A textbook of Computer Based Numerical and Statiscal Techniques part 5 ppsx
A textbook of Computer Based Numerical and Statiscal Techniques part 5 ppsx
... 0 .55 55e1, b = 0. 454 5e1, c = 0. 453 5e1 then (b – c) = 0.0010e1 = 0.1000e – l a( b – c) = (0 .55 55e1) × (0.1000e – 1) = (0. 055 5e0) = 0 .55 50e – 1 ab = (0 .55 55e1) × (0. 454 5e1) = 0. 252 4e2 ac = (0 .55 55e1) ... roots of the equation, obtained by x 1 = 2 22 44 and 22 bb ac bb ac aa −+ − −− − =x These are called closed form solution. Similar formulae are also available for cubic...
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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
... 0.00 05 because − ×= 3 1 10 0.00 05 2 14 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ∂ ×100 u u = − δ× − 5 6 12 5 100 25 V V VV = − ××=−×=− − (12 5) 7 0. 05 100 5 11.667% (2 5) ... second decimal places. Example 4. If 0.333 is the approximate value of 1 3 , then find its absolute, relative and percentage errors. Sol. Given that True value () 1 3...
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A textbook of Computer Based Numerical and Statiscal Techniques part 14 ppsx
... 122 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 22. Evaluate: ∆ n [sin (ax + b)] Sol. We know f∆ (x) = f(x + h) – f (x) therefore ∆ sin (ax + b) = sin [a (x + h)+b] – sin)(ax ... hD hD hD 116 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 14. Evaluate the following: I. ∆ 2 (cos 2x) II. ∆ 2 (3e x ) III. ∆ tan –1 x IV. ∆(x + cos x) the inte...
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A textbook of Computer Based Numerical and Statiscal Techniques part 16 ppsx
... given a set of equidistant values of arguments and its corresponding value of f(x). Suppose for n + 1 equidistant argument values x = a, a + h, a + 2h, , a + nh, are given. 140 COMPUTER BASED NUMERICAL ... 0f−+×−×+= 138 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES i.e., 4f(3) = 124 i.e., f(3) = 31 (function values are 3 n type and this is not a pol...
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A textbook of Computer Based Numerical and Statiscal Techniques part 17 ppsx
... R.H.S. 3.9 FACTORIAL NOTATIONS The product of n consecutive factors each at a constant difference and the first factor being x is called a factorial function or a factorial polynomial of degree n and ... L.H.S. 150 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES CorCor CorCor Cor ollary:ollary: ollary:ollary: ollary: To show that, () n n x ∆ = n! h n and 1n+ ∆...
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A textbook of Computer Based Numerical and Statiscal Techniques part 25 ppsx
... − 2 .5 1 0 .5 4 1 0 .5 4 2 .5 0 .5 4 2 .5 1 4 7 .5 13 17 .5 1 .5 3 4 .5 1 .5 1 .5 3 3 1 .5 1 .5 4 .5 3 1 .5 f (5) = 5 15 65 1 75 20. 25 6. 75 6. 75 20. 25 −++ = 0.246913 – 2.2222 + 9.62962 + 8.6419 75 f (5) = 18 .51 850 831 ... 4, and x 3 = 5. 5 f(x 0 ) = 4, f(x 1 ) = 7 .5, f(x 2 ) = 13 and f(x 3 ) = 17 .5 also, find the value of f (5) 232 COMPUTER BASE...
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A textbook of Computer Based Numerical and Statiscal Techniques part 26 ppsx
... . 65 .39 .342 .39 .423 .39 .5 30 35 .5 .342 .5 .423 .5 . 65 . 65 .342 . 65 .423 . 65 .5 −−− −−− + −−− − − − = 22.84 057 797. Ans. 238 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. sin 6 π ... using Lagrange’s inverse interpolation formula. () 0 5 10 15 16. 35 14.88 13 .59 12.46 x fx [Ans. 8.337] 242 COMPUTER BASED NUMERICAL AND STATISTICAL...
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A textbook of Computer Based Numerical and Statiscal Techniques part 29 ppsx
... co-efficients a 1 , a 2 a n will produce little error for small x near zero. But probably substantial error near the ends of the interval 268 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply ... approximation to a given function. For evaluating a function f(x) on a computer it is generally more efficient of space and time to have an analytic approxi...
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A textbook of Computer Based Numerical and Statiscal Techniques part 51 ppsx
... 0 .52 7 1.407 0 .54 5 1. 455 3. 858 1.710 6.006 0.443 1 .55 7 24 0.612 0.632 0. 157 0.9684 0 .53 8 1.399 0 .55 5 1.4 45 3.3 95 1. 759 6.031 0. 452 1 .54 8 25 0.600 0.619 0. 153 0.9696 0 .54 8 1.392 0 .56 5 1.4 35 3.931 ... juice drained from cans immediately after filling for 20 samples are taken by a random method (at an interval of every 30 minutes). Each of the samples includes...
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