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Let \(f(x)=1 /(1+x)\) and let \(x_{0}=0, x_{1}=1, x_{2}=2 .\) Calculate the divided differences \(f\left[x_{0}, x_{1}\right]\) and \(f\left[x_{0}, x_{1}, x_{2}\right] .\) Using these divided differences, give the quadratic polynomial \(P_{2}(x)\) that interpolates \(f(x)\) at the given node points \(\left\\{x_{0}, x_{1}, x_{2}\right\\} .\) Graph the error \(f(x)-P_{2}(x)\) on the interval \([0,2]\).

Short Answer

Expert verified
The quadratic polynomial is \( P_2(x) = 1 - \frac{2}{3}x + \frac{1}{6}x^2 \). Graph the error \( f(x) - P_2(x) \) over the interval [0, 2].

Step by step solution

01

- Calculate Values of f(x) at Given Points

Evaluate the function at the given node points: \[ f(x_0) = f(0) = \frac{1}{1+0} = 1 \] \[ f(x_1) = f(1) = \frac{1}{1+1} = \frac{1}{2} \] \[ f(x_2) = f(2) = \frac{1}{1+2} = \frac{1}{3} \]
02

- Calculate Divided Difference f[x0, x1]

Use the formula for divided differences to compute \( f[x_0, x_1] \): \[ f[x_0, x_1] = \frac{f(x_1) - f(x_0)}{x_1 - x_0} = \frac{\frac{1}{2} - 1}{1 - 0} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \]
03

- Calculate Divided Difference f[x1, x2]

Use the formula for divided differences to compute \( f[x_1, x_2] \): \[ f[x_1, x_2] = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{\frac{1}{3} - \frac{1}{2}}{2 - 1} = \frac{\frac{1}{3} - \frac{2}{6}}{1} = \frac{-\frac{1}{6}}{1} = -\frac{1}{6} \]
04

- Calculate Second-Order Divided Difference f[x0, x1, x2]

Use the formula for higher-order divided differences to compute \( f[x_0, x_1, x_2] \): \[ f[x_0, x_1, x_2] = \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} = \frac{-\frac{1}{6} - (-\frac{1}{2})}{2 - 0} = \frac{\frac{-1}{6} + \frac{1}{2}}{2} = \frac{\frac{1}{3}}{2} = \frac{1}{6} \]
05

- Construct Quadratic Polynomial P2(x)

Using the divided differences, construct the quadratic polynomial \( P_2(x) \): \[ P_2(x) = f(x_0) + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) \] By substituting the values, we get: \[ P_2(x) = 1 - \frac{1}{2}x + \frac{1}{6}x(x - 1) \] Simplify the polynomial: \[ P_2(x) = 1 - \frac{1}{2}x + \frac{1}{6}x^2 - \frac{1}{6}x = 1 - \frac{2}{3}x + \frac{1}{6}x^2 \]
06

- Graph the Error

The error function is given by \( f(x) - P_2(x) \). Calculate the error over the interval [0, 2] and graph it. Observing the error function graph can help visualize how well the polynomial interpolation approximates the function over this interval.

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Key Concepts

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

Polynomial Interpolation
Polynomial interpolation is a method used to estimate values between known data points. It involves finding a polynomial that passes through a given set of points. This method is useful in various fields, including numerical analysis and computer graphics.

In this exercise, we use divided differences to build up a quadratic polynomial that interpolates the function \( f(x) = \frac{1}{1+x} \).

Divided differences help us systematically derive the coefficients of the interpolating polynomial. The base form of the polynomial grows progressively with each step, adding terms that ensure it passes through all given data points.

The process starts with calculating the function values at specific points, then computing the first and second divided differences. These values are then used to form the polynomial. Understanding this method allows you to effectively estimate unknown values within a given range based on known data.
Error Analysis
Error analysis in polynomial interpolation helps us measure how accurately the polynomial approximates the target function. It gives insights into the reliability of the interpolation, especially over different intervals.

The error in polynomial interpolation depends on the chosen points and the function's nature. A key concept here is that the error between the actual function \( f(x) \) and the interpolating polynomial \( P_2(x) \) can be analyzed by computing \( f(x) - P_2(x) \) over the defined interval, in this case, [0, 2].

The error is often visualized using a graph that shows how the difference between \( f(x) \) and \( P_2(x) \) varies. Minimizing this error is crucial for accurate interpolation.

In our exercise, graphing the error function highlights the deviations and helps identify where the polynomial closely follows the true function and where it doesn't. This insight can lead to improvements in interpolation methods or the selection of better points for interpolation.
Quadratic Polynomial
A quadratic polynomial is a second-degree polynomial of the form \( ax^2 + bx + c \). In our interpolation exercise, we derive the quadratic polynomial \( P_2(x) \) which approximates the function \( f(x) = \frac{1}{1+x} \) over three points: \( x_0 = 0 \), \( x_1 = 1 \), and \( x_2 = 2 \).

The quadratic polynomial we obtained is:

\( P_2(x) = 1 - \frac{2}{3}x + \frac{1}{6}x^2 \)

This polynomial was constructed using the divided differences method. Each divided difference calculated at each step contributes to the coefficients in our final polynomial.

Polynomials of higher degree, like quadratics, can capture more complexity and curvature in data compared to linear polynomials. They provide a better fit, especially for functions with non-linear behavior. However, increasing the degree also increases the risk of overfitting if there are too many data points.

Understanding quadratic polynomials and their behavior is crucial for effectively applying polynomial interpolation techniques in forecasts, modeling, and approximating unknown values from given data points.

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Most popular questions from this chapter

Write a computer program to do linear interpolation and to checkits accuracy. Input \(x_{0}\) and \(x_{1}\) and then generate the data values using \(y=e^{x}\) (or some other commonly available function). For a variety of values of \(x\), both inside and outside \(\left[x_{0}, x_{1}\right]\), compute \(P_{1}(x), e^{x}\), and their difference \(E(x)=e^{x}-P_{1}(x)\). Plot the values of \(E(x)\), to see how the error varies with \(x\).

When the nodes \(\left\\{x_{j}\right\\}\) are evenly spaced, the formulas for the divided difference and the interpolation polynomial \(P_{n}(x)\) take simpler forms. Let \(x_{j}=x_{0}+j h\) for \(j=0, \pm 1, \pm 2, \ldots\), and let \(f_{j} \equiv f\left(x_{j}\right) .\) Define the first-order forward difference of \(f(x)\) by $$ \Delta f\left(x_{j}\right) \equiv \Delta f_{j}=f_{j+1}-f_{j} $$ Define the second-order forward difference by $$ \begin{aligned} \Delta^{2} f\left(x_{j}\right) &=\Delta f_{j+1}-\Delta f_{j}=\left[f_{j+2}-f_{j+1}\right]-\left[f_{j+1}-f_{j}\right] \\ &=f_{j+2}-2 f_{j+1}+f_{j} \end{aligned} $$ Define higher-order forward differences by $$ \Delta^{k} f_{j}=\Delta^{k-1} f_{j+1}-\Delta^{k-1} f_{j}, \quad k \geq 2 $$ Show that (a) \(f\left[x_{j}, x_{j+1}\right]=\frac{1}{h} \Delta f_{j}\) (b) \(f\left[x_{j}, x_{j+1}, x_{j+2}\right]=\frac{1}{2 ! h^{2}} \Delta^{2} f_{j}\) (c) \(f\left[x_{j}, x_{j+1}, \ldots, x_{j+k}\right]=\frac{1}{k ! h^{k}} \Delta^{k} f_{j}, \quad k \geq 1\)

Find the solution to the interpolation problem of finding a polynomial \(q(x)\) with \(\operatorname{deg}(q) \leq 2\) and such that $$ q\left(x_{0}\right)=y_{0}, \quad q\left(x_{1}\right)=y_{1}, \quad q^{\prime}\left(x_{1}\right)=y_{1}^{\prime} $$ with \(x_{0} \neq x_{1}\). Hint: Write \(q(x)=y_{0} M_{0}(x)+y_{1} M_{1}(x)+y_{1}^{\prime} M_{2}(x)\) where \(\operatorname{deg}\left(M_{i}\right) \leq 2, i=\) \(0,1,2\) and each \(M_{i}(x)\) satisfies suitable interpolating conditions at the points \(x_{0}\) and \(x_{1}\). For example, \(M_{0}(x)\) should satisfy $$ M_{0}\left(x_{0}\right)=1, \quad M_{0}\left(x_{1}\right)=M_{0}^{\prime}\left(x_{1}\right)=0 $$

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}

(a) Let \(f(x)\) be a polynomial of degree \(m .\) For \(x \neq x_{0}\), define $$ g_{1}(x)=f\left[x_{0}, x\right]=\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}} $$ Show \(g(x)\) is a polynomial of degree \(m-1\). This is another justification for regarding the divided difference as an analog of the derivative. Hint: The numerator \(f(x)-f\left(x_{0}\right)\) is a polynomial of degree \(m\) with a zero at \(x=x_{0}\). Use the fundamental theorem of algebra, noting that \(x_{0}\) is a root of \(f(x)-f\left(x_{0}\right)\) (b) For \(x_{0}, x_{1}, x\) distinct and for \(f(x)\) a polynomial of degree \(m\), define $$ g_{2}(x)=f\left[x_{0}, x_{1}, x\right] $$ Show \(g_{2}(x)\) is a polynomial of degree \(m-2\).

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