Numerical-methods Question And Answer 6w6t32

www.Vidyarthiplus.in SELVAM COLLEGE OF TECHNOLOGY NAMAKKAL-3 DEPARTMENT OF MATHEMATICS SUB: NUMERICAL METHODS UNIT I 1) Write the Descartes rule of signs Sol: 1) An equation f (x ) = 0 cannot have more number of positive roots than there are changes of sign in the of the polynomial f (x ) . 2)An equation f (x ) = 0 cannot have more number of positive roots than there are changes of sign in the of the polynomial f (x ) . 2) What is the order of convergence of Newton Raphson method if the multiplicity of the root is one. Sol: Order of convergence of N.R method is 2. 3) Newton Raphson method is also known as the method of ……….. Sol: Iteration ( Newton’s iteration method) Derive newtons algorithm to derive p th root of a number N. 1 p

Sol: If x = N then x p − N = 0 is the equation to be solved. Let f ( x) = x p − N , f ′( x) = px p −1 By NR rule ,if xr is the r th iterate f ( xr ) x r +1 = x r − f ′( x r ) p

p

x − N ( p − 1) xr + N = xr − r p −1 = p −1 pxr pxr 4) When would we not use N-R method . Sol: If x1 is the exact root and x 0 is its approximate value of the equation f ( x0 ) f (x ) = 0.we know that x1 = x 0 f ′( x0 ) f ( x0 ) will be large and the computation of the root by If f ′( x 0 ) is small,the error f ′( x0 ) this,method will be a slow process or may even be 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 convergence in NR method? Sol: The rate of convergence in NR method is of order 2. 6) Write the iterative formula of NR method.

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Sol: x n +1 = x n -

f ( xn ) f ′( x n )

7) State the order of convergence and convergence condition for NR method? Sol: The order of convergence is 2 2 Condition of convergence is f ( x ) f ′′( x ) < f ′( x ) 8) If g(x) is continuous in [a , b] then under what condition the iterative method x = g(x) has a unique solution in [a , b]. Sol: Let x = r be a root of x = g(x) .Let I = [a , b] be the given interval combining the point x = r.if g ′( x) < 1 for all x in I, the sequence of approximation x0 , x1 ,......x n will converge to the root r,provided that the initial approximation x 0 is chosen in r. 9) Write a sufficient condition for Guass siedel method to converge .(or) State a sufficient condition for Guass Jacobi method to converge. Sol: The process of iteration by Guass siedel method will converge if in each equation of the system the absolute value of the largest coefficient is greater than the sum of the absolute values of the remaining coefficients. 10) Give two indirect method to solve a system of linear equations? Sol: (i) Guass Jacobi method (ii) Guass Siedel method 11) State True or False : “Guass siedel iteration converges only if the coefficient matrix is diagonally dominant” Sol: True. 12) Compare Guass Siedel and Guass elimination method? Sol: Guass Jacobi method Guass siedel method. The rate of convergence of Guass 1. Convergence method is slow Siedel method is roughly twice that of Guass Jacobi. 2.

Direct method

Indirect method

3.

Condition for convergence is the coefficient matrix diagonally dominant

Condition for convergence is the coefficient matrix diagonally dominant

13) Is the iteration method a self correcting method always? Sol: In general iteration is a self correcting method since the round off error is smaller. 14) State the principle used in Gauss Jordan method? Sol: Coefficient matrix is transformed into diagonal matrix.

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www.Vidyarthiplus.in 15) For solving a linear system, compare Gauss elimination method and Gauss Jordan method. Sol: Gauss elimination method Gauss Jordan method. 1. Coefficient matrix is Coefficient matrix is transformed transformed into upper triangular into diagonal matrix matrix

2.

Direct method

Direct method

3.

We obtain the solution by back substitution method

No need for substitution method

16) The numerical methods of solving linear equations are of two types : one is direct and the other is …………. Sol: iterative. 17) Define round off error? Sol: The round off error is the quantity R which must be added to the finite representation of a computed number in order to make it the true representation of that number. 18) Explain the term pivoting. Sol: In the elimination process if any one of the pivot elements a11 , a 22 ,........a nn vanishes are become very small compared to other elements in that column ,then we attempt to rearrange the remaining rows so as to obtain a non vanishing pivot or to avoid the multiplication by a large number.this strategy is called pivoting. The pivoting is of two types 1) Partial pivoting 2) Complete pivoting. 19) Why Guass siedel method is better than Jacobi’s iterative method? Sol: since the current values of the unknowns at each stage of iteration are used in proceeding to the next stage of iteration,the convergence in Guass siedel method will be more rapid than in Guass Jacobi method. 20) 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 vertical. Sol:True 21) In the case of fixed point iteration method ,the convergence is ……….. Sol: Linear. 22) If the eigen values of A are 1,3,4 then the dominant eigen value of A is …… Sol:4 23) If the eigen values of A are 1,3,-4 then the dominant eigen value of A is …… Sol:-4 24) If the eigen values of A are 1,3,-3 then the dominant eigen value of A is …… www.Vidyarthiplus.in

www.Vidyarthiplus.in Sol:No Dominant eigen value. 25) The power method will work satisfactory only if A has a …….. Sol:Dominant eigen value.

26) Say true or false: The convergence in the Gauss –Siedel method is thrice as fast as in jacobi’s method. Sol: False.The rate of convergence of Guass siedel methodb is roughly twicw that of the Jacobi’s method. 27) Distinguish between direct and iterative method of solving simultaneous equation. Sol: Direct method Iterative method. 1. We get exact solution Approximjate solution

2.

Simple take less time

Time consuming laborious

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www.Vidyarthiplus.in SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL-3 MA1251-NUMERICAL METHODS UNIT-IV INTERPOLATION AND NUMERICAL INTEGRATION AND DIFFERENTIATION Two Marks Q&A 1. State Lagrange’s interpolation formula. Let f ( x ) = 0 be a function which takes the values y 0 , y1 , y 2 , y 3 ,............, y n corresponding to

x = x0 , x1 , x 2 , x3 ,............, x n . Thus Lagrange’s interpolation formula is (x − x1 )(x − x2 )..............( x − x n ) (x − x0 )(x − x2 )..............( x − x n ) y = f ( x) = y0 + y1 + ............ ( x0 − x1 )( x0 − x 2 ).........( x0 − x n ) ( x1 − x0 )( x1 − x 2 ).........( x1 − x n ) (x − x0 )(x − x1 )..............( x − x n−1 ) yn . ........................ + ( x n − x0 )( xn − x1 ).........( xn − xn −1 ) 2. What is the Lagrange’s formula to find y, if three sets of values (x0 , y0 ), (x1 , y1 ) & (x2 , y 2 ) are given. The Lagrange’s interpolation formula is (x − x1 )(x − x2 ) y + (x − x0 )(x − x2 ) y + (x − x0 )(x − x1 ) y . y = f ( x) = 2 1 0 ( x 0 − x1 )( x 0 − x 2 ) ( x1 − x 0 )( x1 − x 2 ) ( x 2 − x0 )(x 2 − x1 ) 3. Using Lagrange’s interpolation, find the polynomial through (0,0),(1,1) and (2,2). The polynomial through the given points is given by (x − x1 )(x − x 2 ) (x − x0 )(x − x2 ) (x − x0 )(x − x1 ) y = f ( x) = y0 + y1 + y2 . ( x0 − x1 )( x0 − x 2 ) ( x1 − x0 )( x1 − x2 ) ( x 2 − x0 )( x 2 − x1 )

=

(x − 1)(x − 2 ) (0) + (x − 0)(x − 2 ) (1) + (x − 0 )(x − 1) (2) (0 − 1)(0 − 2 ) (1 − 0)(1 − 2) (2 − 0 )(2 − 1)

(

)

= −( x 2 − 2 x ) + x 2 − x = x 4. When do we apply Lagrange’s interpolation? Lagrange’s interpolation formula can be used when the values of “x” are equally spaced or not. It is mainly used when the values are unevenly spaced. 5. Give the inverse of Lagrange’s interpolation formula. The inverse Lagrange’s interpolation formula is

( y − y1 )( y − y 2 )..............( y − y n ) ( y − y 0 )( y − y 2 )..............( y − y n ) x0 + x1 + ............ ( ) ( ) ( y 0 − y1 ) y 0 − y 2 ......... y 0 − y n ( y1 − y0 )( y1 − y 2 ).........( y1 − y n ) ( y − y 0 )( y − y1 )..............( y − y n −1 ) xn . ........................ + ( y n − y 0 )( y n − y1 ).........( y n − y n −1 ) x=

6. What do you understand by inverse interpolation? The process of finding a value of “x” for the corresponding value of “y” is called inverse interpolation. The Lagrange’s inverse interpolation formula can be obtained by interchanging “x” and “y” in Lagrange’s interpolation formula. 7. Define divided difference. Let the function y = f (x ) take the values

f ( x0 ), f ( x1 ), f ( x 2 ),............... f ( x n ) corresponding to the values x0 , x1 , x 2 ,.............x n of the argument ‘x’ where x1 − x 0 , x 2 − x1 ,............x n − x n −1 need not necessarily be equal. The first divided difference of f (x ) for the arguments

x0 , x1 is

f ( x 0 , x1 ) =

similarly, f ( x , x ) = f ( x 2 ) − f ( x1 ) etc. 1 2 x 2 − x1 The second divided difference of f(x) for three arguments

x0 , x1 , x 2 is defined as

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f ( x1 ) − f ( x 0 ) . x1 − x0

www.Vidyarthiplus.in f ( x1 , x 2 ) − f ( x 0 , x1 ) f ( x 2 , x3 ) − f ( x1 , x 2 ) etc. f ( x 0 , x1 , x 2 ) = , f ( x1 , x 2 , x3 ) = x 2 − x0 x3 − x1

Sol: 8. Show that the divided difference operator ∆ is linear. f ( x1 ) ± g ( x1 )] − [ f ( x0 ) ± g ( x 0 ) ] f ( x1 ) − f ( x 0 ) g ( x1 ) − g ( x 0 ) [ = ∆ ( f ( x ) ) ± ∆ (g ( x) ) ∆[ f ( x ) ± g ( x ) ] = = ± x1 − x0 x1 − x0 x1 − x0 9. Evaluate

∆10 (1 − x )(1 − 2 x )(1 − 3x ).............(1 − 10 x ) ,by taking h =1.

Sol: ∆ (1 − x )(1 − 2 x )(1 − 3 x ).......... ...(1 − 10 x ) = (1.2.3.4.5.6.7.8.9.10)(10!) = (10!) 10. Obtain a divided difference table for the following data: X: 5 7 11 13 17 Y: 150 392 1452 2366 5202. Sol: x y ∆ f(x) ∆2 f ( x) ∆3 f ( x) 10

5

2

150

392

392 − 150 242 = = 121 7−5 2

11

1452

1452 − 392 1060 = = 245 11 − 7 4

13

2366

7

17

5202

2366 − 1452 914 = = 457 13 − 11 2 5202 − 2366 2836 = = 709 17 − 13 4

245 − 121 124 = = 20.67 11 − 5 6 457 − 245 212 = = 35.33 13 − 7 6 709 − 457 252 = = 42 17 − 11 6

35.33 − 20.67 14.66 = = 1.83 13 − 5 8 42 − 35.33 6.67 = = 0.667 17 − 7 10

11. Write the Newton’s divided difference interpolation formula for unequal intervals. f ( x) = f ( x0 ) + (x − x0 ) f ( x0 , x1 ) + ( x − x0 )(x − x1 ) f ( x0 , x1 , x 2 ) + ............

+ ( x − x0 )( x − x1 ).......( x − xn−1 ) f ( x0 , x1 ,.....xn ) . 12. Write the Newton’s forward difference interpolation formula. Let y = f (x ) be a function which takes the values y 0 , y1 , y 2 , y 3 ,............, y n corresponding to the values

x0 , x1 , x 2 , x3 ,............, x n where the values of x are equally spaced. Then the Newton’s forward difference interpolation formula is given by yn = y0 +

u u(u − 1) 2 u(u − 1)(u − 2) 3 ∆ y 0 + .............................. ∆y 0 + ∆ y0 + 1! 2! 3! Where u = x − x0 h

1 ,find f (a, b) & f ( a, b, c) by using divided differences. x2 1 Sol: Given, f ( x ) = 2 . x

13. If f ( x ) =

1 1 − (b + a) f (b) − f (a ) b 2 a 2 a2 − b2 = 2 2 f (a, b) = =− 2 2 . = b−a b−a a b (b − a ) a b

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www.Vidyarthiplus.in c+b b+a −( 2 2 )+ 2 2 2 2 2 2 f (b, c) − f ( a, b) a b = b c (b + a) − a b (b + c) f (a, b, c) = = c b 2 2 2 c−a c−a a b c (c − a )

=

(ab + bc + ca)(c − a) (ab + bc + ca) = a 2b 2 c 2 (c − a ) a 2b 2 c 2 th

14. What is the nature of n divided differences of a polynomial of n Sol: Let f (x ) = a polynomial of degree “n”.

th

degree?

∆n f (x) = nth divided difference = constant. 15. Find the second divided differences with arguments a,b,c if f ( x ) = Sol:Given, f ( x ) =

1 x

1 1 − f (b) − f ( a) b a − 1 f ( a, b ) = = = b−a b − a ab 1 1 − + 1 f (b, c) − f (a, b) bc ab . f (a, b, c) = = = c−a c−a abc

16. Obtain a divided difference table for the following data: X: -1 0 2 3 f(x): -8 3 1 12. x

y

-1

-8

0

3

2

1

3

12

∆ f(x)

∆2 f ( x)

3+8 = 11 0 +1 1− 3 = −1 2−0 12 − 1 = 11 3− 2

− 1 − 11 = −4 2 +1 11 + 1 =4 3−0

∆3 f ( x)

4+4 =2 3 +1

17. Find the polynomial which takes the following values: X: 0 1 2 Y: 1 2 1 Sol: The Lagrange’s interpolation formula is (x − x1 )(x − x2 ) y + (x − x0 )(x − x2 ) y + (x − x0 )(x − x1 ) y . y = f ( x) = 0 1 2 ( x 0 − x1 )( x 0 − x 2 ) ( x1 − x 0 )( x1 − x 2 ) ( x 2 − x0 )(x 2 − x1 ) (x − 1)(x − 2) (1) + (x − 0)(x − 2) (2) + (x − 0)(x − 1) (1). y = f ( x) = (0 − 1)(0 − 2) (1 − 0)(1 − 2) (2 − 0)(2 − 1) x 2 − 3 x + 2 2( x 2 − 2 x ) x 2 − x = + + = −2 x 2 + 4 x + 2 2 ( −1) 2 18. Obtain a divided difference table for the following data: X: 2 3 5 f(x): 0 14 102.

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1 . x

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y

2

0

3

14

5

102

∆ f(x)

∆2 f ( x)

14 − 0 = 14 3−2 102 − 14 = 44 5−3

44 − 14 = 10 5−2

19. Write the Newton’s backward difference interpolation formula. Let y = f (x ) be a function which takes the values y 0 , y1 , y 2 , y 3 ,............, y n corresponding to the values

x0 , x1 , x 2 , x3 ,............, x n where the values of x are equally spaced. Then the Newton’s backward difference interpolation formula is given by

v v (v + 1) 2 v (v + 1)(v + 2) 3 y n = y 0 + ∇y 0 + ∇ y0 + ∇ y 0 + .............................. 1! 2! 3! x − x0 Where v = . h 20. Find the polynomial for the following data by Newton’s backward difference formula. X: 0 1 2 3 f(x): -3 2 9 18. x

f(x) = y

0

-3

1

2

∇y

∇2 y

∇3 y

2-(-3)=5 7-5= 2 9-2 = 7 2

3

2

9 18-9 = 9( ∇y 0 )

3

9-7 =2( ∇ y 0 )

2-2 = 0( ∇ y 0 )

18( y 0 )

x0 = 3, h = 1, y 0 = 18, v =

x − x0 x − 3 = = x−3 h 1

The Newton’s backward interpolation formula is

v v (v + 1) 2 v (v + 1)(v + 2) 3 y n = y 0 + ∇y 0 + ∇ y0 + ∇ y 0 + .............................. 1! 2! 3! ( x − 3)( x − 3 + 1) = 18 + ( x − 3)(9) + ( 2) + ( 0) 2 = 18 + 9 x − 27 + x 2 − 5 x + 6 = x 2 + 4 x − 3 n ax + b ). 21. Evaluate ∆ (e

f ( x) = e ax +b we know that ∆f ( x ) = f ( x + h) − f ( x) ∴ ∆(e ax +b ) = e a ( x + h )+b − e ax +b = e ax +b (e ah − 1)

Sol: let

∴ ∆2 (e ax +b ) = ∆ (∆ (e ax +b )) = (e ah − 1)(e a ( x + h ) +b − e ax +b ) = e ax +b (e ah − 1)

2

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In general, ∴ ∆

(e ) = e (e ax + b

ax +b

ah

− 1) . n

22. What are the advantages of Lagrange’s formula over Newton’s formula? The forward and backward interpolation formulae of Newton can be used only when the values of the independent variable x are equally spaced and 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 variable are equally spaced or not and whether the difference of y become smaller or not. 23. Find the second degree polynomial fitting the following data: X: 1 2 4 f(x): 4 5 13. x Y ∆ f(x) ∆2 f ( x) 1

4

2

5

4

13

5−4 =1 2 −1 13 − 5 =4 4−2

4 −1 =1 4 −1

f ( x) = f ( x0 ) + ( x − x0 ) f ( x0 , x1 ) + ( x − x 0 )( x − x1 ) f ( x0 , x1 , x 2 ) + ........... = 4 + ( x − 1)(1) + ( x − 1)( x − 2)(1) = 4 + x − 1 + x 2 − 3x + 2 = x 2 − 2 x + 5 . 24. What are the disadvantages in practice in applying Lagrange’s interpolation formula? 1. It takes time. 2. It is laborious. 25. State the properties of divided differences. 1. Divided differences are symmetrical in all their arguments. 2. Divided differences operator is linear. ∆ f ( x) ± g ( x) = ∆ f ( x) ± ∆ g ( x)

(

3. ∆

)

(

)

(

)

(cf ( x)) = c∆( f ( x) ) . th

th

4. The n divided differences of a polynomial of the n degree are constant. 26. 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. 27. When Newton’s forward interpolation formula is used. The formula is used mainly to interpolate the values of ‘y’ near the beginnig of a set of tabular values. 28. When do we use Newton’s divided differences formula? This is used when the data are unequally spaced. 29. Newton’s forward interpolation formula used only foe equidistant (or) equal interval. 30. Say true or false: Newton’s interpolation formulae are not suited to estimate the value of a function near the middle of a table. Sol:TRUE 31. Say true or false: Newton’s forward and Newton’s backward interpolation formulae are applicable for interpolation near the beginning and the end respectively of tabulated values. Sol:TRUE. 32. Given f(0) = -2,f(1) = 2 and f(2) = 8 Find the root of Newton’s interpolating polynomial equation f(x) = 0. Sol: X f(x) ∆ f(x) ∆2 f(x) 0 -2 2-(-2)=4 6-4 =2 1 2 8-2 = 6 2 8

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y 0 = −2, h = 1, u =

x − x 0 x − (0) = =x h 1

Newton’s forward interpolation formula is

u u (u − 1) 2 u (u − 1)(u − 2) 3 ∆y 0 + ∆ y0 + ∆ y 0 + .............................. 1! 2! 3! ( x )( x − 1) = −2 + ( x )(4) + ( 2) 2! = −2 + 4 x + x 2 − x = x 2 + 3 x − 2. − 3 ± 9 + 8 − 3 ± 17 2 .) Now, f ( x ) = x + 3 x − 2 = 0 ⇒ x = = 2 2 y = y0 +

33.Forward difference operator. Let y = f (x ) be a function of

x and let y 0 , y1 , y 2 ,..... of the values of y. corresponding to x0 , x0 + h, x0 + 2h,.... of the values of x. Here,the independent variable (or argument), x proceeds at equally spaced intervals and h (constant),the difference between two consecutive values of x is called the interval of

differencing. Now the forward difference operator is defined as

∆y 0 = y1 − y 0 ∆y1 = y 2 − y1 ...................... ∆y n = y n+1 − y n These are called first differences. 34.Forward difference table.

x0

y0

x1 x2 x3 x4

y1 y2 y3 y4

35. Backward difference operator. The backward difference operator

∆y 0 ∆y1 The image part with relationship ID rId315 was not found in the file.

The image part with relationship ID rId316 was not found in the file.

∆2 y 0

∆2 y1 ∆2 y 2

∆3 y 0

∆3 y1

∇ is defined as

∇y n = y n − y n −1 For n = 0,1,2... ∇ y0 = y0 − y −1 ∇ y1 = y1 − y 0 ∇ y 2 = y 2 − y1

and so on.

∇ 2 y n = y n − 2 y n −1 + y n −2 ∇ 3 y n = ∇ 2 y n − ∇ 2 y n−1 = y n − 3 y n −1 + 3 y n −2 − y n −3

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∆4 y 0

www.Vidyarthiplus.in 36 Backward difference table.

x

The image part with relationship ID rId343 was not found in the file.

x−4 x−3 x−2 x −1 x0

y−4 y −3 y −2 y −1 y0

The image part with relationship ID rId344 was not found in the file.

∇y −3

∇y − 2 ∇y −1 ∇y 0

∇2 y

∇3 y

∇4 y

∇ 2 y −2 ∇ 2 y −1 ∇ 2 y0

∇ 3 y −1 ∇ 3 y0

∇ 4 y0

Newton’s backward interpolation formula is

v v(v + 1) 2 v (v + 1)(v + 2) 3 y n = y 0 + ∇y 0 + ∇ y0 + ∇ y 0 + .............................. 1! 2! 3! 37. ∆{ f ( x) ± g ( x)} = ∆{ f ( x)} ± ∆{g ( x)}

The image part with relationship ID rId394 was not found in the file.

(ie)The divided difference (of any order) of the sum (or) difference of two functions is equal to the sum(or)difference of the corresponding separate divided differences. Proof: If f ( x ), g ( x ) are two functions and x0 , x1 be two arguments,

∆{ f ( x) ± g ( x)} =

[ f ( x1 ) ± g ( x1 )] − [ f ( x0 ) ± g ( x0 )] (x1 − x0 )

( x0 , x1 )

[ f ( x1 ) − f ( x0 )] ± [g ( x1 ) − g ( x0 )] (x1 − x0 ) = ∆{ f ( x)} ± ∆{g ( x)} =

( x0 , x1 ) Similarly the result is true for any higher order difference. 38. ∆ cf ( x) = c∆ f ( x) (ie) the divided difference of the product of a constant and a function is equal to the product of the constant and the divided difference of the function. Proof:

{

}

{

}

 f ( x1 ) − f ( x0 )  cf ( x1 ) − cf ( x0 ) = c  = c∆{ f ( x)} − x1 − x0 x x 1 0   ∴ ∆{cf ( x)} = c∆{ f ( x)}

∆{cf ( x)} =

Similarly the result is true for any higher order difference. 39.The divided differences are symmetrical in all their arguments.

f ( x1 ) − f ( x0 ) f ( x0 ) − f ( x1 ) f ( x0 ) f ( x1 ) = f ( x1 , x0 ) = + = x1 − x 0 x 0 − x1 x0 − x1 x1 − x 0 f ( x1 , x 2 ) − f ( x0 , x1 ) f ( x1 , x 2 ) f ( x0 , x1 ) − = Now, f ( x0 , x1 , x 2 ) = x 2 − x0 x 2 − x0 x 2 − x0 f ( x1 ) − f ( x0 ) f ( x2 ) − f ( x1 ) − = (x2 − x0 )(x2 − x1 ) (x2 − x0 )(x1 − x0 ) f ( x0 , x1 ) =

=

f ( x0 ) f ( x2 ) f ( x1 ) f ( x1 ) − − + (x2 − x0 )(x2 − x1 ) (x2 − x0 )(x2 − x1 ) (x2 − x0 )(x1 − x0 ) (x2 − x0 )(x1 − x0 )

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www.Vidyarthiplus.in =

=

f ( x2 ) f ( x1 )  x1 − x0 + x2 − x1  f ( x0 ) −  + (x2 − x0 )(x2 − x1 ) (x2 − x0 )  (x2 − x1 )(x1 − x0 ) (x2 − x0 )(x1 − x0 )  x 2 − x0 f ( x0 ) f ( x2 ) f ( x1 )  −  + (x2 − x0 )(x 2 − x1 ) (x 2 − x0 )  (x2 − x1 )(x1 − x0 ) (x 2 − x0 )(x1 − x0 )

f ( x0 , x1 , x 2 ) =

f ( x0 ) f ( x1 ) f ( x2 ) − + (x0 − x1 )(x0 − x2 ) (x1 − x2 )(x1 − x0 ) (x1 − x0 )(x1 − x2 )

Similarly ,we can prove the result for higher differences Hence the divided differences are symmetrical in their arguments.

40. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the interior points, using forward difference. Sol:

f ′( x) =

(2u − 1) ∆2 y + (3u 2 − 6u + 2) ∆3 y + (4u 3 − 18u 2 + 22u − 6) ∆4 y + .................. dy 1  = ∆y 0 +  0 0 0 2! 3! 4! dx h  

(

)

  2 6u 2 − 18u + 11 4 3 ( 1 ) ∆ y + u − ∆ y + ∆ y 0 + ..................  0 0 12   3 d y 1  (2u − 3) ∆4 y + .................. . f ′′′( x ) = 3 = 3 ∆3 y 0 +  0 2 dx h   f ′′( x) =

d2y 1 = dx 2 h 2

41. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the point Sol: At x

x = x0 , using forward difference.

= x0 ,

dy 1  1 1 1  = ∆y 0 − ∆2 y 0 + ∆3 y 0 − ∆4 y 0 + .................. dx h  2 3 4  2 11 d y 1   f ′′( x ) = 2 = 2 ∆2 y 0 − ∆3 y 0 + ∆4 y 0 + .................. 12 dx h   3 3 d y 1   f ′′′( x ) = 3 = 3 ∆3 y 0 − ∆4 y 0 + .................. 2 dx h  

f ′( x ) =

42. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the interior points, using backward difference. Sol:

f ′( x) =

dy 1  (2v + 1) ∇ 2 y + (3v 2 + 6v + 2) ∇ 3 y + (4v 3 + 18v 2 + 22v + 6) ∇ 4 y + .................. = ∇y n +  n n n dx h  2! 3! 4! 

(

)

 2  6v 2 + 18v + 11 4 3 ∇ + + ∇ + ∇ y n + .................. y ( v 1 ) y  n n 12   3 (2v + 3) ∇ 4 y + .................. . d y 1  f ′′′( x) = 3 = 3 ∇ 3 y n +  n 2 dx h  

f ′′( x ) =

d2y 1 = dx 2 h 2

43. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the point Sol: At x

f ′( x) =

= xn , dy 1  1 1 1  = ∇y n + ∇ 2 y n + ∆3 y 0 + ∇ 4 y n + .................. 2 3 4 dx h  

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x = x n , using backward difference.

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d y 1 = dx 2 h 2 d3y 1 f ′′′( x ) = 3 = 3 dx h

f ′′( x ) =

44. Find

11 4   2 3 ∇ y n + ∇ y n + ∇ y n + .................. 12   3 4  3  ∇ y n + ∇ y n + .................. 2  

dy at x = 1 from the following table: dx

X: Y:

1 1

2 8

3 27

4 64

X:

Y:

∆

∆2

∆3

1

1

2

8

Sol:

7 12 19 3

6 18

27 37

4

64

Here, h = 1& x 0

=1

The Newton’s forward difference formula for

dy at x = x0 = 1 is dx

dy 1  1 1 1  = ∆y 0 − ∆2 y 0 + ∆3 y 0 − ∆4 y 0 + .................. dx h  2 3 4 

1  1  = 7 − (12) + (6 ) = 3 3  2  45.what is cubic Spline? A cubic polynomial which has continuous slope and curvature is called a cubic spline 46.What is a natural cubic spline? A cubic spline fitted to the given data such that the end cubies approach linearity at their entremities is called natural cubic spline 47.Define a cubic spline S(x) which is commonly used for interpolation. We define a cubic, S(x) as follows: i)S(x) is a polynomial of degree one for X <X0 and X>Xn ii)S(x) is at most a cubic polynomial in each interval (xi-1, xi) ,i=1,2,3,…….,n iii)S(x),S’(x) and S’’(x) are continuous at each point (xi,yi), i=0,1,2,…….n and iv)S(xi) = yi i =0,1,2,3………n 48. If y(xi)= yi, i = 0,1,2,…….n write down the formula for the cubic spline polynomial y(x) , valid in Xi-1 ≤X≤Xi Solution: Here h =1 Y(x) = 1/6 [(xi - x)3 Mi -1 + (x –xi-1)3 ]+ (xi - x)[yi-1 -1/6 Mi -1] +(x –xi-1) [yi -1/6 Mi] 49.Write the end conditions on Mi(x) in natural cubic splines Solution: M0(x) = 0 , Mn(x) = 0 50. Write the relation between the second derivatives Mi(x) in cubic splines with equal mesh spacing. Solution : Mi-1 + 4 Mi +Mi+1 = 6/h2 [yi-1 – 2yi +yi+1], i = 1,2,3,…..n-1

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www.Vidyarthiplus.in SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL-3 NUMERICAL METHODS UNIT III- NUMERICAL DIFFERENTIATION AND INTEGRATION

Two Marks Q&A

Branch: III CSE, III IT& II EEE

1. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the interior points, using forward difference. Sol: dy 1  (2u − 1) ∆2 y + 3u 2 − 6u + 2 ∆3 y + 4u 3 − 18u 2 + 22u − 6 ∆4 y + .................. = ∆y 0 + f ′( x) =  0 0 0 2! 3! 4! dx h  

(

(

)

(

)

)

 d2y 1  2 6u 2 − 18u + 11 4 3 ( 1 ) = ∆ y + u − ∆ y + ∆ y 0 + ..................  0 0 2 2 12 dx h   3 d y 1  (2u − 3) ∆4 y + .................. . f ′′′( x ) = 3 = 3 ∆3 y 0 +  0 2 dx h   f ′′( x ) =

2. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the point x = x0 , using forward difference. Sol: At x = x0 , dy 1  1 1 1  = ∆y 0 − ∆2 y 0 + ∆3 y 0 − ∆4 y 0 + .................. dx h  2 3 4  2 11 d y 1   f ′′( x ) = 2 = 2 ∆2 y 0 − ∆3 y 0 + ∆4 y 0 + .................. 12 dx h   3 3 d y 1   f ′′′( x ) = 3 = 3 ∆3 y 0 − ∆4 y 0 + .................. 2 dx h  

f ′( x ) =

3. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the interior points, using backward difference. Sol: dy 1  (2v + 1) ∇ 2 y + 3v 2 + 6v + 2 ∇ 3 y + 4v 3 + 18v 2 + 22v + 6 ∇ 4 y + .................. f ′( x) = = ∇y n +  n n n dx h  2! 3! 4! 

(

(

)

(

)

)

 6v 2 + 18v + 11 4 d2y 1  2 3 = ∇ + + ∇ + ∇ y n + .................. y ( v 1 ) y  n n 2 2 12 dx h   3 d y 1  (2v + 3) ∇ 4 y + .................. . f ′′′( x) = 3 = 3 ∇ 3 y n +  n 2 dx h  

f ′′( x ) =

4. State Newton’s formula to find f ′( x ), f ′′( x ) & f ′′′( x ) at the point x = x n , using backward difference. Sol: At x = x n , dy 1  1 1 1  = ∇y n + ∇ 2 y n + ∆3 y 0 + ∇ 4 y n + .................. dx h  2 3 4  2 11 d y 1   f ′′( x ) = 2 = 2 ∇ 2 y n + ∇ 3 y n + ∇ 4 y n + .................. 12 dx h  

f ′( x) =

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f ′′′( x ) =

3 d y 1  3  = 3 ∇ y n + ∇ 4 y n + .................. 3 2 dx h  

dy at x = 1 from the following table: dx 2 3 4 X: 1 Y: 1 8 27 64

5. Find

Sol: X: Y: 1

∆

∆2

∆3

1 7

2

12

8 19

3

6 18

27 37

4

64

Here, h = 1& x 0 = 1 The Newton’s forward difference formula for

dy at x = x0 = 1 is dx

dy 1  1 1 1  = ∆y 0 − ∆2 y 0 + ∆3 y 0 − ∆4 y 0 + .................. dx h  2 3 4 

1 1   = 7 − (12) + (6 ) = 3 2 3  

6. In Numerical integration, what should be the number of intervals to apply Simpson’s one-third rule and Simpson’s three-eight rule. Sol: To apply Simpon’s 1/3rd rule, the number of subintervals must be EVEN. To apply Simpon’s 1/8th rule, the number of subintervals must be a multiple of 3. 7. Compare Trapezoidal rule and Simpson’s one-third rule for evaluating numerical integration. Sol: Trapezoidal rule Simpson’s one-third rule

Any number of intervals Least accuracy

Number of intervals must be even. More accuracy

8. State the formula of Trapezoidal rule. Sol: The Trapezoidal rule is given by x0 + nh h ∫x f ( x)dx = 2 {y0 + 2( y1 + y 2 + y3 + .......... yn−1 ) + y n } 0 9. State the formula of Simpson’s one-third rule. Sol: Simpson’s one-third rule is given by

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www.Vidyarthiplus.in x0 + nh

h

∫ f ( x)dx = 3 {y

0

+ 4( y1 + y 3 + ..........) + 2( y 2 + y 4 + ........) + y n }. Provided when “n” is

x0

even. . 15. Write down the order of truncation error of trapezoidal rule and Simpson’s 1/3rd rule. Sol: The error in the trapezoidal formula is of the order h 2 . − h3 ( y 0′′ + y1′′ + y ′2′ + ......... y n′′−1 ) The truncation error in the trapezoidal rule is E = 12 1/ 2

16. Compute

∫ ydx using trapezoidal rule if y (0) = 1, y (1/4) = 1.01049, y (1/2) = 1.04291. 0

Sol: Given, Here, h = ¼ The Trapezoidal rule is 1/ 2 h (1 / 4) ∫0 ydx = 2 {y0 + 2 y1 + y 2 } = 2 {1 + 2(1.01049) + 1.04291} = 0.50798 17. Write down the order of truncation error of Simpson’s 1/3rd rule. Sol: The error in the Simpson’s 1/3rd rule formula is of the order h 4 . − h 5 iv y 0 + y iv 3 + .........) . ( The truncation error in the Simpson’s 1/3rd l rule is E = 90 20. A curve es through (0,1),(0.25,0.9412),(0.5,0.8),(0.75,0.64),(1,0.5). 1

Find

∫ f ( x)dx by trapezoidal rule. 0

Sol: X: 0 F(x) 1

( y0 )

0.25 0.5 0.9412 0.8 ( y1 ) ( y2 )

0.75 0.64 ( y3 )

1 0.5 ( y4 )

Here, h = 0.5 The Trapezoidal rule is 1/ 2 h 0 .5 ∫0 ydx = 2 {y0 + 2( y1 + y 2 + y3 ) + y4 } = 2 {0.399 + 2(0.352 + 0.242 + 0.129) + 0.054} = 0.4748 22. When does Simpson’s rule give exact result? Sol: Simpson’s rule will give exact result, if the entire curve y = f (x ) is itself a parabola. 23. State true or false. Whenever Trapezoidal rule is applicable Simpson’s rule can be applied. Sol: False 24. From the following table find the area bounded by the curve and the x axis from x = 2 to x = 7 x 2 3 4 5 6 7

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f (x )

8

27

64

www.Vidyarthiplus.in 125 216 343

Sol: Here h = 1 and only 6 ordinates are given We use trapezoidal rule 2 h Area = ∫ ydx = [( y 0 + y 4 ) + 2 y 2 + 4( y1 + y 2 )] = 53.8733 2 0 26. For what type of functions, Simpson’s rule and direct integration will give exact result? Sol: Simpson’s rule will give exact result, if the entire curve y = f (x ) is itself a parabola. 27. Why is trapezoidal rule so called? Sol: The trapezoidal rule is so called,because it approximates the integral by the sum of n trapezoids. 28. How the accuracy can be increased in trapezoidal rule of evaluating a given definite integral? Sol: If the number of points of the base segment b-a, (the range of integration) is increased,a better approximation to the area given by the definite integral will be obtained. 1 1 29. Evaluate ∫ dx by trapezoidal rule, dividing the range into 4 equal parts. 1 x 2

Sol: h = 1/8 x 4/8 f (x ) 8/4 1

5/8 8/5

6/8 8/6

7/8 8/7

8/8 8/8

1

∫ xdx = 0.6971. 1 2

30. In Numerical integration, what should be the number of intervals to apply Simpson’s one-third rule and Simpson’s three-eight rule. Sol: To apply Simpon’s 1/3rd rule, the number of subintervals must be EVEN. To apply Simpon’s 1/8th rule, the number of subintervals must be a multiple of 3. 31.Compare Trapezoidal rule and Simpson’s one-third rule for evaluating numerical integration. Sol: Trapezoidal rule Simpson’s one-third rule Any number of Number of intervals must be intervals even. Least accuracy More accuracy 32. State the formula of Trapezoidal rule. Sol: The Trapezoidal rule is given by x0 + nh

h

∫ f ( x)dx = 2 {y

0

+ 2( y1 + y 2 + y 3 + .......... y n −1 ) + y n }

x0

33. State the formula of Simpson’s one-third rule. Sol: Simpson’s one-third rule is given by x0 + nh

h

∫ f ( x)dx = 3 {y

0

+ 4( y1 + y 3 + ..........) + 2( y 2 + y 4 + ........) + y n }. Provided when “n” is even.

x0

.

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www.Vidyarthiplus.in 34. Write down the order of truncation error of trapezoidal rule and Simpson’s 1/3rd rule. 2 Sol: The error in the trapezoidal formula is of the order h . The truncation error in the trapezoidal rule is

E=

− h3 ( y 0′′ + y1′′ + y ′2′ + ......... y n′′−1 ) 12

1/ 2

∫ ydx using trapezoidal rule if y (0) = 1, y (1/4) = 1.01049, y (1/2) = 1.04291.

35. Compute

0

Sol: Given, Here, h = ¼ The Trapezoidal rule is 1/ 2

h

∫ ydx = 2 {y

0

+ 2 y1 + y 2 } =

(1 / 4) {1 + 2(1.01049) + 1.04291} = 0.50798

0

2

36. Write down the order of truncation error of Simpson’s 1/3rd rule. 4

Sol: The error in the Simpson’s 1/3rd rule formula is of the order h . The truncation error in the Simpson’s 1/3rd l rule is

E=

− h 5 iv ( y 0 + y iv 3 + .........) . 90

37. A curve es through (0,1),(0.25,0.9412),(0.5,0.8),(0.75,0.64),(1,0.5). 1

Find

∫ f ( x)dx by trapezoidal rule. 0

Sol: X: F(x)

0 1

0.25 0.9412

( y 0 ) ( y1 )

0.5 0.8

0.75 0.64

1 0.5

( y 2 ) ( y3 ) ( y 4 )

Here, h = 0.5 The Trapezoidal rule is 1/ 2

h

∫ ydx = 2 {y

0

+ 2( y1 + y 2 + y 3 ) + y 4 } =

0

0 .5 {0.399 + 2(0.352 + 0.242 + 0.129) + 0.054} = 0.4748 2

38. When does Simpson’s rule give exact result? Sol: Simpson’s rule will give exact result, if the entire curve y = f (x ) is itself a parabola. 39. State true or false. Whenever Trapezoidal rule is applicable Simpson’s rule can be applied. Sol: False 40. From the following table find the area bounded by the curve and the x axis from x = 2 to x = 7 x 2 3 4 5 6 7 f (x ) 8 27 64 125 216 343 Sol: Here h = 1 and only 6 ordinates are given We use trapezoidal rule 2

Area =

h

∫ ydx = 2 [( y

0

+ y 4 ) + 2 y 2 + 4( y1 + y 2 )] = 53.8733

0

41. For what type of functions, Simpson’s rule and direct integration will give exact result? Sol: Simpson’s rule will give exact result, if the entire curve y = f (x ) is itself a parabola. 42. Why is trapezoidal rule so called? Sol: The trapezoidal rule is so called,because it approximates the integral by the sum of n trapezoids. 43. How the accuracy can be increased in trapezoidal rule of evaluating a given definite integral?

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www.Vidyarthiplus.in Sol: If the number of points of the base segment b-a, (the range of integration) is increased,a better approximation to the area given by the definite integral will be obtained. 1

44. Evaluate

1

∫ xdx by trapezoidal rule, dividing the range into 4 equal parts. 1 2

Sol: h = 1/8

x f (x ) 1

4/8 8/4

5/8 8/5

6/8 8/6

7/8 8/7

8/8 8/8

1

∫ xdx = 0.6971. 1 2

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www.Vidyarthiplus.in SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL-3 NUMERICAL METHODS UNIT IV – INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS

Two Marks Q&A

1. Write down Euler algorithm to the differential equation

2.

3.

4. 5.

6.

dy = f ( x, y ) dx

Solution: y n + 1 = y n + hf ( x n , y n ) When n = 0,1,2,... This is Euler’s algorithm. It can also be written as y ( x + h) = y ( x ) + hf ( x, y ) State true or false. In Euler’s method, if “h” is small, the method is too slow and if “h” is large, it gives inaccurate value. Solution: The statement is true. State Modified Euler algorithm to solve y ′ = f ( x, y ), y ( x0 ) = y 0 at x = x0 + h . Solution: h h   y n +1 = y n + hf  x n + , y n + f ( x n , y n ) 2 2   h h   y1 = y 0 + hf  x0 + , y 0 + f ( x0 , y 0 ) 2 2   The Modified Euler method is based on the average of points. Solution: The statement is true. State the disadvantage of Taylor series method. Solution: dy In the differential equation = f ( x, y ), the function f ( x, y ), may have a dx complicated algebraical structure. Then the evaluation of higher order derivatives may become tedious. This is the demerit of this method. Write down the fourth order Taylor Algorithm. Solution: 2 3 4 iv ″′ h ″ h ′ h y m +1 = y m + hy m + ym + ym + ym Γ4 Γ3 Γ2 n th Here y m denotes the r derivative with respect to x at the point ( x m , y m )

7. Write the merits and demerits of the Taylor method of solution. Solution: The method gives a straight forward adaptation of classic to develop the solution as an infinite series. It is a powerful single step method if we are able to find the successive derivatives 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, Milne’s method etc.

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8. Which is better Taylor’s method or R. K. Method? Solution: 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 derivatives may be difficult. Also the R.K formulas involve the computation of f ( x, y ) at various positions, instead of derivatives and this function occurs in the given equation. 9. Taylor series method will be very useful to give some…………for powerful numerical methods such as Runge-kutta method,Milne’s method etc. Solution: Initial starting values. 10. State Taylor series algorithm for the first order differential equation. Solution: dy To find the numerical solution of = f ( x, y ) with the condition y ( x0 ) = y 0 .We dx expand xm in a Taylor series, getting y (x ) at a general point

h ′ h2 ″ ym + y m + .... Γ1 Γ2 denotes the r th derivatives of y w .r .to x at the point ( x m , y m ) .

y m +1 = y m + Here y m

r

11. Write the Runge-Kutta method algorithm of second order for solving y ′ = f ( x, y ), y ( x0 ) = y 0 . Solution: Let h denote the interval between equidistant values of x. If the initial values are ( x0 , y 0 ), the first increment in y is computed from the formulas. k1 = hf ( x0 , y 0 )

k  h  k 2 = hf  x0 + , y 0 + 1 and∆y = k 2 2 2  Then x1 = x0 + h, y1 = 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 x1 , y1 in the place of x 0 , y 0 respectively. 12. State the third order R.K method algorithm to find the numerical solution of the first order differential equation. Solution:

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www.Vidyarthiplus.in To solve the differential equation y ′ = f ( x, y ) by the third order R.K method, we use the following algorithm. k1 = hf ( x, y )

k  h  k 2 = hf  x + , y + 1  2 2  k 3 = hf ( x + h, y + 2k 2 − k1 ) 1 (k1 + 4k 2 + k 3 ) 6 13. Write down the Runge-Kutta formula of fourth order to solve dy = f ( x, y ) withy ( x0 ) = y 0. dx Solution: Let h denote the interval between equidistant values of x. If the initial values are ( x0 , y 0 ) ,the first increment in is computed from the formulas. k1 = hf ( x 0 , y 0 ) and∆y =

The image part with relationship ID rId675 was not found in the file.

k  h  k 2 = hf  x0 + , y 0 + 1  2 2  k h k 3 = hf ( x0 + , y 0 + 2 ) 2 2 k 4 = hf ( x 0 + h, y 0 + k 3 ) 1 ( k1 + 2 k 2 + 2 k 3 + k 4 ) 6 thenx1 = x0 + h, y1 = y 0 + ∆y The increment in y in the second interval is computed in a similar manner using the same four formulas,using the values x1 , y1 in the place of x0 , y0 respectively. and∆y =

14. State the special advantage of Runge-Kutta method over taylor series method. Solution: Runge-Kutta 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 derivatives may be difficult. Also the Runge-Kutta formulas involve the computation of f ( x, y ) at various positions,instead of derivatives and this function occurs in the given equation. 15. Say true or false. Modified Euler’s method is the Runge-Kutta method of second order. Solution: the statement is true. 16. Is Euler’s method formula, a particular case of second order Runge-Kutta method? Solution: Yes, Euler’s modified formula is a particular case of second order Runge-Kutta method.

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www.Vidyarthiplus.in 17. State which is better. Taylor’s method or R-K method? why? (or) What are the advantages of R.K method over Taylor’s method. (or) State the special advantage of R.K method over Taylor method. (or) Why is Runge-Kutta method preferred to Taylor series method. (or) Compare Taylor’s series and R.K method. Solution: R-K methods do not require prior calculation of higher derivatives of y (x ) as the taylor method does. Since the differential equations are using in applications often complicated, the calculation of derivatives may be difficult. Also the R-K formulas involve the computation of f ( x, y ) at various positions, instead of derivatives and this function occurs in the given equation.

18. The fourth order Runge-Kutta methods are used widely in………to differential equations. Solution: Getting numerical solutions. 19. How many prior values are required to predict the next value in Milne’s method? Solution: Four prior values. 20. Pick out the correct answer: The error term in Milne’s Predictor formula is h 4 ′ 14h 4 ′ 14h 4 19h 4 (a) ∆ y 0 (c) − ∆ y 0 (d) ∆ y 0 (b) ∆ y0 45 45 720 90 14h 4 ′ Solution: The error term is ∆ y0 .Correct answer is (a). 45 21. What is the error term in Milne’s corrector formula? h ′ Solution:The error term is − ∆4 y 0 90 22. Say ‘True or False’. Milne’s method is a self starting method. Solution:The statement is false. 23. Say True or False. Predictor-Corrector methods are single –step methods. Solution:The statement is false. 24. Predictor corrector methods are………starting methods. Solution: Not self starting methods. 25. How many prior values are required to predict the next value in Adam’s method? Solution:Four prior values. 26. Say True or False. Adams Bash forth method is a self starting method.

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www.Vidyarthiplus.in Solution:The statement is false. 27. What will you do,if there is a considerable difference between predicted value and corrected value,in predictor corrector methods? Solution: If there is a considerable difference between predicted value and corrected value,we take the corrected value as the predicted value and find out the new corrected value.This process is repeated till there is no great difference between two consecutive corrected values. 28.Compare Runge-Kutta methods and predictor – corrector methods for solution of initial value problem. Solution:Runge-Kutta methods 1.Runge-methods are self starting,since they do not use information from previously calculated points. 2.As mesne are self starting,an easy change in the step size can be made at any stage. 3.Since these methods require several evaluations of the function f ( x, y ), they are time consuming. 4.In these methods,it is not possible to get any information about truncation error.

Predictor Corrector methods: 1.These methods require information about prior points and so they are not self starting. 2.In these methods it is not possible to get easily a good estimate of the truncation error. 29. Say True or False. Predictor-Collector methods are single –step methods. Solution:The statement is false 30.What is a Predictor-collector method of solving a differential equation? Solution: Predictor-collector methods are methods which require the values of y at x n , x n−1 , x n − 2 ,.... for computing the value of y at x n +1 . We first use a formula to find the

value of y at x n +1 . and this is known as a predictor formula.The value of y so got is improved or corrected by another formula known as corrector formula. 31. Write Milne’s Predictor formula. Solution: Milne’s Predictor formula is 4h  ′ ′ ′ 2 y1 − y 2 + 2 y 3  y 4, p = y 0 +  3  ′ Where y 4 = f ( x1 , y1 ) 32. Write Milne’s corrector formula. Solution: Milne’s corrector formula is h ′ ′ ′ y 4 ,c = y 2 +  y 2 + 4 y 3 + y 4    3 ′ Where y 4 = f ( x 4 , y 4, p )

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www.Vidyarthiplus.in 33. Write down Adams-bashforth Predictor formula. Solution: Adams-bashforth Predictor formula is h  ′ ′ ′ ′ 55 y 3 − 59 y 2 + 37 y1 − 9 y 0  y 4, p = y3 +   24  34. Write down Adams-bashforth Corrector formula. Solution: Adams-bashforth Corrector formula is h  ′ ′ ′ ′ 9 y 4 + 19 y 3 − 5 y 2 + y1  y 4 ,c = y 3 +  24  ′ Where y 4 = f ( x 4 , y 4, p ) 35. Using Euler’s method find y (0.2) from

dy = x + y, y (0) = 1 with h =0.2. dx

Solution: By Euler’s method y1 = y 0 + hf ( x0 , y 0 ) = 1 + (0.2) f (0,1) = 1 + (0.2)(0 + 1) = 1+ (0.2) =1.2 y1 = y (0.2) = 1.2

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SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL-3 NUMERICAL METHODS UNIT V – BOUNDARY VALUE PROBLEMS IN ORDINARY AND PARTIAL DIFFERENTIAL EQUATIONS Two Marks Q&A

1. The number of conditions required to solve the Laplace equation is…… Solution: Four 2. What is the purpose of Liebmann’s process? Solution: The purpose of Liebmann’s process is to find the solution of the Laplace equation U xx + U yy = 0 by iteration. 3. 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. Solution: Since u satisfies the laplace equation and u=100 on the boundary of a square 1 u i , j = (100 + 100 + 100 + 100 ) 4 = 100 4. For the following mesh in solving ∇ 2 u = 0 find one set of rough values of u at interior mesh points. Solution: By symmetry, u 2 = u 3 1 Assume u 2 = 3. (Q u 2 is at distance from u = 2 ) 3 Therefore the rough values are 1 u1 = (1 + 1 + 2u 2 ) = 2 4 u2 = 3 1 u 3 = (5 + 5 + 2u 2 ) = 4 4 1 u 4 = (u1 + u 4 + 2 + 4) = 3 4 5. Write the Laplace equations U xx + U yy = 0 in difference quotients. Solution: u i −1, j − 2u i , j + u i +1, j u i , j −1 − 2u i , j + u i , j +1 + =0 h2 k2 ∂ 2u ∂ 2u 6. Write down the standard five point formula to solve Laplace equation 2 + 2 = 0. ∂y ∂x Solution: 1 u i , j = u i −1, j + u i +1, j + u i , j −1 + u i , j +1 4

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www.Vidyarthiplus.in 7. Define a difference quotient. Solution: 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.

8. State Liebmann’s iteration process formulae. Solution: 1 ( n +1) (n) ( n) ( n +1) n +1 u i , j = u i −1, j + u i +1, j + u i , j −1 + u i , j +1 4

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9. Write the diagonal five-point formula to solve the Laplace equation U xx + U yy = 0 Solution: 1 u i , j = u i −1, j −1 + u i −1, j +1 + u i +1, j −1 + u i +1, j +1 4 10. Write the difference scheme for solving the Laplace equation. Solution: The five point difference for ∇ 2φ = 0 is 1 u i , j = u i −1, j + u i +1, j + u i , j −1 + u i , j +1 4

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11. Write down the finite difference form of the equation. ∇ 2 u = f ( x, y ) Solution: u i −1, j + u i +1, j + u i , j −1 + u i , j +1 − 4u i , j = h 2 f (ih, jh) 12. Write the difference scheme for ∇ 2 u = f ( x, y ). Solution: Consider a square mesh with interval of differencing as h. Taking x = ih, y = jh the difference equation reduces to u i −1, j − 2u i , j + u i +1, j u i , j −1 − 2u i , j + u i , j +1 = f (ih, ih) + h2 h2 (i.e)u i −1, j + u i =1, j + u i , j +1 − 4u i , j = h 2 f (ih, jh) 13. State the five point formula to solve the Poisson equation u xx + u yy = 100 . Solution: u i −1, j + u i +1, j + u i , j −1 + u i , j +1 − 4u i , j = h 2 f (ih, jh) = 100 (Since h=1) 14. State the general form of Poisson’s equation in partial derivatives. Solution: ∂ 2u ∂ 2u + = f ( x, y ). ∂x 2 ∂y 2 15. Write a note on the stability and convergence of the solution of the difference equation corresponding to the hyperbolic equation u tt = a 2 u xx .

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www.Vidyarthiplus.in Solution: 1 , the solution of the difference equation is stable and coincides with the a 1 solution of the differential equation. For λ > ,the solution is unstable. a 1 For λ < ,the solution is stable but not convergent. a 16. State the explicit scheme formula for the solution of the wave equation. Solution: The formula to solve numerically the wave equation a 2 u xx − u tt = 0 is

For λ =

u i , j = 2(1 − λ2 a 2 )u i , j + λ2 a 2 (u i +1, j + u i −1, j ) − u i , j −1 The schematic representation is shown below. The solution value at any point (i,j+1) on the ( j + 1) th level is expressed in of solution values on the previous j and (j-1) levels (and not in of values on the same level).Hence this is an explicit difference formula. 17. Write down the general and simplest forms of the difference equation corresponding to the hyperbolic equation u tt = a 2 u xx . Solution: the general form of the difference equation to solve the equation utt = a 2u xx . is

u i , j +1 = 2(1 − λ2 a 2 )u i , j + λ2 a 2 (u i +1, j + u i −1, j ) − u i , j −1 ........(1) If λ2 a 2 = 1, coefficient of u i , j in (1) is =0 The recurrence equation (1) takes the simplified form u i , j +1 = u i +1, j + u i −1, j − u i , j −1 18. State the conditions for the equation. Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G where A,B,C,D,E,F,G are function of x and y to be (i)elliptic,(ii)parabolic(iii)hyperbolic. Solution: The given equation is said to be (i)elliptic at a point ( x, y ) in the plane if B 2 − 4 AC < 0 (ii)Parabolic if B 2 − 4 AC = 0 (iii)hyperbolicif B 2 − 4 AC > 0 19. State the condition for the equation Au xx + 2 Bu xy + Cu yy = f (u x , u y , x, y ) to be (i) elliptic,(ii)parabolic(iii)hyperbolic when A,B,C are functions of x and y Solution: The equation is elliptic if (2 B 2 ) − 4 AC < 0 (i.e) B 2 − AC < 0. It is parabolic if B 2 − AC = 0 and hyperbolic if B 2 − 4 AC > 0

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20. Fill up the blank. The equation yu xx + u yy = 0 is hyperbolic in the region…… Solution: Here A=y,B=0,C=1 2 B − 4 AC = 0 − 4 y = −4 y The equation is hyperbolic in the region (x,y) where B 2 − 4 AC > 0 i.e., − 4 y > 0 or y < 0 It is hyperbolic in the region y < 0 21. What is the classification of f x − f yy = 0 ? Solution: Here A=0,B=0,C=-1 B 2 − 4 AC = 0 − 4 × 0 × −1 = 0 So the equation is parabolic. 22. Give an example of a parabolic equation. Solution: ∂ 2u ∂u The one dimensional heat equation = α 2 2 is parabolic. ∂t ∂x 23. State Schmidt’s explicit formula for solving heat flow equation. Solution: u i , j +1 = λu i +1, j + (1 − 2λ )u i , j + λu i −1, j 1 1 , u i , j +1 = u i +1, j + u i −1, j 2 2 24. Fill up the blank. Bender-Schmidt recurrence scheme is useful to solve………equation. Solution: One dimensional heat 25. Write an explicit formula to solve numerically the heat equation (parabolic equation) u xx − au t = 0 Solution: ui , j +1 = λui +1, j + (1 − 2λ )ui , j + λui −1, j

If λ =

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k (h is the space for the variable x and k is the space in the time direction). h2a The above formula is a relation between the function values at the two levels j+1 and j and is called a two level formula. The solution value at any point (i,j+1) on the (j+1)th level is expressed in of the solution values at the points (i-1,j),(i,j) and (i+1,j) on the j th level.Such a method is called explicit formula. the formula is geometrically represented below. ∂u 1 26. What is the value of k to solve = u xx by Bender-schmidt method with h = 1 if h ∂t 2 and k are the increments of x and t respectively? Solution:

Where λ =

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www.Vidyarthiplus.in ∂u = u xx ∂t α 2 = 2, h = 1

Given : 2

kα 2 k (2) λ= 2 = = 2k 1 h Here 1 1 λ = 2k = Q 0 < λ ≤  2 2 1 k= 4 27. What is the classification of one dimensional heat flow equation. Solution: ∂ 2u 1 ∂u = 2 One dimensional heat flow equation is 2 ∂x α ∂t Here A=1,B=0,C=0 ∴ B 2 − 4 AC = 0 Hence the one dimensional heat flow equation is parabolic. 28. Write down the Crank-Nicholson formula to solve u t = u xx Solution: 1 1 λu i +1, j +1 + λu i −1, j +1 − (λ + 1)u i , j +1 2 2 1 1 = λu i +1, j − λu i −1, j − (λ − 1)u i , j 2 2 λ (λu i +1, j +1 + u i −1, j +1 ) − 2(λ + 1)u i , j +1 (Or) = 2(λ − 1)u i , j − λ (u i +1, j + u i −1, j ) 29. Write down the implicit formula to solve one dimensional heat flow equation. 1 u xx = 2 u t c Solution: Same as before question. 30. Why is Crank Nicholson’s scheme called an implicit scheme? Solution: The Schematic representation of crank Nicholson method is shown below. 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 j th level. Hence it is an implicit method. 31. Fill up the blanks. kα 2 In the parabolic equation. u t = α 2 u xx if λ = 2 h Where k = ∆t and h = ∆x, then

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www.Vidyarthiplus.in (a) explicit method is stable only if λ = ....... (b) implicit method is convergent when λ = ...... Solution: 1 (a)Explicit method is stable only if λ < 2 1 (b)Implicit method is convergent when λ = 2 32. What type of equations can be solved by using Crank-Nicholson’s difference formula? Solution: Crank-Nicholson’s difference formula is used to solve parabolic equations of the form. u xx = aut 33. Write the crank Nicholson difference scheme to solve u xx = aut with u (0, t ) = T0 , u (l , t ) = T1 and the initial condition as u ( x,0) = f ( x ) Solution: The scheme is 1 1 1 1 λu i −1, j +1 + λu i +1, j +1 − (λ + 1)u i , j +1 = λu i −1, j − λu i +1, j − (λ − 1)u i , j 2 2 2 2

34. For what purpose Bender-Schmidt recurrence relation is used? Solution: To solve one dimensional heat equation. 35. What are the methods to solve second order boundary-value problems? Solution: (i)Finite difference method (ii)Shooting method.

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