Question

A secret formula for eternal youth, \(f(x)\), was discovered by Dr. Quick, who has been working in our biotech company. However, Dr. Quick has disappeared and is rumored to be negotiating with a rival organization. From the notes that Dr. Quick left behind in his hasty exit it is clear that \(f(0)=0, f(1)=2\), and that \(f\left[x_{0}, x_{1}, x_{2}\right]=1\) for any three points \(x_{0}, x_{1}, x_{2} .\) Find \(f(x)\).

Short answer

Answer: The secret polynomial formula \(f(x)\) is given by: $$ f(x_n)=\prod_{i=0}^{n-1} (x_i-x_{i-1})\text{, for }n\ge 1 $$

Step-by-step solution

Determine the degree of the polynomial

We know that \(f(x)\) is a polynomial. Since the divided difference \(f\left[x_{0}, x_{1}, x_{2}\right]=1\) for any three points, we can suspect that the degree of the polynomial is at least 2. Let's assume that the degree of the polynomial is \(n\).

Use divided differences properties

We have given \(f\left[x_0, x_1, x_2\right]=1\) for any three points \(x_0, x_1, x_2\). We can rewrite this using the divided differences formula:$$ \frac{f(x_2)-f\left[x_1, x_2\right]}{x_2-x_0} =1 $$Now we express \(f\left[x_1, x_2\right]\) as another divided difference:$$ \frac{f(x_2)-\frac{f(x_1)-f(x_0)}{x_1-x_0}}{x_2-x_0} =1 $$We can see a pattern, and we can use the same technique until we get to the degree of the polynomial, \(n\).

Simplify and search for a pattern

The expression we obtained in step 2 can be simplified:$$ f(x_2) - \frac{f(x_1)-f(x_0)}{x_1-x_0}(x_2-x_0)=x_2(x_1-x_0)+f(x_0) $$By applying the same process multiple times and looking for a pattern, we obtain the expression:$$ f(x_n) = x_n(x_{n-1}-x_{n-2}) \cdots (x_1-x_0)+f(x_0) $$

Use the given values to find the formula

We have given the values of \(f(0)\) and \(f(1)\). Let's plug them in the expression we obtained in step 3. For \(f(0)\), we have:$$ f(0) = 0(x_{-1}-x_{-2})\cdots(x_1-x_0)+f(x_0)=0 $$Since \(f(x_0)=0\), this equation holds true. Now let's try plugging in \(f(1)\). Using our expression from step 3, we get:$$ f(1)=1(x_0-x_{-1})\cdots(x_1-x_0)+0=2 $$This simplifies to:$$ (x_0-x_{-1})\cdots(x_1-x_0)=2 $$Now, we see that the sequence for \(f(x_n)\) and the coefficients are equal to each other (for example, \(f(1)=x_0-x_{-1}\)). So, we can rewrite our expression from step 3 as:$$ f(x_n)=\prod_{i=0}^{n-1} (x_i-x_{i-1})\text{, for }n\ge 1 $$We can now use this formula to find the secret formula \(f(x)\).

Key concepts

Divided Differences

Divided differences provide a powerful way to understand the variations and structure of a function across specified points. They play an essential role in polynomial interpolation, helping to construct a polynomial that fits through given data points. Understanding how divided differences work involves checking how values at different x-coordinates influence each other.

To get a clearer picture, consider the fact that a divided difference of three points \( f[x_0, x_1, x_2] \) equals 1. This statement shows the amount of change per unit in function value relative to the change in input, computed recursively. The formula for a divided difference is given by:

• \( f[x_0, x_1] = \frac{f(x_1)-f(x_0)}{x_1-x_0} \)
• \( f[x_0, x_1, x_2] = \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2-x_0} \)

This recursive approach helps in understanding not just the function’s behavior but also creates the groundwork for building polynomials that satisfy given criteria. Given that the second divided difference equates to a constant (here, 1), it suggests a systematic pattern within the polynomial structure.

Polynomial Degree

The degree of a polynomial is the highest power of its variable, representing its complexity and behavior. When working with polynomial interpolation, determining the degree is crucial as it defines the minimal number of points required to determine the polynomial uniquely.

In the exercise given, the constant divided difference \( f[x_0, x_1, x_2] = 1 \) indicates that the polynomial degree is at least 2. The reason for this is that a second divided difference corresponds to a quadratic behavior, which implies that the underlying function can be expressed as a polynomial of degree 2.

Degrees give insights into how well a polynomial can model specific datasets. Specifically:

• Degree 0: A constant function.
• Degree 1: A linear function fitting two points.
• Degree 2: A quadratic polynomial requiring at least three points, exactly matching our exercise scenario.

Once the polynomial degree is ascertained, it directs subsequent calculations and solution pathways. It determines how divided differences build up into the interpolation formula.

Interpolation Formula

The interpolation formula offers a method to estimate values between known data points using polynomials. Constructing this formula involves arranging the data into a pattern dictated by the degree of the polynomial, using divided differences.

For polynomial interpolation, the Newton's form of the interpolation polynomial uses divided differences directly to form the formula:\[P(x) = f(x_0) + f[x_0, x_1](x-x_0) + f[x_0, x_1, x_2](x-x_0)(x-x_1) + \cdots\]

Here, each term involves a divided difference and product of terms such as \((x-x_0)\), expanding depending on the number of given points. For our specific problem, knowing \( f[x_0, x_1, x_2] = 1 \) constructs a quadratic interpolation polynomial.

• Each new term adds curvature or complexity to the interpolation polynomial.
• The pattern follows a nested multiplication framework, which builds the polynomial progressively.

Ultimately, the interpolation formula makes the problem of finding \( f(x) \) manageable, boiling down the mystery formula to a testable polynomial expression.