136 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. The difference table for the given table is: ∆∆∆∆∆ − − − − − − 4 4 43 43 44 45 10 10 10 10 10 10 1 10000 5191 2 15191 354 5545 24 3 20736 330 24 5875 0 0 4 26611 330 24 6205 24 27 5 32816 354 51 6559 75 135 6 39375 429 84 6988 9 270 7 46363 420 186 7408 177 270 8 53771 597 84 8005 93 135 9 61776 690 51 8695 144 xyyyyyy 10 70471 834 9529 11 80000 Here, sum of fifth differences is small which may be neglected 0.270 and –270 are the adjacent values which are equal in magnitude and opposite in sign. Horizontal lines between these values point out the incorrect functional value 46363 coefficient of first middle term in (1 – p) 5 is +10. ∴ Error is given by 10e = 270 ⇒ e = 27 Hence, correct functional value 46363 27 4.6336. 10000 − = 3.7 TECHNIQUE TO DETERMINE THE MISSING TERM Let, given a set of equidistant values of arguments and its corresponding value of f(x). Suppose for n + 1 equidistant argument values x = a, a + h, a + 2h, , a + nh, are given. CALCULUS OF FINITE DIFFERENCES 137 y= f(x) = f(x 0 ), f(x 1 ), f(x 2 ) , f(a + nh). i.e., f(x n ). Let one of the value of f(x) is missing. Say it f(i). To determine this missing value of f(x), assume that f(x) can be represented by a polynomial of degree (n – 1) since n values of f(x) are known. Hence, 1 () n fx − ∆ = constant and () 0 n fx ∆= Therefore, (E – I) n f(x) = 0 because ∆ = E – 1 ⇒ 12 12 [ ( 1) ] ( ) 0 nn n n n nnnn ECEICEI EIfx −− − −+ −+− = or 12 12 ( ) ( ) ( ) ( 1) ( ) 0 nnn nn n Efx CE fx CE fx fx −− −+−+−= For first tabulated value of x, put x = 0 12 (1) (0) (0) (0) ( 1) (0) 0 2 nnn n n nn Ef E f E f f −− − ⇒− + −+−= or (1) ( ) ( 1) ( 2) ( 1) (0) 0 2 n nn fn nfn fn f − −−+ −− +− = (1) In equation (1), except missing term, each term is known and hence from this way missing term can be obtained. If two values of f(x) are mssing then in that case only (n – 1) values of f(x) can be given by a polynomial of degree (n –2). i.e., 1 () 0 n fx − ∆= or (E –1) n–1 f(x) = 0. This gives for x = 0, (the first tabulated value) and for x = 1, (second tabulated value) and by solving these two we get the two missing values for given function f(x). Similarly method proceeds to find three and more missing terms in given function f(x). Example 6. Estimate the missing term in the following table: = 0123 4 () 1 3 9 ? 81 x yfx Explain why values differ from 3 3 or 27. Sol. Since we have given 4 values, therefore 4 () 0, fx x∆=∀ i.e., 4 (1)()0, Efx x−=∀ i.e., 432 (4641)()0, EEEEfx x−+−+ =∀ i.e., 432 () 4 () 6 () 4 () () 0, Efx Efx Efx Efx fx x−+−+=∀ i.e., (4)4(3)6(2)4(1)()0, fx fx fx fx fx x+− ++ +− ++ = ∀ (on taking interval of differencing being 1) On putting x = 0, we get (4) 4 (3) 6 (2) 4 (1) (0) 0fffff−+−+= (1) Substituting the value of f(0), f(1), f(2), f(4) in (1), we get 81 4 (3) 6 9 4 3 1 0f−+×−×+= 138 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES i.e., 4f(3) = 124 i.e., f(3) = 31 (function values are 3 n type and this is not a polynomial) Example 7. Find the missing value of the data: 12 3 4 5 ( ) 7 ? 13 21 37 x fx Sol. Since 4 values are known, let us assume the fourth order differences being zero. Also since one value is unknown, we assume 4 () 0, fx x∆=∀ i.e., 4 (1)()0, Efx x−=∀ i.e 432 (4641)()0, EEEEfx x−+−+ =∀ i.e., 432 () 4 () 6 () 4 () () 0, Efx Efx Efx Efx fx x−+−+=∀ i.e., (4)4(3)6(2)4(1)()0,fx fx fx fx fx x+− ++ +− ++ =∀ (on taking interval of differencing being 1) On putting x = 0, we get (4) 4 (3) 6 (2) 4 (1) (0) 0fffff−+−+= (1) Substituting the value of f(0), f(1), f(2), f(4) in (1), we get 37 4(21) 6(13) 4 (1) 7 0f−+−+= −=⇒=38 4 (1) 0 (1) 9.5ff Hence, the required missing value is 9.5. Example 8. Find the missing values in the table: −−− 45 50 55 60 65 () 3 2 2.4 x fx Sol. Difference table is as follows: ∆∆ ∆ − − −+− +− −−− −− −− − 23 1 11 113 13 313 33 3 45 3 3 50 5 2 239 55 2 4 23.63 60 0.4 2 2.4 65 2.4 xy y y y y yy yyy yy yyy yy y CALCULUS OF FINITE DIFFERENCES 139 as only three entries y 0 , y 2 , y 4 are given, the function y can be represented by a second degree polynomial. ∴∆ 3 y 0 = 0 and ∆ 3 y 1 = 0 ⇒ 3y 1 + y 3 = 9 and y 1 + 3y 3 = 3.6 On solving these, we get y 1 = 2.925, y 2 = 0.225 Example 9. Obtain the missing terms in the following table: x123 45 6 7 8 f(x) 1 8 ? 64 ? 216 343 512 Sol. Here we have six known values, therefore sixth differences being zero. i.e., ∆ 6() 0fx= For all values of x i.e., 6 (1)()0 Efx−= , ∀x i.e., 65 4 3 2 ( 6 15 20 15 6 1) ( ) 0, EE E E EEfx−+ − + −+ = x∀ i.e., −+ − + −+=∀ 65 4 3 2 ()6()15()20()15()6()()0, Efx Efx Efx Efx Efx Efx fx x i.e., +− ++ +− ++ +− ++ = ∀( 6)6( 5)15( 4)20( 3)15( 2)6( 1) ()0, fx fx fx fx fx fx fx x (1) On putting x = 1 and x = 2 in equation (1), we get (7) 6 (6) 15 (5) 20 (4) 15 (3) 6 (2) (1) 0ff f f fff−+ − + −+= (2) f(8) – 6f(7) + 15f(6) – 20f(5) + 15f(4) – 6f(3) + f(2) = 0 (3) Putting the value of f(8), f(7), f(6), f(4), f(2), f(1) in equation (1) and (2), we get 343 – 6 × 216 + 15f(5) – 20 × 64 + 15f(3) – 6 × 8 + 1 = 0 Also 512 – 6 × 343 +15 × 216 – 20f (5) + 15 × 64 – 6f (3) + 8 = 0 i.e., 15f(5) + 15f(3) = 2280 and 20f (5) + 6f(3) = 2662 i.e., f(5) + f(3) = 152 and 10f(5) + 3f(3) = 1331 On solving these two, we get f(3) = 27 and f(5) = 125. Example 10. Assuming that the following values of y belong to a polynomial of degree 4, compute the next three values: x0 1 2 3 4567 y1-11 11−−−− 140 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. For the given data, difference table is: ∆∆ ∆ ∆ − − − − − − ∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆ 234 3 2 2 3 3 43 2 54 3 54 2 65 6 7 01 2 11 4 28 21 4 16 28 31 4 16 2 41 16 516 6 7 xy y y y y y y yy yy yy yy y y Since values of y belong to a polynomial of degree 4, fourth difference must be constant. But 4 0 16 y∆= Therefore other fourth order differences will be 16. Thus, 4 1 16 y∆= , ∴ 33 21 16 yy∆−∆= ⇒ 3 2 24 y∆= ∴ 22 32 24 yy∆−∆= ⇒ 2 3 28 y∆= 43 28 yy∆−∆= ⇒ 4 30 y∆= 54 30 yy−= ⇒ 5 31 y = Again, ∆ 4 y 2 = 16 then after solving, we get y 6 = 129 and ∆ 4 y 3 = 16 gives y 7 = 351. CALCULUS OF FINITE DIFFERENCES 141 PROBLEM SET 3.2 1. Locate the error in the following table and correct them? 3.60 3.61 3.62 3.63 3.64 3.65 3.66 3.67 3.68 ( ) 0.112046 0.120204 0.128350 0.136462 0.144600 0.152702 0.160788 0.168857 0.176908 x fx [Ans. f(3.63) = 0.136482] 2. Locate the error in the following: –1, 0, 7, 26, 65, 124, 215, 342, 511 [Ans. correct value = 63, Error = 2] 3. Obtain the missing term in the following table: 2.0 2.1 2.2 2.3 2.4 2.5 2.6 ( ) 0.135 ? 0.111 0.100 ? 0.082 0.074 x fx [Ans. f(2.1) = 0.123, f(2.4) = 0.0900] 4. Estimate the production for the year 1964 and 1966 from the following data: 1961 1962 1963 1964 1965 1966 1967 200 220 260 ? 350 ? 430 Year Production [Ans. f(1964) = 306 f(1966 = 390)] 5. Given, log 100 = 2, log 101 = 2.0043, log 103 = 2.0128, log 104 = 2.0170. find log 102. [Ans. log 102 = 2.0086] 6. Estimate the missing term in the following: 1234 5 6 7 2 4 8 ? 32 64 128 x y Explain why the result differs from 16. [Ans. f(4) = 16.1] 7. Find the first term of the series whose second and subsequent terms are 8, 3, 0, –1, 0. [Ans. First term is 15.] 8. Obtain the missing term in the following table: 0 0.1 0.2 0.3 0.4 0.5 0.6 0.135 ? 0.111 0.100 ? 0.082 0.074 x f(x) [Ans. f(0.1)=0.123, f(0.4) = 0.090] 142 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 9. Evaluate the production of wool in the year 1935 from the given data: ( ) 1931 1932 1933 1934 1935 1936 1937 17.1 13 14 9.6 ? 12.4 18.2 Year x Production (y) [Ans. 6.6] 3.8 SEPARATION OF SYMBOLS The relation E = 1 + ∆ ⇒ E n = (1 + ∆) n has been used to express E n y x in terms of y x and its differences. (1+∆) n has been expanded by binomial theorem without using y x in it. Such methods of operations are known as method of separation of symbols. Point to be noted that the operations on symbols has no meaning without operand y x i.e., y x+nh = E n y x = (1 + ∆) n y x = (1 + n C 1 ∆ + n C 2 ∆ 2 + )y x = y x + n C 1 ∆y x + n C 2 ∆ 2 y x + This type of operation in which we separate the operand from operator is called separation of symbols. Example 1. Show that ∆ r y k = ∇ r y k+r Sol. We know ∇ = 1 – E –1 . Therefore ∇ r y k+r = (1 – E –1 ) r y k+r = 1 r E E −    y k+r =(E – 1) r (E –r y k+r ) = (E – 1) r (E –r y k+r ) = ∆ r y k [3 ∆ ≡ E – 1] Example 2. Show that − = ∆ ∑ n1 2 k0 f k = ∆ f n – ∆f 0 Sol. 1 2 0 n k − = ∆ ∑ f k = () 1 2 0 1 n k E − = − ∑ f k = () 1 2 0 21 n k EE − = −+ ∑ f k = () 1 21 0 2 n kkk k fff − ++ = −+ ∑ = f 2 – 2 f 1 + f 0 + f 3 – 2 f 2 + f 1 + f 4 – 2 f 3 + f 2 + f 5 – 2f 4 + f 3 + f n–1 – 2f n–2 + f n–3 CALCULUS OF FINITE DIFFERENCES 143 + f n – 2f n–1 + f n–2 + f n+1 – 2f n + f n–1 = f n+1 – f n + f 0 – f 1 , on adding and canceling the diagonal terms =(f n+1 – f n ) – (f 1 – f 0 ) = ∆ f n – ∆ f 0. Example 3. Prove that + 1 x 2 y = 1 2 (y x + y x+1 ) – 1 16 ( 2 x y ∆ + ∆ 2 y x+1 ); Assuming that, 3 ∆ y x =0. Sol. 1 2 x y + = E 1/2 y x = (1 + ∆) 1/2 y x = 2 11 1 28  +∆−∆   y x (1) Because 3 x y ∆ = 0 Now, 3 x y ∆ = 0 ⇒ 22 1xx yy + ∆−∆ = 0 ⇒ 2 1x y + ∆ = 2 x y ∆ and x y ∆ = 1x y + – y x Therefore from equation (1), we have 1 2 x y + = 1 1 () 2 xxx yyy + +− – 22 1 1 82 2 xx yy +  ∆∆ +    = 22 11 11 ()( ) 216 xx x x yy y y ++ +−∆+∆ . Example 4. Using the method of separation of symbol, show that ∆ −+−+− = −∆ + − 2 01234 0 0 0 11 1 u u u u u u u u 24 8 Sol. On taking R. H. S of given identity 23 0 11 1 1 1 22 2 2 u   −∆+∆−∆+      = 1 00 11 11 1 1 222 1 2 uu −  ⋅=+∆    +∆   = 11 00 (2 ) (1 ) uEu −− +∆ = + = 23 0 (1 ) EE E u−+−+ = 0123 uuuu−+−+ Hence proved. Example 5. Prove by the method of separation of symbols, that  +++++= +∆+∆+   2 234 x 2 3 12 4 0000 uuu u x u x x x x e u x u u 1! 2! 3! 4! 2! Sol. L.H.S. of given identity = 234 312 4 0 1! 2! 3! 4! u uu u uxx x x++ + + + = 23 23 00 0 0 1! 2! 3! xx x uEu Eu Eu++ + + 144 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 23 23 0 1 1! 2! 3! xx x EE E u  ++++   = (1 ) 00 0 xE x x x eu e u e eu +∆ ∆ ==⋅ = 23 23 0 1 2! 3! x xx ex u  +∆+ ∆+ ∆+   = 23 23 0000 2! 3! x xx eu xu u u  +∆+∆+∆+   = R.H.S. Example 6. Prove that: u x = u x–1 + 2x u − ∆ + 2 3x u − ∆ + + 1nn xn xn uu − −− ∆+∆ Sol. To prove u x = u x–1 + ∆u x–2 + 2 3x u − ∆ + + 1n xn u − − ∆ + n xn u − ∆ , shift the last term of right hand side and then solve left hand side. u x – n xn u − ∆ = (1 – ∆ n E –n )u x = 1 n E  ∆  −      u x = 1 n E (E n – n ∆ ) u x = 1 n E nn E E  −∆  −∆  u x 3 1 + ∆ = E ⇒ E – ∆ = 1 = 1 n E [E n–1 + 2n E − ∆ + 23n E − ∆ + + 1n− ∆ ]u x = (E –1 + 2 E − ∆ + 23 E − ∆ + + 1nn E −− ∆ )u x = u x–1 + 2x u − ∆ + 2 3x u − ∆ + + 1n xn u − − ∆ Example 7. Show that: u 2n –n C 1 2u 2n–1 + n C 2 2 2 u 2n–2 – + (–2) n u n = (–1) n (c – 2an) Where u n = an 2 + bn + c Sol. L.H.S = u 2n –n C 1 2u 2n–1 + n C 2 2 2 u 2n–2 – + (–2) n u n = E n u n –n C 1 ·2E n–1 u n + n C 2 .2 2 E n–2 u n – + (–2) n u n = (E n–n n C 1 .2E n–1 + n C 2 .2 2 E n–2 – +(–2) n u n = (E – 2) n u n = (∆ – 1) n u n = (–1) n (1 – ∆) n u n = (–1) n () 2 1 1 1.2 n nn nu −  −∆+ ∆   (On neglecting higher order differences as u n is a polynomial of second degree) = () () 2 1 1 2 n nn n nn unu u −  −−∆+ ∆   = () ()()() 2 22 22 1 2 n nn an bn c n an bn c an bn c  − −++−∆+++∆++   = () (){}() 2 22 22 1 2 n nn an bn c n a n b n a n  − −++−∆+∆+∆   CALCULUS OF FINITE DIFFERENCES 145 = () () () {} () 2 22 111 n an bn c n a n an bn n n  −++−+−−+−   () {} 2 2 2 1 2 nn an n  − +∆+−   = () () () () 2 2 12 21 2 n nn an bn c n an a b a n  − −++−+++⋅∆+   = () () () () 2 2 12{212} 2 n nn an bn c n an a b a n n  − − + +− +++ ⋅ +−    = () () () () 22 12 n an bn c n an a b a n n  −++−+++−  = (–1) n [c –2an] = R.H.S. Example 8. Using the method of separation of symbols, show that: n xn u − ∆ = u x – nu x–1 + () − nn 1 2 u x–2 – + (–1) n u x–n Sol. R.H.S. = u x – nu x–1 + () 1 2 nn − u x–2 – + (–1) n u x–n = −− − − −+ −+− 12 (1) ( 1) 2 nn xx x x nn unEu Eu Eu = 12 (1) 1 ( 1) 2 n xx nn nE E E u u −− − −  −+ −+−   = 1 (1 ) n x Eu − − = 11 1 nn n xxx n E uuu EE E −∆   −= =     = − − ∆=∆= 1 L.H.S. nn n xx Eu u Example 9. Use the method of separation of symbols to prove the following identities: 1. −− +∆+∆+=+∆ +∆ + xx x2x4 01122 x1x12x2 u C u C u u C u C u 2. +++ −+−+ xx1x2x3 uuuu = −− − −    −∆ + ∆ − ∆ +        23 24 6 x (1/ 2) x (3/ 2) x (5/ 2) x (7 / 2) 1 1 1.3 1 1.3.5 1 u u u u 28 2!8 3!8 3. u 0 + n C 1 u 1 x + n C 2 u 2 x 2 + n C 3 u 3 x 3 + = (1 + x) n u 0 + n C 1 (1 + x) n–1 x∆u 0 + n C 2 (1 + x) n–2 x 2 ∆ 2 u 0 + Sol. 1. R.H.S. = −− +∆ +∆ + 24 1122 xx xx x uCu Cu = −− +∆ +∆ + 21 42 12 xx xx x uCEuCEu = −− +∆ +∆ + 21 42 12 [1 ] xx x CE CE u . TERM Let, given a set of equidistant values of arguments and its corresponding value of f(x). Suppose for n + 1 equidistant argument values x = a, a + h, a + 2h, , a + nh, are given. CALCULUS OF FINITE. 0f−+×−×+= 138 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES i.e., 4f(3) = 124 i.e., f(3) = 31 (function values are 3 n type and this is not a polynomial) Example 7. Find the missing value of the data: 12. term, each term is known and hence from this way missing term can be obtained. If two values of f(x) are mssing then in that case only (n – 1) values of f(x) can be given by a polynomial of degree