A textbook of Computer Based Numerical and Statiscal Techniques part 23 pot

206 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply Stirling’s formula to find a polynomial of degree four which takes the values of y as given below: x y −− 12345 11111 [Ans. 42 28 1 33 uu−+ ] 8. Apply Stirling’s formula to interpolate the value of y at x = 1.91 from the following data: x y 1.7 1.8 1.9 2.0 2.1 2.2 5.4739 6.0496 6.6859 7.3851 8.1662 9.0250 [Ans. 6.7531] 4.5 BESSEL’S Example 1. Using Bessel’s formula find the value of y at x = 3.75 for the data given below: x 2.5 3.0 3.5 4.0 4.5 5.0 y 24.145 22.043 20.225 18.644 17.262 16.047 Sol. Difference table for the given data is as: 2345 22.524.145 2.102 1 3.0 22.043 0.284 1.818 0.047 0 3.5 20.225 0.237 0.009 1.581 0.038 0.003 1 4.0 18.644 0.199 0.006 1.382 0.032 2 4.5 17.262 0.167 1.215 3 5.0 16.047 xy ∆∆∆∆∆ − − − −− −−− −− − Here h = 0.5 ∴ u = 3.75 3.5 0.5 0.5 xa h −− == INTERPOLATION WITH EQUAL INTERVAL 207 Now from Bessel’s formula, we have {} 22 1 (1) (0) (1) (0) ( 1) 1(1) 2 () (0) 222! 2 3! uu u ff f f uu fu u f  −−   +∆+∆− −   =+−∆+ +    ∆ 3 f (–1) 44 5 (1) (2) ( 1)( 1)( 2)( 12) (1)(1)(2) ( 2) 4! 2 5! ffuuuuu uuu u f  ∆−+∆− + − − − +−− ++∆−+    [] [] 1 (0.5)(0.5) 1 18.644 20.225 (0.5 0.5)( 1.581) 1.99 2.37 222 =++−−+×+ 16 9 0 (0.5 1)(0.5)( 0.5)(2.5) 0 2 +  ++ + − × +   = 19.407. (Approx.) Example 2. Following table gives the values of e x for certain equidistant values of x. Find the value of e x at x = 0.644 using Bessel’s formula: x x e 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 Sol. Given h = 0.01, take it origin as 0.64 0.644 0.64 0.004 0.4 0.01 0.01 xa u h −− ⇒= = = = u = 0.4 Difference table for the given data is as: xfx ∆∆∆∆∆∆ − − − − 234 5 6 () 0.61 1.840431 0.018497 0.62 1.858928 0.000165 0.018682 0.000004 0.63 1.877610 0.000189 0.000004 0.18871 0 0.000006 0.64 1.896481 0.000189 0.000002 0.000007 0.1906 0.000002 0.000001 0.65 1.915541 0.000191 0. 000001 0.019251 0.000003 0.66 1.934792 0.000144 0.019445 0.67 1.954237 208 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Bessel’s formula {} 22 1 (1) (0) (1) (0) ( 1) 1(1) 2 () (0) 222! 2 3! uu ff f f uu fu u f  −−   +∆+∆− −   =+−∆+ +    ∆ 3 f (–1) () 44 5 1 ( 1)( 2) ( 1) ( 2) (1)(1)(2)(1/2) ( 2) 4! 2 5! uuuu f f uuu u u f  +−−∆−+∆− +−−− ++∆−+    0.4( 0.6) 1.896481 1.915541 0.000191 0.000189 ( 0.1)(0.01906) 222 − ++   =+−+     (0.4)( 0.6)( 0.1)(0.000002) (1.4)(0.4)( 0.6)( 1.6) 0.000001 0.000002 6242 −− −− +  ++   (1.4)(0.4)( 0.6)( 0.1)( 1.6) ( 0.000001) 120 −−− +− = 1.906011–0.001906 – 0.0000228 + 0.000000008 + 0.000000033 + 1.68 × 10 –10 = 1.904082 Example 3. Find the value of y 25 from the following data using Bessel’s formula. Data being y 20 = 2854, y 24 = 3162, y 28 = 3544, y 32 = 3992. Sol. The difference table for the data is as: 23 20 2854 308 24 3162 74 382 8 28 3544 66 448 32 3992 xy y y y∆∆ ∆ − 25 24 0.25 4 u − == (3162 3544) (0.25)( 0.75)( 0.25)( 8) 66 74 (0.25) ( 0.25) 382 (0.25)( 0.75) 226 f + −−− +  =+−×+− +   = 3353 – 95.5 – 6.5625 – 0.0625 = 3250.875 INTERPOLATION WITH EQUAL INTERVAL 209 Example 4. The pressure p of wind corresponding to velocity v is given by following data. Estimate pressure when v = 25. v P 10 20 30 40 1.1 2 4.4 7.9 Sol. The difference table for the given data is as: vP∆∆ ∆ − 23 10 1.1 0.9 20 2 1.5 2.4 0.4 30 4.4 1.1 3.4 40 7.9 Let origin = 20, h = 10, 25 20 0.5 10 u − == Bessel’s formula for interpolation is: 2 23 01 0 1 0 1 1 (1) 11(1) 2 () ( ) 222!23! uuu uu Pu P P u P P P P −−  −−   −∆   =++−∆+ ∆+ + ∆     () 1 1 0.5(0.5 1) 1.5 1.1 1 0.5(0.5 1) 2 4.4 0.5 2.4 0.5 0.4 222226 −+ −    =++−×+ +− ×−       1 6.4 0 0.16250 0 2 =× +− + = 3.2 – 0.16250 P 25 = 3.03750 Example 5. Probability distribution function values of a normal distribution are given as follows: x px 0.2 0.6 1.0 1.4 1.8 ( ) 0.39104 0.33322 0.24197 0.14973 0.07895 Find the value of p(x) for x = 1.2. Sol. Taking the origin at 1.0 and h = 0.4 x = a + uh ⇒ 1.2 = 1.0 + u × 0.4 210 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 1.2 1.0 1 0.4 2 u − == The difference table is: ufufufufufu∆∆ ∆ ∆ − − −− − −− − − 5 5 52 53 54 10 ( ) 10 ( ) 10 ( ) 10 ( ) 10 ( ) 2 39104 5782 1 33322 3343 9125 3244 0 24197 99 999 9224 2245 1 14973 2146 7078 2 7895 Bessel‘s formula is f (u)= {} 22 3 1 (1) (0) (1) (0) ( 1) 1(1) 2 (0) ( 1) 222! 2 3! uu u ff f f uu uf f  −−   +∆+∆− −   +−∆ + + ∆ −    10 5 f (0.5) = 11 1 24197 14973 2146 99 22 00 22!2   −   +−    ++ +     = 19457.0625 ∴ f (0.5) = 0.194570625 Hence, p(1.2) = 0.194570625. Example 6. Given y 0 , y 1 , y 2 , y 3 , y 4 , y 5 (fifth difference constant), prove that 1 2 2 125(cb)3(ac) yc 2 256 −+ − =+ where a = y 0 + y 5 , b = y 1 + y 4 , c = y 2 + y 3 . Sol. Put 1 2 u = in bessel’s formula, we get ()() 22 4 4 1/2 0 1 0 1 1 2 11 3 () 2 16 256 yyy yy yy −−− = + − ∆ +∆ + ∆ +∆ Shifting origin to 2, we have () 22 44 123 21 10 2 2 11 3 () ( ) 2 16 256 yyy yy yy = + − ∆ +∆ + ∆ +∆ = () 321432 543210 13 2 2 (3223 ) 2 16 256 c yyyyyy yyyyyy − −++−++ −++−+ INTERPOLATION WITH EQUAL INTERVAL 211 1 2 2 y = 4321 13 ()(32) 2 16 256 c yyyy abc−−−++−+ = 13 () (32) 2 16 256 c bc a b c−−+ −+ 1 2 2 y = [] 1 25( ) 3( ) 2 256 c cb ac +−−− Example 7. If third differences are constant, prove that 22 1 x x1 x1 x x 2 11 y(yy)(yy) 216 +− + =+−∆+∆ · Sol. Put 1 2 u = in Bessel’s formula, we get 22 101 0 1 2 11 ()( ) 216 yyy yy − =+−∆+∆ Shifting the origin to x. 22 11 1 2 11 ()( ) 216 xx x x x yyy yy +− + =+−∆+∆ Example 8. Given that: x fx 468101214 ( ) 3.5460 5.0753 6.4632 7.7217 8.8633 9.8986 Apply Bessel’s formula to find the value of f(9). Sol. Taking the origin at 8, h = 2, 9 = 8 + 2u or u = 1 2 The difference table is: 4 4 42 43 44 45 10 10 10 10 10 10 2 35460 15293 1 50753 1414 13879 120 0 64632 1294 5 1258 125 24 1 77217 1169 19 11416 106 2 88633 1063 10353 3 98986 uuuuuu uyyyyyy ∆∆∆∆∆ − −− − − −− − 212 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Bessel’s formula is 1 2 22 10 0 0 1 1 (1) 11(1) 2 () ( ) 2 2 2! 3! u uu u uu yyyuy yy −  −−  −   =++−∆+ ∆+∆+   ∆ 3 y –1 44 5 32 2 1 (2)(1) (1) (1)(1)(2) 1 2 () 4! 2 5! uu uu uuu u yy y −− −  −−− +  +−−  +×∆+∆+ ∆ 4 1/2 11 311 3 . 11 1 22 222 2 10 (77217 64632) 0 . ( 1169 1294) 0 . ( 19 15) 0 222 242 y   −−−     =+++ −−++ −++ ⇒ 4 1/2 10 71078.27344 y = ∴ 1/2 7.107827344 y = Hence, f(9) = 7.107827344 Example 9. Find a polynomial for the given data using Bessel’s formula f(2) = 7, f(3) = 9, f(4) = 12, f(5) = 16. Sol. Let us take origin as 3 therefore, u = x – 3 xyyyy∆∆ ∆ 23 27 2 39 1 30 412 1 4 516 Bessel’s formula: 22 10 01 0 11(1) () 222!2 u yy PP yyyu y −  ∆+∆ −  =++−×∆+      = 912 1 ( 1) 3 222 uu u +−  +−×+   = 2 21 3 3 2222 uu u+−+ − = 2 5 9 22 u u++ INTERPOLATION WITH EQUAL INTERVAL 213 Put u = x – 3 y = 2 (3)5 (3)9 22 x x − +−+ = () 2 15 69 (3)9 22 xx x −++ −+ y = 2 11 6 22 xx−+ . Ans. PROBLEM SET 4.6 1. Find y (0.543) from the following values of x and y: 0.1 0.2 0.3 0.4 0.5 0.6 0.7 ( ) 2.631 3.328 4.097 4.944 5.875 6.896 8.013 x yx [Ans. 6.303] 2. Apply Bessel’s formula to find the value of y 2.73 from the data given below: 2.5 2.6 2.7 2.8 0.4938, 0.4953, 0.4965, 0.4974 yyyy==== 2.9 3.0 0.4981, 0.4987. yy== [Ans. 0.496798] 3. Find y 25 by using Bessel’s interpolation formula from the data: 20 24 28 32 24, 32, 35, 40 yyyy==== [Ans. 32.9453125] 4. Apply Bessel’s formula to evaluate y 62.5 from the data: x x y 60 61 62 63 64 65 7782 7853 7924 7993 8062 8129 [Ans. 7957.1407] 5. Apply Bessel’s formula to obtain the value of f (27.4) from the data: x fx 25 26 27 28 29 30 ( ) 4.000 3.846 3.704 3.571 3.448 3.333 [Ans. 3.649678336] 6. Apply Bessel’s formula to find the value of f(12.2) from the following data: 0 5 10 15 20 25 30 ( ) 0 0.19146 0.34634 0.43319 0.47725 0.49379 0.49865 x fx [Ans. 0.39199981] 214 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply Bessel’s interpolation formula, show that tan160° = 0.2867, Given that: 0 5 10 15 20 25 30 tan 0 0.0875 0.1763 0.2679 0.3640 0.4663 0.5774 x x °° °°°°° 8. Apply Bessel’s formula to find a polynomial of degree 3 from the data: 46810 13820 x y [Ans. 32 21347 6 3; 9 6 18 2 x uuuu − +++= ] 9. From the following table find the value of f (0.5437) by Gauss and Bessel’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] 10. Apply Bessel’s formula to obtain a polynomial of degree three: 7891011 ( ) 14 17 19 22 25 x fx [Ans. 3 2 281 5 157 66 x xx−+ − + ] 4.6 LAPLACE EVERETTS Example 1. Find the value of f(27.4) from the following table: u fx 25 26 27 28 29 30 ( ) 4.000 3.846 3.704 3.571 3.448 3.333 Sol. Here, u = 27.4 27.0 0.4, 1 − = origin is at 27.0 h = 1 Also, w = 1 – u = 0.6 INTERPOLATION WITH EQUAL INTERVAL 215 Difference table is: ufufufufufu ∆ ∆∆ ∆ − − − −− − − −− − 4 3 3 32 33 3 10 ( ) 10 ( ) 10 ( ) 10 ( ) 10 ( ) 2 4000 154 1 3846 12 142 3 0 3704 9 4 133 1 1 3571 10 3 123 2 23448 8 115 3 3333 By Laplace Everett’s formula, f (0.4) = () 2.4)(1.4 (0.4)( 0.6)( 1.6) (1.4)(0.4)( 0.6) (0.4)(3571) (10) ( 3) 3! 5! −−  − ++ −+   () {} (1.6)(0.6)( 0.4) (2.6)(1.6)(0.6)( 0.4)( 1.4) (0.6)(3704) 9 (4) 3! 5! −−− ++ + = 3649.678336. Hence f (27.4) = 3649.678336. Example 2. Using Laplace Everett’s formula, find f(30), if f(20) = 2854, f(28) = 3162, f(36) = 7088, f(44) = 7984. Sol. Take origin at 28, h = 8 ∴ a + hu = 30 ⇒ 28 + 8u = 30 ⇒ u = 0.25 Also, w = 1 – u = 1 – .25 = 0.75 . 206 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply Stirling’s formula to find a polynomial of degree four which takes the values of y as given below: x y −− 1234 5 11111 [Ans. 42 28 1 33 uu−+ ] 8 = 1.2. Sol. Taking the origin at 1.0 and h = 0.4 x = a + uh ⇒ 1.2 = 1.0 + u × 0.4 210 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 1.2 1.0 1 0.4 2 u − == The difference table is: ufufufufufu∆∆. formula to find the value of f(12.2) from the following data: 0 5 10 15 20 25 30 ( ) 0 0.19146 0.34634 0.43319 0.47725 0.49379 0.49865 x fx [Ans. 0.39199981] 214 COMPUTER BASED NUMERICAL AND STATISTICAL