# Engineering Math Numerical Techniques-II

### Contents

**Describe the following Operators: Forward Difference, Backward Difference, Shift etc.**3**Relate various Operators**4**Describe Forward and Backward Finite Difference Table**5**Find missing terms using Finite Difference Table**6**Describe the method of Separation of Symbols and apply it to prove the useful Identities**7**Show a Polynomial in its successive differences and also in Factorial Notation**8**Recall the term Interpolation**9**Formulate Lagrange Interpolation**9**Apply Lagrange’s Interpolation Formula for a given set of data**10**Formulate Hermite Interpolation**11**Apply Hermite’s Interpolation Formula for a given set of data**11**Describe Divided Differences**12**Differentiate between Finite Differences and Divided Differences**13**Recall Divided Difference Table**14- Recall Newton’s Divided Difference Method of Interpolation 15
**Apply Newton Divided Difference Method of Interpolation for the given set of data**16**Describe Gregory-Newton Forward Interpolation Formula**18**Apply Gregory-Newton Forward Interpolation Formula for the given set of data**18**Describe Gregory-Newton Backward Interpolation Formula**20**Apply Gregory-Newton Backward Interpolation Formula for the given set of data**21**Describe Gauss’s Forward Formula of Interpolation**22**Apply Gauss’s Formula of Interpolation for the given set of data**23**Describe Gauss’s Backward Formula of Interpolation**26**Apply Gauss’s Backward Formula of Interpolation for the given set of data**27**Describe Stirling’s Formula of Interpolation**28**Apply Stirling’s Formula of Interpolation for the given set of data**29**Find the derivative from the given set of Data using Numerical Differentiation**30**Describe Newton-Cotes Formulas**31**Deduce the Trapezoidal Rule from Newton-Cotes Formulas**32**Apply Trapezoidal Rule to solve the definite Integral**33**Describe Simpson’s 1/3rd Rule**34**Apply the Simpson’s 1/3rd Rule to find the approximate value of the integral**34**Describe Simpson’s 3/8th Rule**35**Apply the Simpson’s 3/8th Rule to find the approximate value of the integral**36**Describe the Gaussian One-Point,Two-Point, and Three- Point Formula**37**Apply the Gaussian One-Point, Two-Point, and Three-Point Formula to find the Definite Integral**38**Describe the Quadrature Formulas: i. Gauss-Legendre Formula, ii. Gauss-Chebyshev Formula, iii. Gauss-Hermite Formula, iv. Gauss-Laguerre Formula**40

Describe the following Operators: Forward Difference, Backward Difference, Shift etc.

In mathematics and computer science, several operators are used to manipulate and analyze functions or data. Here are descriptions of some common operators:

- Forward Difference Operator: The forward difference operator is used to approximate the derivative of a function. Given a function f(x), the forward difference operator Δf(x) is defined as:

Δf(x) = f(x + h) – f(x)

where h is a small constant called the step size. The forward difference operator is used to approximate the derivative at a point x by dividing the difference in function values by the step size h.

- Backward Difference Operator: The backward difference operator is similar to the forward difference operator, but it approximates the derivative using function values at x and x-h. The backward difference operator Δf(x) is defined as:

Δf(x) = f(x) – f(x – h)

- Central Difference Operator: The central difference operator is a more accurate approximation of the derivative than either the forward or backward difference operators. It uses function values on both sides of the point x. The central difference operator Δf(x) is defined as:

Δf(x) = [f(x + h) – f(x – h)] / 2

- Shift Operator: The shift operator is used to shift a sequence of data or function values by a fixed amount. Given a sequence a(n), the shift operator S(k) is defined as:

S(k)n = a(n – k)

where k is the amount by which the sequence is shifted. The shift operator is useful in time series analysis and signal processing.

- Laplacian Operator: The Laplacian operator is used to calculate the second derivative of a function. In two dimensions, the Laplacian operator Δ is defined as:

Δf = ∂²f/∂x² + ∂²f/∂y²

It is often used in physics and engineering to describe the behavior of waves, heat, and other physical phenomena.

These are just a few examples of the many operators used in mathematics and computer science. Each operator has specific uses and properties, and understanding them can help in analyzing and manipulating data and functions.

Relate various Operators

Several operators used in mathematics and computer science are related to each other. Here are some examples:

- Forward Difference and Backward Difference Operators: The forward difference operator Δf(x) and the backward difference operator Δf(x) are related to each other. Using Taylor’s theorem, we can show that the forward difference operator is an approximation of the first derivative of f(x) at x+h, while the backward difference operator is an approximation of the first derivative of f(x) at x-h. Therefore, we can relate the forward and backward difference operators as:

Δf(x) ≈ -Δf(x-h)

- Central Difference and Taylor Series: The central difference operator is related to the Taylor series expansion of a function. Using Taylor’s theorem, we can show that the central difference operator Δf(x) is an approximation of the second derivative of f(x) at x. Therefore, we can relate the central difference operator and Taylor series as:

Δf(x) ≈ (1/2)f”(x)h² + O(h⁴)

where O(h⁴) denotes the error term, which goes to zero as h approaches zero.

- Shift Operator and Fourier Transform: The shift operator S(k) is related to the Fourier transform of a sequence a(n). The Fourier transform of the shifted sequence S(k)a(n) is given by:

FS(k)a = e^{(-ikω)} Fa

where Fa is the Fourier transform of the sequence a(n). This relation is known as the shift property of the Fourier transform and is used in signal processing and image processing.

- Laplacian Operator and Eigenvalues: The Laplacian operator Δ is related to the eigenvalues and eigenvectors of a function or data. The Laplacian operator can be viewed as a linear operator that maps a function to its second derivative. The eigenvalues of the Laplacian operator correspond to the frequencies of the functions that satisfy certain boundary conditions, and the eigenvectors correspond to the shapes of these functions. This relation is used in spectral methods for solving partial differential equations.

These are just a few examples of how various operators used in mathematics and computer science are related to each other. Understanding these relations can help in solving problems and developing new algorithms and methods.

Describe Forward and Backward Finite Difference Table

Forward and backward finite difference tables are used to approximate derivatives of a function at a point using finite differences. A finite difference is the difference between the values of a function at two nearby points.

- Forward Finite Difference Table:

A forward finite difference table is used to approximate derivatives of a function at a given point using forward differences. The table is constructed by calculating the forward differences of a function at a given point, and then using these differences to approximate the higher-order derivatives.

The first column of the table corresponds to the function values at the given point, and the subsequent columns correspond to the forward differences of the previous column. For example, the first forward difference column corresponds to the first difference of the function values, and is calculated as:

Δf(x) = f(x + h) – f(x)

where h is the step size between the two points. The second forward difference column corresponds to the second difference of the function values, and is calculated as:

Δ²f(x) = Δf(x + h) – Δf(x)

and so on. The entries in each column are used to approximate the corresponding derivative of the function at the given point.

- Backward Finite Difference Table:

A backward finite difference table is used to approximate derivatives of a function at a given point using backward differences. The table is constructed in a similar way as the forward difference table, but using backward differences instead.

The first column of the table corresponds to the function values at the given point, and the subsequent columns correspond to the backward differences of the previous column. For example, the first backward difference column corresponds to the first difference of the function values, and is calculated as:

Δf(x) = f(x) – f(x – h)

where h is the step size between the two points. The second backward difference column corresponds to the second difference of the function values, and is calculated as:

Δ²f(x) = Δf(x) – Δf(x – h)

and so on. The entries in each column are used to approximate the corresponding derivative of the function at the given point.

Both forward and backward finite difference tables are useful tools for approximating derivatives of a function at a given point, and can be used in numerical methods for solving differential equations, optimization problems, and other mathematical applications.

Find missing terms using Finite Difference Table

Finite difference tables can be used to approximate derivatives of a function at a given point. They can also be used to find missing terms in a sequence of values, provided that the sequence has a constant step size. Here is an example of how to find missing terms using a finite difference table:

Suppose we have the following sequence of values:

1, ?, 7, 12, ?, 28

We want to find the missing terms in the sequence. To do this, we construct a finite difference table using the given values:

1 6 5 3

7 12 16

5 4

The first column corresponds to the given values. The second column corresponds to the first differences, which are calculated as the differences between consecutive values:

6 = 7 – 1

5 = 7 – 2

3 = 12 – 7

The third column corresponds to the second differences, which are calculated as the differences between consecutive first differences:

5 = 6 – 1

4 = 3 – 1

We notice that the second differences are constant, which indicates that the original sequence is a second-order polynomial. Therefore, we can use the coefficients of the polynomial to find the missing terms.

From the third column of the finite difference table, we know that the second difference is 5. This corresponds to the coefficient of the quadratic term in the polynomial. Therefore, we have:

f(x) = ax^{2 }+ bx + c

where a = 5. To find b and c, we use the values in the first column of the finite difference table. We know that:

f(0) = 1

f(1) = ?

f(2) = 7

f(3) = 12

f(4) = ?

f(5) = 28

Substituting these values into the polynomial equation, we get:

f(0) = a(0)^{2} + b(0) + c = c = 1

f(2) = a(2)^{2} + b(2) + c = 4a + 2b + c = 7

f(3) = a(3)^{2} + b(3) + c = 9a + 3b + c = 12

f(5) = a(5)^{2} + b(5) + c = 25a + 5b + c = 28

Solving these equations simultaneously, we get:

a = 5, b = -4, c = 1

Therefore, the missing terms in the sequence are:

1, 6, 7, 12, 21, 28

Describe the method of Separation of Symbols and apply it to prove the useful Identities

The method of separation of symbols is a technique used in differential calculus to derive useful identities by manipulating expressions symbolically, rather than using specific functions. The basic idea is to write a differential expression in terms of symbols for differentiation (such as d/dx or D), and then manipulate the symbols to arrive at a desired result.

Here is an example of how to use the method of separation of symbols to prove a useful identity:

Prove that (d/dx) ln(x) = 1/x

We start by writing the left-hand side of the equation in terms of symbols:

(d/dx) ln(x) = D ln(x)

where D represents the differentiation symbol.

Next, we use the definition of the natural logarithm:

ln(x) = ∫(1/x) dx

where ∫ represents integration. We can differentiate both sides of this equation with respect to x to get:

(d/dx) ln(x) = d/dx (∫(1/x) dx)

Using the fundamental theorem of calculus, we can write the right-hand side as:

(d/dx) ln(x) = d/dx (∫(1/x) dx) = 1/x

Therefore, we have proven the identity:

(d/dx) ln(x) = 1/x

which is a useful result in differential calculus.

Show a Polynomial in its successive differences and also in Factorial Notation

Given a polynomial P(x) of degree n, we can represent it in terms of its successive differences and factorial notation.

The first difference of P(x) is given by:

ΔP(x) = P(x+1) – P(x)

The second difference of P(x) is given by:

Δ^{2} P(x) = ΔP(x+1) – ΔP(x) = P(x+2) – 2P(x+1) + P(x)

We can continue this process to find higher order differences. In general, the k-th difference of P(x) is given by:

Δ^{k} P(x) = Δ^{(k-1)} P(x+1) – Δ^{(k-1)} P(x)

Using the factorial notation, we can write P(x) in terms of its differences as follows:

P(x) = Σ[k=0 to n] Δ^{k} P(0) / k! x(x-1)…(x-k+1)

where Δ^{k} P(0) represents the k-th difference of P(x) evaluated at x=0.

For example, consider the polynomial P(x) = x^{3}. Its successive differences are:

ΔP(x) = (x+1)^{3} – x^{3} = 3x^{2} + 3x + 1

Δ^{2} P(x) = (x+2)^{3} – 2(x+1)^{3} + x^{3 }= 6x + 6

Δ^{3} P(x) = (x+3)^{3} – 3(x+2)^{3} + 3(x+1)^{3} – x^{3} = 6

Using the above formula, we can write P(x) as:

P(x) = x(x-1)(x-2)/2! Δ2 P(0) + x(x-1)(x-2)(x-3)/3! Δ3 P(0)

Substituting the values of the differences, we get:

P(x) = x^{3}

This verifies that the formula is valid for this polynomial.

Recall the term Interpolation

Interpolation is the process of estimating the value of a function for a given input that lies between the values for which the function is defined. In other words, interpolation is the technique of constructing a function that passes through a given set of data points. The goal of interpolation is to estimate the value of a function at a point where the function is not explicitly known, based on the values of the function at other points where it is known. Interpolation is widely used in various fields, including engineering, science, economics, and finance.

Formulate Lagrange Interpolation

Lagrange interpolation is a method of constructing a polynomial that passes through a given set of data points. Given a set of n+1 data points (x_{0},y_{0}),(x_{1},y_{1}),…(x_{n},y_{n}), where no two xi are equal, the Lagrange interpolation polynomial L(x) of degree n that passes through these points is given by:

L(x) = Σ[i=0 to n] y_{i} l_{i}(x)

where l_{i}(x) is the ith Lagrange basis polynomial, defined as:

l_{i}(x) = Π_{[j}=0 to n, j!=i] (x-x_{j}) / (x_{i}-x_{j})

The Lagrange basis polynomials have the property that l_{i}(x_{i}) = 1 and l_{i}(x_{j}) = 0 for all j ≠ i, which means that the Lagrange interpolation polynomial L(x) passes through all the given data points. The Lagrange interpolation polynomial can be evaluated at any value of x within the range of the given data points to obtain an estimate of the function value at that point.

Apply Lagrange’s Interpolation Formula for a given set of data

Let’s say we are given the following set of data points:

(1, 3), (2, 5), (4, 11), (5, 13)

We want to find the Lagrange interpolation polynomial that passes through these points.

The Lagrange interpolation polynomial of degree n that passes through these n+1 points is given by:

L(x) = Σ[i=0 to n] y_{i }l_{i}(x)

where li(x) is the ith Lagrange basis polynomial, defined as:

l_{i(}x) = Π[j=0 to n, j!=i] (x-x_{j}) / (x_{i}-x_{j})

For our set of data points, we have n=3, so we can write:

L(x) = y_{0 }l_{0}(x) + y_{1} l_{1}(x) + y_{2} l_{2}(x) + y_{3} l_{3}(x)

Where

l_{0}(x) = (x-2)(x-4)(x-5) / [(1-2)(1-4)(1-5)] = -(1/24)(x^{3} – 11x^{2} + 38x – 40)

l_{1}(x) = (x-1)(x-4)(x-5) / [(2-1)(2-4)(2-5)] = (1/12)(x^{3} – 9x^{2} + 20x – 10)

l_{2}(x) = (x-1)(x-2)(x-5) / [(4-1)(4-2)(4-5)] = -(1/12)(x^{3} – 8x^{2} + 13x – 5)

l_{3}(x) = (x-1)(x-2)(x-4) / [(5-1)(5-2)(5-4)] = (1/24)(x^{3} – 7x^{2} + 16x – 10)

Therefore, the Lagrange interpolation polynomial that passes through the given set of data points is:

L(x) = 3 l_{0}(x) + 5 l_{1}(x) + 11 l_{2}(x) + 13 l_{3}(x)

Simplifying this expression, we get:

L(x) = (1/24)(-x_{3} + 11x_{2} – 14x – 1)

This is the Lagrange interpolation polynomial that passes through the given set of data points. We can evaluate this polynomial at any value of x within the range of the given data points to obtain an estimate of the function value at that point.

Formulate Hermite Interpolation

Hermite interpolation is a type of polynomial interpolation that allows for both the function value and the derivative value at each data point to be specified. Suppose we have a set of n+1 data points {(x_{0}, y_{0}, y’_{0}), (x_{1}, y_{1}, y’_{1}), …, (x_{n}, y_{n}, y’_{n})}, where y’i denotes the derivative of y with respect to x evaluated at xi. The Hermite interpolation polynomial H(x) of degree at most 2n+1 that passes through these data points is given by:

H(x) = Σ[i=0 to n] [y_{i}h_{i}(x) + y’_{i}h_{i}‘(x)]

where hi(x) and hi'(x) are the ith Hermite basis polynomials, defined recursively as:

h0(x) = 1

h1(x) = 2(x-x_{0})

hi(x) = (x-x_{0})_{2}Π_{j}=0 to i-1 for i > 1

hi'(x) = 2(x-x_{0})Π_{j}=0 to i-1 + (x-x_{0})_{2}∑[j=0 to i-1]Πk=0 to i-1, k!=j/(x_{j}-x_{k}) for i > 1

The Hermite basis polynomials hi(x) and hi'(x) are used to construct H(x) such that it satisfies the given data points and their derivatives.

Note that the degree of the Hermite interpolation polynomial is at most 2n+1, which is twice the number of data points. This is because we are specifying both the function value and the derivative value at each data point.

Apply Hermite’s Interpolation Formula for a given set of data

Suppose we have the following set of data points:

{(0, 1, 0), (1, 2, 1), (2, 5, 3)}

We want to find the Hermite interpolation polynomial of degree at most 5 that passes through these data points.

First, we construct the Hermite basis polynomials hi(x) and hi'(x) as follows:

h0(x) = 1

h1(x) = 2(x-0) = 2x

h2(x) = (x-0)^{2(x-1)} = x^{2(x-1)}

h3(x) = (x-0)^{2(x-2) }= x^{2(x-2)}

h4(x) = (x-1)^{2(x-2) }= (x-1)^{2(x-2)}

h5(x) = (x-0)^{2(x-1)²⁽ˣ⁺¹⁾} = x^{2(x-1)²⁽ˣ⁺¹⁾}

Next, we use these basis polynomials to construct the Hermite interpolation polynomial:

H(x) = 1h0(x) + 0h0′(x) + 2h1(x) + 1h1′(x) + 5h2(x) + 3h2′(x)

Simplifying, we get:

H(x) = 1 + 2x + 3x^{2 }– x^{2(x-1)} + 2x^{2(x-2)}

Expanding and collecting like terms, we get:

H(x) = 3x^{4} – 8x^{3} + 6x^{2} + 2x + 1

Therefore, the Hermite interpolation polynomial of degree at most 5 that passes through the given data points is

H(x) = 3x^{4} – 8x^{3} + 6x^{2 }+ 2x + 1.

Describe Divided Differences

Divided differences are a way of recursively computing the coefficients of an interpolating polynomial, given a set of data points. The first divided difference is defined as:

f[x_{0}] = f(x_{0})

where f(x_{0}) is the function value at the data point x_{0}. For any two data points xi and x_{j} (where i and j are distinct integers), the second divided difference is defined as:

f[x_{i},x_{j}] = (f[x_{j}] – f[x_{i}]) / (x_{j} – x_{i})

Similarly, the third divided difference is defined as:

f[x_{i},x_{j},x_{k}] = (f[x_{j},x_{k}] – f[x_{i},x_{j}]) / (x_{k} – xi)

And so on, for higher order divided differences.

Using these divided differences, we can recursively compute the coefficients of the interpolating polynomial. For example, for a set of three data points (x_{0}, y_{0}), (x_{1}, y_{1}), and (x_{2}, y_{2}), the interpolating polynomial can be written as:

P(x) = f[x_{0}] + f[x_{0},x_{1}](x – x_{0}) + f[x_{0},x_{1},x_{2}](x – x_{0})(x – x_{1})

where the coefficients f[xi] are just the function values at the data points, and the coefficients f[xi,xj] and f[xi,xj,xk] are the divided differences.

Note that the divided differences depend only on the values of the function at the data points, and not on the specific form of the function itself. This makes them a useful tool for computing interpolating polynomials in practice.

Differentiate between Finite Differences and Divided Differences

Finite differences and divided differences are related concepts used in the computation of interpolating polynomials, but they have some differences.

Finite differences are the differences between consecutive function values at equidistant points. For example, the first-order forward difference at a point x is given by:

f[x + h] – f[x]

where h is the distance between the equidistant points. Higher-order finite differences can be defined recursively using the previous finite differences.

Divided differences, on the other hand, are the ratios of finite differences. For example, the first-order divided difference at two points x0 and x1 is given by:

f[x_{1}] – f[x_{0}] / (x_{1} – x_{0})

Higher-order divided differences can be defined recursively using the previous divided differences.

One key difference between finite differences and divided differences is that finite differences depend on the spacing between the equidistant points, while divided differences do not. This means that divided differences can be used to interpolate non-equidistant data points, while finite differences cannot.

In summary, finite differences are the actual differences between consecutive function values at equidistant points, while divided differences are the ratios of finite differences and can be used to interpolate non-equidistant data points.

Recall Divided Difference Table

A divided difference table is a way to organize the divided differences of a set of data points for use in constructing an interpolating polynomial. The table is usually arranged in a triangular form, with the data points in the first column, the first-order divided differences in the second column, the second-order divided differences in the third column, and so on.

To construct a divided difference table, we start by writing down the data points in the first column. Then, we calculate the first-order divided differences by subtracting the function values of adjacent data points and dividing by the difference in their x-coordinates. The second-order divided differences are calculated similarly, using the first-order divided differences, and so on.

For example, suppose we have the following set of data points:

(x_{0}, y_{0}) = (0, 1)

(x_{1}, y_{1}) = (1, 2)

(x_{2}, y_{2}) = (2, 5)

(x_{3}, y_{3}) = (3, 10)

The divided difference table for this data would look like:

x | y | |||

0 | 1 | |||

1 | 1.5 | 2.5 | 2.5 | |

1 | 2 | |||

3 | 4 | 5 | ||

2 | 5 | |||

5 | 5 | |||

3 | 10 |

Here, the first column contains the x-values of the data points, the second column contains the y-values, and the remaining columns contain the divided differences at various orders. For example, the first-order divided difference between (x_{0}, y_{0}) and (x_{1}, y_{1}) is (y_{1 }– y_{0}) / (x_{1} – x_{0}) = (2 – 1) / (1 – 0) = 1, which is why there is a 1 in the second column of the second row. The second-order divided difference between (x0, y0) and (x2, y_{2}) is ((y_{2} – y_{1}) / (x_{2} – x_{1}) – (y_{1} – y_{0}) / (x_{1 }– x_{0})) / (x_{2} – x_{0}) = (5/1 – 2/1) / (2 – 0) = 1.5, which is why there is a 1.5 in the third column of the third row. The table can be used to construct an interpolating polynomial of the data using divided difference formulas.

## Recall Newton’s Divided Difference Method of Interpolation

Newton’s Divided Difference Method of Interpolation is a numerical method used to interpolate a polynomial function using divided differences. The method is based on the fact that the divided difference of order k is defined as the difference of the divided difference of order k-1 evaluated at successive pairs of points. In this way, we can recursively compute the divided differences of higher order until we obtain the required interpolation polynomial.

The general formula for Newton’s divided difference method is as follows:

f[x_{0}, x_{1}, …, x_{n}] = (f[x_{1}, x_{2}, …, x_{n}] – f[x_{0}, x_{1}, …, x_{n-1}]) / (x_{n} – x_{0})

where f[xi] represents the function value at the point xi and f[x_{0}, x_{1}, …, xn] represents the divided difference of order n.

Using this formula, we can construct a divided difference table to find the coefficients of the interpolation polynomial. The interpolation polynomial in Newton’s divided difference form is given by:

Pn(x) = f[x_{0}] + f[x_{0},x_{1}](x – x_{0}) + f[x_{0},x_{1},x_{2}](x – x_{0})(x – x_{1}) + … + f[x_{0},x_{1},…,x_{n}](x – x_{0})(x – x_{1})…(x – x_{n-1})

where f[x_{0}, x_{1}, …, x_{n}] is the nth divided difference.

Apply Newton Divided Difference Method of Interpolation for the given set of data

Suppose we have the following set of data:

x | f(x) |

0.1 | 1.221 |

0.2 | 1.4918 |

0.3 | 1.6651 |

0.4 | 1.8514 |

We want to find an interpolation polynomial using Newton’s Divided Difference Method.

To start, we construct the divided difference table as follows:

x | f(x) | f[x] | f[x,x] | f[x,x,x] | f[x,x,x,x] |

0.1 | 1.221 | ||||

0.2 | 1.4918 | 2.707 | |||

0.3 | 1.6651 | 1.735 | 0.924 | ||

0.4 | 1.8514 | 2.427 | 1.136 | -0.231 |

Here, f[x] represents the first divided difference, f[x,x] represents the second divided difference, and so on.

The interpolation polynomial in Newton’s divided difference form is given by:

P3(x) = f[x_{0}] + f[x_{0},x_{1}](x – x_{0}) + f[x_{0},x_{1},x_{2}](x – x_{0})(x – x_{1}) + f[x_{0},x_{1},x_{2},x_{3}](x – x0)(x – x_{1})(x – x_{2})

where x_{0}, x_{1}, x_{2}, and x_{3} are the given data points.

Using the divided difference table, we can compute the coefficients of the interpolation polynomial as follows:

f[x_{0}] =_{ }1.221

f[x_{0},x_{1}] = 2.707

f[x_{0},x_{1},x_{2}] = 0.924

f[x_{0},x_{1},x_{2},x_{3}] = -0.231

Substituting these values into the formula for P_{3}(x), we get:

P_{3}(x) = 1.221 + 2.707(x – 0.1) + 0.924(x – 0.1)(x – 0.2) – 0.231(x – 0.1)(x – 0.2)(x – 0.3)

Thus, the interpolation polynomial using Newton’s Divided Difference Method for the given set of data is:

P_{3}(x) = 1.221 + 2.707(x – 0.1) + 0.924(x – 0.1)(x – 0.2) – 0.231(x – 0.1)(x – 0.2)(x – 0.3)

Describe Gregory-Newton Forward Interpolation Formula

The Gregory-Newton Forward Interpolation Formula is a method of polynomial interpolation used to find the approximate value of a function at a point within a given range. It is a modification of Newton’s divided difference method and is used for equally spaced data points.

Suppose we have a set of n+1 equally spaced data points (x_{0}, y_{0}), (x_{1}, y_{1}), …, (x_{n}, y_{n}) with a common difference h = x_{i} – x_{i}-1. The Gregory-Newton Forward Interpolation Formula for finding the value of the function f(x) at a point x within the interval [x_{0}, x_{n}] is given by:

f(x) ≈ y_{0 }+ tΔy_{0} + t(t-1)Δ²y_{0 }/2! + t(t-1)(t-2)Δ³y_{0 }/3! + … + t(t-1)…(t-n+1)Δⁿy_{0 }/n!

where t = (x – x_{0}) / h and Δⁿy_{0} represents the nth forward difference of y_{0}.

The formula can be simplified using the Newton’s notation:

f(x) = y_{0} + (x-x_{0})Δy_{0} + (x-x_{0})(x-x_{1})Δ²y_{0 }/2!_{ }+ (x-x_{0})(x-x_{1})(x-x_{2})Δ³y_{0 }/3!_{ }+ … + (x-x_{0})(x-x_{1).}..(x-x_{n-1})Δⁿy_{0}/n!

where Δⁿy_{0} represents the nth forward difference of y_{0}.

Apply Gregory-Newton Forward Interpolation Formula for the given set of data

Suppose we have the following set of data:

x | y |

0 | 2 |

1 | 5 |

2 | 12 |

3 | 23 |

We want to use the Gregory-Newton Forward Interpolation Formula to find the value of y at x = 1.5.

First, we need to calculate the forward differences:

Δy_{1} = y_{1} – y_{0 }= 5 – 2 = 3

Δy_{2} = y_{2 }– y_{1} = 12 – 5 = 7

Δy_{3} = y_{3} – y_{2} = 23 – 12 = 11

Next, we can use these forward differences to construct the divided difference table:

x | y | Δy_{1} |
Δy_{2} |
Δy_{3} |

0 | 2 | 3 | 4 | 1 |

1 | 5 | 7 | 5 | |

2 | 12 | 11 | ||

3 | 23 |

The first row shows the values of x and y, and the subsequent rows show the forward differences. The empty cells in the table can be filled in using the formula:

Δy_{k} = Δy_{k-1} – Δy_{k-1},k-1

For example, to fill in the cell for Δy2, we use:

Δy_{2} = Δy_{1},k-1 – Δy_{1},1-1 = 4 – 3 = 1

Now that we have the divided difference table, we can use the Gregory-Newton Forward Interpolation Formula:

y(x) = y_{0} + uΔy_{1 }+ u(u-1)/2! Δy_{2} + u(u-1)(u-2)/3! Δy_{3}

where u = (x – x0)/h and h is the common difference between the x values, which in this case is 1.

To find y(1.5), we first calculate u:

u = (1.5 – 0)/1 = 1.5

Then we can substitute into the formula:

y(1.5) = 2 + 1.5(3) + 1.5(0.5)(1) + 1.5(0.5)(-0.5)(1) = 8.25

Therefore, we have estimated the value of y at x = 1.5 to be 8.25 using the Gregory-Newton Forward Interpolation Formula.

Describe Gregory-Newton Backward Interpolation Formula

The Gregory-Newton Backward Interpolation Formula is used to find the value of a function at a point inside a given set of values. It is a modification of Newton’s Divided Difference Method of Interpolation. Let x_{0}, x_{1}, x_{2}, …, xn be n + 1 distinct equally spaced points and let f(x_{0}), f(x_{1}), f(x_{2}), …, f(x_{n}) be the corresponding function values. Let h = x_{1} – x_{0} be the common distance between the adjacent points.

The Gregory-Newton Backward Interpolation Formula is given by:

f(x) = f(x_{n}) + Δf_{1}(u) + Δf_{2}(u)(u – 1) + Δf_{3}(u)(u – 1)(u – 2) + … + Δf_{n}(u)(u – 1)…(u – n + 1)

where u = (x – x_{n}) / h, and the backward differences are defined by:

Δf_{1} = f(x_{n}) – f(x_{n-1})

Δf_{2 }= Δf_{1}(x_{n} – x_{n-1}) / (x_{n} – x_{n-2}) – (f(x_{n-1}) – f(x_{n-2}))

Δf_{3} = Δf_{2}(x_{n} – x_{n-1})(x_{n} – x_{n-2}) / (x_{n} – x_{n-3})(x_{n} – x_{n-2}) – (Δf_{1}(x_{n} – x_{n-2}) / (x_{n} – x_{n-3}) – (f(x_{n-2}) – f(x_{n-3})))

Δf_{4} = Δf_{3}(x_{n} – x_{n-1})(x_{n} – x_{n-2})(x_{n} – x_{n-3}) / (x_{n} – x_{n-4})(x_{n} – x_{n-3}) – (Δf_{2}(x_{n} – x_{n-2})(x_{n} – x_{n-3}) / (x_{n} – x_{n-4}) – (Δf_{1}(x_{n} – x_{n-3}) / (x_{n} – x_{n-4}) – (f(x_{n-3}) – f(x_{n-4}))))

.

.

.

Δf_{n} = Δf_{n-1}(x_{n} – x_{n-1})(x_{n} – x_{n-2})…(x_{n} – x_{n-n+1}) / (x_{n} – x_{n-n})

The formula can be simplified by introducing the coefficients of the divided differences:

a_{0} = f(x_{n}), a_{1} = Δf_{1,} a_{2} = Δf_{2}, a_{3} = Δf_{3}, …, a_{n} = Δf_{n}

Then, the formula can be written as:

f(x) = a_{0} + a_{1}(u – 1) + a_{2}(u – 1)(u – 2) + a_{3}(u – 1)(u – 2)(u – 3) + … + a_{n}(u – 1)(u – 2)…(u – n + 1) ** **

Apply Gregory-Newton Backward Interpolation Formula for the given set of data

To apply Gregory-Newton Backward Interpolation Formula, we can use the same set of data as in the previous example:

x = [0.1, 0.2, 0.3, 0.4, 0.5]

y = [1.221, 1.4918, 1.6651, 1.8437, 2.0385]

Let’s find the value of y at x = 0.25 using the Gregory-Newton Backward Interpolation Formula:

Step 1: Calculate the backward difference table

x | y | Δy_{1} |
Δy_{2} |
Δy_{3} |
Δy_{4} |

0.1 | 1.221 | ||||

0.2 | 1.4918 | 0.2708 | |||

0.3 | 1.6651 | 0.1733 | -0.0975 | ||

0.4 | 1.8437 | 0.1786 | 0.0053 | 0.1028 | |

0.5 | 2.0385 | 0.1948 | 0.0162 | 0.0109 | -0.0919 |

Step 2: Calculate the value of u

u = (x – x^{n}) / h = (0.25 – 0.3) / 0.1 = -0.5

Step 3: Apply the Gregory-Newton Backward Interpolation Formula

y = y^{n} + uΔy^{1} + u(u + 1)Δy^{2}/2! + u(u + 1)(u + 2)Δy^{3}/3! + u(u + 1)(u + 2)(u + 3)Δy^{4}/4!

y^{n} = y^{4 }= 2.0385

Δy^{1} = Δy^{4} = -0.0919

Δy^{2} = -0.0109

Δy^{3} = -0.1028

y = 2.0385 – 0.5(-0.0919) + (-0.5)(-0.5 + 1)(-0.0109)/2! + (-0.5)(-0.5 + 1)(-0.5 + 2)(-0.1028)/3! + (-0.5)(-0.5 + 1)(-0.5 + 2)(-0.5 + 3)(-0.0919)/4!

y = 1.6211

Therefore, the value of y at x = 0.25 using the Gregory-Newton Backward Interpolation Formula is approximately 1.6211.

Describe Gauss’s Forward Formula of Interpolation

Gauss’s Forward Formula of Interpolation is a method for approximating the value of a function at some point between two known values. This formula is a modification of the Gregory-Newton forward interpolation formula and provides a more accurate approximation of the function value. The formula is given as:

f(x) ≈ y_{0} + uΔy_{0} + (u(u-1)/2!)(Δ^{2}y_{0}) + (u(u-1)(u-2)/3!)(Δ^{3}y_{0}) + … + (u(u-1)(u-2)…(u-n+1)/n!)(Δ^{n} y_{0})

where

- f(x) is the approximation of the value of the function at x.
- y0 is the value of the function at the lower limit of the interval.
- u = (x-x
_{0})/h, where x_{0}is the lower limit of the interval, h is the step size, and x is the point of approximation. - Δy
_{0}, Δ^{2}y_{0}, Δ^{3}y_{0}, …, Δ^{n}y_{0}are the finite differences of the function y with respect to the independent variable x, evaluated at x_{0}.

This formula is particularly useful when the interval is equally spaced and the values of the function are tabulated in a table of finite differences.

Apply Gauss’s Formula of Interpolation for the given set of data

As Gauss’s Formula of Interpolation involves higher-order differences, let’s first calculate the necessary differences for the given data:

x | f(x) |

0.1 | 1.2214 |

0.2 | 1.4918 |

0.3 | 1.7782 |

0.4 | 2.0807 |

0.5 | 2.4000 |

First-order differences:

1st Diff | |

0.2 | 0.2704 |

0.3 | 0.2864 |

0.4 | 0.3025 |

0.5 | 0.3193 |

Second-order differences:

1st Diff | 2nd Diff | |

0.3 | 0.2864 | 0.0161 |

0.4 | 0.3025 | 0.0168 |

0.5 | 0.3193 | 0.0168 |

Third-order differences:

1st Diff | 2nd Diff | 3rd Diff | |

0.4 | 0.3025 | 0.0168 | -0.0001 |

0.5 | 0.3193 | 0.0168 | 0.0000 |

Now, let’s use Gauss’s Formula of Interpolation to find the value of f(0.35):

We need to find the value of f(0.35) using the data points x = 0.3, 0.4, and 0.5. Let’s take 0.4 as the central point and write the formula:

f(0.35) = f(0.4) + u * Δ^{1} + u * (u-1) * Δ^{2} / 2! + u * (u-1) * (u+1) * Δ^{3} / 3!

where u = (x – x_{0}) / h = (0.35 – 0.4) / 0.1 = -0.5

Δ_{1} = 0.3025, Δ_{2} = 0.0168, Δ_{3} = -0.0001

Substituting the values:

f(0.35) = 2.0807 + (-0.5) * 0.3025 + (-0.5) * (-1.5) * 0.0168 + (-0.5) * (-1.5) * (-0.5) * (-2.5) * (-0.0001) / 6

f(0.35) = 1.9035

Therefore, the value of f(0.35) using Gauss’s Formula of Interpolation is approximately 1.9035.

Describe Gauss’s Backward Formula of Interpolation

Gauss’s Backward Formula of Interpolation is used to approximate the value of a function at a certain point using a set of equally spaced data points in the reverse order, i.e., from the highest value of the independent variable to the lowest value.

Suppose we have a set of equally spaced data points in the reverse order, i.e., (x_{0}, y_{0}), (x_{0}-h, y1), (x_{0}-2h, y2), …, (x_{n}, y_{n}), where h is the common difference between the consecutive values of x. We want to find the value of y at a point x = x_{0} + mh, where m is a positive integer.

The Gauss’s Backward Formula of Interpolation is given by:

y = y_{n} + u_{1}Δy + u_{1}u_{2}Δ²y + u_{1}u_{2}u_{3}Δ³y + … + u_{1}u_{2}…u_{m}Δ_{m}y

Where

Δy = y_{1} – y_{0}

Δ²y = Δy_{2} – Δy_{1}

Δ³y = Δ²y_{2} – Δ²y_{1}

and so on, where Δⁿy represents the nth forward difference of y.

The coefficients u_{1}, u_{2}, u_{3}, …, um can be calculated as:

u_{1} = (x – x_{n})/h

u_{2 }= (x – x_{n} + h)/h(x_{n} – x_{0})

u_{3} = (x – x_{n} + 2h)/(h²(x_{n} – x_{0})(x_{n }– x_{0} + h))

and so on, where um is given by:

um = (x – x_{n} + (m-1)h)/(hᵐ(x_{n} – x_{0})(x_{n} – x_{0 }+ h)…(x_{n} – x_{0} + (m-1)h))

Once the coefficients are calculated, the value of y at x = x_{0} + mh can be found by substituting the values in the above formula.

The Gauss’s Backward Formula of Interpolation is useful when we have to approximate the value of a function at a point in the past, i.e., for a negative value of the independent variable.

Apply Gauss’s Backward Formula of Interpolation for the given set of data

Let’s consider an example to apply Gauss’s Backward Formula of Interpolation.

Suppose we have the following set of data:

x | f(x) |

0 | 1 |

1 | 3 |

2 | 7 |

3 | 13 |

4 | 21 |

We want to use Gauss’s Backward Formula of Interpolation to estimate the value of f(5). We will use the last four points in the table to interpolate the value of f(5).

First, we calculate the forward differences and arrange them in a table:

x | f(x) | Δf(x) | Δ²f(x) | Δ³f(x) |

1 | 3 | 2 | 4 | 6 |

2 | 7 | 2 | 4 | |

3 | 13 | 2 | ||

4 | 21 |

Next, we use the Gauss’s Backward Formula of Interpolation to get the estimate:

f(5) = 21 + (-1)(2) + (-1)(-3)(4)/(2!) + (-1)(-3)(-5)(6)/(3!) = 34

Therefore, we estimate that f(5) is approximately 34.

Note: The negative signs in the formula alternate for each additional term of the interpolation.

Describe Stirling’s Formula of Interpolation

Stirling’s Formula of Interpolation is a mathematical formula used in numerical analysis to approximate the value of a function at a point using the values of the function at nearby points. It is named after the Scottish mathematician James Stirling.

The formula is given by:

f(x) ≈ f(x_{0}) + f'(x_{0})(x – x_{0}) + (1/2!) f”(x_{0})(x – x_{0})^{2} + … + (1/n!) f(n)(x_{0})(x – x_{0})^{n}

where f(x) is the value of the function at the point x, f(x_{0}) is the value of the function at the point x_{0}, f'(x_{0}) is the first derivative of the function at x_{0}, f”(x_{0}) is the second derivative of the function at x_{0}, and so on, with f^{(n)}(x_{0}) representing the nth derivative of the function at x_{0}.

The formula is an approximation and becomes more accurate as more terms are added to the series. It is particularly useful when the function is difficult or time-consuming to evaluate directly.

Stirling’s Formula of Interpolation is a special case of Taylor’s Series, which is a way of approximating a function with a polynomial series. However, Stirling’s Formula only considers the terms up to the nth derivative, while Taylor’s Series considers all the derivatives of the function.

Apply Stirling’s Formula of Interpolation for the given set of data

To apply Stirling’s Formula of Interpolation to a set of data, we need to have the values of the function and its derivatives at a point x_{0}, as well as the value of x at which we want to approximate the function.

Let’s assume we have the following set of data:

x_{0} = 0

f(x_{0}) = 1

f'(x_{0}) = 2

f”(x_{0}) = -3

f”'(x_{0}) = 4

f””(x_{0}) = -5

Using Stirling’s Formula of Interpolation, we can write:

f(x) ≈ f(x_{0}) + f'(x_{0})(x – x_{0}) + (1/2!) f”(x_{0})(x – x_{0})^{2} + (1/3!) f”'(x_{0})(x – x_{0})^{3} + (1/4!) f””(x_{0})(x – x_{0})^{4}

Substituting the values we have:

f(0.5) ≈ 1 + 2(0.5 – 0) + (1/2!) (-3)(0.5 – 0)^{2} + (1/3!) 4(0.5 – 0)^{3} + (1/4!) (-5)(0.5 – 0)^{4}

Simplifying the expression, we get:

f(0.5) ≈ 1 + 1 + (-0.375) + (0.16666667) + (-0.02604167)

f(0.5) ≈ 1.764625

Therefore, using Stirling’s Formula of Interpolation, we have approximated the value of the function f(x) at x = 0.5 to be approximately 1.764625.

Find the derivative from the given set of Data using Numerical Differentiation

Numerical differentiation is a method used to estimate the derivative of a function at a point using a set of data. There are several numerical differentiation methods, but one common approach is the three-point central difference formula.

he three-point central difference formula can be used to estimate the first derivative of a function f(x) at a point x0, given the values of the function at x0, x1, and x2:

f'(x0) ≈ (f(x2) – f(x1))/(x2 – x1)

In this case, we are given the following set of data:

x0 = 0, f(x0) = 2

x1 = 0.1, f(x1) = 2.2

x2 = 0.2, f(x2) = 2.4

To estimate the derivative of the function f(x) at x0 = 0, we can use the three-point central difference formula:

f'(0) ≈ (f(0.2) – f(0.1))/(0.2 – 0.1)

≈ (2.4 – 2.2)/(0.2 – 0.1)

≈ 2

Therefore, the estimated value of the derivative of the function f(x) at x0 = 0 is approximately 2. Therefore, the estimated value of the derivative of the function f(x) at x0 = 0 is approximately 2.

Describe Newton-Cotes Formulas

Newton-Cotes formulas are a family of numerical integration methods used to approximate the value of a definite integral of a function. They work by approximating the function with a polynomial and then integrating the polynomial over the interval of integration.

The simplest Newton-Cotes formula is the trapezoidal rule, which approximates the area under the curve of a function between two points a and b as the area of the trapezoid formed by the function and the line connecting the two endpoints. The formula is given by:

f(x) dx ≈ (b-a)/2 * (f(a) + f(b))

This formula assumes that the function f(x) is approximately linear between a and b.

The Simpson’s rule is another commonly used Newton-Cotes formula. It approximates the area under the curve of a function between three points a, b, and c as the area of a parabolic segment. The formula is given by:

∫[a,b,c] f(x) dx ≈ (b-a)/6 * (f(a) + 4f((a+b)/2) + f(b)) + (c-b)/6 * (f(b) + 4f((b+c)/2) + f(c))

This formula assumes that the function f(x) is approximately quadratic between a and c.

The Newton-Cotes formulas can be extended to approximate integrals over more intervals and higher degree polynomials. However, higher-degree formulas tend to be less accurate due to the accumulation of errors in the polynomial approximation.

Overall, Newton-Cotes formulas provide a useful and easy-to-implement method for approximating the value of a definite integral of a function. They are particularly useful when the function is difficult or impossible to integrate analytically.

Deduce the Trapezoidal Rule from Newton-Cotes Formulas

The Trapezoidal Rule is a Newton-Cotes formula used to approximate the definite integral of a function f(x) over the interval [a, b].

To deduce the Trapezoidal Rule from the general Newton-Cotes formula, we can start by approximating the function f(x) over the interval [a, b] with a straight line connecting the points (a, f(a)) and (b, f(b)).

The equation of the line can be written as:

y = f(a) + (f(b) – f(a))/(b – a) * (x – a)

We can then integrate the line over the interval [a, b] to get an approximation of the integral of f(x) over the same interval. The result is:

f(x) dx ≈ ∫[a,b] [f(a) + (f(b) – f(a))/(b – a) * (x – a)] dx

Simplifying the right-hand side of the equation, we get:

f(x) dx dx ≈ f(a)*(b-a) + (f(b) – f(a))/(b – a) * [(b^{2} – a^{2})/2]

Rearranging and simplifying the equation further, we arrive at the Trapezoidal Rule:

f(x) dx ≈ (b-a)/2 * [f(a) + f(b)]

Therefore, the Trapezoidal Rule can be deduced from the general Newton-Cotes formula by approximating the function f(x) over the interval [a, b] with a straight line and integrating the line over the same interval.

Apply Trapezoidal Rule to solve the definite Integral

Let’s consider the definite integral:

x^{2} dx

To apply the Trapezoidal Rule, we first need to divide the interval [0,2] into subintervals. In this case, we will use a single subinterval since we only have one interval to integrate over. We can choose the endpoints of the subinterval as a=0 and b=2.

The Trapezoidal Rule can then be written as:

f(x) dx ≈ (b-a)/2 * [f(a) + f(b)]

Applying this formula to our integral, we get:

x^{2} dx ≈ (2-0)/2 * [f(0) + f(2)]

We need to evaluate f(0) and f(2), which are:

f(0) = 0^{2} = 0

f(2) = 2^{2 }= 4

Substituting these values into the Trapezoidal Rule formula, we get:

x^{2} dx ≈ (2-0)/2 * [0 + 4] = 4

Therefore, the Trapezoidal Rule approximation of the definite integral x^{2} dx is 4

Describe Simpson’s 1/3rd Rule

Simpson’s 1/3 Rule is a numerical integration technique used to approximate the definite integral of a function f(x) over an interval [a,b]. This method is based on approximating the integrand by a quadratic function over each subinterval, and then integrating the quadratic function to get an approximation of the integral.

The Simpson’s 1/3 Rule formula can be written as:

f(x) dx≈ (b-a)/6 * [f(a) + 4*f((a+b)/2) + f(b)]

where f(a) and f(b) are the values of the function f(x) at the endpoints of the interval [a,b], and f((a+b)/2) is the value of the function at the midpoint of the interval.

The idea behind Simpson’s 1/3 Rule is to use the quadratic equation that passes through the points (a,f(a)), ((a+b)/2, f((a+b)/2)), and (b,f(b)) to approximate the integrand. This quadratic equation can be written as:

y = Ax^{2} + Bx + C

where A, B, and C are constants that can be solved using the values of f(a), f((a+b)/2), and f(b).

By integrating this quadratic equation over the interval [a,b], we can obtain an approximation of the integral of f(x) over the same interval. The resulting formula is Simpson’s 1/3 Rule.

Simpson’s 1/3 Rule is generally more accurate than the Trapezoidal Rule for smooth functions. However, if the function is not smooth or if there are odd numbers of intervals, then the composite Simpson’s 1/3 Rule or Simpson’s 3/8 Rule can be used instead.

Apply the Simpson’s 1/3rd Rule to find the approximate value of the integral

Let’s consider the definite integral:

x^{3} dx

To apply Simpson’s 1/3 Rule, we first need to divide the interval [0,1] into subintervals. In this case, we will use a single subinterval since we only have one interval to integrate over. We can choose the endpoints of the subinterval as a=0 and b=1.

The Simpson’s 1/3 Rule formula can then be written as:

f(x) dx ≈ (b-a)/6 * [f(a) + 4*f((a+b)/2) + f(b)]

Applying this formula to our integral, we get:

x^{3} dx ≈ (1-0)/6 * [f(0) + 4*f(0.5) + f(1)]

We need to evaluate f(0), f(0.5), and f(1), which are:

f(0) = 0^{3} = 0

f(0.5) = (0.5)^{3} = 0.125

f(1) = 1^{3} = 1

Substituting these values into the Simpson’s 1/3 Rule formula, we get:

x^{3} dx ≈ (1-0)/6 * [0 + 4*0.125 + 1] = 0.25

Therefore, the Simpson’s 1/3 Rule approximation of the definite integral x^{3} dx is 0.25

Describe Simpson’s 3/8th Rule

Simpson’s 3/8 Rule is a numerical integration technique used to approximate the definite integral of a function f(x) over an interval [a,b]. This method is based on approximating the integrand by a cubic function over each subinterval, and then integrating the cubic function to get an approximation of the integral.

The Simpson’s 3/8 Rule formula can be written as:

f(x) dx ≈ (b-a)/8 * [f(a) + 3f((2a+b)/3) + 3f((a+2b)/3) + f(b)]

where f(a) and f(b) are the values of the function f(x) at the endpoints of the interval [a,b], and f((2a+b)/3) and f((a+2b)/3) are the values of the function at the points (2a+b)/3 and (a+2b)/3 respectively.

The idea behind Simpson’s 3/8 Rule is to use the cubic equation that passes through the points (a,f(a)), ((2a+b)/3, f((2a+b)/3)), ((a+2b)/3, f((a+2b)/3)), and (b,f(b)) to approximate the integrand. This cubic equation can be written as:

y = Ax^{3} + Bx^{2} + Cx + D

where A, B, C, and D are constants that can be solved using the values of f(a), f((2a+b)/3), f((a+2b)/3), and f(b).

By integrating this cubic equation over the interval [a,b], we can obtain an approximation of the integral of f(x) over the same interval. The resulting formula is Simpson’s 3/8 Rule.

Simpson’s 3/8 Rule is generally more accurate than Simpson’s 1/3 Rule for smooth functions. However, it requires three function evaluations per subinterval instead of two, which can make it computationally more expensive for large numbers of subintervals.

Apply the Simpson’s 3/8th Rule to find the approximate value of the integral

Example: Find the approximate value of the integral of f(x) = x^{3} – 2x^{2 }+ x + 1 over the interval [0,2] using Simpson’s 3/8 Rule with 3 subintervals.

Solution:

To use Simpson’s 3/8 Rule, we first need to divide the interval [0,2] into subintervals. Since we are using 3 subintervals, each subinterval will have length (2-0)/3 = 0.6667.

The endpoints of the subintervals are:

x_{0} = 0

x_{1} = 0.6667

x_{2} = 1.3333

x_{3} = 2

Using Simpson’s 3/8 Rule, the approximate value of the integral is:

x^{3} dx ≈ (2-0)/8 * [f(0) + 3f(0.6667) + 3f(1.3333) + f(2)]

We need to evaluate the function f(x) at each of these points:

f(0) = 0^{3} – 20^{2} + 0 + 1 = 1

f(0.6667) = 0.6667^{3} – 20.6667^{2} + 0.6667 + 1 = 0.5866

f(1.3333) = 1.3333^{3} – 21.3333^{2 }+ 1.3333 + 1 = -0.3585

f(2) = 2^{3} – 22^{2} + 2 + 1 = -1

Substituting these values into the formula, we get:

f(x) dx ≈ (2-0)/8 * [1 + 30.5866 + 3(-0.3585) + (-1)]

≈ 0.1813

Therefore, the approximate value of the integral of f(x) over the interval [0,2] using Simpson’s 3/8 Rule with 3 subintervals is 0.1813.

Describe the Gaussian One-Point,Two-Point, and Three- Point Formula

Gaussian quadrature is a numerical integration technique that uses weighted sum of function values at certain points in the integration interval. The Gaussian quadrature formulas are derived to approximate the definite integral of a function over a given interval by selecting a set of nodes (points) in the interval along with corresponding weights. These nodes and weights are selected in such a way that the accuracy of the approximation is maximized.

The Gaussian one-point formula:

The Gaussian one-point formula uses one point and weight to approximate the definite integral. The formula is given as:

f(x) dx ≈ w_{0}*f(x0)

where x_{0} and w_{0} are the node and weight respectively. The node and weight are chosen such that the formula gives exact results for polynomial functions of degree up to 2n-1 where n is the number of nodes used.

The Gaussian two-point formula:

The Gaussian two-point formula uses two points and weights to approximate the definite integral. The formula is given as:

f(x) dx ≈ w0f(x0) + w1f(x1)

where x0, x1, w0 and w1 are the nodes and weights respectively. The nodes and weights are chosen such that the formula gives exact results for polynomial functions of degree up to 2n-1 where n is the number of nodes used.

The Gaussian three-point formula:

The Gaussian three-point formula uses three points and weights to approximate the definite integral. The formula is given as:

f(x) dx ≈ w0f(x0) + w1f(x1) + w2*f(x2)

where x_{0}, x_{1}, x_{2}, w_{0}, w_{1} and w_{2} are the nodes and weights respectively. The nodes and weights are chosen such that the formula gives exact results for polynomial functions of degree up to 2n-1 where n is the number of nodes used.

In general, Gaussian quadrature formulas with n nodes and weights can be derived to give exact results for polynomial functions of degree up to 2n-1. The accuracy of Gaussian quadrature formulas makes them very useful in numerical integration.

Apply the Gaussian One-Point, Two-Point, and Three-Point Formula to find the Definite Integral

Example:

Approximate the definite integral of the function f(x) = x^{3} + x^{2} – 1 over the interval [-1, 1] using the Gaussian one-point, two-point, and three-point formulas.

Solution:

First, we need to find the nodes and weights for each formula. The nodes and weights can be found using the Legendre polynomials, which are orthogonal polynomials over the interval [-1, 1]. The Legendre polynomials can be computed using the recurrence relation:

P0(x) = 1

P1(x) = x

Pn(x) = (2n-1)/n * x * Pn-1(x) – (n-1)/n * Pn-2(x)

The roots of the Legendre polynomial of degree n are the nodes for the Gaussian quadrature formula with n points. The weights can be computed using the formula:

wi = 2 / [(1-xi^{2}) * (P’n(xi))_{2}]

where P’n(x) is the derivative of the Legendre polynomial of degree n.

Using the Legendre polynomials, we can find the nodes and weights for the Gaussian quadrature formulas:

Gaussian one-point formula:

Using the Legendre polynomial of degree 1, we can find the node and weight for the Gaussian one-point formula:

x_{0} = 0

w_{0} = 2

Gaussian two-point formula:

Using the Legendre polynomial of degree 2, we can find the nodes and weights for the Gaussian two-point formula:

x_{0} = -0.577350

x_{1} = 0.577350

w_{0} = 1.00000

w_{1} = 1.00000

Gaussian three-point formula:

Using the Legendre polynomial of degree 3, we can find the nodes and weights for the Gaussian three-point formula:

x_{0} = -0.774597

x_{1} = 0

x_{2} = 0.774597

w_{0} = 0.555556

w_{1} = 0.888889

w_{2} = 0.555556

Now, we can use the Gaussian quadrature formulas to approximate the definite integral:

Gaussian one-point formula:

f(x) dx ≈ w_{0}f(x^{0})

≈ 2(0^{3} + 0^{2} – 1)

≈ -2

Gaussian two-point formula:

f(x) dx ≈ w0f(x0) + w1f(x1)

≈ 1*(-0.577350^{3} + (-0.577350)^{2} – 1) + 1*(0.577350^{3} + (0.577350)^{2} – 1)

≈ -0.153887

Gaussian three-point formula:

f(x) dx ≈ w_{0}f(x_{0}) + w_{1}f(x_{1}) + w_{2}f(x_{2})

≈ 0.555556(-0.774597^{3} + (-0.774597)^{2} – 1) + 0.888889*(0^{3} + 0^{2} – 1) + 0.555556*(0.774597^{3} + (0.774597)^{2} – 1)

≈ -0.067598

Therefore, the approximate values of the definite integral using the Gaussian one-point, two-point, and three-point formulas are -2, -0.153887, and -0.067598, respectively

Describe the Quadrature Formulas: i. Gauss-Legendre Formula, ii. Gauss-Chebyshev Formula, iii. Gauss-Hermite Formula, iv. Gauss-Laguerre Formula

Quadrature formulas are used to numerically approximate definite integrals. Some commonly used quadrature formulas are:

i. Gauss-Legendre Formula:

This formula uses Legendre polynomials to approximate the integral. It is accurate for integrals over the interval [-1,1]. The formula is given by:

f(x) dx ≈ (b-a)/2 * Σ wi*f(xi)

where wi are the weights and xi are the roots of the n-th Legendre polynomial.

ii. Gauss-Chebyshev Formula:

This formula is used to approximate integrals over the interval [-1,1] using Chebyshev polynomials. The formula is given by:

f(x) / √(1-x^{2}) dx ≈ π/n * Σ wi*f(xi)

where wi are the weights and xi are the roots of the n-th Chebyshev polynomial.

iii. Gauss-Hermite Formula:

This formula is used to approximate integrals of the form f(x) dx over the interval [-∞,∞]. The formula is given by:

f(x) e^{(-x2)} dx ≈ Σ wi*f(xi)

where wi are the weights and xi are the roots of the n-th Hermite polynomial.

iv. Gauss-Laguerre Formula:

This formula is used to approximate integrals of the form f(x) dx over the interval [0,∞]. The formula is given by:

f(x) e^{(-x)} dx ≈ Σ wi*f(xi)

where wi are the weights and xi are the roots of the n-th Laguerre polynomial.