A textbook of Computer Based Numerical and Statiscal Techniques part 27 ppt

246 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The second divided difference for x 0 , x 1 , x 2 is given by f(x 0 , x 1 , x 2 ) = [x 0 , x 1 , x 2 ] = [][] 12 01 20 ,, xx xx xx − − The third divided difference for x 0 , x 1 , x 2 , x 3 is given by f(x 0 , x 1 , x 2 , x 3 ) = [x 0 , x 1 , x 2 , x 3 ] = [][] 123 012 30 ,, ,, xxx xxx xx − − and so on. f(x 0 , x 1 , x 2 ,x n ) = [x 0 , x 1 , x 2 ,x n ] = [][ ] 12 01 1 0 , , , , , , nn n xx x xx x xx − − − 5.4.1 Properties of Divided Differences 1. Divided differences are symmetric with respect to the arguments i.e., independent of the order of arguments. i.e., [x 0 , x 1 ] = [x 1 , x 0 ] Also, [x 0 , x 1 , x 2 ] = [x 2 , x 0 , x 1 ] or [x 1 , x 2 , x 0 ] 2. The n th divided differences of a polynomial of n th degree are constant. Let f(x) = A 0 x n + A 1 x n–1 + + A n–1 x + A n by a polynomial of degree n provided A 0 ≠ 0 and arguments be equally spaced so that x 1 – x 0 = x 2 – x 1 = x n – x n–1 = h Then first divided difference [x 0 , x 1 ] = 10 0 10 yy y xx h −∆ = − Second divided difference [x 0 , x 1 , x 2 ] = 2 1 2! h = 2 0 y∆ [x 0 , x 1 , x 2 x n–1 , x n ] = 1 ! n nh = 0 n y ∆ Since, function is a n th degree polynomial Therefore, 0 n y ∆ = constant ∴ n th divided difference will also be constant. 3. The n th divided difference can be expressed as the quotient of two determinants of order (n + 1). i.e., [x 0 , x 1 ] = 10 10 yy xx − − = − −− 10 10 10 yy xxxx ⇒ [x 0 , x 1 ] = 1010 1111 yyxx ÷ Similarly, [x 0 , x 1 , x 2 ] = 222 01 2 0 1 2 01 2 0 1 2 111 1 11 yy y x x x xx x x x x÷ and so on. 4. The n th divided difference can be expressed as the product of multiple integral. INTERPOLATION WITH UNEQUAL INTERVAL 247 5.4.2. Relation Between Divided Differences and Ordinary Differences Let the arguments x 0 , x 1 , x 2 , x n , be equally spaced such that x 1 – x 0 = x 2 – x 1 = = x n – x n–1 = h ∴ x 1 = x 0 + h x 2 = x 0 + 2h x n = x 0 + nh Now, first divided difference for arguments (x 0 , x 1 ) be given by () 1 0 x fx ∆ = () () 10 10 fx fx xx − − = () () 00 fx h fx h +− = () 0 fx h ∆ (1) Similarly, second divided difference be given as () 12 2 0 xx fx ∆ = ()() 12 01 20 1 ,, fx x fx x xx −  − = () () () () 1021 20 21 10 1 fx fxfx fx xx xx xx −− −  −− −  = ()()() () 0000 2 1 2 fx h fx h fx h fx hh h  +− + +− −   = 2 1 2 h [] 000 (2)2( )() fx h fxhfx −− ++ = () 2 0 2 2! fx h ∆ (2) () 123 0 xxx fx ∆ = 20 1 xx − [] 123 012 (,,) (,,) fx x x fx x x − = 1 3h () () 22 01 22 22 fxfx hh  ∆∆ −    = () () 22 10 3 6 fx fx h ∆−∆ [From (1) = () 3 0 3 3! fx h ∆ In general, we have 1 n n xx ∆ f(x 0 ) = () 0 ! n n fx nh ∆ which shows the relation between divided difference and ordinary difference. 5.5 NEWTON’S DIVIDED DIFFERENCE FORMULA Let y 0 , y 1 , y n , be the values of y = f(x) corresponding to the arguments x 0 , x 1 , x n , then from the definition of divided differences, we have [] 0 0 0 , = yy xx xx − − So, that, () [] 000 , yy x x xx =+ − (1) Again, [] [][ ] 001 01 1 = 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 1 01 , ,, y y xx xx x x xx xxx =+− + − − (3) Also [] [][ ] 01 012 012 2 ,, ,, ,,, xx x x x x xx x x xx − = − which gives, [][ ] () [] 01 012 2 012 ,, ,, ,,, xx x x x x x x xx x x =+− (4) From (3) and (4) y = () [] () [] () ()() [] 0 0 01 0 1 012 0 1 2 012 , ( ) ,, ,,, y xxxx xxxxxxx xxxxxxxxxx +− +− − +− − − Proceeding in this manner, we get y = f(x) = y 0 + (x – x 0 ) [x 0 , x 1 ] + (x – x 0 ) (x – x 1 ) [x 0 , x 1 , x 2 ] + (x – x 0 ) (x – x 1 ) (x – x 2 ) [x 0 , x 1 , x 2 , x 3 ] + + (x – x 0 ) (x – x 1 ) (x – x 2 ) (x – x n–1 ) [x 0 , x 1 , x n ] + (x – x 0 ) (x – x 1 ) (x – x n ) [x, x 0 , x 1 , x 2 , x n ] This 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 = y 0 + (x – x 0 ) 0 y∆ + (x – x 0 ) (x – x 1 ) 2 0 y∆ + (x – x 0 ) (x – x 1 ) (x – x 2 ) 3 0 y∆ + (x – x 0 ) (x – x 1 ) (x – x 2 ) (x – x 3 ) 4 0 y∆ + + (x – x 0 ) (x – x 1 ) (x – x n–1 ) 0 n y ∆ Example 1. Apply Newton’s divided difference formula to find the value of f(8) if f(1) = 3, f(3) = 31, f(6) = 223, f(10) = 1011, f(11) = 1343, Sol. The divided difference table is given by () 23 13 28 2 14 331 50510 192 3 64 9 9 1 6 223 133 7 19 788 4 197 8 8 1 10 1011 135 5 27 332 1 332 11 1343 xfx ∆ ∆∆ = = == = == = = On, applying Newton’s divided difference formula, we have f(x) or y x = y 0 + (x – x 0 ) 0 y∆ + (x – x 0 ) (x – x 1 ) 2 0 y∆ + (x – x 0 ) (x – x 1 ) (x – x 2 ) 3 0 y∆ + f(x) or y x = 3 + (x – 1) × 14 + (x – 1) (x – 3) × 10 + (x – 1) (x – 3) (x – 6) × 1 for f(8), we put x = 8 in above equation, we get f(8) = 3 + (7) (14) + (7) (5) (10) + (7) (5) (2) INTERPOLATION WITH UNEQUAL INTERVAL 249 f(8) = 3 + 98 + 350 + 70 f(8) = 521. Ans. Example 2. Find the function u x in powers of x – 1 given that u 0 = 8, u 1 = 11, u 4 = 68, u 5 = 123, Sol. Here, x 0 = x 0 , x 1 = 1, x 2 = 4, x 3 = 5, y 0 = 8, y 1 = 11, y 2 = 68, y 3 = 123, The divided difference table is given by () xfx ∆∆ ∆ 23 08 3 111 4 19 1 468 9 55 5 123 From Newton’s divided difference formula, we have y x = 8 + (x – 0) × (3) + (x – 0)(x–1) × (4) + (x – 0) (x – 1) (x – 4) × 1 y x = 8 + 3x + 4x 2 – 4x + x(x 2 – 5x + 4) y x = x 3 – x 2 + 3x + 8 In order to express it in powers 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 STATISTICAL TECHNIQUES Example 3. Find the Newton’s divided difference interpolation polynomial for: () : 0.5 1.5 3.0 5.0 6.5 8.0 : 1.625 5.875 31.0 131.0 282.125 521.0 x fx Sol. The divided difference table is given as, if here () : 0.5 1.5 3.0 5.0 6.5 8.0 : 1.625 5.875 31.0 131.0 282.125 521.0 x fx The table for divided difference is as: () xfx ∆∆∆∆ 234 0.5 1.625 4.25 1.5 5.875 5 16.75 1 3.0 31.0 9.5 0 50 1 5.0 131.0 14.5 0 100.75 1 6.5 282.125 19.5 159.25 8.0 521.0 By using Newton’s divided difference formula, we have y x = 1.625 + (x – 0.5) × 4.25 + (x – 0.5) (x – 1.5) × 5 + (x – 0.5) (x – 1.5) (x – 3) × 1 y x = 1.625 + 4.25x – 2.125 + 5x 2 – 10x + 3.75 + (x – 0.5) (x 2 – 4.5x + 4.5) y x = 1.625 + 4.25x – 2.125 + 5x 2 – 10x + 3.75 + x 3 – 4.5 x 2 + 4.5x – 0.5x 2 + 2.25 x – 3.25 ∴ y x = x 3 + 11x – 10x + 1 y x = x 3 + x + 1. Ans. Example 4. Construct a divided difference table for the following: () :1 2 4 7 12 : 22 30 82 106 216 x fx INTERPOLATION WITH UNEQUAL INTERVAL 251 Sol. The divided difference table is given as, () () () () () 23 4 122 30 22 8 21 26 8 230 6 41 82 30 36 6 26 1.6 42 71 8 26 0.535 1.6 4 82 3.6 0.194 72 121 106 82 1.75 3.6 8 0.535 74 122 22 8 7 106 1.75 12 4 216 106 22 5 12 216 xfx fx fx fx fx ∆∆ ∆ ∆ − = − − = − −−− ==− −− −+ =− = −− −+ == −− − = − − = Example 5. (i) Prove that 3 bcd 1 a  ∆   = – 1 abcd (ii) Show that the nth divided differences [x 0 , x 1 , x n ] for u x = 1 x is () n 01 n 1 x , x , x  −    Sol. (i) () () () () () () () 23 2 3 2 1 11 1 11 1 11 11 1 11 1 11 1 1 xfx fx fx fx a a ba ba ba b babc cb cb bc abcd c cbcd dc dc dc d d ∆∆∆ − =− − − − =− − − − − =− − 252 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From the table, we observe that 3 bcd ∆ 1 a    = – 1 abcd (1) (ii) From (1), we see that 3 bcd ∆ 1 a    = – 1 abcd = (– 1) 3 f(a, b, c, d) In general, 01 ,, n n xx x ∆ 0 1 x    = (–1) n f(x 0 , x 1 , x 2 , x n ) = () [] 012 1 ,,, n n xxx x − Example 6. (i) Find the third divided difference with arguments 2, 4, 9, 10 of the function f(x) = x 3 – 2x. (ii) if f(x) = 2 1 x , find the first divided differences f(a, b), f(a, b, c,), f (a, b, c, d) (iii) If f(x) = g(x) h(x), prove that f(x 1 , x 2 ) = g(x 1 ) h(x 1 , x 2 ) + g (x 1 , x 2 ) h (x 2 ) Sol. (i) () () () () 23 24 56 4 26 42 131 26 456 15 92 711 56 23 15 131 1 94 102 269 131 9711 23 10 4 980 711 269 10 9 10 980 xfx fx fx fx ∆∆∆ − = − − = − −− == −− − = − − = − Hence third divided difference is 1. INTERPOLATION WITH UNEQUAL INTERVAL 253 (ii) () () () () 23 2 22 22 2222 22 222 2 2222 22 2 1 11 1 1 1 xfx fx fx fx a a ab ba ba ab ab bc ca b babc abc acd abd bcd bc bc abcd bc cd bd c cbcd cd cd d d ∆∆ ∆ − +  =−  −  ++    +++ +   −−     ++    +  −   From the above divided difference table, we observe that first divided differences, f(a, b)= – 22 ab ab +    f(a, b, c)= 222 ab bc ca abc ++    and f(a, b, c, d)= – 2222 abc acd abd bcd abcd +++    (iii) R.H.S. = g(x 1 ) () () () () () 21 2 1 2 21 21 hx hx gx gx hx xx xx −− + −− = 21 1 xx − [{g (x 1 )h (x 2 ) – g(x 1 ) h(x 1 )} + {g(x 2 ) h (x 2 ) – g(x 1 ) h (x 2 )}] = ()() ()() 22 11 21 gx hx gx hx xx − − = 2 x ∆ g (x 1 ) h (x 1 ) = 2 x ∆ f(x 2 ) f(x 1 , x 2 ) = L.H.S. Hence the result. Example 7. The following are the mean temperatures (°F) on 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 Day Date Temp. Date Temp. 0 15 June 58.8 16 Dec. 40.7 30 15 July 63.4 15 Jan 38.1 60 14 August 62.5 14 Feb. 39.3 Sol. Divided difference table for summer is: () () () 2 0 58.8 4.6 163.4 2.75 0.9 2 62.5 xfx fx fx ∆∆ − − ∴ f(x) = 58.8 + (x – 0) (4.6) + (x – 0) (x – 1) (–2.75) = – 2.75x 2 + 7.35 x + 58.8 For maximum and minimum of f(x), we have f ′ (x)= 0 ⇒ – 5.5x + 7.35 = 0 ⇒ x = 1.342 Again, f ’’(x) = – 5.5 < 0 ∴ f(x) is maximum at x = 1.342 Since unit ≡ 30 days 1.342 = 30 × 1.342 = 40.26 days ∴ Maximum temperature was on 15 June + 40 days i.e., on 25 July and value of maximum temperature is [f (x)] max = [f(x)] 1.342 = 63.711°F approximately. Divided difference table for winter is as follows: () () () 2 0 40.7 2.6 1 38.1 1.9 1.2 2 39.3 xfx fx fx ∆∆ − INTERPOLATION WITH UNEQUAL INTERVAL 255 ∴ f(x) = 40.7 + (x – 0) (–2.6) + x(x – 1) (1.9) = 1.9x 2 – 4.5x + 40.7 For f(x) to be maximum or minimum, we have f ′ (x) = 0 3.8x – 4.5 = 0 ⇒ x = 1.184 Again, f ′′ (x) = 3.8 > 0 ∴ f(x) is minimum at x = 1.184 Again, unit 1 ≡ 30 days ∴ 1.184 = 30 × 1.184 = 35.52 days ∴ Minimum temperature was on 16 Dec + 35.5 days i.e., on the mid night of 20th Jan. and its value can be obtained similarly. [f(x)] min = [f(x)] 1.184 = 63.647 °F approximately. Example 8. The mode of a certain frequency Curve y = f (x) is very near to x = 9 and the values of frequency density f(x) for x = 8.9, 9.0 and 9.3 are respectively equal to 0.30, 0.35 and 0.25. Calculate the approximate value of mode. Sol. Divide difference table for given frequency density is as () () () 2 100 100 100 8.9 30 50 9 3500 9.0 35 36 100 3 9.3 25 xfx fx fx ∆∆ − − Applying Newton’s divided difference formula 100 f(x) = 30 + (x – 8.9) × 50 9 + (x – 8.9) (x – 9) 3500 36  −   = – 97. 222x 2 + 1745.833x – 1759.7217 ∴ f(x) = – 0.9722x 2 + 17.45833x – 17.597217 f ′ (x) = – 1.9444x + 17.45833 Putting f ′ (x) = 0, we get x = 17.45833 1.9444 = 8.9788 Also, () fx ′′ = – 1.9444 i.e., (–) ve ∴ f(x) is maximum at x = 8.9788 Hence, mode is 8.9788. . result. Example 7. The following are the mean temperatures (°F) on 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. fx fx a a ba ba ba b babc cb cb bc abcd c cbcd dc dc dc d d ∆∆∆ − =− − − − =− − − − − =− − 252 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From the table, we observe that 3 bcd ∆ 1 a    . differences of a polynomial of n th degree are constant. Let f(x) = A 0 x n + A 1 x n–1 + + A n–1 x + A n by a polynomial of degree n provided A 0 ≠ 0 and arguments be equally spaced so that x 1