Numerical-Methods Question and Answer

7/29/2019 Numerical-Methods Question and Answer

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SELVAM COLLEGE OF TECHNOLOGY NAMAKKAL-3DEPARTMENT OF MATHEMATICS

SUB: NUMERICAL METHODSUNIT I

1) Write the Descartes rule of signsSol:

1) An equation )( x f = 0 cannot have more number of positive roots than there arechanges of sign in the terms of the polynomial )( x f .2)An equation )( x f = 0 cannot have more number of positive roots than there arechanges of sign in the terms of the polynomial )( x f .

2) What is the order of convergence of Newton Raphson method if the multiplicity of theroot 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 ( Newtons iteration method)Derive newtons algorithm to derive th p root of a number N.

Sol: If p N x1

= then 0= N x p is the equation to be solved.Let 1)(,)( == p p px x f N x x f

By NR rule ,if r x is thethr iterate

)()(

1r

r r r x f

x f x x

=

+

11

)1(

+=

=

p

r

pr

p

r

pr

r px

N x p px

N x x

4) When would we not use N-R method .Sol: If 1 x is the exact root and 0 x is its approximate value of the equation

)( x f = 0.we know that 1 x = 0 x - )()(

0

0

x f x f

If )( 0 x f is small,the error )()(

0

0

x f x f

will be large and the computation of the root by

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 itcrosses 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.

www.Vidyarthiplus.in



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Sol: 1+n x = n x - )()(

n

n

x f x f

7) State the order of convergence and convergence condition for NR method?Sol: The order of convergence is 2

Condition of convergence is2

)()()( x f x f x f


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15) For solving a linear system, compare Gauss elimination method and Gauss Jordanmethod.Sol:

Gauss elimination method Gauss Jordan method.1.

2.

3.

Coefficient matrix istransformed into upper triangular matrix

Direct method

We obtain the solution by back substitution method

Coefficient matrix is transformedinto diagonal matrix

Direct method

No need for substitution method

16) The numerical methods of solving linear equations are of two types : one is direct and theother is .Sol: iterative.

17) Define round off error?Sol: The round off error is the quantity R which must be added to the finite representationof 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 nnaaa ,........, 2211 vanishesare become very small compared to other elements in that column ,then we attempt torearrange the remaining rows so as to obtain a non vanishing pivot or to avoid themultiplication by a large number.this strategy is called pivoting.The pivoting is of two types1) Partial pivoting2) Complete pivoting.

19) Why Guass siedel method is better than Jacobis 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 bemore rapid than in Guass Jacobi method.

20) Say true or false. Newtons method is useful in cases where the graph of the function when it crosses the xaxis 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




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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 jacobis method.Sol: False.The rate of convergence of Guass siedel methodb is roughly twicw that of the

Jacobis method.

27) Distinguish between direct and iterative method of solving simultaneous equation.Sol:

Direct method Iterative method.1.

2.

We get exact solution

Simple take less time

Approximjate solution

Time consuming laborious




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SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL-3MA1251-NUMERICAL METHODS

UNIT-IVINTERPOLATION AND NUMERICAL INTEGRATION AND DIFFERENTIATION

Two Marks Q&A1. State Lagranges interpolation formula.

Let 0)( = x f be a function which takes the values n y y y y y ...,,.........,,, 3210 corresponding to

....,,.........,,,3210 n

x x x x x x = Thus Lagranges interpolation formula is

( )( )( ) ( )

( )( )( ) ( )

( )( )( ) ( )

..........)(

)....(..................................

.....................)(

)....(...................)(

)....(..........)(

110

110

112101

200

02010

21

nnnnn

n

n

n

n

n

y x x x x x x x x x x x x



x f y

+

+

+

==

2. What is the Lagranges formula to find y, if three sets of values

( ) ( ) ( )221100 ,&,,, y x y x y x are given.The Lagranges interpolation formula is

( )( )( )

( )( )( )

( )( )( )

.)()()(

)( 21202

101

2101

200

2010

21 y x x x x

x x x x y

x x x x x x x x

y x x x x

x x x x x f y

+

+

==

3. Using Lagranges interpolation, find the polynomial through (0,0),(1,1) and (2,2).The polynomial through the given points is given by

( )( )( )

( )( )( )

( )( )( )

.)()()(

)( 21202

101

2101

200

2010

21 y x x x x

x x x x y

x x x x x x x x

y x x x x

x x x x x f y

+

+

==

( )( )( )( )

( )( )( )( )

( )( )( )( )

( ) x x x x x

x x x x x x

=+=

+

+

=

22 )2(

)2(120210

)1(210120

)0(201021

4. When do we apply Lagranges interpolation?Lagranges 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 Lagranges interpolation formula.

The inverse Lagranges interpolation formula is( )( )( ) ( )

( )( )( ) ( )

( )( )( ) ( )

..........)(

)....(..................................

.....................)(

)....(...................)(

)....(..........

110

110

112101

200

02010

21

nnnnn

n

n

n

n

n

x y y y y y y y y y y y y



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

Lagranges inverse interpolation formula can be obtained by interchanging x and y in Lagranges interpolationformula.

7. Define divided difference.

Let the function )( x f y=

take the values )(.......),........(),(),( 210 n x f x f x f x f corresponding to thevalues n x x x x ....,.........,, 210 of the argument x where 11201 ...,........., nn x x x x x x need notnecessarily be equal.

The first divided difference of )( x f for the arguments 10 , x x is01

0110

)()(),(

x x x f x f

x x f

= .

similarly,12

1221

)()(),(

x x x f x f

x x f

= etc.

The second divided difference of f(x) for three arguments 210 ,, x x x is defined as




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13

2132321

02

1021210

),(),(),,(,

),(),(),,(

x x x x f x x f

x x x f x x

x x f x x f x x x f

=

= etc.

8. Show that the divided difference operator is linear. Sol:

[ ] [ ] [ ] ( ) ( ))()()()()()()()()()()()(01

01

01

01

01

0011 x g x f x x

x g x g x x

x f x f x x

x g x f x g x f x g x f =

=

=

9. Evaluate ( )( )( ) ( ) x x x x 101.............3121110 ,by taking h =1.Sol: ( )( )( ) ( ) ( )210 !10)!10)(10.9.8.7.6.5.4.3.2.1(101.............31211 == x x x x

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

1212

24257150392 ==

2454

1060

711

3921452==

4572

914111314522366

==

7094

2836131723665202

==

67.206

124511121245

==

33.35

6212

713245457

==

426

2521117457709

==

83.1866.14

51367.2033.35

==

667.01067.6

71733.3542

==

7 392

11 1452

13 2366

17 5202

11. Write the Newtons divided difference interpolation formula for unequal intervals.( ) ( )( ) ............),,(),()()(

210101000+++= x x x f x x x x x x f x x x f x f

( ) ),.....,()).......(( 10110 nn x x x f x x x x x x + .

12. Write the Newtons forward difference interpolation formula.Let )( x f y = be a function which takes the values n y y y y y ...,,.........,,, 3210 corresponding to the values

n x x x x x ...,,.........,,, 3210 where the values of x are equally spaced.Then the Newtons forward difference interpolation formula is given by

( ) ( )..............................

!3)2(1

!21

!1 03

02

00+

+

++= y

uuu y

uu y

u y yn

Where

h

x xu 0

=

13. If 21

)( x

x f = ,find ),,(&),( cba f ba f by using divided differences.

Sol: Given, 21

)( x

x f = .

( ).

)(11

)()(),( 2222

2222

baab

abbaba

abab

aba f b f

ba f +

=

=

=

=




7/32

( )accba

cbbaabcbac

baab

bcbc

acba f cb f

cba f

++=

++

+

=

=

222

22222222 )()()(

),(),(),,(

( ) 222222)())((

cbacabcab

accbaaccabcab ++=

++=

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

= )( x f n nth divided difference = constant.

15. Find the second divided differences with arguments a,b,c if x

x f 1

)( = .

Sol: Given, x

x f 1

)( =

ababab

aba f b f

ba f 1

11)()(

),(

=

=

=

.1

11),(),(

),,(abcac

abbc

ac

ba f cb f cba f =

+

=

=

16. Obtain a divided difference table for the following data:X: -1 0 2 3

f(x): -8 3 1 12.

x y f(x) )(2 x f )(3 x f -1 -8

111083

=+

+

10231

=

1123112 =

412111

=+

403111 =

+ 213

44=+

+

0 3

2 1

3 12

17. Find the polynomial which takes the following values:X: 0 1 2Y: 1 2 1

Sol : The Lagranges interpolation formula is( )( )

( )( )( )

( )( )( )

( ).

)()()()( 2

1202

101

2101

200

2010

21 y x x x x

x x x x y

x x x x x x x x

y x x x x

x x x x x f y

+

+

==

( )( )( )

( )( )( )

( )( )( )

).1(12)02(10

)2(21)01(20

)1(20)10(21

)(

+

+

== x x x x x x x f y

2422)1(

)2(22

23 2222

++=

+

+

+= x x

x x x x x x

18. Obtain a divided difference table for the following data:X: 2 3 5

f(x): 0 14 102.




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Sol:x y f(x) )(2 x f 2 0

1423014 =

44

35

14102=

10251444

=

3 14

5 102

19. Write the Newtons backward difference interpolation formula .Let )( x f y = be a function which takes the values n y y y y y ...,,.........,,, 3210 corresponding to the values

n x x x x x ...,,.........,,, 3210 where the values of x are equally spaced.Then the Newtons backward difference interpolation formula is given by

( ) ( )..............................

!3)2(1

!21

!1 03

02

00+

+++

+++= y

vvv y

vv y

v y yn

Whereh x x

v 0

= .

20. Find the polynomial for the following data by Newtons backward difference formula.X: 0 1 2 3f(x): -3 2 9 18.

x f(x) = y y y2 y3 0 -3

2-(-3)=5

9-2 = 7

18-9 = 9( 0 y )

7-5= 2

9-7 =2( 02 y )

2-2 = 0( 03 y )

1 2

2 9

3 18( 0 y )

31

3,18,1,3 000 =

=

==== x

xh x x

v yh x

The Newtons backward interpolation formula is( ) ( )

..............................!3

)2(1!2

1!1 0

30

200

+++

++

++= yvvv

yvv

yv

y yn

)0()2(2

)13)(3()9)(3(18 +

+++=

x x x

346527918 22 +=+++= x x x x x

21. Evaluate )(baxn

e+

.Sol: let baxe x f +=)(

we know that )()()( x f h x f x f +=

( ) ( )1)( == +++++ ahbaxbaxbh xabax eeeee

( ) ( )( ) ( ) ( )2)(2 1)(1 === ++++++ ahbaxbaxbh xaahbaxbax eeeeeee




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In general, ( ) ( ).1 nahbaxbaxn eee = ++ 22. What are the advantages of Lagranges formula over Newtons formula?

The forward and backward interpolation formulae of Newton can be used only when the values of theindependent variable x are equally spaced and can also be used when the differences of the dependent variable y

become smaller ultimately. But Lagranges interpolation formula can be used whether the values of x, theindependent 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 4f(x): 4 5 13.

x Y f(x) )(2 x f 1 4

11245

=

424513 =

11414

=

2 5

4 13

( ) ( )( ) ...........),,(),()()( 210101000 +++= x x x f x x x x x x f x x x f x f 522314)1)(2)(1()1)(1(4 22 +=+++=++= x x x x x x x x .

24. What are the disadvantages in practice in applying Lagranges 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. ( ) ( ) ( ))()()()( x g x f x g x f = 3. ( ) ( ))()( x f c xcf = .4. The thn divided differences of a polynomial of the thn degree are constant.

26. When Newtons 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 Newtons 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 Newtons divided differences formula?

This is used when the data are unequally spaced.29 . Newtons forward interpolation formula used only foe equidistant (or) equal interval.30. Say true or false :

Newtons 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 : Newtons forward and Newtons 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 Newtons interpolating polynomialequation f(x) = 0.

Sol:X f(x) f(x) 2 f(x)0 -2

2-(-2)=4

8-2 = 66-4 =21 2

2 8




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x x

h

x xuh y =

=

===

1)0(

,1,2 00

Newtons forward interpolation formula is( ) ( )

..............................!3

)2(1!2

1!1 0

30

200

+

+

++= yuuu

yuu

yu

y y

)2(!2)1)((

)4)((2

++=x x

x

.2342 22 +=++= x x x x x

Now,2

1732

893023)( 2

=

+==+= x x x x f .)

33.Forward difference operator .Let )( x f y = be a function of x and let ,.....,, 210 y y y of the values of . corresponding to

,....2,, 000 h xh x x ++ of the values of . Here,the independent variable (or argument), x proceeds at equallyspaced intervals and h (constant),the difference between two consecutive values of is called the interval of differencing.

Now the forward difference operator is defined as

nnn y y y

y y y y y y

=

=

=

+ 1

121

010

......................

These are called first differences.

34.Forward difference table .

0 x

1 x

2 x

3 x

4 x

0 y

1 y

2 y

3 y

4 y

0 y

1 y 02 y

12 y 2

2 y 0

3 y

13 y 0

4 y

35. Backward difference operator.The backward difference operator is defined as

1=

nnn y y y For ...2,1,0=n

and so on.

3211223

212

33

2

+==

+=

nnnnnnn

nnnn

y y y y y y y

y y y y

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122

011

100

y y y y y y

y y y

=

=

=




11/32

36 Backward difference table. x y2 y3 y4

4 x

3 x

2 x

1 x

0 x

4 y

3 y

2 y

1 y

0 y

3 y

2 y

1 y

0 y

22

y

12

y

0

2

y

13

y

03 y 0

4 y

Newtons backward interpolation formula is( ) ( )

..............................!3

)2(1!2

1!1 0

30

200

+++

++

++= yvvv

yvv

yv

y yn

37. { } { } { })()()()( x g x f x g x f = (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 )(),( x g x f are two functions and 10 , x x be two arguments,

{ } [ ] [ ]( )

),(

)()()()()()(

10

01

0011

x x

x x x g x f x g x f

x g x f

=

=[ ] [ ]

( )010101 )()()()(

x x x g x g x f x f

{ } { }),(

)()(

10 x x

x g x f =

Similarly the result is true for any higher order difference.

38. { } { })()( x f c xcf = (ie) the divided difference of the product of a constant and a function is equal to theproduct of the constant and the divided difference of the function.Proof:

{ } { })()()()()()(01

01

01

01 x f c x x

x f x f c

x x xcf xcf

xcf =

=

=

{ } { })()( x f c xcf = Similarly the result is true for any higher order difference.

39.The divided differences are symmetrical in all their arguments.

01

1

10

001

10

10

01

0110

)()(),(

)()()()(),(

x x x f

x x x f

x x f x x

x f x f x x

x f x f x x f

+

==

=

=

Now,02

1021210

),(),(),,( x x

x x f x x f x x x f

= =02

10

02

21 ),(),( x x x x f

x x x x f

=( )( ) ( )( )0102

01

1202

12 )()()()( x x x x

x f x f x x x x

x f x f

=( )( ) ( )( ) ( )( ) ( )( )0102

0

0102

1

1202

1

1202

2 )()()()( x x x x

x f x x x x

x f x x x x

x f x x x x

x f

+

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=( )( ) ( ) ( )( ) ( )( )0102

0

0112

1201

02

1

1202

2 )()()( x x x x

x f x x x x x x x x

x x x f

x x x x x f

+

+

=( )( ) ( ) ( )( ) ( )( )0102

0

0112

02

02

1

1202

2 )()()( x x x x

x f x x x x

x x x x

x f x x x x

x f

+

),,( 210 x x x f =

( )( ) ( )( ) ( )( )21012

0121

1

2010

0 )()()(

x x x x

x f

x x x x

x f

x x x x

x f

+

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

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

( ) ( ) ( ) ++++++== ..................!4

622184!3

263!2

121)( 0

423

03

2

02

0 yuuu

yuu

yu

yhdx

dy x f

( )+

+

++== ..................1211186

)1(1

)( 04

2

03

02

22

2

yuu

yu yhdx yd

x f

( ) ++== ..................2

321)( 0

40

333

3

yu

yhdx

yd x f .

41. State Newtons formula to find )(&)(),( x f x f x f at the point 0 x x = , using forward difference.

Sol: At 0 x x = ,

++== ..................41

31

211

)( 04

03

02

0 y y y yhdxdy

x f

++== ..................12111

)( 04

03

02

22

2

y y yhdx

yd x f

+== ..................231

)( 04

03

33

3

y yhdx

yd x f

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

( ) ( ) ( ) ++++++++++== ..................!4

622184!3

263!2

121)( 4

233

22

nnnn yvvv

yvv

yv

yhdx

dy x f

( ) ++++++== ..................12

11186)1(

1)( 4

232

22

2

nnn yvv

yv yhdx

yd x f

( ) +++== ..................2

321)( 4333

3

nn yv

yhdx

yd x f .

43. State Newtons formula to find )(&)(),( x f x f x f at the point n x x = , using backward difference.

Sol: At n x x = ,

++++== ..................41

31

211

)( 4032

nnn y y y yhdxdy

x f




13/32

+++== ..................12111

)( 432222

nnn y y yhdx

yd x f

++== ..................231

)( 43333

nn y yhdx

yd x f

44. Finddxdy

at x = 1 from the following table:

Sol:

Here, h = 1& 10 = x

The Newtons forward difference formula for dxdy

at 10 == x x is

++= ..................41

31

211

04

03

02

0 y y y yhdxdy

( ) 3631

)12(21

7 =+=

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 spline47.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 X n

ii)S(x) is at most a cubic polynomial in each interval (x i-1, xi) ,i=1,2,3,.,niii)S(x),S (x) and S (x) are continuous at each point (x i,y i), i=0,1,2,.n andiv)S(x i) = y i i =0,1,2,3n

48. If y(x i)= y i, i = 0,1,2,.n write down the formula for the cubic spline polynomial y(x) , valid inX i-1 XX i

Solution:

Here h =1Y(x) = 1/6 [(x i - x)

3 M i -1 + (x x i-1)3 ]+ (x i - x)[y i-1 -1/6 M i -1 ] +(x x i-1) [y i -1/6 M i]

49.Write the end conditions on M i(x) in natural cubic splinesSolution: M 0(x) = 0 , M n(x) = 0

50. Write the relation between the second derivatives M i(x) in cubic splines with equal mesh spacing.Solution :

M i-1 + 4 M i +M i+1 = 6/h2 [y i-1 2y i +y i+1], i = 1,2,3,..n-1

X: 1 2 3 4Y: 1 8 27 64

X: Y: 2 3

1 17

19

37

12

186

2 8

3 27

4 64




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15/32

SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL-3NUMERICAL METHODS

UNIT III- NUMERICAL DIFFERENTIATION AND INTEGRATIONTwo Marks Q&A Branch: III CSE, III IT& II EEE

1. State Newtons formula to find )(&)(),( x f x f x f at the interior points, using forwarddifference.Sol:

( ) ( ) ( ) ++++++== ..................!4

622184!3

263!2

121)( 04

23

03

2

02

0 yuuu yuu yu y

hdxdy x f

( ) ++++== ..................

1211186

)1(1

)( 04

2

03

02

22

2

yuu

yu yhdx

yd x f

( ) ++== ..................2

321)( 0

40

333

3

yu

yhdx

yd x f .

2. State Newtons formula to find )(&)(),( x f x f x f at the point 0 x x = , using forwarddifference.

Sol: At 0 x x = ,++== ..................

41

31

211

)( 04

03

02

0 y y y yhdxdy

x f

++== ..................12111

)( 04

03

02

22

2

y y yhdx

yd x f

+== ..................231

)( 04

03

33

3

y yhdx

yd x f

3. State Newtons formula to find )(&)(),( x f x f x f at the interior points, usingbackward difference.Sol:

( ) ( ) ( ) ++++++++++== ..................!4

622184!3

263!2

121)( 4

233

22

nnnn yvvv

yvv

yv

yhdx

dy x f

( ) ++++++== ..................

1211186

)1(1

)( 42

3222

2

nnn yvv

yv yhdx

yd x f

( ) +++== ..................2

321)( 4333

3

nn yv

yhdx

yd x f .

4. State Newtons formula to find )(&)(),( x f x f x f at the point n x x = , using backward

difference.Sol: At n x x = ,

++++== ..................41

31

211

)( 4032

nnn y y y yhdxdy

x f

+++== ..................12111

)( 432222

nnn y y yhdx

yd x f




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++== ..................231

)( 43333

nn y yhdx

yd x f

5. Finddxdy

at x = 1 from the following table:

Sol:

Here, h = 1& 10 = x

The Newtons forward difference formula for dxdy

at 10 == x x is

++= ..................41

31

211

04

03

02

0 y y y yhdxdy

( ) 3631

)12(21

7 =+=

6. In Numerical integration, what should be the number of intervals to apply Simpsonsone-third rule and Simpsons three-eight rule.Sol: To apply Simpons 1/3 rd rule, the number of subintervals must be EVEN .

To apply Simpons 1/8 th rule, the number of subintervals must be a multiple of 3 .7. Compare Trapezoidal rule and Simpsons one-third rule for evaluating numericalintegration.

Sol:

8. State the formula of Trapezoidal rule.Sol: The Trapezoidal rule is given by

( ){ }nnnh x

x

y y y y y yh

dx x f +++++= +

13210 ..........22)(0

0

9. State the formula of Simpsons one-third rule.Sol: Simpsons one-third rule is given by

X: 1 2 3 4Y: 1 8 27 64

X: Y: 2 3

1 17

19

37

12

186

2 8

3 27

4 64

Trapezoidal rule Simpsons one-third ruleAny number of intervals

Number of intervals must be even.

Least accuracy More accuracy




17/32

( ) ( ){ }.........2..........43

)( 423100

0

n

nh x

x

y y y y y yh

dx x f +++++++=+

Provided when n is

even..

15. Write down the order of truncation error of trapezoidal rule and Simpsons 1/3rd

rule.Sol: The error in the trapezoidal formula is of the order 2h .

The truncation error in the trapezoidal rule is ( )12103

.........12

+++

=

n y y y yh

E

16. Compute 2/1

0

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

Sol: Given,Here, h = The Trapezoidal rule is

{ }2102/1

022 y y y

h ydx ++=

( ){ } 50798.004291.1)01049.1(212

4/1=++=

17. Write down the order of truncation error of Simpsons 1/3 rd rule.Sol: The error in the Simpsons 1/3 rd rule formula is of the order 4h .

The truncation error in the Simpsons 1/3 rd l rule is ( ).........90

30

5

++

= iviv y yh

E .

20. A curve passes through (0,1),(0.25,0.9412),(0.5,0.8),(0.75,0.64),(1,0.5).

Find 1

0

)( dx x f by trapezoidal rule.

Sol:

Here, h = 0.5The Trapezoidal rule is

{ }432102/1

0

)(22

y y y y yh

ydx ++++= { } 4748.0054.0)129.0242.0352.0(2399.025.0

=++++=

22. When does Simpsons rule give exact result?Sol: Simpsons rule will give exact result, if the entire curve )( x f y = is itself a parabola.23. State true or false.

Whenever Trapezoidal rule is applicable Simpsons rule can be applied.Sol: False

24. From the following table find the area bounded by the curve and the x axis from x = 2to x = 7

2 3 4 5 6 7

X: 0 0.25 0.5 0.75 1F(x) 1

( )0 y 0.9412( )1 y

0.8( )2 y

0.64( )3 y

0.5( )4 y




18/32

)( x f 8 27 64 125 216 343

Sol: Here h = 1 and only 6 ordinates are givenWe use trapezoidal rule

Area = [ ])(42)(2 21240

2

0

y y y y yh

ydx ++++= = 53.8733

26. For what type of functions, Simpsons rule and direct integration will give exact result?Sol: Simpsons rule will give exact result, if the entire curve )( x f y = 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 definiteintegral?

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.

29. Evaluate dx

x1

21

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

Sol: h = 1/84/8 5/8 6/8 7/8 8/8

)( x f 8/4 8/5 8/6 8/7 8/8

dx x

1

21

1= 0.6971.

30. In Numerical integration, what should be the number of intervals to apply Simpsons one-third rule and

Simpsons three-eight rule.Sol: To apply Simpons 1/3 rd rule, the number of subintervals must be EVEN .

To apply Simpons 1/8 th rule, the number of subintervals must be a multiple of 3 .31.Compare Trapezoidal rule and Simpsons one-third rule for evaluating numerical integration.

Sol:

32. State the formula of Trapezoidal rule.Sol: The Trapezoidal rule is given by

( ){ }nnnh x

x

y y y y y yh

dx x f +++++= +

13210 ..........22

)(0

0

33. State the formula of Simpsons one-third rule.Sol: Simpsons one-third rule is given by

( ) ( ){ }.........2..........43

)( 423100

0

n

nh x

x

y y y y y yh

dx x f +++++++=+

Provided when n is even.

.

Trapezoidal rule Simpsons one-third ruleAny number of intervals

Number of intervals must beeven.

Least accuracy More accuracy




19/32

34. Write down the order of truncation error of trapezoidal rule and Simpsons 1/3 rd rule.

Sol: The error in the trapezoidal formula is of the order 2h .

The truncation error in the trapezoidal rule is ( )12103

.........12

+++

=

n y y y yh

E

35. Compute 2/1

0

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

Sol: Given,Here, h = The Trapezoidal rule is

{ }2102/1

0

22

y y yh

ydx ++= ( ){ } 50798.004291.1)01049.1(2124/1

=++=

36. Write down the order of truncation error of Simpsons 1/3 rd rule.

Sol: The error in the Simpsons 1/3 rd rule formula is of the order 4h .

The truncation error in the Simpsons 1/3 rd l rule is ( ).........90

30

5

++

= iviv y yh

E .

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

Find 1

0

)( dx x f by trapezoidal rule.

Sol:

Here, h = 0.5The Trapezoidal rule is

{ }432102/1

0 )(22 y y y y y

h

ydx++++=

{ } 4748.0054.0)129.0242.0352.0(2399.025.0

=++++=

38. When does Simpsons rule give exact result?Sol: Simpsons rule will give exact result, if the entire curve )( x f y = is itself a parabola.

39. State true or false.Whenever Trapezoidal rule is applicable Simpsons rule can be applied.

Sol: False40. From the following table find the area bounded by the curve and the x axis from x = 2 to x = 7

2 3 4 5 6 7)( x f 8 27 64 125 216 343

Sol: Here h = 1 and only 6 ordinates are givenWe use trapezoidal rule

Area = [ ])(42)(2 21240

2

0

y y y y yh

ydx ++++= = 53.873341. For what type of functions, Simpsons rule and direct integration will give exact result?

Sol: Simpsons rule will give exact result, if the entire curve )( x f y = 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?

X: 0 0.25 0.5 0.75 1F(x) 1

( )0 y 0.9412

( )1 y 0.8

( )2 y 0.64

( )3 y 0.5

( )4 y




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

44. Evaluate dx x

1

21

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

Sol: h = 1/84/8 5/8 6/8 7/8 8/8

)( x f 8/4 8/5 8/6 8/7 8/8

dx x

1

21

1= 0.6971.




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UNIT IV INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONSTwo Marks Q&A

1. Write down Euler algorithm to the differential equation ),( y x f dxdy =

Solution:),(1 nnnn y xhf y y +=+ When ,...2,1,0=n This is Eulers algorithm. It can also be

written as ),()()( y xhf x yh x y +=+ 2. State true or false.

In Eulers method, if h is small, the method is too slow and if h is large, itgives inaccurate value.Solution: The statement is true.

3. State Modified Euler algorithm to solve 00 )(),,( y x y y x f y == at h x x += 0 .Solution:

+++=

+++=+

),(2

,2

),(2

,2

000001

1

y x f h

yh

xhf y y

y x f h

yh

xhf y y nnnnnn

4. The Modified Euler method is based on the average of points.Solution: The statement is true.

5. State the disadvantage of Taylor series method.Solution:

In the differential equation ),,( y x f dxdy

= the function ),,( y x f may have a

complicated algebraical structure. Then the evaluation of higher order derivatives may

become tedious. This is the demerit of this method.6. Write down the fourth order Taylor Algorithm.

Solution:iv

mmmmmm yh

yh

yh

hy y y432

432

1 +

+

+

+=

+

Here nm y denotes thethr derivative with respect to at the point ( )mm y x ,

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 thesuccessive derivatives easily. If ).( y x f involves some complicated algebraic structuresthen the calculation of higher derivatives becomes tedious and the method fails.This isthe major drawback of this method. However the method will be very useful for findingthe starting values for powerful methods like Runge - Kutta method, Milnes method etc.




22/32

8. Which is better Taylors method or R. K. Method?Solution:

R.K Methods do not require prior calculation of higher derivatives of )( x y ,as theTaylor method does. Since the differential equations using in applications are oftencomplicated, the calculation of derivatives may be difficult.Also the R.K formulas involve the computation of ),( y x f at various positions, instead

of derivatives and this function occurs in the given equation.

9. Taylor series method will be very useful to give somefor powerfulnumerical methods such as Runge-kutta method,Milnes method etc.Solution: Initial starting values.

10. State Taylor series algorithm for the first order differential equation.Solution:

To find the numerical solution of ),( y x f dxdy

= with the condition 00 )( y x y = .We

expand )( x y at a general point m x in a Taylor series, getting

....21

2

1+

+

+=

+ mmmm yh

yh

y y

Here r m y denotes the r th derivatives of w .r .to x at the point ( )mm y x , .

11. Write the Runge-Kutta method algorithm of second order for solving.)(),,( 00 y x y y x f y ==

Solution :Let h denote the interval between equidistant values of . x If the initial values

are ),,( 00 y x the first increment in is computed from the formulas.

21

002

001

2,

2

),(

k yand k

yh

xhf k

y xhf k

=

++=

=

Then y y yh x x +=+= 0101 ,The increment is in the second interval is computed in a similar manner using thesame three formulas,using the values 11 , y x in the place of 00 , y x respectively.

12. State the third order R.K method algorithm to find the numerical solution of thefirst order differential equation.Solution :




23/32

To solve the differential equation ),( y x f y = by the third order R.K method, we usethe following algorithm.

)4(61

)2,(

2,

2

),(

321

123

12

1

k k k yand

k k yh xhf k

k y

h xhf k

y xhf k

++=

++=

++=

=

13. Write down the Runge-Kutta formula of fourth order to solve

.00 )(),( y xwithy y x f dxdy

==

Solution :Let h denote the interval between equidistant values of . x If the initial values are

),( 00 y x ,the first increment in is computed from the formulas.

y y yh xthenx

k k k k yand

k yh xhf k

k y

h xhf k

k yh xhf k

y xhf k

+=+=

+++=

++=

++=

++=

=

0101

4321

3004

2003

1002

001

,

)22(61

),(

)2

,2

(

2,

2

),(

The increment in y in the second interval is computed in a similar manner using the

same four formulas,using the values 11 , y x in the place of 00 , y x respectively.

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 ),( x y 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 ),( y x f at various

positions,instead of derivatives and this function occurs in the given equation.

15. Say true or false.Modified Eulers method is the Runge-Kutta method of second order.Solution : the statement is true.

16. Is Eulers method formula, a particular case of second order Runge-Kuttamethod?Solution :Yes, Eulers modified formula is a particular case of second order Runge-Kuttamethod.

TheimagepartwithrelationshipID rId675wasnotfound in thefile.




24/32

17. State which is better. Taylors method or R-K method? why?(or) What are the advantages of R.K method over Taylors 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 Taylors series and R.K method.Solution:R-K methods do not require prior calculation of higher derivatives of )( x y as the taylor

method does.Since the differential equations are using in applications often complicated, thecalculation of derivatives may be difficult.Also the R-K formulas involve the computation of ),( y x f at various positions, insteadof derivatives and this function occurs in the given equation.

18. The fourth order Runge-Kutta methods are used widely into differentialequations.Solution : Getting numerical solutions.

19. How many prior values are required to predict the next value in Milnes method?Solution:

Four prior values.

20. Pick out the correct answer:The error term in Milnes Predictor formula is

(a)

0

4

4514

yh

(b) 04

4514

yh (c) 0

4

72019

yh (d)

04

90 y

h

Solution : The error term is

0

4

4514

yh

.Correct answer is (a).

21 . What is the error term in Milnes corrector formula?

Solution: The error term is

0

4

90 y

h

22 . Say True or False.Milnes 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 arestarting methods.Solution: Not self starting methods.

25 . How many prior values are required to predict the next value in Adams method?Solution: Four prior values.

26. Say True or False .Adams Bash forth method is a self starting method.




25/32

Solution :The statement is false.

27. What will you do,if there is a considerable difference between predicted value andcorrected value,in predictor corrector methods?Solution:

If there is a considerable difference between predicted value and correctedvalue,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 consecutivecorrected values.

28.Compare Runge-Kutta methods and predictor corrector methods for solution of initial value problem .Solution: Runge-Kutta methods1.Runge-methods are self starting,since they do not use information from previouslycalculated 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 ),,( y x f they are timeconsuming.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,....,, 21 nnn x x x for computing the value of at . 1+n x We first use a formula to find the

value of y at .1+n x and this is known as a predictor formula.The value of so got isimproved or corrected by another formula known as corrector formula.

31. Write Milnes Predictor formula.Solution:

Milnes Predictor formula is

+

+=3210,4 223

4 y y y

h y y p

Where ),(114

y x f y =

32. Write Milnes corrector formula.

Solution:Milnes corrector formula is

+

+

+=

4322,4 43 y y y

h y y c

Where ),( ,444 p y x f y =




26/32

33. Write down Adams-bashforth Predictor formula.Solution :

Adams-bashforth Predictor formula is

+

+=01233,4 937595524

y y y yh

y y p

34. Write down Adams-bashforth Corrector formula.Solution:

Adams-bashforth Corrector formula is

+

+

+=12343,4 519924

y y y yh

y y c

Where ),( ,444 p y x f y =

35. Using Eulers method find )2.0( y from 1)0(, =+= y y xdxdy

with h =0.2.

Solution:By Eulers method

),( 0001 y xhf y y +=

= )1,0()2.0(1 f +

= )10)(2.0(1 ++ = )2.0(1 + =1.2

2.1)2.0(1 == y y




27/32


UNIT V BOUNDARY VALUE PROBLEMS IN ORDINARYAND 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 Liebmanns process?Solution:The purpose of Liebmanns process is to find the solution of the Laplace equation

0=+ yy xx U U by iteration.

3. If u satisfies Laplace equation and u=100 on the boundary of a square what will be thevalue of u at an interior grid point.

Solution:Since u satisfies the laplace equation and u=100 on the boundary of a square

( )10010010010041

,+++=

jiu

100= 4. For the following mesh in solving 02 = u find one set of rough values of u at interior

mesh points.Solution:

By symmetry, 32 uu =

Assume .32 =u ( 2uQ is at 31

distance from 2=u )

Therefore the rough values are

2)211(41

21=++= uu

32 =u

3)42(41

4)255(41

414

23

=+++=

=++=

uuu

uu

5. Write the Laplace equations 0=+ yy xx U U in difference quotients.

Solution:

022

21,,1,

2,1,,1 =

++

+++

k

uuu

h

uuu ji ji ji ji ji ji

6. Write down the standard five point formula to solve Laplace equation .022

2

2

=

+

yu

xu

Solution:

[ ]1,1,,1,1, 41

+++++=

ji ji ji ji ji uuuuu




28/32

7. Define a difference quotient.Solution:

A difference quotient is the quotient obtained by dividing the difference betweentwo values of a function by the difference between two corresponding values of theindependent variable.

8. State Liebmanns iteration process formulae.

Solution:

[ ])1(1,)(1,)(,1)1(,11, 41 +

++

+

++++= n

jin

jin

jin

jin

ji uuuuu

9. Write the diagonal five-point formula to solve the Laplace equation 0=+ yy xx U U

Solution:

[ ]1,11,11,11,1, 41

+++++++=


10. Write the difference scheme for solving the Laplace equation.Solution:

The five point difference for 02

= is

[ ]1,1,,1,1, 41

+++++=


11. Write down the finite difference form of the equation. ),(2 y x f u = Solution:

),(4 2,1,1,,1,1 jhih f huuuuu ji ji ji ji ji =+++ ++

12. Write the difference scheme for ).,(2 y x f u = Solution:

Consider a square mesh with interval of differencing as h.Taking jh yih x == , the difference equation reduces to

),(4).(

),(22

2,1,,1,1

2

1,,1,

2

,1,,1

jhih f huuuuei

ihih f h

uuu

h

uuu

ji ji ji ji

ji ji ji ji ji ji

=++

=+

++

+=

++

13. State the five point formula to solve the Poisson equation .100=+ yy xx uu

Solution:100),(4 2,1,1,,1,1 ==+++ ++ jhih f huuuuu ji ji ji ji ji (Since h=1)

14. State the general form of Poissons equation in partial derivatives.Solution :

).,(22

2

2

y x f y

u x

u=

+

15. Write a note on the stability and convergence of the solution of the differenceequation corresponding to the hyperbolic equation .2 xxtt uau =




29/32

Solution:

For ,1a

= the solution of the difference equation is stable and coincides with the

solution of the differential equation. For a1> ,the solution is unstable.

For

a

1


30/32

20. Fill up the blank.The equation 0=+ yy xx u yu is hyperbolic in the region

Solution:Here A=y,B=0,C=1

y y AC B 44042 == The equation is hyperbolic in the region (x,y) where 042 > AC B

i.e., 04 > y or 0< y It is hyperbolic in the region 0< y

21. What is the classification of ?0= yy x f f

Solution:Here A=0,B=0,C=-1

0104042 == AC B So the equation is parabolic.

22. Give an example of a parabolic equation.

Solution:The one dimensional heat equation 2

22 u

t u

=

is parabolic.

23. State Schmidts explicit formula for solving heat flow equation.Solution:

ji ji ji ji uuuu ,1,,11, )21( ++ ++=

If [ ] ji ji ji uuu ,1,11, 21

,21

+++==

24. Fill up the blank.Bender-Schmidt recurrence scheme is useful to solveequation.

Solution:One dimensional heat25. Write an explicit formula to solve numerically the heat equation (parabolic equation)

0= t xx auu Solution:

ji ji ji ji uuuu ,1,,11, )21( ++ ++=

Whereah

k 2

= (h is the space for the variable x and k is the space in the time direction).

The above formula is a relation between the function values at the two levels j+1 and j and iscalled a two level formula. The solution value at any point (i,j+1) on the (j+1)th level isexpressed in terms of the solution values at the points (i-1,j),(i,j) and (i+1,j) on the j th level.Sucha method is called explicit formula. the formula is geometrically represented below.

26. What is the value of k to solve xxut u

21=

by Bender-schmidt method with 1=h if h

and k are the increments of and t respectively?Solution:




31/32

Given : xxut u =

2

Here

41

21

021

2

21

)2(

1,2

2

2

2

=


32/32

(a) explicit method is stable only if .......= (b) implicit method is convergent when ......= Solution:

(a)Explicit method is stable only if 21

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Numerical-Methods Question and Answer