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A textbook of Computer Based Numerical and Statiscal Techniques part 20

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To obtain first term, we use Newton’s forward interpolation formula, Here, a = 3, h = 1, x = 1 ∴ u = –2. Similarly, to obtain tenth term, we use Newton’s backward interpolation formula. given the following table:. Now on applying ‘Newton’s backward difference formula, we get. The area A of a circle of diameter d is given for...

A textbook of Computer Based Numerical and Statiscal Techniques part 21

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Gausss forward formula is given by. From the following table, find the value of e 1.17 using Gauss forward formula:. From the following table find y when x = 1.45.. Use Gauss’s forward formula to find a polynomial of degree four which takes the following values of the function f(x):. Gauss forward formula is. Use Gauss’s forward formula to find...

A textbook of Computer Based Numerical and Statiscal Techniques part 22

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If f(x) is a polynomial of degree four find the value of f(5.8) using Gauss’s backward formula from the following data:. From Gauss backward formula. Using Gauss backward interpolation formula, find the population for the year 1936.. Gauss backward formula is. Using Gauss’s backward formula show that . Find the value of cos 51°42 1 by Gauss’s backward formula from...

A textbook of Computer Based Numerical and Statiscal Techniques part 23

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Apply Stirling’s formula to find a polynomial of degree four which takes the values of y as given below:. Apply Stirling’s formula to interpolate the value of y at x = 1.91 from the following data:. Using Bessel’s formula find the value of y at x = 3.75 for the data given below:. Now from Bessel’s formula, we have. Find...

A textbook of Computer Based Numerical and Statiscal Techniques part 24

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By Everett’s formula,. Apply Laplace Everett’s formula to find the value of log 2375 from the data given below:. Find the value of e –x when x = 1.748 from the following data:. Here h = 0.01, take origin as 1.74.. Apply this formula to find the value of y 11 and y 16 if y 0 = 3010, y...

A textbook of Computer Based Numerical and Statiscal Techniques part 25

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Using Lagrange’s formula, find the value of (i) y x if y 1 = 4, y 3 = 120, y 4 = 340, y 5 = 2544,. (ii) y 0 if y –30 = 30, y –12 = 34, y 3 = 38, y 18 = 42, Sol. Now using Lagrange’s interpolation formula, we have. Now from Lagrange’s interpolation formula,...

A textbook of Computer Based Numerical and Statiscal Techniques part 26

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Certain corresponding values of x and log 10 x are given as. Find the log 10 301 by Lagrange’s formula. Find the weight of babies during 5 to 5.6 months of life. Find the value of tan 33° by Lagrange’s formula if tan 30. Apply Lagrange’s formula to find f(5) and f(6) given that f(2. (x – x n )...

A textbook of Computer Based Numerical and Statiscal Techniques part 27

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The second divided difference for x 0 , x 1 , x 2 is given by f(x 0 , x 1 , x 2. The third divided difference for x 0 , x 1 , x 2 , x 3 is given by. x 1 – x 0 = x 2 – x 1 = ...x n – x n–1...

A textbook of Computer Based Numerical and Statiscal Techniques part 28

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Using Newton’s divided difference formula, prove that f(x. the Newton’s divided difference formula is f(x. Using Newton’s divided difference formula, calculate the value of f(6) from the following data:. The divided difference table is. Applying Newton’s divided difference formula, f(x. Find the value of log 10 656 using Newton’s divided difference formula from the data given below:. Divided difference table...

A textbook of Computer Based Numerical and Statiscal Techniques part 29

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Hermite’s interpolation formula is. Apply Hermite formula to find a polynomial which meets the following specifications:. Apply Hermite’s interpolation to find f(1.05) given:. Apply Hermite’s interpolation to find log 2.05 given that:. Determine the Hermite polynomial of degree 5 which fits the following data and hence find an approximate value of log e 2.7. Compute e by Hermite’s formula for...

A textbook of Computer Based Numerical and Statiscal Techniques part 30

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Chebyshev polynomial approximation. (iii) Economization of power series: To describe the process of economization, which is essential due to Lanczos, we first express the given function as a power series in x. Thus we obtain the Chebyshev series expansion of the given continuous function f x. on the interval [–1, 1]. in which the number of terms retained depends on...

A textbook of Computer Based Numerical and Statiscal Techniques part 31

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5.7.4 Cubic Spline Interpolation for Equally and Unequally Spaced Values According to the idea of draftsman spline, it is required that both dy. The cubic spline have possess the following properties:. 3a i (x – x i ) 2 + 2b i (x – x i. 6a i (x – x i ) 4 + 2b i ...(5). 3a i...

A textbook of Computer Based Numerical and Statiscal Techniques part 32

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Applying equations (1), (2) and (3) for y 0, we get. 6.2.2 Derivatives Using Newton’s Backward Difference Formula Newton’s backward interpolation formula is given by. Differentiating both sides of equation (1) with respect to x, we get. Therefore putting u = 0 in (2), we get. x, we get. Therefore putting u = 0 in (3), we get. Applying these...

A textbook of Computer Based Numerical and Statiscal Techniques part 33

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Then, by Newton’s backward formula, we have. 5 from the following table:. So we use Newton’s divide difference formula.. Newton’s divided difference formula is given by:. (5) from the data given below:. Here, the arguments are not equally spaced and therefore we shall apply Newton’s divided difference formula.. Find f′(4) from the following data:. Though this problem can be solved...

A textbook of Computer Based Numerical and Statiscal Techniques part 34

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Putting n =1 in equation (2) and taking the curve y = f(x) through (x 0 , y 0 ) and (x 0 , y 0 ) as a polynomial of degree one so that differences of order higher than one vanish, we get. we get. Adding the above integrals, we get. 6.6 SIMPSON’S ONE-THIRD RULE. x y 2 2...

A textbook of Computer Based Numerical and Statiscal Techniques part 35

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Hence the required area of the cross-section of the river = 708 sq. (ii) By Simpson’s one-third rule,. Hence the required area of the cross-section of the river = 710 sq. Since the interval of integration is divided into an even number of subintervals, we shall use Simpson’s one-third rule.. By Simpson’s one-third rule. 6.10 EULER-MACLAURIN’S FORMULA. Again we have...

A textbook of Computer Based Numerical and Statiscal Techniques part 36

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7.3 PICARD’S METHOD OF SUCCESSIVE APPROXIMATIONS. Consider first order differential equation. The first approximate solution (approximation) y 1 of y is given by. Similarly, the second approximation y 2 is given by y 2 = y 0 + z x x 0 f x y dx b g , 1. for the nth aproximation y n is given by. Use...

A textbook of Computer Based Numerical and Statiscal Techniques part 37

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Find y(.05) and y(.1) correct to 6 decimal places.. Using Euler’s method, we obtain. We improve y 1 by using Euler’s modified method y 1. correct to 6 decimal places. Hence we take y i.e., we have y. Again, using Euler’s method, we obtain. We improve y 2 by using Euler’s modified method. correct to 7 decimal places. Hence, we...

A textbook of Computer Based Numerical and Statiscal Techniques part 38

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we get. 2, find y(0.1) and y(0.2) correct to 4 decimal places.. We have x 0 = 0, y 0 = 2, h = 0.1 Then, we get. 0.2 , we have x 0 = 0.1, y we get k 1 = hf x y ( 0 , 0. 1 and h = 0.2 on the interval [0.1] using Runge-Kutta...

A textbook of Computer Based Numerical and Statiscal Techniques part 39

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2e x – y at x = 0.4 and x = 0.5 by Milne’s method, given their values of the four points.. By Milne’ predictor formula, we have. Now again by Corrector formula, we get. By Corrector formula, we get. Apply Milne’s method to find a solution of the differential equation dy x y 2 dx. in the range 0...