basic components of computer based measurement and control system

... 186 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Take the mean of equation (1) and (2) 22 (0) (1) (0) ( 1) (1) (1) () (0) 22 2!2 ff f ... = 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 ... Gauss forward formula to find the value of U 9 , if u(0) = 14, u(4) = 24, u(8) = 32, u (12) = 35, u(16) = 40 [Ans. 33.1162109] 192 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 5. Apply Gauss

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... 0 = 1, x 1 = 2.5, x 2 = 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) 230 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. ... 3), (8, 1), (9, 1) and (10, 9). Find the value of y for x = 9.5 using Lagrange’s interpolation formula. Sol. We are given () :78910 :311 9 x fx 234 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ... 226 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 1. Using Lagrange’s formula, find the value of (i) y x if y 1 = 4, y 3 = 120, y 4 =

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... y corresponding to argument value of x (for equal and unequal spaced argument). On the other hand the process of Estimating the value of x for a entry value of y (which is not in the table) is ... 0.1541 and fifth approximation is u 5 = 0.1542 Hence u 4 ≈ u 5 (correct to 3 places of decimal) We have to find cube root of 10. Since 10 lies between the value of y corresponding to x =2 and ... 244 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 10. Show that the sum of Lagrangian co-efficients is unity. Sol. Let () x ∏ = () () () 01 n xx xx xx −− − The reciprocal of

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... three days, 30 days apart round the pds. of summer and winter. Estimate the app. dates and values of max. and min. temperature. 254 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Summer Winter ... = x,x x ,x x,x , x xx − − 248 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES which gives, [][ ] () [] 001 101 x,x x ,x x x x,x ,x =+− (2) From (1) and (2), [] ()( ) () [] 0 0 01 0 ... of (x – 1), we use synthetic division method as, −+11 1 3 8 103 11 0 3 11 11 11 1 4 1 12 x 3 – x 2 + 3x + 8 = (x – 1) 3 + 2 (x – 1) 2 + 4 (x – 1) + 11. Ans. 250 COMPUTER BASED NUMERICAL AND

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... (x)y′ i (2) Where u i (x) and vi(x) are polynomials in x of degree ≤ 2n + 1 and satisfy. (i) u i (x i ) = 0, 1, ij ij ≠   =  (3a) 260 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ... = 1 – 4 x 264 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES L 1 (x)= 0 10 0 40 4 xx x x xx −− == −− ∴ L 0 ′ (x)= () 1 11 and 44 Lx ′ −= Hence, L 0 ’ (x 0 )= – 1 4 and L 1 ’ (x 1 ) ... (3a) and (3b) in (5), we get a i x + b i = 1 (6a) and c i x + d i = 0(6b) Since [L i (x i )] 2 = 1 Again, using conditions (3c) and (3d) in (5), we get a i + 2L i ′ (x i ) = 0 (6c) and c

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... (x), as well as to express the powers of x in terms of the Chebyshev polynomials. Some of the Chebyshev polynomials and the expansion for powers of x in terms of T n (x) are given as follows: T ... substantial error near the ends of the interval 272 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (x near ± 1). In particular, it seems reasonable to look for other sets of simple related functions ... f 2 f n be the values of given function and 12 , n φφ φ be the corresponding values of the approximating function. Then the error vector is e where the components of e are given by e i =

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... ++ −≤≤ −≤++≤∈ and Hence P n (x) is a good uniform approximation to ()fx in which the number of terms retained depends on the given tolerance of ∈ . However for a large number of functions, ... error after two terms of the series is 1 0.00052 1920 ≤= . Thus we have sin x ≈ () () 13 169 5 192 128 Tx Tx − Now, to get the economized series, we put basic values of T 1 and T 3 sin x ≈ () ... 0.1562x 3 Which gives sin x to 3 significant digit accuracy and therefore, it is the economized series. 278 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 6. Prove that () () ()

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... most n . 3. S(x) and its (n –1) derivatives are continuous on [a, b]. 4. S(x) is a polynomial of degree one for x < a and x > b. The process of constructing such type of polynomial is called ... . (11) and after using a i–1 , b i–1 , and c i–1 S’ (x i )= 3 [] () () [] 11 2 11 1 11 2. 2 62 6 i i ii i ii i i ii Sx Sx h PP h Ph P P hh − −− − + −+ + −− 288 COMPUTER BASED NUMERICAL AND STATISTICAL ...  290 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ⇒ () () () 123 1 2 3 462 MMM fx fxfx ++= − +   but M 0 = 0 M 3 then it becomes 12 1 2 412418 MM andM M+= = = M 1 = 2 and M

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... 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 ... fxdx y y y y y + + =++++ ∫ and so on. 318 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Adding all these integrals from x 0 to x 0 + nh, where n is a multiple of 4, we get 0 0 012345678 ... taking 1 6 h = Hence compute an approximate value of π in each case. 320 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. (i) The values of 2 1 () 1 fx x = + at 123 0, , , ,1 444 x =

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... 326 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES [] 2(10) 7(0) 32(4) 12(7) 32(9) 7(12) 32(15) 12(14) 32(8) 7(3) 45 = +++++ + ++ = 708 Hence the required area of the cross-section of ... () () () 212 720 hh fxdx fx f x f x h ′′′′ =−+− ∫ (2) 328 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Between limits x = x 0 and x = x n from equation (2), we have {}{} 0 000 11 () ... 0.0490291] GGG +0)26-4 % Numerical Solution of Ordinary Differential Equation 7.1 INTRODUCTION In the fields of Engineering and Science, we come across physical and natural phenomena which, when represented

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... 338 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 7. Find the solution of dy =1+ xy, y(0) = 1 dx which passess through (0, 1) in the interval (0, 0.5) such that the value of ... 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 x in the neighbourhood of x = 0, when 22 , ... Picard’s method of successive approximations. y=y 0 + fxydx x x , bg 0 z . (1) Sol. The first approximation y 1 of y is obtained by substituting y = 1 2 In the right hand member of (1) i.e.,

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... present book Computer 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 all Engineering ... feel no difficulty to understand the subject A unique feature of this book is to provide with an algorithm and computer program in Clanguage to understand the steps and methodology used in writing ... presented in a very systematic and logical manner In each chapter, all concepts, definitions and large number of examples in the best possible way have been discussed in detail and lucid manner...

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... computation, it is to COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES be cut-off to a manageable size such as 0.29, 0.286, 0.2857, etc The process of cutting off super-flouts digits and retaining ... reliability of the numerical result will depend on an error estimate or bound, therefore the analysis of error and the sources of error in numerical methods is also a critically important part of the ... known as rounding off a number or we can say that process of dropping unwanted digits is called rounding-off Number are rounded-off according to the following rules: To round-off the number to...

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... ∆ABC, a = 2.3 cm, b = 5.7 cm and ∠ B = 90o If possible errors in the computed value of b and a are mm and mm respectively, find the possible error in the measurement of angle A Sol Given δb = mm ... 31 The error in the measurement of the area of a circle is not allowed to exceed 0.1% How accurately should the diameter be measured? = Sol Let d is the diameter of a circle, and then its area ... 8.5 cm and A = 45o, find allowable errors in b, c, and A such that the area of ∆ABC may be determined nearest to a square centimeter Sol Let area of the ∆ABC be given by, = X = bc sin A 18 COMPUTER...

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... numbers of digits If we assume computer memory store digits in each location and also store one or more signs then to represent real number, computer assumed a fixed position for the decimal point and ... division of mantissa of the numerator by that of the denominator and denominator exponent is subtracted from the numerator exponent The resultant exponent is obtained by adjusting it appropriately and ... value of the number a = 4.568 and b = 6.762 using the four-digit  b − a arithmetic and compare the result by taking c = a +     Sol Since a = 4568el , b = 6762e1 and c be the middle value of...

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... iterations, the root of f ( x) = x3 − x − = up to three places of decimals is 1.325, which is of desired accuracy Example Find the root of the equation x3 – x – = between and to three places of decimal ... 2(0.5) − = −1 , which is negative Thus f(0.5) is negative and f(1) is positive Then the root lies between 0.5 and 40 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The ... ALGEBRAIC AND TRANSCENDENTAL EQUATION Here ei and ei + are the errors in ith and (i + 1)th iterations respectively Comparing the above equation with lim i →∞ ei +1 e k i ≤A We get k = and A = 0.5...

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...  0.10203  Since, x3 and x4 are approximately the same upto four places of decimal, hence the required root of the given equation is 1.4036 54 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ... real root of the equation f(x) = x3 – 2x – = by the method of false position up to three places of decimal Sol Given that f(x) = x3 – 2x – = So that f(2) = (2)3 – 2(2) – = – 50 COMPUTER BASED NUMERICAL ... (α ) In order to find the order of convergence, it is necessary to find a formula of the type ei + = Ae k with an appropriate value of k i .(6) With the help of (6), we can write ei = Ae k or...

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... (or Rate) of Convergence of Newton-Raphson Method Let α be the actual root of equation f(x) = i.e., f(a) = Let xn and xn+1 be two successive approximations to the actual root α If en and en+1 ... is small, we can neglect second and higher degree terms in h and therefore, we get f(x0) + hf ′ (x0) = 68 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES f ( x0 ) From which we have, h = – ... PROBLEM SET 2.3 Use the method of Iteration to find a positive root between and of the equation xex – [Ans 0.5671477] Find the Iterative method, the real root of the equation 3x – log10 x = correct...

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... + 240 = 5( 2.99) 78 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 4( 238.977) + 240 = 2.9925 399.627 Hence, the required value of (240)1/5 correct to three places of decimal is 2.993 ... process when the prescribed accuracy is obtained 80 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 2.8.2 Rate or Order of Convergence of Secant Method On substituting xn = ỵ + ån , etc in ... lemmas (1) 81 ALGEBRAIC AND TRANSCENDENTAL EQUATION Lemma 2.1: Under the assumptions and notations of the theorem: f ′′( x∞ ) Ek Ek f ′( x∞ ) −1 Proof Using the definition of xk+1, we find Ek +...

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... extract the quadratic factors that are the products of the pairs of 92 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES complex roots, and then complex arithmetic can be avoided because such ... approximated values of p and q Since R and S are both functions of the two parameters p and q then the improved values are given by R(p + ∆p, q + ∆q) = .(4) S(p + ∆p, q + ∆q) = Expand equation (4) ... values of p and q First approximation: Let p0 and q0 be the initial approximations, then the first approximation can be obtained by p1 = p0 + ∆p and q1 = q0 + ∆q Because given equation is of the...

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