176 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. The difference table for the given date is as follows: xy y y y∇∇ ∇ 23 34.8 3.6 4 8.4 2.5 6.1 0.5 514.5 3 9.1 0.5 623.6 3.5 12.6 0.5 7 36.2 4 16.6 0.5 852.8 4.5 21.1 9 73.9 To obtain first term, we use Newton’s forward interpolation formula, Here, a = 3, h = 1, x = 1 ∴ u = –2 Hence we have f(1) = 23 33 3 3 (1) (1)(2) 2! 3! uu uu u yuy y y −−− +∆ + ∆ + ∆ On putting the subsequent values, we get f(1) = () () 2(3) (2)(3)(4) 4.8 2 3.6 (2.5) (0.5) 3.1 26 −− −−− +− × + + = Similarly, to obtain tenth term, we use Newton’s backward interpolation formula. So a + nh = 9, h = 1, a + nh + uh = 10 ∴ u = 1 ⇒ f (10) = 23 99 9 9 (1) (1)(2) 2! 3! uu uu u yuy y y +++ +∇ + ∇ + ∇ = 73.9 + 21.1 + 4.5 + 0.5 = 100 Example 4. Find the value of an annuity at 3 5 8 %, given the following table: 11 44 5 5 6 22 172.2903 162.888 153.7245 145.3375 137.6483 Rate Annuity Value INTERPOLATION WITH EQUAL INTERVAL 177 Sol. Difference table, 23 4 4 172.2903 9.4014 1 4 162.888 0.237 2 9.1644 0.537 5 153.7245 0.774 0.6132 8.387 0.0762 1 5 145.3375 0.6978 2 7.6892 6 137.6483 Rate Annuity Values ∇∇ ∇ ∇ − − − −− − x = 343 1 5,6, 88 2 an=== y = () 23 ( 1.25)( 0.25) ( 1.25)( 0.25)(0.75) (6) 1.25 (6) (6) (6) 2! 3! yy y y −− −− +− ∇ + ∇ + ∇ = ( 1.25)( 0.25) 137.6483 ( 1.25)( 7.6892) (0.6978) 2! −− +− − + ( 1.25)( 0.25)(0.75) ( 1.25)( 0.25)(0.75)(1.75) ( 0.0762) (0.6132) 3! 4! −− −− +−+ = 147.2251 Approx. Example 5. In an examination, the number of candidates who obtained marks between certain limits are as follows: Marks No of candidates −−−−−0192039405960798099 . 41 62 65 50 17 Find no. of candidates who obtained fewer than 70 marks. Sol. First, we form the difference table. Marks less than x No of candidates y ∇∇ ∇ ∇ − − − − 234 () . () 19 41 62 39 103 3 65 18 59 168 15 0 50 18 79 218 33 17 99 235 178 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here, we have h = 20, a = 99 ∴ u = 70 99 1.45 20 − =− Now on applying ‘Newton’s backward difference formula, we get f(70) = 23 4 (1)(1)(2)(1)(2)(3) () () () () () 2! 3! 4! uu uu u uu u u fa ufa fa fa fa ++++++ +∇ + ∇ + ∇ + ∇ = 235 + (–1.45)(17) + ( 1.45)( 0.45) ( 1.45)( 0.45)(0.55) ( 33) ( 18) 26 −− −− ×− + ×− = 235 – 24.65 – 10.76625 – 1.076625 = 235 – 36.492875 = 198.507 { 198 ∴ Total no. of candidates who obtained fewer than 70 marks are 198. Example 6. The area A of a circle of diameter d is given for the following values: d A 80 85 90 95 100 5026 5674 6362 7088 7854 Find A for 105. Sol. First of all we form the difference table as follow: dA xfx ∇∇∇∇ − 234 () () 80 5026 648 85 5674 40 688 2 90 6362 38 4 726 2 95 7088 40 766 100 7854 Here, h = 5, a = 100, x = 105 ∴ u = 105 100 1 5 − = Now on applying Newton’s backward difference formula, we have (105)f = 23 4 (1)(1)(2)(1)(2)(3) () () () () () 2! 3! 4! uu uu u uu u u fa ufa fa fa fa ++++++ +∇ + ∇ + ∇ + ∇ INTERPOLATION WITH EQUAL INTERVAL 179 = 40 2 1 2 3 1 2 3 4 7854 1 766 2 4 26 24 × ×× ××× +× + + ×+ × = 7854 + 766 + 46 = 8666 Which is the required area for the given diameter of circle. Example 7. The probability integral 2 1 t 2 0 2 Pe − = ∫ π π dt has the following values: x P 1.00 1.05 1.10 1.15 1.20 1.25 0.682689 0.706282 0.728668 0.749856 0.769861 0.788700 Calculate P for x = 1.235 Sol. First we form the difference table xfx ∇∇ ∇∇ ∇ − − − − − 2345 () 1.00 0.682689 0.023593 1.05 0.706282 0.001207 0.022386 0.000009 1.10 0.728668 0.001198 0.000006 0.021188 0.000015 0.000004 1.15 0.749856 0.001183 0.000002 0.020005 0.000017 1.20 0.769861 0.001166 0.018839 1 .25 0.788700 Here, h = 0.05 a = 1.20 ∴ u = 1.235 1.20 0.3 0.05 − =− ()fx = 23 (1) (1)(2) () () () () 2! 3! uu uu u fa ufa fa fa +++ +∇ + ∇ + ∇ + 45 ( 1)( 2)( 3) ( 1)( 2)( 3)( 4) () () 4! 5! uu u u uu u u u fa fa +++ ++++ ∇+ ∇ = ( 0.3)(0.7) ( 0.3)(0.7)(1.7) 0.788700 ( 0.3)(0.018839) ( 0.001166) (0.000017) 26 −− +− + − + () 0.3 (0.7)(1.7)(2.7) ( 0.3)(0.7)(1.7)(2.7)(3.7) (0.000002) (0.000004) 24 120 − − ++ 180 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 0.788700 – 0.0056517 + 0.0012243 – 0.0000010115 – 0.00000008032 = 0.7888225488 – 0.00566189532 = 0.78316065356 Example 8: Calculate the value of tan 48°15′ from the following table: 45 46 47 48 49 50 1.00000 1.03553 1.07237 1.11061 1.15037 1.19175 x tan x °°°°°°° Sol. Given that a + nh = 50 h = 1 a + nh + uh = 48°15′ = 48.25° ∴ 50 + u = 48.25 ⇒ u = –1.75 The difference table for given data is as follows: xyyyyyy°∇∇∇∇∇ ° ° ° − °− ° ° 5 5 52 53 54 55 10 10 10 10 10 10 45 100000 3553 46 103553 131 3684 9 47 107237 140 3 3824 12 5 48 111061 152 2 3976 10 49 115037 162 4138 50 119175 ()ya nh+ = 23 (1) (1)(2) 2! 3! anh anh anh anh uu uu u yuy y y ++ + + +++ +∇ + ∇ + ∇ + 45 (1)(2)(3) (1)(2)(3)(4) 4! 5! anh anh uu u u uu u u u yy ++ +++ ++++ ∇+ ∇ 5 48.25 10 y = () 1.75)( 0.75 ( 1.75)( 0.75)(0.25) 119175 ( 1.75) 4138 162 10 26 −− −− +− × + × + × ( 1.75)( 0.75)(0.25)(1.25) ( 1.75)( 0.75)(0.25)(1.25)(2.25) (2) (5) 24 120 −− −− +×−+ ×− 5 48.25 10 y = 112040.2867 ⇒ 48.25 y = tan 48°15′ = 1.120402867. INTERPOLATION WITH EQUAL INTERVAL 181 PROBLEM SET 4.2 1. The population of a town is as follows: 1921 1931 1941 1951 1961 1971 ( )202429364651 Year Population in lakhs Estimate the increase in population during the period 1955 to 1961 [Ans. 621036.8 lakhs.] 2. From the following table find the value of tan 17 ° . 04 8 12 16 20 24 tan 0 0.0699 0.1405 0.2126 0.2867 0.3640 0.4402 θ θ [Ans. 0.3057] 3. From the given table find the value of log 5875 40 45 50 55 60 65 log 1.60206 1.65321 1.69897 1.74036 1.77815 1.81291 x x [Ans. 3.7690058] 4. From the following table, find y when x = 1.84 and 2.4 1.7 1.8 1.9 2.0 2.1 2.2 2.3 5.474 6.050 6.686 7.389 8.166 9.025 9.974 x x e [Ans. 6.36, 11.02] 5. From the following table of half yearly premium for policies maturing at different ages, estimate the premium for policy maturing at the age of 63: 45 50 55 60 65 ( .) 114.84 96.16 83.32 74.48 68.48 Age Premium in Rs [Ans. 70.585152] 6. The values of annuities are given for the following ages. Find the value of annuity at the age of 1 27 2 . Age Annuity 25 26 27 28 29 16.195 15.919 15.630 15.326 15.006 [Ans. 15.47996] 7. Show that Newton’s Gregory interpolation formula can be written in the form as 23 4 00 0 0 0 x uuxuxauxabuxabcu =+∆−∆+∆− ∆+ where a = 111 1 ( 1), = 1 ( +1), = 1 ( 1) 234 xb xc x−+ − −+ etc. 182 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 4.4 CENTRAL DIFFERENCE FORMULAE As earlier we study formulae for leading terms and differences. These formulae are fundamental and are applicable to nearly all cases of interpolation, but they do not converge as rapidly as central difference formulae. The main advantage of central difference formulae is that they give more accurate result than other method of interpolation. Their disadvantages lies in complicated calculations and tedious expression, which are rather difficult to remember. These formulae are used for interpolation near the middle of a argument values. In this category we use the following formulae: 4.4.1 Gauss Forward Difference Formula We know Newton’s Gregory forward difference formula is given by ()fa hu+ = 23 (1) (1)(2) () () () () 2! 3! uu uu u fa ufa fa fa −−− +∆ + ∆ + ∆ + 4 ( 1)( 2)( 3) () 4! uu u u fa −−− ∆+ (1) Substitute a = 0, h = 1 in (1), we get ()fu = 23 4 (1)(1)(2)(1)(2)(3) (0) (0) (0) (0) (0) 2! 3! 4! uu uu u uu u u fuf f f f −−−−−− +∆ + ∆ + ∆ + ∆ + (2) Now obtain the values of 234 (0), (0), (0) fff∆∆∆ To get these values, 3 (1) f∆− 22 (0) ( 1) ff=∆ −∆ − ⇒ 2 (0) f∆ 32 (1) (1) ff=∆ − +∆ − Also, 4 (1) f∆− 33 (0) ( 1) ff=∆ −∆ − ⇒ 3 (0) f∆ 43 (1) (1) ff=∆ − +∆ − 5 (1) f∆− 44 (0) ( 1) ff=∆ −∆ − ⇒ 4 (0) f∆ 54 (1) (1) ff=∆ − +∆ − 6 (1) f∆− 55 (0) ( 1) ff=∆ −∆ − ⇒ 5 (0) f∆ 65 (1) (1) ff=∆ − +∆ − and so on. Substituting these values in equation (2) ()fu = 32 43 (1) (1)(2) (0) (0) ( 1) ( 1) [ ( 1) ( 1)] 2! 3! uu uu u fuf f f f f −−−  +∆ + ∆−+∆−+ ∆−+∆−  54 65 ( 1)( 2)( 3) ( 1)( 2)( 3)( 4) (1) (1) (1) (1) 4! 5! uu u u uu u u u ff ff −−− −−−−   + ∆ −+∆ − + ∆ −+∆ − +   ()fu = {} 23 (1) (1) (2) (0) (0) (1) 1 (1) 22!3 uu uu u fuf f f −−− +∆ + ∆ − + + ∆ − + {} {} 45 ( 1)( 2) ( 3) ( 1)( 2)( 3) ( 4) 1 ( 1) 1 ( 1) 6 4 24 5 uu u u uu u u u ff −− − −−− − +∆−+ +∆− 6 ( 1)( 2)( 3)( 4) ( 1) 120 uuuuu f −−−− +∆−+ INTERPOLATION WITH EQUAL INTERVAL 183 ()fu = 23 4 (1) (1)(1) (1)(1)(2) (0) (0) (1) (1) (1) 2! 3! 4! uu u uu u uu u fuf f f f −+−+−− +∆ + ∆ − + ∆ − + ∆ − 5 (1)(1)(2)(3) ( 1) 5! uuu u u f +−−− +∆−+ (3) But ∆ 5 f (–2) = ∆ 4 f (–1) – ∆ 4 f (–2) ⇒∆ 4 f (–1) = ∆ 4 f (–2) + ∆ 5 f (–2) and ∆ 6 f (–2) = ∆ 5 f (–1) + ∆ 5 f (–2) ⇒∆ 5 f (–1) = ∆ 5 f (–2) + ∆ 6 f (–2) The equation (3) becomes ()fu = 23 4 (1) (1)(1) (1)(1)(2) (0) (0) (1) (1) (2) 2! 3! 4! uu u uu u uu u fuf f f f −+−+−− +∆ + ∆ − + ∆ − + ∆ − 55 ( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3) (2) (2) 4! 5! uuu u uuu u u ff +−− +−−− +∆−+ ∆− + 6 (1)(1)(2)(3) (2) 5! uuu u u f +−−− ∆− ()fu∴ = 23 4 (1) (1)(1) (1)(1)(2) (0) (0) ( 1) ( 1) ( 2) 2! 3! 4! uu u uu u uu u fuf f f f −+−+−− +∆ + ∆ − + ∆ − + ∆ − 56 ( 2)( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3) ( 2) ( 2) 5! 5! uuuuu uuuuu ff ++ −− + −−− +∆−+∆−+ This formula is known as Gauss forward difference formula. This formula is applicable when u lies between 1 0and 2 4.4.2 Gauss Backward Difference Formula This formula is also solved by using Newton’s forward difference formula. Now, we know Newton’s formula for forward interpolation is 23 4 (1)(1)(2)(1)(2)(3) ( ) () () () () () 2! 3! 4! uu uu u uu u u fa hu fa ufa fa fa fa −−−−−− += +∆+ ∆ + ∆ + ∆ + (1) Put a = 0, and h = 1, in equation (1), we get 23 4 ( 1) ( 1)( 2) ( 1)( 2)( 3) ( ) (0) (0) (0) (0) (0) 2! 3! 4! uu uu u uu u u fu f uf f f f −−−−−− =+∆+ ∆ + ∆ + ∆ + (2) Now 2 22 3 33 4 44 5 (0) ( 1) ( 1) (0) ( 1) ( 1) (0) ( 1) ( 1) (0) ( 1) ( 1) and so on. ff f ff f ff f ff f ∆=∆−+∆− ∆=∆−+∆− ∆=∆−+∆− ∆=∆−+∆− On substituting these values in (2), we get 223 34 (1) (1)(2) () (0) (1) (1) (1) (1) (1) (1) 2! 3! uu uu u fufuff ff ff −−−  = +∆−+∆−+ ∆−+∆−+ ∆−+∆−  45 ( 1)( 2)( 3) ( 1) ( 1) 4! uu u u ff −−−  +∆−+∆−+  184 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ∴ 23 (1) (1) (2) () (0) (1) (1)1 (1)1 22 3 uuu u fu f uf u f f −− −   = +∆−+∆−+ + ∆−+     + 4 ( 1)( 2) ( 3) (1)1 34 uu u u f −− −  ∆− + +   = 23 (1) (1)(1) (0) ( 1) ( 1) ( 1) 2! 3! uuuu fuf uf f ++− +∆− + ∆ − + ∆ − + 4 (1)(1)(2) ( 1) 4! uuu u f +−− ∆−+ (3) Again 334 (1) (2) (2) fff∆−=∆−+∆− 445 (1) (2) (2) fff∆−=∆−+∆− 556 (1) (2) (2) fff∆−=∆−+∆− and so on. Therefore, equation (3) becomes 234 (1) (1)(1) () (0) (1) (1) (2) (2) 2! 3! uu uuu fu f uf f f f ++−  =+∆−+ ∆−+ ∆−+∆−  + 45 56 ( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3) ( 2) ( 2) ( 2) ( 2) 4! 5! uuu u uuu u u ff ff +−− +−−−   ∆ − +∆ − + ∆ − +∆ − +   23 4 ( 1) ( 1) ( 1) ( 1) ( 1) ( 2) () (0) (1) (1) (2) 1 (2) 2! 3! 3! 4 uu uuu uuu u fu f uf f f f ++−+−−  = +∆−+ ∆−+ ∆−+ + ∆−   5 (1)(1)(2) (3) 1 ( 2) 4! 5 uuu u u f +−− −  ++∆−+   23 (1) (1)(1) (2)(1)(1) () (0) (1) (1) (2) 2! 3! 4! uu uuu u uuu fu f uf f f ++−++− =+∆−+ ∆−+ ∆−+ 45 (2)(1)(1)(2) (2) (2) 5! uuuuu ff ++ −− ∆−+ ∆−+ This is known as Gauss Backward difference formula and useful when u lies between 1 and 0 2 − . 4.4.3 Stirling’s Formula This is another central difference formula and useful when 11 1 || 22 2 uor u<−<< . It gives best estimation when 11 . 44 u−< This formula is obtained by taken mean of Gauss forward and Gauss backward difference formula. Gauss forward formula for interpolating central difference is, 23 4 (1) (1)(1) (1)(1)(2) () (0) (0) (1) (1) (2) 2! 3! 4! uu u uu u uu u fu f u f f f f −+−+−− = +∆ + ∆−+ ∆−+ ∆− 56 ( 2)( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3) (2) (2) 5! 5! uuuuu uuuuu ff ++ −− + −−− +∆−+∆−+ (1) Gauss Backward difference is, 23 (1) (1)(1) (2)(1)(1) () (0) (1) (1) (2) 2! 3! 4! uu uuu u uuu fu f uf f f ++−++− =+∆−+ ∆−+ ∆−+ INTERPOLATION WITH EQUAL INTERVAL 185 45 (2)(1)(1)(2) ( 2) ( 2) 5! uuuuu ff ++ −− ∆−+ ∆−+ (2) Take mean of Equation (1) and (2) [] 2 (0) ( 1) (1)(1) () (0) (1) 22! 2 uf f uuu fu f f ∆+∆− −++  =+ +∆−   33 4 (1) (2) (1)(1) (1)(1) (2)(2 (2) 3! 2 3! 2 ff uuu uuu u u f  ∆−+∆− +− +− −++  ++∆−      5 ( 2)( 1) ( 1)( 2) (2) 5! uuuuu f ++ −− +∆−+ 33 2 2 (0) ( 1) ( 1) ( 2) (1)(1) () (0) (1) 22! 3! 2 ff f f uuu u fu f u f  ∆+∆− ∆−+∆− +−  =+ +∆−+      2 45 ( 1) ( 1) ( 2)( 1) ( 1)( 2) ( 2) ( 2) 3! 5! uuu u uuu u ff +− ++−− +∆−+ ∆−+ This is called Stirling’s formula. 4.4.4 Bessel’s Interpolation Formula This is one of the another type of central difference formula and obtained by (1) shifting the origin by 1 in Gauss backward difference and then (2) replacing u by (u – 1), (3) take mean of this equation with Gauss forward formula. Gauss backward difference formula is, () 23 4 (1) (1)(1) (2)(1)(1) () (0) (1) (1) (2) 2 2! 3! 4! uu uuu u uuu fu f uf f f f ++−++− =+∆−+ ∆−+ ∆−+ ∆− 5 (2)(1)(1)(2) (2) 5! uuuuu f ++ −− ++∆−+ Now shift the origin by one, we get () 23 4 (1) (1)(1) (2)(1)(1) ( ) (1) (0) (0) ( 1) 1 2! 3! 4! uu uuu u uuu fu f uf f f f ++−++− =+∆+ ∆ + ∆−+ ∆− 5 ( 2)( 1) ( 1)( 2) ( 1) 5! uuuuu f ++ −− +∆−+ On replacing u by (u – 1) () 23 4 ( 1) ( 1) ( 2) ( 1) ( 1)( 2) () (1) ( 1) (0) (0) (1) 1 2! 3! 4! uu uuu uuu u fu f u f f f f −−−+−− =+−∆+ ∆ + ∆−+ ∆− 5 (1)((1)(2)(3) (1) 5! uuuuu f +− −− +∆−+ …(1) Gauss forward difference formula is, 23 4 (1) (1)(1) (1)(1)(2) () (0) (0) (1) (1) (2) 2! 3! 4! uu u uu u uu u fu f uf f f f −+−+−− = +∆ + ∆−+ ∆−+ ∆− 56 (2)(1)(1)(2) (1)(1)(2)(3) ( 2) ( 2) 5! 5! u u uu u u uu u u ff ++ −− + −−− +∆−+∆−+ …(2) . etc. 182 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 4.4 CENTRAL DIFFERENCE FORMULAE As earlier we study formulae for leading terms and differences. These formulae are fundamental and are applicable. applicable to nearly all cases of interpolation, but they do not converge as rapidly as central difference formulae. The main advantage of central difference formulae is that they give more accurate. −− +−+ = 147.2251 Approx. Example 5. In an examination, the number of candidates who obtained marks between certain limits are as follows: Marks No of candidates −−−−−01 9203 9405960798099 . 41