Answered: cos(x? – x) = x*

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

Concept explainers

Question

Use Newton’s method to find all the roots of the
equation correct to eight decimal places. Start by drawing a
graph to find initial approximations.

Expert Solution

Step 1

Newton's method: This method is also known as Newton Raphson method and it is a technique used to

find the approximation of the root of the function.

It is given that $\cos (x^{2} - x) = x^{4}$ . Rewriting it, we get, $x^{4} - \cos (x^{2} - x) = 0$ . Let us define a function $f (x)$

such that $f (x) = x^{4} - \cos (x^{2} - x)$ . Then, the derivative of $f (x)$ is $f' (x) = 4 x^{3} + (2 x - 1) \sin (x^{2} - x)$ .

Step 2

The graph of the function $f (x) = x^{4} - \cos (x^{2} - x)$ is given below:

Since the curve intersects the x-axis at two points, there are two roots of the equation $f (x) = 0$ . Let

us consider the initial approximations $x_{0} = - 0.5$ and $x_{0} = 1.2$ .

Step 3

First, we will use the initial approximation $x_{0} = - 0.5$ .

First iteration:

We have $f (x_{0}) = - 0.66918887$ and $f' (x_{0}) = - 1.86327752$ . So we get,

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & - 0.5 - \frac{- 0.66918887}{- 1.86327752} \\ = & - 0.85914611 \end{array}$

Hence, $x_{1} = - 0.85914611$ .

Second iteration:

We have $f (x_{1}) = 0.57131762$ and $f' (x_{1}) = - 5.25399218$ . So we get,

$\begin{array}{rcl} x_{2} & = & x_{1} - \frac{f (x_{1})}{f' (x_{1})} \\ = & - 0.85914611 - \frac{0.57131762}{- 5.25399218} \\ = & - 0.75040639 \end{array}$

Hence, $x_{2} = - 0.75040639$ .

Step 4

Third iteration:

We have $f (x_{2}) = 0.06264139$ and $f' (x_{2}) = - 4.10874469$ . So we get,

$\begin{array}{rcl} x_{3} & = & x_{2} - \frac{f (x_{2})}{f' (x_{2})} \\ = & - 0.75040639 - \frac{0.06264139}{- 4.10874469} \\ = & - 0.73516052 \end{array}$

Hence, $x_{3} = - 0.73516052$ .

Fourth iteration:

We have $f (x_{3}) = 0.00119097$ and $f' (x_{3}) = - 3.9527852$ . So we get,

$\begin{array}{rcl} x_{4} & = & x_{3} - \frac{f (x_{3})}{f' (x_{3})} \\ = & - 0.73516052 - \frac{0.00119097}{- 3.9527852} \\ = & - 0.73485922 \end{array}$

Hence, $x_{4} = - 0.73485922$ .

Step by step

Solved in 7 steps with 1 images

Knowledge Booster

$Application of Integration$

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