A textbook of Computer Based Numerical and Statiscal Techniques part 24 pdf
... advantage of these formulas is, they can also be used in case of equal intervals but the formulae for equal intervals cannot be used in case of unequal intervals. 5.2 LAGRANGE’S INTERPOLATION FORMULA Let ... interpolation have the disadvantages of being applicable only to equally spaced argument values. So it is required to develop interpolation formulae for unequally spaced argument...
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... polynomials, if the coefficients are all real valued then the complex roots occurs in conjugate pair. Therefore we extract the quadratic factors that are the products of the pairs of 88 COMPUTER BASED ... default polynomial. If the initial approximation of p and q are not known then the last three terms of given polynomial a n – 2 x 2 + a n–1 x + a n = 0 can be used to...
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... given intermediate or argument values of x, three types of differences are useful: 104 96 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now to obtain the values of b i and C i we use ... roots of cubic equations. The disadvantage of this method is that additional computation is also necessary. This method can be applied to find the complex roots and multiple ro...
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A textbook of Computer Based Numerical and Statiscal Techniques part 44 pdf
... coefficient of correlation? [Ans. 0.4r = ] 7. x and y are two random variables with the same standard deviation and correlation coefficient r. Show that the coefficient of correlation between x and ... Similarly, (2) may be proved. 424 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM SET 9.2 1. Find the equation of the lines of regression on the basi...
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A textbook of Computer Based Numerical and Statiscal Techniques part 48 pdf
... causes of variation. There are some advantages, when a manufacturing process is operating in a state of statistical control. 1. The important use and advantage of SQC is the control, maintenance and ... It primarily aims at the isolation of the chance and assignable causes of variation and consequently helps in the detection, identification and elimination of the as...
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A textbook of Computer Based Numerical and Statiscal Techniques part 61 pdf
... METHOD Step 1. Start of the program to interpolate the given data Step 2. Input the value of n (number of terms) Step 3. Input the array ax for data of x Step 4. Input the array ay for data of y Step ... value of y6 – 3.4903 Enter the value of y7 – 3.6693 594 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES float p, q; float x, y = 0; float nr, dr; float diff[20][20...
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A textbook of Computer Based Numerical and Statiscal Techniques part 62 docx
... the Range Lower limit a = 0 Upper limit b = 6 Enter the number of subintervals = 6 Value of the integral is: 1.3571 598 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Enter the value of y2 ... ALGORITHM FOR TRAPEZOIDAL RULE Step 1. Start of the program for numerical integration Step 2. Input the upper and lower limits a and b Step 3. Obtain the number of sub...
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A textbook of Computer Based Numerical and Statiscal Techniques part 1 ppt
... This page intentionally left blank This page intentionally left blank A TEXTBOOK OF COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Anju Khandelwal M.Sc., Ph.D. Department of Mathematics SRMS ... Based Numerical and Statistical Techniques is primarily written according to the unified syllabus of Mathematics for B. Tech. II year and M.C .A. I year students of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 2 ppsx
... problems. A major advantage for numerical technique is that a numerical answer can be obtained even when a problem has no analytical solution. However, result from numerical analysis is an approximation, in ... significant figures at each step of computation. At each step of computations, retain at least one more significant figure than that given in the data, perform the last...
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A textbook of Computer Based Numerical and Statiscal Techniques part 3 ppsx
... ERRORS AND FLOATING POINT 9 Example 11. Find the relative error in calculation of 7.342 0 .241 . Where numbers 7.342 and 0 .241 are correct to three decimal places. Determine the smallest interval ... 16(0.001) = 0.036 Also, Max. Relative Error = = 0.036 0.009 4 (Because E r(max) = δ ; u u u = 4 at x = y = z = 1). 10 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES T...
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