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Let \(x_{0}

Short Answer

Expert verified
The polynomial is: \[ q(x) = ax^2 + bx + c \]. Ensure x-values create a non-singular system for uniqueness.

Step by step solution

01

Polynomial Assumption

Assume the polynomial to be of the form: \[ q(x) = ax^2 + bx + c \] where a, b, and c are constants to be determined.
02

Apply Initial Conditions

Use the given conditions to create a system of equations: 1. \[ q(x_0) = y_0 \rightarrow a(x_0)^2 + b(x_0) + c = y_0 \]2. \[ q^{\text{'}}(x_1) = y_1^{\text{'}} \rightarrow 2ax_1 + b = y_1^{\text{'}} \]3. \[ q(x_2) = y_2 \rightarrow a(x_2)^2 + b(x_2) + c = y_2 \]
03

Form System of Equations

Equations derived from Step 2:1. \[ ax_0^2 + bx_0 + c = y_0 \]2. \[ 2ax_1 + b = y_1^{\text{'}} \]3. \[ ax_2^2 + bx_2 + c = y_2 \]
04

Solve for Constants

Solve the system of equations to find values of a, b, and c. Use substitution or matrix methods to solve the linear systems.
05

Check Uniqueness

Check if the polynomial is uniquely determined by these equations. The system is uniquely determined if the determinant of the coefficient matrix does not equal zero. Here, ensure the x-values and conditions do not lead to a singular matrix.
06

Summary

Compile the found coefficients into the polynomial expression: \[ q(x) = ax^2 + bx + c \]

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Key Concepts

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

polynomial interpolation
Polynomial interpolation involves finding a polynomial that passes through a given set of points.
In this exercise, we need a polynomial of degree at most 2, meaning a quadratic polynomial. We generally represent a quadratic polynomial by: ax^2 + bx + c
Given points or conditions tell us where the polynomial should pass through or what slope it should have at specific points.
For this exercise, the conditions are given by: q(x0) = y0, q'(x1) = y1', and q(x2) = y2.
We use these conditions to form a system of equations, which we will then solve to find the coefficients 'a', 'b', and 'c' of our polynomial. This process ensures that our polynomial perfectly fits the given conditions.
derivative constraint
A derivative constraint is a condition involving the derivative of the polynomial.
In our example, we are given q'(x1) = y1', which means the first derivative of q(x) at x = x1 equals y1'. The derivative constraint affects the slope of our polynomial at a specific point. For a quadratic polynomial q(x) = ax^2 + bx + c, its derivative is q'(x) = 2ax + b.
So the condition q'(x1) = y1' translates to: 2ax1 + b = y1'. This equation helps us set up our system of equations and is critical in solving for a particularly fitting polynomial.
system of equations
A system of equations consists of multiple equations that we need to solve simultaneously.
Our quadratic polynomial interpolation involves setting up 3 equations based on given conditions: q(x0) = y0, q'(x1) = y1', q(x2) = y2.
We write these conditions in terms of 'a', 'b', and 'c': 1. ax0^2 + bx0 + c = y0 2. 2ax1 + b = y1' 3. ax2^2 + bx2 + c = y2
These three equations make up our system of equations. Solving this system will give us specific values for 'a', 'b', and 'c', which define our quadratic polynomial.
uniqueness of solutions
Uniqueness of solutions indicates whether our system of equations has exactly one solution.
This is important because multiple solutions or no solutions mean we can't find a definite polynomial fitting the given conditions.
To ensure uniqueness, we need to check if the determinant of the system's coefficient matrix is non-zero. In other words, the matrix based on our (x0, x1, x2) values should not be singular.
If the determinant is zero, then the system is either dependent (infinite solutions) or inconsistent (no solution). But if the determinant is non-zero, it means there is a unique solution, and thus a unique polynomial that meets our constraints.

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Most popular questions from this chapter

To visualize the change in the values of $$ \Psi_{n}(x)=\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right) $$ as \(n\) increases and as \(x\) varies over \(\left[x_{0}, x_{n}\right]\), graph the special case $$ \Psi_{n}(x)=x(x-1) \cdots(x-n) $$ for \(0 \leq x \leq n\). Do this for \(n=3,4, \ldots, 8\).

Let \(w(x)=1 / \sqrt{1-x^{2}}\) for \(-1

Given the data below, find \(f\left[x_{0}, x_{1}\right]\) and \(f\left[x_{0}, x_{1}, x_{2}\right] .\) Then calculate \(P_{1}(0.15)\) and \(P_{2}(0.15)\), the linear and quadratic interpolates evaluated at \(x=0.15\). $$ \begin{array}{ccc} \hline & & \\ n & x_{n} & f\left(x_{n}\right) \\ \hline 0 & 0.1 & 0.2 \\ 1 & 0.2 & 0.24 \\ 2 & 0.3 & 0.30 \\ \hline \end{array} $$

Consider the proof of the error formula for linear interpolation $$ f(x)-P_{1}(x)=\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)}{2} f^{\prime \prime}(c) $$ with \(\min \left\\{x_{0}, x_{1}, x\right\\} \leq c \leq \max \left\\{x_{0}, x_{1}, x\right\\} .\) From the construction of \(P_{1}(x)\), the error formula is clearly true if \(x=x_{0}\) or \(x=x_{1}\). Thus, we consider only the case \(x \neq x_{0}, x_{1} .\) Introduce $$ E(t)=f(t)-P_{1}(t) $$ and $$ G(t)=E(t)-\frac{\left(t-x_{0}\right)\left(t-x_{1}\right)}{\left(x-x_{0}\right)\left(x-x_{1}\right)} E(x) $$ with \(t\) varying and \(x\) fixed. (a) Show \(G\left(x_{0}\right)=G\left(x_{1}\right)=G(x)=0\). (b) Using Rolle's theorem or the mean value theorem, show that \(G^{\prime}(t)\) has at least two zeros; and then show that \(G^{\prime \prime}(t)\) has at least one root, calling it \(c\). (c) Calculate \(E^{\prime \prime}(t)\) and \(G^{\prime \prime}(t) .\) Then evaluate \(G^{\prime \prime}(c)\) and solve for \(E(x)\) to conclude the derivation of the error formula. This derivation generalizes to a proof of the general result in Theorem \(4.2 .1\) for any \(n \geq 1\).

Given the data points \((0,2),(1,1)\), find the following: (a) The straight line interpolating this data. (b) The function \(f(x)=a+b e^{x}\) interpolating this data. Hint: Find \(a\) and \(b\) so that \(f(0)=2, f(1)=1\). (c) The function \(f(x)=a /(b+x)\) interpolating this data. In each case, graph the interpolating function.

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