Sunday, April 1, 2012
NUMERICAL METHIODS
Unit I :
Solution of equations and eigen value problems
Part A
1.
If g(x) is continuous in [a,b] then
under what condition the iterative method x = g(x) has unique solution in
[a,b].
2.
By
Newton’s method find an iterative formula to find 1/N.
3.
Find the positive root of x3 + 5x – 3 =
o usind fixed point iteration method with 0.6 as first approximation.
æ1 3 ö
|
|||||
4.
|
Find inverse of A = ç
|
÷ by Gauss –
Jordan method.
|
|||
ç
|
2
|
7
|
÷
|
||
è
|
ø
|
||||
5.
State the convergence and order of
convergence for method of false position.
6.
Why
Gauss Seidel iteration is a method of successive corrections.
7.
Compare Gauss Jacobi and Gauss Siedel
methods for solving linear system of the form AX = B.
8. State
the conditions for convergence of Gauss Siedel method for solving a system of
equations.
9.
Find an iterative formula to find
N where N is a positive
number.
10. Compare Gaussian
elimination method and Gauss-Jordan method.
11. What type of
eigen value can be obtained using power method.
12. State
the order of convergence and convergence condition for Newton Raphson,s method.
13. Find the
dominant eigen value of A = é1 2 ù
|
by power
method.
|
|||
ê
|
3 4
|
ú
|
||
ë
|
û
|
|||
14.
How is the numerically smallest eigen
value of A obtained.
15.
State two difference between direct and
iterative methods for solving system of equations.
Part B
1. Consider
the non – linear system x2
– 2x – y +0.5 =0 and x2+4y2
– 4 = 0. Use Newton – Raphson method with the starting value (x0,
y0)
= (2.00, 0.25) and compute (x1,
y1),
(x2,
y2)
and (x3, y3)
.
2.
Fins
the real root of xex – 3 =
0 by regula - falsi method.
3. Find
the real root using method of false position for x3
– 2x – 5 = 0 coreect to three decimal places.
é 2
|
-1
|
0
ù
|
|||||
4.
|
Find all the eigen values of
the matrix
|
ê
|
ú
|
by power
method (Apply
|
|||
-1
|
2
|
-1
|
|||||
ê
|
ú
|
||||||
ê 0
|
-1
|
2 ú
|
|||||
ë
|
û
|
||||||
only 3
iterations).
|
|||||||
5. Use
Newton’s backward difference formula to construct an interpolating polynomial
of degree 3 for the data:
f( - 0.75) = - 0.0718125, f( - 0.5) = -
0.02475, f( - 0.25) = - 0.3349375 and f(0) = 1.101. Hence find f (- 13
).
6.
Solve the system of equations using
Gauss Seidel iterative methods.
20x – y – 2z = 17, 3x + 20y – z = -18, 2x – 3y +20z
= 25. 7.Find the largest eigen values and its corresponding vector of the
matrix
é 1
|
3
|
-1ù
|
||||
ê
|
3
|
2
|
4
|
ú
|
by power
method.
|
|
ê
|
ú
|
|||||
ê-1
|
4
|
10 ú
|
||||
ë
|
û
|
|||||
8. Using Gauss- Jordan obtain the inverse of the
matrix
é2
|
2
|
3ù
|
|||
ê
|
2
|
1
|
1
|
ú
|
|
ê
|
ú
|
||||
ê1
|
3
|
5ú
|
|||
ë
|
û
|
||||
9.
Using
Gauss Seidel method solve the system of equations starting with the values
x = 1 , y = -2 and
z = 3,
x
+
3y + 5z = 173.61, x – 27y + 2z = 71.31, 41x – 2y + 3z = 65.46
10. Solve the
following equations by Jacobi’s iteration method
x
+
y + z = 9, 2x – 3y + 4z = 13, 3x + 4y +
5z = 40.
Unit II :
Interpolation and Approxiamtion
Part A
- Construct a linear
interpolating polynomial given the points (x0,y0) and (x1,y1).
- Obtain the
interpolation quadratic polynomial for the given data by using
Newton’s
forward difference formula.
X :
|
0
|
2
|
4
|
6
|
Y :
|
-3
|
5
|
21
|
45
|
3.
Obtain the divided difference table for the following data.
X : -1 0
|
2
|
3
|
|||
Y : -8 3
|
1
|
12
|
|||
4.
|
Find the
polynomial which takes the following values.
|
||||
X
|
:
|
0
|
1
|
2
|
|
Y
|
:
|
1
|
2
|
1
|
|
5.
Define
forward, backward, central differences and divided differences.
6.
Evaluate D10 (1-x) (1-2x)
(1-3x)--------(1-10x), by taking h=1.
7.
Show that the divided difference
operator D
is linear.
8.
State the order of convergence of cubic
spline.
9.
What
are the natural or free conditions in cubic spline.
10. Find the cubic spline for the
following data
X : 0
|
2
|
4
|
6
|
|||||
Y : 1
|
9
|
21
|
41
|
|||||
11.
|
State the
properties of divided differences.
|
|||||||
3
|
1
|
-1
|
||||||
12.
|
Show that D (
|
)
|
=
|
.
|
||||
a
|
abcd
|
|||||||
bcd
|
||||||||
13.
Find the divided differences of f(x) = x3 + x + 2 for the arguments 1,3,6,11.
14.
State Newton’s forward and backward
interpolating formula.
15.
Using
Lagranges find y at x = 2 for the following
X : 0
|
1
|
3
|
4
|
5
|
Y : 0
|
1
|
81
|
256
|
625
|
Part B
1.
Using Lagranges interpolation formula
find y(10) given that y(5) = 12, y(6) = 13, y(9) = 14 and y(11) = 16.
2.
Find
the missing term in the following table
x : 0
|
1
|
2
|
3
|
4
|
y : 1
|
3
|
9
|
-
|
81
|
3. From the data
given below find the number of students whose weight is between
60 to 70.
Wt
(x) : 0-40 40-60 60-80 80-100 100-120
No of
students
: 250 120 100 70 50
4. From the
following table find y(1.5) and y’(1) using cubic spline.
X : 1 2 3
Y : -8 -1 18
5. Given sin 450 = 0.7071, sin 500 = 0.7660, sin
550 = 0.8192, sin 600 = 0.8660, find
sin 520 using Newton’s forward interpolating formula.
6. Given log 10 654 = 2.8156, log 10 658 = 2.8182, log 10 659 = 2.8189, log 10 661 =
2.8202, find using Lagrange’s formula
the value of log 10 656.
7. Fit a Lagrangian
interpolating polynomial y = f(x) and find f(5)
x : 1
|
3
|
4
|
6
|
||||
y : -3
|
0
|
30
|
132
|
||||
8.
|
Find y(12)
using Newton’ forward interpolation formula given
|
||||||
x :
|
10
|
20
|
30
|
40
|
50
|
||
y :
|
46
|
66
|
81
|
93
|
101
|
||
9.
Obtain the root of f(x) = 0 by
Lagrange’s inverse interpolation given that f(30) = -30, f(34) = -13, f(38) =
3, f(42) = 18.
10. Fit a natural
cubic spline for the following data
x : 0
|
1
|
2
|
3
|
y : 1
|
4
|
0
|
-2
|
11. Derive Newton’s
divided difference formula.
12. The following
data are taken from the steam table:
Temp0 c :
|
140
|
150
|
160
|
170
|
180
|
Pressure :
|
3.685
|
4.854
|
6.502
|
8.076
|
10.225
|
Find the pressure at temperature t = 1420
|
and at t = 1750
|
||||
13.
Find
the sixth term of the sequence 8,12,19,29,42.
14.
From
the following table of half yearly premium for policies maturing at different
ages, estimate the premium for
policies maturing at the age of 46.
Age x :
|
45
|
50
|
55
|
60
|
65
|
|
Premium y : 114.84
|
96.16
|
83.32
|
74.48
|
68.48
|
||
15. Form the
divided difference table for the
following data
|
||||||
x :
|
-2
|
0
|
3
|
5
|
7
|
8
|
y : -792
|
108
|
-72
|
48
|
-144
|
-252
|
|
Unit III
Differentiation
and Integration
Part A
1.
What
the errors in Trapezoidal and Simpson’s rule.
2.
Write
Simpson’s 3/8 rule assuming 3n intervals.
1
dx
3.
Evaluate
ò1+ x
4 using Gaussian quadrature with two points.
-1
4.
In Numerical integration what should be
the number of intervals to apply
Trapezoidal,
Simpson’s 1/3 and Simpson’s 3/8.
5.
|
Evaluate
|
1
|
x 2 dx
|
using Gaussian three point
quadrature formula.
|
|||
ò
|
|||||||
1+ x 4
|
|||||||
-1
|
|||||||
1
|
|||||||
6.
|
State two point Gaussian
quadratue formula to evaluate
|
ò
|
f (x)dx .
|
||||
-1
|
|||||||
7. Using
Newton backward difference write the formula for first and second order
derivatives at the end value x = x0
upto fourth order.
8.
|
Write down the expression for
|
dy
|
and
|
d 2 y
|
at x = x0 using Newtons
forward
|
||
dx
|
dx 2
|
||||||
difference
formula.
|
|||||||
9.
|
State Simpson’s 1/3 and
Simpson’s 3/8 formula.
|
||||||
P
|
|||||||
10.
|
Using trapezoidal rule evaluate
|
ò
|
sin xdx by dividing into six equal parts.
|
||||
0
|
|||||||
Part B
1.
Using Newton’s backward difference
formula construct an interpolating polynomial of degree three and hence find
f(-1/3) given f(-0.75) = - 0.07181250, f(-0.5) =
-
0.024750, f(-0.25) = 0.33493750, f(0) = 1.10100.
2.
|
Evaluate
|
òò
|
dxdy
|
by Simpson’s 1/3 rule with Dx =Dy = 0.5 where
0
|
||||
1+ x + y
|
||||||||
3.
|
Evaluate I =
|
2 2 dx dy by using
Trapezoidal rule, rule taking h= 0.5 and h=0.25.
|
||||||
òò
|
||||||||
1 1 x + y
|
||||||||
Hence the
value of the above integration by Romberg’s method.
|
||||||||
4.
|
From the following data find
y’(6)
|
|||||||
X : 0
|
2
|
3
|
4
|
7
|
9
|
|||
Y: 4
|
26
|
58
|
112
|
466
|
922
|
|||
5.
|
Evaluate
|
2 2 dx dy
|
numerically
with h= 0.2 along x-direction and k = 0.25 along y
|
||||
òò x 2 + y 2
|
|||||||
1 1
|
|||||||
direction.
|
|||||||
6.
|
Find the value of sec (31) from
the following data
|
||||||
q(deg ree)
: 31
|
32
|
33
|
34
|
||||
Tan q
|
: 0.6008
|
0.6249
|
0.6494
|
0.6745
|
|||
7. Find
the value of x for which f(x) is maxima in the range of x given the following
table, find also maximum value of f(x).
X:
|
9
|
10
|
11
|
12
|
13
|
14
|
Y : 1330
|
1340
|
1320
|
1250
|
1120
|
930
|
|
8. The
following data gives the velocity of a particle for 20 seconds at an interval
of five seconds. Find initial acceleration using the data given below
Time(secs) :
|
0
|
5
|
10
|
15
|
20
|
||||||||||
Velocity(m/sec):
0
|
3
|
14
|
69
|
228
|
|||||||||||
7
|
dx
|
||||||||||||||
9.
|
Evaluate
|
using
Gaussian quadrature with 3 points.
|
|||||||||||||
ò
|
2
|
||||||||||||||
3 1+ x
|
|||||||||||||||
10. For a
given data
|
find
|
dy
|
and
|
d 2 y
|
at x = 1.1
|
||||||||||
dx 2
|
|||||||||||||||
dx
|
|||||||||||||||
X :
|
1.0
|
1.1
|
1.2
|
1.3
|
1.4
|
1.5
|
1.6
|
||||||||
Y:
|
7.989
|
8.403
|
8.781
|
9.129
|
9.451
|
9.750
|
10.031
|
||||||||
UNIT – IV :
INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL
EQUATIONS
PART – A
|
|
1.
|
By Taylor
series, find y(1.1) given y¢ = x + y, y(1)
= 0.
|
2.
|
Find the
Taylor series upto x3 term satisfying 2 y¢ + y = x +1, y(0) = 1.
|
dy
|
|
3.
|
Using Taylor series method find y at x = 0.1 if dx = x 2 y -1, y(0) = 1 .
|
4.
State Adams – Bashforth
predictor and corrector formula.
State Adams – Bashforth
predictor and corrector formula.
5.
What
is the condition to apply Adams – Bashforth method ?
dy
|
|||||
6.
|
Using modified
Euler’s method, find
|
y(0.1)
|
if
|
dx = y 2
|
+ x 2 , y(0) = 1.
|
7.
Write down the formula to solve 2nd
order differential equation using Runge-Kutta method of 4th
order.
8.
In the derivation of fourth order
Runge-Kutta formula, why is it called fourth order.
9.
Compare R.K. method and Predictor
methods for the solution of Initial value problems.
10.
|
Using Euler’s
method find the
solution of the
|
IVP
|
dydx = log( x + y), y(0) = 2
|
||||||||||||||||||
at x = 0.2 taking
h = 0.2 .
|
|||||||||||||||||||||
PART-B
|
|||||||||||||||||||||
11.
|
The
|
differential
|
equation
|
dy
|
=
y -
x2
|
is
|
satisfied
|
||||||||||||||
dx
|
|||||||||||||||||||||
by y(0)= 1, y(0.2)= 1.12186,
y(0.4)=
1.46820,
y(0.6)=
1.7379 .Compute the
value
|
|||||||||||||||||||||
of y(0.8) by Milne’s predictor - corrector
formula.
|
|||||||||||||||||||||
12.
|
By means
of Taylor’s series
expension, find y at
|
x = 0.1,and
x = 0.2
correct to
|
|||||||||||||||||||
three decimals places, given
|
dy
|
- 2 y = 3e x , y(0) =
|
0.
|
||||||||||||||||||
dx
|
|||||||||||||||||||||
13.
|
Given
|
y¢¢ + xy¢ + y = 0, y(0)
=
1,
y¢(0)
=
0,
|
find
|
the
|
value
|
of
|
y(0.1)
|
by using
|
|||||||||||||
R.K.method of fourth order.
|
|||||||||||||||||||||
dy
|
|||||||||||||||||||||
14.
|
Using Taylor;s
series method find y at x = 0.1,
|
if
|
dx = x 2 y -1 , y(0)=1.
|
||||||||||||||||||
dy
|
|||||||||||||||||||||
15.
|
Given
|
dx = x 2 (1 + y) , y(1) =
1, y(1.1) =
1.233, y(1.2) =
1.548, y(1.3)=1.979,
|
|||||||||||||||||||
evaluate
y(1.4) by Adam’s- Bashforth method.
|
|||||||||||||||||||||
16.
|
Using Runge-Kutta
method of 4th order, solve
|
dy
|
=
|
y 2
|
- x 2
|
with y(0)=1
at
|
|||||||||||||||
dx
|
y 2
|
+ x 2
|
|||||||||||||||||||
x=0.2.
|
|||||||||||||||||||||
17.
|
Using
|
Milne’s method
to
|
find
|
y(1.4) given
|
that
|
5xy¢ = y 2 - 2 = 0 given that
|
|||||||||||||||
y(4) = 1, y(4.1)
=
1.0049,
y(4.2) = 1.0097, y(4.3) = 1.0143.
|
|||||||||||||||||||||
18.
|
Given
|
||||||||||
dy
|
= x3 + y, y(0)
=
2,
y(0.2) = 2.443214, y(0.4) = 2.990578, y(0.6)
=
3.823516
|
||||||||||
dx
|
|||||||||||
find y(0.8) by Milne’s predictor-corrector
method taking h = 0.2.
|
|||||||||||
19.
|
Using
|
R.K.Method
|
of
|
order 4,
find
|
y for
x = 0.1,
0.2,
|
0.3
|
given
|
that
|
|||
dy
|
= xy + y 2 , y(0) = 1 also find the solution at x = 0.4 using Milne’s method.
|
||||||||||
dx
|
|||||||||||
20. Solve dydx = y - x 2 , y(0) = 1.
Find y(0.1) and y(0.2) by R.K.Method of
order 4.
Find
y(0.3) by Euler’s method.
Find
y(0.4) by Milne’s predictor-corrector method.
21.
|
Solve
|
y¢¢ - 0.1(1 - y 2 ) y¢ + y = 0 subject to
|
y(0) = 0, y¢(0) = 1 using fourth
order
|
|||
Runge-Kutta
Method.
|
||||||
Find y(0.2) and
y¢(0.2) . Using step
size
|
Dx = 0.2 .
|
|||||
22.
|
Using 4th order RK Method
compute y for x = 0.1 given y¢ =
|
xy
|
given
y(0) =
|
|||
1 + x 2
|
||||||
1 taking
h=0.1.
|
||||||
dy
|
||||||
23.
|
Determine the value of y(0.4) using Milne’s method given dx = xy + y 2 , y(0) = 1,
|
|||||
use
Taylors series to get the value of y at x = 0.1, Euler’s method for y at x =
0.2 and RK 4th order method for
y at x=0.3.
24. Consider the
IVP dydx = y - x 2 +1, y(0) = 0.5
(i)
Using the modified Euler method, find
y(0.2).
(ii)
Using
R.K.Method of order 4, find y(0.4) and y(0.6).
(iii)
Using
Adam- Bashforth predictor corrector method, find y(0.8).
25. Consider the second order IVP y¢¢ - 2 y¢ + 2 y
= e2t
S
int,
with y(0) = -0.4 and y’(0)=-0.6.
(i)
Using
Taylor series approximation, find y(0.1).
(ii)
Using
R.K.Method of order 4, find y(0.2).
UNIT-5
PART-A
- Define the local truncation
error.
- Write down the
standard five point formula used in solving laplace equation U xx
+ U yy
= 0 at the point ( iDx,
jDy ).
- Derive Crank-Niclson scheme.
- State Bender Schmidt’s
explicit formula for solving heat flow equations
5.
Classify
x 2 f xx + (1-y 2 ) f yy = 0
6.
What is the truncation error of the
central difference approximation of
y ' (x)?
7. What is the error for
solving Laplace and Poissson’s equation by finite difference method.
8. Obtain the finite difference scheme fore the
difference equations 2 d
2
y
+ y = 5. dx 2
9. Write dowm the
implicit formula to solve the one dimensional heat equation.
- Define the diagonal five
point formula .
PART B
1. Solve the equqtion U t
= U xx subject to condition u(x,0) = sin px ; 0 £ x £ 1,u(0,t)
=
u(1,t) =0 using Crank- Nicholson method
taking h = 1/3 k = 1/36(do on time
step)
- Solve U xx + U yy =
0 for the following square mesh with boundary values
1
2
u 1
|
u 2
|
1
|
4
|
||
u 3
|
u 4
|
||||
2
|
5
|
||||
4
|
5
|
|||||||
3. Solve U xx = U tt with boundary condition u(0,t)
= u(4,t) and the initial condition
|
||||||||
u t (x,0) = 0 ,
u(x,0)=x(4-x) taking h =1, k = ½
|
(solve one period)
|
|||||||
4.
|
Solve xy II + y = 0 , y(1)
=1,y(2) = 2, h = 0.25
|
by finite difference method.
|
||||||
5.
|
Solve the
boundary value problem xy II -2y + x = 0,
subject to y(2) = 0 =y(3).Find
|
|||||||
y(2.25),y(2.5),y(2.75).
|
||||||||
6 . Solve the vibration problem ¶y = 4
|
¶2 y
|
subject to the boundary
conditions
|
||||||
¶t
|
¶x 2
|
|||||||
y(0,t)=0,y(8,0)=0
and y(x,0)= 12 x(8-x).Find y at x=0,2,4,6.Choosing
D x = 2, D t =
|
||||||||
1
|
up
|
|||||||
2
|
||||||||
compute to 4 time steps.
|
||||||||
7.
|
Solve D2 u = -4(x + y) in the region given 0 £ x £ 4, 0 £ y £ 4. With all boundaries
|
|||||||
kept
|
||||||||
at
|
0 0
|
and
choosing
|
D x = D y = 1.Start
with zero vector and do 4 Gauss- Seidal
|
|||||||||
iteration.
|
||||||||||||
0
0
|
0
0
|
0
0
|
0
0
|
0 0
|
||||||||
0 0
|
||||||||||||
0 0
|
||||||||||||
0
0
|
||||||||||||
0 0
|
0 0
|
0 0
|
0 0
|
0 0
|
||||||||
8.
|
Solve u xx + u yy = 0
|
over the square mesh of
|
sid
|
e 4 units,
satisfying the
|
|||
following
|
|||||||
conditions .
|
|||||||
u(x,0)
=3x
|
for
|
0 £ x £ 4
|
|||||
u(x,
4)
|
=
x 2
|
for
|
0 £ x £ 4
|
||||
u(0,y)
= 0,
|
for
|
0 £ y £ 4
|
|||||
u(4,y)
|
=
12+y
|
for 0 £ y £ 4
|
|||||
9.
|
Solve ¶2u
|
- 2 ¶u = 0, given that
u(0,t)=0,u(4.t)=0.u(x,0)=x(4-x).Assume
|
|||||
¶x 2
|
¶t
|
||||||
h=1.Find
|
|||||||
the
|
values of u upto t =5.
|
||||||
10. Solve y tt = 4y xx subject to the condition y(0,t) =0, y(2,t)=o, y(x,o) = x(2-x),
¶y
(x,0)
=
0
. Do 4steps and find the values upto 2 decimal
accuracy.
¶t
Sunday, April 1, 2012 1
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