**Contents**

Problem 1

1.1 Problem definition

1.2 Solution of the problem

1.2.1 Linear interpolation

1.2.2 Method of least squares interpolation

1.2.3 Lagrange interpolating polynomial

1.2.4 Cubic spline interpolation

1.3 Results and discussion

1.3.1 Lagrange polynomial

Problem 2

2.1 Problem definition

2.2 Problem solution

2.2.1 Rectangular method

2.2.2 Trapezoidal rule

2.2.3 Simpsons rule

2.2.4 Gauss-Legendre method and Gauss-Chebyshev method

Problem 3

3.1 Problem definition

3.2 Problem solution

Problem 4

4.1 Problem definition

4.2 Problem solution

References

**Problem 1**

*1.1 Problem definition*

For the following data set, please discuss the possibility of obtaining a reasonable interpolated value at , , and via at least 4 different interpolation formulas you are have learned in this semester.

*1.2 Solution of the problem*

Interpolation is a method of constructing new data points within the range of a discrete set of known data points.

In engineering and science one often has a number of data points, as obtained by sampling or experimentation, and tries to construct a function which closely fits those data points. This is called curve fitting or regression analysis. Interpolation is a specific case of curve fitting, in which the function must go exactly through the data points.

First we have to plot data points, such plot provides better picture for analysis than data arrays

Following four interpolation methods will be discussed in order to solve the problem:

- Linear interpolation
- Method of least squares interpolation
- Lagrange interpolating polynomial

**Fig 1. Initial data points**

- Cubic spline interpolation

**1.2.1 Linear interpolation**

One of the simplest methods is linear interpolation (sometimes known as lerp). Generally, linear interpolation tales two data points, say and , and the interpolant is given by:

at the point

Linear interpolation is quick and easy, but it is not very precise/ Another disadvantage is that the interpolant is not differentiable at the point .

**1.2.2 Method of least squares interpolation**

The method of least squares is an alternative to interpolation for fitting a function to a set of points. Unlike interpolation, it does not require the fitted function to intersect each point. The method of least squares is probably best known for its use in statistical regression, but it is used in many contexts unrelated to statistics.

**Fig 2. Plot of the data with linear interpolation superimposed**

Generally, if we have data points, there is exactly one polynomial of degree at most going through all the data points. The interpolation error is proportional to the distance between the data points to the power n. Furthermore, the interpolant is a polynomial and thus infinitely differentiable. So, we see that polynomial interpolation solves all the problems of linear interpolation.

However, polynomial interpolation also has some disadvantages. Calculating the interpolating polynomial is computationaly expensive compared to linear interpolation. Furthermore, polynomial interpolation may not be so exact after all, especially at the end points. These disadvantages can be avoided by using spline interpolation.

Example of construction of polynomial by least square method

Data is given by the table:

Polynomial is given by the model:

In order to find the optimal parameters the following substitution is being executed:

, , …,

Then:

The error function:

It is necessary to find parameters , which provide minimums to function :

It should be noted that the matrix must be nonsingular matrix.

For the given data points matrix become singular, and it makes impossible to construct polynomial with order, where - number of data points, so we will use polynomial

**Fig 3. Plot of the data with polynomial interpolation superimposed**

Because the polynomial is forced to intercept every point, it weaves up and down.

**1.2.3 Lagrange interpolating polynomial**

The Lagrange interpolating polynomial is the polynomial of degree that passes through the points , , …, and is given by:

,

Where

Written explicitly

When constructing interpolating polynomials, there is a tradeoff between having a better fit and having a smooth well-behaved fitting function. The more data points that are used in the interpolation, the higher the degree of the resulting polynomial, and therefore the greater oscillation it will exhibit between the data points. Therefore, a high-degree interpolation may be a poor predictor of the function between points, although the accuracy at the data points will be "perfect."

**Fig 4. Plot of the data with Lagrange interpolating polynomial interpolation superimposed**

One can see, that Lagrange polynomial has a lot of oscillations due to the high order if polynomial.

**1.2.4 Cubic spline interpolation**

Remember that linear interpolation uses a linear function for each of intervals . Spline interpolation uses low-degree polynomials in each of the intervals, and chooses the polynomial pieces such that they fit smoothly together. The resulting function is called a spline. For instance, the natural cubic spline is piecewise cubic and twice continuously differentiable. Furthermore, its second derivative is zero at the end points.

Like polynomial interpolation, spline interpolation incurs a smaller error than linear interpolation and the interpolant is smoother. However, the interpolant is easier to evaluate than the high-degree polynomials used in polynomial interpolation. It also does not suffer from Runges phenomenon.

**Fig 5. Plot of the data with Lagrange interpolating polynomial interpolation superimposed**

It should be noted that cubic spline curve looks like metal ruler fixed in the nodal points, one can see that such interpolation method could not be used for modeling sudden data points jumps.

*1.3 Results and discussion*

The following results were obtained by employing described interpolation methods for the points ; ; :

Linear interpolationLeast squares interpolationLagrange polynomialCubic splineRoot mean square0.1480.2090.0150.140.1460.6780.6640.6120.6410.6491.5691.6491.4791.5621.566**Table 1. Results of interpolation by different methods in the given points.**

In order to determine the best interpolation method for the current case should be constructed the table of deviation between interpolation results and root mean square, if number of interpolations methods increases then value of RMS become closer to the true value.

Linear interpolationLeast squares interpolationLagrange polynomialCubic splineAverage deviation from the RMS**Table 2. Table of Average deviation between average deviation and interpolation results.**

One can see that cubic spline interpolation gives the best results among discussed methods, but it should be noted that sometimes cubic spline gives wrong interpolation, especially near the sudden function change. Also good interpolation results are provided by Linear interpolation method, but actually this method gives average values on each segment between values on it boundaries.

**Problem 2**

*2.1 Problem definition*

For the above mentioned data set, if you are asked to give the integration of between two ends and ? Please discuss the possibility accuracies of all the numerical integration formulas you have learned in this semester.

*2.2 Problem solution*

In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral.

There are several reasons for carrying out numerical integration. The integrand may be known only at certain points, such as obtained by sampling. Some embedded systems and other computer applications may need numerical integration for this reason.

A formula for the integrand may be known, but it may be difficult or impossible to find an antiderivative which is an elementary function. An example of such an integrand is , the antiderivative of which cannot be written in elementary form.

It may be possible to find an antiderivative symbolically, but it may be easier to compute a numerical approximation than to compute the antiderivative. That may be the case if the antiderivative is given as an infinite series or product, or if its evaluation requires a special function which is not available.