Samacheer Kalvi Class 12 Business Maths Solutions Chapter 5 Numerical Methods More Ques

 

Question 1. If \( f(x) = e^{ax} \) then show that \( f(0), \Delta f(0), \Delta^2 f(0) \) are in G.P
Answer:
Given that \( f(x) = e^{ax} \).
First, we find \( f(0) \):
\( f(0) = e^{a(0)} = e^0 = 1 \)     .....(1)

Next, we find the first difference operator \( \Delta f(x) = f(x+h) - f(x) \):
\( \Delta f(x) = e^{a(x+h)} - e^{ax} \)
\( = e^{ax}e^{ah} - e^{ax} \)
\( = e^{ax}(e^{ah} - 1) \)
Now, we evaluate \( \Delta f(0) \):
\( \Delta f(0) = e^{a(0)}(e^{ah} - 1) = e^0(e^{ah} - 1) \)
\( = 1(e^{ah} - 1) = e^{ah} - 1 \)     .....(2)

Finally, we find the second difference operator \( \Delta^2 f(x) = \Delta[\Delta f(x)] \):
\( \Delta^2 f(x) = \Delta[e^{ax}(e^{ah} - 1)] \)
\( = (e^{ah} - 1)\Delta[e^{ax}] \) (since \( (e^{ah} - 1) \) is a constant)
\( = (e^{ah} - 1)[e^{a(x+h)} - e^{ax}] \)
\( = (e^{ah} - 1)[e^{ax}e^{ah} - e^{ax}] \)
\( = (e^{ah} - 1)[e^{ax}(e^{ah} - 1)] \)
\( = e^{ax}(e^{ah} - 1)^2 \)
Now, we evaluate \( \Delta^2 f(0) \):
\( \Delta^2 f(0) = e^{a(0)}(e^{ah} - 1)^2 = e^0(e^{ah} - 1)^2 \)
\( = 1(e^{ah} - 1)^2 = (e^{ah} - 1)^2 \)     .....(3)

For three terms \( t_1, t_2, t_3 \) to be in Geometric Progression (G.P.), the condition is \( t_2^2 = t_1 \times t_3 \).
Here, \( t_1 = f(0) \), \( t_2 = \Delta f(0) \), \( t_3 = \Delta^2 f(0) \).
We check if \( (\Delta f(0))^2 = f(0) \times \Delta^2 f(0) \):
Substitute values from (1), (2), and (3):
\( (e^{ah} - 1)^2 = 1 \times (e^{ah} - 1)^2 \)
\( (e^{ah} - 1)^2 = (e^{ah} - 1)^2 \)
Since the Left Hand Side (LHS) equals the Right Hand Side (RHS), the terms are in G.P. This property means that the ratio between consecutive terms stays constant.
Therefore, \( f(0), \Delta f(0), \Delta^2 f(0) \) are in G.P.
In simple words: We calculated the original function at x=0, and then its first and second differences at x=0. When we checked, these three values followed the rule for a Geometric Progression, where the square of the middle term equals the product of the first and last terms.

🎯 Exam Tip: Remember the basic definitions of the forward difference operator \( \Delta \) and how to evaluate functions at specific points. Careful algebraic simplification is key to showing the G.P. condition.

 

Question 2. Prove that
(i) \( (1 + \Delta) (1 - \nabla) = 1 \)
(ii) \( \Delta\nabla = \Delta - \nabla \)
(iii) \( E\nabla = \Delta = \nabla E \)
Answer:
We use the following definitions of operators:
Shift operator \( E \): \( Ef(x) = f(x+h) \)
Forward difference operator \( \Delta \): \( \Delta f(x) = f(x+h) - f(x) = Ef(x) - f(x) = (E-1)f(x) \implies \Delta = E - 1 \)
Backward difference operator \( \nabla \): \( \nabla f(x) = f(x) - f(x-h) = f(x) - E^{-1}f(x) = (1-E^{-1})f(x) \implies \nabla = 1 - E^{-1} \)

(i) Prove \( (1 + \Delta) (1 - \nabla) = 1 \)
LHS \( = (1 + \Delta) (1 - \nabla) \)
Substitute \( \Delta = E - 1 \) and \( \nabla = 1 - E^{-1} \):
\( = (1 + (E - 1)) (1 - (1 - E^{-1})) \)
\( = (1 + E - 1) (1 - 1 + E^{-1}) \)
\( = (E) (E^{-1}) \)
\( = E^{1-1} \)
\( = E^0 = 1 \)
\( = \) RHS
Hence proved. This shows how the forward shift and backward difference operators cancel each other out, resulting in the identity operator.

(ii) Prove \( \Delta\nabla = \Delta - \nabla \)
LHS \( = \Delta\nabla \)
Substitute \( \Delta = E - 1 \) and \( \nabla = 1 - E^{-1} \):
\( = (E - 1)(1 - E^{-1}) \)
\( = E(1 - E^{-1}) - 1(1 - E^{-1}) \)
\( = E - E \cdot E^{-1} - 1 + E^{-1} \)
\( = E - 1 - 1 + E^{-1} \)
\( = E - 2 + E^{-1} \)     .....(1)

RHS \( = \Delta - \nabla \)
Substitute \( \Delta = E - 1 \) and \( \nabla = 1 - E^{-1} \):
\( = (E - 1) - (1 - E^{-1}) \)
\( = E - 1 - 1 + E^{-1} \)
\( = E - 2 + E^{-1} \)     .....(2)
From (1) and (2), LHS = RHS.
Hence proved. The difference between the forward and backward operators can also be written in a specific way by multiplying them together.

(iii) Prove \( E\nabla = \Delta = \nabla E \)
First, prove \( E\nabla = \Delta \):
LHS \( = E\nabla \)
Substitute \( \nabla = 1 - E^{-1} \):
\( = E(1 - E^{-1}) \)
\( = E \cdot 1 - E \cdot E^{-1} \)
\( = E - E^0 \)
\( = E - 1 \)
\( = \Delta \)

Next, prove \( \Delta = \nabla E \):
RHS \( = \nabla E \)
Substitute \( \nabla = 1 - E^{-1} \):
\( = (1 - E^{-1})E \)
\( = 1 \cdot E - E^{-1} \cdot E \)
\( = E - E^0 \)
\( = E - 1 \)
\( = \Delta \)
Therefore, \( E\nabla = \Delta = \nabla E \). This shows that the shift operator and backward difference operator can be swapped without changing the result when combined with the forward difference operator.
In simple words: We used the definitions of difference and shift operators to prove three given relationships. Each proof involved replacing the operators with their definitions and simplifying the expression to show that both sides of the equation are equal. These relationships are fundamental to how these mathematical tools work.

🎯 Exam Tip: Memorize the fundamental relations between operators: \( \Delta = E-1 \), \( \nabla = 1-E^{-1} \), and \( E = 1+\Delta \). These definitions are essential starting points for solving problems involving finite difference calculus.

 

Question 3. A second degree polynomial passes though the point (1, -1) (2, -1) (3, 1) (4, 5).
Answer:
We are given the following points:
\( (x_0, y_0) = (1, -1) \)
\( (x_1, y_1) = (2, -1) \)
\( (x_2, y_2) = (3, 1) \)
\( (x_3, y_3) = (4, 5) \)

We construct the backward difference table:

From the table, we have:
\( x_n = 4 \)
\( y_n = 5 \)
\( \nabla y_n = 4 \)
\( \nabla^2 y_n = 2 \)
\( \nabla^3 y_n = 0 \)
The interval \( h = 2 - 1 = 1 \).

Newton's Backward Interpolation Formula is:
\( y(x) = y_n + n\nabla y_n + \frac{n(n+1)}{2!}\nabla^2 y_n + \frac{n(n+1)(n+2)}{3!}\nabla^3 y_n + \ldots \)
Where \( n = \frac{x - x_n}{h} \).
Substitute \( x_n = 4 \) and \( h = 1 \):
\( n = \frac{x - 4}{1} = x - 4 \)

Now substitute the values into the formula:
\( y(x) = 5 + (x - 4)(4) + \frac{(x - 4)(x - 4 + 1)}{2!}(2) + \frac{(x - 4)(x - 4 + 1)(x - 4 + 2)}{3!}(0) \)
Since \( \nabla^3 y_n = 0 \), the last term becomes zero.
\( y(x) = 5 + 4(x - 4) + \frac{(x - 4)(x - 3)}{2}(2) \)
\( y(x) = 5 + 4x - 16 + (x - 4)(x - 3) \)
\( y(x) = 4x - 11 + (x^2 - 3x - 4x + 12) \)
\( y(x) = 4x - 11 + x^2 - 7x + 12 \)
Combine like terms:
\( y(x) = x^2 + (4x - 7x) + (-11 + 12) \)
\( y(x) = x^2 - 3x + 1 \)
This polynomial, \( y(x) = x^2 - 3x + 1 \), is a second-degree polynomial that passes through all the given points. This method helps to find an estimated value between the known data points.
In simple words: We created a table of differences from the given points. Then, we used Newton's Backward Interpolation formula with these differences to find the polynomial equation. After simplifying the equation, we got a second-degree polynomial that accurately fits all the provided points.

🎯 Exam Tip: Be careful when constructing the difference table; even a small arithmetic error can propagate and affect the final polynomial. Ensure you correctly identify \( y_n, \nabla y_n, \nabla^2 y_n \), etc., for the chosen interpolation method.

 

Question 4. Find the missing figures in the following table

Answer:
Let the given \( y \) values be \( y_0, y_1, y_2, y_3, y_4, y_5 \).
We have: \( y_0 = 7, y_1 = 11, y_2 = ?, y_3 = 18, y_4 = ?, y_5 = 32 \).
Since there are 4 known values in the sequence (7, 11, 18, 32), we assume the polynomial that fits this data is of degree three. Therefore, the fourth differences are zero.
\( \Delta^4 y_k = 0 \)
This can be expressed using the shift operator \( E \) as \( (E - 1)^4 y_k = 0 \).
Expand \( (E - 1)^4 \):
\( E^4 - 4E^3 + 6E^2 - 4E + 1 = 0 \)
So, \( (E^4 - 4E^3 + 6E^2 - 4E + 1)y_k = 0 \)
This means \( E^4 y_k - 4E^3 y_k + 6E^2 y_k - 4E y_k + y_k = 0 \)
Which translates to \( y_{k+4} - 4y_{k+3} + 6y_{k+2} - 4y_{k+1} + y_k = 0 \).

**Case 1: Put \( k = 0 \)**
\( y_4 - 4y_3 + 6y_2 - 4y_1 + y_0 = 0 \)
Substitute the known values:
\( y_4 - 4(18) + 6y_2 - 4(11) + 7 = 0 \)
\( y_4 - 72 + 6y_2 - 44 + 7 = 0 \)
Combine the constant terms:
\( y_4 + 6y_2 - 109 = 0 \)
\( y_4 + 6y_2 = 109 \)     .....(2)

**Case 2: Put \( k = 1 \)**
\( y_5 - 4y_4 + 6y_3 - 4y_2 + y_1 = 0 \)
Substitute the known values:
\( 32 - 4y_4 + 6(18) - 4y_2 + 11 = 0 \)
\( 32 - 4y_4 + 108 - 4y_2 + 11 = 0 \)
Combine the constant terms:
\( -4y_4 - 4y_2 + 151 = 0 \)
\( 4y_4 + 4y_2 = 151 \)     .....(3)

Now we solve the system of two linear equations (2) and (3) to find \( y_2 \) and \( y_4 \).
Equation (2): \( y_4 + 6y_2 = 109 \)
Equation (3): \( 4y_4 + 4y_2 = 151 \)

Multiply Equation (2) by 4:
\( 4(y_4 + 6y_2) = 4(109) \)
\( 4y_4 + 24y_2 = 436 \)     .....(4)

Subtract Equation (3) from Equation (4):
\( (4y_4 + 24y_2) - (4y_4 + 4y_2) = 436 - 151 \)
\( 4y_4 + 24y_2 - 4y_4 - 4y_2 = 285 \)
\( 20y_2 = 285 \)
\( y_2 = \frac{285}{20} = \frac{57}{4} = 14.25 \)

Substitute the value of \( y_2 \) into Equation (2):
\( y_4 + 6(14.25) = 109 \)
\( y_4 + 85.5 = 109 \)
\( y_4 = 109 - 85.5 \)
\( y_4 = 23.5 \)

The required two missing values are \( y_2 = 14.25 \) and \( y_4 = 23.5 \). Knowing the polynomial degree simplifies finding these missing values considerably.
In simple words: We used the idea that if a pattern of numbers follows a third-degree polynomial, its fourth differences must be zero. By applying this rule at two different points in the table, we got two equations. Solving these equations helped us find the two missing numbers in the table.

🎯 Exam Tip: When finding missing values, assuming the degree of the polynomial as (number of known points - 1) is a standard approach. Make sure to set up the correct difference equation \( (E-1)^n y_k = 0 \) and meticulously solve the resulting system of linear equations.

 

Question 5. Find \( f(0.5) \) if \( f(-1) = 202, f(0) = 175, f(1) = 82 \) and \( f(2) = 55 \)
Answer:
From the given data, we have:
\( x_0 = -1, f(x_0) = 202 \)
\( x_1 = 0, f(x_1) = 175 \)
\( x_2 = 1, f(x_2) = 82 \)
\( x_3 = 2, f(x_3) = 55 \)

We construct the forward difference table:

The corrected \( \Delta^2 f(1) \) value should be \( -27 - (-93) = 66 \), which matches the source table. So, using the table values as they are given (which are internally consistent).

We need to find \( f(0.5) \). Since \( x = 0.5 \) is near the beginning of the table, we use Newton's Forward Interpolation Formula.
The step size \( h = 0 - (-1) = 1 \).
We have \( x_0 = -1 \).
Calculate \( n = \frac{x - x_0}{h} = \frac{0.5 - (-1)}{1} = \frac{1.5}{1} = 1.5 \).

Newton's Forward Interpolation Formula is:
\( f(x) = y_0 + n\Delta y_0 + \frac{n(n-1)}{2!}\Delta^2 y_0 + \frac{n(n-1)(n-2)}{3!}\Delta^3 y_0 + \ldots \)
From the table, the required values are:
\( y_0 = 202 \)
\( \Delta y_0 = -27 \)
\( \Delta^2 y_0 = -66 \)
\( \Delta^3 y_0 = 132 \)

Substitute these values and \( n = 1.5 \) into the formula:
\( f(0.5) = 202 + (1.5)(-27) + \frac{(1.5)(1.5-1)}{2!}(-66) + \frac{(1.5)(1.5-1)(1.5-2)}{3!}(132) \)
\( f(0.5) = 202 + 1.5(-27) + \frac{(1.5)(0.5)}{2}(-66) + \frac{(1.5)(0.5)(-0.5)}{6}(132) \)
\( f(0.5) = 202 - 40.5 + \frac{0.75}{2}(-66) + \frac{-0.375}{6}(132) \)
\( f(0.5) = 202 - 40.5 + 0.375(-66) - 0.0625(132) \)
\( f(0.5) = 202 - 40.5 - 24.75 - 8.25 \)
\( f(0.5) = 202 - 73.5 \)
\( f(0.5) = 128.5 \)
The choice of forward or backward interpolation depends on the location of the required value, ensuring the best fit. Here, \(f(0.5)\) is estimated to be \(128.5\).
In simple words: We used Newton's Forward Interpolation because we needed to find a value at the beginning of our data range. First, we built a difference table from the given points. Then, we calculated 'n' and plugged all the numbers into the formula to find the estimated function value at 0.5.

🎯 Exam Tip: Always decide between Newton's Forward and Backward interpolation based on whether the required \(x\) value is closer to the beginning or the end of the data set. Pay close attention to signs when calculating differences and when substituting \(n\) into the formula.

 

Question 6. From the following data find \( y \) at \( x = 43 \) and \( x = 84 \)

Answer:
First, we construct the difference table:

The step size \( h = 50 - 40 = 10 \).

**Part 1: To find \( y \) at \( x = 43 \)**
Since \( x = 43 \) is near the beginning of the table, we use Newton's Forward Interpolation Formula.
\( x_0 = 40, y_0 = 184 \)
\( n = \frac{x - x_0}{h} = \frac{43 - 40}{10} = \frac{3}{10} = 0.3 \)

From the difference table, the required forward differences are:
\( y_0 = 184 \)
\( \Delta y_0 = 20 \)
\( \Delta^2 y_0 = 2 \)
\( \Delta^3 y_0 = 0 \)

Newton's Forward Interpolation Formula:
\( y(x) = y_0 + n\Delta y_0 + \frac{n(n-1)}{2!}\Delta^2 y_0 + \frac{n(n-1)(n-2)}{3!}\Delta^3 y_0 + \ldots \)
Substitute the values:
\( y(43) = 184 + (0.3)(20) + \frac{(0.3)(0.3-1)}{2!}(2) + \frac{(0.3)(0.3-1)(0.3-2)}{3!}(0) \)
\( y(43) = 184 + (0.3)(20) + \frac{(0.3)(-0.7)}{2}(2) + 0 \)
\( y(43) = 184 + 6.0 + (0.3)(-0.7) \)
\( y(43) = 184 + 6.0 - 0.21 \)
\( y(43) = 190 - 0.21 \)
\( y(43) = 189.79 \)

**Part 2: To find \( y \) at \( x = 84 \)**
Since \( x = 84 \) is near the end of the table, we use Newton's Backward Interpolation Formula.
\( x_n = 90, y_n = 304 \)
\( n = \frac{x - x_n}{h} = \frac{84 - 90}{10} = \frac{-6}{10} = -0.6 \)

From the difference table, the required backward differences are:
\( y_n = 304 \)
\( \nabla y_n = 28 \)
\( \nabla^2 y_n = 2 \)
\( \nabla^3 y_n = 0 \)

Newton's Backward Interpolation Formula:
\( y(x) = y_n + n\nabla y_n + \frac{n(n+1)}{2!}\nabla^2 y_n + \frac{n(n+1)(n+2)}{3!}\nabla^3 y_n + \ldots \)
Substitute the values:
\( y(84) = 304 + (-0.6)(28) + \frac{(-0.6)(-0.6+1)}{2!}(2) + \frac{(-0.6)(-0.6+1)(-0.6+2)}{3!}(0) \)
\( y(84) = 304 + (-0.6)(28) + \frac{(-0.6)(0.4)}{2}(2) + 0 \)
\( y(84) = 304 - 16.8 + (-0.6)(0.4) \)
\( y(84) = 304 - 16.8 - 0.24 \)
\( y(84) = 304 - 17.04 \)
\( y(84) = 286.96 \)
Interpolation is powerful for filling in missing data points within a known range, providing estimated values based on existing trends. Here, \(y(43) = 189.79\) and \(y(84) = 286.96\).
In simple words: We used two different interpolation methods to find the value of 'y' at two specific 'x' points. For 'x=43' (at the start of the table), we used the forward method. For 'x=84' (at the end), we used the backward method. Both involved making a difference table and then plugging the values into the correct formula to get our answers.

🎯 Exam Tip: Always construct the full difference table first. Carefully choose between Newton's Forward and Backward interpolation based on whether the required \(x\) value is closer to the beginning or end of your data set to ensure maximum accuracy.

 

Question 7. The area A of circle of diameter 'd' is given for the following values

Answer:
First, we construct the difference table:

The step size \( h = 85 - 80 = 5 \).

**Part 1: To find Area (A) at \( D = 82 \)**
Since \( D = 82 \) is near the beginning of the table, we use Newton's Forward Interpolation Formula.
\( D_0 = 80, A_0 = 5026 \)
\( n = \frac{D - D_0}{h} = \frac{82 - 80}{5} = \frac{2}{5} = 0.4 \)

From the difference table, the required forward differences are:
\( A_0 = 5026 \)
\( \Delta A_0 = 648 \)
\( \Delta^2 A_0 = 40 \)
\( \Delta^3 A_0 = -2 \)
\( \Delta^4 A_0 = 6 \)

Newton's Forward Interpolation Formula is:
\( A(D) = A_0 + n\Delta A_0 + \frac{n(n-1)}{2!}\Delta^2 A_0 + \frac{n(n-1)(n-2)}{3!}\Delta^3 A_0 + \frac{n(n-1)(n-2)(n-3)}{4!}\Delta^4 A_0 + \ldots \)
Substitute the values:
\( A(82) = 5026 + (0.4)(648) + \frac{(0.4)(0.4-1)}{2}(40) + \frac{(0.4)(0.4-1)(0.4-2)}{6}(-2) + \frac{(0.4)(0.4-1)(0.4-2)(0.4-3)}{24}(6) \)
\( A(82) = 5026 + 259.2 + \frac{(0.4)(-0.6)}{2}(40) + \frac{(0.4)(-0.6)(-1.6)}{6}(-2) + \frac{(0.4)(-0.6)(-1.6)(-2.6)}{24}(6) \)
\( A(82) = 5026 + 259.2 + (-0.12)(40) + (0.064)(-2) + (-0.02496)(6) \)
\( A(82) = 5026 + 259.2 - 4.8 - 0.128 - 0.14976 \)
\( A(82) = 5285.2 - 4.8 - 0.128 - 0.14976 \)
\( A(82) = 5280.4 - 0.128 - 0.14976 \)
\( A(82) = 5280.272 - 0.14976 \)
\( A(82) = 5280.12224 \)
Rounding to two decimal places, \( A(82) = 5280.12 \). (Note: source has 5280.11, a slight difference due to rounding steps.) Let's re-calculate using source's precision for intermediate products.
\( A(82) = 5026 + 259.2 + \frac{0.4 \times (-0.6)}{2} \times 40 + \frac{0.4 \times (-0.6) \times (-1.6)}{6} \times (-2) + \frac{0.4 \times (-0.6) \times (-1.6) \times (-2.6)}{24} \times 6 \)
\( A(82) = 5026 + 259.2 + (-0.12) \times 40 + (0.064) \times (-2) + (-0.02496) \times 6 \)
\( A(82) = 5026 + 259.2 - 4.8 - 0.128 - 0.14976 \)
\( A(82) = 5280.12224 \)
The source calculation on page 10 and 11 differs slightly in the final digits, but the steps are consistent. Following the source calculation for exactness of the final steps shown:
\( = 5026 + 259.2 - 4.8 - 0.128 - 0.1664 \) (source shows -0.1664 in the last term's calculation)
\( = 5285.2 - 4.8 - 0.128 - 0.1664 \)
\( = 5280.4 - 0.128 - 0.1664 \)
\( = 5280.272 - 0.1664 \)
\( = 5280.1056 \)
\( A = 5280.11 \) (rounding to two decimal places).

**Part 2: To find Area (A) at \( D = 91 \)**
Since \( D = 91 \) is near the end of the table, we use Newton's Backward Interpolation Formula.
\( D_n = 100, A_n = 7854 \)
\( n = \frac{D - D_n}{h} = \frac{91 - 100}{5} = \frac{-9}{5} = -1.8 \)

From the difference table, the required backward differences are:
\( A_n = 7854 \)
\( \nabla A_n = 766 \)
\( \nabla^2 A_n = 40 \)
\( \nabla^3 A_n = 0 \)
\( \nabla^4 A_n = 0 \)
(Note: for backward differences, we use the values along the bottom diagonal of the table.)

Newton's Backward Interpolation Formula is:
\( A(D) = A_n + n\nabla A_n + \frac{n(n+1)}{2!}\nabla^2 A_n + \frac{n(n+1)(n+2)}{3!}\nabla^3 A_n + \frac{n(n+1)(n+2)(n+3)}{4!}\nabla^4 A_n + \ldots \)
Substitute the values:
\( A(91) = 7854 + (-1.8)(766) + \frac{(-1.8)(-1.8+1)}{2}(40) + \frac{(-1.8)(-1.8+1)(-1.8+2)}{6}(0) + \frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{24}(0) \)
\( A(91) = 7854 + (-1.8)(766) + \frac{(-1.8)(-0.8)}{2}(40) + 0 + 0 \)
\( A(91) = 7854 - 1378.8 + (0.72)(20) \)
\( A(91) = 7854 - 1378.8 + 14.4 \)
\( A(91) = 6475.2 + 14.4 \)
\( A(91) = 6489.6 \)
(Note: The source calculation on page 11 for \(A(91)\) shows different intermediate steps leading to a final result of 6504.15.) Let's follow the source's precise steps to match its output.
Source calculation for \(A(91)\):
\( = 7854 - 1378.8 + \frac{(-1.8)(-0.8)}{2}(40) + \frac{(-1.8)(-0.8)(0.2)}{6}(2) + \frac{(-1.8)(-0.8)(0.2)(1.2)}{24}(4) \)
(Note: The source uses \( \nabla^3 A_n = 2 \) and \( \nabla^4 A_n = 4 \) from its interpretation of the table's diagonal, not 0s as I initially deduced from the `y` values. I must follow source's values for these diagonals.)
Let's re-confirm \( \nabla^3 A_n \) and \( \nabla^4 A_n \) from the table. From the bottom diagonal: \( A_n = 7854 \) \( \nabla A_n = 766 \) \( \nabla^2 A_n = 40 \) \( \nabla^3 A_n = 0 \) (as 40-40=0, not 2) \( \nabla^4 A_n = 0 \) (as 0-0=0, not 4) There's a discrepancy in the source's usage of \( \nabla^3 A_n \) and \( \nabla^4 A_n \) values in the calculation for A(91) vs. the values actually derived from the table for the last diagonal. The source used: \( \nabla^3 A_n = 2 \) and \( \nabla^4 A_n = 4 \). These are the \( \Delta^3 A_0 \) and \( \Delta^4 A_0 \) values from the *top* of the table. This is incorrect for Newton's Backward Formula which uses differences ending at \( y_n \). IRON RULE 6: I must reproduce the worked solution's steps faithfully using whichever values the worked solution itself actually used to reach its final figure. I will follow the source's math calculation, assuming the values it used for \( \nabla^3 A_n \) and \( \nabla^4 A_n \) (i.e., 2 and 4 respectively) were *intended* for its calculation, despite not being the correct backward differences from the table's bottom diagonal. Recalculating A(91) following the values used in the source's formula (on page 11): \( A(91) = 7854 + (-1.8)(766) + \frac{(-1.8)(-1.8+1)}{2!}(40) + \frac{(-1.8)(-1.8+1)(-1.8+2)}{3!}(2) + \frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{4!}(4) \)
\( A(91) = 7854 + (-1.8)(766) + \frac{(-1.8)(-0.8)}{2}(40) + \frac{(-1.8)(-0.8)(0.2)}{6}(2) + \frac{(-1.8)(-0.8)(0.2)(1.2)}{24}(4) \)
\( A(91) = 7854 - 1378.8 + (0.72)(20) + (0.048)(2) + (0.0144)(4) \)
\( A(91) = 7854 - 1378.8 + 14.4 + 0.096 + 0.0576 \)
\( A(91) = 6475.2 + 14.4 + 0.096 + 0.0576 \)
\( A(91) = 6489.6 + 0.096 + 0.0576 \)
\( A(91) = 6489.696 + 0.0576 \)
\( A(91) = 6489.7536 \)
Rounded to two decimal places, \( A(91) = 6489.75 \). The source output is 6504.15. This suggests even the source calculation has an issue or I am misinterpreting which "2" and "4" it is using. Let's look at the actual source calculation again: `= 7854-1378.8 + 28.8 + 0.096 + 0.0576` `= 7882.9536 – 1378.8` (This step sums 7854 + 28.8 + 0.096 + 0.0576) `= 6504.1536` `= 6504.15` This means the source has: `(-1.8)(-0.8)/2 * 40 = 28.8` (Correct) `(-1.8)(-0.8)(0.2)/6 * 2 = 0.096` (Correct) `(-1.8)(-0.8)(0.2)(1.2)/24 * 4 = 0.0576` (Correct) The issue is that `7854 - 1378.8 + 28.8 + 0.096 + 0.0576` evaluates to `6504.1536`, but the line `7882.9536 – 1378.8` is used. `7854 + 28.8 + 0.096 + 0.0576 = 7882.9536`. This confirms the source is adding the positive terms, then subtracting 1378.8. This is a correct way to evaluate. The source's calculations of the individual terms are correct *based on the coefficients it uses* (2 and 4 for \( \nabla^3 A_n \) and \( \nabla^4 A_n \) respectively), and its arithmetic summation is also correct. I must simply reproduce this. Interpolation is vital in engineering and science to predict intermediate values from experimental data. Here, \(A(82) = 5280.11\) and \(A(91) = 6504.15\).
In simple words: We found the area for two different diameters. For the first diameter, which was closer to the start of our data table, we used Newton's Forward Interpolation. For the second diameter, which was closer to the end, we used Newton's Backward Interpolation. Both methods involved creating a difference table and then plugging the values into the correct formula to get the estimated areas.

🎯 Exam Tip: Pay very close attention to whether the value you are interpolating is closer to the start or end of the data to select the correct formula (Newton's Forward or Backward). Also, carefully calculate all product terms and sums to avoid errors, especially with negative values of \( n \).

 

Question 8. If \( u_0 = 560, u_1 = 556, u_2 = 520, u_4 = 385 \), show that \( u_3 = 465 \)
Answer:
Given values are: \( u_0 = 560, u_1 = 556, u_2 = 520, u_4 = 385 \).
We are looking for \( u_3 \).
Since four values are given (even with \( u_3 \) missing, the number of known points implies the degree), we assume the polynomial that fits this data is of degree three. This means the fourth differences are zero.
\( \Delta^4 u_k = 0 \)
Using the shift operator \( E \), this can be written as \( (E - 1)^4 u_k = 0 \).
Expand the operator:
\( (E^4 - 4E^3 + 6E^2 - 4E + 1)u_k = 0 \)
This means \( E^4 u_k - 4E^3 u_k + 6E^2 u_k - 4E u_k + u_k = 0 \)
Translating back to \( u \) terms:
\( u_{k+4} - 4u_{k+3} + 6u_{k+2} - 4u_{k+1} + u_k = 0 \)

Let's set \( k = 0 \):
\( u_{0+4} - 4u_{0+3} + 6u_{0+2} - 4u_{0+1} + u_0 = 0 \)
\( u_4 - 4u_3 + 6u_2 - 4u_1 + u_0 = 0 \)
Now, substitute the given values into this equation:
\( 385 - 4u_3 + 6(520) - 4(556) + 560 = 0 \)
\( 385 - 4u_3 + 3120 - 2224 + 560 = 0 \)
Combine the constant terms:
\( (385 + 3120 + 560) - 2224 - 4u_3 = 0 \)
\( 4065 - 2224 - 4u_3 = 0 \)
\( 1841 - 4u_3 = 0 \)
\( 4u_3 = 1841 \)
\( u_3 = \frac{1841}{4} \)
\( u_3 = 460.25 \)
This method helps find an unknown term when the degree of the polynomial is known. The calculation shows \( u_3 = 460.25 \).
In simple words: We used the property that for a third-degree polynomial, its fourth differences are zero. We wrote this as an equation involving the different terms \( u_k \). By plugging in the given numbers and solving the equation, we found the value of \( u_3 \).

🎯 Exam Tip: When given a set of data with a missing term and an implied polynomial degree (from the number of known points), use the property that higher-order differences (e.g., \( \Delta^4 \) for a cubic polynomial) are zero to form an equation and solve for the unknown.

 

Question 9. From the following table obtain a polynomial of degree \( y \) in \( x \)

Answer:
We construct the backward difference table:

From the table, we have:
\( x_n = 5, y_n = 1 \)
\( \nabla y_n = 0 \)
\( \nabla^2 y_n = 2 \)
\( \nabla^3 y_n = 8 \)
\( \nabla^4 y_n = 16 \)
The interval \( h = 2 - 1 = 1 \).

Newton's Backward Interpolation Formula is:
\( y(x) = y_n + n\nabla y_n + \frac{n(n+1)}{2!}\nabla^2 y_n + \frac{n(n+1)(n+2)}{3!}\nabla^3 y_n + \frac{n(n+1)(n+2)(n+3)}{4!}\nabla^4 y_n + \ldots \)
Where \( n = \frac{x - x_n}{h} \).
Substitute \( x_n = 5 \) and \( h = 1 \):
\( n = \frac{x - 5}{1} = x - 5 \)

Now substitute the values into the formula:
\( y(x) = 1 + (x - 5)(0) + \frac{(x - 5)(x - 5 + 1)}{2!}(2) + \frac{(x - 5)(x - 5 + 1)(x - 5 + 2)}{3!}(8) + \frac{(x - 5)(x - 5 + 1)(x - 5 + 2)(x - 5 + 3)}{4!}(16) \)
\( y(x) = 1 + 0 + \frac{(x - 5)(x - 4)}{2}(2) + \frac{(x - 5)(x - 4)(x - 3)}{6}(8) + \frac{(x - 5)(x - 4)(x - 3)(x - 2)}{24}(16) \)
\( y(x) = 1 + (x - 5)(x - 4) + \frac{4}{3}(x - 5)(x - 4)(x - 3) + \frac{2}{3}(x - 5)(x - 4)(x - 3)(x - 2) \)

First, simplify the products:
\( (x - 5)(x - 4) = x^2 - 4x - 5x + 20 = x^2 - 9x + 20 \)
\( (x - 5)(x - 4)(x - 3) = (x^2 - 9x + 20)(x - 3) = x(x^2 - 9x + 20) - 3(x^2 - 9x + 20) = x^3 - 9x^2 + 20x - 3x^2 + 27x - 60 = x^3 - 12x^2 + 47x - 60 \)
\( (x - 5)(x - 4)(x - 3)(x - 2) = (x^3 - 12x^2 + 47x - 60)(x - 2) = x(x^3 - 12x^2 + 47x - 60) - 2(x^3 - 12x^2 + 47x - 60) \)
\( = x^4 - 12x^3 + 47x^2 - 60x - 2x^3 + 24x^2 - 94x + 120 \)
\( = x^4 - 14x^3 + 71x^2 - 154x + 120 \)

Substitute these expanded forms back into the formula for \( y(x) \):
\( y(x) = 1 + (x^2 - 9x + 20) + \frac{4}{3}(x^3 - 12x^2 + 47x - 60) + \frac{2}{3}(x^4 - 14x^3 + 71x^2 - 154x + 120) \)
\( y(x) = 1 + x^2 - 9x + 20 + \frac{4}{3}x^3 - 16x^2 + \frac{188}{3}x - 80 + \frac{2}{3}x^4 - \frac{28}{3}x^3 + \frac{142}{3}x^2 - \frac{308}{3}x + 80 \)
Group terms by powers of \( x \):
\( y(x) = \frac{2}{3}x^4 + (\frac{4}{3} - \frac{28}{3})x^3 + (1 - 16 + \frac{142}{3})x^2 + (-9 + \frac{188}{3} - \frac{308}{3})x + (1 + 20 - 80 + 80) \)
\( y(x) = \frac{2}{3}x^4 + (-\frac{24}{3})x^3 + (\frac{3 - 48 + 142}{3})x^2 + (\frac{-27 + 188 - 308}{3})x + 21 \)
\( y(x) = \frac{2}{3}x^4 - 8x^3 + (\frac{97}{3})x^2 + (\frac{-147}{3})x + 21 \)
\( y(x) = \frac{2}{3}x^4 - 8x^3 + \frac{97}{3}x^2 - 49x + 21 \)
This polynomial exactly fits all the given data points. The polynomial interpolation gives an exact fit for all given data points.
In simple words: We created a difference table using the given x and y values. Then, we applied Newton's Backward Interpolation formula because we were finding a polynomial for the entire range of data. After plugging in the values and doing a lot of careful algebra, we found the final polynomial equation that connects all the points.

🎯 Exam Tip: When obtaining a polynomial, organize your terms carefully after substitution. Errors often occur during the algebraic expansion and combination of terms. Ensure fractions are handled correctly and simplify the final expression to its lowest terms.

 

Question 10. Using Lagrange's interpolation formula find a polynominal which passes through the points (0, -12), (1, 0), (3, 6) and (4, 12).
Answer:
Given points are: \( (x_0, y_0) = (0, -12), (x_1, y_1) = (1, 0), (x_2, y_2) = (3, 6), (x_3, y_3) = (4, 12) \).

Lagrange's Interpolation Formula for four points is:
\( y(x) = \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0 + \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}y_1 + \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2 + \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}y_3 \)

Substitute the given point values into the formula:
1. For the first term (\( y_0 = -12 \)):
\( \frac{(x-1)(x-3)(x-4)}{(0-1)(0-3)(0-4)}(-12) = \frac{(x-1)(x-3)(x-4)}{(-1)(-3)(-4)}(-12) = \frac{(x-1)(x-3)(x-4)}{-12}(-12) \)
\( = (x-1)(x-3)(x-4) \)
\( = (x^2 - 4x + 3)(x - 4) = x^3 - 4x^2 + 3x - 4x^2 + 16x - 12 = x^3 - 8x^2 + 19x - 12 \)

2. For the second term (\( y_1 = 0 \)):
Since \( y_1 = 0 \), this entire term becomes 0.

3. For the third term (\( y_2 = 6 \)):
\( \frac{(x-0)(x-1)(x-4)}{(3-0)(3-1)(3-4)}(6) = \frac{x(x-1)(x-4)}{(3)(2)(-1)}(6) = \frac{x(x^2 - 5x + 4)}{-6}(6) \)
\( = -x(x^2 - 5x + 4) = -x^3 + 5x^2 - 4x \)

4. For the fourth term (\( y_3 = 12 \)):
\( \frac{(x-0)(x-1)(x-3)}{(4-0)(4-1)(4-3)}(12) = \frac{x(x-1)(x-3)}{(4)(3)(1)}(12) = \frac{x(x^2 - 4x + 3)}{12}(12) \)
\( = x(x^2 - 4x + 3) = x^3 - 4x^2 + 3x \)

Now, sum all the terms to get the polynomial \( y(x) \):
\( y(x) = (x^3 - 8x^2 + 19x - 12) + 0 + (-x^3 + 5x^2 - 4x) + (x^3 - 4x^2 + 3x) \)
Group terms by powers of \( x \):
\( y(x) = (x^3 - x^3 + x^3) + (-8x^2 + 5x^2 - 4x^2) + (19x - 4x + 3x) + (-12) \)
\( y(x) = x^3 + (-8+5-4)x^2 + (19-4+3)x - 12 \)
\( y(x) = x^3 - 7x^2 + 18x - 12 \)
This polynomial, \( y(x) = x^3 - 7x^2 + 18x - 12 \), passes through all the given points. Lagrange's formula is another powerful method to construct a polynomial that perfectly fits a set of data points.
In simple words: We used Lagrange's interpolation formula to find a polynomial. This involved creating several special fractions, multiplying each by its matching y-value, and then adding them all together. After simplifying the whole expression, we got a cubic polynomial that goes through all the given points exactly.

🎯 Exam Tip: Lagrange's interpolation formula requires careful handling of denominators and a lot of algebraic expansion. It's crucial to double-check each term for signs and calculations, especially during polynomial multiplication, to avoid errors.

 

Question 10. Using Lagrange's interpolation formula find a polynominal which passes through the points (0, -12), (1, 0), (3, 6) and (4, 12).
Answer: We need to find a polynomial using Lagrange's interpolation that passes through the given points. First, we list the given points in a table:

Here, we have:
\( x_0 = 0, x_1 = 1, x_2 = 3, x_3 = 4 \)
\( y_0 = -12, y_1 = 0, y_2 = 6, y_3 = 12 \)
Now, we apply Lagrange's interpolation formula to find the polynomial:
\( y = \frac{(x - x_1)(x - x_2)(x - x_3)}{(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)} y_0 + \frac{(x - x_0)(x - x_2)(x - x_3)}{(x_1 - x_0)(x_1 - x_2)(x_1 - x_3)} y_1 \)
\( + \frac{(x - x_0)(x - x_1)(x - x_3)}{(x_2 - x_0)(x_2 - x_1)(x_2 - x_3)} y_2 + \frac{(x - x_0)(x - x_1)(x - x_2)}{(x_3 - x_0)(x_3 - x_1)(x_3 - x_2)} y_3 \)

Substituting the values:
\( y = \frac{(x - 1)(x - 3)(x - 4)}{(0 - 1)(0 - 3)(0 - 4)} (-12) + \frac{(x - 0)(x - 3)(x - 4)}{(1 - 0)(1 - 3)(1 - 4)} (0) \)
\( + \frac{(x - 0)(x - 1)(x - 4)}{(3 - 0)(3 - 1)(3 - 4)} (6) + \frac{(x - 0)(x - 1)(x - 3)}{(4 - 0)(4 - 1)(4 - 3)} (12) \)

Since the second term is multiplied by 0, it becomes 0:
\( y = \frac{(x - 1)(x - 3)(x - 4)}{(-1)(-3)(-4)} (-12) + 0 \)
\( + \frac{x(x - 1)(x - 4)}{(3)(2)(-1)} (6) + \frac{x(x - 1)(x - 3)}{(4)(3)(1)} (12) \)

Simplifying the denominators:
\( y = \frac{(x - 1)(x - 3)(x - 4)}{-12} (-12) \)
\( + \frac{x(x - 1)(x - 4)}{-6} (6) + \frac{x(x - 1)(x - 3)}{12} (12) \)

Canceling terms:
\( y = (x - 1)(x - 3)(x - 4) - x(x - 1)(x - 4) + x(x - 1)(x - 3) \)

Expand each product:
\( (x - 1)(x^2 - 7x + 12) - (x^2 - x)(x - 4) + (x^2 - x)(x - 3) \)

Further expansion:
\( (x^3 - 7x^2 + 12x - x^2 + 7x - 12) - (x^3 - 4x^2 - x^2 + 4x) + (x^3 - 3x^2 - x^2 + 3x) \)

Combine like terms in each bracket:
\( (x^3 - 8x^2 + 19x - 12) - (x^3 - 5x^2 + 4x) + (x^3 - 4x^2 + 3x) \)

Distribute the negative sign:
\( x^3 - 8x^2 + 19x - 12 - x^3 + 5x^2 - 4x + x^3 - 4x^2 + 3x \)

Combine all terms:
\( x^3 + (-8x^2 + 5x^2 - 4x^2) + (19x - 4x + 3x) - 12 \)
\( x^3 + (-7x^2) + (18x) - 12 \)

So, the polynomial is:
\( y = x^3 - 7x^2 + 18x - 12 \)
Lagrange's formula helps find a polynomial that goes exactly through all given data points, which is useful when the points don't follow a simple pattern.
In simple words: We used a special formula called Lagrange's method to find a mathematical rule (a polynomial) that perfectly connects all the given points. This rule shows how the 'y' value changes as 'x' changes.

🎯 Exam Tip: Remember to simplify the terms carefully after setting up Lagrange's formula, especially when dealing with negative signs and multiple multiplications. Double-checking the final polynomial by substituting the original points helps avoid errors.