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Chapter 5: Problem 10

Find sup \(\left\\{\epsilon \mid N_{\epsilon}\left(\mathbf{X}_{0}\right) \subset S\right\\}\) (a) \(\mathbf{X}_{0}=(1,2,-1,3) ; S=\) the open 4 -ball of radius 7 about (0,3,-2,2) (b) \(\mathbf{X}_{0}=(1,2,-1,3) ; S=\left\\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right)|| x_{i} \mid \leq 5,1 \leq i \leq 4\right\\}\) (c) \(\mathbf{X}_{0}=\left(3, \frac{5}{2}\right) ; S=\) the closed triangle with vertices \((2,0),(2,2),\) and (4,4)

Short Answer

Expert verified
In each scenario, the supremum of the set created by the radii of neighborhoods 饾憗_饾湒(饾憢鈧) 鈯 S was calculated as follows: (a) sup{\(\epsilon |饾憗_饾湒(饾憢鈧) 鈯 S\)} = 5 (b) sup{\(\epsilon |饾憗_饾湒(饾憢鈧) 鈯 S\)} = 3 (c) sup{\(\epsilon |饾憗_饾湒(饾憢鈧) 鈯 S\)} 鈮 0.47

Step by step solution

01

Case (a)

The given set S is an open 4-ball with radius 7 and center 饾憥 = (0,3,-2,2). The vector 饾憢鈧 is given as (1,2,-1,3). We want to find the largest 饾湒 such that 饾憗_饾湒(饾憢鈧) 鈯 S. By considering the distance between 饾憢鈧 and 饾憥, we can find the maximum 饾湒 in this case. Calculate the distance between 饾憢鈧 and 饾憥 using the Euclidean norm: \(d\left(\mathbf{X}_{0},\mathbf{a}\right)=\left\Vert\ \mathbf{X}_{0}-\mathbf{a}\ \right\Vert=\left\Vert\ (1,2,-1,3)-(0,3,-2,2)\ \right\Vert=\left\Vert\ (1,-1,1,1)\ \right\Vert=\sqrt{1^{2}+(-1)^{2}+1^{2}+1^{2}}=\sqrt{4}=2\) The radius of the open ball S is 7, hence, the largest possible 饾湒 is 7 - d(饾憢鈧, 饾憥) = 7 - 2 = 5.
02

Supremum for Case (a)

In this case, the supremum is 5, with sup{\(\epsilon |饾憗_饾湒(饾憢鈧) 鈯 S\)} = 5.
03

Case (b)

In this case, the set S is given by the coordinates of the vector satisfying the inequality \(|x_i| \leq 5\). We are given the same 饾憢鈧 as in part (a). We want to find the largest 饾湒 such that 饾憗_饾湒(饾憢鈧) 鈯 S. Notice that the maximum coordinate change from 饾憢鈧 for each dimension is 5 - (-5) = 10. The largest change in any single coordinate will determine our 饾湒 in this case. Our 饾憢鈧 = (1,2,-1,3). For each coordinate: - x_1: 5 - 1 = 4 - x_2: 5 - 2 = 3 - x_3: -1 - (-5) = 4 - x_4: 5 - 3 = 2
04

Supremum for Case (b)

The largest change in the coordinates is related to x_2, so the supremum for this case is 3, with sup{\(\epsilon |饾憗_饾湒(饾憢鈧) 鈯 S\)} = 3.
05

Case (c)

For this last case, our set S is a closed triangle made up of vertices 饾憥 = (2,0), 饾憦 = (2,2), and 饾憪 = (4,4). The vector 饾憢鈧 is given as (3, 5/2). We want to find the largest 饾湒 such that 饾憗_饾湒(饾憢鈧) 鈯 S. The maximum possible 饾湒 will be limited by the distance from 饾憢鈧 to the closest vertex or side of the triangle. Calculate the distances from 饾憢鈧 to each vertex: - d(饾憢鈧, 饾憥) = 鈭(3, 5/2) - (2, 0)鈭 = 鈭(1, 5/2)鈭 = 鈭(1+6.25) = 鈭7.25 - d(饾憢鈧, 饾憦) = 鈭(3, 5/2) - (2, 2)鈭 = 鈭(1, 1/2)鈭 = 鈭(1+0.25) = 鈭1.25 - d(饾憢鈧, 饾憪) = 鈭(3, 5/2) - (4, 4)鈭 = 鈭(-1, -3/2)鈭 =鈭(1+2.25) = 鈭3.25 The minimum distance from 饾憢鈧 to the vertices is d(饾憢鈧, 饾憦) = 鈭1.25. Now, find the distance from 饾憢鈧 to the sides of the triangle. If start by finding the normal vector of the line containing each side and the corresponding unit normal, we can find the distance from 饾憢鈧 to the side by using dot product. For example, let's analyze the side between vertex 饾憥 and 饾憦 (whose line can be described by \(x_2 = 2x_1 - 4\)): 1. Find the normal vector of the side 饾憥饾憦: 饾憶 = (2, -1) 2. Normalize the normal vector 饾憶: 饾憿 = 饾憶/||饾憶|| = (2/鈭(2虏+1虏), -1/鈭(2虏+1虏)) = (2鈭2/3, -1鈭2/3) 3. Calculate the distance from 饾憢鈧 to the line of side 饾憥饾憦 using ||(饾憢鈧 - 饾憥)路饾憿||: \(||((-1/2,5/2)-鉄2,0鉄)路(2\sqrt{2}/3,-\sqrt{2}/3 )||\approx0.47\) Since the distance from 饾憢鈧 to any vertex is larger than the distance from 饾憢鈧 to the side 饾憥饾憦, the 饾湒 will be limited by the 饾憢鈧's distance to the side 饾憥饾憦.
06

Supremum for Case (c)

In this case, the supremum is determined by the distance to the side 饾憥饾憦, with sup{\(\epsilon |饾憗_饾湒(饾憢鈧) 鈯 S\)} 鈮 0.47

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Supremum
Supremum is a fundamental concept in real analysis referred to as the smallest upper bound of a set. For example, if you have a set of numbers and you're trying to find the highest value that you can approach without actually reaching or going over, that's the supremum. It's particularly crucial when dealing with sets that have no maximum element.

In our exercise, when we talk about finding the supremum of \(\epsilon\) such that \(N_{\epsilon}(\mathbf{X}_{0})\subset S\), we're essentially searching for the largest radius of a neighborhood around \(\mathbf{X}_{0}\) that still fits entirely within the set S. This concept helps to characterize the borders of a set in a precise form which can be quite handy in many areas of mathematics.
Euclidean Norm
Euclidean norm, often represented as \(||\cdot||\), is a measure of length in a space. It's derived from the Pythagorean theorem and computes the 'straight-line' distance from the origin to a point in Euclidean space.

In real analysis, calculating the Euclidean norm corresponds to determining the distance between two points in a multi-dimensional space. For instance, the distance between point \(\mathbf{X}_{0}\) and the center \(\mathbf{a}\) in the first part of the exercise is given by \(||\mathbf{X}_{0}-\mathbf{a}||\). Here, although we're working in a four-dimensional space, the principle is the same as in everyday three-dimensional geometry.
Open and Closed Sets
Open and closed sets are vital when determining the nature of a set with respect to its boundary points. An open set does not include its boundary; you can think of it like a circle that doesn't include the circle's edge itself. Conversely, a closed set does include all its boundary points 鈥 it's like drawing a circle and including the lining as part of the circle.

Examples from our textbook exercise illustrate this well: the open 4-ball of radius 7 around point \((0,3,-2,2)\) is an example of an open set, whereas the closed triangle in part (c) of the exercise is an example of a closed set. In real-world terms, understanding open and closed sets helps to define spaces that are complete or not鈥攊mpacting various fields, including physics and computer graphics.
Distance in Metric Spaces
In the context of metric spaces, distance is a way of defining the space between elements within that set. A metric space is a set equipped with a metric, which is a function that quantitatively defines the distance between two points.

This concept allows us to measure how 'far apart' elements are in a more abstract sense, and it's foundational in real analysis because it introduces precise notions of limits and convergence. For instance, part of our exercise involves calculating the distance of a point from the vertices of a triangle, which requires understanding that the space is equipped with a standard metric, the Euclidean distance.

Practically speaking, these measurements play a fundamental role in areas like geometry, data analysis, and complex physics simulations, which rely on precise calculations of distances.

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Most popular questions from this chapter

Let \(h(\mathbf{U})=f(\mathbf{G}(\mathbf{U}))\) and find \(d_{\mathbf{U}_{0}} h\) by Theorem 5.4.3, and then by writing \(h\) explicitly as a function of \(\mathbf{U}\). (a) \(\begin{aligned} f(x, y) &=3 x^{2}+4 x y^{2}+3 x \\ g_{1}(u, v) &=v e^{u+v-1}, \\ g_{2}(u, v) &=e^{-u+v-1} \end{aligned} \quad\left(u_{0}, v_{0}\right)=(0,1)\) (b) \(\begin{aligned} f(x, y, z) &=e^{-(x+y+z)} \\ g_{1}(u, v, w) &=\log u-\log v+\log w \\ g_{2}(u, v, w) &=-2 \log u-3 \log w \\ g_{3}(u, v, w) &=\log u+\log v+2 \log w \end{aligned}\) \(\left(u_{0}, v_{0}, w_{0}\right)=(1,1,1)\) (c) \(\begin{aligned} f(x, y) &=(x+y)^{2}, \\ g_{1}(u, v) &=u \cos v, \quad\left(u_{0}, v_{0}\right)=(3, \pi / 2) \\ g_{2}(u, v) &=u \sin v, \end{aligned}\) (d) \(\begin{aligned} f(x, y, z) &=x^{2}+y^{2}+z^{2} \\ g_{1}(u, v, w) &=u \cos v \sin w \\ g_{2}(u, v, w) &=u \cos v \cos w, \\ g_{3}(u, v, w) &=u \sin v \end{aligned}\) \(\left(u_{0}, v_{0}\right)=(3, \pi / 2)\)

Prove: If $$ u(x, t)=f(x-c t)+g(x+c t) $$ then \(u_{t t}=c^{2} u_{x x}\)

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Suppose that \(p\) is a homogeneous polynomial of degree \(r\) in \(\mathbf{Y}\) and \(p(\mathbf{Y})>0\) for all nonzero \(\mathbf{Y}\) in \(\mathbb{R}^{n}\). Show that there is a \(\rho>0\) such that \(p(\mathbf{Y}) \geq \rho|\mathbf{Y}|^{r}\) for all \(\mathbf{Y}\) in \(\mathbb{R}^{n}\). HINT: \(p\) assumes a minimum on the set \(\\{\mathbf{Y}|| \mathbf{Y} \mid=1\\}\). Use this to establish the inequality in Eqn. (5.4.41).

Find the equation of the line segment from \(\mathbf{X}_{0}\) to \(\mathbf{X}_{1}\). (a) \(\mathbf{X}_{0}=(1,-3,4,2), \quad \mathbf{X}_{1}=(2,0,-1,5)\) (b) \(\mathbf{X}_{0}=(3,1-2,1,4), \quad \mathbf{X}_{1}=(2,0,-1,4,-3)\) (c) \(\mathbf{X}_{0}=(1,2,-1), \quad \mathbf{X}_{1}=(0,-1,-1)\)
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