2d-Solved problems for Newton-divided differences.

Last Updated on January 14, 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 have two solved problems as an application of Newton’s divided differences.

The first solved problem.

Three points where x and y values are given are required to get the expression for the polynomial based on Newton-divided differences, and the value of a new point with x=2.70.

The next equation will yield the polynomial, but we need to estimate the values of 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 the 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 the b0, b1, and b2 values?

The first coefficient, b0, is equal to the y coordinate of the first point, so b0=y0.

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

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), f[x1,x2]=(y2-y1)/(x2-x1)=(8-5)/(3-2)=3, b2=f[x1,x2)-f[x1,x2]/(x3-x0)=1/2.

Once we have obtained the three values of b₀, b1, and b2, we can write the polynomial as shown in the next slide.

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

We can create a Table for b0,, and b2, and we start filling in the data for the given points; hence, 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 for a new point with x = 2.70. Before we can substitute x=2.7 into the polynomial expression, it is preferred to check the y-values of the given three points in the table using the polynomial expression evaluated at (x₀,x₁,x₂).

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

The second solved problem.

Four points where x and y values are given require obtaining the polynomial expression using Newton’s divided differences. Since we have four points, we have to determine the values of four b’s: b0, b1, b₂, and b₃.

The equation will now be expanded to account for the four values of b’s.

How do we estimate the value of b₁?

The value of b₀ is the same value of y₀ for the first point, which equals 1. For the value of b₁, it will be estimated as the first divided difference between points (x₀,y₀) and (x₁,y₁). The value of b1 is equal to zero.

How do we estimate the value of b₂?

The b₂ values can be estimated in two stages. The first stage is to 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 in the case of a cubic polynomial, a polynomial of degree 3. The equations for b0, b1, b₂, and b₃ are shown in the slide image.

How do we estimate the value of b₃?

The calculation for the value of b3 is shown in the next slide for more details.

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

A table was used to compute the Newton divided differences.

It is much easier to estimate the divided differences and obtain the values of b0, b1, b2, and b3 using the table.

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 the polynomial expression, and shows they match.

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

The PDf file used for this post can be viewed or downloaded from the following document.

This is a link to Holistic Numerical Methods-Newton divided Differences.

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