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.

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.

Discuss in Forum

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.

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