Interpolation Techniques Questions and Answers

Q: The value of \(\left(\dfrac{\Delta^{2}}{E}\right) x^{2}\) (taking \(h = 1\)) is :where :\(\Delta\) - forward difference operatorE - shift operator.

The value is 2.The shift operator acts by \(E f(x) = f(x + h)\), so \(E^{-1}f(x) = f(x - 1)\) with \(h = 1\), and \(\Delta\) commutes with E. Hence\[ \frac{\Delta^{2}}{E}x^{2} = \Delta^{2}(x - 1)^{2} \]The second forward difference of any quadratic \(ax^{2} + bx + c\) with step 1 is the constant \(2a\). Here \(a = 1\), so the result is 2, whatever the shift. Directly: \(\Delta x^{2} = 2x + 1\) and \(\Delta(2x + 1) = 2\).0 would be the third difference of a quadratic, and \(\pm 1\) come from forgetting the factor 2.

Q: If \(u_{0} = 1\), \(u_{1} = 5\), \(u_{2} = 8\), \(u_{3} = 3\), \(u_{4} = 7\), \(u_{5} = 0\), then the value of \(\Delta^{5} u_{0}\) is :where \(\Delta\) forward difference operator.

\(\Delta^{5}u_{0} =\) -61.Expand \(\Delta^{5} = (E - 1)^{5}\) with binomial coefficients:\[ \begin{aligned} \Delta^{5}u_{0} &= u_{5} - 5u_{4} + 10u_{3} \\ &\quad - 10u_{2} + 5u_{1} - u_{0} \\ &= 0 - 35 + 30 - 80 + 25 - 1 \\ &= -61 \end{aligned} \]A difference table gives the same result. Option B, \(-60\), is what you get by dropping the final \(-u_{0}\), and 60 and 24 come from sign errors in the alternating sum.

Q: Which of the following is not correct ?

The incorrect relation is \(\delta = E^{1/2} + E^{-1/2}\).The central difference is \(\delta = E^{1/2} - E^{-1/2}\), with a minus sign. The plus combination is \(2\mu\), twice the averaging operator.The others are correct. With \(\Delta = E - 1\) and \(\nabla = 1 - E^{-1}\): \(\Delta - \nabla = E - 2 + E^{-1} = (E - 1)(1 - E^{-1}) = \Delta\nabla\); \((1 + \Delta)(1 - \nabla) = E \cdot E^{-1} = 1\); and with \(\mu = \tfrac{1}{2}(E^{1/2} + E^{-1/2})\),\[ \begin{aligned} \mu^{2} &= \tfrac{1}{4}(E + 2 + E^{-1}) \\ &= 1 + \tfrac{1}{4}\delta^{2} \end{aligned} \]

Q: If \(u_{1} = 1\), \(u_{3} = 17\), \(u_{4} = 43\) and \(u_{5} = 89\), then value of \(u_{2}\) is :

\(u_{2} = \) 5.With four known values the missing one is found by assuming the data fit a cubic, so the fourth difference over five consecutive values is 0:\[ \begin{aligned} \Delta^{4}u_{1} &= u_{5} - 4u_{4} + 6u_{3} - 4u_{2} + u_{1} = 0 \\ 89 - 172 + 102 - 4u_{2} + 1 &= 0 \\ 20 &= 4u_{2},\ u_{2} = 5 \end{aligned} \]Check: the values 1, 5, 17, 43, 89 have differences 4, 12, 26, 46, then 8, 14, 20, then 6, 6, constant third differences as a cubic requires. The other options break that pattern.

Q: The third divided difference of the function \(f(x) = 1/x\) with arguments \(p, q, r, s\) is :

The third divided difference is \(-\dfrac{1}{pqrs}\).For \(f(x) = 1/x\):\[ \begin{aligned} f[p, q] &= \frac{1/q - 1/p}{q - p} = -\frac{1}{pq} \\ f[p, q, r] &= \frac{-\frac{1}{qr} + \frac{1}{pq}}{r - p} = \frac{1}{pqr} \\ f[p, q, r, s] &= -\frac{1}{pqrs} \end{aligned} \]The pattern is \(f[x_{0}, \ldots, x_{n}] = (-1)^{n}/(x_{0}\cdots x_{n})\). \(1/pqr\) is the second divided difference, \(1/pq\) the first with its sign dropped, and \(-s/pqr\) does not follow the pattern.

Q: Using Newton's forward interpolation formula, the cubic polynomial which takes the following value\(x\) :0123\(f(x)\) :12110is :

The polynomial is \(2x^{3} - 7x^{2} + 6x + 1\).Forward differences of 1, 2, 1, 10: first differences 1, -1, 9; second -2, 10; third 12. Newton's formula with \(h = 1\):\[ \begin{aligned} f(x) &= 1 + x - 2\frac{x(x-1)}{2} + 12\frac{x(x-1)(x-2)}{6} \\ &= 2x^{3} - 7x^{2} + 6x + 1 \end{aligned} \]Check: \(f(1) = 2 - 7 + 6 + 1 = 2\), \(f(2) = 16 - 28 + 12 + 1 = 1\), \(f(3) = 54 - 63 + 18 + 1 = 10\). Option B gives \(f(1) = 4\), option C gives \(f(1) = 2\) but \(f(2) = 17\), and option D gives \(f(1) = 8\).

Q: Which of the following is not correct ?

The incorrect relation is \(1 + \Delta = (E + 1)\nabla^{-1}\).With \(Ef(x) = f(x + h)\), \(\Delta = E - 1\) and \(\nabla = 1 - E^{-1}\):\(E = 1 + \Delta\): true by definition.\(1 - \nabla = E^{-1}\), so \(\nabla(1 - \nabla)^{-1} = \nabla E = E - 1 = \Delta\): true.\(E^{-1}\Delta = E^{-1}(E - 1) = 1 - E^{-1} = \nabla\): true.But \(\nabla^{-1} = E(E - 1)^{-1}\), so \((E + 1)\nabla^{-1} = E(E + 1)/(E - 1)\), which is not \(1 + \Delta = E\). So option D is the false one.

Q: Let \(f(x) = e^{x}\), then \(\Delta^{n} e^{x} = \ldots\)

\(\Delta^{n} e^{x} = e^{x}(e^{h} - 1)^{n}\).With step \(h\):\[ \begin{aligned} \Delta e^{x} &= e^{x + h} - e^{x} = e^{x}(e^{h} - 1) \\ \Delta^{2} e^{x} &= (e^{h} - 1)\Delta e^{x} = e^{x}(e^{h} - 1)^{2} \end{aligned} \]and by induction each application multiplies by the constant \(e^{h} - 1\). In operator form, \(\Delta = E - 1\) and \(E e^{x} = e^{h} e^{x}\).The factor \((e^{h} + 1)\) confuses \(\Delta\) with the averaging operator, \(e^{x}\) alone would be the derivative rather than the difference, and \(e^{-x}\) has no source in the computation.

Q: Everett's formula truncated after second differences is equivalent to ..................... truncated after third differences.

Everett's formula to second differences is equivalent to Bessel's formula to third differences.Both are central-difference formulas for interpolating between \(x_{0}\) and \(x_{1}\). Everett's formula uses only even-order differences (\(y_{0}, y_{1}, \delta^{2}y_{0}, \delta^{2}y_{1}, \ldots\)), while Bessel's uses means of even differences and odd differences. Rewriting the odd differences as differences of even ones turns Bessel's formula up to third differences term for term into Everett's formula up to second differences. Both reduce to the same cubic through the four points \(x_{-1}, x_{0}, x_{1}, x_{2}\).Stirling's formula (printed in option A as 'String's') is centred on a single tabulated point and is not equivalent. Lagrange's formula works with unequal spacing and is not a difference formula at all.

The value of \(\left(\dfrac{\Delta^{2}}{E}\right) x^{2}\) (taking \(h = 1\)) is :

where :
\(\Delta\) - forward difference operator
E - shift operator.

A.-1
B.0
C.1
D.2

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The value is 2.

The shift operator acts by \(E f(x) = f(x + h)\), so \(E^{-1}f(x) = f(x - 1)\) with \(h = 1\), and \(\Delta\) commutes with E. Hence

\[ \frac{\Delta^{2}}{E}x^{2} = \Delta^{2}(x - 1)^{2} \]

The second forward difference of any quadratic \(ax^{2} + bx + c\) with step 1 is the constant \(2a\). Here \(a = 1\), so the result is 2, whatever the shift. Directly: \(\Delta x^{2} = 2x + 1\) and \(\Delta(2x + 1) = 2\).

0 would be the third difference of a quadratic, and \(\pm 1\) come from forgetting the factor 2.

Discuss in Forum

If \(u_{0} = 1\), \(u_{1} = 5\), \(u_{2} = 8\), \(u_{3} = 3\), \(u_{4} = 7\), \(u_{5} = 0\), then the value of \(\Delta^{5} u_{0}\) is :

where \(\Delta\) forward difference operator.

A.24
B.-60
C.60
D.-61

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

\(\Delta^{5}u_{0} =\) -61.

Expand \(\Delta^{5} = (E - 1)^{5}\) with binomial coefficients:

\[ \begin{aligned} \Delta^{5}u_{0} &= u_{5} - 5u_{4} + 10u_{3} \\ &\quad - 10u_{2} + 5u_{1} - u_{0} \\ &= 0 - 35 + 30 - 80 + 25 - 1 \\ &= -61 \end{aligned} \]

A difference table gives the same result. Option B, \(-60\), is what you get by dropping the final \(-u_{0}\), and 60 and 24 come from sign errors in the alternating sum.

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Which of the following is not correct ?

A.\((\Delta - \nabla) = \Delta\nabla\)
B.\((1 + \Delta)(1 - \nabla) = 1\)
C.\(\mu^{2} = 1 + \dfrac{1}{4}\delta^{2}\)
D.\(\delta = E^{\frac{1}{2}} + E^{-\frac{1}{2}}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The incorrect relation is \(\delta = E^{1/2} + E^{-1/2}\).

The central difference is \(\delta = E^{1/2} - E^{-1/2}\), with a minus sign. The plus combination is \(2\mu\), twice the averaging operator.

The others are correct. With \(\Delta = E - 1\) and \(\nabla = 1 - E^{-1}\): \(\Delta - \nabla = E - 2 + E^{-1} = (E - 1)(1 - E^{-1}) = \Delta\nabla\); \((1 + \Delta)(1 - \nabla) = E \cdot E^{-1} = 1\); and with \(\mu = \tfrac{1}{2}(E^{1/2} + E^{-1/2})\),

\[ \begin{aligned} \mu^{2} &= \tfrac{1}{4}(E + 2 + E^{-1}) \\ &= 1 + \tfrac{1}{4}\delta^{2} \end{aligned} \]

Discuss in Forum

If \(u_{1} = 1\), \(u_{3} = 17\), \(u_{4} = 43\) and \(u_{5} = 89\), then value of \(u_{2}\) is :

A.3
B.5
C.13
D.15

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

\(u_{2} = \) 5.

With four known values the missing one is found by assuming the data fit a cubic, so the fourth difference over five consecutive values is 0:

\[ \begin{aligned} \Delta^{4}u_{1} &= u_{5} - 4u_{4} + 6u_{3} - 4u_{2} + u_{1} = 0 \\ 89 - 172 + 102 - 4u_{2} + 1 &= 0 \\ 20 &= 4u_{2},\ u_{2} = 5 \end{aligned} \]

Check: the values 1, 5, 17, 43, 89 have differences 4, 12, 26, 46, then 8, 14, 20, then 6, 6, constant third differences as a cubic requires. The other options break that pattern.

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The third divided difference of the function \(f(x) = 1/x\) with arguments \(p, q, r, s\) is :

A.\(\dfrac{1}{pq}\)
B.\(\dfrac{1}{pqr}\)
C.\(-\dfrac{1}{pqrs}\)
D.\(\dfrac{-s}{pqr}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The third divided difference is \(-\dfrac{1}{pqrs}\).

For \(f(x) = 1/x\):

\[ \begin{aligned} f[p, q] &= \frac{1/q - 1/p}{q - p} = -\frac{1}{pq} \\ f[p, q, r] &= \frac{-\frac{1}{qr} + \frac{1}{pq}}{r - p} = \frac{1}{pqr} \\ f[p, q, r, s] &= -\frac{1}{pqrs} \end{aligned} \]

The pattern is \(f[x_{0}, \ldots, x_{n}] = (-1)^{n}/(x_{0}\cdots x_{n})\). \(1/pqr\) is the second divided difference, \(1/pq\) the first with its sign dropped, and \(-s/pqr\) does not follow the pattern.

Discuss in Forum

Using Newton's forward interpolation formula, the cubic polynomial which takes the following value

\(x\) :	0	1	2	3
\(f(x)\) :	1	2	1	10

is :

A.\(2x^{3} - 7x^{2} + 6x + 1\)
B.\(2x^{3} + 7x^{2} - 6x + 1\)
C.\(2x^{3} + x^{2} - 2x + 1\)
D.\(2x^{3} + 2x^{2} + 3x + 1\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The polynomial is \(2x^{3} - 7x^{2} + 6x + 1\).

Forward differences of 1, 2, 1, 10: first differences 1, -1, 9; second -2, 10; third 12. Newton's formula with \(h = 1\):

\[ \begin{aligned} f(x) &= 1 + x - 2\frac{x(x-1)}{2} + 12\frac{x(x-1)(x-2)}{6} \\ &= 2x^{3} - 7x^{2} + 6x + 1 \end{aligned} \]

Check: \(f(1) = 2 - 7 + 6 + 1 = 2\), \(f(2) = 16 - 28 + 12 + 1 = 1\), \(f(3) = 54 - 63 + 18 + 1 = 10\). Option B gives \(f(1) = 4\), option C gives \(f(1) = 2\) but \(f(2) = 17\), and option D gives \(f(1) = 8\).

Discuss in Forum

Which of the following is not correct ?

A.\(E = 1 + \Delta\)
B.\(\Delta = \nabla(1 - \nabla)^{-1}\)
C.\(\nabla = E^{-1}\Delta\)
D.\(1 + \Delta = (E + 1)\nabla^{-1}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The incorrect relation is \(1 + \Delta = (E + 1)\nabla^{-1}\).

With \(Ef(x) = f(x + h)\), \(\Delta = E - 1\) and \(\nabla = 1 - E^{-1}\):

\(E = 1 + \Delta\): true by definition.
\(1 - \nabla = E^{-1}\), so \(\nabla(1 - \nabla)^{-1} = \nabla E = E - 1 = \Delta\): true.
\(E^{-1}\Delta = E^{-1}(E - 1) = 1 - E^{-1} = \nabla\): true.

But \(\nabla^{-1} = E(E - 1)^{-1}\), so \((E + 1)\nabla^{-1} = E(E + 1)/(E - 1)\), which is not \(1 + \Delta = E\). So option D is the false one.

Discuss in Forum

Let

\[ \begin{aligned} x^{(n)} &= x(x - h)(x - 2h) \cdots (x - (n - 2)h) \\ &\quad (x - (n - 1)h), \end{aligned} \]

where \(h\) is the step size. Then which of the following is correct ?

A.\(\Delta^{n} x^{(n)} = n! h^{n}\) and \(\Delta^{n+1} x^{(n)} = 0\)
B.\(\Delta^{n} x^{(n)} = n!\) and \(\Delta^{n+1} x^{(n)} \neq 0\)
C.\(\Delta^{n} x^{(n)} = n! h\) and \(\Delta^{n+1} x^{(n)} = (n + 1)!\)
D.\(\Delta^{n} x^{(n)} = n!\) and \(\Delta^{n+1} x^{(n)} = (n + 1)!\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The correct pair is \(\Delta^{n} x^{(n)} = n!\,h^{n}\) and \(\Delta^{n+1} x^{(n)} = 0\).

The factorial polynomial behaves under \(\Delta\) (step \(h\)) the way \(x^{n}\) behaves under differentiation:

\[ \begin{aligned} \Delta x^{(n)} &= nh\, x^{(n-1)} \\ \Delta^{2} x^{(n)} &= n(n-1)h^{2}\, x^{(n-2)} \\ \Delta^{n} x^{(n)} &= n!\, h^{n} \end{aligned} \]

The \(n\)th difference is the constant \(n!\,h^{n}\), so one more difference gives 0. This matches the general rule that the \((n+1)\)th difference of any polynomial of degree \(n\) vanishes.

\(n!\) alone is the case \(h = 1\), and options claiming \(\Delta^{n+1} x^{(n)} \neq 0\) or \((n+1)!\) contradict that rule.

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Let \(f(x) = e^{x}\), then \(\Delta^{n} e^{x} = \ldots\)

A.\(e^{x}(e^{h} - 1)^{n}\)
B.\(e^{x}(e^{h} + 1)^{n}\)
C.\(e^{x}\)
D.\(e^{-x}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

\(\Delta^{n} e^{x} = e^{x}(e^{h} - 1)^{n}\).

With step \(h\):

\[ \begin{aligned} \Delta e^{x} &= e^{x + h} - e^{x} = e^{x}(e^{h} - 1) \\ \Delta^{2} e^{x} &= (e^{h} - 1)\Delta e^{x} = e^{x}(e^{h} - 1)^{2} \end{aligned} \]

and by induction each application multiplies by the constant \(e^{h} - 1\). In operator form, \(\Delta = E - 1\) and \(E e^{x} = e^{h} e^{x}\).

The factor \((e^{h} + 1)\) confuses \(\Delta\) with the averaging operator, \(e^{x}\) alone would be the derivative rather than the difference, and \(e^{-x}\) has no source in the computation.

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10.

Everett's formula truncated after second differences is equivalent to ..................... truncated after third differences.

A.String's formula
B.Bessel's formula
C.Stirling's formula
D.Lagrange's formula

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

Everett's formula to second differences is equivalent to Bessel's formula to third differences.

Both are central-difference formulas for interpolating between \(x_{0}\) and \(x_{1}\). Everett's formula uses only even-order differences (\(y_{0}, y_{1}, \delta^{2}y_{0}, \delta^{2}y_{1}, \ldots\)), while Bessel's uses means of even differences and odd differences. Rewriting the odd differences as differences of even ones turns Bessel's formula up to third differences term for term into Everett's formula up to second differences. Both reduce to the same cubic through the four points \(x_{-1}, x_{0}, x_{1}, x_{2}\).

Stirling's formula (printed in option A as 'String's') is centred on a single tabulated point and is not equivalent. Lagrange's formula works with unequal spacing and is not a difference formula at all.

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