Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R,(x) → 0.]f(x) = In(x),a = 8(x) = In(8) + (|n = 1Find the associated radius of convergence R.R =

Answered: Find the Taylor series for f(x)…

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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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R,(x) → 0.]
f(x) = In(x),
a = 8
(x) = In(8) + (|
n = 1
Find the associated radius of convergence R.
R =

Expert Solution

Step 1

Given function is $f (x) = \ln (x)$ and centered at $a = 8$

The Taylor series expansion of the function $f (x)$ centered at $a$ is given by

$f (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{''} (a)}{2!} {(x - a)}^{2} + \frac{f^{'''} (a)}{3!} {(x - a)}^{3} + \frac{f^{(I V)} (a)}{4!} {(x - a)}^{4} + .....$

Now, the Taylor series expansion of the function $f (x)$ centered at $a = 8$ is given by

$f (x) = f (8) + f^{'} (8) (x - 8) + \frac{f^{''} (8)}{2!} {(x - 8)}^{2} + \frac{f^{'''} (8)}{3!} {(x - 8)}^{3} + \frac{f^{(I V)} (8)}{4!} {(x - 8)}^{4} + .....$ ....(1)

Substitute $x = 8$ in $f (x) = \ln (x)$

Hence, $f (8) = \ln (8)$

Differentiate $f (x) = \ln (x)$ with respect to $x$

Hence, $f^{'} (x) = \frac{1}{x}$

Now, substitute $x = 8$ in $f^{'} (x) = \frac{1}{x}$

Hence, $f^{'} (8) = \frac{1}{8}$

Differentiate $f^{'} (x) = \frac{1}{x}$ with respect to $x$

Hence, $f^{''} (x) = - \frac{1}{x^{2}}$

Now, substitute $x = 8$ in $f^{''} (x) = - \frac{1}{x^{2}}$

Hence, $f^{''} (8) = - \frac{1}{{(8)}^{2}}$

Differentiate $f^{''} (x) = - \frac{1}{x^{2}}$ with respect to $x$

Hence, $f^{'''} (x) = \frac{2}{x^{3}}$

Now, substitute $x = 8$ in $f^{'''} (x) = \frac{2}{x^{3}}$

Hence, $f^{'''} (8) = \frac{2}{{(8)}^{3}}$

Differentiate $f^{'''} (x) = \frac{2}{x^{3}}$ with respect to $x$

Hence, $f^{(I V)} (x) = - \frac{6}{x^{4}}$

Now, substitute $x = 8$ in $f^{(I V)} (x) = - \frac{6}{x^{4}}$

Hence, $f^{(I V)} (4) = - \frac{6}{{(8)}^{4}}$

Step 2

Now, substitute $f (8) = \ln (8)$ , $f^{'} (8) = \frac{1}{8}$ , $f^{''} (8) = - \frac{1}{{(8)}^{2}}$ , $f^{'''} (8) = \frac{2}{{(8)}^{3}}$ and $f^{(I V)} (4) = - \frac{6}{{(8)}^{4}}$ ... in equation (1)

$\begin{matrix} f (x) = \ln (8) + \frac{1}{8} (x - 8) + \frac{- \frac{1}{{(8)}^{2}}}{2!} {(x - 8)}^{2} + \frac{\frac{2}{{(8)}^{3}}}{3!} {(x - 8)}^{3} + \frac{- \frac{6}{{(8)}^{4}}}{4!} {(x - 8)}^{4} + ..... \\ = \ln (8) + \frac{1}{8} (x - 8) - \frac{1}{2! \cdot {(8)}^{2}} {(x - 8)}^{2} + \frac{2}{3! \cdot {(8)}^{3}} {(x - 8)}^{3} - \frac{6}{4! \cdot {(8)}^{4}} {(x - 8)}^{4} + ..... \\ = \ln (8) + \frac{1}{8} (x - 8) - \frac{1}{2! \cdot {(8)}^{2}} {(x - 8)}^{2} + \frac{2!}{3! \cdot {(8)}^{3}} {(x - 8)}^{3} - \frac{3!}{4! \cdot {(8)}^{4}} {(x - 8)}^{4} + ..... \\ = \ln (8) + \frac{1}{8} (x - 8) - \frac{1}{2 \cdot {(8)}^{2}} {(x - 8)}^{2} + \frac{1}{3 \cdot {(8)}^{3}} {(x - 8)}^{3} - \frac{1}{4 \cdot {(8)}^{4}} {(x - 8)}^{4} + ..... \\ = \ln (8) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n \cdot 8^{n}} {(x - 8)}^{n} \end{matrix}$

Therefore, $f (x) = \ln (8) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n \cdot 8^{n}} {(x - 8)}^{n}$

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