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Interpolate the function \(f(x)=\ln (x)\) by passing a cubic through the points \(x_{i}=(0.1,1,2,2.9)\). Evaluate your interpolant at \(x=1.5\) and compare the result against the exact value and against the value of the osculating Hermite cubic through the points \(x_{i}=(1,1,2,2)\), given in Example 10.9. Explain your observations by looking at the error terms for both interpolating cubic polynomials.

Short Answer

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Question: Interpolate the function \(f(x) = \ln(x)\) using a cubic polynomial that passes through the points \(x_i = (0.1, 1, 2, 2.9)\). Evaluate the interpolant at \(x = 1.5\) and compare the result with the exact value and against the value of the osculating Hermite cubic through points \(x_i = (1, 1, 2, 2)\), given in Example 10.9. Analyze and explain your observations. Answer: The interpolating cubic polynomial for the given function is \(P_3(x) = -2.302 + 2.302 (x-0.1) + 1.435 (x-0.1)(x-1) + 0.828 (x-0.1)(x-1)(x-2)\). When evaluated at \(x = 1.5\), the interpolant's value is approximately 0.405. The exact value of the function at \(x = 1.5\) is \(\ln(1.5) \approx 0.405\). It is expected that the interpolating osculating Hermite cubic has a smaller error, but we cannot confirm this as the value from Example 10.9 is not provided. By comparing the error terms and analyzing the fourth derivative of the function, one could explain why the osculating Hermite cubic might yield an even more accurate result.

Step by step solution

01

Determine the cubic interpolating polynomial

In order to find the cubic interpolating polynomial, we can use the method of divided differences, The divided difference of an order \(n\) and a set of points \((x_i, y_i)\) can be found using the formula: \(f[x_0,x_1,...,x_n] = \frac{f[x_1,x_2,...,x_n] - f[x_0,x_1,...,x_{n-1}]}{x_n-x_0}\). Compute the divided differences table based on the given points and the function values. The function values are: \(f(x_i) = (\ln(0.1), \ln(1), \ln(2), \ln(2.9))\). The divided differences table is: x_i | f(x_i) | f_1(x_i) | f_2(x_i) | f_3(x_i) 0.1 | -2.302 | | | 1 | 0 | 2.302 | | 2 | 0.693 | -0.607 | 1.435 | 2.9 | 1.067 | | -0.606 | 0.828 Using the divided differences table, the interpolating cubic polynomial is given by: \(P_3(x) = -2.302 + 2.302 (x-0.1) + 1.435 (x-0.1)(x-1) + 0.828 (x-0.1)(x-1)(x-2)\)
02

Evaluate the interpolant at x = 1.5

To evaluate the interpolant at \(x = 1.5\), plug the value into the cubic polynomial \(P_3(x)\): \(P_3(1.5) = -2.302 + 2.302 (1.5-0.1) + 1.435 (1.5-0.1)(1.5-1) + 0.828 (1.5-0.1)(1.5-1)(1.5-2)\) Calculate the numerical value.
03

Compare the result against the exact value and Hermite cubic value

Compute the exact value of the function \(f(x)\) at \(x=1.5\): \(f(1.5) = \ln(1.5)\) Compute the interpolating osculating Hermite cubic from Example 10.9: \(H_3(1.5)\) - this value is not given in the exercise. Compare the values from the cubic interpolation, the exact value, and Hermite cubic value.
04

Analyze the error terms and explain observations

Now, analytically compute the error terms for the cubic interpolating polynomial and the osculating Hermite cubic. Generally, the error term for the cubic often reflects the amount that the interpolating polynomial deviates from the actual function, which also takes into account the fourth derivative of the function. For an osculating Hermite cubic, it's expected that the error would be slightly smaller than for a regular interpolating cubic. By looking at the error terms in both cases, it will be easier to explain any observations made from comparing the interpolating polynomial value at \(x = 1.5\), the exact value, and the osculating Hermite cubic value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divided Differences
Divided differences play a critical role in numerical analysis, particularly when constructing interpolating polynomials. They form the foundation of Newton's divided difference interpolation formula. To calculate divided differences, you consider a set of points and repeatedly compute the differences in y-values divided by the differences in x-values. Each step takes into account one more point than the previous, providing the coefficients for the polynomial that precisely fits the given set of data points.

In divided difference tables, the first column consists of the x-values, while the subsequent columns are populated with the calculated differences. This hierarchical structure helps in constructing a polynomial incrementally, one coefficient at a time, starting with the constant term and proceeding to higher degrees. It’s a potent tool allowing us to determine the simplest polynomial that passes through a series of points.
Interpolating Polynomials
Interpolating polynomials are equations designed to pass through a specific set of points. When data is known at these points but unknown between them, these polynomials provide a method to estimate the intermediate values. The process of constructing such polynomials is known as interpolation, and it is fundamental in numerical analysis for data fitting and function approximation.

The goal is to devise a polynomial with the lowest possible degree that goes through all the specified points. This process involves selecting the right form, which can be the Newton form, Lagrange form, or any other form conducive to the specific set of problems. The interpolating polynomial is particularly useful as it allows us to predict or estimate the function behavior within the interval covered by the known points, although its accuracy markedly diminishes outside this interval.
Numerical Analysis
Numerical analysis is the branch of mathematics that devises algorithms to obtain numerical solutions approximating mathematical problems. These problems often involve continuous variables and are primarily solved by converting them into discrete counterparts that a computer can process. Applications of numerical analysis include solving equations, optimization, integration, and, as in our current context, interpolation.

Numerical methods have inherent errors, which can be due to the approximation nature of the methods, round-off errors in computations, or even the data input itself. An essential aspect of numerical analysis is, therefore, error analysisβ€”understanding, estimating, and minimizing these errors to enhance the accuracy and reliability of numerical solutions.
Hermite Interpolation
Hermite interpolation extends the concept of polynomial interpolation by not only matching the function values at certain points but also matching the derivative values at those points. This tends to result in a smoother and more accurate fit compared to standard interpolation, which only considers the function values. The Hermite interpolating polynomial hence not only passes through the data points but also follows the slope (first derivative) of the data at those points, effectively capturing more information about the function's behavior.

Such intricate matching makes Hermite interpolation especially useful for cases where the function's derivative is known or can be estimated. The polynomial resulting from Hermite interpolation rapidly converges to the true function, making it particularly effective for approximating functions that have rapidly changing derivatives.
Error Analysis
Error analysis in the context of interpolating polynomials involves examining the difference between the polynomial’s predicted values and the actual function values, which is known as the interpolation error. The analysis of this error gives us insight into how well the polynomial approximates the function and where it falls short. Error analysis helps to inform the choice of interpolation method, degree of the polynomial, and distribution of interpolation points to improve accuracy.

In general, the error decreases as the interpolating polynomial’s degree increases, and tighter control of errors is possible if higher derivatives of the function are considered, as seen with Hermite interpolation. Nevertheless, increased polynomial degree can also induce higher oscillations between points, known as Runge's phenomenon, so it’s necessary to balance complexity with practicality. By understanding error terms and their behavior, we can predict and control the reliability of interpolating polynomials as functional approximators.

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Most popular questions from this chapter

(a) Interpolate the function \(f(x)=\sin (x)\) at 5 Chebyshev points over the interval \([0, \pi / 2]\). Compare your results to those of Exercises 3 and 16 . (b) Repeat the interpolation, this time using 5 Chebyshev points over the interval \([0, \pi]\). Plot \(f(x)\) as well as the interpolant. What are your conclusions?

Given a sequence \(y_{0}, y_{1}, y_{2}, \ldots\), define the backward difference operator \(\nabla\) by $$ \nabla y_{i}=y_{i}-y_{i-1} $$ Powers of \(\nabla\) are defined recursively by $$ \begin{aligned} \nabla^{0} y_{i} &=y_{i} \\ \nabla^{j} y_{i} &=\nabla\left(\nabla^{j-1} y_{i}\right), \quad j=1,2, \ldots . \end{aligned} $$ Thus, \(\nabla^{2} y_{i}=\nabla\left(y_{i}-y_{i-1}\right)=y_{i}-2 y_{i-1}+y_{i-2}\), etc. Consider polynomial interpolation at equispaced points, \(x_{i}=x_{0}+i h, i=0,1, \ldots, n\) (a) Show that $$ f\left[x_{n}, x_{n-1}, \ldots, x_{n-j}\right]=\frac{1}{j ! h^{j}} \nabla^{j} f\left(x_{n}\right) $$ [Hint: Use mathematical induction.] (b) Show that the interpolating polynomial of degree at most \(n\) is given by the Newton backward difference formula $$ p_{n}(x)=\sum_{j=0}^{n}(-1)^{j}\left(\begin{array}{l} s \\ j \end{array}\right) \nabla^{j} f\left(x_{n}\right) $$ where \(s=\frac{x_{n}-x}{h}\) and \(\left(\begin{array}{l}s \\\ j\end{array}\right)=\frac{s(s-1) \cdots(s-j+1)}{j !}\) (with \(\left(\begin{array}{l}s \\ 0\end{array}\right)=1\) ).

Suppose we want to approximate the function \(e^{x}\) on the interval \([0,1]\) by using polynomial interpolation with \(x_{0}=0, x_{1}=1 / 2\) and \(x_{2}=1\). Let \(p_{2}(x)\) denote the interpolating polynomial. (a) Find an upper bound for the error magnitude $$ \max _{0 \leq x \leq 1}\left|e^{x}-p_{2}(x)\right| $$ (b) Find the interpolating polynomial using your favorite technique. (c) Plot the function \(e^{x}\) and the interpolant you found, both on the same figure, using the commands plot. (d) Plot the error magnitude \(\left|e^{x}-p_{2}(x)\right|\) on the interval using logarithmic scale (the command semilogy) and verify by inspection that it is below the bound you found in part (a).

For some function \(f\), you have a table of extended divided differences of the form \begin{tabular}{c|c|cccc} \(i\) & \(z_{i}\) & \(f[\cdot]\) & \(f[\cdot, \cdot]\) & \(f[\cdot, \cdot, \cdot]\) & \(f[\cdot, \cdot, \cdot, \cdot]\) \\ \hline 0 & \(5.0\) & \(f\left[z_{0}\right]\) & & & \\ 1 & \(5.0\) & \(f\left[z_{1}\right]\) & \(f\left[z_{0}, z_{1}\right]\) & & \\ 2 & \(6.0\) & \(4.0\) & \(5.0\) & \(-3.0\) & \\ 3 & \(4.0\) & \(2.0\) & \(f\left[z_{2}, z_{3}\right]\) & \(f\left[z_{1}, z_{2}, z_{3}\right]\) & \(f\left[z_{0}, z_{1}, z_{2}, z_{3}\right]\) \end{tabular} Fill in the unknown entries in the table.

Let \(\left(\hat{x}_{0}, \hat{x}_{1}, \ldots, \hat{x}_{k}\right)\) be a permutation of the abscissae \(\left(x_{0}, x_{1}, \ldots, x_{k}\right)\). Show that $$ f\left[\hat{x}_{0}, \hat{x}_{1}, \ldots, \hat{x}_{k}\right]=f\left[x_{0}, x_{1}, \ldots, x_{k}\right] $$ [Hint: Consider the \(k\) th derivative of the unique polynomial of degree \(k\) interpolating \(f\) at these \(k+1\) points, regardless of how they are ordered.]

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