A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx

166 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ∆∆ ∆ ∆ 234 () 01 1 12 1 21 24 2 1 42 38 4 8 416 nPn From Newton’s formula, P(n) 23 4 (1)(1)(2)(1)(2)(3) (0) (0) (0) (0) (0) 2! 3! 4! xx xx x xx x x PxP P P P +−−−−− =+∆+ ∆+ ∆+ ∆ (1)(1)(2)(1)(2)(3) 1 26 24 xx xx x xx x x x −−−−−− =++ + + 232 432 311 1 2 2 6 6 3 24 4 24 4 xxx xx xx xx x   =++−+−++−+ −     = 43 2 11 7 1. 24 12 12 12 xx xx−+ + + Ans. Example 6. Find the value of sin 52 ° from the given table: θ°°°° θ 45 50 55 60 sin 0.7071 0.7660 0.8192 0.8660 Sol. Here a = 45 ° , h = 5, x = 52 therefore u = 1.4 °∆∆∆ ° °− − °− ° 4 4 42 43 10 10 10 10 45 7071 589 50 7660 57 532 7 55 8192 64 468 60 8660 xyyyy INTERPOLATION WITH EQUAL INTERVAL 167 By forward interpolation formula 2 (1) ( ) () () () 2! uu fa hu fa ufa fa − += +∆+ ∆ + ⇒ 4 10 ( ) fx = () 1.4 (0.4)( 0.6) (1.4)(0.4) 7071 (1.4)589 ( 57) ( 7) 26 − ++ −+ − = 7880 ∴ f(52) = 0.7880 or sin 52 ° = 0.7880 Example 7. Following are the marks obtained by 492 candidates in a certain examination: 04040454550505555606065 . 2104354743279 Marks No of Candidates −−−−−− Find out (a) No. of candidates, if they secure more than 48 but less than 50 marks. (b) Less than 48 but not less than 45 marks. Sol. First we form the difference table as follows: ∆∆ ∆ ∆ ∆ − − − 2345 . 40 210 43 45 253 11 54 9 50 307 20 71 74 62 222 55 381 42 151 32 89 60 413 47 79 65 492 Marks less than No of Candidates Here, h = 5, For (a), x = 48, a = 40, 48 40 1.6 5 u − ∴= = Now on applying Newton’s forward difference formula, we have 1.6 0.6 1.6 (0.6)( 0.4) 1.6 (0.6)( 0.4)( 1.4) (48) 210 1.6 4.3 11 9 ( 71) 26 24 f ××−×−− =+×+ ×+ ×+ ×− 1.6 (0.6)( 0.4)( 1.4)( 2.4) 222 120 ×−−− +× 168 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 210 + 68.8 + 5.28 – 1.5904 – 0.576 – 2.38694 = 284.08 – 4.553344 = 279.52 ≅ 280 ∴ No. of students secured more than 48 marks but less than 50 marks = 307 – 280 = 27 for (b), x = 48, a = 40, ∴ u = 1.6 1.6 0.6 1.6 (0.6)( 0.4) 1.6 (0.6)( 0.4)( 1.4) (48) 210 1.6 4.3 11 9 ( 71) 26 24 f ××−×−− =+×+ ×+ ×+ ×− 1.6 (0.6)( 0.4)( 1.4)( 2.4) 222 120 ×−−− +× 279.52 280=≅ ∴ No. of students secured more than 45 marks but less than 48 marks = 280 –253 = 27 Example 8. Use Newton’s forward difference formula to obtain the interpolating polynomial f(x) satisfying the following data: 1234 ( ) 26 18 4 1 x fx If another point x = 5, f(x) = 26 is added to the above data, will the interpolating polynomial be the same as before or different. Explain why? Sol. The difference table is ∆∆∆∆ − − − − 234 126 8 218 6 14 17 3 4 11 0 317 41 28 25 526 xy As we seen here that its third difference is being constant either on taking fifth entry or on not taking fifth entry (i.e., 26). So the interpolating polynomial should be same as before, and no change will occur. Here, h = 1, x = 5, a = 1 ∴ u = 4 xa h − = f(x) = 26 + (x – 1)(x – 8) () 1)( 2 (1)(2)(3) (6) 17 2! 3! xx xxx −− −−− +×−+ × INTERPOLATION WITH EQUAL INTERVAL 169 = 26 – 8x + x – 3x 2 + 9x – 6 – () 32 17 6116 6 xx x −+− = 232 17 187 28 3 17 17 66 xx x x x+−+−+− = 32 17 193 20 11 66 xx x−+ + which is the required interpolating polynomial. Example 9. If p , q, r, s be the successive entries corresponding to equidistant arguments in a table. Show that when third differences are taken into account, the entry corresponding to the argument half way between the arguments at q and r is + B A, 24 where A is the arithmetic mean of q and r and B is arithmetic mean of 3q – 2p – s and 3r – 2s – p. Sol. Given A is the arithmetic mean of q and r ⇒ A = 2 qr+ ⇒ q + r = 2A Also, B is the arithmetic means of 3q – 2p – s and 3r – 2s – p. ⇒Β= 32 32 3333 22 qpsrsp qpsr−−+−− −−+ = Β= 3( ) 3( ) 22 qr ps++ − Let quantities p, q, r and s corresponds to argument a, a + h, a + 2h and a + 3h respectively then the value of the argument lying half way between a + h and a + 2h will be a + h + 2 h i.e., a + 3 2 h Hence,a + mh = a + 3 2 h ⇒ m = 3 2 Now, difference table as follows: ∆∆ ∆ − +−+ −−+− +−+ − + 23 () () () () 2 33 22 3 xfxfx fx fx ap qp ah q r qp rq s r qp ah r srq sr ah s Using Newton’s forward interpolation formula up to third difference only and taking m = 3/2, we get 170 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 3 2 fa h  +   = 23 33 33 3 112 3 22 22 2 () () () () 22 6 fa fa fa fa   −−−     +∆ + ∆ + ∆ = 33 1 ()(2) (33) 28 16 pqp rqp srqp+−+−+− −+− = 19 (99) () 16 16 16 ps pqrs qr +  −+ + − = + −   = 923 91 (2 ) 16 3 16 8 8 24 AB B AAA −  −=−+   = 24 B A + Example 10. The table below shows value of tan x for 0.10 ≤ x ≤ 0.30. 0.10 0.15 0.20 0.25 0.30 tan 0.1003 0.1511 0.2027 0.2553 0.3093 x x Evaluate tan 0.12 using Newton’s forward difference table. Sol. Let us, first form the difference table. ∆∆ ∆ ∆ 234 tan 0.10 0.1003 0.0508 0.15 0.1511 0.0008 0.0516 0.0002 0.20 0.2027 0.0010 0.0002 0.0526 0.0004 0.25 0.2553 0.0014 0.0540 0.30 0.3093 xx Here, h = 0.05 a = 0.10, x = 0.12 ∴ u = 0.12 0.10 0.02 0.4 0.05 0.05 xa h −− === ∴ By using Newton’s forward difference formula, we get f(u) = tan (0.12) = f(a) + 23 4 (1)(1)(2)(1)(2)(3) () () () () 2! 3! 4! uu uu u uu u u fa fa fa fa −−−−−− ∆+ ∆+ ∆+ ∆ INTERPOLATION WITH EQUAL INTERVAL 171 = 0.4( 0.6) 0.4( 0.6)( 1.6) 0.1003 0.4 0.0508 (0.0008) (0.0002) 26 −−− +× + × + × + 0.4( 0.6)( 1.6)( 2.6) (0.0002) 24 −−− × = 0.1003 + 0.02032 – 0.000096 + 0.0000128 – 0.00000832 = 0.1206328 – 0.00010432 = 0.12052848 Approx. PROBLEM SET 4.1 1. The population of a town in the decimal census was as given below. Estimate the population for the year 1895. 93 1891 1901 1911 1921 1931 ( ) 46 66 81 101 Year x Population y in thousands [Ans. 54.8528 thousands] 2. If l x represents the number of persons living at age x in a life table, find as accurately as the data will permit l x for values of x = 35, 42 and 47. Given l 20 = 512, l 30 = 390, l 40 = 360, l 50 = 243. [Ans. 394, 326, 274] 3. From the following table, find the value of e 0.24 0.1 0.2 0.3 0.4 0.5 1.10517 1.22140 1.34986 1.49182 1.64872 x x e [Ans. e 0.24 =1.271249] 4. From the table, Estimate the number of students who obtained marks between 40 and 45. 35 30 40 40 50 50 60 60 70 70 80 . 31 42 51 31 Marks No of students −−−−− [Ans. 17] 5. Find the cubic polynomial which takes the following values 10 012 3 ()121 x fx [Ans. 2x 3 -7x 2 +6x+1] 6. Find the number of men getting wages between Rs. 10 and Rs.15 from following table: 42 −−−− ( .) 010 1020 2030 3040 93035 Wages in Rs Frequency [Ans. 15] 172 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. The following are the numbers of deaths in four successive ten year age groups. Find the number of deaths at 45-50 and 50-55. 31496 Age Deaths −−−− 25 35 35 45 45 55 55 65 13229 18139 24225 [Ans. 11278, 12947] 8. The following table give the marks secured by 100 students in Mathematics: 11 Range of marks No of students −−−−− 30 40 40 50 50 60 60 70 70 80 . 25 35 22 7 Use Newton’s forward difference interpolation formula to find (1) the number of students who got more than 55 marks. (2) the number of students who secured marks in the range from 36 to 45. [Ans. (i) 28, (ii) 36] 9. From the following table of half-yearly premium for policies maturing at different ages, estimate the premium for policies maturing at age of 46. 74.48 45 50 55 60 65 ( ) 114.84 96.16 83.32 64.48 Age Premium in Rupees [Ans. Rs. 110.52] 4.3 NEWTON’S GREGORY FORMULA FOR BACKWARD INTERPOLATION Let y = f(x) be a function of x which takes the values f(a), f(a + h), f(a + 2h), f(a+ nh) for (n + 1) equally spaced values a, a + h, , a + nh of the independent variable x. Let us assume f(x) be a nth degree polynomial given by 01 2 () ()()( 1) ()( 1) n fx A Axanh Axanhxan h Axahnxan h =+−−+−−−−−+ −−−−− (x – a – h) (1) Where A 0 , A 1 , A 2 , A n are to be determined. Put x = a + nh, a + 1, nh − a in (1) respectively. Put x = a + nh, then f(a + nh) = A 0 (2) Put x = a+ (n – 1)h, then 01 1 (1) () fa n h A hA fa nh hA+− = − = + − (from 2) ⇒ 1 () fa nh A h ∇+ = (3) Put x = a + (n – 2)h, then 01 2 (2) 2(2)() fa n h A hA h hA+− = − +− − ⇒ 22 2 2! ( 2)()2() () hA fa n h fa nh fa nh fa nh=+−−++∇+=∇+ 2 2 2 () 2! fa nh A h ∇+ = (4) INTERPOLATION WITH EQUAL INTERVAL 173 Proceeding in similar way, () ! n n n fa nh A nh ∇+ = (5) Substituting the vaues in (1), we get () ( ) ( ) ( ) ( )( 1 ) fa nh f x f a nh x a nh x a nh x a n h h ∇+  =++−− + +−− −−−   () ( ) ! n n fa nh xah nh  ∇+ −−    (6) Put x = a +nh + uh, then x – a – nh = uh and x – a – (n – 1)h = (u +1)h x – a – h = () 1un h +− ∴ Equation (6) becomes + + =++∇++ ∇+++ + +−∇ 2 () (1) ()()() () (1) ( 1) 2! ! n n fa nh uu fx fa nh u fa nh fa nh uhu h u n h nh or 2 (1) (1)( 1) ( )() () () () 2! ! n uu uu u n h fanhuh fanh ufanh fanh fanh n +++− ++=++∇++ ∇+++ ∇+ Which in required Newton’s Gregory formular for backward interpolation. This formula is used when we want to interpolate the value near the end of the table. Example 1. The table below gives the value of tan x for 0.10 ≤ x ≤ 0.30 x x : 0.10 0.15 0.20 0.25 0.30 tan : 0.1003 0.1511 0.2027 0.2553 0.3093 Find (a) tan 0.50 (b) tan 0.26 (c) tan 0.40 Sol. First of all we construct the difference table: xx∇∇ ∇ ∇ 234 tan 0.10 0.1003 0.0508 0.15 0.1511 0.0008 0.0516 0.0002 0.20 0.2027 0.0010 0.0002 0.0526 0.0004 0.25 0.2553 0.0014 0.0540 0.30 0.3093 174 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (a) Here, h = 0.05, a = 0.30, x = 0.50 ∴ 0.50 0.30 4 0.05 u − == tan (0.50) = 23 4 ( 1) ( 1)( 2) ( 1)( 2)( 3) () () () () () 2! 3! 4! uu uu u uu u u fa u fa fa fa fa ++++++ +∇ + ∇ + ∇ + ∇ = 45 456 4567 0.3093 4 0.0540 0.0014 0.0004 0.0002 26 24 ×××××× +× + × + × + × = 0.3093 + 0.216 + 0.014 + 0.008 + 0.007 = 0.5543 (b) Here, h = 0.05, a = 0.30, x = 0.26 ∴ u = 0.26 0.30 0.8 0.05 − =− = () 0.8 (0.2) ( 0.8) (0.2) (1.2) 0.3093 ( 0.8) 0.054 0.0014 0.0004 26 −× −× × +− × + × + × ( 0.8) (0.2) (1.2) (2.2) 0.0002 24 −××× +× = 0.3093 – 0.0432 – 0.000112 – 0.0000128 – 0.00000352 = 0.3093 – 0.04332882 ≅ 0.2662 (c) Here, h = 0.05, a = 0.30, x = 0.40 ∴ u = 0.40 0.30 2 0.05 − = 23 (1) (1)(2) (0.40) tan(0.40) ( ) ( ) ( ) ( ) 2! 3! uu uu u f fa u fa fa fa +++ ==+∇+∇+ ∇ 4 (1)(2)(3) () 4! uu u u fa +++ +∇ On putting the subsequent values, we get f(0.40) = 0.3093 + 2 × 0.054 + 23 234 2345 0.0014 0.0004 0.0002 26 24 ×××××× ×+ ×+ = = 0.3093 + 0.108 + 0.0042 + 0.0016 + 0.0001 = 0.4241. Example 2. Using Newton’s backward difference formula find the value of e –1.9 from the following table of value of e –x . x x e − 1 1.25 1.50 1.75 2.00 0.3679 0.2865 0.2231 0.1738 0.1353 INTERPOLATION WITH EQUAL INTERVAL 175 Sol. Difference table for the given data as follows: 234 1 0.3679 0.0814 1.25 0.2865 0.0180 0.0634 0.0039 1.50 0.2231 0.0141 0.0006 0.0493 0.0033 1.75 0.1738 0.0108 0.0385 2.00 0.1353 x xe − ∇∇ ∇ ∆ − −− −− − Here, u = 1.9 2 0.4 0.25 − =− using Newton’s backward difference formula () x fe − = 23 4 (1) (1)(2) (1)(2)(3) () () () () () 2! 3! 4! uu uu u uu u u fa u fa fa fa fa ++++++ +∇ + ∇ + ∇ + ∇ On putting the subsequent values, we get () x fe − = 0.1353 + (–0.4) × (–0.0385) + () −×− × 0.4 ( 0.441) 0.0108 2 − ×−+−+ +× ( 0.4) ( 0.4 1)( 0.4 2) (0.0033) 6 ( 0.4) ( 0.4 1)( 0.4 2)( 0.4 3) 0.0006 24 − ×−+−+−+ +× = 0.1353+0.0154 – 0.001296 + 0.0002112 + 0.000024 = 0.14959 Example 3. In the following table, values of y are consecutive terms of a series of which 23.6 is the 6th term. Find the first and tenth terms of the series. x y 34 5 6 7 8 9 4.8 8.4 14.5 23.6 36.2 52.8 73.9 . the argument half way between the arguments at q and r is + B A, 24 where A is the arithmetic mean of q and r and B is arithmetic mean of 3q – 2p – s and 3r – 2s – p. Sol. Given A is the arithmetic. 0.7880 Example 7. Following are the marks obtained by 492 candidates in a certain examination: 04040454550505555606065 . 2104354743279 Marks No of Candidates −−−−−− Find out (a) No. of candidates,. r and s corresponds to argument a, a + h, a + 2h and a + 3h respectively then the value of the argument lying half way between a + h and a + 2h will be a + h + 2 h i.e., a + 3 2 h Hence,a