Answered: na fn(x) = 1+ nx?

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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Question

Let

(a) Find the pointwise limit of (fn) for all x ∈ (0,∞).
(b) Is the convergence uniform on (0,∞)?
(c) Is the convergence uniform on (0, 1)?
(d) Is the convergence uniform on (1,∞)?

Expert Solution

Step 1

The given sequence of functions is,

$f_{n} (x) = \frac{n x}{1 + n x^{2}}$ .

We have to:

(a) Find the pointwise limit of $f_{n} (x)$ for all x ∈ (0,∞).
(b) Is the convergence uniform on (0,∞)?
(c) Is the convergence uniform on (0, 1)?
(d) Is the convergence uniform on (1,∞)?

Step 2

The given sequence of functions is,

$f_{n} (x) = \frac{n x}{1 + n x^{2}}$ .

(a) Pointwise limit:

$\lim_{n \to \infty} f_{n} (x) = \lim_{n \to \infty} \frac{n x}{1 + n x^{2}} = \lim_{n \to \infty} \frac{x}{\frac{1}{n} + x^{2}} = \frac{1}{x}$ .

Pointwise limit of the function is $\frac{1}{x}$ .

(b) Uniform convergence on $(0, \infty)$ :

Take $x = \frac{1}{n}$ .

$|f_{n} (\frac{1}{n}) - f (\frac{1}{n})| = |\frac{n \times \frac{1}{n}}{1 + n \times \frac{1}{n^{2}}} - n| = |\frac{n}{n + 1} - n| = |\frac{n - n^{2} - n}{n + 1}| = |\frac{n^{2}}{n + 1}|$ >1.

Thus, $|f_{n} (\frac{1}{n}) - f (\frac{1}{n})| > 1$ .

$f_{n} (x)$ is not uniformly convergent on $(0, \infty)$ .

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