# Results for: “Manish Goyal”

1 eBook

## ALLC4-9 |
Manish Goyal | Laxmi Publications | |||||

290 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ Proceeding in this manner, we get y = f(x) = y0 + (x – x0) [x0, x1] + (x – x0) (x – x1) [x0, x1, x2] + (x – x0) (x – x1) (x – x2) [x0, x1, x2, x3] + ..... + (x – x0) (x – x1) (x – x2) ..... (x – xn–1) [x0, x1, x2, x3, ......, xn] + (x – x0) (x – x1) (x – x2) ..... (x – xn) [x, x0, x1, x2, ......, xn] which is called Newton’s general interpolation formula with divided differences, the last term being the remainder term after (n + 1) terms. Newton’s divided difference formula can also be written as y = y0 + (x – x0) ∆| y0 + (x – x0) (x – x1) ∆| 2y0 + (x – x0) (x – x1) (x – x2) ∆| 3y0 + (x – x0) (x – x1) (x – x2) (x – x3) ∆| 4y0 + ..... + (x – x0) (x – x1) ..... (x – xn–1) ∆| ny0 4.31. RELATION BETWEEN DIFFERENCES DIVIDED DIFFERENCES AND ORDINARY Let the arguments x0, x1, x2, ....., xn be equally spaced such that x1 – x0 = x2 – x1 = ... = xn – xn–1 = h ∴ x1 = x0 + h x2 = x0 + 2h ......... xn = x0 + nh ∆| f(x0) = Now x1 ∆ f ( x0 ) f ( x1 ) − f ( x0 ) f ( x0 + h) − f ( x0 ) See All Chapters |
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## ALLC5-4 |
Manish Goyal | Laxmi Publications | |||||

360 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ (i) By Boole’s rule, z 2h [7y0 + 32y1 + 12y2 + 32y3 + 7y4 + 7y4 + 32y5 + 12y6 + 32y7 + 7y8] 45 2 (10) = [7(0) + 32(4) + 12(7) + 32(9) + 7(12) + 7(12) + 32(15) 45 + 12(14) + 32(8) + 7(3)] = 708 Hence the required area of the cross-section of the river = 708 sq. m. 1 (ii) By Simpson’s rd rule 3 80 h y dx = [(y0 + y8) + 4(y1 + y3 + y5 + y7) + 2(y2 + y4 + y6)] 0 3 10 = [(0 + 3) + 4(4 + 9 + 15 + 8) + 2(7 + 12 + 14)] = 710 3 Hence the required area of the cross section of the river = 710 sq. m. 80 0 y dx = z Example 15. Evaluate z 1.4 0.2 (sin x – loge x + ex) dx approximately using Weddle’s rule correct to 4 decimals. Sol. Let f(x) = sin x – log x + ex. We shall divide the given interval of integration into 12 equal parts so that the arguments are : 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4. The corresponding entries are y0 = f(0.2) = 3.0295, y1 = f(0.3) = 2.8494, y2 = f(0.4) = 2.7975, y3 = f(0.5) = 2.8213, y4 = f(0.6) = 2.8976, y5 = f(0.7) = 3.0147 y6 = f(0.8) = 3.1661, y7 = f(0.9) = 3.3483, y8 = f(1) = 3.5598, y9 = f(1.1) = 3.8001, y10 = f(1.2) = 4.0698, y11 = f(1.3) = 4.3705 y12 = f(1.4) = 4.7042 See All Chapters |
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## ALLC4-6 |
Manish Goyal | Laxmi Publications | |||||

243 INTERPOLATION = 1330 + (– .2)(682) + (.8)(− .2) (.8)(− .2)(− 1.2) (436) + (132) 2 6 = 1162.944 f(5.8) = 1162.944. ∴ ASSIGNMENT 1. 2. 3. 4. The population of a town in the years 1931, ......, 1971 are as follows : Year : 1931 1941 1951 1961 1971 Population : 15 20 27 39 52 (in thousands) Find the population of the town in 1946 by applying Gauss’ backward formula. Apply Gauss’ backward formula to find the value of (1.06)19 if (1.06)10 = 1.79085, (1.06)15 = 2.39656, (1.06)20 = 3.20714, (1.06)25 = 4.29187 and (1.06)30 = 5.74349. Given that x : 50 51 52 53 54 tan x : 1.1918 1.2349 1.2799 1.3270 1.3764 Using Gauss’ backward formula, find the value of tan 51° 42′. Interpolate by means of Gauss’ backward formula, the population of a town KOSIKALAN for the year 1974 given that : Year : 1939 1949 1959 1969 1979 1989 Population : 12 15 20 27 39 52 (in thousands) 5. 6. Apply Gauss’ backward formula to find sin 45° from the following table : θ° : 20 30 40 50 60 70 80 sin θ : 0.34202 0.502 0.64279 0.76604 0.86603 0.93969 0.98481 Using Gauss’ backward formula, estimate the no. of persons earning wages between Rs. 60 and See All Chapters |
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## ALLC4-10 |
Manish Goyal | Laxmi Publications | |||||

302 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ (i) ui(xj) = i≠ j i= j RS0, T1, i≠ j i= j (ii) vi(xj) = 0 ∀ i, j (iii) ui′(xj) = 0 ∀ i, j (iv) vi′(xj) = UV W RS0, T1, ...(3 (i)) ...(3 (ii)) ...(3 (iii)) UV W ...(3 (iv)) Using the Lagrange fundamental polynomials Li(x), we choose ui(x) = Ai(x) [Li(x)]2 and vi(x) = Bi(x) [Li(x)]2 where Li(x) is defined as UV W Li(x) = ...(4) ( x − x0 )( x − x1 ) ... ( x − xi −1 )( x − xi + 1 ) ... ( x − xn ) ( xi − x0 )( xi − x1 ) ... ( xi − xi −1 )( xi − xi + 1 ) ... ( xi − xn ) Since Li2(x) is a polynomial of degree 2n, Ai(x) and Bi(x) must be linear polynomials. Let Ai(x) = aix + bi and Bi(x) = cix + di so that from (4), ui(x) = (aix + bi) [Li(x)]2 vi(x) = (cix + di) [Li(x)]2 ...(5) using conditions 3(i) and 3(ii) in (5), we get UV W aix + bi = 1 cix + di = 0 and ...(6 (i)) ...(6 (ii)) | since [Li(xi)]2 = 1 Again, using conditions 3 (iii) and 3 (iv) in (5), we get ai + 2Li′(xi) = 0 and ...(6 (iii)) ci = 1 From equations 6 (i), 6 (ii), 6 (iii) and 6 (iv), we deduce ai = – 2Li′(xi) bi = 1 + 2xiLi′(xi) ci = 1 di = – xi See All Chapters |
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## ALLC9-2 |
Manish Goyal | Laxmi Publications | |||||

501 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS pe slo The eqn. of the tangent at P0(x0, y0) is y – y0 ⇒ F dy I = GH JK dx P0 Q2 ) , y0 (x 0 f e p slo ( x – x0 ) = f(x0, y0) (x – x0) y = y0 + (x – x0) f(x0, y0) ) ,y1 (f x 1 g(x ) Y y= In this method, we use the property that in a small interval, a curve is nearly a straight line. Thus at the point (x0, y0), we approximate the curve by the tangent at the point (x0, y0). Q1 P0 y0 ...(2) O x0 y2 y1 x1 x2 X This gives the y-coordinate of any point on the tangent. Since the curve is approximated by the tangent in the interval (x0, x1), the value of y on the curve corresponding to x = x1 is given by the above value of y in eqn. (2) approximately. Putting x = x1(= x0 + h) in eqn. (2), we get y1 = y0 + hf(x0, y0) Thus Q1 is (x1, y1) Similarly, approximating the curve in the next interval (x1, x2) by a line through Q1(x1, y1) with slope f(x1, y1), we get y2 = y1 + hf(x1, y1) In general, it can be shown that, yn+1 = yn + hf(xn, yn) This is called Euler’s formula. dy changes rapidly over an dx interval, its value at the beginning of the interval may give a poor approximation as compared to its average value over the interval and thus the value of y calculated from Euler’s method may be in much error from its true value. These errors accumulate in the succeeding intervals and the value of y becomes much erroroneous ultimately. See All Chapters |