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13.26 ALGORITHM FOR LAGRANGE’S INTERPOLATION METHOD. Start of the program to interpolate the given data Step 2. Input the number of terms n. End of the program.. 13.27 PROGRAMMING FOR LAGRANGE’S INTERPOLATION METHOD. printf(“enter the no. scanf(“%d”, &. printf(“enter the value in the form of x–”);. printf(“enter the value of x%d”, i+1);. scanf(“%f”, &. printf(“enter the value in the form...
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C OMPUTER B ASED N UMERICAL AND. C OMPUTER B ASED N UMERICAL. Anju Khandelwal M.Sc., Ph.D.. Department of Mathematics SRMS College of Engineering &. Technical University, Lucknow. M.Sc., Ph.D.. Department of Mathematics St. Copyright © 2009, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved.. No part of this ebook may...
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The reliability of the numerical result will depend on an error estimate or bound, therefore the analysis of error and the sources of error in numerical methods is also a critically important part of the study of numerical technique.. For example, each number contains four significant figures while the numbers contains only three significant figures since zeros only help to...
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This form is useful where error in dependent variable is given and also we are to find errors in both independent variables.. if a number is correct to n decimal places. Also Relative error is less than 1 1. if number is correct to n significant digits and l is the first significant digit of a number.. 1.4.6 Error in...
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Percentage Error in r . On substituting r = 4.5 and value of δR from (1). Percentage Error in h = δ. Find the possible error in the area of a triangle if the error in sides is correct to a millimeter and the angle is measured correct to one degree.. Assume that the area of the triangle ABC. X...
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This means that the floating point result fl (a op b) of the exact operation (a op b) is the nearest floating point number to (a op b), breaking ties by rounding to the floating point number whose bottom bit is zero (the “even” one). When the result of floating point operation is not representable as a normalized floating point...
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Then, we bisect the interval and continue the process till the root is found to be desired accuracy. therefore, the root lies in between a and x 1 . The second approximation to the root now is x 2 = 1. 2.4.1 Procedure for the Bisection Method to Find the Root of the Equation f (x. Find the root of...
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Find the smallest root lying in the interval (1, 2) up to four decimal places for the equation. x 6 – x 4 – x 3 – 1 = 0 by Bisection Method [Ans. Find the smallest root of x 3 – 9x + 1 = 0, using Bisection Method correct to three decimal. Find the real root of e...
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Thus, the root lies between 1.0352 and 1.5. Second approximation: The next approximation to the root is x 3 = x 0 – 1 0 0. Thus, the root lies between 1.0456 and 1.5. Third approximation: The next approximation to the root is 4 0 1 0 0. 0.0071 Thus, the root lies between 1.0487 and 1.5.. Then the root...
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Putting i = 2 in (1) x Therefore reciprocal of 41 is 0.0244.. Find the square root of 20 correct to 3 decimal places by using recursion formula. Therefore correct to three decimal places.. Find the smallest root of the equation. Omitting x 2 and higher powers of x, we get x = 1 approximately.. Values of x 7 and...
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The first approximation is given by. Since x 3 = x 4 hence the required root is 2.798 correct to three places of decimal.. Find the real root of the equation x = e –x using the Newton-Raphson’s method.. We have f(x. Hence the required root is 0.5671 correct to 4 decimal places.’. Using the starting value 2(1 + i),...
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Thus, the approximation value to the root is –1.93375, correct up to five decimals.. Even if all coefficients of a non-linear equation are real, the equation can have complex roots. Which can be solved using the methods discussed in previous section.. Find all roots of the equation f(x. Since the second and third approximations are same for five decimals hence...
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Now to obtain the values of b i and C i we use the following procedure:. Here from the table b b C 1 = –3, C 2 = 4.75, C 3 = 6 therefore after substituting the values of b i and C i in equations (1) and (2) we get. Therefore the first approximation are given by. After...
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Construct a forward difference table for the following values:. Forward difference table for given data is:. If y = x 3 + x 2 – 2x + 1, calculate values of y for x and form the difference table. Therefore, difference table for these data is as:. From the difference table and find ∆ 5 f(0).. The difference table for...
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∆(x + cos x) the interval of differencing being h.. We have ∆ 2 (cos 2x. (E – 1) 2 cos 2x because. (E 2 – 2 E + 1) cos 2x. cos (2x + 4h. 2 cos (2x + 2h. cos 2x. cos (2x + 2h. 2 (cos h – 1) {sin (x + h. the interval of differencing...
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a{b cx b c – b cx. 1) b cx = a b ( c − 1) 2 b cx Proceeding in the same manner, we get. On taking h being the interval of differencing the difference table is as:. Find the value of. Form the forward difference table for given set of data:. Construct the difference table for the...
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3.7 TECHNIQUE TO DETERMINE THE MISSING TERM. Estimate the missing term in the following table:. (on taking interval of differencing being 1) On putting x = 0, we get. Substituting the value of f(0), f(1), f(2), f(4) in (1), we get f. Find the missing value of the data:. Substituting the value of f(0), f(1), f(2), f(4) in (1), we...
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On substitution x = 0, we get B = 5 and for x = 1, we get A = 20.. Therefore we have ∆ f x. 5 On integrating, we have. We have ∆ f x = x(x – 1. We have E n f(a. Substitute a = 0, n = x we have for h = 1. φ(x) is...
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Now, to obtain approximate values of y(x) at some points other than those given by (a), let y(x. x x − n ) and L is to be determined such that equation (b) holds for any intermediate values of x, say x = x′, x 0 <. i.e., F(x) vanishes (n + 2) times in the interval x 0. must...
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From Newton’s formula,. Find the value of sin 52 ° from the given table:. Now on applying Newton’s forward difference formula, we have. of students secured more than 45 marks but less than 48 marks Example 8. Use Newton’s forward difference formula to obtain the interpolating polynomial f(x) satisfying the following data:. Using Newton’s forward interpolation formula up to third...