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Chapter 6: Problem 22

The equation \(\ln x=x-2\) has two solutions. Approximate them using Newton's Method. What happens if \(x_{1}:=\frac{1}{2}\) is the initial point?

Short Answer

Expert verified
The equation \(\ln x = x - 2\) has an approximate solution of \(2.15842\) using Newton's method with an initial guess of \(x_1 = 3\) . If we take \(x_{1} = \frac{1}{2}\) as an initial guess, Newton's method fails as it leads to division by zero.

Step by step solution

01

Identify function and its derivative

Given that \(\ln x = x - 2\) rewriting this makes our function \(f(x) = \ln x - x + 2\). The derivative of this, \(f'(x)\) is calculated using basic differentiation rules. The derivative of \(\ln x\) is \(\frac{1}{x}\) and the derivative of \(x\) is \(1\). So, \(f'(x) = \frac{1}{x} - 1\).
02

Apply Newton's method

Let's take an initial guess. For example, \(x_1 = 1\). Inserting this into the Newton's method gives us the next approximation:\[x_{2} = x_1 - \frac{f(x_1)}{f'(x_1)} = 1 - \frac{\ln 1 - 1 + 2}{1/1 - 1} = 1 - 1 = 0\] We see that the approximation leads us to zero, which is not a valid result since the logarithm of zero is undefined. Thus, Newton's method does not work with the initial guess of \(x_1 = 1\). Therefore, we need to take another initial guess. Let's consider \(x_1 = 3\).
03

Repeat Newton's method with a new initial value

With \(x_1 = 3\), we repeat the procedure:\[x_{2} = x_1 - \frac{f(x_1)}{f'(x_1)} = 3 - \frac{\ln 3 - 3 + 2}{1/3 - 1} \approx 2.15842\].We can repeat this process until the difference between the current approximation and the previous one is less than a predetermined value, say \(1E-10\). The approximations will be one solution.
04

Examine what happens with \(x_{1} = \frac{1}{2}\) as an initial value

We apply Newton's method using an initial guess of \(x_1 = \frac{1}{2}\):\[x_{2} = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.5 - \frac{\ln 0.5 - 0.5 + 2}{1/0.5 - 1} \]The value of the denominator is zero, so we are unable to calculate the next approximation as division by zero is undefined.

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Most popular questions from this chapter

Show that if \(x>0\), then \(1+\frac{1}{2} x-\frac{1}{8} x^{2} \leq \sqrt{1+x} \leq 1+\frac{1}{2} x\).

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