Numerical Methods
Exam Duration: 45 Mins Total Questions : 30
Equation \({ e }^{ x }-1=0\) is required to be solved using Newton's method with an initial guess \({ X }_{ o }=-1\) . Then after one step of Newton;s method, estimate x1 of the solution will be given by
- (a)
0.71828
- (b)
0.36784
- (c)
0.2.587
- (d)
0.000
Let \(f(x)={ e }^{ x }-1=0\\ \)
Now, by Newton-Raphson method
\({ x }_{ k+1 }={ x }_{ k }-\frac { f({ x }_{ k }) }{ f'({ x }_{ k }) } \\ \quad { x }_{ k+1 }={ x }_{ k }-\frac { { e }^{ { x }_{ k } }-1 }{ { e }^{ { x }_{ k } } } \quad \quad \)
\(\quad =\frac { { e }^{ { x }_{ k } }.{ x }_{ k }-{ e }^{ { x }_{ k } }+1 }{ { e }^{ { x }_{ k } } } \quad \quad \quad ...(i)\)
But at (k=0), xo = -1
\({ x }_{ 1 }=\frac { ({ x }_{ o }{ e }^{ { x }_{ 0 } }-{ e }^{ { x }_{ 0 } }+1) }{ { e }^{ { x }_{ 0 } } } \)
\(=\frac { { -e }^{ -1 }-{ e }^{ -1 }+1 }{ { e }^{ -1 } } \)
\(\Rightarrow \quad \frac { -2+e }{ 1 } =0.71828\) \((\because e=271828)\)
The square root of a number N is to be obtained by applying the Newton-Raphson iteration to the equation \({ X }^{ 2 }-N=0\) . If I denotes the iteration index the correct iterative scheme will be
- (a)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }_{ i }+\frac { N }{ { X }_{ i } } \right) \)
- (b)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }^{ 2 }_{ i }+\frac { N }{ { X }^{ 2 }_{ i } } \right) \)
- (c)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }_{ i }+\frac { N^{ 2 } }{ { X }_{ i } } \right) \)
- (d)
\({ X }_{ i+1 }=\frac { 1 }{ 2 } \left( { X }_{ i }+\frac { N }{ { X }_{ i } } \right) \)
Given \({ x }^{ 2 }-N=0\)
\(f(x)={ x }^{ 2 }-N\\ f'(x)=2x\\ f_{ x+i }=x_{ i }-\frac { f({ x }_{ i }) }{ f'({ x }_{ i }) } \\ \quad =x_{ i }-\frac { { x }_{ i }^{ 2 }-N }{ 2x_{ i } } =\frac { 1 }{ 2 } \left[ x_{ i }+\frac { N }{ x_{ i } } \right] \)
The convergence of the bisection method is
- (a)
cubic
- (b)
quadratic
- (c)
linear
- (d)
None of these
Linear, because the order of convergence of bisection method is 1/2.
For the differential equation \(\frac { dy }{ dx } =x-{ y }^{ 2 }\) is given that
X | 0 | 0.2 | 0.4 | 0.6 |
Y | 0 | 0.02 | 0.0795 | 0.1762 |
Using predictor-correction method, the y at next value of x is
- (a)
0.5114
- (b)
0.4648
- (c)
0.3046
- (d)
0.2498
We have, x = 0, 0.2, 0.4, 0.6
\(f(x)=x-{ y }^{ 2 }\)
We get,
\({ f }_{ 1 }(x)=0.1996,\quad { f }_{ 2 }(x)=0.3937,\quad { f }_{ 3 }(x)=0.5689\)
Using predictor formula
\({ y }^{ p }_{ 4 }=\quad { y }_{ o }+\frac { 4 }{ 3 } h(2{ f }_{ 1 }-{ f }_{ 2 }+2{ f }_{ 3 })\)
\(=0+\frac { 0.8 }{ 3 } [2(0.1996)-(0.3937)+2(0.5689)]\\ =0.3049\)
\({ y }^{ C }_{ 4 }=\quad { y }_{ 2 }+\frac { h }{ 3 } ({ f }_{ 2 }-4{ f }_{ 3 }+{ f }^{ p }_{ 4 })\)
\({ f }^{ (p) }_{ 4 }=f({ x }_{ 4 },{ y }^{ p }_{ 4 })=f\left( 0.8,0.3049 \right) \)
\({ y }^{ C }_{ 4 }=0.0795+\frac { 0.02 }{ 3 } [0.3937+4(0.5689)+0.7070]\\ \quad \quad =\quad 0.3046\\ \)
Which one of the following is correct?
- (a)
Bisection method is used for iteration
- (b)
Regula-falsi method is direct method
- (c)
Secant method is direct method
- (d)
Newton-Raphson method is not iterative method
Bisection method, Regula-Falsi method, Secant method, Newton-Raphson method are iterative methods.
If \({ e }^{ o }=1,{ e }^{ 1 }=2.72,{ e }^{ 2 }=7.39,{ e }^{ 3 }=20.09\) and \({ e }^{ 4 }=54.60\) , then by Simpson's \(\frac { 1 }{ 3 } rd\) rule value of \(\int _{ 0 }^{ 4 }{ { e }^{ x } } dx\) is
- (a)
52.78
- (b)
53.87
- (c)
5.278
- (d)
5.387
Here, h=1
By Simpson rule,
\(\int _{ 0 }^{ 4 }{ { e }^{ x } } dx=\frac { h }{ 3 } [({ e }^{ 0 }+{ e }^{ 4 })+4({ e }^{ 1 }+{ e }^{ 3 })+2{ e }^{ 2 }]\\ \int _{ 0 }^{ 4 }{ { e }^{ x } } dx=\frac { 1 }{ 3 } [(55.60)+(91.24)+(14.78)]\\ \quad \quad \quad \quad \quad \quad =\frac { 1 }{ 3 } (161.62)\\ \quad \quad \quad \quad \quad \quad =\quad 53.87\quad \)
Match the items in Columns I and II using the codes given below the columns.
Column I | Column II |
(P) Gauss-Seidel method | (1) Interpolation |
(Q) forward Newton method | (2) Non-linear differential equation |
(R) Runge-Kutta method | (3) Numerical integration |
(S) Trapezoidal rule | (4) Linear algebraic equations |
- (a)
P Q R S 1 2 3 4 - (b)
P Q R S 2 3 4 1 - (c)
P Q R S 3 4 2 1 - (d)
P Q R S 4 1 2 3
Gauss-Seidel method | linear algebraic equation |
Forward Newton-Gauss method | Interpolation |
Runge-Kutta method | Non-linear differential equation |
Trapezoidal rule | Numerical intergration |
Newton-Raphson method is used to compute a root of the equation \({ X }^{ 3 }-13=0\) with 3.5 as the initial value. The approximation after one iteration is
- (a)
3.575
- (b)
3.677
- (c)
3.667
- (d)
3.607
\(f(x)={ x }^{ 2 }-13=0,\quad f'(x)=2x\)
Initial value approximation
\({ x }_{ 0 }=3.5\)
If x1 be the approximation after one iteration. By Newton-Raphson method
\({ x }_{ 1 }={ x }_{ 0 }-\frac { f({ x }_{ 0 }) }{ f'({ x }_{ 0 }) } \\ \quad \quad =3.5-\frac { { (3.5) }^{ 2 }-13 }{ 2\times 3.5 } =3.607\)
The relation to solve x = e -x using Newton-Raphson method is
- (a)
\({ X }_{ n+1 }={ e }^{ { -X }_{ n } }\)
- (b)
\({ X }_{ n+1 }={ X }_{ n }{ -e }^{ { -X }_{ n } }\)
- (c)
\({ X }_{ n+1 }=(1+x_{ n })\frac { { e }^{ { -X }_{ n } } }{ 1+{ e }^{ { -X }_{ n } } } \)
- (d)
\({ X }_{ n+1 }=\frac { { X }^{ 2 }_{ n }-e^{ { -x }_{ n } }(1+{ X }_{ n })-1 }{ { X }_{ n }-{ e }^{ { -X }_{ n } } } \)
Let \(f(x)=x-{ e }^{ -x }\)
\(\therefore \) The given equation can be written as
\(f(x)=x-{ e }^{ -x }=0\\ f'(x)=1+{ e }^{ -x }\)
Newton-Raphson method is
\({ x }_{ n+1 }={ x }_{ n }-\frac { f({ x }_{ n }) }{ f'({ x }_{ n }) } \)
\(\Rightarrow \quad { x }_{ n+1 }={ x }_{ n }-\frac { { x }_{ n }-{ e }^{ { -x }_{ n } } }{ 1+{ e }^{ { -x }_{ n } } } \)
\(=(1+{ x }_{ n })\frac { { e }^{ { -x }_{ n } } }{ 1+{ e }^{ { -x }_{ n } } } \\ \)
The table below gives values of a function F(x) obtained for values of x at intervals 0.25
X | 0 | 0.25 | 0.5 | 0.75 | 1.00 |
f(x) | 1 | 0.9412 | 0.8 | 0.64 | 0.50 |
The value of the integral of the function between the limit 0 to 1 using Simpson's rule is
- (a)
0.7854
- (b)
2.3562
- (c)
3.1416
- (d)
7.5000
Given, h = 0.25
X | 0 | 0.25 | 0.5 | 0.75 | 1.00 |
f(x) | 1 | 0.9412 | 0.8 | 0.64 | 0.50 |
By Simpson's \(\frac { 1 }{ 3 } rd\quad rule\)
\(\int _{ { x }_{ 0 } }^{ { x }_{ 0 }+nh }{ F(x)dx=\frac { h }{ 3 } } [\{ F(0)+F(1)\} +2\{ F(0.5)\} +4\{ F(0.25)\} +F(0.75)\} ]\)
\(=\frac { 0.25 }{ 3 } [(1+0.50)+2(0.8)+4(0.9412+0.64)]\\ =\frac { 0.25 }{ 3 } (9.4248)=0.7854\)
In the solution of the following set of linear equations by Gauss elimination using partial pivoting 5x + y + 2Z = 34; 4y - 3z = 12 and 10x - 2y + z = -4; the pivots for elimination of x and y are
- (a)
10 and 4
- (b)
10 and 2
- (c)
5 and 4
- (d)
5 and -4
The given system of equation
5x + y + 2z = 34 ...(i)
4 y - 3z = 12 ...(ii)
10x - 2y + z = -4 ...(iii)
Now, we follow the Gauss elimination method. First we eliminate the term x in Eqs. (i) and (iii) with the help of Eq. (i) but Eq. (ii) does not have any x term, so we only eliminate x in Eq. (iii) and coefficient of which is called our first pivot.
Ist pivot : \(\quad \quad \quad x-2y+z=-4\\ 2(34-y-2z)-2y+z=-4\\ 68-2y-4z-2y+z=-4\)
II nd pivot : y+3z = 72 ...(iv)
Now, we eliminate the term y in Eqs. (ii) and (iv) and the coefficient of y is called our second pivot.
4y - 3z = 12
4y + 3z = -72
- - +
____________
z = 10
So, system of equation becomes
\(\quad 5x+y+2z=34\\ \quad \quad \quad \quad 4y-3z=12\\ \quad \quad \quad \quad \quad \quad \quad z=10\)
First pivot = 10
Second pivot = 4
The equation x3 +4x-9 = 0 needs to be numerically solved using the Netwon-Raphson method is
- (a)
\({ X }_{ k+1 }=\left( \frac { 2X^{ 3 }_{ k }+9 }{ 3X^{ 2 }_{ k }+4 } \right) .8\)
- (b)
\({ X }_{ k+1 }=\frac { 3{ x }^{ 2 }_{ k }+4 }{ 2{ x }^{ 2 }_{ k }+9 } \)
- (c)
\({ X }_{ k+1 }={ x }_{ k }-3{ x }^{ 2 }_{ k }+4\)
- (d)
\({ X }_{ k+1 }=\frac { 4{ x }^{ 2 }_{ k }+3 }{ 9{ x }^{ 2 }_{ k }+2 } \)
Let \(f(x)={ x }^{ 3 }+4x-9=0\\ f'(x)=3{ x }^{ 2 }+4=0\)
Now, Newton-Raphson method
\({ X }_{ k+1 }={ x }_{ k }-\frac { f({ x }_{ k }) }{ f'{ (x }_{ k }) } \\ { X }_{ k+1 }={ x }_{ k }-\frac { { x }^{ 3 }_{ k }+4{ x }_{ k }-9 }{ 3{ x }^{ 3 }_{ k }+4 } \\ { X }_{ k+1 }=\frac { 3{ x }^{ 3 }_{ k }+4{ x }_{ k }-{ x }^{ 3 }_{ k }-4{ x }_{ k }+9 }{ 3{ x }^{ 2 }_{ k }+4 } \\ \\ { X }_{ k+1 }=8\left( \frac { 2{ x }^{ 3 }_{ k }+9 }{ 3{ x }^{ 2 }_{ k }+4 } \right) \)
A numerical solution of the equation \(f(x)=x+\sqrt { x-3 } \) can be obtained using Newton-Raphson method. If the starting value of x = 2 for ten iterations, the value of x that is to be used in the next step
- (a)
0.306
- (b)
0.739
- (c)
1.694
- (d)
2.306
\({ x }_{ n+1 }={ x }_{ n }-\frac { f({ x }_{ n }) }{ f'({ x }_{ n }) } ,\) given \(f(x)\equiv x+\sqrt { x-3 } =0\)
\(f'(x)=1+\frac { 1 }{ 2\sqrt { x-3 } } \\ f(2)=(2+\sqrt { 2 } -3)=\sqrt { 2 } -1\\ f'(2)=\frac { 2\sqrt { 2 } +1 }{ 2\sqrt { 2 } } \)
\(\Rightarrow \) \({ x }_{ n+1 }=2-\frac { \frac { \sqrt { 2 } -1 }{ 2\sqrt { 2 } +1 } }{ 2\sqrt { 2 } } =1.694\) \(\left( \because { x }_{ 0 }=2 \right) \)
The double (repeated) root of
\(4{ x }^{ 3 }-8{ x }^{ 2 }-3x+9=0\)
by Newton-Raphson method will be
- (a)
1.4
- (b)
1.5
- (c)
1.6
- (d)
1.55
Using trapezoidal rule and the table given below, \(\int _{ 4 }^{ 5.2 }{ In\quad x\quad dx } \) will be
x | 4 | 4.2 | 4.4 | 4.6 | 4.8 | 5 | 5.2 |
In x | 1.39 | 1.44 | 1.48 | 1.53 | 1.57 | 1.61 | 1.65 |
- (a)
1.8277
- (b)
1.9284
- (c)
1.6424
- (d)
0.98795
By trapezoidal rule,
\(=\frac { h }{ 2 } [y_{ 0 }+{ y }_{ n })+2({ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+...+{ y }_{ n-1 })\)
\(\left( \because h=\frac { 5.2-4 }{ 6 } \right) \)
\(=\frac { 0.2 }{ 2 } [(1.39+1.65)+2(1.44+1.48+1.53+1.57+1.61)]\\ =\quad 1.82764\\ \)
Using trapezoidal rule, taking 10 equal interval \(\int _{ 0 }^{ \pi /2 }{ sinx\quad dx } \) will be
- (a)
1.902
- (b)
1.941
- (c)
1.888
- (d)
0.99795
Divide \(\left[ 0,\frac { x }{ 2 } \right] \) into ten parts each of width h =\(\frac { \pi }{ 20 } \)
Then, the value of y= sin x are given in the table as below
x | 0 | \(\frac { \pi }{ 20 } \) | \(\frac { 2\pi }{ 20 } \) | \(\frac { 3\pi }{ 20 } \) | \(\frac { 4\pi }{ 20 } \) | \(\frac { 5\pi }{ 20 } \) |
sin x | 0 | 0.15643 | 0.30902 | 0.45399 | 0.5877 | 0.70711 |
x | \(\frac { 6\pi }{ 20 } \) | \(\frac { 7\pi }{ 20 } \) | \(\frac { 8\pi }{ 20 } \) | \(\frac { 9\pi }{ 20 } \) | \(\frac { 10\pi }{ 20 } \\ \) | |
sin x | 0.80902 | 0.89101 | 0.95106 | 0.9876 | 1 |
Using trapezoidal rule,
\(\int _{ 0 }^{ \pi /2 }{ sinx\quad dx } =\frac { h }{ 2 } [({ y }_{ 0 }+{ y }_{ 10 })+2({ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }+{ y }_{ 5 }+{ y }_{ 6 }+{ y }_{ 7 }+{ y }_{ 8 }+{ y }_{ 9 })]\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { \pi }{ 40 } [12.70624]=0.99795\)
If y1 = 4, y3 = 12, y4 = 19 and yx = 7, then x will be
- (a)
1.42
- (b)
1.68
- (c)
1.86
- (d)
1.98
\(x=\frac { (-5)(-12) }{ (-8)(-15) } (1)+\frac { (3)(-12) }{ (8)(-7) } (3)+\frac { (3)(-5) }{ (15)(7) } (4)\\ \quad =\quad \frac { 1 }{ 2 } +\frac { 27 }{ 14 } -\frac { 4 }{ 7 } =1.86\)
The Newton-Raphson iteration
\({ x }_{ n+1 }=\frac { { x }_{ n } }{ 2 } +\frac { 3 }{ 2{ x }_{ n } }\)
can be used to solve the equation
- (a)
X2 = 3
- (b)
X3 = 3
- (c)
X2 = 2
- (d)
X3 = 2
We Know,
Newton-Raphson Method
xn+1=xn\(-\frac { f({ x }_{ n }) }{ { f }^{ ' }({ x }_{ n }) }\)
From option(a),x2=3
x2 - 3 = 0
f(x)= x2-3
f'(x)= 2x2
Now, \(xn+1 = xn- \frac { { x }_{ n }-3 }{ 2{ x }_{ n } }\)
\(= { x }_{ n }-\frac { { x }_{ n }^{ 2 }-3 }{ 2{ x }_{ n } }\)
\(= \frac { { 2x }_{ n }^{ 2 }-{ x }_{ n }^{ 2 }+3 }{ 2{ x }_{ n } }\)
\({ x }_{ n+1 }=\frac { { x }_{ n }^{ 2 } }{ 2 } +\frac { 3 }{ 2{ x }_{ n } }\)
So, this is right answer.
A real root of equation \({ x }^{ 3 }+{ x }^{ 2 }-1=0 \) by iteration (method of successive approximation) method is
- (a)
0.7548765
- (b)
0.7548756
- (c)
0.7548776
- (d)
0.7548764
Using bisection method, the negative root of \({ x }^{ 3 }-4x+9=0\) correct to three decimal places is
- (a)
-2.506
- (b)
-2.706
- (c)
-2.406
- (d)
None of these
A real root of the equation x-cos x = 0 by the method false position correct to four decimal places is
- (a)
0.7391
- (b)
0.7439
- (c)
0.7347
- (d)
None of these
Using Newton-Raphson method, a root correct to three decimal places of the equation \({ e }^{ x }=1+2x \) is
- (a)
1.256
- (b)
1.255
- (c)
1.286
- (d)
None of these
A root of the equation x3-x-11 = 0 correct to four decimals using bisection method, is
- (a)
2.3737
- (b)
2.3838
- (c)
2.3736
- (d)
None of these
Using bisection method the negative of x3 - x + 11 = 0 is
- (a)
-2.3736
- (b)
-2.3838
- (c)
-2.3737
- (d)
None of these
Which of the following statements applies to the bisection method used for finding roots of functions?
- (a)
coverage within a few iteration
- (b)
guaranteed to work for all continuous functions
- (c)
it is faster than the Newton-Raphson method
- (d)
requires that there will be no error in determining the sign of the function
Bisection method is used for finding the roots of continuous functions. So, option (b) is right.
The Newton-Raphson method is used to find the root of the equation x2 - 2 = 0, if iterations are started from -1, the iterations will be
- (a)
converged to -1
- (b)
converged to \(\sqrt { 2 } \)
- (c)
converged to - \(\sqrt { 2 } \)
- (d)
not converged
Newton-Raphson iteration formula for finding \(\sqrt [ 3 ]{ C } \) where C > 0, is
- (a)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 3 }_{ n }+\sqrt [ 3 ]{ C } }{ 3{ x }^{ 2 }_{ n } } \)
- (b)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 3 }_{ n }-\sqrt [ 3 ]{ C } }{ 3{ x }^{ 2 }_{ n } } \)
- (c)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 2 }_{ n }+C }{ 3{ x }^{ 2 }_{ n } } \)
- (d)
\({ x }_{ n+1 }=\frac { 2{ x }^{ 2 }_{ n }-C }{ 3{ x }^{ 2 }_{ n } } \)
The iteration formula to find square root of a positive real number b using the Newton-Raphson method is
- (a)
\({ x }_{ k+1 }=\frac { 3({ x }_{ k }+b) }{ 2{ x }_{ k } } \)
- (b)
\({ x }_{ k+1 }=\frac { ({ x^{ 2 } }_{ k }+b) }{ 2{ x }_{ k } } \)
- (c)
\({ x }_{ k+1 }=\frac { { x }_{ k }-2{ x }_{ k } }{ ({ x^{ 2 } }_{ k }+b) } \)
- (d)
None of these
Using Newton-Raphson method, a root correct to three decimal places of the equation x3 - 3x -5 = 0 is
- (a)
2.275
- (b)
2.279
- (c)
2.222
- (d)
None of these
Give, f(x) = x3 - 3x - 5 = 0
f(2) = -3 (-ve)
f(3) = 13 (+ve)
Therefore, a real root lies between 2 and 3
f'(x) = 3x2 - 3
Let, the initial approximaqtion be x0 = 2
f'(2) = 12-3 = 9
By Newton-Raphson method
\({ x }_{ 1 }={ x }_{ 0 }-\frac { f({ x }_{ 0 }) }{ f'({ x }_{ 0 }) } =2.333333\\ { x }_{ 2 }={ x }_{ 1 }-\frac { f({ x }_{ 1 }) }{ f'({ x }_{ 1 }) } =2.280555\\ { x }_{ 3 }={ x }_{ 2 }-\frac { f({ x }_{ 2 }) }{ f'({ x }_{ 2 }) } =2.279020\\ { x }_{ 4 }={ x }_{ 3 }-\frac { f({ x }_{ 3 }) }{ f'({ x }_{ 3 }) } =2.279019\)
The correct root is 2.279019.
A real root of equation x3 -5x -7 = 0 by the method of false position correct to three decimal places is
- (a)
2.7472
- (b)
2.0844
- (c)
2.0774
- (d)
None of these
We have
\(f(x)={ x }^{ 3 }-5x-7=0\\ f(2)=8-10=-9\\ f(3)=27-15-7=+5\)
As f(2) and f(3) are of opposite signs. So, root lies between 2 and 3
\({ x }_{ 1 }=\frac { af(b)-b\quad f(a) }{ f(b)-f(a) } \\ { \quad x }_{ 1 }=\frac { 2(5)-3(-9) }{ 5-(-9) } =\frac { 37 }{ 14 } =2.642\\ \quad f(2.64209)=-1.7541\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -ve\\ and\quad f(3)=5\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +ve\quad \)
\(\quad \quad \therefore \quad \) Root lies between 2.6429 and 3
a = 2.6429, b = 3
x2 = 2.7356
f(2.7356) = - 0.2061 -ve
f(3) = 5 +ve
\(\quad \quad \therefore \quad \) Root les between 2.7356 and 3
a = 2.7356, b = 3
x3 = 2.7461
f(2.7461) = -0.02198 -ve
f(3) = 5 +ve
\(\quad \quad \therefore \quad \) Root lies between 2.7461 and 3
a = 2.7461, b = 3
x4 = 2.7472
Hence, the root of the given equation is 2.7472.