A textbook of Computer Based Numerical and Statiscal Techniques part 35 pptx

326 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES [] 2(10) 7(0) 32(4) 12(7) 32(9) 7(12) 32(15) 12(14) 32(8) 7(3) 45 = +++++ + ++ = 708 Hence the required area of the cross-section of the river = 708 sq. m. Ans. (ii) By Simpson’s one-third rule, [] 80 08 1357 246 0 ()4( )2( ) 3 h ydx yy yyyy yyy = ++++++++ ∫ () 10 [ 0 3 4(4 9 15 8) 2(7 12 14)] 3 =++++++++ = 710 Hence the required area of the cross-section of the river = 710 sq. m. Ans. Example 10. Evaluate 1 0 1 dx x+ ∫ by dividing the interval of integration into 8 equal parts. Hence find log e 2 approximately. Sol. Since the interval of integration is divided into an even number of subintervals, we shall use Simpson’s one-third rule. Here, 1 () 1 yfx x == + 0 1 (0) 1, 10 yf== = + 1 118 , 1 89 1 8 yf  ===   + 2 24 , 85 yf  ==   3 38 , 811 yf  ==   4 42 , 83 yf  ==   5 58 , 813 yf  ==   6 64 , 87 yf  ==   7 78 , 815 yf  ==   and 8 1 (1) . 2 yf== Hence, 012345678 1234567 0 1 8888888 84828481 1 9 5113137152 x y yyyyyyyyy By Simpson’s one-third rule [] 1 08 1357 246 0 ()4( )2( ) 13 dx h yy yyyy yyy x = + + +++ + ++ + ∫ = 1 1 8888 424 14 2 24 2 9 11 13 15 5 3 7     ++ +++ + ++         (Hence, h = 1/8) = 0.69315453. Ans. NUMERICAL DIFFERENTIATION AND INTEGRATION 327 Since [] 1 1 0 0 log,(1 ) log 2 1 dx x x + == + ∫ log e 2 = 0.69315452. Ans. 6.10 EULER-MACLAURIN’S FORMULA This formula is based on the expansion of operators. Suppose () (), Fx fx∆= then an operator 1− ∆ , called inverse operator, is defined as 1 () () Fx f x − =∆ Again we have 0 () ( ) Fx f x∆= ⇒ 10 0 () () () Fx Fx f x−= () () 21 1 () Fx F x f x −= 11 ()()() nn n Fx Fx fx −− −= Adding all these, we get 1 0 0 () () () n ni i Fx Fx f x − = −= ∑ (1) where x 0, x 1, x n , are the (n+1) equidistant values of x with interval h. Now 11 () ()(1)() Fx f x E f x −− =∆ = − () 1 1() hD efx − =− 1 22 33 1 1 ( ) 2! 3! hD hD hD f x −    =++ + + −      1 22 33 ( ) 2! 3! hD hD hD f x −  =+ + +   1 22 1 ( ) 1 ( ) 2! 3! hD h D hD f x − −    =+++      () 2 22 22 1 1(1)(2) 1 2! 3! 2! 2! 3! hD h D hD h D Dfx h −   −−  =−+++ +++     22 44 1 1 1 () 2! 12 720 hD h D h D Dfx h −  = −+−+   3 11 () () () () 212 720 hh fxdx fx f x f x h ′′′′ =−+− ∫ (2) 328 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Between limits x = x 0 and x = x n from equation (2), we have {}{} 0 000 11 () () () () () () () 212 n x nnn x h Fx Fx fxdx fx fx f x f x h ′′ −= − − + − ∫ 3 0 { ( ) ( )} 720 n h fx fx ′′′ ′′′ −−+ (3) From eqs. (1) and (3), we have {}{} 1 00 0 1 11 () () () () () () 212 n n x inn i h fx fxdx fx fx f x f x h − = ′′ =−−+− ∑ ∫ {} 3 0 ( ) ( ) 720 n h fx fx ′′′ ′′′ −−+ But 1 00 () () ( ) − == =− ∑∑ nn iin ii fx fx fx and 0 n xxnh =+ . Then the above relation reduces to () {} {} 0 0 00 0 11 () ( ) ( ) ( ) () 212 n xnh in x i h fxdx fx fx f x f x nh f x h + = ′′ =−+− +− ∑ ∫ {} 3 00 ( ) ( ) 720 h fxnh fx ′′′ ′′′ −+−+ (4) ⇒ {} − =+++++ ∫ 0 012 1 () ()2()2() 2( ) () 2 n x nn x h fxdx fx fx fx fx fx () {} {} ′ ′ ′′ ′′ −−+ −+ 24 00 () () () 12 720 nn hh fx fx fx fx ⇒ 0 24 01 2 1 0 0 (2 2)()() 212720 n x nn n n x hhh ydx y y y y y y y y y − ′ ′ ′′′ ′′′ =+++++− −+ −+ ∫ which is known as Euler’s Maclaurin’s summation formula. Example 11. Evaluate 1 0 1 dx x+ ∫ to five places of decimal, using Euler-Maclaurin’ formula. Sol. Let 1 1 y x = + Here, we have x 0 = 0, n = 10 and h = 0.1 Then we want to evaluate 1 0 x x ydx ∫ where 2 1 (1 ) y x ′ = + and 4 6 (1 ) y x ′′′ =− + Using Euler-Maclaurin formula, we get ()() 24 1 012 0 0 0 22 ( ) 1 2 12 720 nn n dx h h h yyy y yy yy x ′ ′ ′′′ ′′′ =++++− −+ −+ + ∫ NUMERICAL DIFFERENTIATION AND INTEGRATION 329 0.112222222221 2 1 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 2  = ++++++++++   24 222 44 (0.1) 1 1 (0.1) 6 6 720 121 21  −−++−+   = 0.693773 − 0.000625+0.000001 = 0.693149. Ans. Example 12. Using Euler-Maclaurin’s formula, obtain the value of log e 2 from 1 0 1 dx x+ ∫ . Sol. Let 1 () 1 yfx x == + 0 1 , 0, 10, 0.1, 1. 1 n yxnhx x ===== + ⇒ 23 12 , (1 ) (1 ) yy xx ′′′ =− = ++ and 4 6 (1 ) y x ′′′ =− + Then by Euler-Maclaurins’s formula, we get 1 0 11222222 2 2 1 0 1 0.1 1 0.2 1 0.3 1 0.4 1 0.5 1 0.6 1 1.0 ydx  = ++++++++  +++++++ +  ∫ () 24 22 4 4 (0.1) 1 1 (0.1) 6 6 12 720 (1 0) (1 1) (1 0) 11   − −+−+   +++ +    = 0.693773 − 0.000625 + 0.0000010 = 0.631149. Ans. Example 13. Find the sum of the series using Euler-Maclaurin formula. 222 2 111 1 . 51 53 55 99 ++++ Sol. Here, we have 0 2 1 ,51,24,2 yxnh x ==== Then 1 35 224 , yy xx − ′′′ ==− and so on. Using Euler-Maclaurin’s formula, we get () ′ ′ ′′′ ′′′ =+++++− −+ −+ ∫ 24 99 012 2324 240 240 2 51 1 (22 2 ) ( ) 212720 hhh dx y y y y y y y y y x 222 22 33 55 1 2 2 2 1 4 2 2 16 24 24 12 720 51 53 55 97 99 99 51 99 51  = +++++ −−+ + −+ +   which gives 99 222 22 2 33 55 51 122 21 1 211 811 315 51 53 55 97 99 51 99 51 55 dx x  +++++= + − − − +   ∫ ⇒ 99 222 2 2 22 33 55 51 111 1 1 11211 811 2 315 51 53 55 99 51 99 51 99 51 55 dx x    ++++= +++ −− −+       ∫ 330 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ⇒ 99 222 2 22 33 55 51 111 111 111111 411 22 3 15 51 53 55 99 51 99 51 99 51 99 x  ++++=−+++−− −+     = 0.00475 + 0.00024 + 0.000002 + = 0.00499. Ans. Example 14. Use Euler-Maclaurin’s formula to prove that 2 (1)(21) 6 n i nn n x ++ = ∑ · Sol. By Euler-Maclaurin’s formula, () () 0 24 6 02 1 0 0 0 ( 2 2 2 ) ( ) 2 12 720 30240 n x vv innnnn x hhhh ydx y y y y y y y y y y y − ′ ′ ′′′ ′′′ =+++++− −+ −− −+ ∫ ⇒ 012 1 11 22 nn yyy y y − +++ + + ()() () 0 35 00 1 12 720 30240 n x vv nn n n x hh h ydx yy yy yy h ′ ′ ′′′ ′′′ = + −− −+ −− ∫ (1) Here, 2 () , () 2 yx x y x x ′ == and 1h = ∴ From (1), Sum () () 22 1 11 122 212 n xdx n n =+++− ∫ 3 1 2 1 2 1 22 0 2 yy n n == F H G I K J , ()() () 32 111 (1)(21) 111 326 6 nn n nnn ++ =−+++−= . (Proved) PROBLEM SET 6.2 1. Use Trapezoidal rule to evaluate 1 3 0 xdx ∫ consisting five sub-intervals. [Ans. 0.26] 2. Calculate an approximate value of integral /2 0 sin xdx π ∫ , by using Trapezoidal rule. [Ans. 0.99795] 3. Evaluate the integral 4 0 x edx ∫ by Simpson’s one-third rule. [Ans. 53.87] 4. Using Simpson’s 3 8 rule, evalute 1 0 1 1 dx x+ ∫ with 1 6 h = ·[Ans. 0.69319] 5. Use Boole’s rule to compute /2 0 sin xdx π ∫ .[Ans. 1.18062] 6. Using Weddle’s rule to evaluate 5 0 45 dx x + ∫ .[Ans. 0.4023] 7. Evaluate the integral /2 0 cos d π θθ ∫ by dividing the interval into six parts. [Ans. 1.1873] NUMERICAL DIFFERENTIATION AND INTEGRATION 331 8. Evaluate using Trapezoidal rule (i) 0 sin ttdt π ∫ (ii) 2 2 52 tdt t − + ∫ [Ans. (i) 3.14, (ii) –0.747] 9. Use Simpson’s rule dividing the range into ten equal parts to show that 2 1 2 0 log(1 ) 1 x dx x + + ∫ = 0.173. 10. A rocket is launched from the ground. Its acceleration is registered during the first 80 seconds and is given in the table below. Using simpson’s one-third rule, find the velocity of the rocket at t = 80 seconds. () () 2 sec 0 10 20 30 40 50 60 70 80 cm sec 30 31.63 33.34 35.47 37.75 40.33 43.25 46.69 50.67 t f [Ans. 30.87 m/sec.] 11. Find by Weddle’s rule the value of the integral I = 1.6 0.4 sin x dx hx ∫ by taking 12 sub-intervals. [Ans. 1.0101996] 12. Evaluate 1.4 0.2 (sin log ) x e xxedx −+ ∫ approximately using Weddle’s rule correct to four decimals. [Ans. 4.051] 13. Evaluate 1/2 2 0 1 dx x− ∫ using Weddle’s rule. [Ans. 0.52359895] 14. Using 3 8 th Simpson’s rule. Evaluate 6 4 0 1 dx x+ ∫ .[Ans. 1.019286497] 15. Evaluate 2 1 2 0 2 1 x x + + ∫ dx, using Weddle’s rule correct to four places of decimals. [Ans. 1.7854] 16. Evaluate 1 0 sin cos xxdx+ ∫ correct to two decimal places using seven ordinates. [Ans. 1.14] 17. Evaluate /2 0 sin xdx π ∫ using the Euler-Maclaurin’s formula. [Ans. 1.000003] 18. Prove that {} 2 3 1 (1) 2 n nn x + = ∑ applying Euler-Maclaurin’s formula. 19. Find the sum of the fourth powers of the first n natural numbers by means of the Euler- Maclaurin’s formula. 243 52330 nnn n  ++−   Ans. 20. Using Euler-Maclaurin’s formula, sum the following series. (i) ++++ 11 11 400 402 498 505 (ii) ++++ 11111 100 101 102 103 104 [Ans. (i) 0.11382114 (ii) 0.0490291] GGG +0)26-4 % Numerical Solution of Ordinary Differential Equation 7.1 INTRODUCTION In the fields of Engineering and Science, we come across physical and natural phenomena which, when represented by mathematical models, happen to be differential equations. For example, simple harmonic motion, equation of motion, deflection of a beam etc., are represented by differ- ential equations. Hence, the solution of differential equation is a necessity in such studies. There are number of differential equations which we studied in Calculus to get closed form solutions. But, all differential equations do not possess closed form of finite form solutions. Even if they possess closed form solutions, we do not know the method of getting it. In such situations, depending upon the need of the hour, we go in for numerical solutions of differential equations. In researches, especially after the advent of computer, the numerical solutions of the differential equations have become easy for manipulations. Hence, we present below some of the methods of numerical solutions of the ordinary differential equations. No doubt, such numerical solutions are approxi- mate solutions. But, in many cases approximate solutions to the required accuracy are quite sufficient. 7.2 TAYLOR’S METHOD Consider the differential equation () , dy fxy dx = (1) with the initial condition y(x 0 ) = y 0 If y(x) is the exact solution of (1) then y(x) can be expanded into a Taylor’s series about the point x = x 0 as () 3 2 0 0 000 0 0 () ( ) ( ) 2! 3! xx xx yx y x y y y y − − ′ ′′ ′′′ =+−+++ (2) where dashed denote diferentiation w.r.t. ‘x’ Differentiating (1) successively w.r.t x, we get ffdyf f yfff xydxxyxy  ∂∂ ∂ ∂ ∂∂ ′′ =+ =+ = +  ∂∂ ∂∂∂∂  (3) 332 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATION 333 ∴ () ff yy f f xxyxy  ∂∂ ∂∂∂ ′′′ ′′ ==+ +  ∂∂∂∂∂  2 222 2 2 22 fff f f f f fff f x y xy xy y xy  ∂∂∂ ∂ ∂ ∂ ∂ =+ + + + +  ∂ ∂ ∂∂ ∂∂ ∂ ∂∂  (4) and so on. Putting x = x 0 and y = y 0 in the expressions for , , yy y ′ ′′ ′′′ and substituting them in equation (2), we can obtain the solution of (1). Note: Taylor’s series method has advantages that it is derived in any order and values of y(x) are easily obtained. However, the method suffers from time consumed in computing higher derivatives. Example 1. Solve the differential equation dy xy dx =+ with (0) 1,y = x ∈ [0,1] by Taylor series expansion to obtain y for x = 0.1. Sol. Here, x 0 = 0, y 0 = 1 (),yxy ′ =+ 0 01 1 y ′ =+= (1 ),yy ′′ ′ =+ 0 11 2 y ′′ =+= (0 ),yy ′′′ ′′ =+ 0 02 2 y ′′′ =+= Using Taylor series expansions about x 0 = 0 is given by () () 2 3 00 0 0 0 0 ( ) ( 0) 2! 3! x x yx y x y y y − − ′′′′′′ =+− + + + at x = 0.1 () () =+++++ 23 4 20.1 20.1 2(0.1) (0.1) 1 0.1 2! 3! 4! y = 1 + 0.1 + 0.1 + 0.000333 + 0.0000083 = 1.11033. Ans. Example 2. Using Taylor series for y(x), find y(0.1) correct to four decimal places if y(x) satisfies 2 (),(0)1 yx y y ′ =+− = where x(0) = 0. Sol. Here, x 0 = 0, y 0 = 1 2 , yxy ′ =− 0 01 1 y ′ =−=− 12 ,yyy ′′ ′ =− 0 121(1)3 y ′′ =−××− = 2 22 yyyy ′′′ ′′ ′ =− − 0 8 y ′′′ =− 26,yyyyy ′′′′ ′′′ ′ ′′ =− − 0 34 y ′′′′ = 2 28 6, yyyyyy ′′′′′ ′′′′ ′ ′′′ ′′ =− − − 0 186 y ′′′′′ =− 334 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The Taylor sereis expansion about x 0 = 0 is given by () () () () () () 2345 000000 0000 2! 3! 4! 5! xx x x yx y x yy y y y y −− − − ′ ′′ ′′′ ′′′′ ′′′′′ =+− + + + + + at x = 0.1, y(0.1) = 1 + 0.1(–1) ++++−+ 01 2 3 01 3 01 4 34 01 5 186 23 4 5 . ! . ! –8 . ! . ! af af af af af af af af = 0.91379 = 0.9138. Ans. Example 3. Using Taylor’s series, find the solution of the differential equation ,xy x y ′ =− y(2) = 2 at x = 2.1 correct to five decimal places. Sol. Here, x 0 = 2, y 0 = 2. Also, 1, y y x ′ =− 0 0 y ′ = 2 , yy y x x ′ ′′ =− + 0 1 0 42 2 y ′′ =− + = 23 22 , yyy y x xx ′′ ′ ′′′ =− + − 0 3 4 y ′′′ =− 234 366 , yyyy y x xxx ′′′ ′′ ′ ′′′′ =− + − + 0 3 4 y ′′′′ = Using Taylor series expansion, we obtain () () () () 23 4 00 0 000 0 0 0 ( ) 2! 3! 4! xx xx xx yx y x x y y y y −− − ′ ′′ ′′′ ′′′′ =+− + + + + at 2.1,x= () () () ()() () 23 4 2.1 2 2.1 2 2.1 2 133 2.1 2 2.1 2 0 2! 2 3! 4 4! 2 y −− −   =+ − + ×+ − +     = 2.00238. (correct to five decimal places) Ans. Example 4. Using Taylor’s series Expansion tabulate the solution 4x = to 4.4x = in steps of 0.1 of differential equation. 5xy′ + y′′ − 2 = 0 with y(4) = 1 Sol. Differentiating successively the differential equation, we obtain 5520xy y yy ′′ ′ ′ ++ = 2 510220 xy yyyy ′′′ ′′ ′′ ′ +++= 515260xy y yy y y ′′′′ ′′′ ′′′ ′ ′′ +++ = 2 5202860 xy y yy y y y ′′′′′ ′′′′ ′′′′ ′ ′′′ ′′ ++++= The values of various derivatives at 0 4, 1 xy== are 0 0.05, y ′ = 0 0.0175, y ′′ =− 0 .01025, y ′′′ = 0 .00845, y ′′′′ =− 0 .008998125. y ′′′′′ = NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATION 335 Then by Taylor series, we obtain () () () () () () 23 4 1 0.05 4 . 00875 4 .0017083 4 .0003521 4 yx x x x x =+ −− −+ −+− − () 5 .00007498 4 x +−+ Tabulating from 4x = to 4.4x = we obtain (4.3) 1.014256y = and (4.4) 1.018701y = . Ans. Example 5. Solve the equation 21,yxy ′ =+ given that 0,y = at 0,x = by the use of Taylor series, taking 0.2h = and going as far as 4.x= Sol. The first few derivatives and their values at 0 0, x = 0 0, y = are 21,yxy ′ =+ 0 1 y ′ = () 2,yxyy ′′ ′ =+ 0 0 y ′′ = () 2,yxyyy ′′′ ′′ ′ ′ =++ 0 4 y ′′′ = () 2,yxyyyy ′′′′ ′′′ ′′ ′′ ′′ =+++ 0 0 y ′′′′ = () 2,y xy yyyy ′′′′′ ′′′′ ′′′ ′′′ ′′′ ′′′ =++++ 0 32 y ′′′′′ = Now by Taylor’s series, we have () () () () () () −− − − ′ ′′ ′′′ ′′′′ ′′′′′ =+− + + + + + 23 4 5 00 0 0 000 0 0 0 0 2! 3! 4! 5! xx xx xx xx yxyxxyyy y y Substituting the values of 0 y and its derivatives, we obtain () ()()()() () () () () () () 345 2 .2 .2 .2 .20.21.204032 3! 4! 5! y =+++++ = .2 + .00533 + .00002133 = .20535466 = .21 Now, with 11 .2, .21, xy== we compute () .4 .y So, we have ()( ) ′ =+= 1 2 .2 .21 1 1.084 y () 1 2{.2 1.084 .21} .8536 y ′′ =+= () 1 2{.2 .8536 2 1.084} 4.67744 y ′′′ =+×= () 1 2{.2 4.67744 3 .8536} 6.992576 y ′′′′ =+×= 1 2{.2(6.992576) 4 4.67744} 40.21655 y ′′′′′ =+×= Substituting the values of 1 y and its derivativies in Taylor series expansion, we obtain () () () () 23 10 10 1101 1 1 2! 3! xx xx yx y x x y y y −− ′ ′′ ′′′ =+ − + + + . numerical solutions of differential equations. In researches, especially after the advent of computer, the numerical solutions of the differential equations have become easy for manipulations. Hence,. % Numerical Solution of Ordinary Differential Equation 7.1 INTRODUCTION In the fields of Engineering and Science, we come across physical and natural phenomena which, when represented by mathematical. Use Trapezoidal rule to evaluate 1 3 0 xdx ∫ consisting five sub-intervals. [Ans. 0.26] 2. Calculate an approximate value of integral /2 0 sin xdx π ∫ , by using Trapezoidal rule. [Ans. 0.99795] 3.