Chapter 5: Problem 13
Develop a second-order method for approximating \(f^{\prime}(x)\) that uses the data \(f(x-h), f(x)\), and \(f(x+3 h)\) only.
Short Answer
Expert verified
Question: Develop a second-order method to approximate the first derivative of a function, \(f^{\prime}(x)\), using the data \(f(x-h), f(x)\), and \(f(x+3h)\).
Answer: \(f^{\prime}(x)=\frac{9[f(x-h)]-[f(x+3h)]+8f(x)}{6h}\)
Step by step solution
01
1. Taylor Series Expansion
To start, we need to expand \(f(x-h), f(x)\), and \(f(x+3h)\) using Taylor series around the point x.
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Taylor series expansions
Using the Taylor series expansion around the point x, we have:
\(f(x-h) = f(x) - hf^{\prime}(x) + \frac{h^2}{2}f^{\prime \prime}(x)-\dots\)
\(f(x) = f(x)\)
\(f(x+3h) = f(x) + 3hf^{\prime}(x) + \frac{9h^2}{2}f^{\prime \prime}(x)+\dots\)
03
2. Combine equations to eliminate higher-order terms
Our goal is to combine these equations in such a way that we eliminate higher-order terms and only have terms up to the second-order. The combination will result in a second-order accurate method.
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Eliminating higher-order terms
We can eliminate \(f^{\prime \prime}(x)\) terms by multiplying the first equation by 9 and then subtracting the third equation:
\(9[f(x-h)]=9[f(x) - hf^{\prime}(x) + \frac{h^2}{2}f^{\prime \prime}(x)]\)
Then, subtract the third equation from the above equation:
\(9[f(x-h)]-[f(x+3h)]=[-2hf^{\prime}(x)+\frac{h^2}{2}f^{\prime \prime}(x)]\)
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3. Solve for \(f^{\prime}(x)\)
Now, all we need to do is to solve for \(f^{\prime}(x)\) from the obtained equation:
\(f^{\prime}(x)=\frac{9[f(x-h)]-[f(x+3h)]+8f(x)}{6h}\)
This is the second-order accurate method to approximate \(f^{\prime}(x)\) using the data \(f(x-h), f(x)\), and \(f(x+3h)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series Expansion
The Taylor series is a mathematical tool used to approximate functions around a specific point.
In our context, we are interested in approximating the derivatives of a function.
This series expresses the function in an infinite sum of terms calculated from the values of its derivatives at a single point.
When we expand a function using Taylor series, we write it as:
By managing the higher-order terms, we can greatly enhance computational efficiency and accuracy.
In our context, we are interested in approximating the derivatives of a function.
This series expresses the function in an infinite sum of terms calculated from the values of its derivatives at a single point.
When we expand a function using Taylor series, we write it as:
- For example, the function at point \( x - h \) can be expressed as \( f(x-h) = f(x) - hf^{\prime}(x) + \frac{h^2}{2} f^{\prime \prime}(x) - \dots \)
- At point \( x \) itself, it's simply \( f(x) \), as no expansion is required.
- For the point \( x + 3h \), the function is expanded to \( f(x+3h) = f(x) + 3hf^{\prime}(x) + \frac{9h^2}{2} f^{\prime \prime}(x) + \dots \).
By managing the higher-order terms, we can greatly enhance computational efficiency and accuracy.
Second-Order Method
The second-order method aims to provide a better approximation of the derivative of a function by taking into account more information and cancelling out higher-order errors.
In this exercise, the data points used are \( f(x-h) \), \( f(x) \), and \( f(x+3h) \).
By carefully combining these, we create an equation that zeroes out higher-order error terms, leaving us with a more accurate estimation.We apply a specific combination approach here to eliminate terms associated with the second derivative.
For instance, multiplying the Taylor expansion of \( f(x-h) \) by 9 and then subtracting the expansion of \( f(x+3h) \), we neutralize unwanted higher-order terms. Resulting in a formula for \( f^{\prime}(x) \) that is more precise due to reduced higher-order errors.
The beauty of the second-order method lies in its ability to strike a balance between complexity and accuracy.
In this exercise, the data points used are \( f(x-h) \), \( f(x) \), and \( f(x+3h) \).
By carefully combining these, we create an equation that zeroes out higher-order error terms, leaving us with a more accurate estimation.We apply a specific combination approach here to eliminate terms associated with the second derivative.
For instance, multiplying the Taylor expansion of \( f(x-h) \) by 9 and then subtracting the expansion of \( f(x+3h) \), we neutralize unwanted higher-order terms. Resulting in a formula for \( f^{\prime}(x) \) that is more precise due to reduced higher-order errors.
The beauty of the second-order method lies in its ability to strike a balance between complexity and accuracy.
Finite Difference Method
The finite difference method is a straightforward numerical technique used to approximate derivatives.
It involves the use of function values at specific points to estimate the derivative at a midpoint or specific location.
This exercise utilizes a particular setup within this method, known as a second-order accurate finite difference.Here's how it works:
As a result, it underscores the power of finite differences in numerical differentiation.
It involves the use of function values at specific points to estimate the derivative at a midpoint or specific location.
This exercise utilizes a particular setup within this method, known as a second-order accurate finite difference.Here's how it works:
- The finite difference is typically set up with points like \( f(x-h) \), \( f(x) \), and \( f(x+3h) \).
- These points are strategically chosen to balance the approximation error via the subtraction and addition of these function values.
- The scheme achieved here results in a derivative approximation formula: \( f^{\prime}(x)=\frac{9[f(x-h)]-[f(x+3h)]+8f(x)}{6h} \).
As a result, it underscores the power of finite differences in numerical differentiation.