Numerical Methods

Question 1
Consider the following recursive iteration scheme for different values of variable P with the initial guess x_1=1:
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right ),\;\;\;n=1,2,3,4,5
For P=2,x_5 is obtained to be 1.414, rounded-off to three decimal places. For P=3,x_5 is obtained to be 1.732, rounded-off to three decimal places.
If P=10, the numerical value of x_5 is _________ . (round off to three decimal places)
A
2.155
B
3.162
C
1.125
D
4.568
GATE CE 2022 SET-1   Engineering Mathematics
Question 1 Explanation: 
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right )
Converges when x_{n+1}=x_{n}=\alpha
\begin{aligned} \alpha &=\frac{1}{2}\left ( \alpha+\frac{P}{\alpha} \right )\\ \frac{\alpha}{2}&=\frac{P}{2\alpha}\\ \alpha&=\sqrt{P} \end{aligned}
When P=2, x_5=\sqrt{2}=1.4124
When P=3, x_5=\sqrt{3}=1.732
When P=10, x_5=\sqrt{10}=3.162
Question 2
Numerically integrate, f(x)=10 x-20 x^{2} from lower limit a=0 to upper limit b=0.5. Use Trapezoidal rule with five equal subdivisions. The value (in units,round off to two decimal places) obtained is ____________
A
0.78
B
0.65
C
0.4
D
0.56
GATE CE 2021 SET-2   Engineering Mathematics
Question 2 Explanation: 
\begin{aligned} y&=10 x-20 x^{2} \\ a&=0, b=0.5, n=5 \\ \text { So, } \qquad \qquad \qquad h&=\frac{b-a}{n}=0.1 \end{aligned}
And

\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
Question 3
The value of \int_{0}^{1} e^{x} d x using the trapezoidal rule with four equal subintervals is
A
1.718
B
1.727
C
2.192
D
2.718
GATE CE 2021 SET-1   Engineering Mathematics
Question 3 Explanation: 
Let \int_{a}^{b}f(x)dx=\int_{0}^{1}e^x \; dx \text{ and }n=4
Then a=0,b=1,f(x)=e^x \text{ and }h=\frac{b-a}{n}=\frac{1-0}{4}=0.25
\begin{array}{|c|c|c|c|c|c|} \hline x&0&0.25&0.50&0.75&1\\ \hline y=f(x=e^x)&1&1.284&1.649&2.117&2.718\\ \hline \end{array}
The formula of trapezoidal rule to the given data is given by
\int_{a}^{b}f(x)dx\simeq \int_{a}^{b}p(x)dx=\frac{h}{2}\left \{ (y_0+y_n)+2(y_1+y_2+y_3) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (1+2.718)+2(1.284+1.640+2.117) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (3.718)+2(5.03) \right \}=1.727
Question 4
The integral
\int_{0}^{1}(5x^3+4x^2+3x+2)dx

is estimated numerically using three alternative methods namely the rectangular, trapezoidal and Simpson's rules with a common step size. In this context, which one of the following statement is TRUE?
A
Simpsons rule as well as rectangular rule of estimation will give NON-zero error.
B
Simpson's rule, rectangular rule as well as trapezoidal rule of estimation will give NON-zero error.
C
Only the rectangular rule of estimation will given zero error.
D
Only Simpson's rule of estimation will give zero error.
GATE CE 2020 SET-2   Engineering Mathematics
Question 4 Explanation: 
Because integral is a polynomial of 3rd degree so Simpson's rule will give error free answer.
Question 5
The value of the function f(x) is given at n distinct values of x and its value is to be interpolated at the point x^*, using all the n points. The estimate is obtained first by the Lagrange polynomial, denoted by I_L, and then by the Newton polynomial, denoted by I_N. Which one of the following statements is correct?
A
I_L is always greater than I_N
B
I_L \; and \; I_N are always equal
C
I_L is always less than I_N
D
No definite relation exists between I_L \; and \; I_N
GATE CE 2019 SET-2   Engineering Mathematics
Question 6
The quadratic equation 2x^{2}-3x+3=0 is to be solved numerically starting with an initial guess as x_{0}=2 . The new estimate of x after the first iteration using Newton-Raphson method is ______
A
0
B
1
C
2
D
3
GATE CE 2018 SET-2   Engineering Mathematics
Question 6 Explanation: 
Given \begin{aligned} f\left ( x \right )&=2x^{2}-3x+3, x_{0}&=2 \\ {f}'\left ( x \right )&=4x-3 \end{aligned}
By Newton-Rapshon
\begin{aligned} x_{1}&=x_{0}-\frac{f\left ( x_{0} \right )}{{f}'\left ( x_{0} \right )} \\ &=2-\frac{2\left ( 2 \right )^{2}-3\left ( 2 \right )+3}{4\left ( 2 \right )-3} \\ &=2-\frac{5}{5}=1 \end{aligned}
Question 7
Consider the equation \frac{\mathrm{d} u}{\mathrm{d} t}=3t^{2}+1 with u=0 at t=0. This is numerically solved by using the forward Euler method with a step size \Delta t= 2. The absolute error in the solution at the end of the first time step is_________
A
0.25
B
6
C
4
D
8
GATE CE 2017 SET-1   Engineering Mathematics
Question 7 Explanation: 
\begin{aligned} \frac{\mathrm{d} u}{\mathrm{d} t}&=3t^{2}+1 \\ f\left ( u,\: t \right )&=3t^{2}+1 \\ u_{0}&=0 \\ t_{0}&=0 \\ \Delta t&=2\\ &\text{By Euler's method} \\ u_{1}&=u_{0}+hf\left ( u_{0}t_{0} \right ) & t_{1}&=t_{0}+h \\ &=u_{0}+h\left ( 3t^{2}_{0}+1 \right ) & &=0+2 \\ &=0+2\left ( 3\left ( 0 \right )^{2}+1 \right ) & &=2 \\ &=2 \end{aligned}
After first iteration u=0 when t=2,
\begin{aligned} \frac{\mathrm{d} u}{\mathrm{d} t}&=3t^{2}+1 \\ du&=\left ( 3t^{2}+1 \right )dt \\ \int du&=\int_{0}^{2}\left ( 3t^{2}+1 \right )dt \\ u&=\left.\begin{matrix} \left ( 3\frac{t^{3}}{3}+t \right ) \end{matrix}\right|_{0}^{2} \\ &=8+2=10 \end{aligned}
Absolute error = Exact value - approx value
=10-2 =8
Question 8
Newton-Raphson method is to be used to find root of equation 3x-e^{x}+\sin x=0. If the initial trial value for the root is taken as 0.333, the next approximation for the root would be _________
A
0.33
B
0.54
C
0.36
D
0.76
GATE CE 2016 SET-1   Engineering Mathematics
Question 8 Explanation: 
According to Newton-Raphson Method:
\begin{aligned} x_{N+1}&=X_{N}-\frac{f\left ( X_{N} \right )}{{F}'\left ( X_{N} \right )} \\ f\left ( x \right )&=3x-e^{x}+\sin x \\ {f}'\left ( x \right )&=3-e^{x}+\cos x \\ \Rightarrow \;\; X_{1}&=X_{0}-\frac{f\left ( 0.333 \right )}{{f}'\left ( 0.333 \right )} \\ &=0.333-\frac{3\times 0.333-e^{0.333}+\sin 0.333}{3-e^{0.333}+\cos 0.333} \\ \therefore \;\; X_{1}&=0.36 \end{aligned}
Question 9
For step-size, \Delta x=0.4, the value of following integral using Simpson's 1/3 rule is _________.
\int_{0}^{0.8}(0.2+25x-200x^{2}+675x^{3}-900x^{4}+400x^{5})dx
A
0.64
B
0.86
C
1.36
D
1.64
GATE CE 2015 SET-2   Engineering Mathematics
Question 9 Explanation: 
a=0, b=0.8, \Delta x=0.4
f\left ( x \right )=0.2+25x-200x^{2}+675x^{3}-900x^{4}+400x^{5}



By Simpson's 1/3 Rule,
\begin{aligned} y\left ( x \right )&=\int_{0}^{0.8}\left ( 0.2+25x-200x^{2}+675x^{3}-900x^{4}+400x^{5} \right )dx \\ &=\frac{4}{3}\left [ y_{0}+4y_{1}+y_{2} \right ] \\ y_{0}&=y\left ( 0 \right )=0.2 \\ y_{1}&=y\left ( 0.4 \right )=2.456 \\ y_{2}&=y\left ( 0.8 \right )=0.232 \\ y\left ( n \right )&=\frac{0.4}{3}\left ( 0.2+4\times 2.456+0.232 \right ) \\ &=1.367 \end{aligned}
Question 10
In Newton-Raphson iterative method, the initial guess value ( x_{ini}) is considered as zero while finding the roots of the euation:f(x)=-2+6x-4x^{2}+0.5x^{3}. The correction, \Delta x, to be added to x_{ini} in the first iteration is____________.
A
0.5
B
0.33
C
0.8
D
0.2
GATE CE 2015 SET-2   Engineering Mathematics
Question 10 Explanation: 
\begin{aligned} f\left ( x \right )&=-2+6x-4x^{2}+0.5x^{3} \\ {f}'\left ( x \right )&=6-8x+1.5x^{2} \\ x_{ini}&=0\\ \text{By Newton }&\text{Raphson Method,}\\ x_{1}&=x_{ini}-\frac{f\left ( x_{ini} \right )}{{f}'\left ( x_{ini} \right )} \\ &= 0-\frac{-2}{6} \\ \Rightarrow \;\;x_{1}&=\frac{1}{3} \\ \therefore \;\; \Delta x&=x_{1}-x_{ini}=\frac{1}{3} \end{aligned}
There are 10 questions to complete.

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