2D-Solved problems for Newton-divided differences.

Last Updated on September 23, 2026 by Maged kamel

Solved problems for Newton-divided differences.

We discussed Newton’s divided-difference interpolation in a previous Post. In this Post, we present two solved problems as applications of Newton’s divided differences.

The first solved Problem.

To find the polynomial expression using Newton’s divided differences, we need three points with given x and y values, plus the Value at a new Point with x = 2.70.

The next equation will yield the polynomial, but we need to estimate b0, b1, and b2. We could start with the first Point and call it (x1, y1), but in these two solved problems I treat the first Point’s coordinates as (X0, y0).

We know that b0 = y0. From the given Table, the first y-value, y0, is 3.
To get b1, we must estimate the first divided difference between x₀, y0, and (x1, y₁).

The graph based on the given points is shown in the next slide. We have three points, meaning we have three coefficients: b0, b1, and b2. the polynomial P(x) =b0+b0*(x-x0)+b2*(x-x0)*(x-x1). Such an Expression will give p(x=x0)=b0, while P(x=x1)=b0+b1*(x1-x0)+b2*(x1-x0)*(x1-x1)=b0+b1*(x1-x0)+0.

How do we estimate b0 and b2?

The first coefficient, b0, equals the y-coordinate of the first Point, so b0 = y0.

The equation shown gives the second divided difference in the image on the next slide

The first divided difference between points (x₁, y₁) and ( x₀,y₀)is estimated as the difference between (y₁ and y₀)/the difference between x₁ and x₀.

After estimating the first two differences, the Value of these differences will be divided by the difference between(x₂-x₀) to get the second divided difference. The coefficient b0=3, while b1=(y1-y0)/(x1-x0)=(5-3)/(2-1)=1, f[x1,x2]=(y2-y1)/(x2-x1)=(8-5)/(3-2)=3, and b2=f[x1,x2)-f[x1,x2]/(x3-x0)=1/2.

Once we have b₀, b1, and b2, we can write the polynomial as shown on the next slide.

The final expression for the polynomial using Newton-divided differences.

We can create a Table for b0, b1, and b2 and start filling in the data for the given points; then we can sketch the first and second Newton divided differences.

The next slide shows the final expression for the polynomial using Newton’s divided differences.

The last requirement is to estimate the y-value at x = 2.70. Before substituting x = 2.7 into the polynomial Expression, it is best to check the y-values of the three points in the Table using the polynomial Expression evaluated at (x₀, x₁, x₂).

From the calculations shown, we can substitute x = 2.70 to obtain P(2.70), which equals 6.995.

The second solved Problem.

Four points with given x and y values require obtaining the polynomial Expression using Newton’s divided differences. Since we have four points, we must determine four b’s: b0, b1, b₂, and b₃.

The equation now expands to account for the four b’s.

How do we estimate b₁?

b₀ equals y₀ for the first Point, which is 1. For b₁, we estimate it as the first divided difference between points (x₀, y₀) and (x₁, y₁). The Value of b1 is equal to zero.

How do we estimate b₂?

We can estimate the b2 values in two stages. In the first stage, compute the first divided difference between the points (x2, y2) and (x1, y₁).

The second stage is to compute the first divided difference between the points (x0, y0) and (x1, y₁). The estimated differences will be divided by (x₂-x₀).

This Table is used to determine the differences for the 4 given points for a cubic polynomial (degree 3). The slide image shows the equations for b0, b1, b₂, and b₃.

How do we estimate b₃?

The next slide shows the calculation of b3 in more detail.

The final Value of b₃ is shown to be=-1/12.

We use a Table to compute the Newton divided differences.

The Table makes it much easier to estimate the divided differences and obtain the values of b0, b1, b2, and b3.

Once we have obtained the three values of b₀, b1, b₂, and b₃.

The final expression for the polynomial using Newton-divided differences.

The Expression of the cubic function is shown. The Table compares the y-values of the four points estimated by plugging them into the polynomial Expression and shows they match.

We can find the polynomial, as shown in the next slide. It is best to check the y-values of the four given points in the Table using the polynomial Expression for (x0, x1, x2, x₃).

You can view or download the PDF file for this Post from the following document.

This links to Holistic Numerical Methods: Newton Divided Differences.

This links to the previous Post: What is Newton-divided difference interpolation?