M.KUMARASAMY COLLEGE OF ENGINEERING, KARUR NUMERICAL METHODS.

UNIT I TWO MARKS

1. What is the order of convergence of Newton – Raphson method if the multiplicity of the root is one. Order of convergence of Newton – Raphson method is 2. 2. Newton – Raphson method is also known as the method of …… Answer: Iteration (Newton iterative method). 3. Derive Newton algorithm for finding the p th root of a number N. Solution: If x = N 1/p, then x p – N = 0 is the equation to be solved. Let f (x) = x p – N = 0 , f 1 (x) = p x p-1 By Newton – Raphson rule, if x r is the r

th itrate f(xr) X r+1 = x r - ------ f1(xr) xr

p - N = xr - -------- Pxp-1

(p-1)xr

p + N = ----------------- Pxp-1

4. When should we not use Newton – Raphson method?

If x1 is the exact root and x0 is its approximate value of

F(x0) the equation f(x) = 0. We know x1 = xo - -------. If f ’(x0) is small F ’(x0) F(x0) the error -------- will be large and the computation of the root by F ’(x0) this, method will be a slow process or may even by impossible. Hence the method should not be used in cases where the graph of the function when it crosses the X axis is nearly horizontal. 5. What is the rate of convergence in Newton – Raphson method? The rate of convergence in Newton – Raphson method is of order 2.

6. What is the criterion for the convergence (convergence condition) in Newton – Raphson method? | f(x) f ”(x) | < [ f ’(x) ] . 7. Write the iterative formula of Newton – Raphson method. f( x r ) X r+1 = x r - --------- f ’( x r ) 8. What are the merits of Newton’s method of iterative? Newton’s method is successfully used to improve the result obtained by other methods. It is applicable to the solution of equations involving algebraical functions as well as transcendental functions. 9. Say true or false: Newton’s method is useful when the graph of the function when it crosses the x – axis is nearly vertical. (True). 10. Say true or false: Newton’s method is useful in cases where the graph of the function when it crosses the x – axis is nearly horizontal. ( False ) . 11. Choose correct answer Newton’s method is convergent. (a) Linearly (b) Quadratically (c) cubically (d) Biquadratically Answer: (b).

12. What is condition for applying the fixed point iteration (successive approximation method) method to find the real root of the equation x=f(x) (or) If g(x) is continuous in [a, b], then under what conditions the iterative method x=g(x) has a unique solution in [a, b]? Let x=r be a root of x=g(x).Let I be an interval combining the point x=r. If |g’(x)| < 1 for all x in I, the sequence of approximation x1, x2, …. X n will converge to the root r, provided that the initial approximation x 0 is chosen in I. 13. What is the order of convergence for fixed point iteration? Soln: The convergence is linear and is of order 1.

14. Say true or false: ’iteration method’ is self correcting method Soln: True, iteration method is a self correcting method, since the round off error is smaller. 15. In case of fixed point iteration method, the convergence is a)Linear b)Quadratic c)Very slow d)h2 soln: linear. 16. What is the condition for the convergence of the iteration method for solving x=ΦΦΦΦ(x)? Soln: Φ’(x) < 1 in the range. 17. True or False : ” Iteration method will always converge”. Soln: False. 18.Explain the term”pivot elements”. Soln: In an augmented matrix a11 a12 ..... a1n b1 a21 a22 ..... a2n b2 …. …. …. …. … an1 an2 ..... ann bn The elements a11,a22,…ann which have been assumed to be non-zero are called pivot elements. 19. State the principle used in Gauss-Jordon method. Soln: Coefficient matrix is transformed into diagonal matrix. 20.For solving a linear system ,compare Gaussian elimination method and Gauss-Jordon method. Gaussian elimination method Gauss-Jordon method 1 2 3

Coefficient matrix is transformed into upper triangle matrix. Direct method We obtain the solution by back substitution method

Coefficient matrix is transformed into diagonal matrix. Direct method No need of back substitution method

21. Give two indirect method to solve a system of linear equation. Soln: (i) Gauss Jocobi method (ii) Gauss-seidal method. 22. Compare Gauss – Jacobi and Gauss – Seidel methods. Gauss - Jacobi method Gauss- Seidel method 1 2 3

Convergence rate is slow Indirect method. Condition for convergence is the coefficient matrix is diagonally dominant.

The rate of convergence of Gauss – Seidel is roughly twice that of Gauss – Jacobi. Indirect method. Condition for convergence is the coefficient matrix is diagonally Dominant.

23. Is the iterative method, a self – correcting method always? In general, Iteration is a self correcting method, since the round off error is smaller. 24. Distinguish between direct and iterative(indirect) method of solving simultaneous equations. Direct method Indirect method 1 2

We get exact solution. Simple, take less time.

Approximate solution. Take consuming time.

25. What do you mean by ‘diagonally dominant’? A matrix is diagonally dominant if the numerical value of the leading diagonal element in each row is greater than or equal to the sum of the numerical values of the other value of the other element.

M.KUMARASAMY COLLEGE OF ENGINEERING, KARUR NUMERICAL METHODS. Unit I University Questions

1. (a) Using Gauss elimination method solve the equation X + 2 y + z = 3; 2 x + 3 y +3 z = 10; 3 x – y – 2 z = 13 (b) Solve the following equations by Gauss-seidel method 28 x + 4 y – z = 32; X + 3y + 10 z = 24; 2 x + 17 y + 4 z = 35. 2. (a) Find the root between ( 2 , 3) of x 3 –2 x - 5=0 by regular false method. (b) Compute real root of f( x , y ) = x 2 + y 2 – 4 = 0, g ( x ,y)= y+ex – 1 =0 correct to three decimal places using Newton Raphson method .

3. (a) Find the positive root of 3 x-√ 1+ sinx = 0 by iteration method (b) Determine the real root of x e x – 3 = 0 correct to five decimal places, using the method of false position. 4. (a) Using Gauss Jacobin iteration method solve the equation X + 17 y - 2z = 48 30 x – 2 y + 3z = 75 2 x + 2 y + 18 z = 30. (b) Solve the following equations by Gauss Jordan method 10 x + y + z = 12 2 x + 10 y + z = 13 X + y + 5 z = 7. 5. (a) Solve the following system by gauss Seidal method x + y + 54 z =110; 27 x + 6y – z = 85; 6x + 15y + 2z = 72. 2 1 1 (b) Find the inverse matrix of 3 2 3 by Jordan method. 1 4 9 6. (a) Find the solution of equation 4 x 2 + 2 x y + y 2 = 30 and 2 x 2 + 3 x y + y 2 = 3 correct to three Places of decimals, using Newtons Raphson method, given that x0 = - 3 and y0 = 2 . (b) Using iterative method, find the root of the equation cosx – x e x =0. 7. (a)Solve for positive root of the equation x-cos x=0, by Newton’s method. (b).Solve by Gauss elimination method, the system of equations 3 x – y + 2 z = 13 , X + 2 y + z = 3, 2 x + 3 y + 3 z = 10. 8. (a) Solve by Gauss seidel method, X + y + 54z = 110, 27 x + 6 y – z = 85, 6 x + 15 y + 3 z = 72. 2 2 3 (b).Using Gauss Jordan method,find inverse of the matrix 2 1 1 1 3 5 9. (a) Using Gauss Jacobin method, solve X + y + 5 z = 16, 2 x + 3 y + z = 4, 4 x + y – z = 4. (b).Using Gauss elimination method solves the following equations. 4 x + 4 y – 3 z = 4; 10 x + 8 y – 6 z = 5; 20 x – 4 y + 22 z = 7. 10.(a) Solve the equation x 6 – 5 x 2 + 136 = 0 and y 4 – 3 x 4 y + 80 = 0 assuming that ( 2 , 3 ) is an approximate solution. Obtain the solution correct to 2 decimals. (b). Solve X + 3 y + 3 z = 16, X + 4 y + 34 z = 18, X + 3 y + 4 z = 19, by Gauss Jordan method.

M.KUMARASAMY COLLEGE OF ENGINEERING THALAVAPALAYAM,KARUR.

UNIT-2 TWO MARKS 1. State Lagrange’s interpolation formula. Let f ( x ) be a function which takes the values y 0, y 1,y 2,…….y n corresponding to x 0,x 1,x 2,……….x n. Then Lagrange’s interpolation formula is y = f ( x ) = ( x – x 1 ) ( x – x 2 )………. ( x – x n ) ------------------------------------------------ y0 ( x 0 – x 1 ) ( x 0 – x 2 )……… ( x 0 – x n ) ( x-x0) (x-x2)………. (x-xn) + ----------------------------- y1 (x1-x0) (x1-x1)…… (x1-xn) (x-x0) (x-x1)………. (x-xn-1) +……. + ----------------------------- yn (xn-x0) (xn-x1)…… (xn-xn-1) 2. What is Lagrange’s formula to find y, if three sets of values (x0,y0), (x1,y1), (x2,y2) are given? Y=f(x)=(x-x1) (x-x2) (x-x0) (x-x2) -------------- y0 + -------------- y1 (x0-x1) (x0-x2) (x1-x0) (x1-x2) (x-x0) (x-x1) + -------------- y2 (x2-x1) (x2-x1) 3. What is the assumption we make when Lagrange’s formula is used Lagrange’s interpolation formula can be used whether the values of x, the independent variable are equally spaced or not whether the difference of y become smaller or not. 4. What advantage has Lagrange’s formula over Newton? The forward and backward interpolation formula of Newton can be used only when the values of the independent variable x are equally spaced can also be used when the differences of the dependent variable y become smaller ultimately. But Lagrange’s interpolation formula can be used whether the values of x, the independent variables are equally spaced or not and whether the differences of y become smaller or not.

5. What is the in practicing Lagrange’s interpolation formula? Though Lagrange’s formula is simple and easy to remember, its applications is not speedy. It requires close attention to sign and there is always a change of committing some error due to a number of positive and negative signs in the numerator and denominator. 6. What is inverse interpolation? Suppose we are given a table of values of x and y. Direct interpolation is the process of finding the values of y corresponding to a value of x, not present in the table. Inverse interpolation is the process of finding the values of x corresponding toa value y, not present in the table. 7. Give the inverse Lagrange’s interpolation formula. (y – y 1) ( y – y 2 )……….( y – y n ) x= x 0 ( y 0 – y 1) ( y 0 – y 2 )……( y 0 – y n ) ( y – y 0 ) ( y – y 2 )…… ( y – y n ) + x 1 ( y 1 – y 0 ) ( y 1 – y 2 )…. ( y 1 – y n ) +………………………………………. ( y – y 0 ) ( y – y 2 )………( y – y n – 1 ) + x n . ( y n – y 0) ( y n – y 2 )……( y n – y n – 1 )

8.Derive NEWTON ‘S backward difference formula by using operator method Pn(x)=Pn(xn+νh)=E

νPn(xn) =(1-∇)-νyn since E=(1-∇)

-1 ν(ν+1) ∇2+ν(ν+1)(ν+2) ∇3 +……..]yn =[1+ν∇+ 2! 3! ν(ν+1) ν(ν+1)(ν+2) = yn+ν∇ yn+ ∇2 yn + ∇3 yn+……… 2! 3! Where ν =(x-xn)/n 9. Derive Newton’s forward difference formula by using operator method. Pn(x)=Pn(x0+uh)=E

uPn(x0)= Euy0 =(1+∆)

uyn

u(u- 1 ) u(u-1)(u-2) = y0+u∆y0+ ∆2 y0 + ∆3 y0+……… 2! 3! Where u=(x-x0)/n 10. State Gregory-Newton forward difference interpolation formula u(u-1) u(u-1)(u-2) Pn(x)= y0+u∆y0+ ∆2 yn + ∆3 yn+……… 2! 3! Where u=(x-x0)/n 11. State Gregory-Newton backward difference interpolation formula. ν(ν+1) ν(ν+1)(ν+2) Pn(x)= yn+ν∇ yn+ ∇2 yn + ∇3 yn+……… 2! 3! Where ν =(x-xn)/n 12. Using Newton’s backward difference formula ,writethe formula for the first and second order derivatives at the end value x=xn upto the fourth order difference terms. ν(ν+1) ν(ν+1)(ν+2) y(x)= yn+ν∇ yn+ ∇2 yn + ∇3 yn+……… 2! 3! Where ν =(x-xn)/n Here x = x n ⇒ν = 0 Y( x ) = y n Y ( x + h ) – y (x) Y ′( x ) = yn′=lim

h→0 h y′′(x)=y n′′ 13. When Newton’s backward interpolation formula is used? The formula is used mainly to interpolate the values of y near the end of a set of tabular values and also for extrapolating the values of y a short distance ahead (to the right) of y0. 14. Newton’s forward interpolation formula used only for ----- intervals. Equidistant or equal intervals.

15.Say true or false: Newton’s interpolation formula are not suited to estimate the value of a function near the middle of a table. Soln: true 16. Say true or false: Newton’s forward and backward interpolation formula are applicable for interpolation near the beginning and end respectively of tabulated values. Soln: true 17. Given f(0)=-2,f(1)=2and f(2)=8.Find the root of the Newton’s interpolating polynomial equation f(x)=0.

x Y=f(x) ∆f(x) ∆2f(x) 0 1 2

-2 2 8

4 6

2

There are only three values are given. Hence the polynomial of degree 2. u(u- 1 ) u(u-1)(u-2) y(x)= y0+u∆y0+ ∆2 y0 + ∆3 y0+……… 2! 3! Where u=(x-x0)/n Here X 0 = 0 , h = 1 – 0 = 1 , u = x x x(x-1) Y(x)=-2+ (4)+ (2) 1! 2! =-2+4x+x(x-1) =-2+4x+x2-x =x2+3x-2 ∴The roots of the equation f( x ) = 0 i.e x 2+ 3x - 2 = 0 are - 3±√9+8 -3±√17 X = = 2 2 18. Obtain the interpolation polynomial for the given data by using Newton’s forward difference formula.

X: 0 2 4 6 Y: -3 5 21 45

x

y

∆y

∆2y

∆3y

0 2 4 6

-3 5 21 45

8 16 24

8 8

0

X – x 0 x - 0 x u = = =

h 2 2 u (u-1) y(x)=y0+n∆y0+ ∆2y0 2 (x/2) (x/2) ((x/2)-1) Y(x)=-3+ (8)+ (8) 1! 2! (x2/4)- (x/2) =-3+4x+ (8) 2! =-3+4x+x2-2x =x2+2x-3 19. State the order of convergence of cubic spline. Order of convergence=4 20. What are the natural (or) free conditions in cubic spline. S′′( X 0 ) = M 0 = 0, S ′ ′( X n ) = M n = 0 are free conditions. 21. State the properties of cubic spline. ( i ) S(Xi)=yi,i=0,1,2,……….n ( ii ) S(x),S′(x),S′′(x) are continuous in [a,b]. ( iii ) S(x) is a cubic polynomial.

M.KUMARASAMY COLLEGE OF ENGINEERING, KARUR NUMERICAL METHODS. Unit II University Questions

1. From the table below, Find the number of students who secured marks between 40 and 45 Marks; 30 - 40 40 - 50 50 - 60 60 - 70 70 – 80 No.of students; 35 48 70 40 22 2. Find the cubic spline given the table x: 0 2 4 6 y: 1 9 41 41 {Answer f(x) = 1+x 3 [0,2],25 – 36 x +18x2–2x 3 [2,4], -103+60x–6x 2 [4,6]}. 3. Find the third degree polynomial f(x) satisfying following data x: 1 3 5 7 y: 24 120 336 720 and find f(4). [Answer f(x) = x3 +6x 2 +11x+6 and f(4) = 1.25] 4. Using Newton’s forward interpolation formula find the polynomial f(x) satisfying the following data. Hence evaluate y at x = 5 x: 4 6 8 10 y: 1 3 8 10. [ f(x) = (1/8) ( - x 3 +19 x2 – 106 x + 192 ; and y(5) = 1.5] 5. Given the data x: - 4 - 1 0 2 5 y: 1245 33 5 9 1335. Find f(x). [Answer: f(x) = 3x 4 + x 3 – 14x +5] 6. The following values of x and y are given: x: 1 2 3 4 y: 1 2 5 11 Find the cubic spline for (1, 2) and evaluate y(1.5). 7. Find the polynomial f(x) by using Lagrange’s formula and hence find f(3) for x : 0 1 2 5 f(x): 2 3 12 147 . 8. Obtain the root of f(x) = 0 by Lagrange Inverse Interpolation given that f(30) = - 30, f(34) = -13, f(38) = 3, f(42) = 18. 9. Find the polynomial f(x) of degree two by using Lagrange’s formula x: 1 2 - 4 f(x): 3 - 5 4

10. Find the missing term in the table: X: 2 4 6 8 10 F(x): 5.6 8.6 13.9 - 35.6. 11. Find the cubic polynomial which takes the following set of values (1,2),(0,1), (2,1), and (3,10). 12. The following are data from the stream table: Temp c : 140 150 160 170 180 Pressure kgf/cm2 : 3.685 4.854 6.302 8.076 10.225 Using Newton’s formula formula, find the pressure of the stream for a temp of 1240. 13. Use Newton’s backward difference formula to construct an interpolating polynomial of degree 3 for the data :f(- 0.75) = - 0.07181250, f( - 0.5) = - 0.024750, f( - 0.25) = 0.33493750 and f(0) = 1.10100. Hence find f( - 1/3). 14. Construct Newton’s forward interpolation polynomial for the following data: X : 4 6 8 10 Y: 1 3 8 16 Use it to find the value of y for x = 5. 15. From the following table, find the value of tan 45015’. xo : 45 46 47 48 49 50 tan x0 : 1 1.03553 1.07237 1.11061 1.15037 1.19175.

M. KUMARASAMY COLEGE OF ENGINEERING, Thalavapalayam , karur.

UNIT –III , NUMERICAL DIFFERENTIATION AND INTEGRATION.

1.Define numerical differentiation. Numerical differentiation is the process of calculating the derivatives of the given function by means of table given values of that function. The process of computing the values of dy/dx, d2y/dx2, d3y/dx3……….. is called numerical differentiation. 2. Define quadrature. The process of evaluating of definite integral from a set of tabulated values of a function of single variable is known as quadrature 3. What is another name of Simpson’s 1/3 rd rule. It is called as closed formula. 4. When Simpson’s 3/8 th rule can be applied. This method can be applied only if the number of sub intervals is a multiple of three. 5. Using Newton’s backward difference formula, write the formula for the first and second order derivatives at the end values x= xn up to the fourth order difference term. (d y / d x ) x = x n = 1 / h [∇ y n + (½)(∇2y n ) + 1/3((∇

3 y n ) +…………. ]

(d 2 y / d 2 x ) x = x n =1 / h 2 [( ∇ 2 y n ) + ( ∇

3 y n) + 11/12 (∇4yn) +…… ]

(d 3 y / d 3 x ) x = x n = 1 / h

3 [∇ 3 y n + 3 / 2 ( ∇ 4 y n ) + …………. ]

1 6. Evaluate ∫∫∫∫ (1/x) dx by trapezoidal rule dividing the range in to 4 equal parts. 0.5 Here h = [ 1 - ( 1 / 2 ) ] / 4 = 1 / 8 ,

Y = 1 / x . x 0 = 4 / 8 = 1 / 2 . X n =1, X 1 = 5 / 8, x 2 = 6 / 8, x 3 = 7 / 8, x 4 = 8 / 8. Y 1 = 1 / 4, y 2 = 1 / 5, y 3 = 8 / 6. y 4 = 8 / 7, y 5 = 8 / 8 A= sum of the first and last ordinates = 8/4 +8/8 = 3.

B = sum of the remaining ordinates. = 8/5 + 8/6 +8/7 = 856/210 1 ∫ (1/x) dx = h/2(A+2B) 0 .5 =1/16( 3+( 856 *2/210)). = 0.6971.

7. What is the order of error in trapezoidal formula? The order of error in trapezoidal formula is h2.

8. What is the order of error in Simpson’s formula?

The order of error in Simpson’s formula is h4.

9. What are the errors in Trapezoidal and Simpson’s rule of numerical integration.? The error in Trapezoidal rule is E < [( b – a ) /12 ] (h 2 M ), in the interval ( a, b),H= (b-a)/M, M= max { y0 “, y1”……….}. The error in Simpson’s rule E < [( b – a ) / 180 ] (h 4 M ), in the interval (a,b),M= max { y0

IV, y1 IV……….}.

x n 10. In order to evaluate ∫∫∫∫ ydx by Simpson’s (1/3) rd rule, Trapezoidal X0

rules well as Simpson’s (3/8) th rule. What is the restriction on the

number of intervals? ` Let n= interval. Rule: Simpson’s 1/3 rule → The number of ordinates is even Rule: Simpson’s 3/8rule → The number of ordinates is multiple of 3. Rule: Trapezoidal rule → Any number of ordinates. ππππ 11. Using Trapezoidal rule evaluate ∫∫∫∫ sinx dx by dividing the range into 6 equal parts. 0 Here y(x) = sinx, h= π/6. X : 0 π / 6 2π / 6 3π /6 4π / 6 5π / 6 π Y : 0 .5 .866 1 0.866 0.5 0 π ∫ sinx dx = h/2[ ( y0 + yn)+ 2( remaining terms)] 0

= ( π / 12 ){ 7.464} =0.622π. 12. State three point Gaussian quadrature . Three point Gaussian Quadrature formula is 1 ∫ f ( x ) d x = 5 / 9 [ f ( - √ 3 / √ 5 ) + f ( √3 / √5 )]+8 / 9 f(0). -1 This formula is exact for polynomial up to 5 degree. 4 13.Using Simpson’s rule find ∫∫∫∫e x dx, given e 0 =1, e 1 = 2.72, e 2 = 7.39, e 3 = 20.09 and e 4 = 54.6 0 Let y = e x and h =1, By Simpson’s rule, 4 ∫ e x dx = h / 3[ ( y0 + y4) + 2y 2 + 4( y 1 + y 3 )] 0 = 1/3 [ ( 1+54.6 )+2( 7.39 ) +4( 2.72+20.09 )] =53.8733 1 14. If I = ∫∫∫∫ e-x2 dx then I1= 0.731 and I2= .7430 with h= 0.5 and h= 0.25. 0 Find I usingRomberg’s method. I= I2+ ( I2-I1) /3 =.7430 + ( .7430- .7314) /3 = 0.7469 15. A curve passes through ( 2,8) (3,27) (4,64) and (5, 125) . Find the area of the curve between X axis and the lines x= 2 and x=5 by trapezoidal rule. 5 ∫ y d x = 1/2[(8+125) +2(27+64)] 2 = 157.5 sq units. 16. Find d y / d x at x=2 from the following data: X: 2 3 4 Y : 26 58 112. ∆y0 = 32, ∆

2y0 =22. ( dy / dx ) x = 2 = 1/h[∆y0- ∆

2y0/2 +∆3y0/3+…..]

=32-( 1 / 2 ) (22) =21.

1 17. Evaluate I= ∫∫∫∫ dt/(1+t), by Gaussian 2 point formula. 0 F(t) = 1/(1+t) ( given range is not in the exact form). 1 1 I= ∫ dt/(1+t) = 1/2 ∫ dt/(1+t)…………..(1) 0 - 1 F ( t) = 1/(1+t) F( -1 / √3 ) =2.366sss F( 1 / √3 ) = 0.634. 1 By Gaussian 2 point formula, ∫f ( t )dt = 3. -1 1

(i) → ∫ d t / ( 1 + t ) = 1.5 -1 1

18. Evaluate ∫∫∫∫ x2/( 1+x4) dx , by using Gaussian 3 pont formula. -1 F(x) = x2/( 1+x4) The Gaussian 3 point formula is 1 ∫ f(x) dx = 5/9[f(-√3/√5)+ f(√3/√5)]+8/9 f(0) -1

f(-√3/√5) = 15/34 f(√3/√5) = 15/34 1 ∫ f(x) dx = 150/ 306 = 0.4902. -1 19. Show that the divided difference operator ∆∆∆∆ is linear. If f(x) and g(x) are 2 function and α ,β are the constants, then ∆[ α f ( x ) + β g ( x )] = [ α f( x 1 ) + β g ( x 1 )] - [ α f ( x 0 ) + β g ( x 0 )] = α[f(x1) - f(x0)] /(x1-x 0)+β [g(x1) - g(x0)]/(x1-x0). = α∆f(x) + β∆g(x) Thus ∆ is linear.

UNIVERSITY QUESTIONS: ( 16 MARKS) 1 1.Evaluate ∫ dx/ (1+x2), using Romberg’s method. Hence find an approximate

0 Value of π. 2.A river is 80 meters wide. The depth “d” in meters at a distance x meters from one bank is given by the following table. Calculate the area of cross- section of the river using Simpson’s rule. X: 1 10 20 30 40 50 60 70 80 Y : 0 4 7 9 12 15 14 8 3.

3. The table given below reveals the velocity v of a body during the time “t” specified. Find its acceleration at time 1.1 T: 1 1.1 1.2 1.3 1.4 V: 43.1 47.7 52.1 56.4 60.8

4. Evaluate the integrals of f(x) = 0.2 +25x-200x2 +675x3 -900x4 +400x5 between the limits of x=0 to 0.8 using two point Gaussian quadrature. 5. From the following table X : 1 1.2 1.4 1.6 1.8 2 2.2 Y: 2.7183 3.3201 4.0552 4.9530 6.0496 7.3891 9.0250 Obtain dy / dx and d2y/dx2 for x= 1.2. π

6. Dividing the range in to 10 equal parts ,find approximate value of ∫ sinx dx by (1) Trapezoidal rule. (ii) Simpson’s rule. 0 1

7. Find ∫ x dx by Gaussian formula. 0 1 1 8. Evaluate ∫ ∫ e( x+y) dxdy by Trapezoidal and Simpson’s rule. 0 0 9. Find the first and second derivatives at x=1.6 for the function represented by the following tabular data : X : 1 1.5 2 3 Y : 0 0.40547 0.69315 1.09861 1 10. Using 3 point Guassian formula , evaluate ∫dx /( 1+x2) 1/2 1 2 0

11. Evaluate ∫ ∫2xydxdy /( 1+x2 )(1+y2) , using Simpson’s method with step 0 1 length of h= k= 0.25. 1 2

12. Evaluate ∫ ∫dxdy /( 1+x+ y) , using Simpson’s method with step length of h=k= 0.5. 0 0

13. Find the value of sec ( 31 ). From the following table. θ ( deg) : 31• 32• 33 • 34 • tanθ :0.6008 0.6249 0.6494 0.6746 6 14. Using Simpson’s 1/3 rd rule ,Evaluate ∫ dx/(1+x2), by dividing the range in to 6 equal parts. 0 15. Find the maximum and minimum value of y . X : -2 -1 0 1 2 3 4 Y : 2 -0.25 0 -0.25 2 15.75 56 16. The following data gives the velocity of a particle for 20 sec at an interval of 5 sec, Find the initial acceleration using the data. Time( sec) : 0 5 10 15 20 Velocity(m/sec) : 0 3 14 69 228 1 17. Using Trapezoidal rule Evaluate ∫ dx( 1+x2) , using 8 intervals and also by -1 3 point Gaussian formula.. 18. Obtain the value of f ′(0.04) using an approximate formula for the given data : X : 0.01 0.02 0.03 0.04 0.05 0.06 F(x) : 0.1023 0.1047 0.1071 0.1096 0.1122 0.1148 2 2 19. Using Trapezoidal rule , Evaluate ∫ ∫ 1/(X +Y) dx dy. Taking 4 sub intervals. 1 1 20. Find first and second derivative at x=1.2 for the data. X : 1 1.1 1.2 1.3 1.4 1.5 1.6 Y : 7.989 8.403 8.781 9.12 9.451 9.750 10.031 5 2 21. Using three point formula Evaluate ∫(1/x)dx & ∫(x2+ 2x +1)dx/( (1+(x+1)4)). 5 1 0 22. Evalute ∫dx/(4x+5) , by Simpson’s rule and hence find the value of log e5 (n=10) 0 2 2 23. Evaluate ∫ ∫ f( x , y ) dx dy by Trapezoidal rule for the following data.

0 0 y x 0 0.5 1 1.5 2 0 2 3 4 5 5 1 3 4 6 9 11 2 4 6 8 11 14

NUMERICAL METHODS UNIT-IV

INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS

1. State the disadvantages of Taylor series method. In the differential equation d y / d x = f ( x , y ), the function f (x, y) may have a complicated algebraically structure. Then the evaluation of higher order derivative may become tedious. This is the demerit of this method. 2. Write down the fourth order Taylor Algorithm. Y n + 1 = Y n + h Y n

’ + h 2 / 2! Y ! ! n + h 3 / 3! Y n

!!! +… Here Y n

m denotes the r th derivative of y w.r. to x at the point ( x n ,y n ). 3. Write the merites and demerits of the Taylor method of solution. The method gives a straightforward adaptation of classic calculus to develop the solution as an infinite series. It is a powerful single step method if we are able to find the successive derivative easily. If f (x, y) involves some complicated algebraic structures.Then the calculation of higher derivatives becomes tedious and the method fails. This is the major drawback of this method. However the method will be very useful for finding the starting values for powerful methods like Runge-Kutta method, Milnes method etc. 4. Which is better Taylor’s method (or) R.K Method.(or) What are the advantages of Runge-Kutta method over Taylor method? R.K. methods do not require prior calculation of higher derivatives of y (x), as the Taylor method does. Since the differential equations using in applications are often complicated, the calculation of derivative may be difficult. Also R.K formulas involve the computation of f (x, y) at various, instead of derivatives and this function occurs in the given equation. 5. Write down Euler algorithm to the differential equation dy/dx=f(x, y). Y n + 1 = Y n + h f ( x n , y n ) when n=0,1,2… This is Euler algorithm. It can also be written as Y ( x + h) = Y ( x ) + h f (x, y). 6. State modified Euler algorithm to solve Y’=f ( x ,y ), Y ( x 0 ) = Y o at

X = x 0 + h. Y n + 1 = Y n + h f [ x n + (h / 2) , (h / 2) f (x n , y n)]

Y1 = Y 0 + h f [ x o + (h / 2) ,y 0 + (h / 2) f ( x 0 , y0)].

7. Write the Runge - Kutta algorithm of second order for solving Y ’= f ( x , y ), y ( x 0 ) = y0 Let h denotes the interval between equidistant values of x. If the initial values are ( x 0 , y 0 ) , the first increment in y is computed from the formulas. K 1 = h f ( x 0 , y 0 ) K 2 = h f [ x 0 + (h / 2) , y 0 + (k 1 / 2)] and ∆y = k 2. Then x 1 = x 0 + h , y 1 = y 0 + ∆y The increment is y in the second interval is computed in a similar manner using the same three formulas, using the values ( x 1, y 1 ) in the place ( x 0 , y 0 ) respectively. 8. State the third order R.K method algorithm to find the numerical solution of the First order differential equation. To solve the differential equation dy / dx= f ( x , y ) by the third order R.K method , we use the following algorithm K 1 = h f ( x , y ) K 2 = h f [ x + h / 2 , y + k 1 / 2 ] K 3 = h f [ x + h , y + 2 k 2 – k 1 ] and ∆y = 1 / 6 [ k 1 + 4 k 2 + k 3 ] . 9. Write down the Runge - Kutta formula for the fourth order to solve dy / dx = f ( x , y ) with y ( x 0 ) = y 0. Let h denotes the interval between equidistant values of x. If the initial values are (x0, y0), the first increment in y is computed from the formulas. K 1 = h f ( x 0 , y 0 ) K 2 = h f [ x 0 + h / 2 , y 0 + k 1 / 2 ] K 3 = h f [ x 0 + h / 2 , y 0 + k 2 / 2 ] K 4 = h f [ x 0 + h , y 0 + k 3 ] ∆ y =1 / 6 [ k 1 + 2 k 2 + 2 k 3 + k 4 ] x 1 = x 0 + h ,y 1 = y 0 + ∆ y . 10. State the special advantage of Runge-Kutta method over Taylor series method. Runge-Kutta methods do not require prior calculation of higher derivatives of y y (x), as the Taylor method does. Since the differential equations using in applications are often complicated, the calculation of derivatives may be difficult.

11. Where the Taylor series method of solving differential equation is powerful? If it is possible to find the successive derivatives in a very easy manner. Then Only Taylor series method in powerful. 12. Is Euler’s method formula, a particular case of second order Runge- Kutta method? Yes, Eulers modified formula is a particular case of second order Runge-Kutta method. 13. What are the distinguished properties of Runge-Kutta method?

(i) These methods do not require the higher order derivatives and requires only the function values at different points. (ii) To evaluate yn+1 we need only y n but not pervious of y’s .

(iii) The solution by these methods agrees with Taylor series solution up to the terms of hr where ‘r’ is the order of the Runge-Kutta method.

14. State the order of error in Runge-Kutta method of fourth order. Error in Runge-Kutta fourth order method is o (h5) where h is the interval of Differencing. 15.In the derivation of fourth order Runge-Kutta formula, why it is called fourth Order. The fourth order Runge-Kutta method agree with Taylor series solution up to the terms of h4. Hence it is called fourth order Runge-Kutta method. 16.What is the predictor-corrector method?(OR) Write Milnes predictor corrector formula Milne’s predictor formula is Yn+1,p=yn-3+4h/3[2y

1n-3-y

1n-1+2y

1n]

Milne’s corrector formula is Yn+1,c=yn-1+h/3[y

1n-1+4y

1n+y

1n+1].

17. Write the predictor-Error and corrector-Error in Milne’s method. The error of term in Milne’s predictor formula is 14/45 ∆4y10 The error of term in Milnes corrector formula is –h/90 ∆4y10 18. What is the condition to apply Adams –Bash forth method?(or) State Adams –Bash forth predictor and corrector formula. In the corrector formula, the value of yn+1 is obtained by using the

predictor Value of yn+1 Adams Bash forth predictor formula is given by Yn+1,p= yn+ h/24[55y

1n-59y

1n-1+37y

1n-2-9y

1n-3]

Adams Bash forth corrector formula is given by Yn+1,c= yn+ h/24[9y

1n+1+19y

1n-5y

1n-1+y

1n-2].

19. Given two examples for single-step and multi step method. Single step methods:

(i) Taylor series method. (ii) Euler method. (iii) Runge-Kutta method.

Multi step Method: (i) Milne’s method. (ii) Adams-Bash forth method.

20. Distinguish single step and multi step method Single step method: To find yn+1 the information at y n is enough Multi step method: To find yn+1, the past four values yn-3, yn-2, yn-1and y

‘ns are needed.

21. Define initial value problem The differential equation together with the initial condition is called an initial value problem. It is given by dy /dx=f(x,y), y(x0)=y0. 22. Write the predictor-Error and corrector-Error in Adams method Predictor error is (251/720) h5yiv(ξ) Corrector error is –(19/720) h5yiv(ξ). 23. State Taylor series algorithm for the first order differential equation dy/dx=f(x,y). To find the numerical solution of dy/dx=f (x,y) with the condition y(x0)=y0. We expand y(x) at a general point x n in a Taylor series , getting Yn+1=yn+h/1!yn

1+h2/2!yn1+……………….

Ynr denotes the rth derivatives of y w.r.to x at the point (x n, y n).

24. Write down the Euler algorithm to the differential equation

dy/dx=f(x,y) Yn+1=y n +h f (x n , y n) where n=0,1,2,3,….. This is Euler algorithm. It can also be written as Y (x + h)=y (x)+h f ( x , y).

25. Say True or False Modified Euler method is the Runge-Kutta method of second order. The statement is true.

M.KUMARASAMY COLLEGE OF ENGINEERING, KARUR NUMERICAL METHODS,UNIT V BOUNDARY VALUE PROBLEMS

1.Define a boundary value problem with an example. The differential equation together with the boundary conditions is called a boundary Value problem Example: xy” +y = 0, y(1) = 1, y(2) = 2

2. Write down the finite difference approximations for y’(x) at x = xi. Also write the error

yi ‘ = yi+1 – yi [forward difference] h yi’ = yi – yi-1 [backward difference] h yi’ = yi+1 – yi-1 [central difference]

2h 3. State the finite approximations for y’ and y” with error terms. Yi = y (xi) and xi+1 = xi +h, I = 0,1,2,………… Then yi’ = yi+1 – yi-1, Error = O(h2) [central difference] 2h yi” = yi-1 – 2yi + yi+1 , Error = O(h2) h2

4. Solve xy” + y = 0, y(1) = 1, y(2) = 2 with h = 0.5 Xiyi” + yi = 0 Xi(yi-1 –2yi + yi+1)/h2 +yi = 0 ⇒4xi(yi+1 –2yi +yi-1) + yi = 0, I=1 For I =1, 4xi(y2 – 2y1 + y0) +y1 = 0

⇒11y1 = 18 ⇒y1 = y(1.5) = 18/11 = 1.6364

5. Write the diagonal five – point formula to solve the laplace equation uxx +uyy = 0 Ui,j = 1[ui-1,j-1 + ui-1,j+1 + ui+1,j-1 + ui+1,j+1] 4

6. Write the standard five point formula to the Laplace equation ð2u + ð2u = 0 (OR) Ðx2 ðy2 Write the difference scheme for solving the Laplace equation Ui,j = 1[ui-1,j-1 + ui-1,j+1 + ui+1,j-1 + ui+1,j+1] 4 7. What is the purpose of Liebmann’s process?

The purpose of Liebmann’s process is to find the solution of the Laplace equation uxx + uyy = 0 by iteration

8. If u satisfies Laplace equation and u = 100 on the boundary of a square what will be the value of u at an interior grid point? Since u satisfies Laplace equation and u = 100 on the boundary of a square, ui,j = 1[100 +100 + 100 + 100] = 100 4

9. For the following mesh in solving ∇2u = 0, find one set of rough values of u at interior mesh points

1 2

U1 U2 U3 U4

4 5 By symmetry, u2 = u3. Assume u2 = 3 The rough values are u1 = (1/4)(1 +1 + 2u2) = 2 U2 = 3 U3 = (1/4)(5 +5+ 2u2) = 4 U4 = (1/4)(u1 + u4 + 2 + 4) = 3

10. Write the Laplace equations uxx + uyy = 0 in difference quotients. uxx + uyy = 0 ⇒ ui-1,j – 2ui,j + ui+1,j + ui,j-1 – 2ui,j + ui,j + h2 k2

11. Define a difference quotient A difference quotient is the quotient obtained by dividing the difference between two values of a function by the difference between two corresponding values of the independent variable

12. State Liebmann’s iteration process formula U(n+1) i,j = (1/4)[u(n+1)i-1,j +u(n) i+1,j+ u(n)i,j-1 + u(n+1)i,j+1].

13. State the conditions for the equation, Auxx + Buxy + Cuyy + Dux + Euy +fu = G where A, B, C, D, E, F, G are functions of x and y, to be (i) elliptic, (ii) parabolic, (iii) hyperbolic The given equation is said to be (i) elliptic at a point (x,y) in the plane if B2 – 4ac < 0(ii) parabolic if B2 – 4ac = 0 (iii) hyperbolic if B2 – 4ac>0

14. At which region, the equation yuxx + uyy = 0 is hyperbolic?

Here A = y, B = 0, C = 1 B2 – 4ac = 0 – 4y = -4y The equation is hyperbolic in the region (x, y) where B2 – 4ac > 0 (ie) –4y >0 ⇒y<0 ⇒ it is hyperbolic in the region y<0

15. What is the classification of fx – fyy = 0? Here A = 0, B = 0, C = -1. B2 – 4ac = 0 –4x0x-1 = 0. So the equation is parabolic

16. Give an example of a parabolic equation The one dimensional heat equation ∂ u/∂t = α2∂2u/∂x2, is parabolic where α2 = (1/a)

17. Write down the finite difference form of the equation ∇2u =f(x,y). [Poisson equation] (or) Write down the five point formula t6o solve the Poisson equation uxx + uyy = f(x,y) ui-1,j +ui,j-1 + ui,j+1 – 4ui,j = h2 f(ih,jh)

18. Derive the difference scheme for ∇2u =f (x,y). Consider a square mesh with interior differencing as h, Taking x = ih, y = jh, the difference equation reduces to

ui-1,j – 2ui,j + ui+1,j + ui,j-1 – 2ui,j + ui,j +1 = f(ih,jh) h2 h2

ie) ui-1,j +ui+1,j + ui,j+1 – 4ui,j = h2f(ih,jh) 19. State the five point formula to solve the Poisson equation uxx+uyy = 100, with h =

1. Ui-1,j +ui+1,j + ui,j-1 + ui,j+1 –4ui,j = h2f(ih,jh)

= 100 [since h = 1] 20. State the general form of Poisson’s equation in partial derivatives.

∂ 2u/∂x2 + ∂2u/∂y2 = f(x,y) 21. State Schmidt’s explicit formula for solving heat flow equation (or) Write down

Bender – Schmidt recurrence equation to solve the heat equation uxx = a ut. ui,j+1 = (1/2)[ui-1,j + ui+1,j]

22, Write an explicit formula formula to solve numerically the heat equation uxx – aut = 0

ui,j+1 = λui+1,j + (1-2λ)ui,j + λui-1,j where λ = k/(ah2) , his the space for the variable x and his the space in the time direction

23. Why the explicit formula to solve the heat equation is so called? The explicit formula is ui,j+1 = λui+1,j + (1-2λ)ui,j + λui-1,j where λ = k/(ah2) (i-1,j+1) (i,j+1) (I+1,j+!)

(i-1,j) (I,j) (I+1,j)

The solution value at any point (I,j+1) on the (j+1)th level is expressed in terms of the `solution values at the points (I-1,j) on the previous jth level, (not on the same level). Hence, this formula is called explicit formulas.

24. What is the value of k to solve ðu/ðt = (½ ) uxx by Bender – Schmidt method with h= 1 if h & k are the increments of x and t respectively? Given ðu/ðt = (1/2) uxx Ie) uxx = 2ut Here a = 2, h = 1 We know that,λ = k/(ah2) = ½ [in Bender Schmidt] Ie) k/(ah2) =1/2 Ie)K/2x1 = ½ Ie)K = 1

25. What is the classification of one dimensional heat flow equation? One dimensional heat flow equation is ∂ 2u/∂x2 = a ∂u/∂t Here A = 1, B = 0, C = 0. B2 – 4ac = 0 ⇒ It is parabolic

26. Write down the Crank – Nicholson formula to solve ut = uxx

Here a =1. Crank – Nicholson formula is(1/2) λui+1,j+1 + (1/2) λui-1,j+1 – (λ+1)ui,j+1 = -(1/2) λui+1,j – (1/2) λui-1,j + (λ-1)ui,j where λ = k/(ah2) ie) λ = k/(h2) [since a =1]

27. Write down the implicit formula to solve one dimensional heat flow equation uxx = (1/c2) ut Here a = (1/c2) Implicit formula (Crank – Nicholson formula) is (1/2) λui+1,j+1 + (1/2) λui-1,j+1 – (λ+1)ui,j+1 = -(1/2) λui+1,j – (1/2) λui-1,j + (λ-1)ui,j where λ = k/(ah2) ie) λ = k/(1/c2)h2 [Since a =(1/c2]] ⇒λ = c2k/h2

28. Why is Crank – Nicholson scheme called an implicit scheme? Crank – Nicholson formula is (1/2) λui+1,j+1 +(1/2) λui-1,j+1 – (λ+1)ui,j+1 = -(1/2) λui+1,j – (1/2) λui-1,j + (λ-1)ui,j where λ = k/(ah2) (i-1,j+1) (I,j+1) (I+1,j+1) (I+j) th level (i-1,j) (I,j) (I+1,j) (jth level) The solution value at any point (I,j+1) on the (j+1)th level is dependent on the solution values at the neighboring points on the same level and on three values on the jth level. Hence it is an implicit formula.

29. What type of equations can be solved by using Crank – Nicholson’s difference formula? It is used to solve parabolic equation of the form, uxx = aut [one dimensional heat equation]

30. Express the implicit form of Crank – Nicholson formula to solve uxx = aut Ui,j+1 = (1/4)[ui-1,j+1 + ui+1,j+1 + ui-1,j + ui+1,j] where λ = 1 ie)k = ah2

31. Write a note on the stability and of the solution of the difference equation corresponding to the hyperbolic equation Utt = a2Uxx

a. For λ = 1/a, the solution of the differential equation is stable and coincides with the solution of the differential equation

b. For λ>1/a, the solution is unstable. c. forλ<1/a, the solution is stable but not convergent

32. Write down one-dimensional wave equation and its boundary conditions. Utt =a2 uxx With boundary conditions U (0,t) =0 U (l,t) =0 And the initial conditions U(x,0) =f(x) Ut(x,0) =0. 33. State the explicit scheme formula for the solution of the wave equation , utt =a2 uxx. Ui,j+1 =2(1-λ2a2)ui,j+λ2a2(ui-1,j +ui+1,j)-ui,j-1, where λ=(k/h). 34.State the explicit formula for the one dimensional wave equation with 1-λ2α2 =0 where λ=(k/h) & α2=(T/m) Ui,j+1=ui-1,j +ui+1,j –ui,j-1 Where 1-λ2a2=0, λ =k/h, α2=(T/m)(Tension/Mass) This formula is known as simplex form of explicit formula. 35 .For What value of λ , the explicit method of solving the hyperbolic equation ∂ 2u/∂x2 = 1/c2 ∂u/∂t2 is stable, where λ =c∆t/∆x.

WKT, for λ=1/a, the equation Utt = a2Uxx ------------(1) is stable, where λ=k/h Given that Uxx = 1/c2 utt. where λ =c∆t/∆x. Comparing (1)&(2) a2=c2 K/h =c∆t/∆x. k/h= ck/h c=1 (2) Becomes uxx=utt for λ =1/a=1/c=1, the equation utt=a2uxx stable ⇒ for λ = 1, iit is stable.

36. Why the explicit formula to solve the wave equation is so called? The explicit formula is Ui,j+1 =2(1-λ2a2)ui,j+λ2a2(ui-1,j +ui+1,j)-ui,j-1, where λ=(k/h).

(I,j+1) (j+1)th level

(i-1,j) (I,j) (I+1,j) j th level (I,j-1) (j-1)th level

The solution value at any point (I,j+1) on the(j+1)th level is expressed in terms of solution values on the previous j and (j-1) levels (and not in terms of values on the same level). Hence this is an explicit difference formula.

37. In one dimensional heat equation,∂u/∂t = α2∂2u/∂x2 Where α= k/ρc, what do the values kρc stand for? K = thermal conductivity ρ = Density c = specific heat