Answered: 1 gRU (7) = E (x – 10)" n²1766" n=1 a)…

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

The “Rutgers” function (the best function!) has Taylor series about the point a = 10

1
fRU (x) = L n²1766"
(x – 10)"
n=1
(a) Find the interval of convergence for the Rutgers function.
(b) Find the interval of convergence for the 1st derivative of the Rutgers function.
(c) Write down
dx5
fR (10)

Expert Solution

Step 1

$\begin{array}{l} Consider the Rutgers function \\ f^{R U} (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{2} 1766^{n}} {(x - 10)}^{n} \\ To find : \\ (a) Find the interval of convergence for the Rutgers function . \\ (b) Find the interval of convergence for the 1 st derivative of the \\ Rutgers function . \\ (c) Write down \frac{d^{5}}{d x^{5}} f^{R U} (10) \end{array}$

Step 2

$\begin{array}{l} (a) \\ Consider the Rutgers function \\ f^{R U} (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{2} 1766^{n}} {(x - 10)}^{n} \\ There we are able to show that there is a number R \\ so the power series converges for |x - a| < R and \\ diverges for |x - a| > R . \\ Let <a_{n}> = <\frac{1}{n^{2} 1766^{n}} {(x - 10)}^{n}> \\ By ratio test : \\ L = \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| \\ = \lim_{n \to \infty} |\frac{\frac{1}{{(n + 1)}^{2} 1766^{n + 1}} {(x - 10)}^{n + 1}}{(\frac{1}{n^{2} 1766^{n}} {(x - 10)}^{n})}| \\ = \lim_{n \to \infty} |\frac{n^{2} (x - 10)}{1766 {(n + 1)}^{2}}| \\ = |\frac{(x - 10)}{1766}| \\ The ratio test tells that if L < 1, then series will be convergent . \\ Now, |\frac{(x - 10)}{1766}| < 1 \\ \Rightarrow - 1 < \frac{(x - 10)}{1766} < 1 \\ \Rightarrow - 1756 < x < 1776 \end{array}$

Step 3

$\begin{array}{l} Now check at the end points . \\ At x = 1776, then \\ f^{R U} (1776) = \sum_{n = 1}^{\infty} \frac{1}{n^{2} 1766^{n}} {(1776 - 10)}^{n} \\ = \sum_{n = 1}^{\infty} \frac{1}{n^{2} 1766^{n}} 1766^{n} \\ = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \\ By p - test, the series \sum_{n = 1}^{\infty} \frac{1}{n^{2}} is convergent . \\ At x = - 1756, then \\ f^{R U} (1776) = \sum_{n = 1}^{\infty} \frac{1}{n^{2} 1766^{n}} {(- 1756 - 10)}^{n} \\ = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \\ By Alternate series test, the series \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} is convergent . \\ Hence, the series \sum_{n = 1}^{\infty} (\frac{1}{n^{2} 1766^{n}} {(x - 10)}^{n}) is convergent in the \\ interval [- 1756, 1776] . \end{array}$

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