origin \((0, 0)\) with a radius of 1. The disk includes the boundary where \( x^2 + y^2 = 1 \). Let \( f : D \to D \) be a function. If \( f \) is continuous, then \( f \) has a fixed point, i.e., there exists some point \((x_0, y_0) \in D\) such that \( f(x_0, y_0) = (x_0, y_0) \). The proof of this theorem is non-constructive, meaning that it does not actually produce a point \((x_0, y_0) \in D\) that is fixed by a given continuous function \( f \); it just guarantees that such a point exists. According to Brouwer, one of the mathematicians credited with the theorem, his pursuit of this result was inspired by observing a stirred cup of coffee—it somehow appears that there is always a region in the swirling cup that is not moving. We have not rigorously defined “continuous” in this class (see Real Analysis for such a definition), but that is not the point of this Problem. All you need to know is that every map \( f : D \to D \) is either continuous or not, but never both. #### Problems: (i) Given a continuous map \( f : D \to D \), what does **this theorem** allow you to conclude? (ii) Given a function \( f : D \to D \) that is not continuous, what does **this theorem** allow you to conclude? (iii) Given a function \( f : D \to D \) for which you are able to verify that \( f(x, y) \neq (x, y) \) for all \((x, y) \in D\), what does **this theorem** allow you to conclude?
### Theorem Let \( D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\} \) be the closed unit disk, visualized below:  ### Explanation of Diagram: The above diagram shows the closed unit disk \( D \) centered at the origin \((0, 0)\) with a radius of 1. The disk includes the boundary where \( x^2 + y^2 = 1 \). Let \( f : D \to D \) be a function. If \( f \) is continuous, then \( f \) has a fixed point, i.e., there exists some point \((x_0, y_0) \in D\) such that \( f(x_0, y_0) = (x_0, y_0) \). The proof of this theorem is non-constructive, meaning that it does not actually produce a point \((x_0, y_0) \in D\) that is fixed by a given continuous function \( f \); it just guarantees that such a point exists. According to Brouwer, one of the mathematicians credited with the theorem, his pursuit of this result was inspired by observing a stirred cup of coffee—it somehow appears that there is always a region in the swirling cup that is not moving. We have not rigorously defined “continuous” in this class (see Real Analysis for such a definition), but that is not the point of this Problem. All you need to know is that every map \( f : D \to D \) is either continuous or not, but never both. #### Problems: (i) Given a continuous map \( f : D \to D \), what does **this theorem** allow you to conclude? (ii) Given a function \( f : D \to D \) that is not continuous, what does **this theorem** allow you to conclude? (iii) Given a function \( f : D \to D \) for which you are able to verify that \( f(x, y) \neq (x, y) \) for all \((x, y) \in D\), what does **this theorem** allow you to conclude?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Explanation of Diagram:
The above diagram shows the closed unit disk \( D \) centered at the origin \((0, 0)\) with a radius of 1. The disk includes the boundary where \( x^2 + y^2 = 1 \).
Let \( f : D \to D \) be a function. If \( f \) is continuous, then \( f \) has a fixed point, i.e., there exists some point \((x_0, y_0) \in D\) such that \( f(x_0, y_0) = (x_0, y_0) \).
The proof of this theorem is non-constructive, meaning that it does not actually produce a point \((x_0, y_0) \in D\) that is fixed by a given continuous function \( f \); it just guarantees that such a point exists. According to Brouwer, one of the mathematicians credited with the theorem, his pursuit of this result was inspired by observing a stirred cup of coffee—it somehow appears that there is always a region in the swirling cup that is not moving.
We have not rigorously defined “continuous” in this class (see Real Analysis for such a definition), but that is not the point of this Problem. All you need to know is that every map \( f : D \to D \) is either continuous or not, but never both.
#### Problems:
(i) Given a continuous map \( f : D \to D \), what does **this theorem** allow you to conclude?
(ii) Given a function \( f : D \to D \) that is not continuous, what does **this theorem** allow you to conclude?
(iii) Given a function \( f : D \to D \) for which you are able to verify that \( f(x, y) \neq (x, y) \) for all \((x, y) \in D\), what does **this theorem** allow you to conclude?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd761fbf2-5ed1-47eb-bdd4-1a6b42441150%2F90e42e68-36d3-4589-be5e-8dd68754ce16%2Fgi4oul4_processed.png&w=3840&q=75)
Transcribed Image Text:### Theorem
Let \( D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\} \) be the closed unit disk, visualized below:
### Explanation of Diagram:
The above diagram shows the closed unit disk \( D \) centered at the origin \((0, 0)\) with a radius of 1. The disk includes the boundary where \( x^2 + y^2 = 1 \).
Let \( f : D \to D \) be a function. If \( f \) is continuous, then \( f \) has a fixed point, i.e., there exists some point \((x_0, y_0) \in D\) such that \( f(x_0, y_0) = (x_0, y_0) \).
The proof of this theorem is non-constructive, meaning that it does not actually produce a point \((x_0, y_0) \in D\) that is fixed by a given continuous function \( f \); it just guarantees that such a point exists. According to Brouwer, one of the mathematicians credited with the theorem, his pursuit of this result was inspired by observing a stirred cup of coffee—it somehow appears that there is always a region in the swirling cup that is not moving.
We have not rigorously defined “continuous” in this class (see Real Analysis for such a definition), but that is not the point of this Problem. All you need to know is that every map \( f : D \to D \) is either continuous or not, but never both.
#### Problems:
(i) Given a continuous map \( f : D \to D \), what does **this theorem** allow you to conclude?
(ii) Given a function \( f : D \to D \) that is not continuous, what does **this theorem** allow you to conclude?
(iii) Given a function \( f : D \to D \) for which you are able to verify that \( f(x, y) \neq (x, y) \) for all \((x, y) \in D\), what does **this theorem** allow you to conclude?
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