Consider the function $ f(x) = \ln(x + 1) $. **Problem Statement:**Compute the approximate value of ln(2) using the Taylor polynomials calculated in (a). How large is the approximation error?---**Explanation:**This problem requires you to use Taylor polynomials to approximate the natural logarithm of 2, denoted as ln(2). You should have already calculated the relevant Taylor polynomials in part (a) of your task. Once you have the polynomial approximation of ln(2), you will need to determine the size of the approximation error. This involves comparing the approximation to the actual value of ln(2) and calculating the difference. This exercise is a practical application of Taylor series in approximating functions and assessing accuracy.

Consider the function, f(x)=ln(x+1) ⋯⋯(1) At x=0, equation (1) becomes, f(0)=ln(0+1)=ln1=0 Taking…

Answered: Compute the approximate value of In(2)…

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

Consider the function $ f(x) = \ln(x + 1) $.

**Problem Statement:**

Compute the approximate value of ln(2) using the Taylor polynomials calculated in (a). How large is the approximation error?

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**Explanation:**

This problem requires you to use Taylor polynomials to approximate the natural logarithm of 2, denoted as ln(2). You should have already calculated the relevant Taylor polynomials in part (a) of your task. Once you have the polynomial approximation of ln(2), you will need to determine the size of the approximation error. This involves comparing the approximation to the actual value of ln(2) and calculating the difference. This exercise is a practical application of Taylor series in approximating functions and assessing accuracy.

Expert Solution

Step 1

Consider the function,

$f (x) = \ln (x + 1) \dots \dots (1)$

At $x = 0,$ equation (1) becomes,

$\begin{array}{rcl} f (0) & = & \ln (0 + 1) \\ = & \ln 1 \\ = & 0 \end{array}$

Taking the derivatives of the function $f (x)$ and using $x = 0,$

$\begin{array}{rcl} f' (x) & = & \frac{1}{(x + 1)} \\ f' (0) & = & 1 \\ f'' (x) & = & \frac{- 1}{{(x + 1)}^{2}} \\ f'' (0) & = & - 1 \\ f''' (x) & = & \frac{2}{{(x + 1)}^{3}} \\ f''' (0) & = & 2 \\ f'''' (x) & = & \frac{- 6}{{(x + 1)}^{4}} \\ f'''' (0) & = & - 6 \dots \end{array}$

Step 2

The Taylor polynomial of $f (x)$ of degree $n$ is given by,

$\begin{array}{rcl} \ln (x + 1) & = & f (0) + \frac{f' (0) (x - 0)}{1!} + \frac{f'' (0) {(x - 0)}^{2}}{2!} + \frac{f''' (0) {(x - 0)}^{3}}{3!} + \frac{f'''' (0) {(x - 0)}^{4}}{4!} + \dots + \frac{f^{(n)} (0) {(x - 0)}^{n}}{n!} + \dots \\ = & 0 + (x \times 1) + \frac{x^{2}}{2} (- 1) + \frac{x^{3}}{6} (2) + \frac{x^{4}}{24} (- 6) + \dots + \frac{f^{(n)} (0) x^{n}}{n!} + \dots \\ = & x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots + \frac{f^{(n)} (0) x^{n}}{n!} + \dots \end{array}$

Then, the Taylor polynomial of $f (x)$ of degree 4 is,

$\ln (x + 1) = \begin{array}{rcl} x \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{x^{2}}{2} \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{x^{3}}{3} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{x^{4}}{4} \end{array} \dots \dots (2)$

Put $x = 1$ in equation (2),

$\begin{array}{rcl} \ln (1 + 1) & = & 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \\ \ln (2) & = & 0.58333334 \dots \dots (3) \end{array}$

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