Math Problem Statement
The following data represents the function x f (x) e , x 1 1.5 2 2.5 3 f (x) 2.7183 4.4817 7.3891 12.1825 20.0855 Estimate the value of f (1.75) by using the Newton’s forward difference interpolation. Obtain a bound on the error at x 1.75 .
Solution
To estimate the value of using Newton's forward difference interpolation, we'll follow these steps:
Step 1: Tabulate the given data
The data provided is:
x & f(x) = e^x \\ \hline 1.0 & 2.7183 \\ 1.5 & 4.4817 \\ 2.0 & 7.3891 \\ 2.5 & 12.1825 \\ 3.0 & 20.0855 \\ \end{array}$$ ### Step 2: Forward differences We construct the forward difference table. Denote the first forward difference by $$\Delta f(x)$$, the second by $$\Delta^2 f(x)$$, and so on. $$\begin{array}{c|c|c|c|c} x & f(x) & \Delta f(x) & \Delta^2 f(x) & \Delta^3 f(x) \\ \hline 1.0 & 2.7183 & 1.7634 & 1.1440 & 0.0486 \\ 1.5 & 4.4817 & 2.9074 & 1.1926 & \\ 2.0 & 7.3891 & 4.7934 & \\ 2.5 & 12.1825 & 7.9030 & \\ 3.0 & 20.0855 & \\ \end{array}$$ ### Step 3: Newton's forward difference formula Newton's forward difference formula for interpolation is: $$P(x) = f(x_0) + p\Delta f(x_0) + \frac{p(p-1)}{2!} \Delta^2 f(x_0) + \frac{p(p-1)(p-2)}{3!} \Delta^3 f(x_0) + \cdots$$ where $$p = \frac{x - x_0}{h}$$, $$h$$ is the difference between consecutive $$x$$-values (in this case $$h = 0.5$$), and $$x_0 = 1.0$$. For $$x = 1.75$$: $$p = \frac{1.75 - 1.0}{0.5} = 1.5$$ Now, we substitute into the interpolation formula: $$P(1.75) = f(1.0) + p \Delta f(1.0) + \frac{p(p-1)}{2!} \Delta^2 f(1.0) + \frac{p(p-1)(p-2)}{3!} \Delta^3 f(1.0)$$ Substituting the values: $$P(1.75) = 2.7183 + 1.5(1.7634) + \frac{1.5(0.5)}{2!} (1.1440) + \frac{1.5(0.5)(-0.5)}{3!} (0.0486)$$ ### Step 4: Calculation $$P(1.75) = 2.7183 + 2.6451 + 0.4290 - 0.0061$$ $$P(1.75) \approx 5.7863$$ ### Step 5: Error bound The formula for the error bound in Newton's forward difference interpolation is: $$E(x) = \frac{f^{(n+1)}(\xi) (x - x_0)(x - x_1) \cdots (x - x_n)}{(n+1)!}$$ Here, the maximum value of $$f^{(4)}(x)$$ (since the 4th forward difference is small) for $$f(x) = e^x$$ can be estimated as the maximum value of $$e^x$$ on the interval $$[1, 3]$$. Since $$e^x$$ is increasing, the maximum value is at $$x = 3$$, which gives $$e^3 \approx 20.0855$$. The error bound can be approximated as: $$E(1.75) \approx \frac{20.0855 (1.75 - 1.0)(1.75 - 1.5)(1.75 - 2.0)(1.75 - 2.5)}{4!}$$ $$E(1.75) \approx \frac{20.0855 \times 0.75 \times 0.25 \times (-0.25) \times (-0.75)}{24}$$ $$E(1.75) \approx \frac{20.0855 \times 0.03515625}{24} \approx 0.0294$$ ### Final result: The estimated value of $$f(1.75)$$ using Newton's forward difference interpolation is approximately **5.7863**. The bound on the error is approximately **0.0294**. --- Would you like more details or clarifications on this method? ### Related Questions: 1. How is Newton's backward difference interpolation different from the forward difference? 2. What are other methods for polynomial interpolation? 3. How does the step size $$h$$ affect the accuracy of Newton’s interpolation? 4. Can you derive the error term for higher-order Newton interpolations? 5. What is the difference between Lagrange interpolation and Newton's method? **Tip:** When working with interpolation, always ensure the spacing between $$x$$-values (step size) is consistent for applying Newton’s method effectively.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Numerical Methods
Interpolation
Exponential Functions
Error Bound
Formulas
Newton’s forward difference interpolation formula: P(x) = f(x_0) + pΔf(x_0) + (p(p-1)/2!) Δ^2f(x_0) + (p(p-1)(p-2)/3!) Δ^3f(x_0)
Error bound formula: E(x) = f^{(n+1)}(ξ) (x-x_0)(x-x_1)...(x-x_n)/(n+1)!
Theorems
Newton's Forward Difference Theorem
Error Bound Theorem in Polynomial Interpolation
Suitable Grade Level
Undergraduate - Numerical Methods Course