Numerical Methods: Numerical Techniques-II

Contents

**Describe the Forward and Backward Finite Difference Table** 4

**Find the missing term using Finite Difference Table** 8

**Describe the method of Separation of Symbols and it to prove the useful Identities** 10

**Show a Polynomial in its successive differences and also in Factorial Notation** 10

**Recall the term Interpolation** 12

**Formulate Lagrange Interpolation** 12

**Apply Lagrange’s Interpolation Formula for a given set of data** 13

**Formulate Hermite Interpolation** 14

**Apply Hermite’s Interpolation Formula for a given set of data** 15

**Describe the Divided Differences** 16

**Differentiate between Finite Differences and Divided Differences** 17

**Recall the Divided Difference Table** 18

**Recall Newton’s Divided Difference Method of Interpolation** 19

**Apply Newton Divided Difference Method of Interpolation for the given set of Data** 20

**Describe Gregory-Newton Forward Interpolation Formula** 21

**Apply Gregory-Newton Forward Interpolation Formula for the given set of Data** 22

**Describe Gregory-Newton Backward Interpolation Formula** 23

**Apply Gregory-Newton Backward Interpolation Formula for the given set of Data** 24

**Describe Gauss’s Forward Formula of Interpolation** 26

**Apply Gauss’s Formula of Interpolation for the given set of Data** 26

**Describe Gauss’s Backward Formula of Interpolation** 28

**Apply Gauss’s Backward Formula of Interpolation for the given set of Data** 29

**Describe Stirling’s Formula of Interpolation** 29

**Apply Stirling’s Formula of Interpolation for the given set of data** 29

**Describe Numerical Differentiation** 29

**Describe Newton-Cotes Formulas** 29

**Deduce the Trapezoidal Rule From Newton-Cotes Formulas** 29

**Describe Simpson’s 1/3rd Rule** 29

**Apply the Simpson’s 1/3 ^{rd} rule to find the approximate value of the integral** 29

**Describe Simpson’s 3/8 ^{th} Rule** 29

**Apply the Simpson’s 3/8th Rule to find the approximate value of the integral** 29

**Describe he Gaussian One-Point, Two-Point, and Three-Point Formula** 29

**Apply the Gaussian One-Point, Two-Point, and Three-Point Formula to find the Definite Integral** 29

**Describe the following Operators: i. Forward Difference, ii. Backward Difference, iii. Shift Operator, iv. Central Difference Operator, v. Average Operator**

i. Forward Difference Operator: The forward difference operator is a mathematical operator that calculates the difference between a function’s values at two consecutive points, where the second point is slightly ahead of the first point. The forward difference operator is defined as:

Δf(x) = f(x+h) – f(x)

where h is the step size, and x is the point where the difference is being calculated. This operator is commonly used in numerical differentiation, where it approximates the derivative of a function at a particular point.

ii. Backward Difference Operator: The backward difference operator is similar to the forward difference operator, but it calculates the difference between a function’s values at two consecutive points, where the first point is slightly behind the second point. The backward difference operator is defined as:

Δf(x) = f(x) – f(x-h)

where h is the step size, and x is the point where the difference is being calculated. Like the forward difference operator, the backward difference operator is commonly used in numerical differentiation.

iii. Shift Operator: The shift operator is a mathematical operator that shifts the values of a sequence by a certain number of places. For example, if we have a sequence {a_{1}, a_{2}, a_{3}, a_{4}}, and we apply the shift operator with a shift value of 1, the resulting sequence would be {a2, a_{3}, a_{4}, 0}. The shift operator is denoted by the symbol E and is defined as:

E[a(n)] = a(n+1)

where a(n) is the nth element of the sequence.

iv. Central Difference Operator: The central difference operator is a mathematical operator that calculates the difference between a function’s values at two points, where the points are equidistant from a central point. The central difference operator is defined as:

δf(x) = f(x+h/2) – f(x-h/2)

where h is the step size, and x is the central point. This operator is commonly used in numerical differentiation because it provides a more accurate estimate of the derivative than the forward or backward difference operators.

v. Average Operator: The average operator is a mathematical operator that calculates the average of a set of values. The average operator is denoted by the symbol ẍ and is defined as:

ẍ = (x_{1} + x_{2} + … + x_{n}) / n

where x_{1}, x_{2}, …, x_{n} are the values being averaged, and n is the number of values. The average operator is used in a wide range of applications, such as in statistics, signal processing, and control systems. ** **

**Relate various Operators **

The forward difference and backward difference operators are related to each other as they both approximate the derivative of a function at a particular point using a finite difference approach. The forward difference operator calculates the difference between a function’s values at two consecutive points, where the second point is slightly ahead of the first point, while the backward difference operator calculates the difference between a function’s values at two consecutive points, where the first point is slightly behind the second point.

- The central difference operator is related to both the forward and backward difference operators as it also approximates the derivative of a function at a particular point using a finite difference approach. However, unlike the forward and backward difference operators, the central difference operator calculates the difference between a function’s values at two points, where the points are equidistant from a central point.
- The shift operator is related to all the other operators as it is often used to generate a sequence of values that can be differentiated or averaged using the other operators. For example, we can apply the shift operator to a sequence of function values to generate a sequence of forward or backward differences.
- The average operator can be used to smooth a sequence of values before applying the finite difference operators. This can help reduce noise in the signal and improve the accuracy of the differentiation. Alternatively, the average operator can be used to compute the average value of a set of central difference approximations at multiple points to estimate the derivative of a function over a range of points.

Here are some examples of how each operator could be applied: Forward Difference Operator:

Suppose we have a function f(x) = x^{2}, and we want to approximate its derivative at x = 1 using the forward difference operator with a step size of h = 0.1. The forward difference approximation would be:

Δf(1) = f(1+0.1) – f(1)

= (1.1)^{2} – 1^{2}

= 0.21

f'(1) ≈ Δf(1)/h = 2.1

- Backward Difference Operator:

Using the same function and point as the example above, but with the backward difference operator, we would have:

Δf(1) = f(1) – f(1-0.1)

= 1^{2} – (0.9)^{2}

= 0.19

f'(1) ≈ Δf(1)/h = 1.9

- Shift Operator:

Suppose we have the sequence {1, 2, 3, 4}, and we want to shift it to the right by two places. The resulting sequence would be {0, 0, 1, 2, 3}.

- Central Difference Operator:

Suppose we have a function f(x) = x^{2}, and we want to approximate its derivative at x = 1 using the central difference operator with a step size of h = 0.1. The central difference approximation would be:

δf(1) = f(1+0.05) – f(1-0.05)

= (1.05)^{2} – (0.95)^{2}

= 0.1

f'(1) ≈ δf(1)/h = 1

- Average Operator:

Suppose we have the sequence {1, 2, 3, 4, 5, 6}, and we want to compute its average value. The average operator would be:

ẍ = (1 + 2 + 3 + 4 + 5 + 6) / 6

= 3.5

These are just a few examples of how each operator could be applied, and there are many other possible applications depending on the problem at hand.

**Describe the Forward and Backward Finite Difference Table**

The forward and backward finite difference tables are used to approximate the derivatives of a function at a specific point using finite difference methods. They involve calculating the differences between function values at different points and using these differences to approximate the derivative.

The forward difference table is used to approximate the derivative of a function at a point x, using values of the function at x and at some points x+h, x+2^{h}, x+3h, and so on. The table is constructed as follows:

i | xi | fi | Δf | Δ^{2}f |
Δ^{3}f |
Δ^{4}f |
… |

0 | x | f(x) | |||||

1 | x + h | f(x+h) | f(x+h)-f(x) | ||||

2 | x + 2h | f(x+2h) | f(x+2h)-f(x+h) | f(x+2h)-2f(x+h) | |||

3 | x + 3h | f(x+3h) | f(x+3h)-f(x+2h) | f(x+3h)-2f(x+2h)+f(x+h) | f(x+3h)-3f(x+2h)+3f(x+h) | … |

In this table, Δf represents the first-order forward difference, Δ^2f represents the second-order forward difference, and so on. The notation Δ^nf represents the nth order forward difference.

An example of using the forward difference table would be to approximate the derivative of the function f(x) = sin(x) at x = π/4, using a step size of h = π/12. The table would look like:

i | xi | fi | Δf | Δ^{2}f |
Δ^{3}f |
Δ^4f |

0 | π/4 | sin(π/4) | ||||

1 | π/4 + π/12 | sin(π/4 + π/12) | sin(π/4+π/12)-sin(π/4) | |||

2 | π/4 + 2π/12 | sin(π/4 + 2π/12) | sin(π/4+2π/12)-sin(π/4+π/12) | sin(π/4+2π/12)-2sin(π/4+π/12) | ||

3 | π/4 + 3π/12 | sin(π/4 + 3π/12) | sin(π/4+3π/12)-sin(π/4+2π/12) | sin(π/4+3π/12)-2sin(π/4+2π/12)+sin(π/4+π/12) | sin(π/4+3π/12)-3sin(π/4+2π/12)+3sin(π/4+π/12) |

Using the values in the table, we can approximate the derivative of sin(x) at x = π/4 as follows:

f'(π/4) ≈ Δf/ h = [sin(π/4 + π/12) – sin(π/4)] / (π/12)

≈ [sin(π/4 + π/12) – sin(π/4)] * 12/π

≈ [sin(π/4 + π/12) – sin(π/4)] * 3.8197

≈ [sin(7π/12) – sin(π/4)] * 3.8197

≈ [0.9659 – 0.7071] * 3.8197

≈ 0.6984

Therefore, the forward difference table can be used to approximate the derivative of a function at a specific point, using values of the function at that point and at some other points with a step size h.

The backward difference table is similar to the forward difference table, but instead of using values of the function at x+h, x+2h, and so on, it uses values at x-h, x-2h, and so on. The table is constructed as follows:

i | xi | fi | Δf | Δ^{2}f |
Δ^{3}f |
Δ^{4}f |
… |

0 | x | f(x) | |||||

1 | x – h | f(x-h) | f(x)-f(x-h) | ||||

2 | x – 2h | f(x-2h) | f(x-h)-f(x-2h) | f(x)-2f(x-h)+f(x-2h) | |||

3 | x – 3h | f(x-3h) | f(x-2h)-f(x-3h) | f(x)-2f(x-h)+2f(x-2h)-f(x-3h) | f(x)-3f(x-h)+3f(x-2h)-f(x-3h) | … |

An example of using the backward difference table would be to approximate the derivative of the function f(x) = e^{x} at x = 0, using a step size of h = 0.1. The table would look like:

i | xi | fi | Δf | Δ^{2}f |
Δ^{3}f |
Δ^{4}f |

0 | 0 | 1 | ||||

1 | -0.1 | 0.9048 | 1 – 0.9048 | |||

2 | -0.2 | 0.8187 | 0.9048 – 0.8187 | 1.0000 – 2.7038 | ||

3 | -0.3 | 0.7408 | 0.8187 – 0.7408 | 2.7038 – 3.0000 | 1.7388 – 2.7038 | |

4 | -0.4 | 0.6703 | 0.7408 – 0.6703 | 3.0000 – 2.4536 | 2.7038 – 1.7388 | 0.9634 – 2.7038 |

To approximate the derivative of e^{x} at x=0 using this table, we can use the first row (i=0) and the second row (i=1) to compute the first-order backward difference of f(x) at x=0, which is:

Δf/ h = [f(0) – f(-h)]/h = [e^{0} – e^{-0.1)}]/0.1 = 0.9526

This is a first-order approximation of the derivative of e^{x} at x=0 using the backward finite difference method with a step size of h=0.1. We could obtain higher-order approximations by using more rows of the table and computing higher-order differences. ** **

**Find the missing term using Finite Difference Table**

To find a missing term in a finite difference table, we can use the formula for the corresponding finite difference order and extrapolate from the known terms. Let’s consider the following finite difference table:

i | xi | fi | Δf | Δ^2f | Δ^3f |

0 | 0.0 | 1.000 | |||

1 | 0.2 | 1.221 | 2.105 | ||

2 | 0.4 | 1.491 | 2.493 | 2.184 | |

3 | 0.6 | 1.822 | 2.778 | 2.175 | 0.422 |

4 | 0.8 | 2.225 | 2.956 | 0.449 | 2.009 |

In this table, we can see that the first-order forward difference is known for all rows except the first one. To find this missing value, we can use the formula for the first-order forward difference:

Δf = f(i+1) – f(i)

Using this formula and the values in the second row, we can compute the missing value in the first row as:

Δf = f(1) – f(0) = 1.221 – 1.000 = 0.221

Therefore, the missing term in the first row of the table is 0.221.

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

The method of separation of symbols is a technique used in algebra and calculus to simplify complex expressions and prove useful identities. The basic idea of this method is to manipulate the symbols of an expression without evaluating them until we obtain the desired result.

To use the method of separation of symbols, we first identify the terms in an expression that can be simplified or manipulated using known algebraic or calculus identities. We then separate the symbols in each term into two or more groups and rearrange them in a way that facilitates the simplification or manipulation. This can often involve factoring, expanding, or rearranging terms to obtain a desired form.

For example, we can use the method of separation of symbols to prove the following useful trigonometric identity:

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

To prove this identity using the method of separation of symbols, we start with the left-hand side of the identity:

sin(a + b)

We then separate the symbols using the angle addition formula for sine:

sin(a) cos(b) + cos(a) sin(b)

Now we can see that this expression matches the right-hand side of the identity we want to prove, so we have shown that sin(a + b) = sin(a) cos(b) + cos(a) sin(b). This is an example of how the method of separation of symbols can be used to prove useful identities in algebra and calculus.

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

To demonstrate a polynomial in its successive differences and factorial notation, let’s consider the quadratic polynomial:

f(x) = x^{2} + 3x + 2

First, we can compute its successive differences by taking the difference between adjacent terms in the sequence of polynomial values. This process is repeated until we reach a constant sequence:

x | f(x) |

0 | 2 |

1 | 6 |

2 | 12 |

x | f(x) | Δf(x) | Δ^{2}f(x) |

0 | 2 | 4 | 6 |

1 | 6 | 8 | |

2 | 12 |

We can see that the second-order differences are constant, which indicates that the original polynomial is a quadratic function. We can also express the polynomial in factorial notation using the binomial theorem:

f(x) = (x + 1)^{2} + 1

Expanding this expression using the binomial theorem, we get:

f(x) = (x + 1)(x + 1) + 1 = x^{2 }+ 2x + 1 + 1 = x^{2} + 3x + 2

Therefore, we have shown that the polynomial f(x) = x^{2} + 3x + 2 can be expressed in both its successive differences and factorial notation. ** **

**Recall the term Interpolation**

Interpolation is a mathematical technique used to estimate a value of a function at a point between the known data points. It involves constructing a function that passes through a given set of points or data, and then using this function to approximate the value of the function at a new point within the interval of the known data.

In other words, interpolation is a method of constructing a smooth curve or surface that fits the given data points, and then using this curve or surface to estimate the values of the function at other points.

Interpolation is widely used in many fields of science and engineering, such as physics, engineering, economics, and finance, to name a few. It is used to model and predict real-world phenomena based on incomplete or incomplete data. Some common interpolation methods include linear interpolation, polynomial interpolation, spline interpolation, and kriging interpolation. ** **

**Formulate Lagrange Interpolation**

Lagrange interpolation is a method of polynomial interpolation used to approximate the value of a function at a given point based on its known values at several other points. The Lagrange interpolation formula is:

f(x) = ∑[yᵢ * ℓᵢ(x)]

where:

- f(x) is the value of the function at the point x we want to estimate
- yᵢ is the value of the function at the i-th known point
- ℓᵢ(x) is the i-th Lagrange basis polynomial, which is defined as:

ℓᵢ(x) = ∏[(x – xⱼ) / (xᵢ – xⱼ)] for j ≠ i

where:

- xᵢ and yᵢ are the i-th known data point
- xⱼ and yⱼ are the j-th known data point, and j ≠ i

In other words, we first compute the Lagrange basis polynomials for each known data point, and then use them to compute the value of the function at the point x we want to estimate by summing up the product of each known function value with its corresponding Lagrange basis polynomial evaluated at x.

The Lagrange interpolation formula is named after Joseph-Louis Lagrange, a French mathematician who first introduced the concept of interpolation in the 18th century. It is widely used in numerical analysis and scientific computing to approximate the value of a function at intermediate points based on known data. ** **

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

An example of how to apply Lagrange’s interpolation formula for a given set of data.

Suppose we are given the following set of data points:

{(0, 1), (1, 2), (2, 5), (3, 10)}

and we want to approximate the value of the function at x = 1.5.

To do this, we can use Lagrange’s interpolation formula:

f(x) = ∑[yᵢ * ℓᵢ(x)]

where:

- f(x) is the value of the function at x = 1.5 we want to estimate
- yᵢ is the value of the function at the i-th known point
- ℓᵢ(x) is the i-th Lagrange basis polynomial, which is defined as:

ℓᵢ(x) = ∏[(x – xⱼ) / (xᵢ – xⱼ)] for j ≠ i

where:

- xᵢ and yᵢ are the i-th known data point
- xⱼ and yⱼ are the j-th known data point, and j ≠ i

First, we need to compute the Lagrange basis polynomials:

ℓ₀(x) = [(x – 1)(x – 2)(x – 3)] / [(0 – 1)(0 – 2)(0 – 3)] = -(1/6)x³ + (1/2)x² – (1/3)x + 1

ℓ₁(x) = [(x – 0)(x – 2)(x – 3)] / [(1 – 0)(1 – 2)(1 – 3)] = (1/2)x³ – (5/2)x² + 4x – 2

ℓ₂(x) = [(x – 0)(x – 1)(x – 3)] / [(2 – 0)(2 – 1)(2 – 3)] = -(1/2)x³ + (5/2)x² – (7/2)x + 5/2

ℓ₃(x) = [(x – 0)(x – 1)(x – 2)] / [(3 – 0)(3 – 1)(3 – 2)] = (1/6)x³ – (1/2)x² + (1/3)x

Next, we can plug in the known function values and the Lagrange basis polynomials into the formula:

f(1.5) = (1 * ℓ₀(1.5)) + (2 * ℓ₁(1.5)) + (5 * ℓ₂(1.5)) + (10 * ℓ₃(1.5))

= (1 * (-0.125)) + (2 * 0.75) + (5 * (-0.875)) + (10 * 0.25)

= 2.125

Therefore, the approximate value of the function at x = 1.5 is 2.125 using Lagrange’s interpolation formula.

**Formulate Hermite Interpolation**

Hermite interpolation is an extension of Lagrange interpolation that allows for the interpolation of a function and its derivatives. The formula for Hermite interpolation is as follows:

Given a set of n+1 data points {(x_{0}, y_{0}), (x_{1}, y_{1}), …, (x_{n}, y_{n})}, and a set of derivatives of the function at these points {(y_{0}‘), (y_{1}‘), …, (yn’)}, we want to find a polynomial p(x) of degree at most 2n+1 that satisfies the following conditions:

p(xi) = yi for i = 0, 1, …, n

p'(xi) = yi’ for i = 0, 1, …, n

The formula for p(x) is:

p(x) = ∑[yihi(x) + yi’hi'(x)]

where:

hi(x) and hi'(x) are the i-th Hermite basis polynomials, which are defined as:

hi(x) = [1 – 2(x – xi)li'(xi)]li²(x)

hi'(x) = (x – xi)li²(x)

where:

li(x) is the i-th Lagrange basis polynomial that satisfies li(xi) = 1 and li(xj) = 0 for j ≠ i

The first term in the polynomial p(x) interpolates the function values yi at xi, and the second term interpolates the derivative values yi’ at xi.

The Hermite interpolation formula is useful in cases where we not only need to interpolate a function but also its derivatives at a given set of points.

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

Let’s consider the following set of data:

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

and the corresponding derivatives:

y_{0}‘ = 2, y1′ = 4, y2’ = 6

We want to find a polynomial p(x) that satisfies:

p(0) = 1, p'(0) = 2

p(1) = 3, p'(1) = 4

p(2) = 5, p'(2) = 6

To apply the Hermite interpolation formula, we first need to compute the Lagrange basis polynomials li(x) for i = 0, 1, 2. These are:

l_{0}(x) = ((x – 1)(x – 2))/2

l_{1}(x) = -(x(x – 2))

l_{2}(x) = ((x – 1)x)/2

Next, we need to compute the Hermite basis polynomials hi(x) and hi'(x) for i = 0, 1, 2. These are:

h_{0}(x) = (1 – 2x)l0²(x)

h_{0}‘(x) = x l0²(x)

h_{1}(x) = (1 – 2(x – 1))l1²(x)

h_{1}‘(x) = (x – 1) l1²(x)

h_{2}(x) = (1 – 2(x – 2))l2²(x)

h_{2}‘(x) = x – 2 l2²(x)

Now, we can apply the Hermite interpolation formula to obtain p(x):

p(x) = y_{0} h_{0}(x) + y_{0′} h_{0′}(x) + y_{1} h_{1}(x) + y_{1}‘ h_{1}‘(x) + y_{2} h_{2}(x) + y_{2′} h_{2′}(x)

Substituting the values for the data and derivatives, we get:

p(x) = 1*(1 – 2x)(x – 2)² + 2x*(x – 2)² + 3*(-x(x – 2)²) – 4(x – 1)²(x – 2) + 5*((x – 1)x)² – 6(x – 1)x(x – 2)

Simplifying, we get:

p(x) = -x³ + 4x² – 3x + 1

Therefore, the polynomial that interpolates the given data and derivatives is:

p(x) = -x³ + 4x² – 3x + 1

**Describe the Divided Differences**

Divided differences are a recursive method used to interpolate a set of data points using a polynomial. They are used to find the coefficients of the polynomial in an efficient way, by computing differences between the data points and using those differences to construct the coefficients. The divided differences can be calculated using a table called the divided difference table.

The divided difference table is constructed by arranging the data points in a diagonal pattern. The first column of the table contains the y-values of the data points. The second column contains the differences between adjacent y-values. The third column contains the differences between adjacent values in the second column, and so on. The divided differences in each column are used to compute the coefficients of the polynomial.

Once the divided difference table has been constructed, the coefficients of the polynomial can be found using the formula:

f[x_{0}, x_{1}, …, xn] = Δ_{^0 }y_{0} / 0! + Δ_{^1} y_{0},1 / 1! (x – x_{0}) + Δ_{^2 }y_{0},1,2 / 2! (x – x_{0}) (x – x1) + … + Δn y_{0},1,2,…,n / n! (x – x_{0}) (x – x_{1}) … (x – xn-1)

where f[x_{0}, x_{1}, …, xn] is the polynomial that interpolates the data points (x_{0}, y_{0}), (x_{1}, y_{1}), …, (xn, yn), Δ^{0} y_{0} = y_{0}, Δ^{1} y_{0},1 = (y_{1} – y_{0}) / (x_{1} – x_{0}), Δ^{2 }y_{0},1,2 = (Δ_{^1} y1,2 – Δ_{^1} y_{0},1) / (x_{2} – x_{0}), and so on. The coefficients of the polynomial are the divided differences in the first row of the divided difference table.

Divided differences are useful because they allow for the interpolation of data with arbitrary intervals between the data points, whereas other interpolation methods may require the intervals to be evenly spaced.

**Differentiate between Finite Differences and Divided Differences**

Finite differences and divided differences are both methods used in numerical analysis for approximating the derivatives of a function or for interpolating a set of data points using a polynomial. However, there are some key differences between these two methods:

- Definition: Finite differences refer to the differences between values of a function calculated at equally spaced points. Divided differences, on the other hand, refer to the differences between values of a function calculated at any arbitrary points.
- Calculation: Finite differences are calculated by taking the difference between values of a function at two equally spaced points, while divided differences are calculated recursively by taking the difference between values of a function at any two points.
- Application: Finite differences are often used for numerical differentiation, while divided differences are primarily used for polynomial interpolation.
- Interpolation: Finite differences can only be used for interpolation when the data points are equally spaced. Divided differences, on the other hand, can be used for interpolation with any set of data points.

In summary, finite differences are used for numerical differentiation and interpolation of equally spaced data points, while divided differences are primarily used for polynomial interpolation with arbitrary data points.

**Recall the Divided Difference Table**

The divided difference table is a table used to compute the divided differences of a set of data points. It is typically used in polynomial interpolation to find the coefficients of the polynomial.

The table has two columns: the first column contains the x-values of the data points, while the second column contains the corresponding y-values. The remaining columns contain the divided differences, which are calculated recursively using the following formula:

f[x_{0}, x_{1}] = (f(x_{1}) – f(x_{0})) / (x_{1} – x_{0})

f[x_{0}, x_{1}, …, x_{n}] = (f[x_{1}, x_{2}, …, x_{n}] – f[x_{0}, x1, …, n-1]) / (x_{n} – x_{0})

where f[x_{0}, x_{1}] represents the divided difference of the function f(x) at the points x_{0} and x_{1}.

The table is constructed by computing the divided differences recursively for each set of points, starting from the first-order divided differences and working up to higher orders. The diagonal entries of the table represent the function values at the corresponding x-values, while the entries in the last column represent the coefficients of the interpolating polynomial.

Here’s an example of a divided difference table for the data points (1, 2), (2, 5), (4, 3), and (5, 1):

x | f(x) | f[x, x+1] | f[x, x+1, x+2] | f[x, x+1, x+2, x+3] |

1 | 2 | 3 | -1 | -0.5 |

2 | 5 | -1 | -2 | |

4 | 3 | -0.5 | ||

5 | 1 |

To use this table for polynomial interpolation, we can use the coefficients in the last column to construct the interpolating polynomial. For example, using the table above, the interpolating polynomial for the given data points can be written as:

f(x) = 2 + 3(x-1) – (x-1)(x-2) – 0.5(x-1)(x-2)(x-4)

Note that this polynomial passes through all four data points, and is of degree 3, since there are four data points. ** **

**Recall Newton’s Divided Difference Method of Interpolation**

Newton’s divided difference method of interpolation is a numerical technique used to construct a polynomial that passes through a given set of data points. It is similar to Lagrange interpolation, but instead of using Lagrange basis polynomials, it uses divided differences to construct the interpolating polynomial. The method can be applied to both equally spaced and unequally spaced data points.

The divided difference of order k of a set of data points (x_{0}, y_{0}), (x_{1}, y_{1}), …, (x_{k}, y_{k}) is defined recursively as:

f[x_{i}] = y_{i}, for i = 0, 1, …, k

f[xi, xi+1, …, xi+j] = (f[xi+1, xi+2, …, xi+j] – f[xi, xi+1, …, xi+j-1]) / (xi+j – xi), for j = 1, 2, …, k, and i = 0, 1, …, k-j.

where f[xi, xi+1, …, xi+j] is the divided difference of order j for the data points xi, xi+1, …, xi+j.

Using these divided differences, the interpolating polynomial can be written in Newton’s form as:

Pn(x) = f[x_{0}] + (x-x_{0})f[x_{0},x1] + (x-x_{0})(x-x1)f[x_{0},x_{1},x_{2}] + … + (x-x_{0})(x-x_{1})…(x-xn-1)f[x_{0},x_{1},…,x_{n}]

where f[xi, xi+1, …, xj] is the divided difference of order j-i for the data points xi, xi+1, …, xj.

Newton’s divided difference method of interpolation can be used to find an approximate value of a function at a point x outside the given set of data points, by evaluating the interpolating polynomial at x.

**Apply Newton Divided Difference Method of Interpolation for the given set of Data**

Suppose we have the following set of data points:

(x_{0}, y_{0}) = (1, 2)

(x_{1}, y_{1}) = (3, 5)

(x_{2}, y_{2}) = (5, 1)

(x_{3}, y_{3}) = (7, 6)

To apply Newton’s divided difference method of interpolation, we first need to calculate the divided differences. We start by calculating the first-order divided differences:

f[x_{0}, x_{1}] = (y_{1} – y_{0}) / (x1 – x0) = (5 – 2) / (3 – 1) = 1.5

f[x_{1}, x_{2}] = (y_{2} – y_{1}) / (x2 – x1) = (1 – 5) / (5 – 3) = -2

f[x_{2}, x_{3}] = (y_{3} – y_{2}) / (x3 – x2) = (6 – 1) / (7 – 5) = 2.5

Next, we calculate the second-order divided differences:

f[x_{0}, x_{1}, x_{2}] = (f[x_{1}, x_{2}] – f[x_{0}, x_{1}]) / (x_{2} – x_{0}) = (-2 – 1.5) / (5 – 1) = -0.875

f[x_{1}, x_{2}, x_{3}] = (f[x_{2}, x_{3}] – f[x_{1}, x_{2}]) / (x_{3} – x_{1}) = (2.5 – (-2)) / (7 – 3) = 1.125

Finally, we calculate the third-order divided difference:

f[x_{0}, x_{1}, x_{2}, x_{3}] = (f[x_{1}, x_{2}, x_{3}] – f[x_{0}, x_{1}, x_{2}]) / (x_{3} – x_{0}) = (1.125 – (-0.875)) / (7 – 1) = 0.25

Using these divided differences, we can write the interpolating polynomial in Newton’s form:

P(x) = 2 + 1.5(x – 1) – 0.875(x – 1)(x – 3) + 0.25(x – 1)(x – 3)(x – 5)

To approximate the value of the function at a point x = 4, we simply evaluate the polynomial at that point:

P(4) = 2 + 1.5(4 – 1) – 0.875(4 – 1)(4 – 3) + 0.25(4 – 1)(4 – 3)(4 – 5) = 3.875

Therefore, the approximate value of the function at x = 4 is 3.875.

**Describe Gregory-Newton Forward Interpolation Formula**

Gregory-Newton Forward Interpolation Formula is a method for finding the approximate value of a function at some point within the range of the given data points. It is an extension of the forward difference method of interpolation.

The formula is given as follows:

f(x) ≈ f(x_{0}) + (x − x_{0})Δf(x_{0}) + (x − x_{0})(x − x1)Δ^{2}f(x_{0}) + ⋯ + (x − x_{0})(x − x_{1})⋯(x − xn−1)Δnf(x0)

where:

- f(x) is the value of the function being interpolated at the point x
- x
_{0}, x_{1}, …, x_{n}are the given data points - Δf(x
_{0}) = f(x_{1}) – f(x_{0}), Δ^{2}f(x_{0}) = Δf(x_{1}) – Δf(x_{0}), and so on, are the forward differences of the function at x0 - Δnf(x0) is the nth forward difference of f(x) at x
_{0}.

This formula is based on the idea that the forward differences of a function can be used to construct a polynomial that approximates the function. The polynomial is constructed using a table of forward differences, where each column is the difference of the previous column. This table is known as the forward difference table.

The advantage of the Gregory-Newton Forward Interpolation Formula is that it only requires knowledge of the values of the function at the beginning of the interval, while the Lagrange interpolation formula requires knowledge of the function at all points in the interval.

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

Suppose we have the following set of data points:

x_{0} = 0, f(x_{0}) = 2

x_{1} = 1, f(x_{1}) = 5

x_{2} = 2, f(x_{2}) = 12

x_{3} = 3, f(x3) = 23

We want to use the Gregory-Newton Forward Interpolation Formula to approximate the value of f(1.5).

First, we construct the forward difference table:

x f(x) Δf Δ^{2}f Δ^{3}f

0 2 3 3 1

1 5 6 4

2 12 8

3 23

We can see that the first column corresponds to the given data points, and the subsequent columns are the forward differences. The Δf column is the first forward difference, the Δ^2f column is the second forward difference, and so on.

Using the formula, we have:

f(1.5) ≈ f(x_{0}) + (x – x_{0})Δf(x_{0}) + (x – x_{0})(x – x1)Δ^{2}f(x_{0}) / 2! + (x – x_{0})(x – x1)(x – x2)Δ^{3}f(x_{0}) / 3!

Substituting the values, we get:

f(1.5) ≈ 2 + (1.5 – 0) * 3 + (1.5 – 0)(1.5 – 1) * 4 / 2! + (1.5 – 0)(1.5 – 1)(1.5 – 2) * 1 / 3!

f(1.5) ≈ 2 + 4.5 + 1.5 + 0.25

f(1.5) ≈ 8.25

Therefore, using the Gregory-Newton Forward Interpolation Formula, we have approximated the value of f(1.5) to be 8.25.

**Describe Gregory-Newton Backward Interpolation Formula**

Gregory-Newton Backward Interpolation Formula is a method used to find the approximate value of a function at a point x based on a set of discrete data points. It is similar to the forward interpolation formula, but instead of using the forward differences, it uses the backward differences.

The formula for Gregory-Newton Backward Interpolation is given by:

f(x) = y_{n} + Δy_{n},p / p! – Δ^{2} y_{n}, p-1 / (p-1)! + Δ^{3} y_{n}, p-2 / (p-2)! – … + (-1)^{p} Δ^{p} y_{n},0 / p!

where y_{n} is the last value in the given set of data points, Δy_n,p is the nth backward difference of order p, and Δ^{k} y_{n},p is the kth difference of the nth backward difference of order p.

To use this formula, we first need to calculate the backward differences of the given set of data points. We can do this using the divided difference table. Once we have the backward differences, we can substitute them in the above formula to find the approximate value of the function at the desired point x.

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

Suppose we have the following set of data points:

x | y |

0.0 | 1.000 |

0.2 | 1.221 |

0.4 | 1.491 |

0.6 | 1.822 |

0.8 | 2.225 |

We want to find the value of y at x = 0.1 using the Gregory-Newton Backward Interpolation formula.

First, we need to calculate the backward differences of the given set of data points:

x | y | Δy | Δ^{2}y |
Δ^{3}y |
Δ^{4}y |

0.8 | 2.225 | ||||

0.6 | 1.822 | 0.403 | |||

0.4 | 1.491 | 0.331 | 0.072 | ||

0.2 | 1.221 | 0.270 | 0.061 | 0.011 | |

0.0 | 1.000 | 0.221 | 0.049 | 0.012 | -0.001 |

Using this table, we can apply the Gregory-Newton Backward Interpolation formula to find the approximate value of y at x = 0.1. Let p = 3 since we need to use the third order backward difference.

f(0.1) = y_{n} + Δy_{n},p / p! – Δ^{2} y_{n}, p-1 / (p-1)! + Δ^{3} y_n, p-2 / (p-2)!

f(0.1) = 2.225 + Δy_{0},3 / 3! – ^{2} y_{0},2 / 2! + Δ3 y_{0},1 / 1!

f(0.1) = 2.225 + (-0.001) / 6 – 0.012 / 2 + 0.011 / 1

f(0.1) = 2.225 – 0.00016667 – 0.006 + 0.011

f(0.1) = 2.229833

Therefore, the approximate value of y at x = 0.1 using Gregory-Newton Backward Interpolation formula is 2.229833.

**Describe Gauss’s Forward Formula of Interpolation**

Gauss’s Forward Formula of Interpolation is a method used to approximate the value of a function at some point within an interval, given a set of data points within that interval. The formula involves constructing a forward-difference table and using the coefficients of the first few terms to calculate the desired value.

The formula is given by:

f(x + kh) = f(x) + kΔf(x) + k(k-1)Δ²f(x)/2! + k(k-1)(k-2)Δ³f(x)/3! + …

where:

- f(x) is the value of the function at x
- k is the number of intervals forward of x
- h is the size of the interval
- Δf(x), Δ²f(x), Δ³f(x), … are the first, second, third, … forward differences of the function at x.

To use Gauss’s Forward Formula, we first construct a forward-difference table, similar to that used in the previous interpolation methods. Once we have the table, we can use the first few coefficients to calculate the value of the function at a desired point within the interval.

One potential disadvantage of this method is that it requires the values of higher-order differences, which may not be readily available or may be subject to significant rounding error. Additionally, the formula may converge slowly for some functions, requiring many terms to achieve accurate results.

**Apply Gauss’s Formula of Interpolation for the given set of Data**

Gauss’s Formula of Interpolation is given by:

f(x) ≈ f(x0) + pf1(x)

where p is the product of the differences between x and the data points, divided by the product of the differences between x0 and the data points. The first forward difference table is used to find the values of pf1(x).

Let’s apply Gauss’s Formula of Interpolation to the following set of data, where we want to approximate f(0.5):

x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 |

f(x) | 1.221 | 1.4918 | 1.6651 | 1.8273 | 1.9820 |

We will take x0 = 0.1 as the starting point, and calculate the first forward difference table:

x | f(x) | 1st FD | 2nd FD | 3rd FD | 4th FD |

0.1 | 1.221 | 0.2708 | 0.0783 | -0.0157 | |

0.2 | 1.4918 | 0.3574 | 0.0626 | ||

0.3 | 1.6651 | 0.4201 | |||

0.4 | 1.8273 | 0.5447 | |||

0.5 | 1.982 |

We can now use the formula to approximate f(0.5):

f(0.5) ≈ f(0.1) + p * f1(0.1,0.2,0.3,0.4,0.5)

where p = (0.5 – 0.1) / ((0.2 – 0.1)(0.3 – 0.1)(0.4 – 0.1)(0.5 – 0.1)) = 75.6

and f1(0.1,0.2,0.3,0.4,0.5) = 0.2708 + 0.3574(0.5 – 0.1) + 0.4201(0.5 – 0.2)(0.5 – 0.1) + 0.5447(0.5 – 0.3)(0.5 – 0.2)(0.5 – 0.1) = 0.9863

Therefore, f(0.5) ≈ 1.221 + 75.6 * 0.9863 = 78.98.

**Describe Gauss’s Backward Formula of Interpolation**

Gauss’s Backward Formula of Interpolation is used to find the value of a function at a point x0, based on a set of equally spaced data points with a step size h. The formula is given by:

f(x0) = y[n] + u*Δy[n] + u(u-1)*Δ²y[n] + u(u-1)(u-2)*Δ³y[n] + …

where:

- f(x0) is the value of the function at x0 that we want to find.
- y[n] is the last data point in the table.
- u = (x0 – xn) / h is the normalized distance between x0 and the last data point xn, where h is the step size between data points.
- Δy[n] = y[n] – y[n-1], Δ²y[n] = Δy[n] – Δy[n-1], Δ³y[n] = Δ²y[n] – Δ²y[n-1], and so on, are the finite differences of the data points.

Note that the formula involves only the backward differences, unlike the forward formula which involves only the forward differences.

Gauss’s Backward Formula is useful when we need to interpolate a function at a point that is closer to the end of the interval of the given data points.

To apply Gauss’s Backward Formula of Interpolation, we need to have a set of equally spaced data points and the value of x0 that we want to interpolate. We also need to compute the backward differences of the data points. Then, we can plug the values into the formula and compute the interpolated value of the function at x0.

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

Gauss’s Backward Formula of Interpolation is used to approximate the value of a function at a given point using a set of equally spaced data points in the backward direction. The formula is given as:

f(x) = y[n] + Δy[n-1, n](x – x[n]) + Δ^{2}y[n-2, n](x – x[n])(x – x[n-1]) + Δ^{3}y[n-3, n](x – x[n])(x – x[n-1])(x – x[n-2]) + …

where y[n] is the last value in the data set, Δy[n-1, n] is the first divided difference, Δ^{2}y[n-2, n] is the second divided difference, and so on.

Let’s apply Gauss’s Backward Formula of Interpolation to the following data set:

x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 |

f(x) | 1.2 | 1.6 | 1.8 | 1.9 | 2.0 |

We want to approximate f(0.15).

First, we need to compute the divided differences:

Δy[4, 5] = f(0.5) – f(0.4) = 2.0 – 1.9 = 0.1

Δy[3, 4] = f(0.4) – f(0.3) = 1.9 – 1.8 = 0.1

Δy[2, 3] = f(0.3) – f(0.2) = 1.8 – 1.6 = 0.2

Δy[1, 2] = f(0.2) – f(0.1) = 1.6 – 1.2 = 0.4

Next, we can use the formula to approximate f(0.15):

f(0.15) = y[n] + Δy[n-1, n](x – x[n]) + Δ^{2}y[n-2, n](x – x[n])(x – x[n-1]) + Δ^{3}y[n-3, n](x – x[n])(x – x[n-1])(x – x[n-2]) + …

= f(0.5) + Δy[4, 5](0.15 – 0.5) + Δ^{^2}y[3, 5](0.15 – 0.5)(0.15 – 0.4) + Δ^{3}y[2, 5](0.15 – 0.5)(0.15 – 0.4)(0.15 – 0.3)

= 2.0 + 0.1(-0.35) + 0.0333(-0.35)(-0.25) + 0.00444(-0.35)(-0.25)(-0.15)

= 1.7343

Therefore, the approximate value of f(0.15) using Gauss’s Backward Formula of Interpolation is 1.7343.

**Describe Stirling’s Formula of Interpolation**

Stirling’s Formula of Interpolation is another method of interpolation that is used to approximate the value of a function at a point based on a set of given data points. It is a modification of Gauss’s Forward Interpolation Formula and is used when the given data points are evenly spaced.

Stirling’s Formula is given by:

f(x) ≈ f(x0) + Δf(Δx/2) + (Δx^{2}/8) Δ^{2}f(Δx/2) + (Δx^{2}/8)(Δx^{2}/6) Δ^{3}f(Δx/2) + …

Where,

Δx = x1 – x0 = x2 – x1 = … = xn – xn-1 (constant spacing)

Δf(x) = f(x) – f(x-Δx)

Δ^{2}f(x) = Δf(x) – Δf(x-Δx)

Δ^{3}f(x) = Δ^{2}f(x) – Δ^{2}f(x-Δx)

and so on.

In this formula, the first term is the value of the function at x0, and the second term is the forward difference of the function at x0 with a step size of Δx/2. The third term involves the second order divided difference of the function at x0 with a step size of Δx/2, and so on.

Stirling’s Formula is useful because it can be used to approximate the value of a function at a point even when the point is not one of the given data points, as long as the spacing between the data points is constant.

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

Stirling’s formula of interpolation is used to interpolate a value at a point x within a given set of data points. The formula is given by:

f(x) = y_{0} + [(x-x_{0})/h]Δy_{0} + [(x-x_{0})(x-x1)/h^{2}]Δ^{2}y_{0} + … + [(x-x_{0})(x-x1)…(x-xn-1)/hn]Δ^{n}-1y_{0}

where:

- x
_{0}, x_{1}, …, x_{n}are the data points - y
_{0}, y_{1}, …, y_{n}are the corresponding function values - h is the step size between the data points
- Δy
_{0}, Δ^{2}y0, …, Δ^{n-1}y_{0}are the divided differences of the function values

To apply Stirling’s formula, we first need to compute the divided differences of the function values. The divided differences can be calculated using the divided difference table. For example, let’s consider the following set of data:

x_{0} = 0, x1 = 1, x_{2} = 2, x_{3} = 3

y_{0} = 1, y1 = 2, y_{2} = 3, y_{3} = 5

Using the divided difference table, we can calculate the divided differences as follows:

Δy_{0} = y_{1} – y_{0} = 2 – 1 = 1

Δ^{2}y_{0} = (y_{2} – y_{1}) – (y_{1 }– y_{0}) = (3 – 2) – (2 – 1) = 0

Δ^{3}y_{0} = ((y_{3} – y_{2}) – (y_{2} – y_{1})) – ((y_{2} – y_{1}) – (y_{1} – y_{0})) = ((5 – 3) – (3 – 2)) – ((3 – 2) – (2 – 1)) = 1

Now, using Stirling’s formula, we can interpolate the value of the function at x = 1.5 as follows:

f(1.5) = y0 + [(1.5-0)/1]Δy0 + [(1.5-0)(1.5-1)/1^{2}]Δ^{2}y0 + [(1.5-0)(1.5-1)(1.5-2)/1^{3}]Δ^{3}y0

= 1 + 1(1.5-0) + 0(1.5-0)(1.5-1)/2 + 1(1.5-0)(1.5-1)(1.5-2)/6

= 1 + 1.5 + 0 + 0.125

= 2.625

Therefore, the interpolated value of the function at x = 1.5 is approximately 2.625.

**Describe Numerical Differentiation**

Numerical differentiation is a technique used to approximate the derivative of a function at a given point using numerical methods. The derivative of a function at a point is defined as the limit of the slope of a tangent line to the function at that point. However, in many cases, it is not feasible to evaluate the limit analytically, especially for complex functions or when the function is only known through a set of discrete data points.

Numerical differentiation approximates the derivative of a function at a point by using finite difference formulas that involve evaluating the function at nearby points. The most common finite difference formulas used for numerical differentiation are the forward, backward, and central difference formulas.

The forward difference formula estimates the derivative of a function at a point using the function values at the current and next points, while the backward difference formula uses the function values at the current and previous points. The central difference formula uses the function values at the current point and at points equidistant on either side of the current point.

Numerical differentiation is widely used in scientific computing, engineering, and other fields that require the analysis of complex functions or systems. It is useful in applications such as numerical optimization, curve fitting, and solving differential equations.

**Apply the concept of Numerical differentiation to find the derivatives at a point from the given set of data**

Here’s an example of how to apply numerical differentiation to find the derivative of a function at a point using a set of data:

Suppose we have the following set of data points for the function f(x) = x^{2} – 3x + 2:

x | f(x) |

0.0 | 2.00 |

0.5 | 1.375 |

1.0 | 0.00 |

1.5 | -0.625 |

2.0 | -1.00 |

We want to find the derivative of f(x) at x = 1.5 using numerical differentiation.

One approach is to use the central difference method, which approximates the derivative using values of the function on either side of the point of interest:

f'(x) ≈ [f(x+h) – f(x-h)] / (2h)

where h is the step size. In this case, since we want to approximate the derivative at x = 1.5, we can take h to be, say, 0.5. Then we have:

f'(1.5) ≈ [f(2.0) – f(1.0)] / (2*0.5) = (-1.00 – 0.00) / 1.0 = -1.00

So the approximation to f'(1.5) using the central difference method with h=0.5 is -1.00.

**Describe Newton-Cotes Formulas**

Newton-Cotes formulas are numerical integration methods that approximate the definite integral of a function over a given interval by using equally spaced points and weights. These formulas are based on the idea of approximating the integrand by a polynomial and integrating the polynomial exactly.

The most common Newton-Cotes formulas are the trapezoidal rule, Simpson’s 1/3 rule, and Simpson’s 3/8 rule.

The trapezoidal rule uses linear interpolation to approximate the integral, and the formula is:

∫(a to b) f(x)dx ≈ (b-a) * [f(a) + f(b)] / 2

Simpson’s 1/3 rule uses quadratic interpolation to approximate the integral, and the formula is:

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

Simpson’s 3/8 rule uses cubic interpolation to approximate the integral, and the formula is:

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

These formulas are useful for approximating integrals of functions that are difficult or impossible to integrate analytically. The accuracy of the approximations depends on the number of points used and the degree of the polynomial used for interpolation.

**Deduce the Trapezoidal Rule From Newton-Cotes Formulas**

The Trapezoidal Rule is a numerical integration technique that is derived from the Newton-Cotes formulas.

The Newton-Cotes formulas are a family of numerical integration techniques that use polynomial interpolation to approximate the value of a definite integral. The Trapezoidal Rule is a specific case of the Newton-Cotes formulas, where the integral is approximated by a linear polynomial.

To derive the Trapezoidal Rule, we start with the first-order Newton-Cotes formula, which uses a linear polynomial to approximate the integrand over the interval [a, b]. The formula is given by:

f(x) dx ≈ (b-a)/2 [f(a) + f(b)]

We can see that this formula is essentially the area of a trapezoid with bases f(a) and f(b) and height (b-a)/2. Hence, the formula is called the Trapezoidal Rule.

The error of the Trapezoidal Rule can be derived using Taylor’s theorem, and it can be shown that the error is proportional to (b-a)^3 times the second derivative of f evaluated at some point c in the interval [a, b].

**Describe Simpson’s 1/3rd Rule**

Simpson’s 1/3rd Rule is a numerical integration technique used to approximate the value of a definite integral. It approximates the integrand by a quadratic polynomial over each subinterval of the integration interval and then integrates the polynomial to approximate the integral.

The formula for Simpson’s 1/3rd Rule is as follows:

∫(a to b) f(x)dx ≈ (b-a)/6 * (f(a) + 4f((a+b)/2) + f(b))

where f(a), f((a+b)/2) and f(b) are the values of the integrand at the endpoints and midpoint of the interval [a, b] respectively.

The rule is called 1/3rd Rule because it uses every third interval to approximate the value of the integral. If the number of intervals is odd, the last interval is approximated using the Trapezoidal Rule.

Simpson’s 1/3rd Rule provides a more accurate approximation than the Trapezoidal Rule for smoother functions with less oscillations.

**Apply the Simpson’s 1/3**^{rd} rule to find the approximate value of the integral

^{rd}rule to find the approximate value of the integral

To apply Simpson’s 1/3 rule, we first need to have the data points of the function. Let’s say we have the following data points:

x = [0, 1, 2, 3, 4, 5]

y = [1, 2.7183, 7.3891, 20.0855, 54.5982, 148.4132]

To find the approximate value of the integral of this function using Simpson’s 1/3 rule, we can use the following formula:

Integral ≈ (h/3) * [y[0] + 4sum(y[1:n:2]) + 2sum(y[2:n-1:2]) + y[n]]

where,

- h = (b – a)/n is the step size, and here a=0, b=5, and n=5 is the number of segments.
- y[0], y[n] are the values of y at the first and last points respectively.
- y[1:n:2] is a slice of y starting from the second element and taking every other element up to n-1.
- y[2:n-1:2] is a slice of y starting from the third element and taking every other element up to n-2.

Substituting the values, we get:

h = (5-0)/5 = 1

Integral ≈ (1/3) * [1 + 4*(2.7183 + 54.5982 + 148.4132) + 2*(7.3891 + 20.0855)]

≈ 143.661

Therefore, the approximate value of the integral using Simpson’s 1/3 rule is 143.661.

**Describe Simpson’s 3/8**^{th} Rule

^{th}Rule

Simpson’s 3/8th rule is a numerical integration technique used to approximate the value of a definite integral. It is an extension of the Simpson’s 1/3rd rule and is used when the number of intervals is a multiple of 3.

The formula for Simpson’s 3/8^{th} rule is given as:

∫[a,b]f(x)dx ≈ (3h/8) [f(a) + 3f(a + h) + 3f(a + 2h) + 2f(a + 3h) + f(b)]

where:

- h = (b – a) / 3
- f(a), f(a + h), f(a + 2h), f(a + 3h), and f(b) are the values of the function f(x) at the equally spaced points a, a + h, a + 2h, a + 3h, and b, respectively.

Note that Simpson’s 3/8th rule requires the number of intervals between a and b to be a multiple of 3. If the number of intervals is not a multiple of 3, then the trapezoidal rule can be used to approximate the remaining interval.

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

Simpson’s 3/8th Rule is a numerical method used for approximating definite integrals. It is based on approximating the integrand using a cubic polynomial and then integrating this polynomial over the given interval.

The formula for Simpson’s 3/8th Rule is:

∫[a,b]f(x)dx ≈ (3h/8) [f(a) + 3f(a + h) + 3f(a + 2h) + 2f(a + 3h) + f(b)]

where:

- h = (b – a) / 3
- f(a), f(a + h), f(a + 2h), f(a + 3h), and f(b) are the values of the function f(x) at the equally spaced points a, a + h, a + 2h, a + 3h, and b, respectively.

To apply Simpson’s 3/8th Rule, we need to divide the given interval into sub-intervals of equal width. In this case, we divide the interval [1, 2] into three sub-intervals of width h = (2 – 1)/3 = 1/3.

Then, we apply Simpson’s 3/8th Rule formula to each of these sub-intervals and sum up the results to get the approximate value of the integral.

Let’s assume that we have the following values of the function f(x) at the endpoints and midpoints of the sub-intervals:

f(1) = 0.5, f(4/3) = 0.1, f(5/3) = 0.2, f(2) = 0.3

Then, using Simpson’s 3/8th Rule, we have:

\int_{1}^{2} f(x)dx \approx \frac{1}{3}\left[f(1) + 3f\left(\frac{4}{3}\right) + 3f\left(\frac{5}{3}\right) + f(2)\right]

\approx \frac{1}{3}\left[0.5 + 3(0.1) + 3(0.2) + 0.3\right]

\approx 0.3667

Therefore, the approximate value of the integral is 0.3667.

**Describe he Gaussian One-Point, Two-Point, and Three-Point Formula**

Gaussian integration formulas are a set of numerical integration techniques that use weighted sums of function values at specific points within the interval of integration to approximate the definite integral of the function. The Gaussian formulas are named after the mathematician Carl Friedrich Gauss, who developed them.

The Gaussian One-Point Formula uses a single point within the interval of integration to approximate the definite integral. The formula is as follows:

f(x)dx ≈ f((a+b)/2) * (b-a)

The Gaussian Two-Point Formula uses two points within the interval of integration to approximate the definite integral. The formula is as follows:

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

The Gaussian Three-Point Formula uses three points within the interval of integration to approximate the definite integral. The formula is as follows:

f(x)dx ≈ (b-a)/2 * [5/9f((a+b)/2) + 8/9f((a+b)/2 – (b-a)/(2√15)) + 8/9*f((a+b)/2 + (b-a)/(2√15))]

The advantage of Gaussian integration formulas over other numerical integration techniques is that they can achieve a high degree of accuracy using relatively few function evaluations. However, they require the evaluation of the function at specific points within the interval of integration, which can be computationally expensive for highly oscillatory or singular functions.

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

Let’s consider an example to apply the Gaussian One-Point, Two-Point, and Three-Point Formula to find the definite integral:

Example:

Estimate the value of the integral ∫[0,1] e^x dx using Gaussian One-Point, Two-Point, and Three-Point Formula.

Solution:

First, let’s calculate the exact value of the integral:

∫[0,1] e^{x} dx = e^1 – e^0 = e – 1

Now, let’s apply the Gaussian formulas to approximate the value of the integral:

- Gaussian One-Point Formula:

The formula for the Gaussian One-Point Formula is:

∫[-1,1] f(x) dx ≈ f(0) * w0

where w0 = 2

We need to transform the given integral from [0,1] to [-1,1] by the following substitution:

t = 2x – 1 => dt/dx = 2 => dx = dt/2

∫[0,1] e^{x} dx = ∫[-1,1] e^{(t+1)}/2 * 1/2 dt

Now, we can use the Gaussian One-Point Formula as follows:

∫[-1,1] e^{(t+1)}/2 * 1/2 dt ≈ e * w0 / 2 = e

Therefore, the approximation of the integral using the Gaussian One-Point Formula is e.

- Gaussian Two-Point Formula:

The formula for the Gaussian Two-Point Formula is:

∫[-1,1] f(x) dx ≈ (1/2) [ f(-1/√3) * w0 + f(1/√3) * w1 ]

where w0 = w1 = 1

Again, we need to transform the given integral from [0,1] to [-1,1] by the substitution:

t = 2x – 1 => dt/dx = 2 => dx = dt/2

∫[0,1] e^{x} dx = ∫[-1,1] e^{(t+1)/}2 * 1/2 dt

Now, we can use the Gaussian Two-Point Formula as follows:

∫[-1,1] e^{(t+1)}/2 * 1/2 dt ≈ (1/2) [ e^{(√3/2)} + e^{(-√3/2)} ]

Therefore, the approximation of the integral using the Gaussian Two-Point Formula is approximately 2.04.

3. Gaussian Three-Point Formula:

The formula for the Gaussian Three-Point Formula is:

∫[-1,1] f(x) dx ≈ (5/9) [ f(-√(3/5)) * w0 + f(0) * w1 + f(√(3/5)) * w2 ]

where w0 = w2 = 5/9 and w1 = 8/9

Once again, we need to transform the given integral from [0,1] to [-1,1] by the substitution:

t = 2x – 1 => dt/dx = 2 => dx = dt/2

∫[0,1] e^{x} dx = ∫[-1,1] e^{(t+1)}/2 * 1/2 dt

Now, we can use the Gaussian Three-Point Formula as follows:

∫[-1,1] e^{(t+1)}/2 * 1/2 dt ≈ (5/9) [ e^{(-√(3/5)}) * f(-√(3/5)) + 8/9 * f(0) + 5/9 * e^{(√(3/5)}) * f(√(3/5)) ]

where f(x) = e^{(x+1)}/2 * 1/2

Substituting the function f(x) in the above formula, we get

∫[-1,1] e^{(t+1)}/2 * 1/2 dt ≈ (5/9) [ e^{(-√(3/5)}) * e^{(-√(3/5)}+1)/2 * 1/2 + 8/9 * e^{(0+1)}/2 * 1/2 + 5/9 * e^{(√(3/5)}+1)/2 * 1/2 ]

Simplifying this expression, we get

∫[-1,1] e^{(t+1)}/2 * 1/2 dt ≈ 0.6839

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

Numerical integration, or quadrature, is the process of approximating the definite integral of a function using a finite number of function evaluations. There are several quadrature formulas available, each designed to integrate a specific class of functions. Four of the most commonly used quadrature formulas are:

i. Gauss-Legendre Formula: This formula is designed to integrate a wide range of functions over the interval [-1, 1]. It uses the roots of the Legendre polynomial to determine the nodes and weights of the formula. The Gauss-Legendre formula is particularly useful for functions with sharp peaks or singularities within the integration interval.

ii. Gauss-Chebyshev Formula: This formula is designed to integrate functions over the interval [-1, 1] that can be written in the form f(x) = (1/sqrt(1 – x^{2})) g(x), where g(x) is a smooth function. The Gauss-Chebyshev formula uses the roots of the Chebyshev polynomial to determine the nodes and weights of the formula.

iii. Gauss-Hermite Formula: This formula is designed to integrate functions of the form f(x) e^{(-x2)} over the entire real line. The Gauss-Hermite formula uses the roots of the Hermite polynomial to determine the nodes and weights of the formula. It is particularly useful for functions that are rapidly decaying and have a Gaussian shape.

iv. Gauss-Laguerre Formula: This formula is designed to integrate functions of the form f(x) e^{(-x)} over the interval [0, ∞). The Gauss-Laguerre formula uses the roots of the Laguerre polynomial to determine the nodes and weights of the formula. It is particularly useful for functions that are rapidly decaying and have an exponential shape.

Here are some examples of each of the quadrature formulas:

i. Gauss-Legendre Formula:

Consider the integral ∫[0,1] x^{2} dx. We can use the Gauss-Legendre formula with n=2 to approximate this integral. The roots and weights for n=2 are given by:

x_{1} = -1/√3, x2 = 1/√3

w_{1} = w_{2} = 1

Using these roots and weights, we have:

∫[0,1] x^{2} dx ≈ (1/2) [f(x1) + f(x2)]

= (1/2) [f(-1/√3) + f(1/√3)]

= (1/2) [(1/3) + (1/3)]

= 1/3

So, the Gauss-Legendre formula with n=2 gives an approximation of 1/3 for the integral ∫[0,1] x^{2 }dx.

ii. Gauss-Chebyshev Formula:

Consider the integral ∫[-1,1] e^{x} dx. We can use the Gauss-Chebyshev formula with n=2 to approximate this integral. The roots and weights for n=2 are given by:

x_{1} = -√(1/2), x_{2} = √(1/2)

w_{1 }= w_{2} = π/2

Using these roots and weights, we have:

∫[-1,1] e^{x} dx ≈ (π/2) [f(x1) + f(x2)]

= (π/2) [f(-√(1/2)) + f(√(1/2))]

= (π/2) [e^{(-√2)} + e^{√2}]

≈ 3.336

So, the Gauss-Chebyshev formula with n=2 gives an approximation of 3.336 for the integral ∫[-1,1] e^{x} dx.

iii. Gauss-Hermite Formula:

Consider the integral ∫[-∞,∞] x^{2} e^{(-x2)} dx. We can use the Gauss-Hermite formula with n=2 to approximate this integral. The roots and weights for n=2 are given by:

x1 = -√(2), x2 = √(2)

w1 = w2 = π/2

Using these roots and weights, we have:

∫[-∞,∞] x^{2} e^{(-x2)} dx ≈ (π/2) [f(x1) + f(x2)]

= (π/2) [f(-√2) + f(√2)]

≈ 1.764

So, the Gauss-Hermite formula with n=2 gives an approximation of 1.764 for the integral ∫[-∞,∞] x^{2} e^{(-x2) }dx.

iv. Gauss-Laguerre Formula:

Consider the integral ∫[0,∞) x^{2} e^{(-x)} dx. We can use the Gauss-Laguerre formula with n=2 to approximate this integral. The roots and weights for n=2 are given by:

x1 = 0.5858, x2 = 3.4142

w1 = 0.8536, w2 = 0.1464

Using these roots and weights, we have:

∫[0,∞) x^{2} e^{(-x)} dx