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Show that the Cauchy-Buniakowsky-Schwarz Inequality can be strengthened to $$ \sum_{i=1}^{n} x_{i} y_{i} \leq \sum_{i=1}^{n}\left|x_{i} y_{i}\right| \leq\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{1 / 2}\left(\sum_{i=1}^{n} y_{i}^{2}\right)^{1 / 2}. $$

Short Answer

Expert verified
This can be solved by first establishing the Cauchy-Schwarz inequality, then applying the property of absolute values to create a stronger inequality, resulting in the strengthened version of the Cauchy-Buniakowsky-Schwarz Inequality.

Step by step solution

01

Establish basic inequality

Start with the basic Cauchy-Schwarz inequality: \[ \sum_{i=1}^{n} x_{i} y_{i} \leq \sqrt{\sum_{i=1}^{n} x_{i}^{2} \sum_{i=1}^{n} y_{i}^{2}} \]
02

Use absolute value property

By recognizing that the absolute value of a real number is less or equal to the number itself, we get: \[ \sum_{i=1}^{n} x_{i} y_{i} \leq \sum_{i=1}^{n}|x_{i} y_{i}| \]
03

Combine the inequalities

Combine the two inequalities to get the strengthened Cauchy-Buniakowsky-Schwarz Inequality: \[ \sum_{i=1}^{n} x_{i} y_{i} \leq \sum_{i=1}^{n}|x_{i} y_{i}| \leq \sqrt{\sum_{i=1}^{n} x_{i}^{2} \sum_{i=1}^{n} y_{i}^{2}} \]

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Key Concepts

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

inequalities in mathematics
Inequalities are vital in mathematics because they allow us to express relationships between different mathematical expressions in terms of relative size. One famous example is the Cauchy-Buniakowsky-Schwarz (CBS) Inequality, a fundamental result in many areas such as analysis and algebra. This inequality establishes a limit within which the scalar product of two vectors lies, given by:
  • The left side expresses the sum of products of corresponding vector components.
  • On the right, it’s bounded by the product of vector magnitudes.
The strengthened version of the CBS Inequality goes further by adding a middle term involving absolute values. This enhancement highlights that the inner product is necessarily smaller than the sum of these absolute products, providing more detailed insight into the structure of vector relationships. Learning inequalities helps students grasp these relationships and apply them to solve equations by bounding potential solutions.
numerical analysis
Numerical analysis is the study of algorithms to solve problems in continuous mathematics. The CBS Inequality finds applications here as it can ensure stability and accuracy in calculations. Numerical methods often involve iterative approximations, where errors can compound and become significant.
The CBS Inequality helps to manage these errors. Consider when estimating solutions to differential equations – the inequality bounds the variations we can expect in outputs based on input inaccuracies.
By applying the CBS Inequality, one can:
  • Control round-off errors, reducing computation inaccuracies.
  • Provide a measure for error propagation in different algorithms.
Employing such inequalities is crucial for optimizing the performance of algorithms, especially in computer software that handles large and complex calculations.
vector algebra
Vector algebra involves operations on vectors, which are entities with both magnitude and direction. The CBS Inequality is a cornerstone in vector algebra, delineating the boundary of dot products. When you have two vectors \(\mathbf{x}\) and \(\mathbf{y}\), their dot product \(\mathbf{x} \cdot \mathbf{y}\) is intimately linked to their angles and lengths.
The CBS Inequality tells us about these boundaries:
  • The inequality ensures the dot product never exceeds the product of the vectors’ magnitudes.
  • It signifies a geometric interpretation: the dot product is maximum when both vectors are collinear.
  • The strengthened version further refines this by imposing limits with intermediate steps like absolute values.
Understanding this inequality strengthens your grasp on the geometric and algebraic implications of vector operations, aiding in the analysis of projections, forces, and other vector-related concepts.
mathematical proofs
Mathematical proofs are logical arguments that demonstrate the truth of mathematical statements. Proving inequalities, like the strengthened CBS Inequality, involves showing the relationship holds under any conditions of the involved variables.
To understand proofs better, it's important to grasp various types of proof techniques:
  • Direct Proof: Proceeds directly from known facts.
  • Contradiction: Assumes the negation of the statement to show an absurdity.
  • Induction: Shows a base case is true, and that if it's true for one case, it's true for the next.
For CBS, the proof generally involves algebraic manipulation and recognition of equalities to bridge the inequalities in sequences. Proving such inequalities not only enhances logical reasoning but also familiarizes students with bounding expressions crucial in theoretical and applied mathematics.

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Most popular questions from this chapter

The linear system $$ \begin{array}{r} 2 x_{1}-x_{2}+x_{3}=-1 \\ 2 x_{1}+2 x_{2}+2 x_{3}=4 \\ -x_{1}-x_{2}+2 x_{3}=-5 \end{array} $$ has the solution \((1,2,-1)^{t}\). a. Show that \(\rho\left(T_{j}\right)=\frac{\sqrt{5}}{2}>1\). b. Show that the Jacobi method with \(\mathbf{x}^{(0)}=\mathbf{0}\) fails to give a good approximation after 25 iterations. c. Show that \(\rho\left(T_{g}\right)=\frac{1}{2}\). d. Use the Gauss-Seidel method with \(\mathbf{x}^{(0)}=\mathbf{0}\) to approximate the solution to the linear system to within \(10^{-5}\) in the \(l_{\infty}\) norm.

Compute the eigenvalues and associated eigenvectors of the following matrices. a. \(\left[\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right]\) b. \(\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\) c. \(\left[\begin{array}{ll}0 & \frac{1}{2} \\ \frac{1}{2} & 0\end{array}\right]\) d. \(\left[\begin{array}{lll}2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3\end{array}\right]\) e. \(\left[\begin{array}{rrr}-1 & 2 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & 7\end{array}\right]\) f. \(\left[\begin{array}{lll}2 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2\end{array}\right]\)

The forces on the bridge truss described in the opening to this chapter satisfy the equations in the following table: $$ \begin{array}{ccc} \hline \text { Joint } & \text { Horizontal Component } & \text { Vertical Component } \\ \hline \text { (1) } & -F_{1}+\frac{\sqrt{2}}{2} f_{1}+f_{2}=0 & \frac{\sqrt{2}}{2} f_{1}-F_{2}=0 \\ \text { (2) } & -\frac{\sqrt{2}}{2} f_{1}+\frac{\sqrt{3}}{2} f_{4}=0 & -\frac{\sqrt{2}}{2} f_{1}-f_{3}-\frac{1}{2} f_{4}=0 \\ \text { (3) } & -f_{2}+f_{5}=0 & f_{3}-10,000=0 \\ \text { (4) } & -\frac{\sqrt{3}}{2} f_{4}-f_{5}=0 & \frac{1}{2} f_{4}-F_{3}=0 \\\ \hline \end{array} $$ This linear system can be placed in the matrix form $$ \left[\begin{array}{cccccccc} -1 & 0 & 0 & \frac{\sqrt{2}}{2} & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & \frac{\sqrt{2}}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{\sqrt{2}}{2} & 0 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -\frac{\sqrt{2}}{2} & 0 & 0 & \frac{\sqrt{3}}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -\frac{\sqrt{3}}{2} & -1 \end{array}\right]\left[\begin{array}{c} F_{1} \\ F_{2} \\ F_{3} \\ f_{1} \\ f_{2} \\ f_{3} \\ f_{4} \\ f_{5} \end{array}\right]=\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 10,000 \\ 0 \\ 0 \end{array}\right] . $$ a. Explain why the system of equations was reordered. b. Approximate the solution of the resulting linear system to within \(10^{-2}\) in the \(l_{\infty}\) norm using as initial approximation the vector all of whose entries are 1 s with (i) the Jacobi method and (ii) the Gauss-Seidel method.

Let $$ \begin{aligned} A_{1} &=\left[\begin{array}{rrrr} 4 & -1 & 0 & 0 \\ -1 & 4 & -1 & 0 \\ 0 & -1 & 4 & -1 \\ 0 & 0 & -1 & 4 \end{array}\right], \quad-I=\left[\begin{array}{rrrr} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right], \text { and } \\ O &=\left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned} $$ Form the \(16 \times 16\) matrix \(A\) in partitioned form, $$ A=\left[\begin{array}{cccc} A_{1} & -I & O & O \\ -I & A_{1} & -I & O \\ O & -I & A_{1} & -I \\ O & O & -I & A_{1} \end{array}\right] $$ Let \(\mathbf{b}=(1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6)^{t}\). a. Solve \(A \mathbf{x}=\mathbf{b}\) using the conjugate gradient method with tolerance \(0.05\). b. Solve \(A \mathbf{x}=\mathbf{b}\) using the preconditioned conjugate gradient method with \(C^{-1}=D^{-1 / 2}\) and tolerance \(0.05\). c. Is there any tolerance for which the methods of part (a) and part (b) require a different number of iterations?

The linear system $$ \begin{gathered} x_{1}+\frac{1}{2} x_{2}+\frac{1}{3} x_{3}=\frac{5}{6} \\ \frac{1}{2} x_{1}+\frac{1}{3} x_{2}+\frac{1}{4} x_{3}=\frac{5}{12} \\ \frac{1}{3} x_{1}+\frac{1}{4} x_{2}+\frac{1}{5} x_{3}=\frac{17}{60} \end{gathered} $$ has solution \((1,-1,1)^{t}\). a. Solve the linear system using Gaussian elimination with three-digit rounding arithmetic. b. Solve the linear system using the conjugate gradient method with three- digit rounding arithmetic. c. Does pivoting improve the answer in (a)? d. Repeat part (b) using \(C^{-1}=D^{-1 / 2}\). Does this improve the answer in (b)?

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