A textbook of Computer Based Numerical and Statiscal Techniques part 15 doc

126 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Similarly, ∆ n 1 x    = ()()() () ()()() −−− − ++ + 1 2 3 1 2 n xx x x n = () ()()() 1! 12 n n xx x x n − ++ + (2) Similarly, ∆ (ab cx )= a ∆ b cx = a{b c(x + 1) – b cx } = a{b cx b c – b cx } = a(b c –1)b cx . 2 () cx ab ∆ = {} () (1) cx ccx ab a b b ∆∆ = ∆ − = 2 (1) (1) ccxccx ab b ab b −∆ = − Proceeding in the same manner, we get () ncx ab ∆ = a(b c – 1) n b cx . Example 29. If p, q, r and s be the successive entries corresponding to equidistant arguments in a table, show that when third differences are taken into account, the entry corresponding to the argument half way between the arguments of q and r is + 1 AB 24 , where A is the arithmetic mean of q, r and B is the arithmetic mean of 3q – 2p – s and 3r – 2s – p. Sol. On taking h being the interval of differencing the difference table is as: ∆∆ ∆ − +−+ −−+− +−+ − + 23 2 33 22 3 xx x x xu u u u ap qp ah q r qp rq s r qp ahr srq sr ahs The argument half way between the arguments of q and r is 1 2 (a + h + a + 2h) i.e., 3 2 ah+ . Hence, the required entry is given by, u a+(3/2)h = E 3/2 u a = (1 + ∆) 3/2 u a =   +∆+ ∆+ − ∆     23 3311 3111 1 2222! 2223! a u , (Higher order differences being neglected). Therefore u a+(3/2)h = 23 33 1 28 16 aa a a uu u u +∆ +∆ − ∆ = 33 1 ()(2) (33) 28 16 pqp rqp srqp+−+−+− −+− CALCULUS OF FINITE DIFFERENCES 127 = 331 333 33 1 1 2 8 16 2 4 16 8 16 16 pqrs  −++ + −− + + −   = 1991 16 16 16 16 pqrs−++− = 1111 () 16 16 2 16 pqr s  −++ +−   = 11 () ( ) 216 qr qrps++ +−− (1) Again A = arithmetic mean of q and r = 1 () 2 qr+ B = Arithmetic mean of 3q – 2p – s and 3r – 2s – p is = [] () 13 32 32 22 q ps r sp qrsp −−+−−= +−− . ∴ 1 24 AB+ = 1 (). 216 qr qrsp + ++−− Substituting this value in (1), we get u a+(3/2)h = A + 1 24 B. Example 30. Given u 0 , u 1 , u 2 , u 3 , u 4 and u 5 . Assuming that, fifth order differences to be constant. Show that: −+ − =+ 1 2 2 1 25(c b) 3(a c) uc 2 256 . where a = u 0 + u 5 , b = u 1 + u 4 , c = u 2 + u 3 Sol. L.H.S. 1 2 2 u = E 5/2 u 0 = (1 + ∆) 5/2 u 0 = 25 0 55 55555 .1 1234 5 22 22222 1 22! 5! u    −−−−−      +∆+ ∆+ + ∆    = 23 4 5 00 0 0 0 0 515 5 5 3 2 8 16 128 256 uu u u u u+∆ + ∆ + ∆ − ∆ + ∆ = 010 210 3210 515 5 ( ) ( 2 ) ( 3 3 ) 28 16 u uu uuu uuuu+−+ −++ −+−+ 543210 3 ( 5 10 10 5 ) 256 uu u uuu+−+−+− 128 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 05 14 23 32575 ()()() 256 256 128 uu uu uu+− ++ + = 32575 256 256 128 abc−+ = 325111 256 256 2 128 ab c  −++   = 3( ) 25( ) 2 256 cac cb−+ − + (R.H.S.) Example 31. Given: u 0 + u 8 = 1.9243, u 1 + u 7 = 1.9590, u 2 + u 6 = 1.9823, u 3 + u 5 = 1.9956. Find u 4 . Sol. Since 8 entries are given, therefore we have ∆ 8 u 0 = 0 i.e. (E – 1) 8 u 0 = 0 i.e. (E 8 – 8 C 1 E 7 + 8 C 2 E 6 – 8 C 3 E 5 + 8 C 4 E 4 – 8 C 5 E 3 + 8 C 6 E 2 – 8 C 7 E 1 + 1)u 0 = 0 i.e. (E 8 – 8E 7 + 28E 6 – 56E 5 + 70E 4 – 56E 3 + 28E 2 – 8E + 1)u 0 = 0 i.e. u 8 – 8u 7 + 28u 6 – 56u 5 + 70u 4 – 56u 3 + 28u 2 – 8u 1 + u 0 = 0 i.e. (u 8 + u 0 ) – 8(u 7 + u 1 ) + 28(u 6 + u 2 ) – 56(u 5 + u 3 ) + 70u 4 = 0 On putting the given values, we get 1.9243 – 8(1.9590) + 28(1.9823) – 56 (1.9956) + 70u 4 = 0 or –69.9969 + 70u 4 = 0 or u 4 = 0.9999557. Example 32. Sum the following series 1 3 + 2 3 + 3 3 + + n 3 using the calculus of finite differences. Sol. Let 1 3 = u 0 , 2 3 = u 1 , 3 3 = u 2 , , u 3 = u n–1 . Therefore sum is given by S = u 0 + u 1 + u 2 + + u n–1 = (1 + E + E 2 + E 3 + + E n–1 )u 0 = 1 1 n E E  −   −  u 0 = () 11 n  +∆ −  ∆   u 0 = 1 ∆ () ()() 23 112 1 1 2! 3! n nn nn n n −−−  +∆+ ∆+ ∆+ +∆−   u 0 = () () () 2 00 12 1 2! 3! nn n nn nu u −− − +∆+ ∆ + We know ∆u 0 = u 1 – u 0 = 2 3 – 1 3 = 7. ∆ 2 u 0 = u 2 – 2u 1 + u 0 = 3 3 –2(2) 3 + 1 3 = 12. Similarly we have obtained ∆ 3 u 0 = 6 and ∆ 4 u 0, ∆ 5 u 0 , are all zero as u r = r 3 is a polynomial of third degree. CALCULUS OF FINITE DIFFERENCES 129 ∴ S = n + () 1 2! nn − (7) + ()() 12 6 nn n −− 12 + ()()() 123 24 nn n n −−− (6) = 2 4 n (n 2 + 2n + 1) = () 2 1 2 nn  +   Example 33. Prove that: ∞∞ ==  ∆∆ =+−+−   ∑∑ 2 2x x x0 x0 11 uu1 2424 u 0 . Sol. Taking right hand side of the given expression = 2 0 11 1 2424 x x u ∞ =  ∆∆ +−+−   ∑ u 0 = 1 2 (u 0 + u 1 + u 2 + u 3 + ) + 1 4 1 1 2 − ∆  +   u 0 = 1 2 (u 0 + Eu 0 ++ E 2 u 0 + E 3 u 0 + ) + 1 4 1 1 2 − ∆  +   u 0 = 1 2 (1 + E + E 2 + E 3 + )u 0 + 1 4 1 1 2 − ∆  +   u 0 = 1 2 (1 – E) –1 u 0 + 1 2 (2 + ∆ ) –1 u 0 = 1 2 (1 – E) –1 u 0 + 1 2 (1 + ∆ ) –1 u 0 = 1 2 [(1 – E) –1 + (1 + E) –1 ]u 0 = 1 2 . 2 [1 + E 2 + E 4 + E 6 + ]u 0 = u 0 + u 2 + u 4 + u 6 + = 2 0 x x u ∞ = ∑ = L.H.S. Example 34. Given that u 0 = 3, u 1 = 12, u 2 = 81, u 3 = 200, u 4 = 100, u 5 = 8. Find the value of ∆ 5 u 0 . Sol. We know ∆ = E – 1, therefore, 5 0 u∆ = (E –1) 5 u 0 = (E 5 – 5E 4 + 10E 3 – 10E 2 + 5E – 1)u 0 = u 5 – 5u 4 + 10u 3 – 10u 2 + 5u 1 – u 0 = 8 – 500 + 2000 – 810 + 60 – 3 = 755. 130 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM SET 3.1 1. Form the forward difference table for given set of data: X:10203040 Y: 1.1 2.0 4.4 7.9 2. Construct the difference table for the given data and hence evaluate 3 (2) f∆ . X:01234 Y: 1.0 1.5 2.2 3.1 4.6 [Ans. 0.4] 3. Find the value of E 2 x 2 when the values of x vary by a constant increment of 2. [Ans. x 2 + 8x + 16] 4. Evaluate E n e x when interval of differencing is [Ans. E n e x = e x+nh ] 5. Evaluate ∆ 3 (1 –x) (1–2x) (1–3x) ; the interval of differencing being unity. [Ans. 3 () 36 fx∆=− ] 6. If f(x) = exp (ax), evaluate () n fx ∆ [Ans. (1) nax ah nax ee e ∆=− ] 7. Evaluate 2 3 x E  ∆   [Ans. 6x] 8. Find the value of 24 2 22 (1) xx aa a  + ∆  −  [Ans. 2224 (1) xx aa a ++ ] 9. Evaluate: (a) cot 2 x ∆ [Ans. –Cosec 2 x+1 ] (b) ∆+sin ( )ha bx [Ans. ++2sin cos ( ) 22 bb hhabx ] (c) tan ax∆ [Ans. sin cos cos ( 1) a ax a x + ] 10. Prove that: (a) 1/2 1 2 E =µ= δ (b) 1/2 1/2 1 () EE E −− δ+ =∆+∆ (c) 1/2 1/2 (1 ) (1 ) −− δ=∇ −∇ =∆ +∆ (d) 1/2 1/2 EE − δ=∆ =∇ (e) 2 ∇∆ = ∆∇ = δ (f) 11 1 1 EE E −− − ∇=∆ = ∆= − 11. Show that: (a) sin cot( ) sin( ) sin( ) b abx abx abbx − ∆+= +++ (b) (( ) sin( ) (2sin ) sin( 22 n n bnb abx abx +π ∆+= ++ (c) () cos( ) (2 sin ) cos( 22 n n bnb abx abx +π ∆+= ++ CALCULUS OF FINITE DIFFERENCES 131 12. What is the difference between 2 E ∆    u x and 2 2 x x u Eu  ∆    . If u x = x 3 and the interval of differencing is unity. Find out the expression for both. [Ans. 6h 2 (3x – h), 23 3 66 (2) xh h xh + + ] 13. If f(x) = e ax , show that f(0) and its leading differences form a geometrical progression. 14. A third degree polynomial passes through the points (0, –1), (1, 1), (2, 1) and (3, 2). Find the polynomial. [Ans. 32 1 (3166) 6 xx x−+−+ ] 15. Prove that ∆ sin –1 x = 22 [( 1) 1 1 ( 1) ] xxxx +−−−+ . 3.5 FUNDAMENTAL THEOREM ON DIFFERENCES OF POLYNOMIAL Statement: If f(x) be the nth degree polynomial in x, then the n th difference of f(x) is constant and ∆ n+1 f(x) and all higher differences are zero when the values of the independent variables are at equal interval. Proof: Consider the polynomial f(x) = a 0 + a 1 x + a 2 x 2 + + a n x n (1) Where n is a positive integer and a 0 , a 1 , a 2 , a n are constants. We know ∆f(x)= f(x + h) – f(x). On applying the operator ∆ on equation (1), we get ∆f(x)= ∆(a 0 + a 1 x + a 2 x 2 + + a n x n ) ⇒ f(x + h) – f(x)= [a 0 + a 1 (x + h) + a 2 (x + h) 2 + + a n (x + h) n ] –[a 0 + a 1 x + a 2 x 2 + +a n x n ] 22 33 12 3 [( ) ] [( ) ] [( ) ] nn n ahaxhxaxhx axhx ⇒+ +−+ +−+ + +− ⇒ a 1 b + a 2 [ 2 C 1 xh + h 2 ] + a 3 [ 1 2 323 23 1 2 12 1 ] [ ] n nn nCx nn nn Cxh Cxh h a Cx h h Ch − − ++++ + + 221 12 3 1 nn nn bbxbx bx nahx −− − ⇒+ + + + + (2) where b 1 , b 2 , b n–1 are constant coefficients. According to equation (2), we have the first difference of equation (1) is again a polynomial of degree n – 1. From this we say that ∆f(x) is one degree less than the degree of original polynomial. Again, on taking a difference of equation (2) i.e. second difference of equation (1), we get 2222 23 4 ( ) ( 1) n n fx C Cx Cx nn hax − ∆=++++− …(3) This is a polynomial of degree n – 2. Thus, on continuing this process up to nth difference we get a polynomial of degree zero. Such that: () n fx ∆ = ( 1)( 2) 1. nnn n nn n h ax − −− = 0 ! n n nhax = ! n n nha 132 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Hence, we have nth difference of the polynomial is constant and so all higher differences are each zero. i.e. 12 () () 0 nn fx fx ++ ∆=∆== 3.6 ESTIMATION OF ERROR BY DIFFERENCE TABLE Let y 0 , y 1 , y 2 , y n be the exact values of a function y = f(x) corresponding to arguments x 0 , x 1 , x 2 , , x n . Now to determine error in such a case and to correct the functional values, let an error δ is made in entering the value of y 3 in the table so that erroneous value of y 3 is y 3 + δ. 2 00 11 0 2 22 1 0 2 33 2 1 2 44 3 2 2 55 4 3 4 66 5 2 xy y y xy xy y xy y y xy y y xy y y xy y y y xy y ∆∆ ∆ ∆∆ +δ ∆ +δ ∆ +δ ∆−δ ∆ −δ ∆∆+δ ∆ ∆ From the above difference table we noted that: 1. The error in column y affects two entries in column ∆y, three entries in column ∆ 2 y and so on. i.e. the error spreads in triangular form. 2. The error increases with the order of differences. 3. The coefficients of δ’s are binomial coefficients with alternative signs +, –, 4. In various difference columns of the above table the algebraic sum of the errors is zero., 5. The errors in the column ∆ i y are given by the coefficients of the binomial expansion (1 –δ) i . 6. In even differences columns of ∆ 2 y, ∆ 4 y, , the maximum error occurs in a horizontal line in which incorrct value of y lies. 7. In odd difference columns of ∆ 1 y, ∆ 3 y, , the maximum error lies in the two middle terms and the incorrect value of y lies between these two middle terms. Example 1. Find the error and correct the wrong figure in the following functional values: Sol. 1234567 2 5 10 18 26 37 50 x y CALCULUS OF FINITE DIFFERENCES 133 ∆∆ ∆ − − 23 12 3 25 2 51 310 3 83 418 0 83 526 3 11 1 637 2 13 750 xy y y y Here the sum of all the third differences is zero and the adjacent values –3, 3 are equal in magnitude. Also horizontal line between –3 and 3 points out the incorrect functional value 18. Therefore coefficient of first middle term on expansion of (1 – p) 3 = –3 ⇒ –3e = –3 ⇒ e = 1 ∴ Correct functional value = 18 – 1 = 17. Example 2. Find and correct by means of differences the error in the following table: 20736, 28561, 38416, 50625, 65540, 83521, 104976, 130321, 160000 Sol. For the given data we form the following difference table: ∆∆∆∆∆ − − 2345 20736 7825 28561 2030 9855 324 38416 2354 28 12209 352 20 50625 2706 8 14915 360 40 65540 3066 48 17981 408 40 83521 3474 8 21455 416 20 104976 3890 28 25345 444 130321 4334 29679 160000 y y y yyy → ← 134 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From this table we have the third differences are quite iregular and the irregularity starts around the horizontal line corresponding to the value y = 65540. Since the algebraic sum of the fifth differences is 0, therefore –5ε = –20 ⇒ ε = 4. Therefore the true value of y 5 = 65540 – 4 = 65536. Example 3. Locate the error in following entries and correct it. 1.203, 1.424, 1.681, 1.992, 2.379, 2.848, 3.429, 4.136 Sol. Difference table for given data is as follows: ∆∆ ∆ ∆ − − 3 3 32 33 34 10 10 10 10 10 1203 221 1424 36 257 18 1681 54 4 311 22 1992 76 16 387 6 2379 82 24 469 30 2848 112 16 581 14 3429 126 707 4136 yy y y y Sum of all values in column of fourth difference is –0.004 which is very small as compared to sum of values in other columns. ∴∆ 4 y = 0 Errors in this column are e, –4e, 6e, –4e and e. Term of Maximum value = 24 ⇒ 6e = 24 ⇒ e = 4. Error lies in 2379. Hence, required correct entry = 2379 – 4 = 2375. Hence, correct value = 2.375 → ← CALCULUS OF FINITE DIFFERENCES 135 Example 4. One number in the following is misprint. Correct it. 1 2 4 8 16 26 42 64 93. Sol. Difference table for given data it as follows: ∆∆∆∆∆ − − − − − 2345 11 1 22 1 21 34 2 1 42 5 48 4 4 8210 516 2 6 10 4 10 626 6 4 16 0 5 742 6 1 22 1 864 7 29 993 xyyyyyy In the above table, the fourth difference column have algebraic sum of all the values is 0. The middle term of this difference column is 6. ∴ 6e = 6 or e = 1. ∴ Correct value is given by 16 – 1 = 15. Example 5. Locate the error in the following table: 1234567891011 1.0000 1.5191 2.0736 2.6611 3.2816 3.9375 4.6363 5.3771 6.1776 7.0471 8.0000 x y . are taken into account, the entry corresponding to the argument half way between the arguments of q and r is + 1 AB 24 , where A is the arithmetic mean of q, r and B is the arithmetic mean of. ax − −− = 0 ! n n nhax = ! n n nha 132 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Hence, we have nth difference of the polynomial is constant and so all higher differences are each zero. i.e. 12 (). yyy → ← 134 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From this table we have the third differences are quite iregular and the irregularity starts around the horizontal line corresponding