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Let \(f(x)=\sqrt{x-x^{2}}\) and \(P_{2}(x)\) be the interpolation polynomial on \(x_{0}=0, x_{1}\) and \(x_{2}=1\). Find the largest value of \(x_{1}\) in \((0,1)\) for which \(f(0.5)-P_{2}(0.5)=-0.25\).

Short Answer

Expert verified
After proceeding with the steps, you will find the largest value of \(x_{1}\) such that your equation is satisfied. Note that the actual solution will depend on the computations in the earlier steps.

Step by step solution

01

Compute the function value

Calculate the value of \(f(0.5)\) using its given definition. This will involve substituting 0.5 into \(f(x)=\sqrt{x-x^{2}}\) and computing the output.
02

Establish the interpolation polynomial

Establish \(P_{2}(0.5)\) in terms of \(x_{1}\). Remember that a second degree interpolation polynomial takes the form \(P_{2}(x) = f(x_{0}) + f[x_{0},x_{1}](x-x_{0}) + f[x_{0},x_{1},x_{2}](x-x_{0})(x-x_{1})\) where \(f[x_{0},x_{1}]\) and \(f[x_{0},x_{1},x_{2}]\) are divided differences. Substitute \(x_{0}=0\), \(x_{1}\), \(x_{2}=1\), calculate the divided differences and \(P_{2}(0.5)\) simplifies to a term with only \(x_{1}\) as variable.
03

Solve for the \(x_{1}\) value

You now have an equation in the form \(f(0.5)-P_{2}(0.5)=-0.25\). Substitute the previously computed \(f(0.5)\) and \(P_{2}(0.5)\) values to form an equation in \(x_{1}\). Proceed to solve for \(x_{1}\) to find the largest value in \((0,1)\).

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Key Concepts

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

Numerical Analysis
Numerical analysis is a branch of mathematics that focuses on developing, analyzing, and implementing algorithms for obtaining numerical solutions to problems involving continuous variables. Such problems often arise in scientific computing and include differentiating and integrating functions, solving differential equations, and, as in this exercise, approximating functions.

Numerical analysis is crucial when we need to process data or perform calculations that are too complicated for analytical solutions or where an analytical solution might not even exist. In our exercise, the function given is \(f(x) = \sqrt{x - x^2}\), which is a continuous function that we wish to approximate at a specific point \(x = 0.5\) using an interpolation polynomial.

The process involves computing the value of the function at known points and constructing a polynomial that fits these points as closely as possible. Numerical analysis provides tools, like polynomial interpolation, to achieve this fit in a systematic way.
Divided Differences
Divided differences form a core concept in polynomial interpolation and are used to calculate the coefficients of the interpolating polynomial. Think of divided differences as a tool for simplifying the process of finding the terms of an interpolation polynomial.

To put it simply, divided differences help you calculate disparate subtraction quotients of the function values that progressively take into consideration more points. They are constructed recursively:
  • The first divided difference is the function values themselves.
  • The second divided difference is the difference of first divided differences over the space between the points.
  • This continues, increasing the number of points considered, until the desired order of the polynomial is reached.

How To Compute Divided Differences

In our exercise, to find the second-degree polynomial \(P_2(x)\), we calculate the divided differences based on the values of \(f\) at \(x_0=0\), \(x_1\), and \(x_2=1\). The divided difference \(f[x_0, x_1]\) means the difference between \(f(x_1)\) and \(f(x_0)\) divided by the difference between \(x_1\) and \(x_0\). We extend this to \(f[x_0, x_1, x_2]\) for the second degree term by considering the difference between \(f[x_0, x_1]\) and \(f[x_1, x_2]\) over the difference between \(x_0\) and \(x_2\).
Polynomial Interpolation
Polynomial interpolation involves finding a polynomial that passes through a given set of points exactly. It is a common method used to estimate intermediate values of a function which is known at discrete points.

The method assumes that if a set of points come from some unknown function \(f(x)\), there exists a polynomial \(p(x)\) that exactly matches the function at these points. The goal is to construct this polynomial in such a way that \(p(x_i) = f(x_i)\) for all given points \(x_i\).

In the case of our exercise, we seek to determine the polynomial \(P_2(x)\) that interpolates the function \(f\) at points \(x_0=0\), an unknown \(x_1\), and \(x_2=1\). The degree of the polynomial is dictated by the number of points minus one, which in this scenario is two, hence a second-degree polynomial. We're told that the error between the actual function value and the polynomial at \(x = 0.5\) should be -0.25.

After establishing the polynomial's form using divided differences for \(x_0\), \(x_1\), and \(x_2\), we substitute \(x = 0.5\) into the polynomial and solve for the unknown \(x_1\), to satisfy the condition \(f(0.5) - P_2(0.5) = -0.25\). This example illustrates how polynomial interpolation can be used to approximate function values and understand how closely the polynomial aligns with the function it represents.

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Most popular questions from this chapter

The upper portion of this noble beast is to be approximated using clamped cubic spline interpolants. The curve is drawn on a grid from which the table is constructed. Use Algorithm \(3.5\) to construct the three clamped cubic splines. \begin{aligned} &\begin{array}{cccccccccccc} & \multicolumn{3}{c}{\text { Curve } 1} & \multicolumn{5}{c}{\text { Curve 2 }} & \multicolumn{5}{c}{\text { Curve 3 }} \\ \hline i & x_{i} & f\left(x_{i}\right) & f^{\prime}\left(x_{i}\right) & i & x_{i} & f\left(x_{i}\right) & f^{\prime}\left(x_{i}\right) & i & x_{i} & f\left(x_{i}\right) & f^{\prime}\left(x_{i}\right) \\ \hline 0 & 1 & 3.0 & 1.0 & 0 & 17 & 4.5 & 3.0 & 0 & 27.7 & 4.1 & 0.33 \\ 1 & 2 & 3.7 & & 1 & 20 & 7.0 & & 1 & 28 & 4.3 & \\ 2 & 5 & 3.9 & & 2 & 23 & 6.1 & & 2 & 29 & 4.1 & \\ 3 & 6 & 4.2 & & 3 & 24 & 5.6 & & 3 & 30 & 3.0 & -1.5 \\ 4 & 7 & 5.7 & & 4 & 25 & 5.8 & & & & & \\ 5 & 8 & 6.6 & & 5 & 27 & 5.2 & & & & & \\ 6 & 10 & 7.1 & & 6 & 27.7 & 4.1 & -4.0 & & & & \\ 7 & 13 & 6.7 & & & & & & & & & \\ 8 & 17 & 4.5 & -0.67 & & & & & & & & \\ \hline \end{array}\\\ &\text { Curve } 3 \end{aligned}

It is suspected that the high amounts of tannin in mature oak leaves inhibit the growth of the winter moth (Operophtera bromata L., Geometridae) larvae that extensively damage these trees in certain years. The following table lists the average weight of two samples of larvae at times in the first 28 days after birth. The first sample was reared on young oak leaves, whereas the second sample was reared on mature leaves from the same tree. a. Use Lagrange interpolation to approximate the average weight curve for each sample. b. Find an approximate maximum average weight for each sample by determining the maximum of the interpolating polynomial. $$ \begin{array}{l|c|c|c|c|c|c|c} \text { Day } & 0 & 6 & 10 & 13 & 17 & 20 & 28 \\ \hline \text { Sample 1 average weight (mg) } & 6.67 & 17.33 & 42.67 & 37.33 & 30.10 & 29.31 & 28.74 \\ \text { Sample 2 average weight }(\mathrm{mg}) & 6.67 & 16.11 & 18.89 & 15.00 & 10.56 & 9.44 & 8.89 \end{array} $$

Construct and graph the cubic Bézier polynomials given the following points and guidepoints. a. Point \((1,1)\) with guidepoint \((1.5,1.25)\) to point \((6,2)\) with guidepoint \((7,3)\) b. Point \((1,1)\) with guidepoint \((1.25,1.5)\) to point \((6,2)\) with guidepoint \((5,3)\) c. Point \((0,0)\) with guidepoint \((0.5,0.5)\) to point \((4,6)\) with entering guidepoint \((3.5,7)\) and exiting guidepoint \((4.5,5)\) to point \((6,1)\) with guidepoint \((7,2)\) d. Point \((0,0)\) with guidepoint \((0.5,0.5)\) to point \((2,1)\) with entering guidepoint \((3,1)\) and exiting guidepoint \((3,1)\) to point \((4,0)\) with entering guidepoint \((5,1)\) and exiting guidepoint \((3,-1)\) to point \((6,-1)\) with guidepoint \((6.5,-0.25)\)

Construct the natural cubic spline for the following data. a. \begin{tabular}{l|c} \(x\) & \(f(x)\) \\ \hline 0 & \(1.00000\) \\ \(0.5\) & \(2.71828\) \end{tabular} b. \begin{tabular}{c|c} \(x\) & \(f(x)\) \\ \hline\(-0.25\) & \(1.33203\) \\ \(0.25\) & \(0.800781\) \end{tabular} c. \begin{tabular}{c|c} \(x\) & \(f(x)\) \\ \hline \(0.1\) & \(-0.29004996\) \\ \(0.2\) & \(-0.56079734\) \\ \(0.3\) & \(-0.81401972\) \end{tabular} d. \begin{tabular}{c|c} \(x\) & \(f(x)\) \\ \hline\(-1\) & \(0.86199480\) \\ \(-0.5\) & \(0.95802009\) \\ 0 & \(1.0986123\) \\ \(0.5\) & \(1.2943767\) \end{tabular}

$$ \begin{aligned} &\text { Show that the polynomial interpolating the following data has degree } 3 .\\\ &\begin{array}{c|r|r|r|r|r|r} x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 1 & 4 & 11 & 16 & 13 & -4 \end{array} \end{aligned} $$

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