A textbook of Computer Based Numerical and Statiscal Techniques part 25 ppsx

226 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 1. Using Lagrange’s formula, find the value of (i) y x if y 1 = 4, y 3 = 120, y 4 = 340, y 5 = 2544, (ii) y 0 if y –30 = 30, y –12 = 34, y 3 = 38, y 18 = 42, Sol. (i) Here, x 0 = 1, x 1 = 3, x 2 = 4, x 3 = 5, f(x 0 ) = 4, f(x 1 ) = 120, f(x 2 ) = 340, f(x 3 ) = 2544, Now using Lagrange’s interpolation formula, we have f(x) = ()()() ()()() () ()()() ()()() () 123 023 01 010103 101213 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −−− −−− + −−− −−− + ()( )( ) ()()() () ()()() ()()() () 013 012 23 202123 303132 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −− − − − − + −−− −−− f(x) = ()()() ()()() ()( )( ) ()( )( ) 345 145 4 120 131415 313435 xxx xxx −−− −−− ×+ × −−− −−− ()()() ()()() ()()() ()()() 135 134 340 2544 414345 515354 xxx xxx −−− −−− +×+× −−− −−− y x = f(x) = – 1 6 (x – 3) (x – 4) (x – 5) + 30 (x – 1) (x – 4) (x – 5) – 340 3 (x – 1) (x – 3) (x – 5) + 318 (x – 1) (x – 3) (x – 4) (ii) Here, x 0 = – 30, x 1 = – 12, x 2 = 3, x 3 = 18, y 0 = 30, y 1 = 34, y 2 = 38, y 3 = 42, Now from Lagrange’s interpolation formula, we have f(x) = ()()() ()()() () ()()() ()()() () 123 023 01 010203 101213 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −−− −−− + −− − −−− + ()()() ()()() () ()()() ()()() () 01 3 0 12 23 202123 303132 xx xx xx xx xx xx fx fx xxxxxx x xxxxx −−− −−− + −−− −−− y x = ()()() ()()() ()()() ()()() 12 3 18 30 3 18 30 34 30 12 30 3 30 18 12 30 12 3 12 18 xxx xxx +−− +−− ×+ × −+ −− −− −+ −− −− + ()()() ()()() ()()() ()()() 30 12 18 30 12 3 38 42 330312318 18301812183 xxx xxx ++− ++− ×+ × ++− + +− y x = – 0.001052188 (x + 12) (x – 3) (x – 18) + 0.00419753 (x + 30) (x – 3) (x – 18) – 0.005117845 (x + 30) (x + 12) (x – 18) + 0.001944444 (x + 30) (x + 12) (x – 3) for x = 0 y 0 = – 0.001052188 (12) (–3) (–18) + 0.00419753 (30) (–3) (–18) – 0.005117895 (30) (12) (–18) + 0.001944444 (30) (12) (–3) INTERPOLATION WITH UNEQUAL INTERVAL 227 y 0 = – 0.681817824 + 6.7999986 + 33.1636356 – 2.09999952 y 0 = 39.9636546 – 2.781817344 y 0 = 37.1818. Ans. Example 2. If y 0 , y 1 , y 2 , y 3 y 9 are consecutive terms of a series. Prove that y 5 = 1 70 [56(y 4 + y 6 ) – 28 (y 3 + y 7 ) + 8 (y 2 + y 8 ) – (y 1 + y 9 )] Sol. Here, the arguments are 1, 2, 3, 9 so for these values Lagrange’s formula is given by ()()()()()()()() 12346789 x y xxxxxxxx −−−−−−−− = () ()()()()()()() 1 11235678 y x − −−−−−−− + () ()()()()()() − −−−−−− 2 21124567 y x + () ()()()()()() − −−−−− 3 32113456 y x + () ()()()() 4 43.2.12345 y x − −−−− + () ()()() 6 6 5.4.3.2 1 2 3 y x −−−− + 7 ( 7)6.5.4.3.1( 2)( 1) y x −−− + () () 8 8 7.6.5.4.2.1 1 y x −− () ()()()()()()()() 9 9 8.7.6.5.3.2.1 1 2 3 4 6 7 8 9 x yy x xxxxxxxx + − −−−−−−−− = () () () () () () () 12346 1 10080 2 1680 720 3 720 4 720 6 yyyyy xx xxx ++++ −− −−−−−− () () ++ −− − 78 720 7 1680 8 yy xx + () 9 10080 9 y x − Now for y 5 , put x = 5 ()()()() () 5123 4. 3. 2. 1 1 2 3 4 4 10080 3 1680 720 2 yyyy =++ −−− − −× × − × ()( )() () ()()( ) 46 7 8 9 720 1 1 720 2 720 1680 3 4 10080 yy y y y ++ + + + × − ×− − × − ×− − ×− 5 576 y = 12346789 40320 5040 1440 720 720 1440 5040 40320 y y yyyy y y −+−++−+− 5 576 y = ()()() () 46 37 28 19 11 1 1 720 1440 5040 40320 yy yy yy yy +− −+ +− + y 5 = () ()() () 46 37 28 19 576 576 576 576 720 1440 5040 40320 yy y y y y y y +− ++ +− − y 5 = () () () () 46 37 28 19 70 576 576 70 576 70 576 70 1 70 720 1440 5040 40320 yy yy yy yy ××××   +− ++ +− +     y 5 = 1 70 [56 (y 4 + y 6 ) – 28 (y 3 + y 7 ) + 8 (y 2 + y 8 ) – (y 1 + y 9 )]. Proved. 228 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 3. Find f(x) as a polynomial of x if x :–10367 f(x) : 3 –6 39 822 1611 Sol. Now from Lagrange’s interpolation formula, P(x) = f(x) = ()()()() ()()( )( ) ()()()() () ()() () 0367 1367 36 10 13 16 17 01(03)0607 xxxx xxxx −−−− +−−− ×+ ×− −− −− − − − − + − − − ()()()() ()()()() () ()()() ()()() () 1037 1067 39 822 31303637 61606367 xx x x xxxx +−− − +−−− +×+ × +−−− +−−− + ()()()() ()()()() 1063 1611 71707376 xxxx +−−− × +−−− P(x) = 3 224 x(x – 3) (x – 6) (x – 7) + 6 126 (x + 1) (x – 3) (x – 6) (x – 7) + 39 144 (x + 1)x (x – 6) (x – 7) – 822 126 (x + 1)x (x – 3) (x – 7) + 1611 224 (x + 1)x (x – 6) (x – 3) P(x) = 3 224 (x 4 – 16x 3 + 81x 2 – 126x) + 39 144 (x 4 – 12x 3 + 29x 2 + 42x) + 6 126 (x 4 – 15x 3 + 23x 2 – 45x – 126) – 822 126 (x 4 – 9x 3 + 11x 2 + 21x) + 1611 224 (x 4 – 8x 3 + 9x 2 + 18x) P(x)= x 4 3 16 3 12 39 15 6 822 9 1611 8 3 39 6 822 1611 224 144 126 126 224 224 144 126 126 224 x −× × × × ×   ++−+ + − − + −     + x 2 3 81 39 29 23 6 822 11 9 1611 224 144 126 126 224 ×××××  ++− +   + x 45 6 126 3 39 42 822 21 1611 18 6 126 224 144 126 224 −× × × × ×  −+− − −   P(x) = x 4 (1) + x 3 (–3) + x 2 (5) – 6 P(x) = x 4 – 3x 3 + 5x 2 – 6 which is the required polynomial. Example 4. Value of f(x) are given at a, b, c. Show that the maximum is obtained by x = () ( ) () ( ) () ( ) () () () () () () fa b c fb c a fc a b fa b c fb c a fc a b −+ −+ −  −+ −+ −  22 22 22 Sol. Here, x 0 = a, x 1 = b, x 2 = c, f(x 0 ) = f(a), f(x 1 ) = f(b), f(x 2 ) = f(c) By applying Lagrange’s formula, we have f(x) = ()() ()() () ()() ()() () ()() ()() () xbxc xaxc xaxb fa fb fc abac babc cacb −− −− −− ++ −− −− −− INTERPOLATION WITH UNEQUAL INTERVAL 229 f(x)= () () () () () ()() () () ()() () 222 x b cx bc x a cx ac x a bx ab fa fb fc ab a c b a bc c acb −+ + −+ + −+ + ++ −− −− −− For maximum, we have f’(x)= 0 i.e., f’(x)= () () ()() () () ()() () () ()() 222xbcfa xacfb xabfc abac babc cacb −+ −+ −+ ++ −− −− −− ={2x – (b + c)} (b – c) f(a) + {2x – (a + c)} (c – a) f(b) + {2x – (a + b)}(a–b) f (c) = 0 =2x [(b – c) f(a) + (c – a) f(b) + (a – b) f(c)] = (b 2 – c 2 ) f(a) + (c 2 – a 2 ) f(b) + (a 2 – b 2 ) f(c) (3 2 ≠ 0) ∴ x = () () () () () () () ()() ()() () 22 22 22 b c fa c a fb a b fc fa b c fb c a fc a b −+−+− −+ −+ − . Proved. Example 5. Applying Lagrange’s formula, find a cubic polynomial which approximate the following data x : –2 –1 2 3 y(x) : –12 –8 3 5 Sol. Now, using Lagrange’s formula, we have f(x)= ()()() () () ()() ()()() () ()()() 12312 2238 21 22 23 12 12 13 xxx xxx +−−×− +−−×− + −+ −− −− −+ −− −− + ()()() ()()() ()()() ()()() 2133 2125 222123 323132 xxx xxx ++−× ++−× + ++− ++− f(x)= 32 32 3 32 32 11 4 6 3 412 76 44 53 44 xxx xxx xx xxx   −+−− −−+− −−+ +−−   f(x)= x 3 2 3211 12 1 387 18243 211 5344 5 4 534 5 3 2 xx  −−+ + − ++ + ++− + − +−   ∴ f(x)=– 32 1 3 241 3.9 15 20 60 xx x−+− Example 6. (i) Determine by Lagrange’s formula, the percentage number of criminals under 35 years: Age % no. of Criminals Under 25 year 52 Under 30 years 67.3 Under 40 years 84.1 Under 50 years 94.4 (ii) Find a Lagrange’s interpolating polynomial for the given data: x 0 = 1, x 1 = 2.5, x 2 = 4, and x 3 = 5.5 f(x 0 ) = 4, f(x 1 ) = 7.5, f(x 2 ) = 13 and f(x 3 ) = 17.5 also, find the value of f(5) 230 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. (i) Here x 0 = 25, x 1 = 30, x 2 = 40 and x 3 = 50 f(x 0 ) = 52, f(x 1 ) = 67.3, f(x 2 ) = 84.1 and f(x 3 ) = 94.4 By using Lagrange’s interpolation formula, we have f(x) = ()()() ()()() ()()() ()()() 30 40 50 25 40 50 52 67.3 25 30 25 40 25 50 30 25 30 40 30 50 xxx xxx −−− −−− ×+ × −−− −−− + ()()() ()()() ()()() ()()() 30 25 50 25 30 40 84.1 94.4 40 25 40 30 40 50 50 25 50 30 50 40 xxx xxx −−− −−− ×+ × −−− −−− Now for f(35), put x = 35 f(35) = ()( )( ) ()()() ()()( ) ()( )( ) ()( )( ) ()()( ) ()()() ()()() 5515 10515 51015 1055 52 67.3 84.1 94.4 5 15 25 5 10 20 15 10 10 25 20 10 −− −− − − ×+ × + × + × −− − − − − f(35) = – 10.5 + 50.475 + 42.05 + 4.72 f(35) = 77.405 (ii) By using Lagrange’s formula, we have f(x) = ()()() ()()() ()()( ) ()()( ) 2.545.5 145.5 47.5 12.51415.5 2.512.542.55.5 xxx xxx −−− −−− ×+ × −−− −−− + ()()() ()( )( ) ()( )() ()( ) () 12.55.5 12.54 13 17.5 4 1 4 2.5 4 5.5 5.5 1 5.5 2.5 5.5 4 xx x xx x −− − −− − ×+ × −− − − − − Put x = 5 f(5) = ()()( ) ()()() ()()( ) ()( )() ()( )( ) ()( )( ) ()( )() ()()() −− − ×+ × + × + × −−− −− − 2.5 1 0.5 4 1 0.5 4 2.5 0.5 4 2.5 1 4 7.5 13 17.5 1.5 3 4.5 1.5 1.5 3 3 1.5 1.5 4.5 3 1.5 f(5) = 5 15 65 175 20.25 6.75 6.75 20.25 −++ = 0.246913 – 2.2222 + 9.62962 + 8.641975 f(5) = 18.51850831 – 2.222222 = 16.296296. Ans. f(x)= ()()() ()()( ) 47.5 2.5 4 5.5 1 4 5.5 20.25 6.75 xxx xxx − −−−+ −−− + () 13 6.75 − (x – 1) (x – 2.5) (x – 5.5) + 17.5 20.25 (x – 1) (x – 2.5) (x – 4) P(x) = – 0.1481x 3 + 1.5555x 2 – 1.6666x + 4.2592 which is a required polynomial At x = 5, f(x) = 16.3012 (Approx.) Example 7. By means of Lagrange’s formula, show that (i) y 0 = 1 2 (y 1 – y –1 ) – 1 8 ()() 31 1 3 11 yy y y 22 −−  −− −   INTERPOLATION WITH UNEQUAL INTERVAL 231 (ii) y 3 = 0.05 (y 0 + y 6 ) – 0.3 (y 1 + y 5 ) + 0.75 (y 2 + y 4 ) (iii) y 1 = y 3 – 0.3 (y 5 – y –3 ) + 0.2 (y –3 – y –5 ) Sol. (i) For the arguments – 3, –1, 1, 3, the Lagrange’s formula is y x = ()()() ()() () ()()() ()()( ) 31 113 313 31 31 33 13 11 13 xxx x xx yy −− +−− + −− + − + −− −− −+ −− − − + ()()() ()()() ()()() ()()() 13 313 311 131113 333131 xxx xxx yy ++− ++− + ++− ++− y x = ()()() () ()()() 31 113 313 48 16 xxx x xx yy −− +−− +−− + − + ()()() () ()()() 13 313 311 16 48 xxx xxx yy ++− ++− + − (1) Putting x = 0 in (1), we get y 0 = – 1 16 y –3 + 9 16 y -1 + 9 16 y 1 – 1 16 y 3 = 1 2 (y 1 + y –1 ) – 1 8 ()() 31 1 3 11 22 yy y y −−  −− −   (ii) For the arguments 0, 1, 2, 4, 5, 6, the Lagrange’s formula is y x = ()()()()() ()()()()() ()()()()() () ()()()() 01 12456 02456 0102040506 1012141516 xxxxx xxxxx yy −−−−− −−−−− + −− − − − −−−−− + ()() ()()() ()()() ()() ()()()()() () ()()()() 24 01456 01256 2021242526 4041424546 xxxxx xxxxx yy −−−−− −−−−− + −−−− − −−−−− + ()()()()() ()() ()()() ()()()()() () ()()()() 56 01246 01245 50 515254 56 60616264 65 xxxxx xxxxx yy −−−−− −−−−− + − −−−− − −−−− (2) Putting x = 3 in (2), we get y 3 = 0.05 y 0 – 0.3 y 1 + 0.75 y 2 + 0.75 y 4 – 0.3 y 5 + 0.05y 6 y 3 = 0.05 (y 0 + y 6 ) – 0.3 (y 1 + y 5 ) + 0.75 (y 2 + y 4 ) (iii) For the arguments –5, –3, 3, 5, the Lagrange’s formula is y x = ()()() ()() () ()()() () ()() 53 335 535 5 3 5 3 55 35 33 35 xxx xxx yy −− +−− +−− + −+ −− −− −+ −− −− + ()()() ()()() ()()() ()()() 35 535 533 533335 555353 xxx xxx yy ++− ++− + ++− ++− (3) Putting x = 1 in equation (3), we get y 1 = (–0.2) y –5 + 0.5 y –3 + y 3 – 0.3y 5 = y 3 – 0.3 (y 5 – y –3 ) + 0.2 (y –3 – y –5 ) 232 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 8. Prove that Lagrange’s formula can be expressed in the () () () () 2n n 2n 0000 2n 1111 2n nnnn P x | x x x f x | x x x f x | x x x f x | x x x = 0 where P n (x) = f(x). Sol. Let P n (x) = a 0 + a 1 x + a 2 x 2 + a n x n (1) Given that P n (x) = f(x) ⇒ P n (x i ) = f(x i ); i = 0, 1, 2 , n (2) Substitute x = x 0 , x 1 , x 2 , x n Successively in equation (2) ⇒ f(x 0 ) = a 0 + a 1 x 0 + a 2 x 2 0 + + a n x 0 n f(x 1 ) = a 0 + a 1 x 1 + a 2 x 1 2 + + a n x 1 n f(x n ) = a 0 + a 1 x n + a 2 x 2 n + + a n x n n Now Eliminating a 0 , a 1 , a 2 , a n from above equations, we get () () () () 2 2 0000 2 1111 2 | | | | n n n n n nnnn Px xx x fx x x x fx x x x fx x x x − − − − = 0 or () () () () 2 2 0000 2 1111 2 | | | | n n n n n nnnn Px xx x fx x x x fx x x x fx x x x = 0 Example 9. Four equidistant values u –1 , u 0 , u 1 and u 2 being given, a value is interpolated by Lagrange’s formula show that it may be written in the form. u x = yu 0 + xu 1 + ( ) 2 yy 1 3! − ∆ 2 u –1 + ( ) 2 xx 1 3! − ∆ 2 u 0 where x + y = 1. Sol. ∆ 2 u 1 = (E – 1) 2 u –1 = (E 2 – 2E + 1) u –1 = u 1 – 2u 0 + u –1 ∆ 2 u 0 = (E 2 – 2E + 1) u 0 = u 2 – 2u 1 + u 0 R.H.S. = (1 – x) u 0 + xu 1 + ()() {} 2 11 1 3! xx −−− (u 1 – 2u 0 + u –1 ) + () 2 1 3! xx − (u 2 – 2u 1 + u 0 ) where y = 1 – x = – ()() ()()() ()() 101 12 212 1 2 622 xx x x x x x xx uuu − −− +−− + − ++ ()() 2 11 6 xxx u +− + (1) INTERPOLATION WITH UNEQUAL INTERVAL 233 Applying Lagrange’s formula for the arguments – 1, 0, 1 and 2. u x = () () ()()() ()()() ()( )( ) ()() ()()( ) ()() ()()() 1012 12 211 1 2 1 1 123 112 211 321 xx x x x x x xx x xx uuuu − −− −−+ + − + − +++ −−− −− − = ()() ()()() ()() ()() 1012 12 211 1 2 1 1 6226 xx x x x x x xx x xx uuuu − −− −−+ + − + − −+ −+ (2) From (1) and (2), we observe that R.H.S. = L.H.S. Hence the result. Example 10. Find the cubic Lagrange’s interpolating polynomial from the following data () :01 2 5 :2312147 x fx Sol. Here, () () () () 012 3 012 3 0, 1, 2, 5, 2, 3, 12, 147, xxx x fx fx fx fx === = === = Lagrange’s formula is f(x)= ()()( ) ()()() () ()()() ()()() () 12 3 023 01 010203 101213 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −−− −−− + −−− −−− ()()( ) ()()() () ()()() ()()() () 01 3 012 23 202123 303132 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −−− −−− ++ −−− −−− f(x)= ()()() () ()() () ()()() ()() () () 125 025 23 010205 101215 xxx xxx −−− −−− + −−− −−− ()()() ()()() () ()()() ()()() () 015 012 12 147 202125 505152 xxx xxx −−− −−− ++ −−− −−− = 1 5 − (x – 1) (x – 2) + 3 4 x (x – 2) (x – 5) – 2x (x – 1) (x – 5) + 49 20 x (x – 1) (x – 2) = – 1 5 (x 3 – 8x 2 + 17x – 10) + 3 4 (x 3 – 7x 2 + 10) – 2 (x 3 – 6x 2 + 5x) + 49 20 (x 3 – 3x 2 + 2x) ⇒ f(x)= x 3 + x 2 – x + 2 which is the required Lagrange’s interpolating polynomial. Example 11. The function y = f(x) is given at the points (7, 3), (8, 1), (9, 1) and (10, 9). Find the value of y for x = 9.5 using Lagrange’s interpolation formula. Sol. We are given () :78910 :311 9 x fx 234 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here, () () () () 0123 0123 7, 8, 9, 10, 3, 1, 1, 9, xxxx fx fx fx fx ==== ==== Lagrange’s interpolation formula is f(x)= ()()() ()()() () ()()() ()()() () 123 023 01 010203 101213 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −−− −−− + −−− −−− + ()()() ()()() () ()()() ()()() () 013 012 23 202123 303132 xx xx xx xx xx xx fx fx xxxxxx xxxxxx −−− −−− + −−− +−− = ()()( ) () () () () ()()( ) ()()( ) () ()()( ) ()()( ) () 8 9 10 7 9 10 7 8 10 311 7879710 8789810 9798910 xxx xxx xxx −−− −−− −−− ++ −−− −−− −−− ()()() ()()() () 789 9 10 7 10 8 10 9 xxx −−− + −−− = – 1 2 (x – 8) (x – 9) (x – 10) + 1 2 (x – 7) (x – 9) (x – 10) – 1 2 (x – 7) (x – 8) (x – 10) + 3 2 (x – 7) (x – 8) (x – 9) (1) Putting x = 9.5 in eqn. (1), we get f (9.5) = – 1 2 (9.5 – 8) (9.5 – 9) (9.5 – 10) + 1 2 (9.5 – 7) (9.5 – 9) (9.5 – 10) – 1 2 (9.5 – 7) (9.5 – 8) (9.5 – 10) + 3 2 (9.5 – 7) (9.5 – 8) (9.5 – 9) = 3.625 Example 12. Find the unique polynomial P(x) of degree 2 such that: P(1) = 1 P(3) = 27 P(4) = 64 Use Lagrange’s method of interpolation. Sol. Here, () () () 01 2 01 2 1, 3, 4, 1, 27, 64, xx x fx fx fx == = == = Lagrange’s interpolation formula is P(x)= ()() ()() () ()() ()() () ()() ()() () 12 02 01 012 0102 1012 1021 xx xx xx xx xx xx fx fx fx xxxx xxxx xxxx −− −− −− ++ −− −− −− = ()() ()() () ()() ()() () ()() ()() () 34 14 13 12764 1314 3134 4143 xx xx xx −− −− −− ++ −− −− −− = 1 6 (x 2 – 7x + 12) – 27 2 (x 2 – 5x + 4) + 64 3 (x 2 – 4x + 3) = 8x 2 – 19x + 12 Hence the required polynomial is, P(x) = 8x 2 – 19x + 12 INTERPOLATION WITH UNEQUAL INTERVAL 235 PROBLEM SET 5.1 1. Using Lagrange’s interpolation formula, find y(10) from the following table. :5 6 911 :12131416 X Y [Ans. 14.6666667] 2. Use Lagrange’s interpolation formula to fit a polynomial for the data :1023 : 83112 x x u − − Hence or otherwise find the value of u 1 .[Ans. u x = 2x 3 – 6x 2 + 3x + 3, u 1 = 2] 3. Compute the value of f(x) for x = 2.5 from the following data: () :12 3 4 :182764 x fx Using Lagrange’s interpolation method. [Ans. 15.625] 4. If y(1) = – 3, y(3) = 9, y(4) = 30 and y(6) = 132, find the four point Lagrange’s interpolation polynomial which takes the same values as the function y at the given points. [Ans. x 3 – 3x 2 + 5x – 6] 5. Find the polynomial of degree three which takes the values given below: :0124 :1125 x y [Ans. – 1 12 x 3 + 3 4 x 2 – 2 3 x + 1] 6. Find f(x) by using Lagrange’s interpolation formula: () :0234 : 659 705 729 804 x fx Also find maximum value of f(x)[Ans. 151 24 x 3 – 249 8 x 2 + 721 12 x + 659] [No real value of x exist for which f(x) is max.] 7. Given log 10 654 = 2.8156, log 10 658 = 2.8182, log 10 659 = 2.8189 and log 10 661 = 2.8202. Find log 10 656 by Lagrange’s interpolation formula [Ans. 2.8169] 8. Compute Sin 15° by Lagrange’s Method from the data given below: : 0 30 45 60 90 : 0.0000 0.50000 0.70711 0.86603 1.0000 x y °°°°° [Ans. 0.25859] 9. The percentage of Criminals for different age group are given below: Age less than :25304050 Percentage of Criminals :52678494 Apply Lagrange’s formula to find the percentage of criminals under 35 years of age. [Ans. 77] . Criminals for different age group are given below: Age less than :253 04050 Percentage of Criminals :52678494 Apply Lagrange’s formula to find the percentage of criminals under 35 years of age. [Ans Proved. 228 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 3. Find f(x) as a polynomial of x if x :–10367 f(x) : 3 –6 39 822 1611 Sol. Now from Lagrange’s interpolation formula, P(x). = () () () () () () () ()() ()() () 22 22 22 b c fa c a fb a b fc fa b c fb c a fc a b −+−+− −+ −+ − . Proved. Example 5. Applying Lagrange’s formula, find a cubic polynomial which approximate the following data x : –2 –1 2 3 y(x)