a) Set up Newton- Raphson's sheme of finding the square root of a positive number N. Hence evaluate 12 to four decimal places.


b) Given that f(0)=1, (1)=3, (3)=55.find the unique polynomial of degree 2 or less which fits the given data. Hence evaluate (2)

Let's break down the problems:

### Part (a): Newton-Raphson Scheme for Finding the Square Root of a Positive Number \( N \)
The Newton-Raphson method is an iterative method used to find successively better approximations to the roots (or zeroes) of a real-valued function. To find the square root of a positive number \( N \), we can apply the Newton-Raphson method as follows:

1. **Set up the equation**: To find \( \sqrt{N} \), we solve the equation \( f(x) = x^2 - N = 0 \).

2. **Apply Newton-Raphson iteration**:
   The iteration formula is given by:
   \[
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   \]
   For \( f(x) = x^2 - N \), the derivative \( f'(x) \) is \( 2x \). Therefore, the iteration formula becomes:
   \[
   x_{n+1} = x_n - \frac{x_n^2 - N}{2x_n}
   \]
   Simplifying this, we get:
   \[
   x_{n+1} = \frac{1}{2} \left(x_n + \frac{N}{x_n}\right)
   \]
   This is the Newton-Raphson iteration for finding the square root of \( N \).

3. **Evaluate \( \sqrt{12} \) to four decimal places**:
   Start with an initial guess \( x_0 \), for example, \( x_0 = 3.5 \) (since \( \sqrt{12} \approx 3.464 \)).

   Iterating:
   \[
   x_1 = \frac{1}{2} \left(3.5 + \frac{12}{3.5}\right) = 3.4643
   \]
   \[
   x_2 = \frac{1}{2} \left(3.4643 + \frac{12}{3.4643}\right) \approx 3.4641
   \]
   \[
   x_3 = \frac{1}{2} \left(3.4641 + \frac{12}{3.4641}\right) \approx 3.4641
   \]
   So, \( \sqrt{12} \approx 3.4641 \) (correct to four decimal places).

### Part (b): Finding the Unique Polynomial of Degree 2 or Less
Given points are \( f(0) = 1 \), \( f(1) = 3 \), and \( f(3) = 55 \).

1. **General form of the polynomial**:
   A polynomial of degree 2 can be written as:
   \[
   f(x) = ax^2 + bx + c
   \]

2. **Set up the system of equations**:
   Using the given points:
   - For \( x = 0 \): \( f(0) = c = 1 \) 
   - For \( x = 1 \): \( f(1) = a(1)^2 + b(1) + c = a + b + 1 = 3 \)
   - For \( x = 3 \): \( f(3) = a(3)^2 + b(3) + c = 9a + 3b + 1 = 55 \)

   Simplifying these, we get:
   \[
   a + b = 2 \quad \text{(Equation 1)}
   \]
   \[
   9a + 3b = 54 \quad \text{(Equation 2)}
   \]

3. **Solve the system of equations**:
   - From Equation 1: \( b = 2 - a \).
   - Substitute \( b = 2 - a \) into Equation 2:
     \[
     9a + 3(2 - a) = 54
     \]
     Simplifying:
     \[
     9a + 6 - 3a = 54 \implies 6a = 48 \implies a = 8
     \]
   - Substituting \( a = 8 \) into Equation 1: \( b = 2 - 8 = -6 \).

   So, \( a = 8 \), \( b = -6 \), and \( c = 1 \).

   Therefore, the polynomial is:
   \[
   f(x) = 8x^2 - 6x + 1
   \]

4. **Evaluate \( f(2) \)**:
   \[
   f(2) = 8(2)^2 - 6(2) + 1 = 32 - 12 + 1 = 21
   \]

Thus, \( f(2) = 21 \).

---

Would you like further details or have any questions? 

Here are some related questions:

1. How would you generalize Newton-Raphson's method for other functions?
2. How do you determine a good initial guess for the Newton-Raphson method?
3. Can the Newton-Raphson method fail? If so, under what circumstances?
4. How would you extend the polynomial interpolation method to a cubic polynomial?
5. What is the significance of using a second-degree polynomial for interpolation?

**Tip:** When using Newton-Raphson, always verify that the function and its derivative are well-behaved near the root to ensure convergence.

Learn how to use the Newton-Raphson method to find the square root of a positive number and polynomial interpolation to find a unique polynomial of degree 2 or less. Includes step-by-step solutions and examples. Suitable for advanced high school students and beyond.

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