Samacheer Kalvi Class 12 Business Maths Solutions Chapter 5 Numerical Methods Exercise 5.3

 

Question 1. \( \Delta^2 y_0 = \)
(a) \( y_2 – 2y_1 + y_0 \)
(b) \( y_2 + 2y_1 - y_0 \)
(c) \( y_2 + 2y_1 + y_0 \)
Answer: (a) \( y_2 – 2y_1 + y_0 \)
\( \Delta^2 y_0 = (E - 1)^2 y_0 \)
\( = (E^2 – 2E + 1) y_0 \)
\( = E^2 y_0 – 2E y_0 + y_0 \)
\( = y_2 - 2y_1 + y_0 \) This shows how the second difference of \( y_0 \) can be expressed using the shift operator \( E \).
In simple words: The second difference of \( y_0 \) is found by taking the value of \( y \) at the second step, subtracting twice the value at the first step, and then adding the initial value of \( y \). This is a standard formula in numerical analysis.

🎯 Exam Tip: Remember the relationship between the difference operator \( \Delta \) and the shift operator \( E \), which is \( \Delta = E - 1 \). This conversion is key to solving many problems involving these operators.

 

Question 2. \( \Delta f(x) = \)
(a) \( f(x + h) \)
(b) \( f(x) – f(x + h) \)
(c) \( f(x + h) – f(x) \)
(d) \( f(x) – f(x - h) \)
Answer: (c) \( f(x + h) – f(x) \)
The forward difference operator \( \Delta \) is defined as the change in the function value from \( x \) to \( x + h \).
\( \Delta f(x) = f(x + h) – f(x) \). This definition is fundamental in understanding finite differences.
In simple words: The forward difference of a function \( f(x) \) is simply the value of the function at the next step \( (x+h) \) minus its current value at \( x \). It tells you how much the function changes over an interval \( h \).

🎯 Exam Tip: Always recall the basic definition of the forward difference operator \( \Delta \): it signifies the difference between the next value and the current value of a function.

 

Question 3. \( E = \)
(a) \( 1 + \Delta \)
(b) \( 1 – \Delta \)
(c) \( 1 + \nabla \)
(d) \( 1 – \nabla \)
Answer: (a) \( 1 + \Delta \)
The shift operator \( E \) moves the function value one step forward. It is directly related to the forward difference operator \( \Delta \).
We know that \( \Delta f(x) = f(x + h) – f(x) \).
Also, \( E f(x) = f(x + h) \).
Substituting \( f(x + h) = E f(x) \) into the \( \Delta \) definition:
\( \Delta f(x) = E f(x) – f(x) \)
\( \Delta f(x) = (E - 1) f(x) \)
\( \implies \Delta = E - 1 \)
\( \implies E = 1 + \Delta \). This relationship is a cornerstone in difference calculus.
In simple words: The shift operator \( E \) means you move to the next value of the function. This is the same as taking the current value (1) and adding the forward change \( (\Delta) \) to it.

🎯 Exam Tip: Mastering the relationships between \( E \), \( \Delta \), and \( \nabla \) (backward difference) is essential. Remember \( E = 1 + \Delta \) and \( E^{-1} = 1 - \nabla \).

 

Question 4. If \( h = 1 \), then \( \Delta(x^2) = \)
(a) \( 2x \)
(b) \( 2x – 1 \)
(c) \( 2x + 1 \)
(d) \( 1 \)
Answer: (c) \( 2x + 1 \)
We are asked to find the forward difference of \( x^2 \) when the interval \( h \) is 1.
The definition of \( \Delta f(x) \) is \( f(x + h) – f(x) \).
Given \( f(x) = x^2 \) and \( h = 1 \).
So, \( \Delta(x^2) = (x + 1)^2 – x^2 \)
\( = (x^2 + 2x + 1) – x^2 \)
\( = 2x + 1 \). This calculation demonstrates how the difference operator works on a polynomial function.
In simple words: When you find the change in \( x^2 \) for a step of 1, you compare \( (x+1)^2 \) with \( x^2 \). After doing the math, you find the difference is \( 2x + 1 \).

🎯 Exam Tip: When dealing with difference operators, always expand terms like \( (x+h)^2 \) carefully and simplify. Pay close attention to the value of \( h \) provided in the question.

 

Question 5. If \( c \) is a constant then \( \Delta c = \)
(a) \( c \)
(b) \( \Delta \)
(c) \( \Delta^2 \)
(d) \( 0 \)
Answer: (d) \( 0 \)
The forward difference operator \( \Delta f(x) \) is defined as \( f(x + h) – f(x) \).
If \( f(x) = c \) (a constant), then \( f(x + h) = c \).
So, \( \Delta c = c – c \)
\( = 0 \). This shows that constants do not change with the forward difference operation.
In simple words: If a number never changes (it's a constant), then the difference between its current value and its next value will always be zero. It's like finding the change in something that stays exactly the same.

🎯 Exam Tip: Remember that the difference of a constant is always zero, similar to how the derivative of a constant is zero in calculus.

 

Question 6. If \( m \) and \( n \) are positive integers then \( \Delta^m \Delta^n f(x) = \)
(a) \( \Delta^{m+n} f(x) \)
(b) \( \Delta^m f(x) \)
(c) \( \Delta^n f(x) \)
(d) \( \Delta^{m-n} f(x) \)
Answer: (a) \( \Delta^{m+n} f(x) \)
The difference operators follow the index law of exponents when applied consecutively. Just like in algebra where \( x^m \cdot x^n = x^{m+n} \), applying \( \Delta^m \) and then \( \Delta^n \) to a function is equivalent to applying \( \Delta \) a total of \( m+n \) times.
So, \( \Delta^m \Delta^n f(x) = \Delta^{m+n} f(x) \). This property simplifies repeated difference calculations.
In simple words: If you take the difference of a function \( m \) times, and then take the difference of that result \( n \) more times, it's the same as just taking the difference a total of \( m+n \) times from the start.

🎯 Exam Tip: The index rule applies to difference operators, meaning you can add the powers when operators are applied in sequence. This is a common simplification technique.

 

Question 7. If 'n' is a positive integer \( \Delta^n[\Delta^{-n} f(x)] \)
(a) \( f(2x) \)
(b) \( f(x + h) \)
(c) \( f(x) \)
(d) \( \Delta f(x) \)
Answer: (c) \( f(x) \)
Using the index law for difference operators, when you multiply powers with the same base, you add the exponents.
Here, the base is \( \Delta \), and the exponents are \( n \) and \( -n \).
\( \Delta^n [\Delta^{-n} f(x)] = \Delta^{n + (-n)} f(x) \)
\( = \Delta^0 f(x) \)
\( = f(x) \). This shows that applying an operator and its inverse returns the original function.
In simple words: When you apply a difference operator \( n \) times and then apply its inverse \( -n \) times, you undo all the changes. It brings the function back to its original state, \( f(x) \).

🎯 Exam Tip: Any operator raised to the power of zero acts as the identity operator, meaning it leaves the function unchanged. This is a crucial property for simplifying expressions.

 

Question 8. \( E f(x) = \)
(a) \( f(x – h) \)
(b) \( f(x) \)
(c) \( f(x + h) \)
(d) \( f(x + 2h) \)
Answer: (c) \( f(x + h) \)
The shift operator \( E \) is defined to increase the argument of a function by \( h \), which is the interval of differencing. Applying \( E \) to \( f(x) \) means shifting the function one step forward.
Therefore, \( E f(x) = f(x + h) \). This is the fundamental definition of the shift operator.
In simple words: The shift operator \( E \) simply moves the function one step forward by adding \( h \) to the \( x \) value. It tells you the value of the function at the next point.

🎯 Exam Tip: Understand that \( E \) shifts forward, \( E^{-1} \) shifts backward. This basic definition is key to manipulating difference equations.

 

Question 9. \( \nabla = \)
(a) \( 1 + E \)
(b) \( 1 – E \)
(c) \( 1 - E^{-1} \)
(d) \( 1 + E^{-1} \)
Answer: (c) \( 1 - E^{-1} \)
The backward difference operator \( \nabla \) is defined as \( \nabla f(x) = f(x) - f(x - h) \).
We know that the inverse shift operator \( E^{-1} \) means \( E^{-1} f(x) = f(x - h) \).
Substitute \( f(x - h) = E^{-1} f(x) \) into the \( \nabla \) definition:
\( \nabla f(x) = f(x) - E^{-1} f(x) \)
\( \nabla f(x) = (1 - E^{-1}) f(x) \)
\( \implies \nabla = 1 - E^{-1} \). This relationship helps connect backward differences to the shift operator.
In simple words: The backward difference \( \nabla \) finds the difference between the current value and the previous value of a function. This is like starting with the current value (1) and subtracting the effect of shifting backward \( (E^{-1}) \).

🎯 Exam Tip: Distinguish between forward \( (\Delta = E - 1) \) and backward \( (\nabla = 1 - E^{-1}) \) difference operators. They are inverses in a sense, but their formulas with \( E \) are distinct.

 

Question 10. \( \nabla f(a) = \)
(a) \( f(a) + f(a – h) \)
(b) \( f(a) – f(a + h) \)
(c) \( f(a) – f(a – h) \)
(d) \( f(a) \)
Answer: (c) \( f(a) – f(a – h) \)
The backward difference operator \( \nabla \) at a point \( a \) is defined as the current function value minus the function value at the previous point. The interval of differencing is \( h \).
So, \( \nabla f(a) = f(a) - f(a - h) \). This is the direct definition of the backward difference for a specific point.
In simple words: The backward difference of a function at point \( a \) is simply its value at \( a \) minus its value at the point just before it, \( a-h \).

🎯 Exam Tip: Remember that \( \nabla \) always involves subtracting a previous value from the current value, while \( \Delta \) involves subtracting the current value from the next value.

 

Question 11. For the given points \( (x_0, y_0) \) and \( (x_1, y_1) \) the Lagrange's formula is
(a) \( y(x) = \frac { x-x_1 }{x_0-x_1} y_0 + \frac { x-x_0 }{x_1-x_0} y_1 \)
(b) \( y(x) = \frac { x_1-x}{x_0-x_1} y_0 + \frac { x-x_0 }{x_1-x_0} y_1 \)
(c) \( y(x) = \frac { x-x_1 }{x_0-x_1} y_0 + \frac { x-x_0 }{x_1-x_0} y_0 \)
(d) \( y(x) = \frac { x_1-x }{x_0-x_1} y_0 + \frac { x-x_0 }{x_1-x_0} y_0 \)
Answer: (a) \( y(x) = \frac { x-x_1 }{x_0-x_1} y_0 + \frac { x-x_0 }{x_1-x_0} y_1 \)
Lagrange's interpolation formula is used to find a polynomial that passes through a given set of points. For two points \( (x_0, y_0) \) and \( (x_1, y_1) \), the formula is a weighted average of the \( y \) values.
The weights are polynomials in \( x \) such that the weight for \( y_0 \) is 1 when \( x=x_0 \) and 0 when \( x=x_1 \), and vice-versa for \( y_1 \). This elegant formula constructs the unique polynomial.
In simple words: Lagrange's formula helps you find a curve that goes exactly through a set of given points. For two points, it's like blending the two y-values using special fractions that make sure the curve hits each point perfectly.

🎯 Exam Tip: Pay close attention to the subscripts in Lagrange's formula. The numerator \( (x-x_i) \) excludes the current point \( x_i \), and the denominator \( (x_j-x_i) \) ensures the term becomes 1 when \( x=x_j \).

 

Question 12. Lagrange's interpolation formula can be used for
(a) equal intervals only
(b) unequal intervals only
(c) both equal and unequal intervals
(d) None of the options
Answer: (c) both equal and unequal intervals
Lagrange's interpolation formula is highly versatile because its construction does not depend on the spacing between the data points. Unlike Newton's formulas which require equally spaced intervals, Lagrange's formula works for any set of distinct x-values. This flexibility makes it widely applicable in various numerical methods. It's often preferred when data points are unevenly distributed.
In simple words: Lagrange's formula is good because it can work with any set of points, whether they are spaced evenly apart or not. This means you can use it for many different types of data.

🎯 Exam Tip: Understand the key advantage of Lagrange's interpolation formula: its ability to handle unequally spaced data points, which is a limitation for other interpolation methods like Newton's forward/backward difference formulas.

 

Question 13. If \( f(x) = x^2 + 2x + 2 \) and the interval of differencing is unity then \( \Delta f(x) \)
(a) \( 2x - 3 \)
(b) \( 2x + 3 \)
(c) \( x + 3 \)
(d) \( x - 3 \)
Answer: (b) \( 2x + 3 \)
We need to find the forward difference of the function \( f(x) = x^2 + 2x + 2 \).
The interval of differencing is unity, which means \( h = 1 \).
The definition of \( \Delta f(x) \) is \( f(x + h) – f(x) \).
Substitute \( h = 1 \): \( \Delta f(x) = f(x + 1) – f(x) \)
First, find \( f(x + 1) \):
\( f(x + 1) = (x + 1)^2 + 2(x + 1) + 2 \)
\( = (x^2 + 2x + 1) + (2x + 2) + 2 \)
\( = x^2 + 4x + 5 \)
Now, subtract \( f(x) \):
\( \Delta f(x) = (x^2 + 4x + 5) – (x^2 + 2x + 2) \)
\( = x^2 + 4x + 5 – x^2 – 2x – 2 \)
\( = (x^2 - x^2) + (4x - 2x) + (5 - 2) \)
\( = 0 + 2x + 3 \)
\( = 2x + 3 \). This step-by-step expansion helps avoid errors in calculation.
In simple words: To find the forward difference, we calculate the function's value one step ahead \( (x+1) \) and subtract its current value at \( x \). When we do the math for \( x^2 + 2x + 2 \), the change comes out to be \( 2x + 3 \).

🎯 Exam Tip: When dealing with polynomial functions, remember to carefully expand all terms like \( (x+h)^2 \) before combining like terms. A small error in expansion can lead to an incorrect final answer.

 

Question 14. For the given data find the value of \( \Delta^3 y_0 \) is

(a) 1
(b) 0
(d) -1
Answer: (b) 0
To find \( \Delta^3 y_0 \), we use the relationship between the difference operator \( \Delta \) and the shift operator \( E \):
\( \Delta = E - 1 \)
So, \( \Delta^3 y_0 = (E - 1)^3 y_0 \)
Expand \( (E - 1)^3 \) using the binomial theorem: \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \).
\( = (E^3 - 3E^2 + 3E - 1) y_0 \)
Applying the operators to \( y_0 \):
\( = E^3 y_0 - 3E^2 y_0 + 3E y_0 - y_0 \)
Using the definition of the shift operator \( E^k y_0 = y_k \):
\( = y_3 - 3y_2 + 3y_1 - y_0 \)
From the given data in the table:
\( y_0 = 12 \)
\( y_1 = 13 \)
\( y_2 = 15 \)
\( y_3 = 18 \)
Substitute these values into the formula:
\( \Delta^3 y_0 = 18 - 3(15) + 3(13) - 12 \)
\( = 18 - 45 + 39 - 12 \)
Now, perform the addition and subtraction:
\( = (18 + 39) - (45 + 12) \)
\( = 57 - 57 \)
\( = 0 \). This shows that the third difference for this data set is zero, indicating the data might follow a quadratic pattern.
In simple words: To find the third difference of \( y_0 \), we use a formula that combines the \( y \) values from \( y_0 \) to \( y_3 \). After plugging in the numbers from the table and doing the calculations, the final answer is 0.

🎯 Exam Tip: When calculating higher-order differences, remember to use the binomial expansion of \( (E-1)^n \) to derive the correct coefficients for \( y_k \) terms. Always substitute the values carefully to avoid arithmetic errors.