Question

Consider a function \(f(x)\) such that \(f(2)=1.5713, f(3)=1.5719, f(5)=\) \(1.5738,\) and \(f(6)=1.5751 .\) Estimate \(f(4)\) using a second degree interpolating polynomial (interpolating the first three data points) and a third degree interpolating polynomial (interpolating the first four data points). Round the final results to four decimal places. Is there any advantage here in using a third degree interpolating polynomial?

Short answer

In this exercise, we estimated the value of a function, \(f(x)\), at \(x=4\) using two interpolation methods: second-degree (quadratic) and third-degree (cubic) interpolating polynomials. Using the given data points, we constructed each interpolating polynomial and evaluated it at \(x=4\). The quadratic interpolating polynomial resulted in an estimated value of \(f(4) \approx 1.5724\) and the cubic interpolating polynomial resulted in an estimated value of \(f(4) \approx 1.5725\). Although the cubic interpolating polynomial may provide a closer approximation to the function's curve, especially with more data points, the difference in the estimates was insignificant in this case.

Step-by-step solution

1. Second Degree Interpolating Polynomial (Quadratic)

We will use the first three data points to create a quadratic interpolating polynomial \(P_2(x) = ax^2 + bx + c\). First, we will substitute the values of the points into the polynomial equation and solve for the coefficients \(a, b,\) and \(c\). The system of equations is: \begin{align*} 4a + 2b + c &= 1.5713 \\ 9a + 3b + c &= 1.5719 \\ 25a + 5b + c &= 1.5738 \end{align*} Solving this system of equations for \(a, b,\) and \(c\) gives us the quadratic interpolating polynomial \(P_2(x)=0.0011x^2 - 0.0055x + 1.5757\).

2. Evaluate the Second Degree Polynomial at \(x=4\)

Now that we have the quadratic interpolating polynomial, we can estimate \(f(4)\) by evaluating \(P_2(4)\): \(P_2(4) = 0.0011(4)^2 - 0.0055(4) + 1.5757 \approx 1.5724\)

3. Third Degree Interpolating Polynomial (Cubic)

Now we will use all four data points to create a cubic interpolating polynomial \(P_3(x) = ax^3 + bx^2 + cx + d\). We will substitute the values of the four points into the polynomial equation and solve for the coefficients \(a, b, c,\) and \(d\). The system of equations is: \begin{align*} 8a + 4b + 2c + d &= 1.5713 \\ 27a + 9b + 3c + d &= 1.5719 \\ 125a + 25b + 5c + d &= 1.5738 \\ 216a + 36b + 6c + d &= 1.5751 \end{align*} Solving this system of equations for \(a, b, c,\) and \(d\) gives us the cubic interpolating polynomial \(P_3(x)=-0.0002x^3 + 0.0018x^2 - 0.0032x + 1.573\)

4. Evaluate the Third Degree Polynomial at \(x=4\)

Now that we have the cubic interpolating polynomial, we can estimate \(f(4)\) by evaluating \(P_3(4)\): \(P_3(4) = -0.0002(4)^3 + 0.0018(4)^2 - 0.0032(4) + 1.573 \approx 1.5725\)

5. Compare the Results and Discuss Advantages

The second-degree interpolating polynomial gave an estimate of \(f(4) \approx 1.5724\), while the third-degree interpolating polynomial gave an estimate of \(f(4) \approx 1.5725\). The two estimates are very close to each other. The advantage of using a third-degree interpolating polynomial is that it can more closely approximate the function's curve, particularly when more data points are used. However, in this case, the difference in the estimates is too small to give any significant advantage to the third-degree polynomial.

Key concepts

Numerical Analysis

Numerical analysis is a branch of mathematics that deals with algorithms and techniques for finding approximate solutions to complex mathematical problems. Often in science and engineering, we encounter functions that are difficult or impossible to describe analytically with a formula. Numerical analysis provides tools to analyze these functions and come up with numerical solutions.

Numerical methods are particularly useful for interpolation, which is the process of estimating unknown values that fall within a range of known data points. Interpolating polynomials, which you may encounter in problems like estimating the value of a function at a certain point given a set of values, are just one example of the powerful techniques numerical analysis offers.

Polynomial Interpolation

Polynomial interpolation is a fundamental concept underpinning numerical analysis where a polynomial function is fitted through a set of discrete points. The goal is to construct a polynomial that passes exactly through these data points and use it to estimate values at other points within the domain. This is particularly useful when we want to approximate the value of a function we don't have the explicit form for, or when working with discrete data from experiments and measurements.

Polynomial interpolation involves solving for the coefficients of the polynomial that will make the polynomial pass through the given points. The degree of the polynomial corresponds to the number of data points minus one; for example, with three points, you would use a quadratic (second-degree) polynomial, and with four points, a cubic (third-degree) polynomial.

Quadratic Interpolating Polynomial

A quadratic interpolating polynomial is a second-degree polynomial used to approximate functions based on three data points. This polynomial can be of the form \(P_2(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are the coefficients that need to be solved for.

Once the coefficients are determined by solving a system of linear equations derived from the input data points, the quadratic polynomial can be used to estimate function values at other points within the interval spanned by the data. A quadratic polynomial is often sufficient for simple curves and in cases where the data doesn't have extreme changes in slopes or curvatures. However, it may not capture the detailed behavior of more complex functions.

Cubic Interpolating Polynomial

Cubic interpolating polynomials offer a higher level of approximation accuracy by utilizing four data points. With the general form \(P_3(x) = ax^3 + bx^2 + cx + d\), a cubic polynomial can accommodate the subtle curvatures and features of more complex datasets.

The process of deriving a cubic interpolating polynomial involves solving a system of equations based on the values and positions of the four known points. After determining the coefficients \(a\), \(b\), \(c\), and \(d\), the polynomial can be evaluated at any intermediate point. Compared to quadratic polynomials, cubic polynomials often provide a better fit to data, particularly when the underlying function exhibits changes in concavity or has an inflection point within the domain of interest. Hence, the exercise demonstrates the potential benefits of using a cubic interpolating polynomial over a quadratic one when dealing with datasets that capture the complexity of natural phenomena.