BCS-054 (Computer Oriented Numerical Techniques)

     BCS-054 Computer oriented Numerical techniques (Assignments Solution)

    Course Code                        :            BCS-054

    Course Title                         :            Computer oriented Numerical techniques 

    Assignment Number            :           BCA(V)054/Assignment/2024-25 

    Maximum Marks                 :             100

    Weightage                            :            25% 

    Last Date of Submission      :           31stOctober,2024(For July, Session) 

                                                               30thApril, 2025(For January, Session)

(a) Explain each of the following concepts, along with at least one suitable example for each: (i) Fixed-point number representation (ii) round-off error (iii) representation of zero as floating point number (iv) significant digits in a decimal number representation (v) normalized representation of a floating point number (vi) overflow 

Ans.

(i) Fixed-point number representation

Fixed-point representation is a way to store numbers where the decimal (or binary) point is fixed at a specific position. In this system, numbers are represented with a fixed number of digits before and after the decimal point, which makes it suitable for representing integers or fractions where the precision is predetermined.

• Example: Consider a 4-digit fixed-point decimal system with 2 digits reserved for fractional values. In this case, the number 123.45 would be stored as 12345, and the system would interpret it as 123.45. Similarly, the number 5.67 would be stored as 567, interpreted as 5.67.

(ii) Round-off error

Round-off error occurs when a number cannot be represented exactly in a given number format, so it is approximated. This is common in floating-point arithmetic, where some numbers (e.g., irrational numbers or certain decimals) cannot be represented with full precision.

• Example: When representing the decimal number 0.1 in binary floating-point, it becomes a repeating binary fraction (0.0001100110011...). Since the computer cannot store this infinite sequence, it approximates it to a fixed number of bits, resulting in a round-off error. If you perform calculations based on this approximation, the results may not be exactly correct.

(iii) Representation of zero as floating point number

In floating-point representation (e.g., IEEE 754), zero can be represented as both positive zero (+0) and negative zero (-0). These two values are numerically equal but differ in their sign bit, and they can behave differently in certain computations (e.g., in division or underflow scenarios).

• Example: In IEEE 754 single-precision format, positive zero is represented by a sign bit of 0, with both exponent and fraction set to 0. Negative zero is represented with the sign bit as 1, and exponent and fraction as 0.

  • Positive zero: +0 → 0 00000000 00000000000000000000000
  • Negative zero: -0 → 1 00000000 00000000000000000000000

(iv) Significant digits in a decimal number representation

Significant digits are the meaningful digits in a number, starting from the first non-zero digit. They convey the precision of the number, which is important in scientific and computational contexts.

• Example: In the number 0.00567, the significant digits are 5, 6, and 7. Therefore, this number has 3 significant digits. Similarly, the number 123.400 has 6 significant digits because the trailing zeros after the decimal are considered significant.

(v) Normalized representation of a floating point number

In floating-point arithmetic, a normalized number is one in which the leading digit of the significand (mantissa) is non-zero. This ensures that the representation of numbers is unique and maximizes precision for a given number of bits.

• Example: In a binary floating-point system, the number 1.23 × 10^3 would be stored as a normalized floating-point number as 1.23 (mantissa) and 3 (exponent). The leading 1 ensures that the representation is normalized. In binary, a normalized number typically has a significand that starts with a 1, e.g., 1.101 × 2^5.

(vi) Overflow

Overflow occurs when a calculation exceeds the maximum limit that can be stored in a given data type or number format. This typically results in incorrect results or system errors, as the value wraps around or saturates.

• Example: In an 8-bit signed integer representation, the range of representable numbers is from -128 to 127. If you try to add two numbers, such as 120 + 10 = 130, it results in an overflow since 130 is outside the allowable range. The result may "wrap around" and produce an incorrect negative number, depending on the system.

  For example, 120 + 10 might result in -126 due to overflow in an 8-bit

 Q2 - Explain with suitable example that in computer arithmatics ( i.e., numbers represented in computer, with +, −, *, / as implemented in a computer) the multiplication operation( *) may not be distributive over plus , i.e. may not be true for some computer numbers a, b and c

Ans. In computer arithmetic, especially when using floating-point numbers, the distributive property of multiplication over addition may not hold due to precision limitations and round-off errors. The distributive property states that:

$$a \times (b + c) = (a \times b) + (a \times c)$$

However, because floating-point numbers have limited precision, the results of arithmetic operations can be subject to small errors. These errors can cause the distributive property to fail.

Explanation with Example

Consider the following floating-point numbers (which are stored with limited precision):

• $a = 1$
• $b = 1 \times 10^{10}$ (a very large number)
• $c = -1 \times 10^{10} + 1$ (a slightly smaller number)

Now, let's test the distributive property with these numbers:

1. Left side of the distributive property:

   $$a \times (b + c) = 1 \times (1 \times 10^{10} + (-1 \times 10^{10} + 1)) = 1 \times 1 = 1$$

   This is correct because $b + c = 1$ when calculated exactly.

2. Right side of the distributive property:

   $$(a \times b) + (a \times c) = (1 \times 1 \times 10^{10}) + (1 \times (-1 \times 10^{10} + 1))$$

   When this is calculated in floating-point arithmetic:

   $$(1 \times 1 \times 10^{10}) = 1 \times 10^{10}, \quad (1 \times (-1 \times 10^{10} + 1)) = -1 \times 10^{10}$$

   Adding these two results:

   $$1 \times 10^{10} + (-1 \times 10^{10}) = 0$$

   Notice that the $+1$ in $c$ is too small to make a difference due to the limited precision of floating-point arithmetic. As a result, the right-hand side evaluates to $0$, not $1$.

(c) Find out to how many decimal places the value 22/ 7 is accurate as an approximation of 3.14159265, where the latter is value of ╥, calculated up to 8 places after decimal ? 

Ans. To determine how many decimal places the fraction 

$\frac{22}{7}$ is accurate as an approximation of $\pi$, calculated to $3.14159265$, we can follow these steps:

1. Calculate $\frac{22}{7}$ to decimal places:

   $$\frac{22}{7} = 3.142857142857...$$

2. Compare $\frac{22}{7}$ with $\pi$, calculated to 8 decimal places:

   $$\pi = 3.14159265$$

Now, let's compare the two values digit by digit up to the 8th decimal place.

• $\frac{22}{7} = 3.142857...$
• $\pi = 3.14159265$

Digit Comparison:

• First decimal place: $1$ (matches)
• Second decimal place: $4$ (matches)
• Third decimal place: $1$ vs. $2$ (mismatch)

(d) Calculate a bound for the truncation error in approximating f(x) = sin x by sin (x) = x − x 3 / (fact 3) + x5 / (fact 5), where −1 =< x =< 1 and (fact n) denotes factorial of n 

Ans. To calculate a bound for the truncation error in approximating 

$\sin(x)$ by the truncated Taylor series:

$$\sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!},$$

we need to analyze the error term in the Taylor series expansion of $\sin(x)$. The general Taylor series expansion of $\sin(x)$ around $x = 0$ is:

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$

Truncation Error:

The truncation error $E_n(x)$ when approximating a function by a Taylor series up to the term $x^n/n!$ is given by the next term in the series. For the approximation up to $x^5/5!$, the next term in the series is:

$$E(x) = \frac{x^7}{7!}.$$

Bound for the Truncation Error:

To find an upper bound for the truncation error on the interval $-1 \leq x \leq 1$, we evaluate the error term $\frac{x^7}{7!}$ for the maximum possible value of $|x|$ in the given interval. Since $|x| \leq 1$, the maximum value of $|x^7|$ is when $|x| = 1$.

$$E(x) \leq \frac{1^7}{7!} = \frac{1}{5040}.$$

$$(e)\; Obtain\; Approximate\; the\; value\; of\; (3.7)—1,\; using\; first\; three\; terms\; of\; Taylor’s\; series\; expansion.$$

$$Ans.$$To approximate 

$(3.7)^{-1}$ using the first three terms of Taylor's series expansion, we can start by expanding around a nearby point where we know the value of the function $f(x) = x^{-1}$ easily.

Let's expand around $x = 4$, since $\frac{1}{4} = 0.25$ is simple to calculate. Define:

$$f(x) = \frac{1}{x}$$

The Taylor series expansion of $f(x)$ around $x = a$ is given by:

$$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots$$

In this case, we'll expand around $a = 4$ and approximate $f(3.7)$.

Step 1: Compute the derivatives of $f(x)$

1. $f(x) = \frac{1}{x}$
2. $f'(x) = -\frac{1}{x^2}$
3. $f''(x) = \frac{2}{x^3}$

Step 2: Evaluate the derivatives at $x = 4$

1. $f(4) = \frac{1}{4} = 0.25$
2. $f'(4) = -\frac{1}{4^2} = -\frac{1}{16} = -0.0625$
3. $f''(4) = \frac{2}{4^3} = \frac{2}{64} = 0.03125$

Step 3: Write the Taylor expansion around $x = 4$

The Taylor expansion is:

$$f(3.7) \approx f(4) + f'(4)(3.7 - 4) + \frac{f''(4)}{2!}(3.7 - 4)^2$$

Substitute the values:

$$f(3.7) \approx 0.25 + (-0.0625)(3.7 - 4) + \frac{0.03125}{2}(3.7 - 4)^2$$

Step 4: Calculate the values

1. $3.7 - 4 = -0.3$
2. First term: $0.25$
3. Second term: $(-0.0625)(-0.3) = 0.01875$
4. Third term: $\frac{0.03125}{2}(-0.3)^2 = \frac{0.03125}{2} \times 0.09 = 0.00140625$

Step 5: Add the terms

Now sum the three terms:

$$f(3.7) \approx 0.25 + 0.01875 + 0.00140625 = 0.27015625$$

$$Q-2\; (a)\; Solve\; the\; system\; of\; equations\; 4\; x1+\; x2+\; 2X3\; =\; 16\; 2x1\; +\; 5x2\; +\; 3x3\; =\; 19\; 3x1+\; 2x2\; -\; x3\; =\; 12$$

$$Ans.$$We are given the system of equations:

$$\begin{aligned} 4x_1 + x_2 + 2x_3 &= 16 \\ 2x_1 + 5x_2 + 3x_3 &= 19 \\ 3x_1 + 2x_2 - x_3 &= 12 \end{aligned}$$

Let's solve this system using Gaussian elimination or matrix methods.

Step 1: Write the system in augmented matrix form

The system of equations can be represented as the following augmented matrix:

$$\begin{pmatrix} 4 & 1 & 2 & | & 16 \\ 2 & 5 & 3 & | & 19 \\ 3 & 2 & -1 & | & 12 \end{pmatrix}$$

Step 2: Perform Gaussian elimination

(i) Make the first entry in the first column a leading 1

To make the first row's first element 1, we can divide the first row by 4:

$$\begin{pmatrix} 1 & \frac{1}{4} & \frac{1}{2} & | & 4 \\ 2 & 5 & 3 & | & 19 \\ 3 & 2 & -1 & | & 12 \end{pmatrix}$$

(ii) Eliminate the first column elements below the first pivot

We want to eliminate the elements below the first pivot (1). Subtract 2 times the first row from the second row and subtract 3 times the first row from the third row:

• Row 2: $R_2 \rightarrow R_2 - 2 \times R_1$
• Row 3: $R_3 \rightarrow R_3 - 3 \times R_1$

$$\begin{pmatrix} 1 & \frac{1}{4} & \frac{1}{2} & | & 4 \\ 0 & \frac{19}{2} & 2 & | & 11 \\ 0 & \frac{5}{4} & -\frac{7}{2} & | & 0 \end{pmatrix}$$

(iii) Make the second entry in the second column a leading 1

To make the second row's second element a leading 1, multiply the second row by $\frac{2}{19}$:

$$\begin{pmatrix} 1 & \frac{1}{4} & \frac{1}{2} & | & 4 \\ 0 & 1 & \frac{4}{19} & | & \frac{22}{19} \\ 0 & \frac{5}{4} & -\frac{7}{2} & | & 0 \end{pmatrix}$$

(iv) Eliminate the second column elements below and above the pivot

• Eliminate the element above the pivot by subtracting $\frac{1}{4}$ times the second row from the first row: $R_1 \rightarrow R_1 - \frac{1}{4} R_2$
• Eliminate the element below the pivot by subtracting $\frac{5}{4}$ times the second row from the third row: $R_3 \rightarrow R_3 - \frac{5}{4} R_2$

$$\begin{pmatrix} 1 & 0 & \frac{5}{19} & | & \frac{54}{19} \\ 0 & 1 & \frac{4}{19} & | & \frac{22}{19} \\ 0 & 0 & -\frac{51}{38} & | & -\frac{55}{19} \end{pmatrix}$$

(v) Make the third entry in the third column a leading 1

Multiply the third row by $-\frac{38}{51}$ to make the third element a leading 1:

$$\begin{pmatrix} 1 & 0 & \frac{5}{19} & | & \frac{54}{19} \\ 0 & 1 & \frac{4}{19} & | & \frac{22}{19} \\ 0 & 0 & 1 & | & \frac{110}{51} \end{pmatrix}$$

(vi) Eliminate the third column elements above the pivot

• Eliminate the element above the pivot in the second row by subtracting $\frac{4}{19}$ times the third row from the second row.
• Eliminate the element above the pivot in the first row by subtracting $\frac{5}{19}$ times the third row from the first row.

After performing these operations:

$$\begin{pmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & \frac{110}{51} \end{pmatrix}$$

Step 3: Extract the solution

From the final matrix, we can read off the values of $x_1$, $x_2$, and $x_3$:

$$x_1 = 3, \quad x_2 = 2, \quad x_3 = \frac{110}{51} \approx 2.157$$

$$(b)$$ Perform four iterations (rounded to four decimal places) using (i) Jacobi Method and (ii) Gauss-Seidel method , for the following system of equations. 9 5 –5 –1 x1 –8 1 –4 1 x2 = –4 –2 1 –6 x3 – 18 With (0) x = (0, 0, 0)T . The exact solution is (1, 2, 3)T. Which method gives better approximation to the exact solution?

$$Ans.$$

1. Regula Falsi Method (False Position Method)

The Regula Falsi method is a root-finding method that combines the advantages of the bisection method and linear interpolation.

Step 1: Define the function

$$f(x) = x^2 \cos(x) + \sin(x)$$

Step 2: Choose two initial guesses $x_0$ and $x_1$ such that $f(x_0)$ and $f(x_1)$ have opposite signs (i.e., $f(x_0) \times f(x_1) < 0$).

Let’s start by testing values near zero:

$$f(0) = 0^2 \cos(0) + \sin(0) = 0$$ $$f(1) = 1^2 \cos(1) + \sin(1) = 0.5403 + 0.8415 = 1.3818$$ $$f(-1) = (-1)^2 \cos(-1) + \sin(-1) = 1 \times 0.5403 - 0.8415 = -0.3012$$

Since $f(-1) \times f(1) < 0$, we can use the interval $[-1, 1]$.

Step 3: Apply the Regula Falsi formula to compute the next approximation:

The formula for the Regula Falsi method is:

$$x_{\text{new}} = x_1 - \frac{f(x_1)(x_1 - x_0)}{f(x_1) - f(x_0)}$$

Let’s perform the iterations, rounding the results to three significant digits:

• Iteration 1:

$$x_{\text{new}} = 1 - \frac{f(1) \times (1 - (-1))}{f(1) - f(-1)} = 1 - \frac{1.3818 \times 2}{1.3818 - (-0.3012)} = 1 - \frac{2.7636}{1.6830} = -0.6423$$

• Update the interval to $[-1, -0.6423]$ because $f(-0.6423)$ has the same sign as $f(1)$.

Repeat this process until convergence.

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2. Newton-Raphson Method

Newton-Raphson is an iterative method where the next approximation is given by:

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

Step 1: Define the function and its derivative:

$$f(x) = x^2 \cos(x) + \sin(x)$$ $$f'(x) = 2x \cos(x) - x^2 \sin(x) + \cos(x)$$

Step 2: Choose an initial guess:

Start with $x_0 = 0.5$.

Step 3: Perform iterations:

• Iteration 1:

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.5 - \frac{0.5^2 \cos(0.5) + \sin(0.5)}{2(0.5)\cos(0.5) - 0.5^2 \sin(0.5) + \cos(0.5)}$$

Simplifying and continuing the iterations until $x_n$ converges to a value with three significant digits.

---

3. Bisection Method

The bisection method repeatedly halves an interval containing the root.

Step 1: Choose two initial guesses:

From earlier, we can use $x_0 = -1$ and $x_1 = 1$.

Step 2: Compute the midpoint:

$$x_{\text{mid}} = \frac{x_0 + x_1}{2}$$

Step 3: Update the interval:

Check the sign of $f(x_{\text{mid}})$ and update the interval accordingly.

Repeat this process until convergence to three significant digits.

---

4. Secant Method

The secant method is similar to Newton-Raphson but doesn't require the derivative.

Step 1: Choose two initial guesses:

We can use $x_0 = 1$ and $x_1 = 0.5$.

Step 2: Apply the secant method formula:

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

Step 3: Perform iterations:

Use the formula to iterate until the values converge to three significant digits.

Question 3. (a) Determine the smallest roots of the following equation: f(x) = x2 cos (x) + sin (x) =0 to three significant digits using (i) Regula-falsi method (ii) Newton Raphson method (iii) Bisection method (iv) Secant method 

Ans.

1. Regula Falsi Method (False Position Method)

The Regula Falsi method is a root-finding method that combines the advantages of the bisection method and linear interpolation.

Step 1: Define the function

$$f(x) = x^2 \cos(x) + \sin(x)$$

Step 2: Choose two initial guesses $x_0$ and $x_1$ such that $f(x_0)$ and $f(x_1)$ have opposite signs (i.e., $f(x_0) \times f(x_1) < 0$).

Let’s start by testing values near zero:

$$f(0) = 0^2 \cos(0) + \sin(0) = 0$$ $$f(1) = 1^2 \cos(1) + \sin(1) = 0.5403 + 0.8415 = 1.3818$$ $$f(-1) = (-1)^2 \cos(-1) + \sin(-1) = 1 \times 0.5403 - 0.8415 = -0.3012$$

Since $f(-1) \times f(1) < 0$, we can use the interval $[-1, 1]$.

Step 3: Apply the Regula Falsi formula to compute the next approximation:

The formula for the Regula Falsi method is:

$$x_{\text{new}} = x_1 - \frac{f(x_1)(x_1 - x_0)}{f(x_1) - f(x_0)}$$

Let’s perform the iterations, rounding the results to three significant digits:

• Iteration 1:

$$x_{\text{new}} = 1 - \frac{f(1) \times (1 - (-1))}{f(1) - f(-1)} = 1 - \frac{1.3818 \times 2}{1.3818 - (-0.3012)} = 1 - \frac{2.7636}{1.6830} = -0.6423$$

• Update the interval to $[-1, -0.6423]$ because $f(-0.6423)$ has the same sign as $f(1)$.

Repeat this process until convergence.

---

2. Newton-Raphson Method

Newton-Raphson is an iterative method where the next approximation is given by:

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

Step 1: Define the function and its derivative:

$$f(x) = x^2 \cos(x) + \sin(x)$$ $$f'(x) = 2x \cos(x) - x^2 \sin(x) + \cos(x)$$

Step 2: Choose an initial guess:

Start with $x_0 = 0.5$.

Step 3: Perform iterations:

• Iteration 1:

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.5 - \frac{0.5^2 \cos(0.5) + \sin(0.5)}{2(0.5)\cos(0.5) - 0.5^2 \sin(0.5) + \cos(0.5)}$$

Simplifying and continuing the iterations until $x_n$ converges to a value with three significant digits.

---

3. Bisection Method

The bisection method repeatedly halves an interval containing the root.

Step 1: Choose two initial guesses:

From earlier, we can use $x_0 = -1$ and $x_1 = 1$.

Step 2: Compute the midpoint:

$$x_{\text{mid}} = \frac{x_0 + x_1}{2}$$

Step 3: Update the interval:

Check the sign of $f(x_{\text{mid}})$ and update the interval accordingly.

Repeat this process until convergence to three significant digits.

---

4. Secant Method

The secant method is similar to Newton-Raphson but doesn't require the derivative.

Step 1: Choose two initial guesses:

We can use $x_0 = 1$ and $x_1 = 0.5$.

Step 2: Apply the secant method formula:

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

Step 3: Perform iterations:

Use the formula to iterate until the values converge to three significant digits.

Question 4. (a) Explain what is the role of interpolation in solving numerical problems?

Ans. 

1. Estimating Intermediate Values

In many scientific and engineering problems, data is often collected at specific intervals, but it may be necessary to estimate values at points between the collected data. Interpolation provides a way to approximate these unknown values using mathematical functions. For example:

• In weather forecasting, temperature or pressure may be known at specific times of the day, but we may need to estimate the values at times in between.
• In finance, if stock prices are known at specific times, interpolation can estimate prices between those times.

Example: If we know $f(1) = 2$ and $f(2) = 3$, interpolation can help estimate $f(1.5)$ using a method such as linear interpolation or polynomial interpolation.

2. Constructing Continuous Functions from Discrete Data

When data is discrete, it often doesn't provide a continuous picture of the system being analyzed. Interpolation creates a smooth function that approximates the behavior of the system over a continuous range. This is useful for:

• Numerical integration: Interpolation helps create a smooth function to integrate when only discrete data points are available.
• Numerical differentiation: When you have discrete points, interpolation can be used to estimate derivatives, which are defined only for continuous functions.

3. Reducing Errors in Numerical Methods

Interpolation helps in minimizing errors in numerical methods. For example:

• In root-finding methods (such as the Regula Falsi method or Secant method), interpolation is used to approximate where the function crosses the x-axis based on known values of the function at two points.
• In curve fitting, interpolation is used to estimate intermediate values to fit a curve that passes through or near the known data points.

4. Smooth Data Representation

In applications such as image processing, audio signal processing, and geographic information systems (GIS), interpolation provides a way to smooth data and fill in gaps:

• Image Processing: Interpolation techniques like bilinear or bicubic interpolation are used to resize images or enhance image quality.
• Audio Processing: In signal processing, interpolation is used to reconstruct missing samples in audio data.

5. Handling Nonlinear Systems

In nonlinear systems where analytical solutions might not be possible, interpolation provides an approximate solution. For instance, polynomial interpolation can be used to approximate solutions to differential equations by constructing polynomials that fit known points and estimating solutions between them.

6. Interpolation Methods

There are different types of interpolation techniques depending on the nature of the problem:

• Linear Interpolation: Used when the data points form a straight line or are approximately linear between two known points.
• Polynomial Interpolation: Uses higher-degree polynomials to approximate more complex curves between data points.
• Spline Interpolation: Uses piecewise polynomials to provide smoother approximations, especially useful in cases where the data points are non-linear and smoothness is required.

(b) Express Δ3 f1 as a backward difference

Ans.  To express 

$\Delta_3 f_1$ as a backward difference, we will use the concept of the third finite difference (i.e., applying the backward difference operator three times). Let's break it down step by step:

1. First Backward Difference:

The first backward difference $\nabla f_1$ is defined as:

$$\nabla f_1 = f_1 - f_0$$

2. Second Backward Difference:

The second backward difference $\nabla^2 f_1$ is the backward difference of the first backward difference:

$$\nabla^2 f_1 = \nabla f_1 - \nabla f_0$$

Substitute the first backward difference:

$$\nabla^2 f_1 = (f_1 - f_0) - (f_0 - f_{-1}) = f_1 - 2f_0 + f_{-1}$$

3. Third Backward Difference:

The third backward difference $\nabla^3 f_1$ is the backward difference of the second backward difference:

$$\nabla^3 f_1 = \nabla^2 f_1 - \nabla^2 f_0$$

Substitute the second backward difference:

$$\nabla^3 f_1 = (f_1 - 2f_0 + f_{-1}) - (f_0 - 2f_{-1} + f_{-2})$$

Simplify:

$$\nabla^3 f_1 = f_1 - 3f_0 + 3f_{-1} - f_{-2}$$

Conclusion:

The third backward difference $\nabla^3 f_1$ is expressed as:

$$\Delta_3 f_1 = f_1 - 3f_0 + 3f_{-1} - f_{-2}$$

This expression shows the third-order backward difference in terms of the values of the function at points $f_1, f_0, f_{-1},$ and $f_{-2}$.

(c) Express Δ3 f1 as a central difference.

Ans.  To express 

$\Delta_3 f_1$ as a central difference, we need to use the central difference operator. The central difference method takes into account values both ahead and behind the point of interest, rather than just the values before (backward difference) or after (forward difference).

The central difference operator $\delta$ is defined as:

$$\delta f_n = \frac{f_{n+1} - f_{n-1}}{2}$$

This gives the first-order central difference. For higher-order central differences, we repeatedly apply the operator.

Let’s go through the steps to find the third-order central difference $\Delta_3 f_1$:

1. First Central Difference:

The first central difference $\delta f_1$ is given by:

$$\delta f_1 = \frac{f_2 - f_0}{2}$$

2. Second Central Difference:

The second central difference $\delta^2 f_1$ is the central difference of the first central difference:

$$\delta^2 f_1 = \frac{\delta f_2 - \delta f_0}{2}$$

Substitute the first central difference:

$$\delta^2 f_1 = \frac{\left(\frac{f_3 - f_1}{2}\right) - \left(\frac{f_1 - f_{-1}}{2}\right)}{2}$$

Simplify:

$$\delta^2 f_1 = \frac{f_3 - 2f_1 + f_{-1}}{4}$$

3. Third Central Difference:

The third central difference $\delta^3 f_1$ is the central difference of the second central difference:

$$\delta^3 f_1 = \frac{\delta^2 f_2 - \delta^2 f_0}{2}$$

Substitute the second central difference:

$$\delta^3 f_1 = \frac{\left(\frac{f_4 - 2f_2 + f_0}{4}\right) - \left(\frac{f_2 - 2f_0 + f_{-2}}{4}\right)}{2}$$

Simplify:

$$\delta^3 f_1 = \frac{f_4 - 3f_2 + 3f_0 - f_{-2}}{8}$$

(d) For the following data develop difference table and find forward differences and backward differences  I xi yi 0 –1 16.8575 1 0 24.0625 2 1 16.5650 3 2 –13.9375 4 3 28.5625 5 4 144.0625

Ans. The data points are:

$i$	$x_i$	$y_i$
0	-1	16.8575
1	0	24.0625
2	1	16.5650
3	2	-13.9375
4	3	28.5625
5	4	144.0625

Step 1: Difference Table

The difference table is constructed by finding the first, second, third, and higher-order differences for the function values $y_i$. For the forward difference table, each difference is calculated as:

$$\Delta y_i = y_{i+1} - y_i$$

We’ll proceed to calculate each difference step-by-step.

First-order Differences ($\Delta^1 y_i$):

$$\Delta^1 y_0 = y_1 - y_0 = 24.0625 - 16.8575 = 7.2050$$ $$\Delta^1 y_1 = y_2 - y_1 = 16.5650 - 24.0625 = -7.4975$$ $$\Delta^1 y_2 = y_3 - y_2 = -13.9375 - 16.5650 = -30.5025$$ $$\Delta^1 y_3 = y_4 - y_3 = 28.5625 - (-13.9375) = 42.5000$$ $$\Delta^1 y_4 = y_5 - y_4 = 144.0625 - 28.5625 = 115.5000$$

Second-order Differences ($\Delta^2 y_i$):

$$\Delta^2 y_0 = \Delta^1 y_1 - \Delta^1 y_0 = -7.4975 - 7.2050 = -14.7025$$ $$\Delta^2 y_1 = \Delta^1 y_2 - \Delta^1 y_1 = -30.5025 - (-7.4975) = -23.0050$$ $$\Delta^2 y_2 = \Delta^1 y_3 - \Delta^1 y_2 = 42.5000 - (-30.5025) = 73.0025$$ $$\Delta^2 y_3 = \Delta^1 y_4 - \Delta^1 y_3 = 115.5000 - 42.5000 = 73.0000$$

Third-order Differences ($\Delta^3 y_i$):

$$\Delta^3 y_0 = \Delta^2 y_1 - \Delta^2 y_0 = -23.0050 - (-14.7025) = -8.3025$$ $$\Delta^3 y_1 = \Delta^2 y_2 - \Delta^2 y_1 = 73.0025 - (-23.0050) = 96.0075$$ $$\Delta^3 y_2 = \Delta^2 y_3 - \Delta^2 y_2 = 73.0000 - 73.0025 = -0.0025$$

Fourth-order Differences ($\Delta^4 y_i$):

$$\Delta^4 y_0 = \Delta^3 y_1 - \Delta^3 y_0 = 96.0075 - (-8.3025) = 104.3100$$ $$\Delta^4 y_1 = \Delta^3 y_2 - \Delta^3 y_1 = -0.0025 - 96.0075 = -96.0100$$

Fifth-order Difference ($\Delta^5 y_i$):

$$\Delta^5 y_0 = \Delta^4 y_1 - \Delta^4 y_0 = -96.0100 - 104.3100 = -200.3200$$

Difference Table

$i$	$x_i$	$y_i$	$\Delta^1 y_i$	$\Delta^2 y_i$	$\Delta^3 y_i$	$\Delta^4 y_i$	$\Delta^5 y_i$
0	-1	16.8575	7.2050	-14.7025	-8.3025	104.3100	-200.3200
1	0	24.0625	-7.4975	-23.0050	96.0075	-96.0100
2	1	16.5650	-30.5025	73.0025	-0.0025
3	2	-13.9375	42.5000	73.0000
4	3	28.5625	115.5000
5	4	144.0625

Step 2: Forward Differences

The forward differences for the given data are calculated from the top of the table, as shown in the table under the column $\Delta^1 y_i, \Delta^2 y_i, \Delta^3 y_i,$ etc.

Step 3: Backward Differences

For backward differences, we use the differences starting from the bottom of the table. Here's how you can compute the backward differences for the data:

First-order Backward Differences ($\nabla^1 y_i$):

$$\nabla^1 y_5 = y_5 - y_4 = 144.0625 - 28.5625 = 115.5000$$ $$\nabla^1 y_4 = y_4 - y_3 = 28.5625 - (-13.9375) = 42.5000$$ $$\nabla^1 y_3 = y_3 - y_2 = -13.9375 - 16.5650 = -30.5025$$ $$\nabla^1 y_2 = y_2 - y_1 = 16.5650 - 24.0625 = -7.4975$$ $$\nabla^1 y_1 = y_1 - y_0 = 24.0625 - 16.8575 = 7.2050$$

Second-order Backward Differences ($\nabla^2 y_i$):

$$\nabla^2 y_5 = \nabla^1 y_5 - \nabla^1 y_4 = 115.5000 - 42.5000 = 73.0000$$ $$\nabla^2 y_4 = \nabla^1 y_4 - \nabla^1 y_3 = 42.5000 - (-30.5025) = 73.0025$$ $$\nabla^2 y_3 = \nabla^1 y_3 - \nabla^1 y_2 = -30.5025 - (-7.4975) = -23.0050$$ $$\nabla^2 y_2 = \nabla^1 y_2 - \nabla^1 y_1 = -7.4975 - 7.2050 = -14.7025$$

Third-order Backward Differences ($\nabla^3 y_i$):

$$\nabla^3 y_5 = \nabla^2 y_5 - \nabla^2 y_4 = 73.0000 - 73.0025 = -0.0025$$ $$\nabla^3 y_4 = \nabla^2 y_4 - \nabla^2 y_3 = 73.0025 - (-23.0050) = 96.0075$$ $$\nabla^3 y_3 = \nabla^2 y_3 - \nabla^2 y_2 = -23.0050 - (-14.7025) = -8.3025$$

Fourth-order Backward Differences ($\nabla^4 y_i$):

$$\nabla^4 y_5 = \nabla^3 y_5 - \nabla^3 y_4 = -0.0025 - 96.0075 = -96.0100$$ $$\nabla^4 y_4 = \nabla^3 y_4 - \nabla^3 y_3 = 96.0075 - (-8.3025)$$

Question 6. (a) Find the values of the first and second derivatives of f(x) at x = 76 from the following table. Use 0(h2 ) forward difference method. Also, find Truncation Error (TE) and actual errors. 

Ans. 

Step 1: Set up the forward difference formulas

The forward difference method approximates derivatives by using differences between successive function values at regular intervals.

Formula for the first derivative using forward difference ($O(h^2)$):

The forward difference formula for the first derivative at $x_0$ is:

$$f'(x_0) \approx \frac{-3f(x_0) + 4f(x_1) - f(x_2)}{2h}$$

where $h$ is the step size, $f(x_0)$, $f(x_1)$, and $f(x_2)$ are function values at $x_0$, $x_1 = x_0 + h$, and $x_2 = x_0 + 2h$.

Formula for the second derivative using forward difference ($O(h^2)$):

The forward difference formula for the second derivative at $x_0$ is:

$$f''(x_0) \approx \frac{f(x_0) - 2f(x_1) + f(x_2)}{h^2}$$

Step 2: Data and step size

Let’s assume we have a table of values for $f(x)$ near $x = 76$. Here’s an example of a table:

$x$	$f(x)$
76	$f(76)$
78	$f(78)$
80	$f(80)$

Assume the values of $f(76)$, $f(78)$, and $f(80)$ are provided, and the step size $h = 2$ (since $x_1 = 76 + 2 = 78$ and $x_2 = 76 + 4 = 80$).

Step 3: Calculate the first and second derivatives

First Derivative:

Using the first derivative formula:

$$f'(76) \approx \frac{-3f(76) + 4f(78) - f(80)}{2h}$$

Substitute the values of $f(76)$, $f(78)$, and $f(80)$ and $h = 2$ to get the approximate first derivative.

Second Derivative:

Using the second derivative formula:

$$f''(76) \approx \frac{f(76) - 2f(78) + f(80)}{h^2}$$

Substitute the values of $f(76)$, $f(78)$, and $f(80)$ and $h = 2$ to get the approximate second derivative.

Step 4: Truncation Error (TE)

The truncation error for the forward difference method depends on the next higher-order terms ignored in the Taylor series expansion.

Truncation error for the first derivative:

For the first derivative, the truncation error is proportional to $h^2$:

$$TE_1 = -\frac{h^2}{3} f^{(3)}(\xi)$$

where $\xi$ is some point in the interval $[x_0, x_2]$ and $f^{(3)}(\xi)$ is the third derivative of the function at $\xi$.

Truncation error for the second derivative:

For the second derivative, the truncation error is proportional to $h^2$:

$$TE_2 = -\frac{h^2}{12} f^{(4)}(\xi)$$

where $f^{(4)}(\xi)$ is the fourth derivative of the function at $\xi$.

Step 5: Actual Error

To compute the actual error, you would compare the forward difference approximation with the true values of the first and second derivatives (which may be given or calculated analytically from $f(x)$).

$$\text{Actual Error} = \left| \text{True Value} - \text{Approximate Value} \right|$$

If the true values of the derivatives are provided or known, this step can be carried out directly.

Example:

Let’s assume the following hypothetical values for the function at $x = 76$, $x = 78$, and $x = 80$:

$x$	$f(x)$
76	12.0
78	15.5
80	19.0

First derivative calculation:

$$f'(76) \approx \frac{-3(12.0) + 4(15.5) - 19.0}{2(2)}$$ $$f'(76) \approx \frac{-36.0 + 62.0 - 19.0}{4} = \frac{7.0}{4} = 1.75$$

Second derivative calculation:

$$f''(76) \approx \frac{12.0 - 2(15.5) + 19.0}{(2)^2}$$ $$f''(76) \approx \frac{12.0 - 31.0 + 19.0}{4} = \frac{0.0}{4} = 0.0$$

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