106 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 1. Construct a forward difference table for the following values: 0 5 10 15 20 25 ()71114182432 x fx Sol. Forward difference table for given data is: ∆∆∆∆∆ − − − 2345 07 4 511 1 32 10 14 1 1 410 15 18 2 1 60 20 24 2 8 25 32 xyyyyyy Example 2. If y = x 3 + x 2 – 2x + 1, calculate values of y for x = 0, 1, 2, 3, 4, 5 and form the difference table. Also find the value of y at x = 6 by extending the table and verify that the same value is obtained by substitution. Sol. For x = 0, 1, 2, 3, 4, 5, we get the values of y are 1, 1, 9, 31, 73, 141. Therefore, difference table for these data is as: 23 01 0 11 8 86 29 14 22 6 331 20 42 6 473 26 68 6 5141 32 100 6241 xyy y y∆∆ ∆ Because third differences are zero therefore ∆ 3 y 3 = 6 ⇒∆ 2 y 4 – ∆ 2 y 3 = 6 ⇒∆ 2 y 4 –26 = 6 ⇒∆ 2 y 4 = 32 Now, ∆ 2 y 4 = 32 ⇒∆y 5 – ∆y 4 = 32 CALCULUS OF FINITE DIFFERENCES 107 ⇒∆y 5 – 68 = 32 ⇒∆y 5 = 100 Further, ∆y 5 = 100 ⇒ y 6 – y 5 = 100 ⇒ y 6 – 141 = 100 ⇒ y 6 = 241 Verification: For given function x 3 + x 2 – 2x + 1, at x = 6, y(6) = (6) 3 + (6) 2 – 2(6) + 1 = 241 Hence Verified. Example 3. Given f(0) = 3, f(1) = 12, f(2) = 81, f(3) = 200, f(4) = 100 and f(5) = 8. From the difference table and find ∆ 5 f(0). Sol. The difference table for given data is as follows: 2345 () () () () () () 03 9 112 60 69 10 2 81 50 259 119 269 755 3 200 219 496 100 227 4100 8 92 58 xfxfxfxfxfxfx∆∆ ∆ ∆ ∆ − − − − − − Hence, 5 (0) f∆ = 755. Example 4. Construct the forward difference table, given that: 51015202530 9962 9848 9659 9397 9063 8660 x y and point out the values of ∆ 2 y 10 , ∆ 4 y 5 . Sol. For the given data, forward difference table is as: 234 5 9962 114 10 9848 75 189 2 15 9659 73 1 262 1 20 9397 72 2 334 3 25 9063 69 403 30 8660 x y yyyy∆∆ ∆ ∆ − − − −− − − − − − From the table, ∆ 2 y 10 ,∆ 4 y 5 is as ∆ 2 y 10 = –73 and ∆ 4 y 5 = –1. 108 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 5. Find f(6) given that f(0) = –3, f(1) = 6, f(2) = 8, f(3) = 12, the third differences being constant. Sol. For given data we construct the difference table: ∆∆ ∆ − − 23 () () () () 03 9 16 7 29 28 2 4 312 xfx fx fx fx We have, f(6) = f(0+6) = E 6 f(0) = (1+ ∆) 6 f(0) = 23 (1 6 15 20 ) (0) f+∆+ ∆+ ∆ [Higher differences being zero] = 23 (0) 6 (0) 15 (0) 20 (0) ff f f+∆ + ∆ + ∆ = –3 + 6 × 9 + 15 × (–7) + 20 × 9 = –3 + 54 – 105 + 180 = 126. Example 6. Prove that: (a) f(4) = f(3) + ∆f(2) + ∆ 2 f(1) + ∆ 3 f(1). (b) f(4) = f(0) + 4∆f(0) + 6∆ 2 f(–1) + 10∆ 3 f(–1) Sol. (a) We have, f(4) – f(3) = ∆f(3) = ∆[f(2) + ∆f(2)] [Because ∆f(2) = f(3) – f(2)] = ∆f(2) + ∆ 2 f(2) = ∆f(2) + ∆ 2 [f(1) + ∆f(1)] [Because ∆f(1) = f(2) – f(1)] = ∆f(2) + ∆ 2 f(1) + ∆ 3 f(1) Therefore, f(4) = f(3) + ∆f(2) + ∆ 2 f(1) + ∆ 3 f(1) (b) We have, f(4) = E 5 f(–1) = (1 + ∆) 5 f(–1) = {1 + 5 C 1 ∆+ 5 C 2 ∆ 2 + 5 C 3 ∆ 3 } f(–1) (On taking up to third differences) = f(–1) + 5∆f(–1) + 10∆ 2 f (–1) + 10∆ 3 f(–1) = [ f(–1) + ∆f(–1)] +4[∆f(–1)+ ∆ 2 f(–1)] + 6∆ 2 f(–1) + 10∆ 3 f(–1) = −+∆− +∆ −+∆− +∆ −+ ∆ − 23 [(1) (1)] 4[(1) (1)] 6 (1) 10 (1) ff ff f f = +∆ +∆ − + ∆ − 23 (0) 4 (0) 6 ( 1) 10 ( 1) ff f f Because, −+∆−= −+ − −=(1) (1) (1) (0) (1) (0)ffffff. CALCULUS OF FINITE DIFFERENCES 109 Example 7. Find the function whose first difference is e x . Sol. We know that − ∆= − = −(1) xxhxxh ee eee , where h is the interval of differencing. Therefore,  =∆=∆  −−  1 11 x xx hh e ee ee Hence, required function is given by 1 x h e e − . Example 8. Find the first term of the series whose second and subsequent terms are 8, 3, 0, –1, and 0. Sol. If the interval of differencing is unity, then f(1) = E –1 f(2) = (1+ ∆) –1 f(2) = (1 – ∆ + ∆ 2 – ∆ 3 + )f(2). Since we have five observations, therefore the 4th differences will be constant and 5th differences will be zero. 2 () () () 28 5 33 3 2 40 2 1 51 2 1 60 xfx fx fx∆∆ − − − − Hence, f(1) = f(2) – ∆f(2) + ∆ 2 f(2) [Higher order differences are 0] f(1) = 8 – (–5) + 2 = 15 3.4.2 Backward or Ascending Differences If we subtract from each value of y except y 0 , the previous value of y, we get y 1 – y 0 , y 2 –y 1 , y 3 – y 2 , y n – y n–1 . These differences are called first backward differences of y and are denoted by ∇y. The symbol ∇ denotes the backward difference operator. That is, ∇y 1 = 10 yy− ∇y 2 = 21 yy− ∇y n = y n – y n-1 Also it can be written as, ∇+()fx h = ()()fx h fx+− Similarly, second forward difference is given by, ∇+ 2 () fx h = ∇+−∇()()fx h fx 110 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES In general, + ∇ 1 n r y = −− + ∇−∇ 11 1 , or nn rr yy ∇+() n fx h = −− ∇+−∇ 11 () () nn fx h fx Backward Difference Table: 234 00 1 2 11 2 3 23 24 22 3 4 3 34 2 33 4 4 44 xy xy y xy y yy xy y y yy xy y y xy ∇∇∇∇ ∇ ∇ ∇∇ ∇∇ ∇∇ ∇ ∇ Example 9. Construct the backward difference table for y = log x given that: 10 20 30 40 50 1 1.3010 1.4771 1.6021 1.6990 x y and find the values of ∇ 3 log 40 and 4 ∇ log 50. Sol. For the given data, backward difference table as: ∇∇∇ ∇ − −− − 23 4 10 1 0.3010 20 1.3010 0.1249 0.1761 0.0738 30 1.4771 0.0511 0.0508 0.1250 0.0230 40 1.6021 0.0281 0.0969 50 1.6990 xyy yy y Hence,Hnc 3 log 40 = 0.0738 ∇ and 4 log 50 = 0.0508 ∇− Example 10. Given that: 12345678 1 8 27 64 125 216 343 512 x y Construct backward difference table and obtain 4 () f8∇ . CALCULUS OF FINITE DIFFERENCES 111 Sol. Backward difference table for given data is as: 234 () () () () () 11 7 28 12 19 6 327 18 0 37 6 464 24 0 61 6 5 125 30 0 91 6 6 216 36 0 129 6 7 343 42 169 8 512 xfx fx fx fx fx∇∇ ∇ ∇ Hence, 4 (8) 0. f∇= Example 11. Construct the backward difference table from the data: sin 30 o = 0.5, sin 35 o = 0.5736, sin 40 o = 0.6428, sin 45 o = 0.7071 Assuming third difference to be constant, find the value of sin 25°. Sol. Backward difference table for given data is as: ∇∇ ∇ − − − − − 23 25 0.4225 0.0775 30 0.5000 0.0039 0.0736 0.0005 35 0.5736 0.0044 0.0692 0.0005 40 0.6428 0.0049 0.0643 45 0.7071 xyy y y Since third differences are constant therefore ∇ 3 y 40 = – 0.0005 ⇒∇ 2 y 40 – ∇ 2 y 35 = 0.0005 ⇒ –0.0044 – ∇ 2 y 35 = –0.0005 112 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ⇒∇ 2 y 35 = –0.0039 Again, ∇y 35 – ∇y 30 = –0.0039 ⇒ 0.0736 – ∇y 30 = –0.0039 ⇒∇y 30 = 0.0775 Again, y 30 – y 25 = 0.0775 ⇒ 0.50 – y 25 = 0.0775 ⇒ y 25 = 0.4225 Therefore, sin 25 ° = 0.4225 3.4.3 Central Differences The central difference operator is denoted by the symbol δ and central differences is given by, () ( ) ( ) or 22 hh fx fx fxδ=+−− x y δ = +− − , 22 hh xx yy or 1/2 yδ = y 1 – y 0 3/2 yδ = y 2 – y 1 − δ 1 2 n y = 1nn yy − − Central Difference Table: δδ δ δ δ δ δδ δδ δδ δ δ 234 00 1/2 2 11 1 3 3/2 3/2 24 22 2 2 3 5/2 5/2 2 33 3 7/2 44 xy xy y xy y yy xy y y yy xy y y xy 3.4.4 Other Difference Operators (a) The Operator E: The operator E is called shift operator or displacement or translation operator. It shows the operation of increasing the argument value x by its interval of differencing h so that. CALCULUS OF FINITE DIFFERENCES 113 Ef (x) = f(x + h) or Ey x = y x+h Similarly, Ef(x + h)= f(x + 2h) In general, n x Ey = or ( ) ( ) n xnh y Efx fx nh + =+ In the same manner, E –1 f(x)= f(x – h) Also, E –2 f(x)= f(x – 2h) E –n f(x)= f(x – nh) This is called inverse of shift operator. (b) Differential Operator D: The differential operator for a function y = f(x) is defined by ()Df x = () d fx dx 2 () Dfx = 2 2 () d fx dx and so on. The operator ∆ is an analogous to the operator D of differential calculus. In finite differences, we deal with ratio of simultaneous increments of mutually dependent quantities where as in differential calculus, we find the limit of such ratios when the increment tends to 0. (c) The Unit Operator 1: The unit operator 1 has a property that 1. f(x) = f(x). It is also called identity operator. (d) Averaging Operator µµ µµ µ: The operator µ is a averaging operator and is defined by, µy x = +−  +    22 1 2 hh xx yy i.e., ()fxµ =  ++ −   1 ()() 22 2 hh fx fx 3.4.5 Properties of Operators 1. The operators ,,,, and ED∆∇ δµ are all linear operators. i.e., ∇ (af (x + h) + bφ(x + h)= [af (x + h)+bφ(x + b)] – [af (x) + bφ(x)] = a[f(x + h) – f(x)] + b[φ(x + h) – φ(x)] = a ∇ f(x + h) + b ∇ φ(x+h) Hence, ∇ is a linear operator. On substituting a = 1, b = 1, we get ∇ [f(x + h) + φ(x + h)] = ∇ f(x + h) + ∇ φ(x + h) Also on substituting b = 0, we get ∇ [af(x + h)], = a ∇ f(x + h) 2. The operator is distributive over addition. 3. All the operators follows the law of indices. i.e., 114 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ∆ p ∆ q f(x)= ∆ p+q f(x) = ∆ q ∆ p f(x) Also, ∆[ f(x) + φ(x)] = ∆[φ(x) + f(x)] 4. E and ∆ are not commutative with respect to variables. 5. If f(x) = 0, then it does not mean that either ∆ = 0 or f(x) = 0. 6. Operators E and ∆ cannot stand without operands. 3.4.6 Relation between Different Operators There are few relations defined between these operators. Some of them are: 1. ∇ = 1 – E –1 or E = (1 – ∇ ) –1 2. ∆ = E – 1 or E = 1 + ∆ 3. E ∇ = ∇ E = ∆ 4. E = e hD = 1 + ∆ , where D is the differential operator. 5. δ = E 1/2 – E –1/2 6. µ = 1 2 (E 1/2 + E –1/2 ) 7. δ E 1/2 = ∆ Proof: 3. (E ∇ ) f(x)= E{ ∇ f(x)} = E{f(x) – f (x – h)} = Ef (x) – Ef (x – h) = f (x + h) – f(x) = ∆ f (x) (1) Also, ( ∇ E) f(x)= ∇ {Ef (x)} = ∇ f (x + h) = f(x + h) – f(x) = ∆f(x) (2) From (1) and (2), we get E ∇ = ∆ and ∇ E = ∆ ⇒ E ∇ = ∇ E = ∆ . 4. Ef (x)= f (x + h) = f(x) + h ′ f (x) + 2 2! h ′′ f (x) + (By using Taylor’s theorem) = 1.f(x) + hDf (x) + 2 2! h D 2 f (x) + = e hD f (x) Ef (x)= e hD f(x) or E = e hD Since, E = 1 + ∆ , therefore ∆ = e hD – 1. 5. δ y x = y + 2 h x – y x – 2 h = E 1/2 y x – E –1/2 y x = (E 1/2 – E –1/2 )y x Therefore, δ = E 1/2 – E –1/2 6. µ y x = 1 2 22 hh xx yy +−  +   = 1 2 (E 1/2 y x + E –1/2 y x ) = 1 2 (E 1/2 + E –1/2 )y x CALCULUS OF FINITE DIFFERENCES 115 Therefore, µ = 1 2 (E 1/2 + E –1/2 ) 7. δ E 1/2 y x = δ y x+h = 2 h x y + – y x = ∆ y x Therefore, δ E 1/2 = ∆ . Example 12. Show that: (a) (E 1/2 + E –1/2 ) (1 + ∆) 1/2 = 2 + ∆ (b) ∆ = 1 2 δ 2 + δ +δ 2 14 Sol. (a) Since 1 + ∆ = E therefore (E 1/2 + E –1/2 ) E 1/2 = E + 1 = 1 + ∆ + 1 = ∆ + 2. (b) 22 1 1/4 2 δ+δ +δ = 1 2 (E 1/2 – E –1/2 ) 2 + (E 1/2 – E –1/2 ) () 2 1/2 1/2 1 1 4 EE − +− = 1 2 (E + E –1 –2) + (E 1/2 – E –1/2 ) 1/2 1/2 2 EE −  −    = 1 2 (2E – 2) = E – 1 = ∆ Example 13. Prove that (1) ∆ + ∆∇ ∇= − ∇∆ (2) (1 + ∆) (1 – ∇ ) ≡ 1 Where ∆ and ∇ are forward and backward difference operators respectively. Sol. (1) ∆∇  −  ∇∆  y x = 1 1 11 1 1 EE E E − −  −− −   − −  y x = 1 1 1 1 E E E E E E  −    −   − − −      y x = 1 E E  −   y x = (E – E –1 )y x = {(1 + ∆ ) – (1 – ∇ )} y x = ( ∆ + ∇ ) y x Hence, ∆∇ − ∇∆ = ∆ + ∇ . (2) (1 + ∆ ) (1 – ∇ ) y x = (1 + ∆ ) [y x – ∇ y x ] = (1 + ∆ ) [y x –{y x – y x–h }] = (1 + ∆ ) [y x –h ] = E(y x – h ) = EE –1 y x = 1. y x (the interval of differencing being 1) Hence, (1 + ∆) (1 – ∇ ) ≡ 1. . 106 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 1. Construct a forward difference table for the following values: 0 5 10 15 20 25 ()71114182432 x fx Sol. Forward difference table. Difference Operators (a) The Operator E: The operator E is called shift operator or displacement or translation operator. It shows the operation of increasing the argument value x by its interval of differencing h. 2 2 () d fx dx and so on. The operator ∆ is an analogous to the operator D of differential calculus. In finite differences, we deal with ratio of simultaneous increments of mutually dependent quantities