A textbook of Computer Based Numerical and Statiscal Techniques part 22 pps

196 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 5. If f(x) is a polynomial of degree four find the value of f(5.8) using Gauss’s backward formula from the following data: f(4) = 270, f(5) = 648, ∆f(5) = 682, ∆ 3 f(4) = 132. Sol. Given (5) 682f∆= ⇒ f (6) – f (5) = 682 ⇒ f (6) = 682 + 648 ⇒ f (6) = 1330 Also, ∆ 3 f (4) = 132 ⇒ (E – 1) 3 f (4) = 132 ⇒ f (7) – 3f (6) + 3f (5) – f (4) = 132 ⇒ f (7) = 3 × 1330 – 3 × 648 + 270 + 132 f (7) = 2448 Now form difference table as: xfxfxfxfx∆∆ ∆ 23 () () () () 4270 378 5 648 304 682 132 6 1330 436 1118 72448 Take a = 6, h = 1 a + hu = 5.8 ∴ 0.2u =− From Gauss backward formula () 23 1) ( 1) (1) (0.2) (0) (1) (1) (2) 2! 3! uuu uu ffuf f f +− + −= +∆−+ ∆−+ ∆− ( 0.2)(0.8) ( 0.2)(0.8)( 1.2) 1330 ( 0.2) 682 436 132 26 −−− =+−×+ ×+ × = 1330 – 136.4 – 34.88 + 4.224 = 1162.944 Hence f (5.8) = 1162.944. INTERPOLATION WITH EQUAL INTERVAL 197 Example 6. Using Gauss backward interpolation formula, find the population for the year 1936. Given that Year Population in thousands 1901 1911 1921 1931 1941 1951 ( )121520273952 Sol. Here h = 10. Take origin at 1941 to evaluate population in 1936 1936 1941 5 0.5 10 10 xa u h −−− ⇒= = ==− Difference table for given data is as: ufufufufufufu∆∆ ∆ ∆ ∆ − − − − −− − 2345 () () () () () () 412 3 315 2 50 220 2 3 73 10 127 5 7 12 4 039 1 13 152 Gauss backward formula is 23 (1) (1)(1) () (0) (1) (1) (2) 2! 3! uu uuu fu f uf f f ++− =+∆−+ ∆−+ ∆− (2)(1)(1) 4! uuuu++ − + ∆ 4 f (–2) 5 ( 2)( 1) ( 1)( 2) (3) 5! uuuuu f ++ −− +∆−+ = 39 + ( 0.5)(0.5) ( 0.5)(0.5)( 1.5) (0.5) 12 1 (4) 26 −−− −×+ ×+ ×− = 39 – 6.0 – 0.125 – 0.25 = 32.625 thousands Hence, the population in 1936 is 32625 thousand. 198 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM SET 4.4 1. Given that 12500 = 111.803399, 12510 = 111.848111, 12520 = 111.892806, 12530 = 111.937483. Using Gauss’s backward formula show that 12516 = 111.8749301 2. Find the value of cos 51°42 1 by Gauss’s backward formula from the following data: 50 51 52 53 54 cos 0.6428 0.6293 0.6157 0.6018 0.5878 x x °°°°° [Ans. 0.61981013] 3. The population of a town in the years are as follows: () 1931 1941 1951 1961 1971 15 20 27 39 52 Year Population in thousands Find the population of the town in 1946 by applying Gauss’s backward formula. [Ans. 22898] 4. Interpolate by means of Gauss’s backward formula, the population of a town KOSIKALAN for the year 1974, given that: () 1939 1949 1959 1969 1979 1989 12 15 20 27 39 52 Year Population in thousands [Ans. 32.345 thousands appros] 5. Use Gauss interpolation formula to find y 41 from the following data: y 30 = 3678.2, y 35 = 2995.1, y 40 = 2400.1, y 45 = 1876.2, y 50 = 1416.3 [Ans. y 41 = 2290.1] 6. Use Gauss’s backward formula to find the value of y when x = 3.75, given the following table: 2.5 3.0 3.5 4.0 4.5 5.0 24.145 22.043 20.225 18.644 17.262 16.047 x x y [Ans. 19.704] INTERPOLATION WITH EQUAL INTERVAL 199 Stirling’s Formula Example 1. Evaluate sin (0.197) from the table given: 0.15 0.17 0.19 0.21 0.23 0.14944 0.16918 0.18886 0.20846 0.22798 x sin x Sol. The difference table is given by 23 45 2 0.15 0.15 0.14944 0.01974 1 0.17 0.17 0.16918 0.00006 0.01968 0.00002 0 0.19 0.19 0.18886 0.00008 0.00002 0.0196 0 1 0.21 0.21 0.20846 0.00008 0.01952 2 0.23 0.23 0.22798 xxsinx ∆∆ ∆ ∆ ∆ − −− − − − ∴ u = xa h − = 0.197 0.19 0.02 − = 0.35 From Stirling formula, we have f(u)= f(0) + () ( ) 01 2 uf f ∆+∆−   + 2 2! u () 2 1 f ∆− + ()() 11 3! uuu +− () () 33 12 2 ff  ∆−+∆−    + ()() 2 11 3! uuu +− () 4 2 f ∆− = 0.18886 + 0.35 () 2 0.35 0.0196 0.01968 22 +  +   × (–0.00008) + () () () 0.35 1 0.35 0.35 1 6 +− × 0.00002 2 −    + (0.35) 2 () 2 0.35 1 0.00002 24  − ×    = 0.18886 + 0.0068741 – 0.0000049 + 0.0000005 – 0.00000009 = 0.19573 (Approx.) Example 2. Find the value of e x = at x = 0.644 using by Stirling’s formula. The following data given below: x x ye = 0.61 0.62 0.63 0.64 0.65 0.66 0.67 1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237 200 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. For the given data difference table is as: x xe ∆∆ ∆ ∆ ∆ ∆ − − −− − − 23456 3 0.61 1.840431 0.018497 2 0.62 1.858928 0.000185 0.018682 0.000004 1 0.63 1.877610 0.000189 0.000004 0.018871 0 0.000006 0 0.64 1.896481 0.000189 0.000002 0.000007 0.01906 0.000002 0.000001 1 0.65 1.915541 0.0 0091 0.000001 0.019251 0.000003 2 0.66 1.934792 0.000194 0.019445 3 0.67 1.954237 Here h = 0.01 ∴ u = 0.644 0.64 0.4 0.01 xa h −− == By stirling formula, we get, 33 3 3 [ (0) (1) (1) (2) (1)(1) () (0) (1) 22! 3! 2 uf f f f uuu u fu f f  ∆+∆− ∆−+∆− +− =+ +∆−+    + 2 45 ( 1) ( 1) ( 2)( 1) ( 1)( 2) ( 2) ( 2) 3! 5! uuu u uuu u ff +− ++−− ∆−+ ∆−+ 2 (0.4) (1.4)(0.4)( 0.6) 0.01906 0.018871 0.000004 1.896481 0.4 0.000189 22 62 − +   =+ +×+     22 (0.4) (0.4) 1 24  −  + (1.4)(2.4)(0.4)( 0.6)( 1.6) 0.000006 0.000001 0.000002 120 2 −− −  ×+   2 (0.4 2)(1.4)(0.4)( 0.6)( 0.4 2) ( 0.000007) 720 +−−− +×− = 1.896481 + 0.0075862 + 0.00001512–0.000000112 – 0.0000000112 + 0.000000026 – 0.0000002 = 1.904081. (Approx.) INTERPOLATION WITH EQUAL INTERVAL 201 Example 3. Employ Stirling’s formula to evaluate y 12.2 from the data given below (y x = 1+ log 10 sin x). 5 10 11 12 13 14 10 23967 28060 31788 35209 38368 x x y ° Sol. For the given data difference table is as: 5234 10 10 23967 4093 11 28060 365 3728 58 12 31788 307 13 3421 45 13 35209 262 3159 14 38368 x xy °∆∆∆∆ − −− − ∴ u = 12.2 – 12 = 0.2 From Stirling formula [] {} () {} 2 2 2 2 (0.2) 0.2 1 0.2 (0.2) 1 0.2 (0.2) 1 ( ) 31788 3421 3728 ( 307) (45 58) ( 13) 22624! fu − − =+ ++ −+ ++ − = 31788 – 714.9 – 6.14 – 1.648 + 0.0208 y 12.2 = 32495 10 5 y 12.2 = 32495 ∴ y 12.5 = 0.32495. Ans. 202 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 4. Apply Stirling’s formula to find a polynomial of degree three which takes the following values of x and y: x y − 2 46810 213820 Sol. Let u = 6 2 x − . Now, we construct the following difference table: uuuuu xuy y y y y ∆∆∆∆ −− −− 234 222 3 411 1 24 603 3 0 54 818 7 12 10220 Stirling’s formula is 33 2 2 [ (0) (1)] (1) (2) (1)(1) () (0) (1) 22! 3! 2 uf f f f uuu u fu f f  ∆+∆− ∆−+∆− +− =+ +∆−+    + 2 4 (1)(1) (2) 3! uuu f +− ∆− 22 25 ( 1)44 33 0 22 6 2 u uuu yu +−+    =+ + ×+ +      23 73 2 3() 22 3 uu uu=+ + + − 32 2372 3 3223 uuuu=+++− 32 2317 3 32 6 uu u=+ + + 32 32 2317 2636176 33 32 6 32 22 62 xx x uu u −− −   =+++= + + +     = 0.0833x 3 – 1.125x 2 + 8.9166x – 19. INTERPOLATION WITH EQUAL INTERVAL 203 Example 5. Apply Stirling’s formula to find the value of f(1.22) from the following table which gives the value of 2 x x 2 0 1 f(x) e 2 π = ∫ dx at intervals of x = 0.5 from x = 0 to 2. () 0 0.5 1.0 1.5 2.0 0 0.191 0.341 0.433 0.477 x fx Sol. Let the origin be at 1 and h = 0.5 ∴ x = a + hu, u = 1.22 1.00 0.44 0.5 xa h −− == Applying Stirling’s formula [] 33 2 2 (0) ( 1) (1) (2) (1)(1) () (0) (1) 22! 3! 2 uf f ff uuu u fu f f  ∆+∆− ∆−+∆− +− =+ +∆−+    + 2 4 (1)(1) (2) 3! uuu f +− ∆− ∴ f(0.44) = f(0) + (0.44) 2 2 1 (0.44) [ (0) (1)] (1) 22 ff f∆+∆−+ ∆− {} {} 222 33 4 (0.44) (0.44) 1 (0.44) (0.44) 1 1 ( 1) ( 2) ( 2) 62 24 ff f −−  +∆−+∆−+∆−  = f(0) + (0.22) [∆f(0) + ∆f(–1)] + 0.0968 ∆ 2 f(–1) – 0.029568 [∆ 3 f(–1) + ∆ 3 f(–2)] – 0.06505∆ 4 f(–2) (1) The difference table is as follows: u x fx fx fx fx fx∆∆∆∆ − −− − − − 33 323334 10 ( ) 10 ( ) 10 ( ) 10 ( ) 10 ( ) 20 0 191 10.5 191 41 150 17 01 341 58 27 92 10 1 1.5 433 48 44 2 2 477 204 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES f(0) and the differences are being multiplied by 10 3 ∴ 10 3 f(0.44) ≈ 341 + 0.22 × (150 + 92) + 0.0968 × (–58) – 0.029568 × (–17 + 10) – 0.006505 × 27 ≈ 341 + 0.22 × 242 – 0.0968 × 58 + 0.029568 × 7 – 0.006505 × 27 ≈ 341 + 53.24 – 5.6144 + 0.206276 – 0.175635 ≈ 388.66 f(0.44) = 0.389 at x = 1.22. Ans. Example 6. Use Stirling’s formula to find y 28 given. y 20 = 49225, y 25 = 48316, y 30 = 47236, y 35 = 45926, y 40 = 44306 Sol. Let the origin be at 30 and h = 5 a + hu = 28 ⇒ 30 + 5u = 28 ⇒ u = –0.4 The difference table is as follows: 234 2 20 49225 909 1 25 48316 171 1080 59 0 30 47236 230 21 1310 80 1 35 45926 310 1620 2 40 44306 ux y y y y y∆∆∆∆ − − −− −− −− −− − − By Stirling’s formula, 2 (0.4) 1080 1310 ( .4) 47236 ( .4) ( 230) 22! f − −−  −= +− + −   (.6)( .4)( 1.4) 3! −− + () {} 2 2 .4 ( 4.4) 1 59 80 (21) 24! −−− −−  +−   Hence, y 28 = 47691.8256. INTERPOLATION WITH EQUAL INTERVAL 205 PROBLEM SET 4.5 1. Use Stirling’s formula to find y 32 from the following table: x y 20 25 30 35 40 35 14.035 13.674 13.257 12.734 12.089 11.309 [Ans. 13.062] 2. Use the following table to evaluate tan 16° by Stirling’s formula: 0 5 10 15 20 25 30 0 0.0875 0.1763 0.2679 0.364 0.4663 0.5774tan θ°°° ° °°° ° θ° [Ans. 0.2866980499] 3. Use Stirling’s formula to find the value of f(1.22) from the following data: xfx() 1.0 0.84147 1.1 0.89121 1.2 0.93204 1.3 0.96356 1.4 0.98545 1.5 0.99749 1.6 0.99957 1.7 0.99385 1.8 0.97385 [Ans. 0.9391002] 4. Find f(0.41) using Stirling’s formula if, x fx 0.30 0.35 0.40 0.45 0.50 ( ) 0.1179 0.1368 0.1554 0.1736 0.1915 [Ans. 0.15907168] 5. Use Stirling’s formula to find y 35 , data being: y 20 = 512, y 30 = 439, y 40 = 346, and y 50 = 243, [Ans. 394.6875] 6. From the following table find the value of f(0.5437) by Stirling’s formula: 0.51 0.52 0.53 0.54 0.55 0.56 0.57 ( ) 0.529244 0.537895 0.546464 0.554939 0.663323 0.571616 0.579816 x fx [Ans. 0.558052] . 196 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 5. If f(x) is a polynomial of degree four find the value of f(5.8) using Gauss’s backward formula from the following data: f(4). in 1946 by applying Gauss’s backward formula. [Ans. 228 98] 4. Interpolate by means of Gauss’s backward formula, the population of a town KOSIKALAN for the year 1974, given that: () 1939 1949 1959. 0.125 – 0.25 = 32.625 thousands Hence, the population in 1936 is 32625 thousand. 198 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM SET 4.4 1. Given that 12500 = 111.803399, 12510