A textbook of Computer Based Numerical and Statiscal Techniques part 17 ppsx

146 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 22 21 (1) (1 ) xx x xx x EEE Eu u u EE −   +∆ + − +∆ = =     = 2 1 [1 ( 1)] x xx xx EE uEEEu E −  −+ =+ −   = (1 + ∆E) x u 0 = (1 + x C 1 ∆E + x C 2 ∆ 2 E 2 + )u 0 = +∆+∆ + 2 2 0102 0 xx uCEuCEu = +∆+∆+ 2 01122 xx uCuCu = L.H.S. 2. R.H.S. = −− − −    −∆ + ∆ − ∆ +        23 24 6 (1/2) (3/2) (5/2) (7/2) 1 1 1.3 1 1.3.5 1 28 2!8 3!8 xx x x uu u u = 2 1/2 2 3/2 4 5/2 111 (1/2)(3/2)1 224 2!4 xx x Eu Eu Eu −− −   −⋅∆ + ∆      3 67/2 (1/2)(3/2)(5/2) 1 3! 4 x Eu −   −∆+      = ()() 2 1/2 2 1 2 1 1/2 3/2 111 1 1 224 2!4 EE E −− −  −−     +− ∆ + ∆            3 21 (1/2)(3/2)(5/2) 1 3! 4 x Eu −  −−−  +∆+      = 1/2 2 1 1/2 11 [1 ] 24 x EEu −−− +∆ = 1/2 2 1/2 1/2 1/2 2 1/2 14 1 2[4(1) ] 24 2 xx E EuEE u E − −−−  +∆ =+∆+∆   = 21/2 1 1 [(2)] (2) (1) xxx uuEu −−− +∆ = +∆ = + = 2345 [1 ] x EEEEE u −+−+−+ = +++++ −+−+−+ 12345 = L.H.S. xxxxxx uuuuuu 3. R.H.S. = (1 + x) n u 0 + n C 1 (1 + x) n–1 x 0 u∆ + n C 2 (1 + x) n–2 x 2 2 0 u∆ + = {(1 + x) + x∆} n u 0 = (1 + x (1 + ∆)} n u 0 = (1 + xE) n u 0 = [1 + n C 1 xE + n C 2 x 2 E 2 + n C 3 x 3 E 3 + ]u 0 = u 0 + n C 1 u 1 x + n C 2 u 2 x 2 + n C 3 u 3 x 3 + = L.H.S. CALCULUS OF FINITE DIFFERENCES 147 Example 10. Prove that n2n 3n 4n 1 13 135 xx x x 224246 ∆−∆+∆− ∆ + =  +−−   nn 11 xx 22 . Sol. L.H.S. = 2 13 1 22 1 21.2 n x   −−    ∆−∆+ ∆+    = () 1/2 1/2 1 1 2 n nn xEx x − −  ∆+∆ =∆ =∆ −   = 11 11 1 22 22 nn nn xxxx   +− −− =+ −−     = R.H.S. 3.9 FACTORIAL NOTATIONS The product of n consecutive factors each at a constant difference and the first factor being x is called a factorial function or a factorial polynomial of degree n and is defined by x (n) = x(x – h) (x – 2h) (x – 3h) (x – (n – 1)), n > 0 If interval of differencing being unity then x (n) = x(x – 1) (x – 2) (x – 3) (x–(n – 1)), n > 0 Because of their properties, this function play an important role in the theory of finite differences and also it helps in finding the various order differences of a polynomial directly by simple rule of differentiation. Example 11. Obtain the function whose first difference is 9x 2 + 11x + 5. Sol. Let f(x) be the required function so that () fx ∆ = 9x 2 + 11x + 5 Let 9x 2 + 11x + 5 = 9 [x] 2 + A [x] + B = 9x (x – 1) + Ax + B On substitution x = 0, we get B = 5 and for x = 1, we get A = 20. Therefore we have () fx ∆ = 9[x] 2 + 20 [x] + 5 On integrating, we have f (x)= [] [] [] +++ 32 9205 32 xx xc = 3x(x – 1) (x – 2) + 10x (x – 1) + 5x + c = 3x 3 + x 2 + x + c, where c is the constant of integration. Example 12. Find a function f x for which x f ∆ = x(x – 1). Sol. We have x f ∆ = x(x – 1) = x (2) Therefore f x = () 3 3 x + C, where C is an arbitrary constant = ()() 1 12 3 xx x C −−+ 148 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 13. Prove that if m is a positive integer then () () () () () m mm1 x1 xx . m! m! m 1 ! − + =+ − Sol. To prove this we start from R.H.S. and use factorial notations to simplify. () () () 1 !1! mm xx mm − + − = ()()( )()()( ) () 12 1 12 2 !1! xx x x m xx x x m mm − − −+ − − −+ + − = ()()( ) () 12 2 1 ! xx x x m xm m m −− −+  −++  = ()()()( ) 112 2 ! xxx x xm m +−− −+ = () 1 ! m x m + . Example 14. Prove that 2() ( 2) (1) mm xmmx − ∆=− ; where m is a positive integer and interval of differencing being unity. Sol. We know, ()m x = ( 1)( 2) ( 1)xx x x m−− −− . Therefore, ()m x ∆ = {}{} ( 1) ( 1)( 2) ( 1 1) ( 1) ( 1)xxx x x m xx xm + − − +−− − − −− = {} ( 1)( 2) ( 2) 1 ( 1)xx x x m x x m −− −− +−−− = mx (m–1) Also, 2()m x ∆ = {} () ( 1) ( mm xmx − ∆∆ =∆ = (1)m mx − ∆ = (2) (1) m mm x − − . Hence proved Example 15. Evaluate 1 () nn n ax bx − ∆+ Sol. 1 () nn n ax bx − ∆+ = 1 () ( ) nn nn ax bx − ∆+∆ = !(0) !an b an+= Example 16. Denoting () x n x(x 1) (x n 1) , n! −−+ = prove that for any polynomial φ(x) of degree k, () = = ∑ k xi i i0 (x) (0) φ∆φ . Sol. We have E n f(a)= f(a + nh) = 2 12 () () () () nn nn n fa C fa C fa C fa +∆+∆ ++∆ Substitute a = 0, n = x we have for h = 1 f(x)= 2 12 (0) (0) (0) (0) xx xx x fCfCf Cf +∆+∆ ++∆ Again f(x) = φ(x) is the given polynomial of degree k therefore we have () k xL ∆φ = (constant). Therefore higher order differences become zero. CALCULUS OF FINITE DIFFERENCES 149 ∴ 2 12 ( ) (0) (0) (0) (0) xx xk k xCC C φ=φ+ ∆φ+ ∆ ++ ∆φ = () 1 (0) k xi i i = ∆φ ∑ 3.10 RECIPROCAL FACTORIAL NOTATION The reciprocal factorial is denoted by ()n x − and is defined by ()n x − = () 11 ()(2) ( ) () n xhxhxnh xnh = ++ + + If the interval of differencing being unity, then ()n x − = () 11 ( 1)( 2) ( ) () n xx xn xn = ++ + + Example 17. Prove that −−− ∆=+ 2(m) (m2) xm(m1)x Sol. We have, ()m x − = 1 ( 1)( 2) ( )xx xm ++ + Therefore, ()m x − ∆ = 11 ( 2)( 3) ( 1) ( 1)( 2) ( )x x xm x x xm − ++ ++ ++ + = 111 ( 2)( 3) ( ) ( 1) ( 1)x x xm xm x  −  ++ + ++ +  = ( 1)( 2) ( )( 1) m xx xmxm − ++ + ++ = −+ −− −=− [( 1)] 1 () mm mx mx Therefore, 2( )m x − ∆ = () ( 1) ()( ) mm xmx −−− ∆∆ =∆− = (1) (2) (1) mm mx m m x −− −− −∆ =− − − = m (m + 1) x (–m–2) . Example 18. Express f(x) = ()() − ++ x1 x1x3 in terms of negative factorial polynomials. Sol. Here, f(x)= ()() 1 13 x xx − ++ Multiply and divide by (x + 2) = ()()() (1)(2) 123 xx xxx −+ +++ = ()()()()() 14 4 112 123xxx xxx −+ +++ +++ = x (–1) – 4x (–2) + 4x (–3) 150 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES CorCor CorCor Cor ollary:ollary: ollary:ollary: ollary: To show that, () n n x ∆ = n! h n and 1n+ ∆ x (n) = 0 We know that, () n x ∆ = (x + h) (n) – (x) (n) (1) Since, (x + h) (n) = (x + h) .x. (x – h) (x – 2h) {x – (n–2)h} (x) (n) = x(x – h) (x – 2h) {x – (n – 1)h} Therefore from (1), we have ⇒ (x + h) (n) – (x) (n) = (x + h) .x. (x – h) (x – 2h) {x – (n – 2)h} – x (x – h) ( x – 2h) {x – (n – 1)h} or (x + h) (n) – (x) (n) = x (x – h) (x – 2h) {x – (n – 2)h} [(x + h) –{x – (n – 1)h}] or (x + h) (n) – (x) (n) = x (n – 1) nh (a) This is a polynomial of degree (n – 1) in factorial notation. Similarly, () 2 n x ∆ = () n x  ∆∆  = () 1n nhx −  ∆  [using (a)] or nh () 1n x − ∆ = nh () () () 1 1 n n xh x − −  +−   = nh[(x + h) x (x – h) {x – (n – 3)h} – x (x – h) {x – (n – 2)h}] = nh[x(x–h) {x – (n – 3) h}] (nh – h) = nh (n – 1) h [x (x – h) (x – 2h) {x – (n – 3)h}] = n(n – 1) h 2 x (n–2) This is a polynomial of degree (n – 2) in factorial notation. Proceeding in the same way, we get () n n x ∆ = n (n – 1) (n – 2) 3.2.1.h n x n–n = n! h n x 0 = n! h n . This term is a constant Again, () 1 n n x + ∆ = () n n x  ∆∆  = ! n nh  ∆  = 0. Because n!h n is constant 3.11 METHOD OF REPRESENTING POLYNOMIAL IN FACTORIAL NOTATIONS 3.11.1 Direct Method Example 19. Express 2x 3 – 3x 2 + 3x – 10 and its differences in factorial notation, consider the interval of differencing being unity. Sol. Let 2x 3 – 3x 2 + 3x – 10 = Ax (3) + Bx (2) + Cx (1) + D = Ax (x – 1) (x – 2) + Bx (x – 1) + Cx + D (1) Where A, B, C and D are constants to be found. Now substitute x = 0 on both the sides of equation (1), we get D = – 10. Again, substituting x = 1 on both sides of equation (1), we get 2 – 3 + 3 – 10 = C + D i.e., C = 2 CALCULUS OF FINITE DIFFERENCES 151 Again substituting x = 2, we get 16 – 12 + 6 – 10 = 2B + 2C + D or 0 = 2B – 6 i.e., B= 3. By equating the coefficient of x 2 on both sides of (1), we get A = 2 Hence, the required polynomial in factorial notation will be f(x)= 2x (3) + 3x (2) + 2x (1) – 10 Again by the simple rule of differentiation, we have (2) () 6 6 2 fx x x∆= ++ 2 ( ) 12 6 fx x∆=+ 3 () 12 fx∆= Example 20. Find the relation between α, β and γ in order that ++ 2 xx αβ γ may be expressible in one term in the factorial notation. Sol. Let, ()fx = 2(2) () xx abxα+β +γ = + where a and b are certain unknown constants. Now, (2) () abx+ = ()[(1)]abxabx++− = 22 ()( )() abxabbx abx abbx+−+=+−− = 22222 ()(2) a ab abbxbx x x−+ − + =α+β+γ Comparing the coefficients of various powers of x we get 222 ,2 , aab abb bα= − β= − γ= Eliminating a and b from the above equations, we get 22 4 γ+αγ=β This is the required relation. 3.11.2 Method of Synthetic Division It is also called the method of detached coefficient. Example 21. Represent −+− 32 2x 3x 3x 10 in factorial notation using synthetic division method. Sol. Let, 32 (3)(2)(1) ()23310 fxxxx AxBxCxD=−+−= +++ = (1)(2) (1)Ax x x Bx x Cx D−−+ −++ In this case we divide the function ()fx by x. Then the remainder will be –10 and the quotient will be 2 233. xx−+ ∴ D = –10 152 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now divide the quotient by (x – 1) i.e., − − −+ − −+ −+ 2 2 21 (1) 233 22 3 1 2 x x xx xx x x In this case quotient is (2x – 1) and remainder is 2 = C. Again divide (2x – 1) by (x – 2), 2 (2)21 24 3 xx x −− − Therefore the quotient 2 = A and the remainder are 3 = B. Hence the required polynomial will be (3) (2) 23210 xxx++− We can also simplify the above in the following way: Taking the coefficients of various powers of x in f(x), we have 1 2 –3 3 –10 = D (a) 02–1 22–12= C (b) 04 323= B (c) 0 2 = A The following steps are to be followed in the method of synthetic division: 1. Put the coefficients of different powers of x in order beginning with the coefficients of higher power of x. 2. Put 1 in the left hand side column and write zero below the coefficients of highest power of x (in this case we have written zero below 2 which is coefficient of x 3 ). 3. Now multiply 2 by 1 and 0 by 1, add them to get the sum 2, put 2 below –3 as given in (a). Now multiply –3 by 1 and 2 by 1 and add them to get the sum –1 which is to be written below 3. The remainder –10 is the value of D. 4. Add the terms of corresponding columns of (a) and get 2, –1 and 2 of (b) 5. Now again apply the steps (1) and (3): in this way we get (2 × 2) + (0 × 2) = 4 which is to be written below –1. The remainder 2 of (b) is equal to C. 6. Apply step (4) on (b) and get 2 and 3 of (c). 7. Again apply the steps (2), (3) and (4) to get 2 which will be equal to A and remainder 3 of (c) which will be equal to B. CALCULUS OF FINITE DIFFERENCES 153 Example 22. Express f(x) = x 4 – 12x 3 + 24x 2 – 30x + 9, and its successive differences in factorial notation. Hence show that () 5 fx ∆ = 0. Sol. Let f(x) = A[x] 4 + B [x] 3 + C[x] 2 + D[x] + E. Use method of synthetic division, we divide by x, x – 1, x – 2, x – 3 etc. successively, then 1 1 –12 24 –30 9 = E 1–11 13 2 1 –11 13 –17 = D 2 –18 3 1 –9 –5 = C 3 41 –6= B 1 = A Hence, f(x) = [x] 4 – 6[x] 3 – 5[x] 2 – 17[x] + 9 ∴ () fx ∆ = 4[x] 3 – 18[x] 2 – 10[x] – 17 () 2 fx ∆ = 12[x] 2 – 36[x] – 10 () 3 fx ∆ = 24[x] – 36 ⇒ () 4 fx ∆ = 24 And () 5 fx ∆ = 0. Example 23: Find the lowest degree polynomial which takes the following values: () x 012345 fx 038152435 Sol. We know that f (a + nh)= f (a) + n C 1 () fa ∆ + n C 2 () 2 fa ∆ + + n C n () n fa ∆ (1) Putting a = 0, h = 1, n = x, we get f(x)= f(0) + x C 1 () 0f ∆ + x C 2 () 2 0 f ∆ + x C 3 () 3 0 f ∆ = f(0) + x (1) () 0f ∆ + () 2 2! x () 2 0 f ∆ + () 3 3! x () 3 0 f ∆ + (2) 154 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now form the difference table for the given data to find ∆ f (0), 2 ∆ f(0), 3 ∆ f(0) etc. ∆∆ ∆ 23 () () () () 00 3 13 2 50 28 2 70 315 2 90 424 2 11 535 xfxfxfxfx Substituting the values of 23 (0), (0), (0), (0) ffff∆∆ ∆ in (2), we get f(x)= (2) (1) 03 2 0 2! x x++⋅+ = 2 3(1) 2 xxx x x+−=+ . Example 24. A second degree polynomial passes through the points (0,1), (1,3), (2,7), (3,13). Find the polynomial. Sol. Let f(x) = Ax 2 + Bx + C. Now form the difference table for given points: ∆∆ 2 () () () 01 2 13 2 4 27 2 6 313 xfx fx fx Since ()fx∆ = 2 Ax Bx C∆+∆+∆ Therefore, ()fx∆ = 22 {( 1) } ( 1 ) 0 Ax x Bx x+−+ +−+ = (2 1)Ax B++ Put x = 0, (0)f∆ = 2AB AB+⇒ += CALCULUS OF FINITE DIFFERENCES 155 Again, 2 () fx∆ = ⇒∆ = = ⇒ = 2 2(0)221 Af AA Also, we have B = 1 Therefore polynomial for the given point is f(x) = x 2 + x + 1. Example 25. Write down the polynomial of lowest degree which satisfies the following set of numbers 0, 7, 26, 63, 124, 215, 342, 511. Sol. We know (2) (3) (1) 2 2 ( ) (0) (0) (0) (0) 2! 3! xx fxfxfff=+∆+∆ +∆ + By forming the difference table we calculate 23 (0), (0), (0), (0) ff f f∆∆ ∆ etc. ∆∆∆∆ 234 () () () () () 00 7 17 12 19 6 226 18 0 37 6 363 24 0 61 6 4 124 30 0 91 6 5 215 36 0 127 6 6342 42 169 7511 xfxfxfxfxfx Now substituting the values of () ∆∆ ∆ 23 (0), (0), 0 , (0) ff f f from the difference table, we obtain f(x)= (1) (2) (3) 12 6 07 2! 3! xxx++ + = 76(1)(1)(2)xxx xx x+−+−− = 232 76 6 3 2 xxxxxx+ −+− + = 32 33 xxx++ . R.H.S. 3.9 FACTORIAL NOTATIONS The product of n consecutive factors each at a constant difference and the first factor being x is called a factorial function or a factorial polynomial of degree n and. 4x (–3) 150 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES CorCor CorCor Cor ollary:ollary: ollary:ollary: ollary: To show that, () n n x ∆ = n! h n and 1n+ ∆ x (n) = 0 We know that, () n x ∆ =. in the factorial notation. Sol. Let, ()fx = 2(2) () xx abxα+β +γ = + where a and b are certain unknown constants. Now, (2) () abx+ = ()[(1)]abxabx++− = 22 ()( )() abxabbx abx abbx+−+=+−− =