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Chapter 1: Problem 13

Use induction on \(k\) to prove that if \(\mathbf{x}_{1}, \ldots, \mathbf{x}_{k} \in \mathbb{R}^{n},\) then $$\left\|\mathbf{x}_{1}+\cdots+\mathbf{x}_{k}\right\|\leq\left\|\mathbf{x}_{1}\right\|+\cdots+\left\|\mathbf{x}_{k}\right\|$$.

Short Answer

Expert verified
The inequality holds for all integers \( k \) by induction.

Step by step solution

01

Base Case (k=1)

For the base case, consider \( k = 1 \). Then, the statement \( \left\| \mathbf{x}_1 \right\| \leq \left\| \mathbf{x}_1 \right\| \) holds true as both sides are equal, thereby demonstrating the inequality for \( k = 1 \).
02

Inductive Hypothesis

Assume that the inequality holds for some integer \( k = m \), i.e., \( \left\| \mathbf{x}_1 + \cdots + \mathbf{x}_m \right\| \leq \left\| \mathbf{x}_1 \right\| + \cdots + \left\| \mathbf{x}_m \right\| \). This is our inductive hypothesis.
03

Inductive Step (k = m + 1)

We need to prove the inequality for \( k = m + 1 \). Consider:\[\left\| \mathbf{x}_1 + \cdots + \mathbf{x}_{m+1} \right\| = \left\| (\mathbf{x}_1 + \cdots + \mathbf{x}_m) + \mathbf{x}_{m+1} \right\|\]By the triangle inequality, we have:\[\left\| (\mathbf{x}_1 + \cdots + \mathbf{x}_m) + \mathbf{x}_{m+1} \right\| \leq \left\| \mathbf{x}_1 + \cdots + \mathbf{x}_m \right\| + \left\| \mathbf{x}_{m+1} \right\|\]According to the inductive hypothesis:\[\left\| \mathbf{x}_1 + \cdots + \mathbf{x}_m \right\| \leq \left\| \mathbf{x}_1 \right\| + \cdots + \left\| \mathbf{x}_m \right\|\]Thus:\[\left\| \mathbf{x}_1 + \cdots + \mathbf{x}_{m+1} \right\| \leq (\left\| \mathbf{x}_1 \right\| + \cdots + \left\| \mathbf{x}_m \right\|) + \left\| \mathbf{x}_{m+1} \right\|\]Hence, the inequality holds for \( k = m + 1 \).

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Key Concepts

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

Triangle Inequality
In vector calculus, the triangle inequality is a fundamental concept that helps us understand distances in vector spaces. It's named after the geometric property of triangles, where the sum of the lengths of any two sides must be greater than or equal to the length of the third side.
In mathematical terms, the triangle inequality states that for any vectors \(\mathbf{a}\) and \(\mathbf{b}\), you have:
  • \(\|\mathbf{a} + \mathbf{b}\| \leq \|\mathbf{a}\| + \|\mathbf{b}\|\)
This idea is not just important for geometry, but crucial for proving inequalities involving sums of vectors in real vector spaces. It helps to establish boundaries for how long a combination of vectors can be. This principle is pivotal in the process of proving our exercise statement by induction.
Basis Step
The basis step, also known as the base case, is the starting point of any mathematical induction proof. This first step involves verifying that the statement to be proven is true for the initial value. In our problem, this is when \(k = 1\).

At this stage, we check if the inequality holds when there is just a single vector. This is typically quite straightforward. We see if \(\|\mathbf{x}_1\| \leq \|\mathbf{x}_1\|\), which holds true because they are obviously equal. This confirmation forms a solid foundation on which the rest of the induction process builds its argument. It's crucial because if the base case fails, the entire induction proof can't proceed.
Inductive Hypothesis
The inductive hypothesis is a key step in the induction process. In this step, we assume that the statement holds true for some arbitrary integer \(k = m\). This assumption is not about proving \(m\), but about laying the groundwork for showing the truth across all iterations.

In our exercise, we presume the inequality \(\|\mathbf{x}_1 + \cdots + \mathbf{x}_m\| \leq \|\mathbf{x}_1\| + \cdots + \|\mathbf{x}_m\|\) holds for \(m\) vectors. This strategic assumption underpins the next steps, as it provides the necessary condition to prove the statement for \(k = m + 1\). This step is crucial for bridging the base case to the more general case.
Norms in Real Vector Spaces
Norms in real vector spaces are mathematical functions that assign a non-negative length or size to vectors. This concept is important for understanding distances and magnitudes of vectors. The commonly used norm is the Euclidean norm, expressed as \(\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}\) for a vector \(\mathbf{x}\) in \(\mathbb{R}^n\).

Norms satisfy several properties:
  • Non-negativity: \(\|\mathbf{x}\| \geq 0\)
  • Zero Vector: \(\|\mathbf{x}\| = 0\) if and only if \(\mathbf{x} = \mathbf{0}\)
  • Vector Scaling: \(\|c\mathbf{x}\| = |c| \|\mathbf{x}\|\)
  • Triangle Inequality: \(\|\mathbf{x} + \mathbf{y}\| \leq \|\mathbf{x}\| + \|\mathbf{y}\|\)
These properties help establish the behavior of vectors within the space, which is necessary to understanding vector addition and the inequalities involving them. Understanding norms allows us to apply the triangle inequality to sums of vectors, which was essential in the induction proof of our exercise.

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Most popular questions from this chapter

The wind velocity \(\mathbf{v}_{1}\) is 40 miles per hour \((\mathrm{mi} / \mathrm{h})\) from east to west while an airplane travels with air speed \(\mathbf{v}_{2}\) of \(100 \mathrm{mi} / \mathrm{h}\) due north. The speed of the airplane relative to the ground is the vector sum \(\mathbf{v}_{1}+\mathbf{v}_{2}\) (a) Find \(\mathbf{v}_{1}+\mathbf{v}_{2}\) (b) Draw a figure to scale.

Interpret these results geometrically in terms of the parallelogram formed by \(x\) and \(y\). Verify the Cauchy-Schwarz inequality and the triangle inequality for the vectors. $$\mathbf{x}=(2,0,-1), \mathbf{y}=(4,0,-2)$$

Use vector methods to describe the given configurations.The plane determined by the three points \(\left(x_{0}, y_{0}, z_{0}\right),\left(x_{1}, y_{1}, z_{1}\right),\) and \(\left(x_{2}, y_{2}, z_{2}\right)\).

What restrictions must be made on the scalar \(b\) so that the vector \(2 \mathbf{i}+b \mathbf{j}\) is orthogonal to \((\mathrm{a})-3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}\) and (b) \(\mathbf{k} ?\)

(This exercise assumes a knowledge of integration of continuous functions of one variable.) Note that the proof of the Cauchy-Schwarz inequality (Theorem 4) depends only on the properties of the inner product listed in Theorem \(1 .\) Use this observation to establish the following inequality for continuous functions \(f, g:[0,1] \rightarrow \mathbb{R}\): $$\left|\int_{0}^{1} f(x) g(x) d x\right| \leq \sqrt{\int_{0}^{1}[f(x)]^{2} d x} \sqrt{\int_{0}^{1}[g(x)]^{2} d x}$$. Do this by (a) verifying that the space of continuous functions from [0,1] to \(\mathbb{R}\) forms a vector space; that is, we may think of functions \(f, g\) abstractly as "vectors" that can be added to each other and multiplied by scalars. (b) introducing the inner product of functions $$f \cdot g=\int_{0}^{1} f(x) g(x) d x$$ and verifying that it satisfies conditions (i) to (iv) of Theorem 3.
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