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 – 1ab = (0 .55 55e1) × (0. 454 5e1) = 0. 252 4e2ac = (0 .55 55e1) ... roots of the equation, obtained byx1= 22244 and 22bb ac bb acaa−+ − −− −=xThese are called closed form solution. Similar formulae are also available for cubic and biquadratic polynomial ... of the normalized floating-point representation theassociative and the distributive laws of arithmetic are not always valid. The example given belowproves the above statement:Let a = 0 .55 55e1,...
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A textbook of Computer Based Numerical and Statiscal Techniques part 2 ppsx
... problems. A majoradvantage 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 figurethan that given in the data, perform the last operation, and then round off.3. ... in numerical methods is also a critically important part of the study of numerical technique.1.2 ACCURACY OF NUMBERS(i) Exact Number: Number with which no uncertainly is associated to no approximation...
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A textbook of Computer Based Numerical and Statiscal Techniques part 3 ppsx
... 0.00 05 because −×=3110 0.00 05 214 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ∂×100uu= −δ×− 5 612 5 100 25 VVVV= −××=−×=−−(12 5) 70. 05 100 5 11.667%(2 5) ... seconddecimal places.Example 4. If 0.333 is the approximate value of 13, then find its absolute, relative and percentageerrors.Sol. Given that True value ()13x=, and its Approximate value ... 11 11 (1)12 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 20. It is required to obtain the roots of X2 – 2X + log10 2 = 0 to four decimal places. Towhat accuracy should log10...
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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 hDhD116 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 14. Evaluate the following:I. ∆2 (cos 2x)II. ∆2 (3ex)III. ∆ tan–1 xIV. ∆(x + cos x)the interval of differencing ... f(x)∴∆(e a + bx)= e a( x + 1)+b – eax + b = eax + b (e a – 1)∴∆2 (e a + bx)= ∆ (∆e a + bx) = ∆ {eax + b (e a –1)}= (e a – 1) (∆ eax + b)= (e a – 1) eax + b (e a –1)= (e a –1)2...
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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). Supposefor 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) = 124i.e., f(3) = 31(function values are 3n type and this is not a polynomial)Example 7. Find the missing value of the data:12 ... 2212nnnan bn c n a n b n a n−−++−∆+∆+∆142 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 9. Evaluate the production of wool in the year 19 35 from the given data: ( ) 1931...
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A textbook of Computer Based Numerical and Statiscal Techniques part 17 ppsx
... R.H.S.3.9 FACTORIAL NOTATIONSThe product of n consecutive factors each at a constant difference and the first factor being x iscalled a factorial function or a factorial polynomial of degree n and ... L.H.S. 150 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES CorCorCorCorCorollary:ollary:ollary:ollary:ollary:To show that,()nnx∆= n! hn and 1n+∆ x(n) = 0We know that,()nx∆= ... 1 and 2 by 1 and add them to get the sum –1 which is to bewritten below 3. The remainder –10 is the value of D.4. Add the terms of corresponding columns of (a) and get 2, –1 and 2 of (b) 5. ...
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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 14 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 x3 = 5. 5f(x0) = 4, f(x1) = 7 .5, f(x2) = 13 and f(x3) = 17 .5 also, find the value of f (5) 232 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 8. Prove that Lagrange’s ... −f( 35) = – 10 .5 + 50 .4 75 + 42. 05 + 4.72f( 35) = 77.4 05 (ii) By using Lagrange’s formula, we havef(x) = ()()()()()()()()( )()()( )2 .54 5 .5 1 45. 547 .5 12 .51 4 15. 5 2 .51 2 .54 2 .55 .5 xxx xxx−−−...
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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. sin6π ... using Lagrange’s inverse interpolation formula.()0 5 10 15 16. 35 14.88 13 .59 12.46xfx[Ans. 8.337]242 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 7. Find the value of x correct ... life. [Ans. 15. 67]13. Find the value of tan 33° by Lagrange’s formula if tan 30° = 0 .57 74, tan 32° = 0.6249, tan 35 = 0.7002, tan 38° = 0.7813. [Ans. 0.64942084]14. Apply Lagrange’s formula to...
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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 littleerror for small x near zero. But probably substantial error near the ends of the interval268 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 timeto have an analytic approximation to f(x) rather than to store a table ... polynomial. Therefore all the interpolation formulae shouldgive the same functional value.274 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 2. Obtain the best lower degree approximation...
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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 takenby a random method (at an interval of every 30 minutes). Each of the samples includes 4 cans. Thesamples are tabulated ... concern has an average sale of Rs. 10000/- daily estimated over a longperiod. A salesman claims that he will increase the average sales by Rs. 700/- a day. The concernis interested in an increased...
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