116 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 14. Evaluate the following: I. ∆ 2 (cos 2x) II. ∆ 2 (3e x ) III. ∆ tan –1 x IV. ∆(x + cos x) the interval of differencing being h. Sol. I. We have ∆ 2 (cos 2x)= (E – 1) 2 cos 2x because ∆ = E – 1. =(E 2 – 2 E + 1) cos 2x = E 2 cos 2x – 2E cos 2x + cos 2x = cos (2x + 4h) – 2 cos (2x + 2h) + cos 2x = cos (2x + 4h) – cos (2x + 2h) – cos (2x + 2h) + cos 2x = 2 sin (2x + 3h) sin (– h) – 2 sin (2x + h) sin h = – 2 sin h [sin (2x + 3h) – sin (2x + h)] = – 2 sin h [2 cos (2x + 2h) sin h] = – 4 sin 2 h cos (2x + 2h). II. We have ∆ (3e x ) = 3( ∆ e x ) = 3 (e x +h – e x ) =3e x (e h –1) ∴ ∆ 2 (3e x )=∆ (∆3e x ) = ∆{3e x (e h –1)} =3 (e h –1) (∆e x ) = 3(e h –1) (e x+h – e x ) =3 (e h –1) e x (e h –1) = 3e x (e h –1) 2 . III. We have ∆ tan –1 x = tan –1 (x + h) – tan –1 x = tan –1 () () 1 xh x xhx +− ++ = tan –1 2 1 h xh x   ++   . IV. We have ∆ (x + cos x) = ∆ x + ∆ cos x ={(x + h) – x} + {cos (x + h) – cos x} = h + 2 sin 2 2 xh+ sin 2 h  −   = h – 2 sin 2 h x  +   sin . 2 h Example 15. Evaluate ∆ 2 E sin (x + h) + () () 2 sin x h Esin x h ∆+ + , where h being the interval of differencing. Sol. To evaluate the given problem we use the operator property that is, ∆ = E – 1 Now 2 E ∆ sin (x + h) + () () 2 sin sin xh Exh ∆+ + = () 2 1 E E − sin (x + h) + ()() () 2 1sin sin 2 Exh xh −+ + CALCULUS OF FINITE DIFFERENCES 117 = (E – 2 + E –1 ) sin (x + h) + −+ + + 2 (21)sin() (2) EE xh sin x h = [sin (x + 2h) – 2 sin (x + h) + sin x] + sin( 3 ) 2 sin( 2 ) sin( ) sin( 2 ) xh xh xh xh +− ++ +  +  = 2 sin (x + h) [cos h – 1] + 2sin( 2 )[cos 1] sin( 2 ) xh h xh +− + = 2 (cos h – 1) {sin (x + h) – 1}. Example 16. Show that B(m + 1, n) = (–1) m ∆ m    1 n where m is a positive integer. Sol. We know that 0 nx e ∞ − ∫ dx = 1 n . Therefore, 0 mnx e ∞ − ∆ ∫ dx = ∆ m 1 n    or 0 mnx e ∞ − ∆ ∫ dx = ∆ m 1 n    , where for m ∆ e –nx , n is to be regarded variable and x is to be regarded as constant. Now, ∆ m e –nx = ∆ m–1 [e –(n+1)x – e –nx ] = ∆ m–1 e –nx (e –x – 1) = (e –x – 1) ∆ m–1 e –nx = (e –x – 1) 2 ∆ m–2 e –nx = = (e –x – 1) m e –nx Therefore, () 0 1 nx x ee ∞ −− − ∫ m dx = ∆ m 1 n    Put e –x = z, so that –e –x dx = dz or dx = – (1/z) dz. Then, 0 1 ∫ n z (z–1) m (–1/z) dz = ∆ m 1 n    or (–1) m 1 1 0 n z − ∫ (1 – z) m dz = ∆ m 1 n    or 1 1 0 n z − ∫ (1 – z) (m + 1)–1 dz = (–1) m ∆ m 1 n    or B (m + 1, n) = (–1) m ∆ m 1 n    118 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 17. Show that e x =  ∆   2 E e x . ∆ x 2x Ee e ; the interval of differencing being h. Sol. Let f(x) = e x , then Ef (x) = f(x + h), therefore Ee x = e x+h . Now, f∆ (x)= f (x + h) – f (x) ∴ x e ∆ = e x+h – e x = e x (e h – 1) ∴ 2 x e ∆ = ∆ ( ∆ e x ) = ∆ {e x (e h – 1)} 2 x e ∆ = (e h – 1) ∆ e x = (e h – 1) 2 e x ∴ 2 E  ∆   e x = ( ∆ 2 E –1 ) e x = ∆ 2 (E –1 e x ) = ∆ 2 (e x–h ) = ∆ 2 (e x e –h ) = e –h ∆ 2 e x = e –h (e h – 1) 2 e x . ∴ 2 E  ∆   e x 2 x x Ee e ∆ = e –h (e h – 1) 2 e x () 2 1 xh hx e ee + − = e –h e x+h = e x . Example 18. Evaluate ∆ 2  +  ++  2 5x 12 x5x6 ; the interval of differencing being unity. Sol. We have ∆ 2 2 512 56 x xx  +   ++  Therefore, ∆ 2 2 512 56 x xx  +   ++  = ∆ 2 ()() 512 23 x xx  +   ++   = ∆ 2   +=∆∆ +∆   ++ + +    23 2 3 23 2 3xx x x = ∆ 11 11 23 32 43xx xx   −+ −   ++ ++    = – 2 ∆ ()() ()() 11 3 23 3 4xx xx   −∆  ++ ++   = – 2 ()()()() 11 34 23xx xx  −  ++ ++   – 3 ()()()() 11 45 34xx xx  −  ++ ++   = ()()()()()() 46 234 345xxx xxx + +++ +++ = () ()()()() 25 16 2345 x xxxx + ++++ . CALCULUS OF FINITE DIFFERENCES 119 Example 19. Evaluate ∆ n e ax+b ; where the interval of differencing taken to be unity? Sol. Given ∆ n e ax+b ; which shows that f(x) = e ax+b . Now ∆f (x)= f(x + 1) – f(x) ∴ ∆(e a + bx )= e a(x + 1)+b – e ax + b = e ax + b (e a – 1) ∴ ∆ 2 (e a + bx )= ∆ ( ∆ e a + bx ) = ∆ {e ax + b (e a –1)} = (e a – 1) (∆ e ax + b ) = (e a – 1) e ax + b (e a –1) = (e a –1) 2 e ax + b . Proceeding in the same way, we get ∆ n e ax + b = (e a –1) n e ax + b Example 20. With usual notations, prove that, ∆ n    1 x = (–1) n . () ( ) ! n nh x x h x nh ++ Sol. 1 n x  ∆   = 1 1 n x −  ∆∆   = 1 11 n xhx −  ∆−  +  = 1 () n h xx h −  − ∆  +  = −  −∆ ∆  +  2 1 () () n h xx h = −   −∆ ∆ −   +   2 11 (1) n xxh = 2 11 1 1 (1) 2 n xhx x hxh −    −∆ − − −    +++    = 2 21 1 (1) 2 n xh x x h −  −∆ − −  ++  = 2 2 2 (1) ()(2) n h xx h x h −  − −∆  ++  = −  −∆  ++  2 22 2! (1) ()(2) n h xx h x h = 3 33 ! (1) ()(2)(3) n h xx hx hx h −  3 −∆  ++ +  . . . . . = ! (1) ( ) ( ) n n nh xx h x nh − ++ Example 21. Prove that: (a) µµ −δδ  µ=   −+ 1 f(x) g(x) f(x) g(x) f(x) 4 11 g(x) g(x )g(x ) 22 Here, interval of differencing being unity. 120 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (b) ∇ 2 = h 2 D 2 – h 3 D 3 + 7 12 h 4 D 4 – (c) ∇ – ∆ = – ∇ ∆ (d) 1 +    2 2 δ = + 22 1 δµ (e) µδ = 1 2 ()∆+∇ Sol. (a) Here R.H.S. is = 1 () () () () 4 11 ()() 22 fx gx fx gx gx gx µµ −δδ −+ Now numerator of R.H.S. is given by: = 1 2 [E 1/2 + E –1/2 ] f(x) . 1 2 (E 1/2 + E –1/2 ) g(x) – 1 4 (E 1/2 – E –1/2 ) f(x) (E 1/2 – E –1/2 )g(x) = 1 4 [ f (x + 1 2 ) + f(x – 1 2 )] [g(x + 1 2 ) + g(x – 1 2 )] – 1 4 [ f (x + 1 2 ) – f(x– 1 2 ] [g(x + 1 2 ) – g(x – 1 2 )] = 1 4 [f (x + 1 2 )g(x + 1 2 ) + f(x + 1 2 ) g(x – 1 2 ) + f (x – 1 2 ) g(x + 1 2 ) f(x – 1 2 )g(x – 1 2 ) – 1 4 [f(x + 1 2 ) g(x + 1 2 ) – f (x + 1 2 ) g(x – 1 2 ) – f (x – 1 2 ) g(x + 1 2 ) + f(x – 1 2 ) f(x– 1 2 )] = 1 2 [f (x + 1 2 ) g(x – 1 2 ) + f (x – 1 2 ) g(x + 1 2 ) ] Therefore right hand side is = 111 11 222 22 11 22 fx gx fx gx gx gx   +−+−+      −+   = () () () () 1/2 1/2 11 1 22 11 22 22 fx fx fx fx EE gx gx gx gx −   +−    +  += =µ      +−     (b) We know E = e hD and ∇ = 1 – E –1 , therefore ∇ 2 = (1 – e –hD ) 2 . = () () () 2 234 1 1 2! 3! 4! hD hD hD hD     −− + − + −      = () () ()   −+−+    2 234 2! 3! 4! hD hD hD hD CALCULUS OF FINITE DIFFERENCES 121 = h 2 D 2 ()     −− +      2 2 1 26 hD hD = h 2 D 2 () ()       +− + − − +        2 22 1 2 26 26 hD hD hD hD = h 2 D 2 ()   −++ −     2 11 1 43 hD hD = h 2 D 2  −+ −   22 7 1 12 hD h D = h 2 D 2 – h 3 D 3 + 7 12 h 4 D 4 – (c) ∇−∆ = (1 – E –1 ) – (E – 1) = 1E E −    – (E – 1) = (E – 1) (E –1 – 1) = – (E – 1) (1 – E –1 ) = – ∇∆ (d) L.H.S. = 2 1 2 x y   δ  +      = () 2 1/2 1/2 1 2 x EE y −  −  +    = 1 2 1 2 EE −   +−  +       y x = 1 2 (E + E –1 )y x . R.H.S. = ( ) 22 1 +δ µ y x = ()() {} 1/2 22 1/2 1/2 1/2 1/2 1 1. 4 x EE EE y −−  +− +   = () 1/2 2 1 1 4 EE −   −   +       y x = 1/2 22 2 4 EE −  ++    y x = 1 2 EE −  +    y x Hence, L.H.S. = R.H.S. (e) x y µδ = µ (E 1/2 – E –1/2 )y x = µ 22 hh xx yy +−  −    = µ 2 h x y +     – µ 2 h x y −     = 1 2 (E 1/2 + E –1/2 ) 2 h x y +     – 1 2 (E 1/2 + E –1/2 ) 2 h x y −     = 1 2 (y x+h + y x ) – 1 2 (y x + y x–h ) = 1 2 (y x+h – y x ) + 1 2 (y x – y x–h ) = 1 2 () x y ∆ + 1 2 () x y ∇ = 1 2 () ∆+∇ y x Hence, µδ = 1 2 () ∆+∇ . 122 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 22. Evaluate: ∆ n [sin (ax + b)] Sol. We know f∆ (x) = f(x + h) – f (x) therefore ∆ sin (ax + b) = sin [a (x + h)+b] – sin)(ax + b) = 2 sin 2 ah cos 2 h ax b   ++     = 2 sin 2 ah sin 2 ah ax b +π  ++   Therefore, ∆ 2 sin (ax + b)= ∆ 2sin sin 22 ah ah ax b +π   ++     = (2 sin 2 ah ) (2 sin 2 ah ) sin [ax + b + 2 ah +π + 2 ah +π ] = 2 2sin 2 ah    sin 2 2 ah ax b π +  ++     On continuing in the same manner, we get ∆ 3 sin (ax + b) = 3 2sin 2 ah    sin 3( ) 2 ah ax b π +  ++     ∆ n sin (ax + b) = 2sin 2 n ah    sin () 2 nah ax b π +  ++     Example 23. Show that: u 0 – u 1 + u 2 – = 1 2 u 0 – 1 4 ∆ u 0 + 1 8 ∆ 2 u 0 – Sol. On taking left hand side = u 0 – u 1 + u 2 – u 3 + = u 0 – Eu 0 + E 2 u 0 – E 3 u 0 + = (1 – E + E 2 – E 3 + )u 0 = () 1 1 E   −−   u 0 = 1 1 E   +  u 0 = 1 11   ++∆  u 0 = 1 2   +∆  u 0 = 1 2 1 1 2 − ∆  +   u 0 = 1 2  ∆∆ ∆ −+−+   23 1 24 8 u 0 = 1 2 u 0 – 1 4 ∆ u 0 – 1 8 ∆ 2 u 0 – 1 16 ∆ 3 u 0 + (R.H.S.) CALCULUS OF FINITE DIFFERENCES 123 Example 24. Prove that: (1) () () δ   fxgx = fµ (x) () gx δ + () gx µ () fx δ (2) () () () () () ()  µδ−µδ δ=    −+   fx gx fx fx gx 11 gx gx g x 22 Interval of differencing is unity. Sol. (1) R.H.S. = fµ (x) () gx δ + () gx µ () fx δ = 1/2 1/2 2 EE − + f(x). (E 1/2 – E –1/2 ) g(x) + 1/2 1/2 2 EE − + g(x). (E 1/2 – E –1/2 ) f(x) = 1 2 [{f(x + 1 2 ) + f(x – 1 2 )} {g (x + 1 2 ) – g (x – 1 2 )} + {g (x + 1 2 ) + g (x – 1 2 )} {f (x + 1 2 ) – f (x – 1 2 )}] = 1 2 [{f(x + 1 2 ) g (x + 1 2 )– f (x+ 1 2 ) g (x – 1 2 ) + f(x – 1 2 ) g (x + 1 2 ) – f (x – 1 2 ) g (x – 1 2 )} + {f(x + 1 2 ) g (x + 1 2 ) + f(x + 1 2 ) g (x – 1 2 ) – f (x – 1 2 ) g (x + 1 2 ) – f (x – 1 2 ) g (x – 1 2 )}] = 1 4 f(x + 1 2 ) g(x + 1 2 ) – f (x – 1 2 ) g (x – 1 2 ) = E 1/2 f (x) g(x) – E –1/2 f(x) g(x) = (E 1/2 – E –1/2 ) f (x) g(x) = fδ (x) g(x). (2) R.H.S. = () () () () 11 22 gx f x f x gx gx gx µδ−µδ  −+   Now first we solve the numerator of right hand side. = 1/2 1/2 2 EE − + g(x) (E 1/2 – E –1/2 ) f(x) – 1/2 1/2 2 EE − + f(x) (E 1/2 – E –1/2 ) g(x) = 1 2 [{g(x + 1 2 ) + g(x – 1 2 )} {f(x + 1 2 ) – f(x – 1 2 )} – {f(x + 1 2 ) + f(x – 1 2 )} {g(x + 1 2 ) – g(x – 1 2 )}] = 111 11 11 1 1 [( )( ) ( )( ) ( )( ) ( ) ( )] 222 22 22 2 2 f x gx f x gx fx gx f x gx++++−−−+−− − – 111 11 11 1 1 [( )( ) ( )( ) ( )( ) ( ) ( )] 222 22 22 2 2 fx gx fx gx fx gx f x gx++−+−+−+−− − = f (x + 1 2 ) g(x – 1 2 ) – f(x – 1 2 ) g(x + 1 2 ) 124 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Therefore right hand side as = 11 11 ()()()() 22 22 11 ()() 22 fx gx fx gx gx g x +−−−+ −+ = 1 () 2 1 () 2 fx gx + + – 1 () 2 1 () 2 fx gx − − = E 1/2 () () fx gx    – E –1/2 () () fx gx    = (E 1/2 – E –1/2 ) () () fx gx    = () () fx gx  δ   . Example 25. Evaluate: (a) x 2 (x 1)! ∆ + ; differencing 1. (b) ∆(log) ax ebx . Sol. (a) Let f(x)= 2 x , g(x) = (x + 1)!, therefore 1 () 2 2 2 xxx fx + ∆= −= and g∆ (x) = (x + 1 + 1)! – (x + 1)! = (x + 1) (x + 1)! () () fx gx  ∆   = ∆− ∆ + () () () () ( )() gx fx fx gx gx hgx = (1)!.22.(1)(1)! ( 1 1)!( 1)! xx xxx xx +−++ ++ + (Because h = 1) = 2 ( 1) (!(1 1) 2 (2)!(1)! (2)! x x xx x xx x +−− =− ++ + . (b) Again let, f(x)= e ax , g(x) = log bx, therefore ∆f(x)= () (1) ax h ax ax ah eeee + −= − ∆g(x)= +− = +log ( ) log log(1 ) h bx h bx x We know that ()()fxgx∆ = ( ) () () ()fx h gx gx fx+∆ + ∆ Therefore (log) ax ebx ∆ = () +  ++ −   ) log 1 (log ) ( 1) ax h ax ah h ebxee x ∆(e ax log bx)= e ax [e ah log 1 h x  +   + (e ah –1) log bx]. Example 26. Prove that, hD =– log (1 – ∇ ) = sin h –1 (µδ). Sol. Because, E –1 =1 – ∇ therefore, hD = log E = – log (E –1 ) = – log (1 – ∇ ) CALCULUS OF FINITE DIFFERENCES 125 Also, µ = 1/2 1/2 1 () 2 EE − + δ = E 1/2 – E –1/2 Therefore, µδ = −− −= − = 1 11 ()( )sin() 22 hD hD EE e e hhD hD = sin h –1 (µδ). Example 27. Show that ∆ log f(x) =  ∆ +   f(x) log 1 f(x) Sol. L.H.S. log ( )fx∆ = log ( ) log ( )fx h fx+− = () () log log () () fx h Efx fx fx  + =   = (1 ) ( ) log () fx fx  +∆   = () () log () fx fx fx  +∇   = () log 1 () fx fx  ∆ +   Example 28. Evaluate (1)  ∆   n 1 x (2) n ∆ (ab cx ). Sol. (1) We have, 1 n x  ∆   = ∆n –1 ∆ 1 x    . Now, ∆ 1 x    = 1 1x + – 1 x = () () 1 1 xx xx −+ + = () () 1 1xx − + ∆ 2 1 x    = ∆∆ 1 x    = ∆ () () 1 1xx  −   +   = (–1) ∆ () 1 1xx    +   = (–1) ()()() 11 12 1xx xx   −  ++ +   = (–1) () ()() 2 12 xx xx x −+ ++ = ()() ()() −− ++ 12 12xx x ∆ 3 1 x    = ()()() ()()() 123 123xx x x −−− +++ . 116 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 14. Evaluate the following: I. ∆ 2 (cos 2x) II. ∆ 2 (3e x ) III. ∆ tan –1 x IV. ∆(x + cos x) the interval of differencing. () ∆+∇ . 122 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 22. Evaluate: ∆ n [sin (ax + b)] Sol. We know f∆ (x) = f(x + h) – f (x) therefore ∆ sin (ax + b) = sin [a (x + h)+b] – sin)(ax. nh − ++ Example 21. Prove that: (a) µµ −δδ  µ=   −+ 1 f(x) g(x) f(x) g(x) f(x) 4 11 g(x) g(x )g(x ) 22 Here, interval of differencing being unity. 120 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (b)