A textbook of Computer Based Numerical and Statiscal Techniques part 32 pps

296 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Aliter: We know that +∆= = =, hD d IEeD dx ∴ log(1 )hD =+∆ ∆∆∆∆∆ =∆−+−+−+ 23456 23456 or  ∆∆∆∆∆ =∆−+−+−+   23456 1 23456 D h (1) ⇒ 2 23456 2 2 1 23456 D h  ∆∆∆∆∆ = ∆−+−+−+    =∆−∆+∆   23 4 2 111 12 h (2) also  ∆∆∆ = ∆−+−+   3 234 3 3 1 234 D h  =∆−∆+   34 3 13 2 h (3) Applying equations (1), (2) and (3) for 0, y we get =   =∆−∆+∆−∆+     0 234 0000 1111 234 xx dy yyyy dx h and =   =∆−∆+∆−      0 2 23 2 00 0 22 111 12 xx dy yy y dx h =   =∆−∆+     0 3 34 00 33 13 2 y xx d yy dx h 6.2.2 Derivatives Using Newton’s Backward Difference Formula Newton’s backward interpolation formula is given by 23 (1) (1)(2) 2! 3! nn n n uu uu u yy uy y y +++ =+∇+ ∇ + ∇ (1) where n xx u n − = . Differentiating both sides of equation (1) with respect to x, we get 2 23 121362 2! 3! nn n dy uuu yy y dx h  +++ =∇+ ∇ + ∇+   (2) NUMERICAL DIFFERENTIATION AND INTEGRATION 297 At ,0 n xxu == . Therefore putting u = 0 in (2), we get 23 111 23 n nnn xx dy yyy dx h =   = ∇ +∇ +∇ +     (3) Again differentiating both sides of equation (2) w.r.t. x, we get 2 2 23 4 22 166123622 624 nn n dy uuu yy y dx h  +++ =∇+ ∇+ ∇+   At , 0 n xxu == . Therefore putting u = 0 in (3), we get 2 23 4 22 111 12 n nn n xx dy yy y dx h =   =∇+∇+∇+      (4) Similarly, 3 34 33 13 2 n nn xx dy yy dx h =   =∇+∇+      and so on. (5) Aliter: We know that 1 1 hD Ee −− −∇= = ∴ −= −∇log (1 )hD 2345 2345  ∇∇∇∇ =−∇+++++   or 2345 1 2345 D h  ∇∇∇∇ =∇+ + + + +   and 2234 2 1111 212 D h  =∇+∇+∇   and 334 3 13 2 D h  =∇+∇+   Applying these identities to n y , we get 234 1111 234 n nnnn xx dy yyyy dx h =   =∇+∇ +∇ +∇     and =   =∇+∇+∇+      2 23 4 22 111 12 n nn n xx dy yy y dx h and 3 34 33 13 2 n nn xx dy yy dx h =   =∇+∇+      298 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.2.3 Derivatives Using Stirling’s Formula If we want to determine the values of the derivatives of the function near the middle of the given set of arguments. We may apply any central difference formula. Therefore using Stirling’s formula, we get. 33 22 22 24 01 1 2 01 2 (1) (1) 22! 3! 2 4! n yy y y uuu uu yyu y y −−− −−  ∆+∆ ∆ +∆ −−  =+ +∆ + + ∆ +      (1) where 0 xx u h − = . Now, differentiating w.r.t. x, we get 33 23 24 01 1 2 12 (3 1) (4 2 ) 23!24! dy y y y y uuudu uy y dx dx −−− −−   ∆+∆ ∆ +∆ −−  =+∆+ +∆+         Since 0 xx u h − = ∴ 1du dx h = ∴ 33 23 24 01 1 2 12 13142 23!24! dy y y y y uuu uy y dx h −−− −−   ∆+∆ ∆ +∆ −− =+∆+ +∆+       (2) At 0, 0, xxu== therefore, putting u = 0 in (2), we get 1 0 33 0 12 11 26 2 xx yy dy y y dx h − −− − ∆+∆   ∆+∆  =− +         Again differentiating, we get −− −−   ∆+∆ − =∆+ + ∆+       233 2 24 12 12 22 16 122 62 4! dy y y uu yy dx h (3) At 0 ,0 xxu== therefore, putting 0u = in (3), we get 0 2 24 12 22 11 12 xx dy yy dx h −− =   =∆−∆+      and so on. 6.2.4 Derivative Using Newton’s Divided Difference Formula Newton’s divided difference formula for finding the successive differentiation at the given value of x. Let us consider a function f(x) of degree n, then 23 0) 00 01 0 012 0 () ( ( ) ( )( )( ) ( )( )( )( ) ( ) yfxfxxxfxxxxx fxxxxxxx fx==+−∆+−−∆ +−−−∆ 01 1 0 ( )( ) ( ) ( ) n n xx xx xx fx − ++− − − ∆ NUMERICAL DIFFERENTIATION AND INTEGRATION 299 Differentiate this equation w.r.t. ‘x’ as many times as we require and put ,i xx = we get the required derivatives. Example 1. Find dy dx at x = 0.1 from the following table: x 0.1 0.2 0.3 0.4 y 0.9975 0.9900 0.9776 0.9604 Sol. Difference table: xy ∆y ∆ 2 y ∆ 3 y 0.1 0.9975 –0.0075 0.2 0.9900 -0.0049 –0.0124 0.0001 0.3 0.9776 -0.0048 –0.0172 0.4 0.9604 Here, 0 0.1, 0.1 xh== and 0 y = 0.9975 we know that, Newton’s forward difference formula. =   = ∆ −∆ +∆     23 000 0.1 111 23 x dy yyy dx h ()() 111 0.0075 0.0049 0.0001 0.1 2 3  =− −− +   =−0.050167. Ans. Example 2. Using following table. x 1.0 1.1 1.2 1.3 1.4 1.5 1.6 y 7.989 8.403 8.781 9.129 9.451 9.750 10.031 Find dy dx and 2 2 dy dx at x = 1.1. Sol. Since the values are at equidistant and we want to find the value of y at x = 1.1. Therefore, we apply Newton’s forward difference formula. 300 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Difference table: xy ∆ y 2 ∆ y ∆ 3 y 4 ∆ y 5 ∆ y 6 y∆ 1.0 7.989 0.414 1.1 8.403 –0.036 0.378 0.006 1.2 8.781 –0.030 –0.002 0.348 0.004 0.001 1.3 9.129 –0.026 –0.001 –0.002 0.322 0.003 –0.001 1.4 9.451 –0.023 –0.002 0.299 0.005 1.5 9.750 –0.018 0.281 1.6 10.031 We have, =   = ∆ −∆ +∆ −∆ +∆ −∆     0 23456 000000 111111 23456 xx dy yyyyyy dx h Putting 2 00 0 1.1, 0.378, 0.030, xy y=∆= ∆= and so on and h = 0.1, we get ()()()() 1.1 11 11 1 0.378 0.030 0.004 0.001 0.001 0.1 2 3 4 5 dy dx   =−−+−−+−     [] 10 0.378 0.015 0.0013 0.00025 0.0002 =+++− [] 10 0.39435 3.9435 == and   =∆−∆+∆−∆+∆      0 2 23 4 5 6 00 0 0 0 22 1 11 5 137 12 6 180 x dy yy y y y dx h ⇒   =−−−×+×      2 22 1.1 1115 0.030 0.004 0.001 0.001 12 6 (0.1) dy dx [] () =− − − + =− 100 0.030 0.004 0.0009 0.0008 100 0.0341 =−0.341 . Ans.Ans. Ans.Ans. Ans. Example 3. Using the given table, find dy/dx at x = 1.2 x 1.0 1.2 1.4 1.6 1.8 2.0 2.2 y 2.7183 3.3201 4.0552 4.9530 6.0496 7.3891 9.0250 NUMERICAL DIFFERENTIATION AND INTEGRATION 301 Sol. xy ∆ y 2 ∆ y 3 y∆ 4 y∆ 5 y∆ 6 y∆ 1.0 2.7183 0.6018 1.2 3.3201 0.1333 0.7351 0.0294 1.4 4.0552 0.1627 0.0067 0.8978 0.0361 0.0013 1.6 4.9530 0.1988 0.0080 0.0001 1.0966 0.0441 0.0014 1.8 6.0496 0.2429 0.0094 1.3395 0.0535 2.0 7.3891 0.2964 1.6359 2.2 9.0250 We have =   = ∆ −∆ +∆ −∆ +∆ −∆     0 23456 000 000 111111 23456 xx dy yyyyyy dx h Here, 2 00 0 1.2, 0.7351, 0.1627, xy y=∆= ∆= and so on and 0.2,h = we get ()()()() 1.2 11111 0.7351 0.1627 0.0361 0.0080 0.0014 0.2 2 3 4 5 dy dx   =−+−+     [] 5 0.7351 0.08135 0.0120 0.002 0.00028 =−+−+ 3.32015= . Ans. Example 4. Find dy dx and 2 2 dy dx of 1/3 yx= at x = 50 from the following table: x 50 51 52 53 54 55 56 1/3 yx= 3.6840 3.7084 3.7325 3.7563 3.7798 3.8030 3.8259 302 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. x 1/3 yx= ∆ y 2 ∆ y 3 y∆ 50 3.6840 0.0244 51 3.7084 –0.0003 0.0241 0 52 3.7325 –0.0003 0.0238 0 53 3.7563 –0.0003 0.0235 0 54 3.7798 –0.0003 0.0232 0 55 3.8030 –0.0003 0.0229 56 3.8259 Here, 0 50, 1. xh== Then, we have Newton’s forward difference formula. 0 23 000 111 23 xx dy yyy dx h =   = ∆ −∆ +∆ −     ()() 50 11 1 0.0244 0.0003 0 12 3 dy dx   =−−+     () =+ = 0.0244 0.00015 0.02455 and () 0 2 23 00 22 1 xx dy yy dx h =  =∆−∆+    [] 2 1 ( 0.0003) (1) =− = 2 2 50 0.0003 dy dx  =−    . Ans. Example 5. The table given below reveals the velocity v of a body during the time t. Find its, acceleration at t = 1.1. t 1.0 1.1 1.2 1.3 1.4 v 43.1 47.7 52.1 56.4 60.8 NUMERICAL DIFFERENTIATION AND INTEGRATION 303 Sol. DifDif DifDif Dif ferfer ferfer fer ence table:ence table: ence table:ence table: ence table: tv ∆ 2 ∆ 3 ∆ 4 ∆ 1.0 43.1 4.6 1.1 47.7 –0.2 4.4 0.1 1.2 52.1 –0.1 0.1 4.3 0.2 1.3 56.4 –0.1 4.4 1.4 60.8 We have, 00 1.1, 4.77 tv== and 0.1h = Then the acceleration at 1.1t = is given by 0 23 000 111 23 tt dv vvv dt h =   =∆−∆+∆    ()() 11 1 4.4 0.1 0.2 0.1 2 3  =−−+   0.2 14.40.5 3  =− + +   45.167= (approx.)Ans. Example 6. Find f′(1.1) and f″(1.1) from the following table : x 1.0 1.2 1.4 1.6 1.8 2.0 () fx 0 0.1280 0.5440 1.2960 2.4320 4.0000 Sol. Difference table x () fx () ∆ fx () ∆ 2 fx () ∆ 3 fx () ∆ 4 fx 1.0 0 0.1280 1.2 0.1280 0.288 0.4160 0.048 1.4 0.5440 0.336 0 0.7520 0.048 1.6 1.2960 0.384 0 1.1360 0.048 1.8 2.4320 0.432 1.5680 2.0 4.0000 304 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here we have to find the derivatives at x=1.1 which lies between given arguments 1.0 and 1.2. So apply Newton’s forward formula, we have 0 1 5( 1) 0.2 xx x ux h − − ===− (1) () () () ()() () 23 00 0 0 12 4(4 1) ( ) 21 3! uu u fx fx ufx fx fx −− − =+∆+ ∆+ ∆ (2) Differentiating w.r.t. x, we get () () () () () 2 23 00 0 21 362 2! 3! u uu du f x fx fx fx dx  − −+ ′ =∆ + ∆ + ∆   Also, from (1) 5 du dx = ∴ () () () () () 2 23 00 0 21 362 5 26 u uu f x fx fx fx  − −+ ′ =∆ + ∆ + ∆   At x =1.1, u = 5(1.1 – 1) = 0.5 ∴ () () () () () () 2 20.5 1 30.5 60.5 2 1.1 5 0.128 0.288 0.048 26 f  −−+ ′ =+ +    [] 5 0.128 0 0.002 =+− () 1.1 0.63.f ′ = Ans. Differentiating equation (2) again w.r.t. ‘x’, we get () () () 23 00 66 5 6 udu f x fx fx dx −  ′′ =∆ + ∆   ∴ () ( ) [] 5 1.1 25 0.288 0.5 1 0.0481 25 0.288 0.024 f ′′ =+−×=−   Hence, () 1.1 6.6.f ′′ = Ans. Example 7. Find f′(1.5) from the following table: x 0.0 0.5 1.0 1.5 2.0 f(x) 0.3989 0.3521 0.2420 0.1245 0.0540 Sol. Here we want to find the derivatives of f (x) at x = 1.5, which is near the end of the arguments. Therefore, apply Newton’s backward formula. NUMERICAL DIFFERENTIATION AND INTEGRATION 305 Difference Table x () fx () fx∇ () 2 fx∇ () 3 fx∇ () 4 fx∇ 0.0 0.3989 –0.0468 0.5 0.3521 –0.0633 –0.1101 0.0609 1.0 0.2420 –0.0024 –0.0215 –0.1125 0.0394 1.5 0.1295 0.037 0.0755 2.0 0.0540 Here, x n = 1.5, h = 0.5 Therefore, 234 1111 234 n nnnn xx dy yyyy dx h =   = ∇ +∇ +∇ +∇ +     () ()() 111 1.5 0.1125 0.0024 0.0609 0.5 2 4 f  ′ =− +− +   [][] 2 0.1125 0.0012 0.0152251 2 0.098475 =− − + =− ⇒ () 1.5 0.19695.f ′ =− Ans. Example 8. From the following table, find the values of dy dx and 2 2 dy dx at x = 2.03. x 1.96 1.98 2.00 2.02 2.04 y 0.7825 0.7739 0.7651 0.7563 0.7473 Sol. Difference table x y y∇ 2 y∇ 3 y∇ 4 y∇ 1.96 0.7825 –0.0086 1.98 0.7739 –0.0002 –0.0088 0.0002 2.00 0.7651 0 –0.0004 –0.0088 –0.0002 2.02 07563 –0.0002 –0.0090 2.04 0.7473 . equidistant and we want to find the value of y at x = 1.1. Therefore, we apply Newton’s forward difference formula. 300 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Difference table: xy ∆ y 2 ∆ y ∆ 3 y 4 ∆ y 5 ∆ y 6 y∆ 1.0. 0.048 1.8 2. 4320 0. 432 1.5680 2.0 4.0000 304 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here we have to find the derivatives at x=1.1 which lies between given arguments 1.0 and 1.2. So apply. ∆ NUMERICAL DIFFERENTIATION AND INTEGRATION 299 Differentiate this equation w.r.t. ‘x’ as many times as we require and put ,i xx = we get the required derivatives. Example 1. Find dy dx at