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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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Using example in photo1, solve the same way showing proof(leftside) and sidework(rightside) of the problem in photo2

The image contains a handwritten mathematical concept related to functions and uniform continuity. Here is a transcription suitable for an educational website: --- **Claim:** Let \( f: [a, +\infty) \to \mathbb{R} \), where \( a > 0 \). The function \( f(x) = \sqrt{x} \) is uniformly continuous on the interval \([a, +\infty)\). --- The notation and symbols are used to express that a function \( f \) defined from the interval \([a, +\infty)\) to the real numbers \(\mathbb{R}\), specifically the square root function \( f(x) = \sqrt{x} \), is uniformly continuous on the interval starting from \( a \) to infinity, where \( a \) is a positive number. The concept of uniform continuity means that for any chosen degree of closeness between function values, the input values can be restricted to ensure they are within that degree of closeness across the entire specified interval.
Transcribed Image Text:The image contains a handwritten mathematical concept related to functions and uniform continuity. Here is a transcription suitable for an educational website: --- **Claim:** Let \( f: [a, +\infty) \to \mathbb{R} \), where \( a > 0 \). The function \( f(x) = \sqrt{x} \) is uniformly continuous on the interval \([a, +\infty)\). --- The notation and symbols are used to express that a function \( f \) defined from the interval \([a, +\infty)\) to the real numbers \(\mathbb{R}\), specifically the square root function \( f(x) = \sqrt{x} \), is uniformly continuous on the interval starting from \( a \) to infinity, where \( a \) is a positive number. The concept of uniform continuity means that for any chosen degree of closeness between function values, the input values can be restricted to ensure they are within that degree of closeness across the entire specified interval.
### Transcription of Mathematical Proof **Topic**: Uniform Continuity of the Function on an Unbounded Interval --- #### Given: The function \( f: [1, \infty) \to \mathbb{R} \) defined as \( f(x) = \frac{1}{x} \). The goal is to prove that \( f \) is uniformly continuous on the interval \([1, \infty)\). --- #### Proof Outline: 1. **Definition and Assumption**: - Let \( \epsilon > 0 \). We need to find \( \delta > 0 \) such that whenever \( |x - u| < \delta \), it follows that \( |f(x) - f(u)| < \epsilon \) for all \( x, u \in [1, \infty) \). 2. **Calculation**: - Assume \( x, u \in [1, \infty) \). - Calculate \( |f(x) - f(u)| = \left| \frac{1}{x} - \frac{1}{u} \right| \). 3. **Simplification**: - This simplifies to \( \left| \frac{u - x}{xu} \right| = \frac{|x - u|}{|x||u|} \). - Since \( x, u \geq 1 \), we have \( |x||u| \geq 1 \). 4. **Conclusion**: - Thus, \( |f(x) - f(u)| \leq |x - u| \). - Choose \( \delta = \epsilon \). Then \( |x - u| < \delta \) implies \( |f(x) - f(u)| < \epsilon \). Therefore, \( f \) is uniformly continuous on \([1, \infty)\). --- **Conclusion:** The function \( f(x) = \frac{1}{x} \) is uniformly continuous on the interval \([1, \infty)\), as demonstrated by the above proof. The key step was showing the bounded nature of \( \left| \frac{1}{xu} \right| \), which ensures that \( |f(x) - f(u)| \) is directly controlled by \( |x - u| \). --- This explanation helps the reader understand the step-by
Transcribed Image Text:### Transcription of Mathematical Proof **Topic**: Uniform Continuity of the Function on an Unbounded Interval --- #### Given: The function \( f: [1, \infty) \to \mathbb{R} \) defined as \( f(x) = \frac{1}{x} \). The goal is to prove that \( f \) is uniformly continuous on the interval \([1, \infty)\). --- #### Proof Outline: 1. **Definition and Assumption**: - Let \( \epsilon > 0 \). We need to find \( \delta > 0 \) such that whenever \( |x - u| < \delta \), it follows that \( |f(x) - f(u)| < \epsilon \) for all \( x, u \in [1, \infty) \). 2. **Calculation**: - Assume \( x, u \in [1, \infty) \). - Calculate \( |f(x) - f(u)| = \left| \frac{1}{x} - \frac{1}{u} \right| \). 3. **Simplification**: - This simplifies to \( \left| \frac{u - x}{xu} \right| = \frac{|x - u|}{|x||u|} \). - Since \( x, u \geq 1 \), we have \( |x||u| \geq 1 \). 4. **Conclusion**: - Thus, \( |f(x) - f(u)| \leq |x - u| \). - Choose \( \delta = \epsilon \). Then \( |x - u| < \delta \) implies \( |f(x) - f(u)| < \epsilon \). Therefore, \( f \) is uniformly continuous on \([1, \infty)\). --- **Conclusion:** The function \( f(x) = \frac{1}{x} \) is uniformly continuous on the interval \([1, \infty)\), as demonstrated by the above proof. The key step was showing the bounded nature of \( \left| \frac{1}{xu} \right| \), which ensures that \( |f(x) - f(u)| \) is directly controlled by \( |x - u| \). --- This explanation helps the reader understand the step-by
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