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Chapter 4: Problem 26

(a) By using function program divdif, calculate the divided differences \(D_{0}=\) \(f\left(x_{0}\right), D_{1}=f\left[x_{0}, x_{1}\right], \ldots, D_{5}=f\left[x_{0}, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right]\), for \(f(x)=e^{x}\) Use \(x_{0}=0, x_{1}=0.2, x_{2}=0.4, \ldots, x_{5}=1.0\) (b) Using the results of (a), calculate \(P_{j}(x)\) for \(x=0.1,0.3,0.5\) and \(j=\) \(1, \ldots, 5 .\) Compare these results to the true values of \(e^{x} .\)

Short Answer

Expert verified
Find divided differences, construct polynomial interpolants, and compare evaluated interpolants at specific points to true values of \( e^x \).

Step by step solution

01

Define the function and nodes

The function given is \( f(x) = e^x \). Define the nodes as \( x_0 = 0, x_1 = 0.2, x_2 = 0.4, x_3 = 0.6, x_4 = 0.8, x_5 = 1.0 \).
02

Calculate function values at the nodes

Evaluate \( f(x) \) at each node: \[ f(x_0) = e^0 = 1 \] \[ f(x_1) = e^{0.2} \approx 1.2214 \] \[ f(x_2) = e^{0.4} \approx 1.4918 \] \[ f(x_3) = e^{0.6} \approx 1.8221 \] \[ f(x_4) = e^{0.8} \approx 2.2255 \] \[ f(x_5) = e^1 \approx 2.7183 \]
03

Calculate first-order divided differences

Calculate \( D_1 = f[x_0, x_1] \): \[ D_1 = \frac{f(x_1) - f(x_0)}{x_1 - x_0} = \frac{1.2214 - 1}{0.2} = 1.107 \]
04

Calculate second-order divided differences

Calculate \( D_2 = f[x_0, x_1, x_2] \): \[ D_2 = \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} = \frac{\frac{1.4918 - 1.2214}{0.2} - 1.107}{0.4} = 0.384 \]
05

Calculate higher-order divided differences

Continue the process to calculate higher-order divided differences using the same formula. For example, \( D_3 = f[x_0, x_1, x_2, x_3] \), \( D_4 = f[x_0, x_1, x_2, x_3, x_4] \), \( D_5 = f[x_0, x_1, x_2, x_3, x_4, x_5] \).
06

Formulate polynomial interpolants

Using the divided differences, formulate the polynomial interpolants \( P_j(x) \) for \( j = 1, \, 2, \, 3, \, 4, \, 5 \). For example, \[ P_1(x) = f(x_0) + D_1(x - x_0) \] \[ P_2(x) = f(x_0) + D_1(x - x_0) + D_2(x - x_0)(x - x_1) \] Continue for higher values of \( j \).
07

Evaluate at specified points

Evaluate \( P_j(x) \) at \( x = 0.1, 0.3, 0.5 \) for each polynomial \( P_j \). Compare these values to the true values of \( e^x \).
08

Compare with true values

The true values of \( e^x \) at \( x = 0.1, 0.3, 0.5 \) are approximately \( 1.1052, 1.3499, 1.6487 \). Compare these with the polynomial approximations to see how closely they match.

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Key Concepts

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

Numerical Methods
Numerical methods are techniques used to solve mathematical problems that cannot be solved analytically. These methods approximate solutions by using algorithms, often implemented on computers. They are essential in engineering, physics, chemistry, and various real-world applications. One key area where numerical methods shine is in approximating the values of functions and solving differential equations. For example, polynomial interpolation, where one uses a polynomial to estimate the values of a function, is a common numerical method.
Polynomial Interpolation
Polynomial interpolation involves approximating a function by a polynomial. Given certain data points, we aim to find a polynomial that exactly fits those points. If we have points \((x_0, y_0), (x_1, y_1), ..., (x_n, y_n)\), we seek a polynomial \(P(x)\) that satisfies \(P(x_i) = y_i\) for all points. One of the main advantages of polynomial interpolation is that it provides a smooth curve that can be easily manipulated. However, high-degree polynomials can lead to oscillations and inaccuracies, especially outside the given data points.
Divided Differences
Divided differences are a recursive way to calculate the coefficients of the interpolating polynomial in Newton's form. For a given function \(f(x)\) evaluated at points \(x_0, x_1, ..., x_n\), the divided difference \(f[x_i, x_j, ..., x_k]\) is computed recursively starting from:
  • Zero order: \(f[x_i] = f(x_i)\)
  • First order: \(f[x_i, x_{i+1}] = \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i}\)
For higher orders, the pattern continues with:
\[ f[x_i, x_{i+1}, ..., x_{i+k}] = \frac{f[x_{i+1}, ..., x_{i+k}] - f[x_i, ..., x_{i+k-1}]}{x_{i+k} - x_i} \]
These differences allow a step-by-step construction of the interpolation polynomial.
Newton's Divided Difference Formula
Newton's Divided Difference Formula provides an efficient method for constructing the interpolation polynomial using calculated divided differences. If \(f(x)\) is evaluated at \(n+1\) points, the Newton form of the interpolation polynomial is:

\[ P_n(x) = f[x_0] + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) + ... + f[x_0, x_1, ..., x_n](x - x_0)(x - x_1)...(x - x_{n-1}) \]
Each new term involves higher-order divided differences and the products of \(x - x_i\) terms. This construction is both straightforward and computationally effective.
Exponential Function Approximation
Approximating the exponential function \(e^x\) is a common task in numerical methods. The exercise you're dealing with uses polynomial interpolation to approximate \(e^x\) at specific points. Starting with points like \(x_0 = 0, x_1 = 0.2, ..., x_5 = 1.0\), you first calculate \(e^x\) at these nodes and then compute the divided differences. Using these, you construct a series of interpolating polynomials \(P_1(x), ..., P_5(x)\).
  • For example, \(P_1(x) = 1 + 1.107(x - 0)\)
  • And \(P_2(x) = 1 + 1.107(x - 0) + 0.384(x - 0)(x - 0.2)\)
Evaluating these polynomials at \(x = 0.1, 0.3, 0.5\) demonstrates how well they approximate the true values of the exponential function. This process highlights both the power and limitations of polynomial interpolation in numerical methods.

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Most popular questions from this chapter

The following table was obtained in solving a differential equation. Using linear interpolation between adjacent nodes \(x_{i}\), produce a continuous graph of this data on the interval \(0 \leq x \leq 6\). \begin{tabular}{c|ccccccc} \(x_{i}\) & \(0.0\) & \(1.0\) & \(2.0\) & \(3.0\) & \(4.0\) & \(5.0\) & \(6.0\) \\ \hline\(y_{i}\) & \(2.0000\) & \(2.1592\) & \(3.1697\) & \(5.4332\) & \(9.1411\) & \(14.406\) & \(21.303\) \end{tabular}

Define $$ s(x)=\left\\{\begin{array}{cl} 2 x^{3}, & 0 \leq x \leq 1 \\ x^{3}+3 x^{2}-3 x+1, & 1 \leq x \leq 2 \\ 9 x^{2}-15 x+9, & 2 \leq x \leq 3 \end{array}\right. $$ Verify that \(s(x)\) is a cubic spline function on \([0,3] .\) Is it a natural cubic spline function on this interval?

(a) Let \([a, b]\) be a given interval and let \(a

(a) Let \(P_{2}^{(0,2)}(x)\) denote the quadratic polynomial that interpolates the data \(\left\\{\left(x_{0}, y_{0}\right),\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right\\}\); let \(P_{2}^{(1,3)}(x)\) denote the quadratic polynomial that interpolates the data \(\left\\{\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)\right\\}\). Finally, let \(P_{3}(x)\) denote the cubic polynomial interpolating the data \(\left\\{\left(x_{0}, y_{0}\right),\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right.\), \(\left.\left(x_{3}, y_{3}\right)\right\\}\). Show that $$ P_{3}(x)=\frac{\left(x_{3}-x\right) P_{2}^{(0.2)}(x)+\left(x-x_{0}\right) P_{2}^{(1,3)}(x)}{x_{3}-x_{0}} $$ (b) How might this be generalized to constructing \(P_{n}(x)\), interpolating \(\left\\{\left(x_{0}, y_{0}\right)\right.\), ..., \(\left.\left(x_{n}, y_{n}\right)\right\\}\), from interpolation polynomials of degree \(n-1\) ?

(a) For \(f(x)=\log x\), bound the minimax approximation error \(\rho_{n}(f)\) on \(1 \leq\) \(x \leq 2 .\) Find a bound for each case \(n \geq 1\), and have the bound converge to zero as \(n \rightarrow \infty\). (b) Find the Taylor polynomial \(t_{3}(x)\) of degree 3 for \(f(x)=\log x\) about \(x=\frac{3}{2}\). What is its maximum error on \(1 \leq x \leq 2 ?\) (c) For \(f(x)=\log (x)\) and the interval \([1,2]\), $$ m_{3}(x)=-1.492776+2.112632 x-0.729104 x^{2}+0.109690 x^{3} $$ Find \(\rho_{3}(f)\) and graph \(f(x)-m_{3}(x)\) on \([1,2]\). Compare \(t_{3}(x)\) and \(m_{3}(x)\).
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