- Solved problems for Newton-divided differences.
- The first solved problem of the two Solved problems for Newton-divided differences.
- How to estimate the value of b1?
- How to estimate the value of b2?
- Using a table to get the Newton-divided differences.
- The final expression for the polynomial using Newton-divided differences.
- The second solved problem of the two Solved problems for Newton-divided differences.

## Solved problems for Newton-divided differences.

We have discussed the Newton-divided difference interpolation in a previous post. In this post, we have two solved problems for Newton-divided differences.

### The first solved problem of the two Solved problems for Newton-divided differences.

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

The next equation will use for obtaining the polynomial, but we need to estimate the values of b_{0},b_{1}, and b_{2}.

We know that ** b_{0}=y_{0}**, from the given table the first point y value or

**y**.

_{0}=3To get the b

_{1}we need to estimate the first divided difference between x

_{0},y

_{0,}and(x

_{1},y

_{1}).

#### How to estimate the value of b_{1}?

The difference between(y_{1} and y_{0}) divided by the difference between x_{1} and x_{0} will give the value of the first divided difference.

#### How to estimate the value of b_{2}?

The second divided difference is given by the shown equation in the next slide image. The first divided difference between points (x_{1},y_{1}) and ( x_{2},y_{2}) is estimated as the difference between (y_{2} and y_{1})/the difference between x_{2} and x_{1}.

While the first divided difference between points (x_{1}, y_{1}) and ( x_{0},y_{0})_{ }is estimated as the difference between (y_{1} and y_{0})/the difference between x_{1} and x_{0}. After estimating the two first differences the value of these differences will be divided by the difference between(x_{2}-x_{0}) to get the second divided difference.

Once we have obtained the three values of b_{0},b_{1},b_{2}. We can find out the polynomial as we can see from the next slide image. We can plug in the values of x_{0}, and x_{1}, together with the values of b_{0},b_{1},b_{2}, The final expression of the polynomial can thus be found.

#### Using a table to get the Newton-divided differences.

We can create a table for the divided differences and perform the necessary calculation as we can see from the next slide image. The same results are obtained.

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

The final expression for the polynomial using Newton-divided differences is shown in the next slide image.

The last requirement is the estimation of the value y value for a new point of a given x value =2.70. Before we can substitute the value of x=2.7 in the polynomial expression, it is preferred to check the y values of the given three points via the table, by the values obtained by using the polynomial expression for (x_{0},x_{1},x_{2}).

From the shown calculations, we can substitute the value of x=2.70, and get the P(2.70), which will be found to be equal to 6.995.

### The second solved problem of the two Solved problems for Newton-divided differences.

Four points where x and y values are given, it is required to get the expression for the polynomial based on Newton-divided differences. since we have four points, then we have to determine the values of 4 b’s, b_{0},b_{1},b_{2}, and b_{3}.

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

#### How to estimate the value of b_{1}?

The value of b_{0} is the same value of y_{0} for the first point which is equal to 1. For the value of b_{1}, it will be estimated as the first divided difference between points (x_{0},y_{0}) and (x_{1},y_{1}).

#### How to estimate the value of b_{2}?

The b_{2} values can be estimated in two stages, the first stage is the get the first divided difference between points (x_{2},y_{2}) and ( x_{1},y_{1}).

The second stage is to get the first divided difference between points (x_{0},y_{0}) and ( x_{1},y_{1}). The estimated differences will be divided by (x_{2}-x_{0}).

This is the table used to get the differences for the 4 given points in the case of a cubic polynomial. A cubic polynomial is a polynomial of degree 3. The equations of b_{0},b_{1},b_{2}, and b_{3} are shown in the slide image.

#### How to estimate the value of b_{3}?

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

The final value of b_{3} is shown to be=** -1/12**.

#### Using a table to get the Newton-divided differences.

As we can see it is much easier to estimate the divided differences and get the different values of b_{0},b_{1},b_{2}, and b_{3} by using the table.As we can see it is much easier to estimate the divided differences and get the different values of b_{0},b_{1},b_{2}, and b_{3} by using the table.

Once we have obtained the three values of b_{0},b_{1},b_{2}, and b_{3}.

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

We can find out the polynomial as we can see from the next slide image. it is preferred to check the y values of the given four points via the table, by the values obtained by using the polynomial expression for (x_{0},x_{1},x_{2},x_{3}).

The expression of the cubic function is shown. All the y values of the given four points, estimated by plugin the polynomial expression, are compared by the given table and found to match.

This is a link for the pdf file used in the illustration of this post.