A textbook of Computer Based Numerical and Statiscal Techniques part 34 docx
... +6(0.3777)] 316 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.5 TRAPEZOIDAL RULE 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 ... 1.3661 7341 3. Ans. 324 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 6. Evaluate the integral 6 3 0 1 dx x+ ∫ by using Weddle’s rule. Sol. Div...
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... 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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... 66 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Putting i = 2 in (1) x 3 = 0.02439 Therefore reciprocal of 41 is 0.0244. Example 8. Find the square root of 20 correct to 3 decimal places ... = x 3 up to four decimal places. So we have 1 2 = 3.4641. 72 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = () () 10 0. 4343 2.8133 1.2 2.4128 log 2.8133 0. 4343...
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A textbook of Computer Based Numerical and Statiscal Techniques part 21 docx
... 0.047875 190 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 4. Use Gauss’s forward formula to find a polynomial of degree four which takes the following values of the function f(x): x 1 2345 f(x) ... No. of Persons earning wages between Rs. 60 to 70 is 423.59375 – 370 = 53.59375 or 54000. (Approx.) 186 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUE...
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A textbook of Computer Based Numerical and Statiscal Techniques part 31 docx
... DIFFERENTIATION The method of obtaining the derivatives of a function using a numerical technique is known as numerical differentiation. There are essentially two situations where numerical differentiation ... Newton forward formula, and if the same is required at a point near the end of the set of given tabular 294 INTERPOLATION WITH UNEQUAL INTERVAL 289 Example 1. Obta...
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A textbook of Computer Based Numerical and Statiscal Techniques part 36 docx
... = . Ans. Here in this example, only I approximation can be obtained and so it gives that approximate value of y for 0.1x= Example 12. Find the series expansion that gives y as a function of ... .00005. Taking logarithm, we obtain 15 log x ≤ log .00005 218295 13 afa f or x ≤ .988. 340 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 10. Obtain y when x = 0.1,...
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A textbook of Computer Based Numerical and Statiscal Techniques part 60 docx
... INTERPOLATION 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 ... 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 Ste...
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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 ... 4008 and 30.4647 + 0.0639 = 30.5286. Example 12. Find the product of 346 .1 and 865.2 and state how many figures of the result are trustworthy, given that the numbers are corr...
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