186 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Take the mean of equation (1) and (2) 22 (0) (1) (0) ( 1) (1) (1) () (0) 22 2!2 ff f f uu uu fu f  +∆+∆− −+ −   =+ ∆+       44 3 (1) (2) ( 1) 2) ( 1) ( 1) ( 1)( 2) (1) 3! 2 4! 2 ff uu u u u uu u f  ∆−+∆− −−++ +−−  +∆−+      + 55 (3)(1)(2)(2) (1)(1)(2) 5! 2 ufuf uuu u  −∆ −+ +∆ − +−−  +    {} () 22 3 1 (1) (0) (1) (0) ( 1) 1(1) 2 () (0) 1 222! 2 3! uu u ff f f uu fu u f f  −−   +∆+∆− −   =+−∆+ + ∆−    () 44 5 1( 1)( 2)( 12) (1) (2) (1)(1)(2) ( 2) 4! 2 5! uuu u u ff uuu u f  +−−− ∆−+∆− +−− ++∆−+    This formula is very useful when 1 2 u = and gives best result when 13 44 u<< . 4.4.5 Laplace-Everett’s Formula Gausss forward formula is given by 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! uuuuu uuuuu ff ++ − + −−− +∆−+∆−+ (1) We know 322 544 (0) (1) (0) ( 1) (0) ( 1) (2) (1) (2) fff fff fff ∆= − ∆−=∆ −∆− ∆−=∆−−∆− Therefore, using this in equation (1), we get [] 222 (1) (1)(1) ( ) (0) (1) (0) ( 1) ( (0) ( 1)) 2! 3! uu u uu fu f uf f f f f −+− =+ − + ∆−+ ∆ −∆− () 444 ( 1) ( 1)( 2) ( 2)( 1) ( 1)( 2) (2) (1) (2) 4! 5! uuu u u uuu u fff +−− ++−− +∆−+ ∆−−∆−+ 22 (1) (1)(1) (1 ) (0) (1) ( 1) ( 1) 2! 3! uu u uu uf uf f f −+− =− + + ∆ −− ∆ − 24 ( 1) ( 1) ( 2)( 1) ( 1)( 2) (0) ( 1) 3! 5! uuu u uuu u ff +− ++−− +∆+ ∆− + 44 ( 2)( 1) ( 1)( 2) (1)(1)(2) ( 2) ( 2) 4! 5! uuuuu uuu u ff ++ −− +−− ∆−− ∆− INTERPOLATION WITH EQUAL INTERVAL 187 22 (1)(2) (1)(1) (1 ) (0) (1) ( 1) (0) 3! 3! uu u u uu uf uf f f −− + − =− + + ∆ −+ ∆ 44 (2)(1)(1)(2) (1)(1)(2)(3) (1) (2) 5! 5! u u uu u u uu u u ff ++ −− + −−− +∆−+∆−+ 24 ( 1)( 1) ( 2)( 1) ( 1)( 2) ( ) (1) (0) ( 1) 3! 5! uu u u u uu u fu uf f f +− ++ −−  =+ ∆+ ∆−+   + {} 24 (1)(2) (1)(1)(2)(3) (1 ) (0) ( 1) ( 2) 3! 5! uu u u uu u u uf f f −− + −−− −+ ∆−+ ∆−+ (2) Substitute 1 – u = w in second part of equation (2) 24 ( 1) ( 1) ( 2)( 1) ( 1)( 2) ( ) (1) (0) ( 1) 3! 5! uuu u uuu u fu uf f f +− ++−−  =+ ∆+ ∆−+   + () 24 1( 1) ( 2)( 1) ( 1)( 2) (0) ( 1) ( 2) 3! 5! www wwwww wf f f −+  ++ −− +∆−+ ∆−+   This is called Laplace-Everett’s formula. It gives better estimate value when 1 2 u > . Example 1. From the following table, find the value of e 1.17 using Gauss forward formula: x x efx = 1 1.05 1.10 1.15 1.20 1.25 1.30 ( ) 2.7183 2.8577 3.0042 3.1582 3.3201 3.4903 3.6693 Sol. The difference table is as given: 23456 1 2.7183 0.1394 1.05 2.8577 0.0071 0.1465 0.0004 1.10 3.0042 0.0075 0 0.154 0.0004 0 1.15 3.1582 0.0079 0 0.0001 1.619 0.0004 0.0001 1.20 3.3201 0.0083 0.0001 0.1702 0.0005 1.25 3.4903 0.0088 0.179 1.30 3.6693 xy ∆∆∆∆∆∆ 188 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now, let taking origin at 1.15, here h = 0.05 Then, 1.17 1.15 0.4 0.05 u − == On applying Gauss’s forward interpolation, we have 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 −+−+−− = +∆ + ∆−+ ∆−+ ∆− 5 (2)(1)(1)(2) (2) 5! uuuuu f ++ −− +∆−+ (0.4)( 0.6) (1.4)(0.4)( 0.6) (0.4) 3.1582 0.4 (0.1619) (0.0079) (0.0004) 26 f −− =+×+ ×+ × (1.4)(0.4)( 0.6)( 1.6)(2.4) 0 0.0001 120 −− ++ × (0.4) 3.1582 0.06476 0.000948 0.0000224 0.0000010752f =+ − − + (0.4) 3.22199 (Approx.)f = Example 2. Given that x x 25 30 35 40 45 log 1.39794 1.47712 1.54407 1.60206 1.65321 log 3.7 = ? Sol. 234 2 1 0 1 2 log log log log log 25 1.39794 0.07918 30 1.47712 0.01223 0.06695 0.00327 35 1.54407 0.00896 0.00115 0.05799 0.00212 40 1.60206 0.00684 0.05115 45 1.65321 xxx xxx x x x x x − − ∆∆ ∆ ∆ − −− − x = a + hu, x = 37, a = 35, h = 5 u = 37 35 0.4 5 xa h −− == INTERPOLATION WITH EQUAL INTERVAL 189 23 4 (1) (1)(1) (1)(1)(2) () (0) (0) (1) (1) (1) 2! 3! 4! uu u uu u uu u fu f uf f f f −+−+−− = +∆ + ∆−+ ∆−+ ∆− (0.4)(0.4 1) (0.4)(0.4 1)(0.4 1) 1.54407 0.4 0.05798 ( 0.00896) (0.00212) 26 −−+ =+×+ ×−+ × (0.4)(0.4 1)(0.4 1)(0.4 2) ( 0.00115) 24 −+− +×− ( ) 1.54407 0.023192 0.0010752 0.00011872 0.00002576fu=+ + − − = 1.56819272 Since log 3.7 = 3.7 10 37 log log 10 10 × = ⇒ log 3.7 = log 37 – log 10 = 1.56819272 – 1 = 0.56819272. Ans. Example 3. From the following table find y when x = 1.45. x y −−++ 1.0 1.2 1.4 1.6 1.8 2.0 0.0 0.112 0.016 0.336 0.992 2.0 Sol. xy y y yy∆∆∆∆ − − − 234 1.0 0.0 0.112 1.2 0.112 0.208 0.096 0.048 1.4 0.016 0.256 0 0.352 0.048 1.6 0.336 0.304 0 0.656 0.048 1.8 0.992 0.352 1.008 2.0 2.0 () 23 4 1( 1) (1) (1)(1)(2) () (0) (0) (1) (1) (2) 2! 3! 4! uuu uu u uu u fu f uf f f f +− −+−− =+∆+ ∆−+ ∆−+ ∆− 1.45 1.4 0.25 0.2 u − == (0.25)( 0.75) (0.25)( 0.75)(1.25) ( ) 0.016 0.25 0.352 0.256 0.048 26 fu −− =− + × + × + × = 0.047875 190 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 4. Use Gauss’s forward formula to find a polynomial of degree four which takes the following values of the function f(x): x 12345 f(x) 1–11–11 Sol. Taking center at 3 i.e., x 0 = 3 and h = 1 u = 0 xx h − ⇒ u = x – 3 Now, for the given data difference table becomes: xfxfxfxfxfx∆∆∆∆ − − − − − − 234 () () () () () 11 2 21 4 28 31 4 16 28 41 4 2 51 Gauss forward 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 −+−+−− =+∆+ ∆−+ ∆−+ ∆− () 3( 4)( 2) (3)(4) (2)(3)(4)(5) 1 ( 3)( 2) ( 4) (8) (16) 26 24 xxx xx xxxx x −−− −− −−−− =+ − − + − + + = 232 432 4 104 2 28 142 308 1 2 6 2 14 24 12 32 80 333333 xxx xx x xx x x−+− + −+ − − −+ − + − + 43 2 2 100 () 8 56 31 33 fx x x x x∴=−+ −+ . Ans. Example 5. Use Gauss’s forward formula to find 30 y for the following data. yyyyy 21 25 29 33 37 18.4708 17.8144 17.1070 16.3432 15.5154 Sol. Let us take the origin at x = 29 then 30 29 1 0.25 44 u − === INTERPOLATION WITH EQUAL INTERVAL 191 Now, for the given data difference table is: xy y y y y∆∆∆∆ − − −− −− −− − − 234 21 18.4708 0.6564 25 17.8144 0.0510 0.7074 0.0054 29 17.1070 0.0564 0.0022 0.7638 0.0076 33 16.3432 0.0640 0.8278 37 15.5154 Putting these values in Gauss forward interpolation formula, we have 0.25 (0.25)( 0.750) 17.1070 (0.25) ( 0.7638) ( 0.0564) 2 y − =+×−+ ×− + (0.25)( 0.750)(1.25) (1.25)(0.25)( 0.750)( 1.75) ( 0.0076) ( 0.0022) 624 −−− ×− + ×− y 0.25 = 17.1070 – 0.19095 + 0.0052875 + 0.00002968 – 0.00000375 y 0.25 = 16.9216. Ans. PROBLEM SET 4.3 1. The values of e –x at x = 1.72 to x = 1.76 are given in the following table: () 1.72 1.73 1.74 1.75 1.76 0.17907 0.17728 0.17552 0.17377 0.17204 x fx Find the values of e –1.7425 using Gauss forward difference formula [Ans. 0.1750816846] 2. Apply Gauss’s forward formula to find the value of f(x) at x = 3.75 from the table: () 2.5 3.0 3.5 4.0 4.5 5.0 24.145 22.043 20.225 18.644 17.262 16.047 x fx [Ans. 19.407426] 3. Apply Central difference formula to obtain f(32). Given that : f(25) = 0.2707, f(30) = 0.3027, f(35) = 0.3386, f(40) = 0.3794. [Ans. 0.316536] 4. Apply Gauss forward formula to find the value of U 9 , if u(0) = 14, u(4) = 24, u(8) = 32, u (12) = 35, u(16) = 40 [Ans. 33.1162109] 192 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 5. Apply Gauss forward formula to find a polynomial of degree three which takes the values of y as given on next page: 2 46810 213820 x y − 23 17 3 2 3 62 3 xx x  +++   Ans. 6. Use Gauss’s forward formula to find the annuity value for 27 years from the following data: 15 20 25 30 35 40 10.3797 12.4622 14.0939 15.3725 16.3742 17.1591 Year Annuity [Ans. 14.643] 7. Use Gauss’s forward formula to find the value of f(x), when 1 , 2 x = given that: () 210 1 100 108 105 110 x fx − GAUSS BACKWARD Example 1. Given that 50 51 52 53 54 tan 1.1918 1.2349 1.2799 1.3270 1.3764 x x ° ° Using Gauss’s backward formula, find the value of tan 51° 42′ Sol. Take the origin at 52° and given h = 1 ∴ 51 42 52 18 0.3 xa uxa h − ′′ ==−=°−°=−=−° Now using Gauss backward formula f(u) = f(0) + u ∆ f (–1) + () () 2 1 1 2! uu f + ∆− ()() () ()()() 3 11 211 2 3! 4! uuu u uuu f +− ++− +∆−+ () 4 2 f ∆− INTERPOLATION WITH EQUAL INTERVAL 193 Difference table for given data is: 23 50 1.1918 0.0431 51 1.2349 0.0019 0.045 0.0002 52 1.2799 0.0021 0.0471 0.0002 53 1.3270 0.0023 0.0494 54 1.3764 xtanx°°∆∆∆ From equation (1) f(–0.3 ° ) = 1.2799 + (–0.3) (0.045) + ()() ()()() 0.3 0.7 0.3 0.7 1.7 0.0021 0.0002 26 −− ×+ × = 1.2799 – 0.0135 – 0.0002205 – 0.0000119 = 1.266167 (Approx.) Example 2. Apply Gauss backward formula to find sin 45 ° from the following table Sin θ° θ° 20 30 40 50 60 70 80 0.34202 0.502 0.64279 0.76604 0.86603 0.93969 0.98481 Sol. Difference table for given data is: 2345 20 0.34202 0.15998 30 0.502 0.01919 0.14079 0.00165 40 0.64274 0.01754 0.00737 0.12325 0.00572 0.01002 50 0.76604 0.02326 0.00265 0.09999 0.00307 0.00179 60 0.86603 0.02633 0.00086 0.07366 0.00221 70 0.93 sin θ° θ° ∆∆∆∆∆ − −− − − −− − − 969 0.02854 0.04512 80 0.98481 − 194 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ∴ u = xa h − = 45 40 10 − = 5 10 = 0.5 Now using Gauss backward formula f(u)= f(0) + u f∆ (–1) + () 1 2! uu + () 2 1 f ∆− + ()() 11 3! uuu +− () 3 2 f ∆− + ()()() 21 1 4! uuuu ++ − () 4 2 f ∆− + ()()()() 21 12 5! uuuuu ++ −− () 5 2 f ∆−+ f(0.5) = 0.64279 + 0.5 × 0.14079 + 0.5 1.5 2 × × (–0.01754) + () 0.5 1.5 0.5 6 ×− × (0.00165) + ()() 0.5 1.5 0.5 2.5 24 ×− × (–0.00737) = 0.64279 + 0.070395 – 0.0065775 – 0.000103125 + 0.00028789 = 0.706792. Ans. Example 3. Apply Gauss backward formula to find the value of (1.06) 19 if (1. 06) 10 = 1.79085, (1.06) 15 = 2.39656, (1. 06) 20 = 3.20714, (1.06) 25 = 4.29187, (1.06) 30 = 5.74349 Sol. The difference table is given by xy ∆∆ ∆ ∆ 234 10 1.79085 0.60571 15 2.39656 0.20487 0.81058 0.06928 20 3.20714 0.27415 0.02346 1.08473 0.09274 25 4.29187 0.36689 1.45162 30 5.74349 ∴ u = xa h − = 19 20 5 − = – 0.2 From Gauss backward formula f(u) = f(0) + u f∆ (–1) + () 1 2! uu + () 2 1 f ∆− + ()() 11 3! uuu +− () 3 2 f ∆− + ()()() 21 1 4! uuuu ++ − () 4 2 f ∆− + f(u) = 3.20714 – 0.2 × 0.81058 – () 0.2 0.8 2 × 0.27415 – ()( ) 0.2 0.8 1.2 6 − × 0.06928 + () () () 0.2 0.8 1.2 1.8 24 − × 0.02346 INTERPOLATION WITH EQUAL INTERVAL 195 f(u) = 3.20714 – 0.162116 – 0.021932 + 0.002216 + 0.00033782 = 3.0256458 (Approx.) Example 4. Using Gauss backward formula, Estimate the no. of persons earning wages between Rs. 60 and Rs. 70 from the following data: () () Wages Rs Below No of Persons in thousands −−− − . 40 40 60 60 80 80 100 100 120 . 250 120 100 70 50 Sol. Difference table for the given data is as: Wages below No of Persons y y y y∆∆∆∆ − − − − 234 . 40 250 120 60 370 20 100 10 80 470 30 20 70 10 100 540 20 50 120 590 ∴ u = xa h − = 70 80 20 − = 10 20 − = – 0.5 From Gauss backward formula f (0.5) = 23 (1) (1)(1) (0) ( 1) ( 1) ( 2) 2! 3! uu uuu fuf f f ++− +∆− + ∆ − + ∆ − 4 (2)(1)(1) (2) 4! uuuu f ++ − +∆−+ = ( 0.5)(0.5) ( 0.5)(0.5)( 1.5) 470 ( 0.5) ( 100) ( 30) ( 10) 26 −−− +− ×+ + ×− + ×− + ( 0.5)(0.5)( 1.5)(1.5) (20) 24 −− × = 470 50 3.75 0.625 0.46875−+ − + = 423.59375 Hence No. of Persons earning wages between Rs. 60 to 70 is 423.59375 – 370 = 53.59375 or 54000. (Approx.) . = 0.047875 190 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 4. Use Gauss’s forward formula to find a polynomial of degree four which takes the following values of the function. formula to find the value of U 9 , if u(0) = 14, u(4) = 24, u(8) = 32, u (12) = 35, u(16) = 40 [Ans. 33.11 6210 9] 192 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 5. Apply Gauss forward formula. 3.6693 xy ∆∆∆∆∆∆ 188 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now, let taking origin at 1.15, here h = 0.05 Then, 1.17 1.15 0.4 0.05 u − == On applying Gauss’s forward interpolation, we have 23