Interpolation Techniques Questions and Answers

Q: The value of\[ \Delta^{5} [(1 - x)(1 - x^{2})(1 + x^{2})] \]is :

The value is \(5!\), taking the usual unit step.Multiply out the product first:\[ \begin{aligned} (1 - x)(1 - x^{2})(1 + x^{2}) &= (1 - x)(1 - x^{4}) \\ &= 1 - x - x^{4} + x^{5} \end{aligned} \]This is a polynomial of degree 5 with leading coefficient 1. With step \(h = 1\), the fifth difference of \(x^{5}\) is \(5!\,h^{5} = 120\), and the fifth difference of every term of lower degree is 0. So the answer is \(5! = 120\).0 would be right for \(\Delta^{6}\), and 1 and 5 confuse the leading coefficient or the degree with the value of the difference.

Q: The missing term in the following tablexf(x)0113293-481using difference operator method is :

The missing value is 31.Five values are tabulated, with one missing, so we assume they lie on a polynomial of degree 3, whose fourth difference is zero:\[ \begin{aligned} \Delta^{4} y_{0} &= y_{4} - 4y_{3} + 6y_{2} - 4y_{1} + y_{0} = 0 \\ 81 - 4y_{3} + 54 - 12 + 1 &= 0 \\ 124 - 4y_{3} &= 0 \\ y_{3} &= 31 \end{aligned} \]The trap is that the given values 1, 3, 9, 81 suggest \(3^{x}\), which would put 27 in the gap. The difference-operator method does not assume an exponential; it fits the lowest-degree polynomial, and that gives 31, not 27, 28, 29 or 30.

Q: Let E be the shift operator and \(\mu\) be the averaging operator. Then :

The averaging operator is μ = (E1/2 + E−1/2)/2.By definition\[ \begin{aligned} \mu f(x) &= \tfrac{1}{2}\left[f\left(x + \tfrac{h}{2}\right) \right. \\ &\quad \left. + f\left(x - \tfrac{h}{2}\right)\right] \end{aligned} \]Since \(E^{1/2}f(x) = f(x + h/2)\) and \(E^{-1/2}f(x) = f(x - h/2)\),\[ \mu = \frac{1}{2}\left(E^{1/2} + E^{-1/2}\right) \]\(E^{1/2} - E^{-1/2}\) is the central difference operator \(\delta\), not \(\mu\); half of it is not a standard operator, and dividing by 3 does not average two values.

Q: If \(f(x) = \dfrac{1}{x^{2}}\), then the first divided difference [a, b] is :

The divided difference is −(a + b)/(a2b2).\[ \begin{aligned} f[a, b] &= \frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{b^{2}} - \frac{1}{a^{2}}}{b - a} \\ &= \frac{a^{2} - b^{2}}{a^{2}b^{2}(b - a)} \\ &= \frac{-(a + b)}{a^{2}b^{2}} \end{aligned} \]since \(a^{2} - b^{2} = -(b - a)(a + b)\). The option with \(+(a + b)\) has the sign wrong. The options \((a + b)/ab\) and \((a - b)/ab\) lack the squares in the denominator that \(1/x^{2}\) produces.

Q: The value of\[ \Delta^{10}[(1 - x)(1 - 2x^{2})(1 - 3x^{3})(1 - 4x^{4})] \]is :

The value is 24 × 10!.The product is a polynomial of degree \(1 + 2 + 3 + 4 = 10\). Its leading term comes from multiplying the leading terms:\[ (-x)(-2x^{2})(-3x^{3})(-4x^{4}) = 24x^{10} \]The tenth forward difference of a polynomial of degree 10 is constant and depends only on the leading term: with unit step, \(\Delta^{10}x^{10} = 10!\). The lower-degree terms vanish under \(\Delta^{10}\). So\[ \Delta^{10}[\ldots] = 24 \times 10! \]12 would come from a sign or product slip in the leading coefficient, and 8! from the wrong degree.

Q: Let \(\Delta\) be the forward difference operator, \(\nabla\) be the backward difference operator and \(\delta\) be the central difference operator. Then :

The correct relation is \(\nabla \Delta = \Delta - \nabla\).Write the operators with the shift E: \(\Delta = E - 1\), \(\nabla = 1 - E^{-1}\), \(\delta = E^{1/2} - E^{-1/2}\). Then\[ \begin{aligned} \nabla\Delta &= (1 - E^{-1})(E - 1) = E - 2 + E^{-1} \\ \Delta - \nabla &= (E - 1) - (1 - E^{-1}) = E - 2 + E^{-1} \\ \delta^{2} &= E - 2 + E^{-1} \end{aligned} \]So \(\nabla\Delta = \Delta - \nabla = \delta^{2}\). That makes option A true and options B, C and D, each of which asserts one of these equalities fails, false.

Q: Let f(x) be a polynomial of degree n. Then the nth forward difference of f(x), that is, \(\Delta^{n}[f(x)]\) is :

The nth forward difference is constant.Each application of \(\Delta\) lowers the degree of a polynomial by one: if \(f(x) = a x^{n} + \ldots\), then \(\Delta f(x) = f(x + h) - f(x) = a n h\, x^{n-1} + \ldots\). After n applications the result is the constant\[ \Delta^{n} f(x) = a\, n!\, h^{n} \]It is not always 0 (that is \(\Delta^{n+1}\)), nor always 1 (it depends on the leading coefficient a and the step h), and it can certainly be determined.

Q: \(\Delta\!\!\!|\,\) is a divided difference operator and \(\Delta\) is a forward difference operator, f(x) and g(x) are two functions and a and b are constants, then :

The correct property is linearity of the divided difference: the operator applied to \(af(x) + bg(x)\) equals a times its value on f plus b times its value on g.A first divided difference is\[ [x_{0}, x_{1}]f = \frac{f(x_{1}) - f(x_{0})}{x_{1} - x_{0}} \]and replacing f by \(af + bg\) splits the numerator into a times the f part plus b times the g part. Higher divided differences are built recursively from the same quotients, so they are linear too.Option C mixes in the forward difference \(\Delta\), a different operator. Options B and D replace f and g by their product, which has nothing to do with a linear combination.

Q: Let f(x) be a function which takes the values \(y_{0}, y_{1}, \ldots, y_{n}\) for \(x = x_{0}, x_{1}, \ldots, x_{n}\) respectively and [a, b] be an interval divided into n subintervals of width h such that :\[ \begin{aligned} &x_{0} = a, x_{1} = x_{0} + n, \ldots, \\ &x_{n} = x_{0} + nh = b. \end{aligned} \]Then the error in the trapezoidal rule for integration is of order :

The error of the trapezoidal rule is of order \(h^{2}\).On one subinterval the trapezoidal rule has error \(-\frac{h^{3}}{12} f''(\xi)\). Over the n subintervals of [a, b], with \(nh = b - a\), the errors add up to\[ E = -\frac{(b - a)h^{2}}{12} f''(\xi) \]so the composite error is \(O(h^{2})\). The \(h^{3}\) in option C is the error on a single strip, and h alone would describe a first-order rule such as the rectangle rule.

Q: The value of \(\Delta^{2} f(x)\), where \(f(x) = ab^{cx}\) is :

\(\Delta^{2} f(x)\) equals \((b^{ch} - 1)^{2}\, ab^{cx}\).With step h:\[ \begin{aligned} \Delta f(x) &= ab^{c(x + h)} - ab^{cx} = (b^{ch} - 1)\, ab^{cx} \\ \Delta^{2} f(x) &= (b^{ch} - 1)\,\Delta\left(ab^{cx}\right) = (b^{ch} - 1)^{2}\, ab^{cx} \end{aligned} \]Each application multiplies by the same factor \(b^{ch} - 1\), because an exponential is an eigenfunction of \(\Delta\).The options with \(b^{ch} + 1\) have the wrong sign (they would come from a sum, not a difference), and \((ab^{cx})^{2}\) squares the function instead of applying \(\Delta\) twice.

11.

The value of

\[ \Delta^{5} [(1 - x)(1 - x^{2})(1 + x^{2})] \]

is :

A.5!
B.0
C.1
D.5

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The value is \(5!\), taking the usual unit step.

Multiply out the product first:

\[ \begin{aligned} (1 - x)(1 - x^{2})(1 + x^{2}) &= (1 - x)(1 - x^{4}) \\ &= 1 - x - x^{4} + x^{5} \end{aligned} \]

This is a polynomial of degree 5 with leading coefficient 1. With step \(h = 1\), the fifth difference of \(x^{5}\) is \(5!\,h^{5} = 120\), and the fifth difference of every term of lower degree is 0. So the answer is \(5! = 120\).

0 would be right for \(\Delta^{6}\), and 1 and 5 confuse the leading coefficient or the degree with the value of the difference.

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12.

The missing term in the following table

x	f(x)
0	1
1	3
2	9
3	-
4	81

using difference operator method is :

A.28
B.29
C.30
D.31

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The missing value is 31.

Five values are tabulated, with one missing, so we assume they lie on a polynomial of degree 3, whose fourth difference is zero:

\[ \begin{aligned} \Delta^{4} y_{0} &= y_{4} - 4y_{3} + 6y_{2} - 4y_{1} + y_{0} = 0 \\ 81 - 4y_{3} + 54 - 12 + 1 &= 0 \\ 124 - 4y_{3} &= 0 \\ y_{3} &= 31 \end{aligned} \]

The trap is that the given values 1, 3, 9, 81 suggest \(3^{x}\), which would put 27 in the gap. The difference-operator method does not assume an exponential; it fits the lowest-degree polynomial, and that gives 31, not 27, 28, 29 or 30.

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13.

Let E be the shift operator and \(\mu\) be the averaging operator. Then :

A.\(\mu = E^{1/2} - E^{-1/2}\)
B.\(\mu = \dfrac{E^{1/2} + E^{-1/2}}{2}\)
C.\(\mu = \dfrac{E^{1/2} - E^{-1/2}}{2}\)
D.\(\mu = \dfrac{E^{1/2} + E^{-1/2}}{3}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The averaging operator is μ = (E^1/2 + E^−1/2)/2.

By definition

\[ \begin{aligned} \mu f(x) &= \tfrac{1}{2}\left[f\left(x + \tfrac{h}{2}\right) \right. \\ &\quad \left. + f\left(x - \tfrac{h}{2}\right)\right] \end{aligned} \]

Since \(E^{1/2}f(x) = f(x + h/2)\) and \(E^{-1/2}f(x) = f(x - h/2)\),

\[ \mu = \frac{1}{2}\left(E^{1/2} + E^{-1/2}\right) \]

\(E^{1/2} - E^{-1/2}\) is the central difference operator \(\delta\), not \(\mu\); half of it is not a standard operator, and dividing by 3 does not average two values.

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14.

If \(f(x) = \dfrac{1}{x^{2}}\), then the first divided difference [a, b] is :

A.\(\dfrac{-(a + b)}{a^{2}b^{2}}\)
B.\(\dfrac{+(a + b)}{a^{2}b^{2}}\)
C.\(\dfrac{a + b}{ab}\)
D.\(\dfrac{a - b}{ab}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The divided difference is −(a + b)/(a²b²).

\[ \begin{aligned} f[a, b] &= \frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{b^{2}} - \frac{1}{a^{2}}}{b - a} \\ &= \frac{a^{2} - b^{2}}{a^{2}b^{2}(b - a)} \\ &= \frac{-(a + b)}{a^{2}b^{2}} \end{aligned} \]

since \(a^{2} - b^{2} = -(b - a)(a + b)\). The option with \(+(a + b)\) has the sign wrong. The options \((a + b)/ab\) and \((a - b)/ab\) lack the squares in the denominator that \(1/x^{2}\) produces.

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15.

The value of

\[ \Delta^{10}[(1 - x)(1 - 2x^{2})(1 - 3x^{3})(1 - 4x^{4})] \]

is :

A.\(24 \times 10!\)
B.\(24 \times 8!\)
C.\(12 \times 10!\)
D.\(12 \times 8!\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The value is 24 × 10!.

The product is a polynomial of degree \(1 + 2 + 3 + 4 = 10\). Its leading term comes from multiplying the leading terms:

\[ (-x)(-2x^{2})(-3x^{3})(-4x^{4}) = 24x^{10} \]

The tenth forward difference of a polynomial of degree 10 is constant and depends only on the leading term: with unit step, \(\Delta^{10}x^{10} = 10!\). The lower-degree terms vanish under \(\Delta^{10}\). So

\[ \Delta^{10}[\ldots] = 24 \times 10! \]

12 would come from a sign or product slip in the leading coefficient, and 8! from the wrong degree.

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16.

Let \(\Delta\) be the forward difference operator, \(\nabla\) be the backward difference operator and \(\delta\) be the central difference operator. Then :

A.\(\nabla \Delta = \Delta - \nabla\)
B.\(\nabla \Delta \neq \delta^{2}\)
C.\(\Delta - \nabla \neq \delta^{2}\)
D.\(\nabla \Delta \neq \Delta - \nabla\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The correct relation is \(\nabla \Delta = \Delta - \nabla\).

Write the operators with the shift E: \(\Delta = E - 1\), \(\nabla = 1 - E^{-1}\), \(\delta = E^{1/2} - E^{-1/2}\). Then

\[ \begin{aligned} \nabla\Delta &= (1 - E^{-1})(E - 1) = E - 2 + E^{-1} \\ \Delta - \nabla &= (E - 1) - (1 - E^{-1}) = E - 2 + E^{-1} \\ \delta^{2} &= E - 2 + E^{-1} \end{aligned} \]

So \(\nabla\Delta = \Delta - \nabla = \delta^{2}\). That makes option A true and options B, C and D, each of which asserts one of these equalities fails, false.

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17.

Let f(x) be a polynomial of degree n. Then the nth forward difference of f(x), that is, \(\Delta^{n}[f(x)]\) is :

A.always 0
B.constant
C.always 1
D.cannot determine

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The nth forward difference is constant.

Each application of \(\Delta\) lowers the degree of a polynomial by one: if \(f(x) = a x^{n} + \ldots\), then \(\Delta f(x) = f(x + h) - f(x) = a n h\, x^{n-1} + \ldots\). After n applications the result is the constant

\[ \Delta^{n} f(x) = a\, n!\, h^{n} \]

It is not always 0 (that is \(\Delta^{n+1}\)), nor always 1 (it depends on the leading coefficient a and the step h), and it can certainly be determined.

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18.

\(\Delta\!\!\!|\,\) is a divided difference operator and \(\Delta\) is a forward difference operator, f(x) and g(x) are two functions and a and b are constants, then :

A.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a \Delta\!\!\!|\,f(x) + b \Delta\!\!\!|\,g(x)\)
B.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a \Delta\!\!\!|\,(f(x)\, g(x)) + b \Delta\!\!\!|\,(f(x) . g(x))\)
C.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a \Delta f(x) + b \Delta g(x)\)
D.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a\Delta(f(x)\, g(x)) + b \Delta (f(x)\, g(x))\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The correct property is linearity of the divided difference: the operator applied to \(af(x) + bg(x)\) equals a times its value on f plus b times its value on g.

A first divided difference is

\[ [x_{0}, x_{1}]f = \frac{f(x_{1}) - f(x_{0})}{x_{1} - x_{0}} \]

and replacing f by \(af + bg\) splits the numerator into a times the f part plus b times the g part. Higher divided differences are built recursively from the same quotients, so they are linear too.

Option C mixes in the forward difference \(\Delta\), a different operator. Options B and D replace f and g by their product, which has nothing to do with a linear combination.

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19.

Let f(x) be a function which takes the values \(y_{0}, y_{1}, \ldots, y_{n}\) for \(x = x_{0}, x_{1}, \ldots, x_{n}\) respectively and [a, b] be an interval divided into n subintervals of width h such that :

\[ \begin{aligned} &x_{0} = a, x_{1} = x_{0} + n, \ldots, \\ &x_{n} = x_{0} + nh = b. \end{aligned} \]

Then the error in the trapezoidal rule for integration is of order :

A.\(h\)
B.\(h^{2}\)
C.\(h^{3}\)
D.cannot determine

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The error of the trapezoidal rule is of order \(h^{2}\).

On one subinterval the trapezoidal rule has error \(-\frac{h^{3}}{12} f''(\xi)\). Over the n subintervals of [a, b], with \(nh = b - a\), the errors add up to

\[ E = -\frac{(b - a)h^{2}}{12} f''(\xi) \]

so the composite error is \(O(h^{2})\). The \(h^{3}\) in option C is the error on a single strip, and h alone would describe a first-order rule such as the rectangle rule.

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20.

The value of \(\Delta^{2} f(x)\), where \(f(x) = ab^{cx}\) is :

A.\((b^{ch} - 1)^{2}\, ab^{cx}\)
B.\((b^{ch} + 1)\, ab^{cx}\)
C.\((ab^{cx})^{2}\)
D.\((b^{ch} + 1)^{2}\, ab^{cx}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

\(\Delta^{2} f(x)\) equals \((b^{ch} - 1)^{2}\, ab^{cx}\).

With step h:

\[ \begin{aligned} \Delta f(x) &= ab^{c(x + h)} - ab^{cx} = (b^{ch} - 1)\, ab^{cx} \\ \Delta^{2} f(x) &= (b^{ch} - 1)\,\Delta\left(ab^{cx}\right) = (b^{ch} - 1)^{2}\, ab^{cx} \end{aligned} \]

Each application multiplies by the same factor \(b^{ch} - 1\), because an exponential is an eigenfunction of \(\Delta\).

The options with \(b^{ch} + 1\) have the wrong sign (they would come from a sum, not a difference), and \((ab^{cx})^{2}\) squares the function instead of applying \(\Delta\) twice.

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