Mathematics: Numerical Analysis: CSIR-NET Previous Year Solved problems.

1.Let $f(x)=\sqrt{x+3}$ for $x\geq -1$ consider the iteration $x_{n+1}=f(x_n),x_0=0;n\geq 0$.

The possible limits of the iterations are:

• -1
• 3
• 0
• $\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}}$

Solution: 

It is given that,

$x_{n+1}=f(x_n),x_0=0;n\geq 0$.

So,

$x_1=f(x_0)=\sqrt{3}$

$x_2=f(x_1)=\sqrt{\sqrt{3}+3}=\sqrt{3+\sqrt{3}}$

$x_3=f(x_2)=\sqrt{3+\sqrt{3+\sqrt{3}}}$

Proceeding in same way..

we get limit:

$\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}}$

So, 4th option is correct.

2. The iteration $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n}),n\geq 0$ for a given $x_0\neq 0$ is an instance of

• Fixed point iteration for $f(x)=x^2-2$
• Newton's method for $f(x)$
• Fixed point iteration for $f(x)=\frac{x^2+2}{2x}$
• Newton's method for $f(x)=x^2+2$

Solution:

Recall,

Fixed point iteration method,

The first step in this method is to rewrite the equation $f(x)=0$ as $x=g(x)$

starting from  $x=x_0$ we find the the first approximation on $x_1$ as $x_1=g(x_0)$.

then we find $x_2=g(x_1)$

we can see here, similar situation is here,

comparing with $x=g(x)$

we get,

$g(x)= \frac{1}{2}\times (x+\frac{2}{x})$

So, 3rd option is correct.

Now Recall, newton's method,

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

so, applying this on $x^2-2$ we get the same form described in the question.

So, 2nd option is correct.

3. The following numerical integration formula is exact for all polynomials of degree less than or

equal to 3.

• Trapezoidal Rule
• Simpson's $\frac{1}{3}$rd rule
• Simpson's $\frac{3}{8}$rd rule
• Gauss -Legendre 2 point formula

Solution:

option 2 is correct:

The truncation error for the Simpson method is proportional to the $f^4(x)$. If the integrand is cubic, then the fourth order derivative is zero. So the quadrature will be exact for this case.

Option 3 is correct:

This method is more accurate than previous.

Option 4 is correct:

Error for this method is:

$\frac{(b-1)^{2n+1}(n!)^4}{(2n+1)[(2n)!]^3}f^{(2n)}(\xi)$          $a<\xi<b$

So, Here $n=2$, so, Sice fourth derivative is zero, so this method is exact for cubic polynomial.

4. Let $f:\mathbb{R}\to \mathbb{R}$ be a smooth function with non-vanishing derivative. the Newton's method's for finding a root of $f(x)=0$ is same as

• fixed point iteration for the map $g(x)=x-f(x)/f'(x)$
• Forward Euler method with unit step length for the differential equation $\frac{dy}{dx}+\frac{f(y)}{f'(y)}=0$
• Fixed point iteration for $g(x)=x+f(x)$
• Fixed point iteration for $g(x)=x-f(x)$

Solution:

Newton's method is:

$x_{n+1}=x_n-\frac{f(x)}{f'(x)}$

So fixed point iteration will be the same.

So, 1st option is correct.

Euler's method:

For,

$\frac{dy}{dx}=f(x,y)$

we have,

$y_{n+1}=y_n+h(f(x_n,y_n))$

here $h=1$

So, we get,

$y_{n+1}=y_n+(f(x_n,y_n))$

here $f(x,y)=\frac{f(y)}{f'(y)}=0$

$\Rightarrow y_{n+1}=y_n+1\times (-\frac{f(y)}{(f'(y)})$

So, 2 option is also correct.

5. Consider the function $f(x)=\sqrt{2+x}$ for $x\geq -1$ and the iteration $x_{n+1}=f(x_n); n\geq 0$ for $x_0=1$ what are the possible limits of the iteration?

• -1
• 2
• 1
• $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$

Solution:

It is given that,

$x_{n+1}=f(x_n),x_0=1;n\geq 0$.

So,

$x_1=f(x_0)=\sqrt{3}$

$x_2=f(x_1)=\sqrt{\sqrt{3}+2}=\sqrt{2+\sqrt{3}}$

$x_3=f(x_2)=\sqrt{2+\sqrt{2+\sqrt{3}}}$

Proceeding in same way..

we get limit:

$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$

4th option is correct.

Note: In CSIR official Key $2$ is also an answer.(I don't know how!) 

6.  The iteration $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n}),n\geq 0$ for a given $x_0\neq 0$ is an instance of

• $x_n$ converges to $\sqrt{2}$ with rate of convergence $1$.
• $x_n$ converges to $\sqrt{2}$ with rate of convergence $2$
• Fixed point iteration for $f(x)=x^2-2$
• Newton's method for $f(x)=x^2-2$

Solution:

See question number $2$.

4th option is correct.

$x_n$ converges to $\sqrt{2}$ , It is evident for the function $f(x)=x^2-2$

Newton Raphson Method is said to have quadratic convergence.

So, 2nd option is correct.

7. which of the following approximations for estimating the derivative of a smooth function $f$ at a point $x$ is of order $2$. (that is the error term is $O(h^2$.

• $f'(x)\approx \frac{f(x+h)-f(x)}{h}$
• $f'(x)\approx \frac{f(x+h)-f(x-h)}{2h}$
• $f'(x)\approx \frac{3f(x)-4f(x-h)+f(x-2h)}{2h}$
• $f'(x)\approx \frac{-3f(x)+4f(x+h)-f(x+2h)}{h}$

solution:

Using forward difference formula, we get

$hf'_k=\frac{1}{2}(-f_{k+2}+4f_{k+1}-3f_k)$

$\Rightarrow f'(x)=\frac{1}{2h}(-3f(x)+4f(x+h)-f(x+2h))$

So, $4$ is correct option.

8.Let $f\in C^3([x_{-1},x_1])$ where $x_{-1}=x_0-h,x_1=x_0+h$with $h\geq 0$$f(x_0)=f_0, f(x_j)=f_j$ for $j=-1,1$ and $f'(x_0)=f'_0$

then for some $\epsilon \in (x_{-1},x_1)$ we have,

• $f'_0=\frac{f_1-f_0}{h}-\frac{h^2}{2} f'''(\epsilon)$
• $f'_0=\frac{f_1-f_{-1}}{h}-\frac{h^3}{3} f'''(\epsilon)$
• $f'_0=\frac{f_1-f_{-1}}{2h}-\frac{h^2}{6} f'''(\epsilon)$
• $f'_0=\frac{f_1-f_{-1}}{2h}-\frac{h^3}{6} f'''(\epsilon)$

Solution:

This question is similar to above question.

3rd option is correct.