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Cont a ER are tig Cont at 9

Cont a ER are tig Cont at 9

Advanced Engineering Mathematics
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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Please revise the solution to the problem, the board provides hints on how to solve the problem but the solution wasn't detailed enough

Here's a transcription of the handwritten mathematical solution: --- **Solution:** Given, \( f \) and \( g \) are continuous at \( a \in \mathbb{R}^p \). So, for \( \epsilon > 0 \), \( \exists \delta_1 > 0 \) such that \[ |f(x) - f(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_1 \] For \( g \), \( \exists \delta_2 > 0 \) such that \[ |g(x) - g(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_2 \] Thus, if we choose \( \delta = \min \{\delta_1, \delta_2\} \), then \[ |f(x) - f(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_1 \leq \delta \] and \[ |g(x) - g(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_2 \leq \delta \] Hence, for our chosen \( \epsilon > 0 \), we get \( \delta > 0 \) such that \[ |f(x) - f(a)| < \epsilon \quad \& \quad |g(x) - g(a)| < \epsilon \quad \text{whenever} \, \|x-a\| < \delta \] **How to show:** \( f \cdot g \) is continuous at \( x = a \). \[ \Rightarrow \text{we have} \] \[ |(f \cdot g)(x) - (f \cdot g)(a)| = |f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)| \] Using the triangle inequality: \[ \leq |f(x)g(x) - f(x)g(a)| + |f(x)g(a) - f(a)g(a)| \] \[ = |f(x)||g(x) - g(a)| + |g(a)||f(x) - f(a)| \] **Now if** \( f \) being continuous, we know \( f \) is bounded on
Transcribed Image Text:Here's a transcription of the handwritten mathematical solution: --- **Solution:** Given, \( f \) and \( g \) are continuous at \( a \in \mathbb{R}^p \). So, for \( \epsilon > 0 \), \( \exists \delta_1 > 0 \) such that \[ |f(x) - f(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_1 \] For \( g \), \( \exists \delta_2 > 0 \) such that \[ |g(x) - g(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_2 \] Thus, if we choose \( \delta = \min \{\delta_1, \delta_2\} \), then \[ |f(x) - f(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_1 \leq \delta \] and \[ |g(x) - g(a)| < \epsilon \quad \forall \, \|x-a\| < \delta_2 \leq \delta \] Hence, for our chosen \( \epsilon > 0 \), we get \( \delta > 0 \) such that \[ |f(x) - f(a)| < \epsilon \quad \& \quad |g(x) - g(a)| < \epsilon \quad \text{whenever} \, \|x-a\| < \delta \] **How to show:** \( f \cdot g \) is continuous at \( x = a \). \[ \Rightarrow \text{we have} \] \[ |(f \cdot g)(x) - (f \cdot g)(a)| = |f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)| \] Using the triangle inequality: \[ \leq |f(x)g(x) - f(x)g(a)| + |f(x)g(a) - f(a)g(a)| \] \[ = |f(x)||g(x) - g(a)| + |g(a)||f(x) - f(a)| \] **Now if** \( f \) being continuous, we know \( f \) is bounded on
The blackboard contains mathematical proofs related to continuity of functions, specifically dealing with the continuity of the product of functions. 1. **Setup and Definitions**: - Let \( f, g : D \subseteq \mathbb{R}^p \) be continuous functions, where \( a \in \mathbb{R}^p \). - We establish that if both \( f \) and \( g \) are continuous at a point \( a \), then their product \( fg \) is also continuous at \( a \). 2. **Proof Structure**: - Begin by stating that for any \( \epsilon > 0 \), there exist \( \delta_1, \delta_2 > 0 \) such that for all \( x \in \mathbb{R}^p \), - if \( ||x - a|| < \delta_1 \), then \( ||f(x) - f(a)|| < \frac{\epsilon}{2} \). - if \( ||x - a|| < \delta_2 \), then \( ||g(x) - g(a)|| < \frac{\epsilon}{2} \). - Choose \( \delta = \min(\delta_1, \delta_2) \). 3. **Conclusions**: - Illustrates the property of continuity for the product \( |f(x) \cdot g(x) - f(a) \cdot g(a)| \). - The expression expands using the distributive property of multiplication over subtraction and absolute values. - The conclusion follows through with bounding \( |f(x) \cdot g(x) - f(a) \cdot g(a)| \) under \( \epsilon \) when \( ||x - a|| < \delta \), proving continuity. 4. **Mathematical Expressions**: - The board uses vectors and norms in a multi-dimensional real space \( \mathbb{R}^p \). - The expressions include manipulatives around the concept \( ||f(x) \cdot g(x) - f(a) \cdot g(a)|| \), illustrating a breakdown into manageable components. This proof is streamlined to validate the operational property of continuity using the structure prescribed by epsilon-delta definitions.
Transcribed Image Text:The blackboard contains mathematical proofs related to continuity of functions, specifically dealing with the continuity of the product of functions. 1. **Setup and Definitions**: - Let \( f, g : D \subseteq \mathbb{R}^p \) be continuous functions, where \( a \in \mathbb{R}^p \). - We establish that if both \( f \) and \( g \) are continuous at a point \( a \), then their product \( fg \) is also continuous at \( a \). 2. **Proof Structure**: - Begin by stating that for any \( \epsilon > 0 \), there exist \( \delta_1, \delta_2 > 0 \) such that for all \( x \in \mathbb{R}^p \), - if \( ||x - a|| < \delta_1 \), then \( ||f(x) - f(a)|| < \frac{\epsilon}{2} \). - if \( ||x - a|| < \delta_2 \), then \( ||g(x) - g(a)|| < \frac{\epsilon}{2} \). - Choose \( \delta = \min(\delta_1, \delta_2) \). 3. **Conclusions**: - Illustrates the property of continuity for the product \( |f(x) \cdot g(x) - f(a) \cdot g(a)| \). - The expression expands using the distributive property of multiplication over subtraction and absolute values. - The conclusion follows through with bounding \( |f(x) \cdot g(x) - f(a) \cdot g(a)| \) under \( \epsilon \) when \( ||x - a|| < \delta \), proving continuity. 4. **Mathematical Expressions**: - The board uses vectors and norms in a multi-dimensional real space \( \mathbb{R}^p \). - The expressions include manipulatives around the concept \( ||f(x) \cdot g(x) - f(a) \cdot g(a)|| \), illustrating a breakdown into manageable components. This proof is streamlined to validate the operational property of continuity using the structure prescribed by epsilon-delta definitions.
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