A textbook of Computer Based Numerical and Statiscal Techniques part 9 docx
A textbook of Computer Based Numerical and Statiscal Techniques part 9 docx
... 66 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Putting i = 2 in (1) x 3 = 0.024 39 Therefore reciprocal of 41 is 0.0244. Example 8. Find the square root of 20 correct to 3 decimal places ... 1.22 39 x 2 = φ(x 1 ) = 1 + () () () () −+− 2345 2222 (1.22 39) (1.22 39) (1.22 39) (1.22 39) 2! 3! 4! 5! + = 1.3263 70 COMPUTER BASED NUMERICAL AND STATISTICAL TECHN...
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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 ... 13.28 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...
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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 ... 423. 593 75 Hence No. of Persons earning wages between Rs. 60 to 70 is 423. 593 75 – 370 = 53. 593 75 or 54000. (Approx.) 186 COMPUTER BASED NUMERICAL AND STATISTICAL TE...
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A textbook of Computer Based Numerical and Statiscal Techniques part 31 docx
... 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 2 89 Example 1. Obtain cubic spline for every subinterval, ... DIFFERENTIATION The method of obtaining the derivatives of a function using a numerical technique is known as numerical differentiation. There are essentially two si...
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A textbook of Computer Based Numerical and Statiscal Techniques part 34 docx
... 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. Divide the interval [0,6] into 6 equal parts each of ... 0. 192 7 0.5820 0 .96 94 6(0.3777) 10 = + + ++++ = 0.075[1.7441 + 5(0.8 392 ) +6(0.3777)] 316 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.5 TRAPEZOIDAL RULE Putting...
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A textbook of Computer Based Numerical and Statiscal Techniques part 36 docx
... obtain () 10 00 , yyhfxy =+ () 1 0.01 1 0 .99 =+ − = () 21 11 , yyhfxy =+ () 0 .99 0.01 0 .99 0 .98 01 =+ − = () 3 0 .98 01 0.01 0 .98 01 0 .97 0 299 y =+− = () 4 0 .97 0 299 0.01 097 0 299 096 0 596 y =+−= Hence, ... approximation of the error, we have. 13 218 295 15 x ≤ .00005. Taking logarithm, we obtain 15 log x ≤ log .00005 218 295 13 afa f or x ≤ .98 8. 340 COMP...
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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
... 0.000202 Similarly, Absolute Error E a = E r X ≤ 0.000202 × 299 446 ≈ 60 So, true value of the product of the given numbers lies between 299 446 – 60 = 299 386 And 299 446 + 60 = 299 506. Hence the mean of ... 0.000333 0.00 099 9 0.333333 a E x == and Percentage Error, E p = 100 0.00 099 9 100 0. 099 % r E ×= ×= . Example 5. Round-off the number 75462 to four significant...
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