Answered: 2n p(x) = E (1+x2n) n=0

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Define

Find the values of x where the series converges and show that we get a continuous function on this set.

Expert Solution

Step 1

Given: $g (x) = \sum_{n = 0}^{\infty} (\frac{x^{2 n}}{1 + x^{2 n}})$
To find: a) The values of x where the given series converges.
b) The function $(\frac{x^{2 n}}{1 + x^{2 n}})$ is continuous.

Step 2

a) We have,
$g (x) = \sum_{n = 0}^{\infty} (\frac{x^{2 n}}{1 + x^{2 n}})$ where $n^{t h}$ term is $b_{n} = (\frac{x^{2 n}}{1 + x^{2 n}})$

Case 1 $|x| > 1$
Using the Divergence test,
$\lim_{n \to \infty} b_{n} = \lim_{n \to \infty} (\frac{x^{2 n}}{1 + x^{2 n}}) = \lim_{n \to \infty} (\frac{1}{\frac{1}{x^{2 n}} + 1}) = (\frac{1}{0 + 1}) [∵ |x| > 1] = 1 \neq 0$
Hence, the given series is divergent for $|x| > 1$ .

Step 3

Case 2 $|x| = 1$
Using the Divergence test,
$\lim_{n \to \infty} b_{n} = \lim_{n \to \infty} (\frac{x^{2 n}}{1 + x^{2 n}}) = \lim_{n \to \infty} (\frac{1^{n}}{1 + 1^{n}}) [∵ |x| = 1, x^{2} = 1] = \lim_{n \to \infty} (\frac{1}{2}) = \frac{1}{2} \neq 0$
Hence, the given series is divergent for $|x| = 1$ .

Case 3 $|x| < 1$
Using the Divergence test,
$\lim_{n \to \infty} b_{n} = \lim_{n \to \infty} (\frac{x^{2 n}}{1 + x^{2 n}}) = (\frac{0}{1 + 0}) [∵ |x| < 1, \lim_{n \to \infty} x^{2 n} = 0] = 0$
Hence, the divergent test is inconclusive for $|x| < 1$ .

Step by step

Solved in 5 steps

Knowledge Booster

$Power Series$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions