**Problem Statement:**Find the first five terms of the Taylor series for the function $ f(x) = \ln(x) $ about the point $ a = 8 $. (Your answers should include the variable x when appropriate.)**Expression:**\[\ln(x) = \boxed{\phantom{fill}} + \boxed{\phantom{fill}} + \boxed{\phantom{fill}} + \boxed{\phantom{fill}} + \cdots\]**Explanation:**The image shows a mathematical problem where the goal is to determine and fill in the first five terms of the Taylor series expansion for the natural logarithm function centered around $ x = 8 $. Each term in the series is represented by an empty box intended for input. The series should be expressed with respect to the variable $ x $.

Answered: Find the first five terms of the Taylor…

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

**Problem Statement:**

Find the first five terms of the Taylor series for the function $ f(x) = \ln(x) $ about the point $ a = 8 $. (Your answers should include the variable x when appropriate.)

**Expression:**

\[
\ln(x) = \boxed{\phantom{fill}} + \boxed{\phantom{fill}} + \boxed{\phantom{fill}} + \boxed{\phantom{fill}} + \cdots
\]

**Explanation:**

The image shows a mathematical problem where the goal is to determine and fill in the first five terms of the Taylor series expansion for the natural logarithm function centered around $ x = 8 $. Each term in the series is represented by an empty box intended for input. The series should be expressed with respect to the variable $ x $.

Expert Solution

Step 1

The given function is $f (x) = \ln x$ about the point a = 8.

Note that, the Taylor series expansion of f(x) centered at x=a is given by,

$f (x) = f (a) + f' (a) \frac{(x - a)}{1!} + f'' (a) \frac{{(x - a)}^{2}}{2!} + f''' (a) \frac{{(x - a)}^{3}}{3!} + . . .$

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

$f (8) = \ln 8$

Obtain the derivative of the function as follows:

$f' (x) = \frac{1}{x} \Rightarrow f' (8) = \frac{1}{8} f'' (x) = - \frac{1}{x^{2}} \Rightarrow f'' (8) = - \frac{1}{64} f''' (x) = 2 \frac{1}{x^{3}} \Rightarrow f''' (8) = \frac{1}{256} f^{(i v)} (x) = - 6 \frac{1}{x^{4}} \Rightarrow f^{(i v)} (8) = - \frac{6}{4096} f^{(v)} (x) = 24 \frac{1}{x^{5}} \Rightarrow f^{(v)} (8) = \frac{24}{32768}$

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

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