Math Problem Statement
Find f(1.75) x 0 1 2 3 f(x) -18 26 11 -6 solve using Newton forward interpolation
Solution
We are tasked with finding using Newton's Forward Interpolation method. Given the data points:
x & f(x) \\ \hline 0 & -18 \\ 1 & 26 \\ 2 & 11 \\ 3 & -6 \\ \end{array}$$ ### Steps for Newton’s Forward Interpolation: 1. **Compute the forward differences** (denoted as $$\Delta f(x)$$): $$\Delta f(x_0) = f(x_1) - f(x_0), \quad \Delta^2 f(x_0) = \Delta f(x_1) - \Delta f(x_0), \quad \text{and so on.}$$ 2. **Use the interpolation formula**: $$f(x) = f(x_0) + \frac{(x - x_0)}{1!} \Delta f(x_0) + \frac{(x - x_0)(x - x_1)}{2!} \Delta^2 f(x_0) + \cdots$$ ### Step 1: Calculate the forward differences Start by calculating the first, second, and third forward differences. #### First forward differences $$\Delta f(x)$$: $$\Delta f(0) = f(1) - f(0) = 26 - (-18) = 44$$ $$\Delta f(1) = f(2) - f(1) = 11 - 26 = -15$$ $$\Delta f(2) = f(3) - f(2) = -6 - 11 = -17$$ #### Second forward differences $$\Delta^2 f(x)$$: $$\Delta^2 f(0) = \Delta f(1) - \Delta f(0) = -15 - 44 = -59$$ $$\Delta^2 f(1) = \Delta f(2) - \Delta f(1) = -17 - (-15) = -2$$ #### Third forward difference $$\Delta^3 f(x)$$: $$\Delta^3 f(0) = \Delta^2 f(1) - \Delta^2 f(0) = -2 - (-59) = 57$$ ### Step 2: Apply the Newton Forward Interpolation formula The formula for Newton’s Forward Interpolation is: $$f(x) = f(x_0) + \frac{(x - x_0)}{1!} \Delta f(x_0) + \frac{(x - x_0)(x - x_1)}{2!} \Delta^2 f(x_0) + \frac{(x - x_0)(x - x_1)(x - x_2)}{3!} \Delta^3 f(x_0)$$ We need to find $$f(1.75)$$, where $$x_0 = 0$$, and we know the values of the differences. - $$f(x_0) = -18$$ - $$\Delta f(x_0) = 44$$ - $$\Delta^2 f(x_0) = -59$$ - $$\Delta^3 f(x_0) = 57$$ Let’s plug in these values for $$x = 1.75$$. 1. First, calculate the term for $$\frac{(x - x_0)}{1!}$$: $$\frac{(1.75 - 0)}{1!} = 1.75$$ 2. Next, calculate the term for $$\frac{(x - x_0)(x - x_1)}{2!}$$: $$\frac{(1.75 - 0)(1.75 - 1)}{2!} = \frac{(1.75)(0.75)}{2} = \frac{1.3125}{2} = 0.65625$$ 3. Then, calculate the term for $$\frac{(x - x_0)(x - x_1)(x - x_2)}{3!}$$: $$\frac{(1.75 - 0)(1.75 - 1)(1.75 - 2)}{3!} = \frac{(1.75)(0.75)(-0.25)}{6} = \frac{-0.328125}{6} = -0.0546875$$ ### Step 3: Calculate $$f(1.75)$$ Now, substitute all the terms into the Newton Forward Interpolation formula: $$f(1.75) = -18 + (1.75 \times 44) + (0.65625 \times -59) + (-0.0546875 \times 57)$$ Calculating each term: - $$1.75 \times 44 = 77$$ - $$0.65625 \times -59 = -38.8125$$ - $$-0.0546875 \times 57 = -3.115234375$$ Thus, $$f(1.75) = -18 + 77 - 38.8125 - 3.115234375$$ $$f(1.75) = 77 - 38.8125 - 3.115234375 - 18$$ $$f(1.75) \approx 17.072265625$$ ### Final Answer: $$f(1.75) \approx 17.07$$ --- Would you like more details or clarification on any of the steps? Here are some related questions to explore further: 1. What is the difference between Newton's Forward and Backward Interpolation methods? 2. How do you choose between using Newton’s Forward or Backward Interpolation in practice? 3. What happens if we have more data points? How does the formula change? 4. How does the degree of the polynomial affect the accuracy of interpolation? 5. Can Newton's interpolation method be applied when data points are unevenly spaced? Tip: The accuracy of interpolation increases with the number of data points, but after a certain degree, the polynomial can oscillate and become less reliable.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Interpolation
Newton's Forward Interpolation
Forward Differences
Formulas
Newton's Forward Interpolation Formula: f(x) = f(x_0) + (x - x_0)/1! * Δf(x_0) + (x - x_0)(x - x_1)/2! * Δ²f(x_0) + ...
Forward Difference Formula: Δf(x_0) = f(x_1) - f(x_0)
Theorems
Newton's Forward Interpolation Theorem
Suitable Grade Level
Grades 11-12