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z . Prove that fn (z) = sin is uniformly convergent to 0 on D(0,2).

z . Prove that fn (z) = sin is uniformly convergent to 0 on D(0,2).

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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need help with proving uniform convergence for complex analysis

**Problem:** Prove that the sequence of functions \( f_n(z) = \sin\frac{z}{n} \) is uniformly convergent to 0 on the interval \( D(0,2) \). **Explanation:** To show uniform convergence, you must demonstrate that for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\) and for all \(z \in D(0,2)\), the inequality \(|f_n(z) - 0| < \epsilon\) holds. **Key Points:** 1. **Function Behavior:** - The function \(f_n(z) = \sin\frac{z}{n}\) tends to 0 as \(n\) increases because \(\frac{z}{n}\) approaches 0 for any fixed \(z\). 2. **Bounding the Sine Function:** - Since \(|\sin x| \leq |x|\) for all \(x\), it follows that \(\left|\sin \frac{z}{n}\right| \leq \left|\frac{z}{n}\right|\). 3. **Choosing \(N\):** - Since \(z \in D(0,2)\), we have \(0 < z < 2\). Therefore, \(\left|\frac{z}{n}\right| \leq \frac{2}{n}\). - We need \(\frac{2}{n} < \epsilon\), which implies \(n > \frac{2}{\epsilon}\). - Choose \(N = \left\lceil \frac{2}{\epsilon} \right\rceil\). For \(n \geq N\), \(\left|\sin \frac{z}{n}\right| < \epsilon\). Thus, the sequence \(f_n(z) = \sin\frac{z}{n}\) converges uniformly to 0 on the interval \(D(0,2)\).
Transcribed Image Text:**Problem:** Prove that the sequence of functions \( f_n(z) = \sin\frac{z}{n} \) is uniformly convergent to 0 on the interval \( D(0,2) \). **Explanation:** To show uniform convergence, you must demonstrate that for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\) and for all \(z \in D(0,2)\), the inequality \(|f_n(z) - 0| < \epsilon\) holds. **Key Points:** 1. **Function Behavior:** - The function \(f_n(z) = \sin\frac{z}{n}\) tends to 0 as \(n\) increases because \(\frac{z}{n}\) approaches 0 for any fixed \(z\). 2. **Bounding the Sine Function:** - Since \(|\sin x| \leq |x|\) for all \(x\), it follows that \(\left|\sin \frac{z}{n}\right| \leq \left|\frac{z}{n}\right|\). 3. **Choosing \(N\):** - Since \(z \in D(0,2)\), we have \(0 < z < 2\). Therefore, \(\left|\frac{z}{n}\right| \leq \frac{2}{n}\). - We need \(\frac{2}{n} < \epsilon\), which implies \(n > \frac{2}{\epsilon}\). - Choose \(N = \left\lceil \frac{2}{\epsilon} \right\rceil\). For \(n \geq N\), \(\left|\sin \frac{z}{n}\right| < \epsilon\). Thus, the sequence \(f_n(z) = \sin\frac{z}{n}\) converges uniformly to 0 on the interval \(D(0,2)\).
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