Chapter 4: Problem 12
Find the slope \(m\) of the line joining \((a, f(a))\) and \((b, f(b))\). Then use the Newton-Raphson method to estimate the values of \(c\) for which \(f^{\prime}(c)=m .\) Continue the process until successive iterations obtained by the calculator are identical. $$ f(x)=\sin (\pi x / 2) ; a=-1, b=1 $$
Short Answer
Step by step solution
Calculate the Function Values
Determine the Slope Formula
Find the Slope
Set Up the Newton-Raphson Method
Differentiate the Function
Choose Initial Guess
Iteration 1
Iteration 2
Check Successive Iterations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope of a Line
To find the slope of a line connecting two points, \((a, f(a))\) and \((b, f(b))\), we use the slope formula:
- \m = \frac{f(b) - f(a)}{b - a}\
- When the slope is positive, the line ascends as you move along; conversely, a negative slope indicates that the line descends.
Differentiation
To apply the Newton-Raphson method effectively, we need both the first and second derivatives of the function. For the function \(f(x) = \sin\left(\frac{\pi x}{2}\right)\), the derivatives are:
- First derivative: \f'(x) = \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right)\
- Second derivative: \f''(x) = -\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi x}{2}\right)\
These derivatives are pivotal in the Newton-Raphson method, as they tell us both how the function is changing and also refine our estimates closer to the actual point where \(f'(x) = m\). Knowing how to differentiate correctly allows us to leverage these tools in the method, helping achieve the estimation faster.
Iteration Process
The Newton-Raphson formula is iteratively applied: \x_{n+1} = x_n - \frac{f'(x_n) - m}{f''(x_n)}\
- Start with an initial guess. Here, choosing \(c = 0.5\) as an initial guess allows computation without encountering indeterminate solutions.
- Apply the formula to get a sequence of values, like getting from \(x_0\) to \(x_1\), \(x_2\) and so forth.
- Continue this process until the estimates converge, meaning \(x_{n+1}\) and \(x_n\) are sufficiently close.