Let {fn(x)} be defined by: fn(x) = Vx for 0 < x< 1. [a] Prove with an e – N argument that {fn(x)} converges pointwise to the function 1 ;if 0 < x < 1 0 ;if x = 0 f(x) = [b] Show that {fn(x)} does not converge uniformly to f(x) by showing that given any n > 0 you can find a point 0 < x < 1 such that | VI – 1| 2. **Note**: The point æ can be a function of n. [c] Using the e – N definition of uniform convergence, prove that {fn(x)} con- verges uniformly to: f(x) = 1, on the interval [}, 1).
Let {fn(x)} be defined by: fn(x) = Vx for 0 < x< 1. [a] Prove with an e – N argument that {fn(x)} converges pointwise to the function 1 ;if 0 < x < 1 0 ;if x = 0 f(x) = [b] Show that {fn(x)} does not converge uniformly to f(x) by showing that given any n > 0 you can find a point 0 < x < 1 such that | VI – 1| 2. **Note**: The point æ can be a function of n. [c] Using the e – N definition of uniform convergence, prove that {fn(x)} con- verges uniformly to: f(x) = 1, on the interval [}, 1).
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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COURSE: Mathematical/
TOPIC: Uniform Convergence
![Let \(\{f_n(x)\}\) be defined by:
\[
f_n(x) = \sqrt[n]{x} \quad \text{for } 0 \leq x \leq 1.
\]
**[a]** Prove with an \(\epsilon - N\) argument that \(\{f_n(x)\}\) converges pointwise to the function
\[
f(x) =
\begin{cases}
1 & ; \text{if } 0 < x \leq 1 \\
0 & ; \text{if } x = 0
\end{cases}
\]
**[b]** Show that \(\{f_n(x)\}\) does not converge uniformly to \(f(x)\) by showing that given any \(n > 0\) you can find a point \(0 < x < 1\) such that \(\left|\sqrt[n]{x} - 1\right| \geq \frac{1}{2}\).
**Note**: The point \(x\) can be a function of \(n\).
**[c]** Using the \(\epsilon - N\) definition of uniform convergence, prove that \(\{f_n(x)\}\) converges uniformly to: \(f(x) = 1\), on the interval \(\left[\frac{1}{2}, 1\right]\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbaa9c753-6b68-41b4-8837-61aa8aa0a19b%2F63922e3f-58ea-4ec4-982f-ad22ff8295d1%2Fujrzu2p_processed.png&w=3840&q=75)
Transcribed Image Text:Let \(\{f_n(x)\}\) be defined by:
\[
f_n(x) = \sqrt[n]{x} \quad \text{for } 0 \leq x \leq 1.
\]
**[a]** Prove with an \(\epsilon - N\) argument that \(\{f_n(x)\}\) converges pointwise to the function
\[
f(x) =
\begin{cases}
1 & ; \text{if } 0 < x \leq 1 \\
0 & ; \text{if } x = 0
\end{cases}
\]
**[b]** Show that \(\{f_n(x)\}\) does not converge uniformly to \(f(x)\) by showing that given any \(n > 0\) you can find a point \(0 < x < 1\) such that \(\left|\sqrt[n]{x} - 1\right| \geq \frac{1}{2}\).
**Note**: The point \(x\) can be a function of \(n\).
**[c]** Using the \(\epsilon - N\) definition of uniform convergence, prove that \(\{f_n(x)\}\) converges uniformly to: \(f(x) = 1\), on the interval \(\left[\frac{1}{2}, 1\right]\).
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