286 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 1. S(x i ) = f(x i ); i = 0, 1, 2, n 2. On each subinterval [x i–1 ,x i ], 1 ≤ i ≤ n, S(x) is a polynomial in n of degree at most n . 3. S(x) and its (n –1) derivatives are continuous on [a, b]. 4. S(x) is a polynomial of degree one for x < a and x > b. The process of constructing such type of polynomial is called spline interpolation. 5.7.4 Cubic Spline Interpolation for Equally and Unequally Spaced Values According to the idea of draftsman spline, it is required that both dy dx and the curvature 2 2 dy dx are the same for the pair of cubic that join at each point. The cubic spline have possess the following properties: 1. S(x i ) = f i , i = 0, 1, 2, ,n. 2. The cubic and their first and second derivatives are continuous i.e., S(x), S I (x) and S II (x) and continuous on [a, b] 3. On each subintervals [x i–1 , x i ] 1 ≤ i ≤ n, S(x) is a third degree polynomial. 4. The third derivatives of the cubics usually have jumps discontinuities at the ducks or the junction points. Y P 0 f(x ) 1 P 1 Ducks f(x ) 2 P 2 x 0 x 1 x 2 x i x i + 1 x i + 2 x n – 1 x n X f(x ) i f(x ) i + 1 P i P i + 1 Spline curve f(x ) i + 2 P i + 2 P n – 1 f(x ) n – 1 f(x ) n P n FIG. 5.4 Where x i = for i = 0, 1, 2 , n may or may not be equally spaced. Let a cubic polynomial for the i th interval is S(x i )= a i (x – x i ) 3 + b i (x – x i ) 2 + c i (x – x i ) + d i (1) Since this polynomial is valid for both the points x i and x i+1 therefore, S(x i )= a i (x i – x i ) 3 + b i (x i – x i ) 2 + c i (x i – x i ) + d i (2) ⇒ S(x i )= d i S(x i+1 )= a i (x i+1 – x i ) 3 + b i (x i+1 – x i ) 2 + c i (x i+1 – x i ) + d i ⇒ () 1i Sx + = 32 1 1l i ii i ii i ah bh ch d + ++ +++ (3) where h i + 1 = x i + 1 – x i . INTERPOLATION WITH UNEQUAL INTERVAL 287 Now, Twice differentiate Equation (1) we get, S’ (x i )= 3a i (x – x i ) 2 + 2b i (x – x i ) + c i (4) S’’(x i )= 6a i (x – x i ) 4 + 2b i (5) Now, Let P i = S’’ (x i ) then equation (5) becomes P i = 6a i (x – x i ) + 2b i at x = x i , P i = 2b i 2 i i P b ⇒= (6) at x = x i+1 , P i + 1 = 6a i (x i + 1 – x i ) + 2b i P i + 1 = 6a i (x i+1 – x i ) + P i [using (6)] P i + 1 = 6a i h i+1 + P i a i = 1 1 6 ii i PP h + + − (7) Now substituting the values of d i , a i and b i from (2), (6) and (7) in (3) () 1i Sx + = () () 1 32 1 11 1 1 62 i i i ii ii i i P PPh h ch sx h + + ++ + −+ ++ () 1i Sx + = () () 2 1 2 1 11 62 i i i ii ii i P h PP h ch sx + + ++ −+++ () 1i Sx + = () 21 1 1 62 ii i i iii PP P Sx h ch + + + −  ++   () 1i Sx + = () () 2 l 11 3 6 i iiiiii h Sx P P P ch + ++ =−++ c i = () () [] 2 1 l 1 1 3 6 ii i iii i Sx Sx h PPP h + + + + − −−+ c i = () () [] 1 l 1 1 2 6 ii i ii i Sx Sx h PP h + + + + − −+ (8) Now, the slope at the point x i (because the curve has equal slope at the point [x i , S(x i )] hence from equation (4). S’ (x i )= 3a i (x i – x i ) 2 + 2b i (x i – x i ) + c i () ii Sx c ′ ⇒= (9) c i = S’ = () () 1 1 1 6 ii i i Sx Sx h h + + + − − 1 2 ii PP +  +  (10) But S’ (x i ) for the last subinterval is, S’ (x i )= 3a i–1 h 2 i + 2b i–1 h i + c i–1 . (11) and after using a i–1 , b i–1 , and c i–1 S’ (x i )= 3 [] () () [] 11 2 11 1 11 2. 2 62 6 i i ii i ii i i ii Sx Sx h PP h Ph P P hh − −− − + −+ + −− 288 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES S’ (x i )= () () 1 1 2 6 ii ii i i Sx Sx PP h h − − + + + (12) For equation (9) and (10) c i = 3a i–1 = h i 2 + 2b i–1 h i + c i–1 On substituting the values of a i–1 , b i–1 , c i–1 and c i () () () () ( ) 1 11 1 1 2 2 666 ii iih iiii ii ii Sx Sx Sx Sx hPhP PP hh + − +− + + − − −+= ++ ()() () () () 1 1 11 1 1 2 2 666 ii ii iiiii ii ii Sx Sx Sx Sx hPhPh PP hh − + +− + + − − −=+++ () () () () () 1 1 11 1 1 1 636 ii ii iii ii ii ii Sx Sx Sx Sx h h P hP hP hh − + ++ − + − +− + −=++ for i = 1, 2, n – 1 ⇒ () ()()() ( ) 11 11 1 1 1 26 iiii ii i ii ii ii Sx Sx Sx Sx hP h hP hP hh +− ++ + − + −− +++= −   (13) Now for equally spaced argument i.e., h i = h Equation (13) becomes [] () ()() 11 1 1 6 42 4 iii i i i hP P P Sx Sx Sx +− + − ++ = − +   or P i+1 + 4P i + P i – 1 = 2 6 h [S(x i + 1 ) – 2S (x i ) + S(x i–1 )] (14) while the S(x) for equally spaced becomes. () () ( ) () 2 33 11 1 11 1) 66 ii ii i i i h Sx x x P x x P x x Sx P hh −− −    =−+− +− −−       + () () 2 1 1 6 iii h xx Sx P h −  −−   (15) Equation (15) gives cubic spline interpolation while equation (14) gives the condition for P i . Remarks: (1) If 0 = P = 0 n P ; it is called free boundary conditions and the spline curve for this condition is called the natural spline because the splines are assumed to take their natural straight line shape outside the interval of approximation. (2) If = 0+11 = P , P nn PP ; 01111 = , , nn n fff f hh ++ == then spline is called periodic splines. (3) For a non-periodic spline we use. 0 '( ) , ( ) n fa f fb f ′′ ′ == ⇒ 0 01 0 6 2 + = i i ff PP f hh  − ′ −   ⇒ 1 1 6 + 2 = nn nn n nn ff PP f hh − −  − ′ −   INTERPOLATION WITH UNEQUAL INTERVAL 289 Example 1. Obtain cubic spline for every subinterval, given in the tubular form. () x0123 f x 1 2 33 244 With the end conditions M 0 = 0 = M 3 Sol. Here, we have equal spaced intervals as h 1 = h 2 = h 3 = 1, hence the condition for M i becomes. () () () 111 1 462 1,2 iiiiiii MMM fx fxfx −+− + ++= − + =  ⇒ () () () 012 2 1 0 462 MMM fx fxfx ++= − +   ⇒ () () () 123 3 2 1 462 MMM fx fxfx ++= − +  Now, after substituting the values of f(x i ) and M 0 = 0 = M 3 we get 12 1 2 4 180 4 1080 MM andM M+= + = 12 24 276 MandM=− = x 0 = 0 x 1 = 1 x 2 = 2 x 3 = 3 h 1 = 1 h 2 = 1 h 3 = 1 t 0 = 1 t 1 = 2 t 2 = 33 t 3 = 244 Now, the corresponding cubic spline can be obtained by having () () ( ) () 33 11 11 6 ii iii fx x x M x x M x x hh −−  =− +− +−  () () () 2 2 11 1 1 , 1,2,3 66 ii iii h h fx M x x fx M i h −− −   −+− − =     Now, for i = 1 (the interval is [0, 1]), f(x) = – 4x 3 + 5x + 1 Similarly, for [1, 2], f(x) = 50x 3 – 162x 2 + 167x – 53 and for [2, 3], f(x) = – 46x 3 + 414x 2 – 985x + 715. Example 2: Find the cube splines for following data: () x:0123 fx :12511 with the end condition M 0 = 0 = M 3 and also calculate f (2.5)and f’(2.5). Sol. Here intervals are equally spaced with difference 1 and n = 3. Now, the condition for M i is () ()() 111 1 462 1,2 iii i ii MMM fx fxfxi −++ − ++ = − + =   ⇒ () () () 012 0 1 2 462 MMM fx fxfx ++= − +   290 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ⇒ () () () 123 1 2 3 462 MMM fx fxfx ++= − +   but M 0 = 0 M 3 then it becomes 12 1 2 412418 MM andM M+= = = M 1 = 2 and M 2 = 4 Now, the corresponding cubic spline can be obtained by having () () ( ) () 33 11 1 6 ii iii fx x x M x x M x x h −−  =− +− +−  () () () 1 11 , 1,2,3 66 ii iii MM fx x x fx i − −−  −+− − =   Now, for i = 1 (the interval is [0, 1]) f(x) = () 3 1 23 3 xx ++ Similarly, for [1, 2] f(x) = 1 3 (x 3 + 2x + 3) and for [2, 3], f(x) = 1 3 () −+ − + 32 2183427 xx x Now, f(2.5) = 7.66 and f ’ (2.5) = 6.16 Example 3. Obtain the cubic spline for the following data: () x:0123 fx : 2 6 8 2 −− Sol. Take initial conditions M 0 = 0 = M 3 for i = 1, 2 n h 2 [M i–1 + 4M i + M i+1 ] = 6 [f i+1 – 2f i + f i–1 ] Here, h = 1; ∴ () 012 210 46201 MMM fffforx ++=−+ ≤≤ () 123321 46212 MMM f ffforx ++=−+ ≤≤ ++=   ++=  210 321 436 472 MMM MMM Using initial conditions, we get M 1 = 4.8,M 2 = 16.8 Hence for 0 ≤ x ≤ 1 spline is given by S (x)= () () ()( )()( ) 33 0100 11 1 101606 6 xM x M x f M x f M  −+−+−−+−−  = () () () () 3 1 4.8 1 12 36 4.8 6 xxx  +− +−−  = 3 0.8 8.8 2 xx−+ Hence for 1 ≤ x ≤ 2 spline is given by S(x)= 2x 3 – 3. 6x 2 – 5.2x + 0.8 Similarly, for 2 ≤ x ≤ 3 S(x) = –2.8x 3 + 25.2x 2 – 62.8x + 39.2. Ans. INTERPOLATION WITH UNEQUAL INTERVAL 291 Example 4. Estimate the function value f at x = 7 using cubic splines from the following data: Given P 2 = P 0 = 0. i i i012 x4916 f234 Sol. h 1 = x 1 – x 0 = 9 – 4 = 5 h 2 = x 2 – x 1 = 16 – 9 = 7 ()() 12 12 0122110 21 11 633 hh hh PPPffff hh + ++=−−− ⇒ 1 1 70 P =− = – 0.0143 Since, n = 3 therefore, there are two cubic splines given by S 1 (x) = x 0 ≤ x ≤ x 1 S 2 (x) = x 1 ≤ x ≤ x 2 () () ( ) {} 33 11 1 6 ii ii i Sx x x P x x P h −− =−+− () () 2 2 11 1 11 66 i i ii i iii ii h h xxf P xx f P hh −− −      +− − +− −            For i = 1 S(x) = () () {} () 2 33 1 10 01 100 11 11 66 h xxP xx P xxf P hh    −+− + −−       () 2 1 01 1 1 1 6 h xx f P h    +−−       () () () 143 27 0.0143 3 25 0.0024 30 5 5 Sx =− +++×  () 7 2.64862S = . Ans. PROBLEM SET 5.5 1. Using the Chebyshev polynomials T n (x), obtain the least square approximation of degree eleven for f(x) = cos –1 x. () () () () () 01 3 5 44 4 2925 fx Tx Tx Tx Tx π  =−− −  ππ π  Ans. () () () 711 44 4 9 49 81 121 Tx T x T x  −−−  ππ π  2. Find the linear least-squares polynomial approximation to the function f(x) = 5 + x 2 on the interval [0, 1]. () 1 29 6 6 yx  =+   Ans. 3. Find the quadratic least squares polynomial approximation to the function f(x) = x 3/2 on the interval [0, 1]. () 2 1 248 60 105 yxx  =−++   Ans. 292 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 4. Using the Chebyshev polynomials T n (x), obtain the least squares approximation of second degree for f(x) = 4x 3 + 2x 2 + 5x – 2 on the interval [1, 1]. [Ans. f(x) = – T 0 (x) + 8T 1 (x) + T 2 (x)] 5. Find the best lower-order approximation to the cubic 9x 3 + 7x 2 , –1 ≤ x ≤ 1. 2 27 9 7 , max.error = in [-1,1] 44 xx  +   Ans. 6. Economize the series e x 2 = 1 + x 2 + 4681012 2 6 24 120 720 xxxx x +++ + + on the interval [–1, 1] allowing for a tolerance of 0.05. [Ans. e x 2 = 1.0075 + 0.869x 2 + 0.8229x 4 ] 7. Economize the series x + 35 7 6 120 5040 xx x ++ on the interval [–1, 1], allowing for a tolerance of 0.0005. 3 383 17 sin 384 96 hx x x  =+   Ans. 8. Find a uniform polynomial approximation of degree 1 to (2x – 1) 3 on the interval [0, 1] so that the maximum norm of the error function is minimized, using Lanczos economization. Also calculate the norm of the error function. Hint: Put x = 1 2 t + , linear approximation = () 3 21, 4 x − maximum error = 1 4 9. Find the lowest order polynomial which approximates the function f(x) =1 – x + x 2 – x 3 + x 4 , 0 ≤ x ≤ 1 with an error less than 0.1. () 2 160 160 131 128 128 128 fx x x  =−+   Ans. 10. Obtain an approximation in the sense of the principle of least squares in the form of a polynomial of second degree to the function f(x) = 2 1 1 x+ in the range –1 ≤ x ≤ 1. [Ans. P (x) = 3 4 (2 π – 5) + 15 4 (3 – p) x 2 ] 11. Find the polynomial of second degree, which is the best approximation in maximum norm to x on the point set {} 41 0, , , 1,0 99 · () 2 19 2 16 8 Px x x  =+−   Ans. 12. Find a polynomial P(x) of degree as low as possible such that () 2 1 max 0.05 x x ePx ≤ −≤ [Ans. 1.0075 + 0.8698x 2 + 0.82292x 4 ] 13. Prove that x 2 = () () 02 1 2 Tx Tx +   · 14. Express T 0 (x) + 2T 1 (x) + T 2 (x) as polynomials in x.[Ans. 2x + 2x 2 ] 15. Economize the series f(x) = 1 – 23 2816 xx x −− · 16. Economize the series cos x = 1 – 24 6 224720 xx x +− · 17. Prove that T n (x) is a polynomial in x of degree n 18. Find the best lower order approximation to the cubic 5x 3 + 4x 2 in the closed interval [–1, 1]. [Ans. 4x 2 + 5 4 x] INTERPOLATION WITH UNEQUAL INTERVAL 293 19. Find cubic spline for the following data: () 012 3 12511 x fx with end conditions P 0 = 0 = P 3 and also calculate f(2.5), f’(2.5). 20. Estimate the function value f at x = 7 using cubic splines from the following data: 01 2 4916 23 4 i i i x f [Ans. S 1 (7) = 2.6229] 21. Fit the following points by the cubic spline: () :1234 :15118 x fx By using the conditions M 0 = 0 = M 3 . Hence find f(1.5) and f ’(2) () 32 1 17 51 94 45 1 2 15 fx x x x x   =−+−≤≤    Ans. () 32 1 55 381 770 53 2 3 15 fx xxx x  =− − − + ≤≤  () 32 1 38 456 1741 1980 3 4 15 fx x x x x  =−+−≤≤  () () 103 94 1.5 , 2 40 15 ff  ′ ==   22. Find the cubic spline corresponding to the interval [2, 3] which means the following representation: () :12345 :3015321825 x fx with the end condition M 1 = 0 = M 5 and also compute f(2.5), f’(3) () () () 32 1 142.9 1058.4 2475.2 1950 2.5 24.03 & 3 2.817 16 fx x x x f f    ′ =− + − + =− =      Ans. 23. Fit the following points by Cubic spline and obtain y(1.5): :1 2 3 :8118 x y −− [Ans. () 3 31412 xx −+− () 1.5 5.625y =− ] 24. Obtain cubic spline approximation valid in the interval [3, 4], Given that 1234 3102965 x y Under the natural spline conditions M(1) = 8 = M(4). () {} 32 1 56 72 2092 2175 15 Sx x x x  =− + − +   Ans. 23 62 112 , 55 mm  ==   GGG CHAPTER 6 Numerical Differentiation and Integration 6.1 INTRODUCTION The differentiation and integration are losely linked processes which are actually inversely related. For example, if the given function y(t) represents an objects position as a function of time, its differentiation provides its velocity, =() () d vt yt dt On the other hand, if we are provided with velocity v(t) as a function of time, its integration denotes its position. 0 () () t yt vtdt = ∫ There are so many methods available to find the derivative and definite integration of a function. But when we have a complicated function or a function given in tabular form, they we use numerical methods. In the present chapter, we shall be concerned with the problem of numerical differentiation and integration. 6.2 NUMERICAL DIFFERENTIATION The method of obtaining the derivatives of a function using a numerical technique is known as numerical differentiation. There are essentially two situations where numerical differentiation is required. They are: 1. The function values are known but the function is unknown, such functions are called tabulated function. 2. The function to be differentiated is complicated and, therefore, it is difficult to differentiate. The choice of the formula is the same as discussed for interpolation if the derivative at a point near the beginning of a set of values given by a table is required then we use Newton forward formula, and if the same is required at a point near the end of the set of given tabular 294 NUMERICAL DIFFERENTIATION AND INTEGRATION 295 values, then we use Newton’s backward interpolation formula. The central difference formula (Bessel’s and Stirling’s) used to calculate value for points near the middle of the set of given tabular values. If the values of x are not equally spaced, we use Newton’s divided difference interpolation formula or Lagrange’s interpolation formula to get the required value of the derivative. 6.2.1 Derivation Using Newton’s Forward Interpolation Formula Newton’s forward interpolation is given by 23 00 0 0 (1) (1)(2) 2! 3! uu uu u yy uy y y −−− =+∆+ ∆ + ∆ (1) where − = 0 xx u h Differentiating equation (1) with respect to u, we get −−+ =∆ + ∆ + ∆ 2 23 00 0 21 3 62 2! 3! dy uuu yy y du (2) Now dy dx = dy du · du dx = 1 h dy du ⋅ Therefore,  −−+−+− =∆+ ∆+ ∆+ ∆   232 23 4 00 0 0 121362418226 2! 3! 4! dy uuuuuu yy y y dx h (3) As == 0, 0 xxu , therefore, putting u = 0 in (3), we get =   =∆−∆+∆−∆     0 234 0000 1111 234 xx dy yyyy dx h Differentiating equation (3) again w.r.t. ‘x’, we get   ==×     2 2 1 dy dy dy ddud du dx dx h du dx dx ()  −+ =∆+−∆+ ∆−   2 23 4 00 0 2 161811 1 12 uu yu y y h (4) Putting u = 0 in (4), we get =   =∆−∆+∆−      0 2 23 2 00 0 22 111 12 xx dy yy y dx h …(5) Similarly, =   =∆−∆+      0 3 34 00 33 13 2 xx dy yy dx h and so on. . by a table is required then we use Newton forward formula, and if the same is required at a point near the end of the set of given tabular 294 NUMERICAL DIFFERENTIATION AND INTEGRATION 295 values,. integration. 6.2 NUMERICAL DIFFERENTIATION The method of obtaining the derivatives of a function using a numerical technique is known as numerical differentiation. There are essentially two situations. backward interpolation formula. The central difference formula (Bessel’s and Stirling’s) used to calculate value for points near the middle of the set of given tabular values. If the values of