236 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 10. If f(x) = 1 ax− show that f(x 0 , x 1 , x 2 , x 3 , x n ) = ()()() 01 1 n ax ax ax −− − and f(x 0 , x 1 , x 2 , x 3 , x n , x) = ()()()() 01 1 n ax ax ax ax −− −− 11. Certain corresponding values of x and log 10 x are given as 10 : 300 304 305 307 log : 2.4771 2.4829 2.4843 2.4871 x x Find the log 10 301 by Lagrange’s formula. [Ans. 2.4786] 12. The following table gives the normal weights of babies during the first 12 months of life: Age in Months :02581012 Weight in lbs : 7.5 10.25 15 16 18 21 Find the weight of babies during 5 to 5.6 months of life. [Ans. 15.67] 13. Find the value of tan 33° by Lagrange’s formula if tan 30° = 0.5774, tan 32° = 0.6249, tan 35° = 0.7002, tan 38° = 0.7813. [Ans. 0.64942084] 14. Apply Lagrange’s formula to find f(5) and f(6) given that f(2) = 4, f(1) = 2, f(3) = 8, f(7) = 128. Explain why the result differs from those obtained by completing the series of powers of 2? [Ans. 38.8, 74; 2 x is not a polynomial] 5.3 ERRORS IN POLYNOMIAL INTERPOLATION Let the function y(x), defined by the (n + 1) points (x i , y i )i = 0, 1, 2, n be continuous and differentiable (n + 1) times, and let y(x) be approximated by a polynomial n φ (x) of degree not exceeding n such that n φ (x i )= yi, i = 1, 2, n. (1) Now use n φ (x) to obtain approximate value of y(x) at some points other than those defined by (1). Since the expression y(x)– n φ (x) vanishes for x = x 0 , x 1 , x n , we put y(x) – n φ (x)= L π n+1 (x) (2) where π n+1 (x) = (x – x 0 ) (x – x 1 ) (x – x n ) (3) and L is to be determined such that equation (2) holds for any intermediate value of x, say x = x’, x 0 < x’ < x n clearly, L = () () () 1 n n yx x x + ′′ −φ ′ π (4) We construct a function F(x) such that F(x)= y(x) – () n x φ – L. () 1n x + π (5) INTERPOLATION WITH UNEQUAL INTERVAL 237 where L is given by (4). It is clear that F(x 0 ) = F(x 1 ) = F(x n ) = F(x’) = 0 that is F(x) vanished (n + 2) times in the interval x 0 ≤ x ≤ x n ; consequently, by the repeated application of Rolle’s theorem, F | (x) must vanish (n + 1) times, F || (x) must vanish n times, etc., in the interval x 0 < x < x n ;. In particular, F (n + 1) (x) must vanish once in the interval. Let this point be given by x = ξ , x 0 < ξ < x n . On differentiating (5) (n + 1) times with respect to x and putting x = ξ , we obtain 0= y (n + 1) ( ξ ) – L. (n + 1)! so that L = () () () 1 1! n y n + ξ + (6) On comparison of (4) and (6), we get y(x’) – n φ (x’) = () () () 1 1! n y n + ξ + π n + 1 (x’) Dropping the prime on x′, we obtain y (x) – n φ (x)= () () 1 1! n x n + π + y (n + 1) ( ξ ), x 0 < ξ < x n (7) Which is the required Expression for error. 5.3.1 Error in Lagrange’s interpolation formula To estimate the error of Lagrange’s interpolation formula for the class of functions which have continuous derivatives of order upto (n + 1) on [a, b] we use above equation 7 (Art. 5.3). Therefore we have, y (x) – L n (x)= R n (x) = () () 1 1! n x n + π + y (n + 1) ( ξ ), a < ξ < b and the quantity E L where E L = [] , max ab |R n (x)| may be taken as an estimate of error. Further, if we assume that () () 1n y + ξ ≤ M n + 1 , a ≤ ξ ≤ b then E L ≤ () 1 1! n M n + + [] , max ab () 1n x + π Example 1. Find the Value of Sin 16 π    from the data given using by Lagrange’s interpolation formula. Hence estimate the error in the solution. :0 4 /2 sin : 0 0.70711 1.0 x yx ππ = 238 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. sin 6 π    = 0 662 0 442 πππ   −−     πππ   −−     (0.70711) + 0 664 0 224 πππ   −−     πππ   −−     (1) = 8 9 (0.70711) – 1 9 = 4.65688 9 = 0.51743 Now, y(x) = sin x, y’(x) = cos x, y’’(x) = sin x, y’’’(x) = – cos x, Hence, () y ′′′ ξ < 1 When x = π/6 () n Rx ≤ 0 66462 3! πππππ   −− −     = 0.02392 where agrees with the actual error in problem. Example 2. Show that the truncation error of quadratic interpolation in an equidistant table is bounded by 3 h 93 max f ξ ′′′ where h is the step size and f is the tabulated function. Sol. Let x i –1 , x i , x i + 1 denote three consecutive equispaced points with step size h. The truncation error of the quadratic Lagrange interpolation is bounded by () 2 ; Efx ≤ 3 6 M max ()()() 11iii xx xx xx −+ −−− where x i – 1 ≤ x ≤ x i + 1 and M 3 = () 11 max x fx −≤ ≤ ′′′ Substitute t = i xx h − then, x – x i – 1 = x – (x i – h)= x – x i + h = th + h = (t + 1) h x – x i + 1 = x – (x i + h)= x – x i – h = th – h = (t – 1) h and (x – x i – 1 ) (x – x i ) (x – x i + 1 )= (t + 1) t(t – 1)h 3 = t(t 2 – 1) h 3 = g(t) Setting g′(t) = 0, we get 3t 2 – 1 = 0 ⇒ t = ± 1 3 For both these values of t, we obtain max () ()( ) 11iii xx xx xx −+ −−− = h 3 11 max t−≤≤ t 2 (1) t − = 3 2 33 h Hence, the truncation error of the quadratic interpolation is bounded by () 3 23 ; 93 h Efx M ≤ () () 3 2 ;max 93 h Efx f ′′′ ≤ξ INTERPOLATION WITH UNEQUAL INTERVAL 239 Example 3. Determine the step size that can be used in the tabulation of f(x) = sin x in the interval 0, 4 π    at equally spaced nodal points so that the truncation error of the quadratic interpolation is less than 5 × 10 –8 . Sol. From Example 2, we know () 3 23 ; 93 h Efx M ≤ For f(x) = sin x, we get f’’’ (x) = – cos x and M 3 = 0/4 max cos x x ≤≤π = 1 Hence the step size h is given by 3 8 5 10 0.009 93 h or h − ≤× = 5.3.2 Inverse Interpolation We know different formulae for obtaining y corresponding to argument value of x (for equal and unequal spaced argument). On the other hand the process of Estimating the value of x for a entry value of y (which is not in the table) is called Inverse Interpolation. In this case when the values of x are unequally spaced, we use Lagrange’s method and when x are equally spaced, then Iterative method should be employed. (a) Lagranges method for inverse interpolation: The only difference of this formula from Lagrange’s method is that x is assumed to be expressible as a polynomial in y. So on interchanging x and y in the Lagrange’s formula we have, x = ()()() ()()() ()()() ()()() 12 02 01 0102 0 1012 1 nn nn yy yy yy yy yy yy xx yyyy yy yyyy yy −− − −− − ++ −− − −− − + ()() () ()()( ) 01 1 01 1 n n nn nn yy yy yy x yyyy yy − − −− − −− − Which is the inverse interpolation formula. (b) Iterative method: Newton’s forward interpolation formula is y u = y 0 + u ∆ y 0 + () ()() 23 00 112 2! 3! uu uu u yy −−− ∆+ ∆+ From this u = () ()() 23 00 0 0 112 1 2! 3! u uu uu u yy y y y −−− −− ∆− ∆−  ∆  (1) On neglecting the second and higher order differences, we get first approximation to u as u 1 = 0 0 u yy y − ∆ (2) 240 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES To find second approximation, retaining the term with second difference in (1) and replace u by u 1 , we get u 2 = 0 1 y ∆ () 11 2 00 1 2! u uu yy y  − −− ∆   Similarly, u 3 = 0 1 y ∆ () ()() 22 22 2 23 00 0 112 2! 3! u uu uu u yy y y −−− −− ∆ − ∆   This process is continued till two successive approximations of u agree with desired accuracy. This technique can equally be also applied by starting with any other interpolation formula. This method is a powerful iterative procedure for finding the roots of an equation to a good degree. Example 4. Using Inverse interpolation find the real root of the equation x 3 + x – 3 which is closed to 1.2. Sol. 234 21 1 0.431 1 1.1 0.569 0.066 0.497 0.006 0 1.2 0.072 0.072 0 0.569 0.006 1 1.3 0.497 0.078 0.647 2 1.4 1.144 xy y y yy∆∆ ∆∆ −− −− − Let the origin be at 1.2. Using Stirling’s formula y = y 0 + u ()() 33 2 01 2 12 1 11 22 6 2 y y uuu yy u y − −− − ∆+∆  +−  ∆+∆ +∆ +      we have, 0 = – 0.072 + u () 2 2 1 0. 569 0.497 0.006 0.006 (0.072) 22 6 2 uu u − ++   ++     or 0 = – 0.072 + 0.533u + 0.036u 2 + u (u 2 – 1) (0.001) or 0 = – 0.072 + 0.532u + 0.036u 2 + 0.00lu 3 (1) From equation (1) u = 0.1353 – 0.0675u 2 – 0.0019u 3 (2) Neglecting all terms beyond the R.H.S of (2), we get u (1) = 0.1353 INTERPOLATION WITH UNEQUAL INTERVAL 241 Substitute u (1) for u in (2), we get second approximation u (2) = 0.1341 u (1) and u (2) are nearly equal up to third decimal place so the required root is x = uh + x 0 = 1.2 + 0.1 × 0.134. Since u = 0 xx h − = 1.2134 Example 5. Values of elliptic integral F( θ ) = 2 0 d 2 1cos θ θ θ + ∫ are given below: () :212325 : 0.3706 0.4068 0.4433F θ°°° θ Find θ for which F( θ ) = 0.3887. Sol. By inverse interpolation formula θ = ()() ()() ()() ()() ()() ()() 12 02 01 012 0102 1012 2021 FF FF FF FF FF FF FFFF FFFF FFFF −− −− −− θ+ θ+ θ −− −− −− = ()( ) ()() () 0.3887 0.4068 0. 3887 0.4433 0.3706 0.3706 0.4068 0.3706 0.4433 −− −− + + = 7.884 + 17.20 – 3.087 = 22° Example 6. From the given table () :20 25 3035 : 0.342 0.423 0.5 0.65 x yx Find the value of x for y(x) = 0.390. Sol. By inverse interpolation formula, x = ()()() ()()() ()()() ()()() 123 023 0 010203 101213 yy yy yy yy yy yy x yyyyyy yyyyyy −−− −−− + −−− −−− x 1 + ()()() ()()() ()()() ()()() 013 012 23 202123 303132 yy yy yy yy yy yy xx yyyyyy yyyyyy −−− −−− + −−− −−− = ()()() ()()() () ()()() ()()() () .39 .423 .39 .5 .39 .65 .39 .342 .39 .5 .39 .65 20 25 .342 .423 .342 .5 .342 .65 .423 .342 .423 .5 .423 .65 −−− −−− + −−− −−− + ()()() ()()() () ()()() ()()() () .39 .342 .39 .423 .39 .65 .39 .342 .39 .423 .39 .5 30 35 .5 .342 .5 .423 .5 .65 .65 .342 .65 .423 .65 .5 −−− −−− + −−− − − − = 22.84057797. Ans. 242 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 7. Find the value of x correct to one decimal place for which y = 7 Given x:1 3 4 y:41219 Sol. Here we use Lagrange’s inverse interpolation formula i.e., x = ()() ()() ()() ()() ()() ()() 12 02 01 012 0102 1012 2021 yy yy yy yy yy yy xxx yyyy yyyy yyyy −− −− −− ++ −− −− −− = ()() ()() ()( ) ()( ) ()( ) ()( ) 712719 7 4719 7 4712 134 4 12 4 19 12 4 12 19 19 4 19 12 −− −− −− ×+ ×+ × −− −− −− = 0.5 + 1.9286 – 0.5714 x = 1.8572 Example 8. Tabulate y = x 3 for x = 2, 3, 4, 5 and calculate the cube root of 10 correct to three decimal places. Sol. For x = 2, y = 8 x = 3, y = 27 x = 4, y = 64 x = 5, y = 125 respectively. Here h = 1 so form forward difference table 23 28 19 327 18 37 6 464 24 61 5125 xyyyy∆∆ ∆ The first approximation is given by (using Newton’s Forward formula) u 1 = 0 1 y ∆ (y u – y 0 ) = 1 19 (10 – 8) = 0.1 The second approximation is u 2 = () 1 2 00 0 1 1 2! u u yy y y  − −− ∆   ∆  = () () 0.1 0.1 1 1 10 8 18 19 2  − −−   = 0.15 INTERPOLATION WITH UNEQUAL INTERVAL 243 The third approximation is u 3 = () ()() 22 2 2 2 23 00 0 0 112 1 2! 3! u uu uu u yy y y y −−− −− ∆ − ∆   ∆  = 1 19 () ()() 0.15 0. 15 1 0.15 0.15 1 0.15 2 10 8 18 6 26 −−− −− −   = 0.1532 Similarly fourth approximation is u 4 = 0 1 y ∆ = () ()() ()()() 33 33 3 33 3 3 23 4 00 0 0 112123 2! 3! 4! u uu uu u uu u u yy y y y −−−−−−   −− ∆ − ∆ − ∆       = () ()() 0.1532 0.1532 1 0.1532 0.1532 1 0.1532 2 1 10 8 18 6 19 2 6  −−− −− −   = 0.1541 and fifth approximation is u 5 = 0.1542 Hence u 4 ≈ u 5 (correct to 3 places of decimal) We have to find cube root of 10. Since 10 lies between the value of y corresponding to x =2 and x = 3, therefore the required value of 3 10 is x = x 0 + uh = 2 + 0.1541 × 1 x = 2.154. Ans. 5.3.3 Expression of Function as a Sum of Partial Fractions Example 9. Let f(x) = 2 32 xx3 x2xx2 +− −−+ Sol. Consider φ (x) = x 2 + x – 3 and tabulate its values for x = 1, –1, 2, we get 2 112 3133 x xx − +− − − Using Lagrange’s formula, we get f(x)= ()() ()() () ()() ()() () ()() ()() () 12 12 11 133 1112 11 12 2121 xx xx xx +− −− −+ −+ −+ +− −−−− −+ = 1 2 (x + 1) (x – 2) – 1 2 (x – 1) (x – 2) + (x – 1) (x + 1) ⇒ f(x)= () ()()() 112 x xxx φ −+− = ()() 111 2121 2xxx −+ −+− . Ans. 244 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 10. Show that the sum of Lagrangian co-efficients is unity. Sol. Let () x ∏ = () () () 01 n xx xx xx −− − The reciprocal of () x ∏ ⇒ () ()()() 01 11 n xxxxx xx = ∏−−− Let ()()() 01 1 n xx xx xx −− − = 0 12 012 n n AAAA xx xx xx xx ++++ −−− − = 0 n i i i A xx = − ∑ (This can be expressible as partial fractions.) Now, A i = ()()()()() 01 1 1 1 i i ii ii in xxxx xx xx xx −+ −− − − − (1) is obtained by taking L.C.M. of (1) and setting x = x i A i can be written as A i = () () 1 i at x x x − ′ ∏ = () 1 x i ′ ∏ From (1) () 1 x Π = () () 0 1 n ii i xxx = ′ Π− ∑ Multiply both sides by () x ∏ , we get 1= () () () 0 n ii i x xxx = ∏ ′ ∏− ∑ = () 0 n i i Lx = ∑ . Proved. PROBLEM SET 5.2 1. Given that y 10 = 1754, y 15 = 2648, y 20 = 3564, find the value of x for y = 3000 by using, iterative method of inverse interpolation. [Ans. 16.935] 2. Given that :1.82.02.22.42.6 : 2.9 3.6 4.4 5.5 6.7 x y Find x when y = 5 using iterative interpolation formula. 3. Using inverse interpolation find the real root of the equation x 3 – 15x + 4 = 0 close to 0.3 correct upto 4 decimal places. [Ans. 0.2679] 4. Find the value of θ if f( θ ) = 0.3887 from the table given below: () 21 23 25 0.3706 0.4068 0.4433f θ°°° θ [Ans. 22°] 5. Find x when f(x) = 14 for the following data using Lagrange’s inverse interpolation formula. () 0 5 10 15 16.35 14.88 13.59 12.46 x fx [Ans. 8.337] INTERPOLATION WITH UNEQUAL INTERVAL 245 6. Using Lagrange’s interpolation formula express the function ()()() 2 31 123 xx xxx ++ −−− as sums of partial fractions. () () 51531 21 223xx x  −+  −− −   Ans. 7. Express the function () ()() 2 2 61 146 xx xxx ++ −−− as a sum of partial factions using Lagrange’s interpolation formula. () ()()() 131371 5 1 35 1 10 4 70 6xxxx  +−+  −+−−   Ans. 8. From the following data find the value of x corresponding to y = 12 using Lagrange’s technique of inverse interpolation. : 1.2 2.1 2.8 4.1 4.9 6.2 : 4.2 6.8 9.8 13.4 15.5 19.6 x y [Ans. 3.55] 9. Obtain the values of t when A = 85 from the following table using Lagrange’s Method of inverse interpolation. :2 5 8 14 : 94.8 87.9 81.3 68.7 t A [Ans. 6.5928] 5.4 DIVIDED DIFFERENCE When the values of the argument are given at unequal spaced interval, then the various differences will also be affected by the changes in the values of the argument. The differences defined by taking into consideration the changes in the values of argument are known as divided differences where as the difference defined earlier are called ordinary differences. Lagrange’s interpolation formula has the disadvantage that if any other interpolation point were added, the interpolation co-efficient will have to be recomputed. So an interpolation polynomial, which has the property that a polynomial of higher degree may be derived from it by simply adding new terms, in Newton’s divided difference formula. Let (x 0 , y 0 ), (x 1 , y 1 ), (x 2 , y 2 ) (x n , y n ) be given (n + 1) points. Let y 0 , y 1 , y 2 , y n be the values of the function corresponding to the values of argument x 0 , x 1 , x 2 , x n which are not equally spaced. Since the difference of the function values with respect to the difference of the arguments are called divided differences, so the first divided difference for the arguments x 0 , x 1 , is given by f(x 0 , x 1 ) = 1 x ∆ y 1 = [x 0 , x 1 ] = 10 10 yy xx − − Similarly f(x 1 , x 2 ) = 2 x ∆ y 1 = [x 1 , x 2 ] = 21 21 yy xx − − and so on. . Interpolation We know different formulae for obtaining y corresponding to argument value of x (for equal and unequal spaced argument). On the other hand the process of Estimating the value of x for a. entry value of y (which is not in the table) is called Inverse Interpolation. In this case when the values of x are unequally spaced, we use Lagrange’s method and when x are equally spaced, then Iterative. second and higher order differences, we get first approximation to u as u 1 = 0 0 u yy y − ∆ (2) 240 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES To find second approximation, retaining