i=1 XVi S Σ. x \i=1 1/2 Σ izl 1/2
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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![**Cauchy-Schwarz Inequality Explanation**
For any vectors \((x_1, \ldots, x_n)\) and \((y_1, \ldots, y_n)\) in \(\mathbb{R}^n\), the Cauchy-Schwarz inequality is demonstrated as follows:
\[
\sum_{i=1}^{n} x_i y_i \leq \left( \sum_{i=1}^{n} x_i^2 \right)^{1/2} \left( \sum_{i=1}^{n} y_i^2 \right)^{1/2}
\]
This inequality states that the absolute value of the dot product of two vectors is less than or equal to the product of their Euclidean norms. It is a fundamental result in linear algebra and has numerous applications in mathematics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2F096e0da5-26e4-42ff-8671-b52075de417c%2F7kaltgc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Cauchy-Schwarz Inequality Explanation**
For any vectors \((x_1, \ldots, x_n)\) and \((y_1, \ldots, y_n)\) in \(\mathbb{R}^n\), the Cauchy-Schwarz inequality is demonstrated as follows:
\[
\sum_{i=1}^{n} x_i y_i \leq \left( \sum_{i=1}^{n} x_i^2 \right)^{1/2} \left( \sum_{i=1}^{n} y_i^2 \right)^{1/2}
\]
This inequality states that the absolute value of the dot product of two vectors is less than or equal to the product of their Euclidean norms. It is a fundamental result in linear algebra and has numerous applications in mathematics.
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