## 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 are introduced as practice problems for Newton-divided differences.

You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.

### The first solved problem.

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},b_{2}.

[latexpage]\begin{equation}

\mathrm{Q}(\mathrm{x})=\mathrm{y}_{0}+f\left[\mathrm{x}_{0}, \mathrm{x}_{1}\right](\mathrm{x}-\mathrm{x}_0)+f\left[\mathrm{x}_{0,}, \mathrm{x}_{1}, \mathrm{x}_{2)}\right]\left(x-x_{0}\right)\left(\mathrm{x}-x_{1}\right)

\end{equation}

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}).

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#### 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},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 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.

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}, b_{3}.

The equation in 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},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},b_{3} by using the table.

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