216 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Difference table is: ufufufufu∆∆ ∆ − − − 23 () () () () 1 2854 308 0 3162 3618 3926 6648 1 7088 3030 896 2 7984 By Everett’s formula, () ( ) ()() ()() () 1.25 (0.25) 0.75 1.75 0.75 ( 0.25) (.25) (0.25)(7088) 3030 0.75 (3162) 3618 3! 5! f −−   =+ −++ +     = 4064 Hence f(30) = 4064. Example 3. Apply Laplace Everett’s formula to find the value of log 2375 from the data given below: x x 21 22 23 24 25 26 log 1.3222 1.3424 1.3617 1.3802 1.3979 1.4150 Sol. Here h = 1 We take origin at 23. Now difference table is given by xx∆∆ ∆ ∆ ∆ − −− −− − − 2345 log 2 21 1.3222 0.0202 1 22 1.3424 0.0009 0.0193 0.0001 0 23 1.3617 0.0008 0.0001 0.0185 0 0.0003 1 24 1.3802 0.0008 0.0002 0.0171 0.0002 2 25 1.3979 0.0006 0.0171 3 26 1.4150 INTERPOLATION WITH EQUAL INTERVAL 217 Here h = 1 ∴ u = 23.75 23 0.75 1 xa h −− == w = 1 – 0.75 = 0.25 From Laplace Everett formula, we have () ()()()() 24 1( 1) 2 1 1 2 () (1) (0) (1) 3! 5! uuu u uuu u fu uf f f +− ++ −−  =+ ∆+ ∆−+   ()() ()()()() 24 1 1 21 12 (0) ( 1) ( 2) 3! 5! www w www w wf f f +− ++−−   ++ ∆−+ ∆−+     = ()() () ()()()( )( ) () 1.75)(0.75 0.25 1.75 2.75 0.75 0.25 1.25 0.75 1.3802 0.0008 0.0002 6 120 −−−  ×+ ×− + ×   + () 2.25 (1.25)(0.25)( 0.75)( 1.75) 0.25(1.25)( 0.75( 0.0008) 0.25 1.3617 ( 0.0001) 6 120 −−  −− ++ + ×−   = 1.035419 + 0.340455 = 1.375874 log 2375 = log (23.75 × 100) = log 23.75 + log 100 ⇒ log 2375 = 1.375872 + 2 = 3.375872 Example 4. Find the value of e –x when x = 1.748 from the following data: x x e − 1.72 1.73 1.74 1.75 1.76 1.77 0.1790 0.1773 0.1755 0.1738 0.1720 0.1703 Sol. Here h = 0.01, take origin as 1.74. The difference table for the given data is as: x xe − ∆∆∆∆∆ − − − − −− − −− − 2345 1.72 0.1790 0.0017 1.73 0.1773 0.0001 0.0018 0.0002 1.74 0.1755 0.0001 0.0004 0.0017 0.0002 0.0008 1.75 0.1738 0.0001 0.0004 0.0018 0.0002 1.76 0.1720 0.0001 0.0017 1.77 0.1703 218 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 1.748 1.74 0.8 0.01 u − == w = 0.2 ()()()()() 2.8 1.8 0.8 0.2 1.2 (0.0004) (0.8) 0.8(0.1738) (0.8)(1.8)( 0.2)( 0.00017) 120 f −− =+−−+ ()()() () ()()()()() () 0.8 0.2 1.2 1.2 2.2 0.2 0.8 1.8 0.2(0.1755) 0.0001 0.0004 6120 −−− ++ ×+ ×− = 0.13904 + 0.0000816 + 0.000003225 + 0.0351 – 0.0000032 + 0.000002534 = 0.174224. Example 5. Prove that if third differences are assumed to be constant y x = () () !! 22 22 100 1 xx 1 uu 1 xy y uy y 33 − −− +∆++∆ where u = 1 – x. Apply this formula to find the value of y 11 and y 16 if y 0 = 3010, y 5 = 2710, y 10 = 2285, y 15 = 1860, y 20 = 1560, y 25 =1510, y 30 = 1835. Sol. 1 11 10 0.2, 5 x − == 2 16 15 0.2. 5 x − == xxxx xy y y y ∆∆ ∆ − − − − − − 25 03010 300 5 2710 125 425 125 10 2285 0 425 125 15 1860 125 300 125 20 1560 250 50 125 25 1510 375 325 30 1835 Using given formula, () 2 2 22 100 1 1 (1) 3! 3! x xx uu yxy yuy y − − − =+ ∆++ ∆ () () () 11 1.2 (0.2)( 0.8) 0.8)(0.64 1 0.2 (1860) (125) (0.8)(2285) (10) 66 y −− =+ ++ = 2196 INTERPOLATION WITH EQUAL INTERVAL 219 ()( ) ()()( ) ()()( ) ()( ) () 16 1.2 0.2 0.8 0.8 0.2 (1.8 0.2 1560 250 0.8 1860 125 66 y −− =+ ++ = 1786 Example 6. Find the compound interest on the sum of Rs. 10,000/- at 7% for the period 16 years if, n x 5 1015202530 (1.07) 1.40255 1.96715 2.75903 3.86968 5.42743 7.61236 Sol. The difference table can be formed as: xy ∆∆ ∆ ∆ ∆ − − − − 23 45 2 5 1.40255 0.5646 1 10 1.96715 0.22728 0.79188 0.09149 0 15 2.75903 0.31877 0.35901 1.11065 0.26752 1.60246 1 20 3.86968 0.05125 1.24345 1.5775 0.97593 2 25 5.42743 1.02718 2.18493 3 30 7.61236 Here, h = 5 ∴ u = 16 15 0.2 5 xa h −− == ∴ w = 110.20.8u−=− = . On applying Laplace Everett formula, we have 0.2(0.2 1)(0.2 1) 0.2(0.2 2)(0.2 1)(0.2 1)(0.2 2) ( ) 0.2 3.86968 0.05125 1.24345 3! 5! fu +− ++−−   =× + × + ×     () 0.8 2.75903 0.8(0.8 1)(0.8 1 0.8(0.8 2)(0.8 1)(0.8 1)(0.8 2) 0.31877 0.35901 6 120 × + +− ++−−   +×+ ×−     = 0.776593 + 2.189027 = 2.96595 (Approx.) 220 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 7. The values of the ellipitc integral () () /2 1/2 2 0 km 1 msin d − =− ∫ π θθ for certain equidistant values of m are given below. Use Everett’s formula to determine k (0.25). m km 0.20 0.22 0.24 0.26 0.28 0.30 ( ) 1.659624 1.669850 1.680373 1.691208 1.702374 1.713889 Sol. Here h = 0.02, take origin as 0.24 u = 0.25 0.24 0.5 0.02 xa h −− == w = 1 – u = 0.5 mkm ∆∆ ∆ ∆ ∆ − − 23 4 5 () 0.20 1.659624 0.010226 0.22 1.669850 0.000297 0.010523 0.000015 0.24 1.680373 0.000312 0.000004 0.010835 0.000019 0.000005 0.26 1.691208 0.000331 0.000001 0.011166 0.000018 0.28 1.702374 0.000349 0.011515 0.30 1.713889 f(u)= ()() () ()() 22 11 1 1 (1) (0) 0 ( 1) 3! 3! uuu www uf f wf f +− + −   +∆++ ∆−     = ()( ) () () ()()()()() () (0.5)(1.5) 0.5 1.5 .5 .5 1.5 2.5 0.5 1.691208 0.000331 0.000001 6120 −−−   ++−     ()( ) ()( )( ) () ()( ) ( )( ) () .5 1.5 .5 .5 1.5 ( .5) 2.5 1.5 0.5 1.680073 0.000312 .000004 6120 −−− ++ + × = 0.845604 – 0.00002069 – 0.00000014 + 0.84018650 – 0.0000195 + 0.00000005 = 1.685750 INTERPOLATION WITH EQUAL INTERVAL 221 Example 8. Find the value of log 337.5 by using Laplace Everett’s formula. Given that: x x 310 320 330 340 350 360 log 2.49136 2.50515 2.51851 2.53148 2.54407 2.55630 Sol. Here h = 10, take origin as 340 337.5 340 0.25 10 xa u h −− == =− w = 1 – u = 1+ 0.25 = 1.25 For the given data difference table is as: x x fx fx fx fx fx∆∆∆∆ ∆ − −− −−− − − − 234 5 log ( ) ( ) ( ) ( ) ( ) 3 310 2.49136 0.01379 2 320 2.50515 0.00043 0.01336 0.00004 1 330 2.51851 0.00039 0.00005 0.01297 0.00001 0.00004 0 340 2.53148 0.00038 0.00001 0.01259 0.00002 1 350 2.54407 0.00036 0.01 223 2 360 2.55630 () () ()() () ()()()() () 24 11 2112 10 1 3! 5! uuu u uuu u fu uf f f +− ++−−  =+ ∆+ ∆−   () ()() () ()()()() () 24 11 2112 01 2 3! 5! www w www w wf f f +− ++−−   ++ ∆−+ ∆−     () ()()() () 1.25 0.25 0.75 0.25 2.54407 0.00036 1.25 2.53148 6 −− =− × + ×− + × ()()() () 0.25 1.25 2.25 0.00038 6 +− ()()()()( ) () 3.25 2.25 1.25 0.25 0.75 0.00001 120 − + = –0.6360175–0.0000140625 + 3.16435 – 0.00004453125 – 0.0000001428 = 2.528273 (Approx.) 222 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM SET 4.7 1. Eliminate odd difference from the Gauss Forward formula to drive Everett’s formula: () ()() ()() 22 01 0 1 12 1 1 1 2! 3! u uu u u uu yufu S S −− + − =− +δ= δ+ δ+ where 0 xx u h − = 2. From the following table of values of x and y = e x , interpolate the value of y when x = 1.91: 1.7 1.8 1.9 2.0 2.1 2.2 5.4739 6.0496 6.6859 7.3891 8.1662 9.0250 x x ye = [Ans. 6.7531] 3. From the following present value annuity a n table: n x a 20 25 30 35 40 11.4699 12.7834 13.7648 14.4982 15.0463 Find the present value of the annuity a 31 , a 33 . [Ans. 13.9186, 14.2306] 4. Find the value of x 1/3 when x = 51 to 54 from the data: x x 1/3 40 45 50 55 60 65 3.4200 3.3569 3.6840 3.8030 3.9149 4.0207 [Ans. 3.7084096, 3.7325079, 3.7563005, 3.7797956] 5. From the following data, find the value of f(31), f(32), f(20) = 3010, f(25) = 3979, f(30) = 4771, f(35) = 5441, f(40) = 6021, f(45) = 6532 [Ans. 4913, 5052] 6. Apply Everett’s formula to find the value of f(26) and f(27) from the data given below: x fx 15 20 25 30 35 40 ( ) 12.849 16.351 19.524 22.396 24.999 27.356 [Ans. 20.121431, 20.707077] 7. The following table gives the values of e x for certain equidistant values of x: 0.61 0.62 0.63 0.64 0.65 0.66 0.67 1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237 x x e Find the value of e x when x = 0.644 using Everett’s formula. [Ans. 1.904082] INTERPOLATION WITH EQUAL INTERVAL 223 8. Apply Everett’s formula to find the values of e –x for x = 3.2, 3.4, 3.6, 3.8 if, 123456 0.36788 0.13534 0.04979 0.01832 0.00674 0.00248 x x e − [Ans. 0.04087, 0.03354, 0.02749, 0.02248] 9. Use Everett’s formula to find the present value of the annuity of n = 36 from the table: 25 30 35 40 45 50 12.7834 13.7648 14.4982 15.0463 15.4558 15.7619 x x a [Ans. 14.620947] 10. Obtain the value of y 25 , given that: y 20 = 2854, y 24 = 3162, y 28 = 3544, y 32 = 3992 [Ans. 3250.875] GGG CHAPTER 5 Interpolation with Unequal Interval 5.1 INTRODUCTION The interpolation formulae derived before for forward interpolation, Backward interpolation and central interpolation have the disadvantages of being applicable only to equally spaced argument values. So it is required to develop interpolation formulae for unequally spaced argument values of x. Therefore, when the values of the argument are not at equally spaced then we use two such formulae for interpolation. 1. Lagrange’s Interpolation formula 2. Newton’s Divided difference formula. The main advantage of these formulas is, they can also be used in case of equal intervals but the formulae for equal intervals cannot be used in case of unequal intervals. 5.2 LAGRANGE’S INTERPOLATION FORMULA Let f(x 0 ), f(x 1 ) f(x n ) be (n + 1) entries of a function y = f(x), where f(x) is assumed to be a polynomial corresponding to the arguments x 0 , x 1 , x 2 , x 0 . So that The polynomial f(x) may be written as f(x) = A 0 (x – x 1 ) (x – x 2 ) (x – x n ) + A 1 (x – x 1 ) (x – x 2 ) (x – x n ) + + A n (x – x 1 ) (x – x 2 ) (x – x n–1 ) (1) where A 0 , A 1 , A n , are constants to be determined. Putting x = x 0 , x 1 , x 2 , , x n in (1) successively, we get For x = x 0 , f(x 0 )=A 0 (x 0 – x 1 ) (x 0 – x 2 ) (x 0 – x n ) ⇒ A 0 = () ()() () 0 0102 0 n fx xxxx xx −− − (2) For x = x 1 , f(x 1 )=A 1 (x 1 – x 0 )(x 1 – x 2 ) (x 1 – x n ) ⇒ A 1 = () ()()( ) 1 10 1 2 0 n fx xx x x x x −− − (3) Similarly, For x = x n ,f(x n )=A n (x n – x 0 )(x n – x 1 ) (x n – x n–1 ) 224 INTERPOLATION WITH UNEQUAL INTERVAL 225 A n = () ()()( ) 02 1 n nn nn fx xxxx xx − −− − (4) Substituting the values of A 0 , A 1 , A n , in equation (1), we get f(x) = ()()() ()()() () ()()() ()()() () 12 02 01 0102 0 1012 1 nn nn xx xx xx xx xx xx fx fx xxxx xx xxxx xx −− − −− − + −− − −− − + + ()()( ) () ()( ) () 01 1 01 1 n n nn nn xx xx xx fx xxxx x x − − −− − −− − (5) This is called Lagrange’s interpolation formula. In equation (5), dividing both sides by (x – x 0 ) (x – x 1 ) (x – x n ), Lagrange’s formula may also to written as ()()()()()()() 0 0 1 0102 0 0 () ( ) nn fx fx xx xx xx x x x x x x xx = −− − − − −− ()()()() ()()( )() 1 1012 1 1 0 1 1 () () 11 n nnnnnn fx fx xxxx xx xx xxxx xx xx − +++ −− −− −− − − Corollary. Show that Lagrange’s formula can be put in the form () ()() ) 0 ()( ' n r n rr r xfx Px xx x = φ = −φ ∑ where () 0 n r x = φ= ∏ (x – x r ) and φ ’(x r ) = () {} d x dx  φ   x=xr Sol. We have, P n (x)= ()()( )( )() ()()( )( )() 01 1 1 01 1 1 0 n rr n rr rr rrn r xx xx xx xx xx xxxx xx xx xx −+ −+ = −− − − − −− − − − ∑ f(x r ) = () () () () ()( )( ) () −+ =  φ   −−−−− −   ∑ 01 1 1 0 n r r r r rr rr rn r fx x xx xxxx xx xx xx (1) Now, () x φ = () 0 n xx r r − ∏ = (given) Therefore, () x φ = ()()() () () () () 012 1 1 rrr n xx xx xx xx xx xx xx −+ −−− − −− − ∴ φ′(x) = () 1 xx − ()()()()()()() 202 rn rn xx xx xx xx xx xx xx −−−+−−−− + ()()()( )()()()()( ) 01 1 01 1 rr n r n xx xx xx xx xx xx xx xx xx +− −− −− −+−− − − ⇒ φ′(x) = [φ′(x)] x = xr = (x r – x 0 ) (x r – x 1 ) (x r – x r–1 ) (x r – x r+ 1 ) (x r – x n ) (2) Hence from equation (1) () () () () 0 () ' n r n rr r xfx Px xx x = φ = −φ ∑ [Using (2)] . interpolation formulae derived before for forward interpolation, Backward interpolation and central interpolation have the disadvantages of being applicable only to equally spaced argument values 2.96595 (Approx.) 220 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 7. The values of the ellipitc integral () () /2 1/2 2 0 km 1 msin d − =− ∫ π θθ for certain equidistant values of. Interpolation formula 2. Newton’s Divided difference formula. The main advantage of these formulas is, they can also be used in case of equal intervals but the formulae for equal intervals cannot