Math Problem Statement

Diberikan data sebagai berikut: xi | f(xi) 1 | 3.5 2 | 4.25 3 | 7 5 | 16.75 6 | 40 (a) Buatlah tabel beda terbagi dari data tersebut. (b) Tentukan polinomial interpolasi dengan menggunakan metode beda terbagi dari data tersebut. (c) Tentukan nilai pendekatan untuk f(2.5) menggunakan interpolasi metode terbagi.

Solution

To solve this problem, let's go through each part of the question in detail.

Given Data:

The data points provided are:

x_i & f(x_i) \\ \hline 1 & 3.5 \\ 2 & 4.25 \\ 3 & 7 \\ 5 & 16.75 \\ 6 & 40 \\ \end{array}$$ ### Solution Steps #### Part (a): Create the Divided Difference Table To create a divided difference table, we use the values of $$x_i$$ and $$f(x_i)$$ and calculate the differences iteratively until we have the complete table. 1. **First Column (Original $$f(x_i)$$ values)**: This is just the given data. 2. **Second Column (First Divided Differences)**: Calculate $$\frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}}$$. 3. **Third Column (Second Divided Differences)**: Calculate $$\frac{\text{First Divided Difference} - \text{Previous First Divided Difference}}{x_i - x_{i-2}}$$, and so on. Let's compute the differences: 1. First Divided Differences: $$f[x_1, x_2] = \frac{4.25 - 3.5}{2 - 1} = 0.75$$ $$f[x_2, x_3] = \frac{7 - 4.25}{3 - 2} = 2.75$$ $$f[x_3, x_5] = \frac{16.75 - 7}{5 - 3} = 4.875$$ $$f[x_5, x_6] = \frac{40 - 16.75}{6 - 5} = 23.25$$ 2. Second Divided Differences: $$f[x_1, x_2, x_3] = \frac{2.75 - 0.75}{3 - 1} = 1.0$$ $$f[x_2, x_3, x_5] = \frac{4.875 - 2.75}{5 - 2} = 0.7083$$ $$f[x_3, x_5, x_6] = \frac{23.25 - 4.875}{6 - 3} = 6.125$$ 3. Third Divided Differences: $$f[x_1, x_2, x_3, x_5] = \frac{0.7083 - 1.0}{5 - 1} = -0.073$$ $$f[x_2, x_3, x_5, x_6] = \frac{6.125 - 0.7083}{6 - 2} = 1.3542$$ 4. Fourth Divided Difference: $$f[x_1, x_2, x_3, x_5, x_6] = \frac{1.3542 - (-0.073)}{6 - 1} = 0.285$$ This completes the divided difference table. #### Part (b): Construct the Interpolating Polynomial Using Newton's divided difference formula, the interpolating polynomial is: \[ P(x) = f(x_1) + f[x_1, x_2](x - x_1) + f[x_1, x_2, x_3](x - x_1)(x - x_2) + \dots \] Substituting the values: $$P(x) = 3.5 + 0.75(x - 1) + 1.0(x - 1)(x - 2) - 0.073(x - 1)(x - 2)(x - 3) + 0.285(x - 1)(x - 2)(x - 3)(x - 5)$$ #### Part (c): Estimate $$f(2.5)$$ Using the Polynomial To find $$f(2.5)$$, substitute $$x = 2.5$$ into $$P(x)$$ and calculate the result. Let me proceed with this calculation. Calculating: 1. First term: $$3.5$$ 2. Second term: $$0.75(2.5 - 1) = 1.125$$ 3. Third term: $$1.0(2.5 - 1)(2.5 - 2) = 1.0 \times 1.5 \times 0.5 = 0.75$$ 4. Fourth term: $$-0.073(2.5 - 1)(2.5 - 2)(2.5 - 3) = -0.073 \times 1.5 \times 0.5 \times -0.5 = 0.027375$$ 5. Fifth term: $$0.285(2.5 - 1)(2.5 - 2)(2.5 - 3)(2.5 - 5) = 0.285 \times 1.5 \times 0.5 \times -0.5 \times -2.5 = 0.2671875$$ Summing these terms: $$f(2.5) \approx 3.5 + 1.125 + 0.75 + 0.027375 + 0.2671875 = 5.6695625$$ So, the approximate value of $$f(2.5)$$ is $$5.67$$. Would you like a more detailed explanation of any of the parts, or do you have any specific questions? --- Here are 5 related questions to expand your understanding: 1. How would the interpolation polynomial change if an additional data point was added? 2. What is the purpose of using divided differences in interpolation? 3. How does Newton’s divided difference interpolation compare to Lagrange interpolation? 4. Can we use the same divided difference table to estimate values outside the given data range? 5. What are some practical applications of polynomial interpolation in real-world data analysis? **Tip:** Divided differences are especially useful for constructing interpolation polynomials efficiently without re-evaluating previous computations.

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Math Problem Analysis

Mathematical Concepts

Numerical Analysis
Polynomial Interpolation
Newton's Divided Differences

Formulas

Divided difference formula: f[x_i, x_{i+1}] = (f(x_{i+1}) - f(x_i)) / (x_{i+1} - x_i)
Newton's Interpolation Polynomial: P(x) = f(x_0) + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) + ...
Estimation formula: f(2.5) using the interpolating polynomial

Theorems

Newton's Divided Difference Theorem

Suitable Grade Level

Undergraduate - Numerical Methods

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