Given f(x) = x²+ 3h
LHS = Δ f(x)
= f(x + h) - f(x)
= [(x + h)² + 3 (x + h)] - [x² + 3x]

= h² + 2xh + 3h
when h = 1,
LHS = 1² + 2x(1) + 3(1)
= 1+2x+3
= 2x+4= RHS
Hence proved.

Δ(log ax) = log (ax + h) -log (ax)
= log\(\left( \frac { ax+h }{ ax } \right) \)
=log \(\left( \frac { ax }{ ax } +\frac { h }{ ax } \right) \)
= log \(\left( 1+\frac { h }{ ax } \right) \)
∴ ∆ (log ax) = log \(\left( 1+\frac { h }{ ax } \right) \)

Δ²U₀ = (E-1)⁴U₀
= (E⁴ − 4E³+ 6E²− 4E+1)U₀
= E⁴U₀ - 4E³U₀ + 6E²U₀− 4EU₀+ U₀
= U₄ − 4U₃ + 6U₂ − 4U₁ + U₀
= 29 − 4(28) + 6(21) − 4(11) + 1.
= 156 – 156 = 0

We know that f(4) − f(3) =  Δf(3)
f(4) − f(3) = Δf(3)
= Δ[f(2) + Δf(2)] ∵[f(3) - f(2) = Δf(2)]
= Δf(2) +Δ²f(2)
= Δf(2) + Δ² [f(1) + Δf(1)]
∴ f(4) = f(4) + Δf(2) +  Δ² f(1) +  Δ³f(1)

y = f(x) = x³+2x+1 for x = 1,2,3,4,5

Since only four values are given,
(E -1)4 u₀ = 0
⇒ (E⁴ - 4E³ + 6E² - 4E + 1) u₀
= u₄ - 4 u₃ + 6 u₂ - 4 u₁ + u₀
⇒ 385 - 4(u₃) + 6(520) - 4 (556) + 560 =0
⇒ 385 - 4u₃ + 3120 - 2224 + 560 = 0
⇒ 1841 - 4u₃ = 0
⇒ 1841 = 4u₃
⇒ u₃ = \(\frac{1841}{4}\) = 460.25
∴ u₃ = 460.25

Given

Since we have to find f(0.5) which is at the beginning of the table, use Newton's forward interpolation formula.
x_n + nh = 0.5 ⇒ -1 + n(1) = 0.5
⇒ n = 0.5 + 1 = 1.5
∴ y_(0.5) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)
The difference table is

∴ y(0.5) = 202+\(\frac { 1.5 }{ 1! } (-27)+\frac { (1.5)(1.5-1) }{ 2! } (-66)+\frac { (1.5)(1.5-1)(1.5-2) }{ 3! } (132)\)
= 202 - 40.5 + (1.5) (.5) (-33) + \(\frac { (1.5)(.5)(-0.5) }{ 6(132) } \) (132)
= 202 - 40.5 - 24.75 - 8.25
= 202 -73.5
= 128.5
Hence f(0.5) = 128.5

Here the intervals are unequal
∴ By Lagrange'sinterpolation formula, we have
x₀ = 15, x₁ = 25, x₂ = 30, x₃ = 35
y₀= 36, y₁ = 40, y₂ = 45, y₃ = 38, and x = 26
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (26-25)(26-30)(26-35) }{ (15-25)(15-30)(15-35) } (36)+\frac { (26-15)(26-1530)(26-35) }{ (25-15)(5-30)(25-35) } (40)+\frac { 26-15)(26-25)(26-35) }{ (30-15)(30-30)(30-35) } (40)+\frac { (26-15)(26-25)(26-30) }{ (35-15)(35-25)(35-30) } (38)\)
= \(\frac { (1)(-4)(-9) }{ (-10)(-15)(-20) } (36)+\frac { (n)(-4)(-9) }{ (10)(-5)(-10) } (40)+\frac { (11)(1)(-9) }{ (15)(5)(-5) } (45)+\frac { (9)(1)(-4) }{ (-20)(10)(5) } (38)\)
= \(-\frac { (36)(36) }{ (150)(20) } +\frac { (44)(9)(40) }{ 500 } +\frac { 99(45) }{ (15)(25) } -\frac { (36)(38) }{ (200)(15) } \)
= \(\frac { 1296 }{ 3000 } +\frac { 15840 }{ 500 } +\frac { 4455 }{ 375 } -\frac { 1368 }{ 1000 } \)
= -0.432 + 31.68 + 11.88 - 1.368
= 41.76.
Hence the number of workers getting income not exceeding Rs. 26 per month is 42.

The forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
Here x₀ + nh = x ⇒ x₀ = 0, h = 1
∴ 0 + n = x ⇒ n = x.
The difference table is

\({ y }_{ (n=x) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\(y=1+\frac { n }{ 1! } (1)+\frac { n(n-1) }{ 2! } (-2)+\frac { n(n-1)(n-2) }{ 6 } (12)\)
⇒ y = 1 + x + (x² - x)(-1) + x(x² - 3x + 2) (2)
⇒ y = 1 + x - x² + x + 2x³ - 6x² + 4x
⇒ y = 1+ 6x - 7x² + 2x³
Hence, the cubic polynomial is 2x³ - 7x²+ 6x + 1.

Here the intervals are unequal. By Lagrange’s interpolation formula we have
x₀ = 5, x₁ = 6, x₂ = 9, x₃ = 11
y₀ = 12, y₁ = 13, y₂ = 14, y₃ = 16
\(y=f(x)=\frac { \left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 0 }-{ x }_{ 1 } \right) \left( x_{ 0 }-{ x }_{ 2 } \right) \left( { x }_{ 0 }-{ x }_{ 3 } \right) } \times { y }_{ 0 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 1 }-{ x }_{ 0 } \right) \left( x_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) } \times { y }_{ 1 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 2 }-{ x }_{ 0 } \right) \left( x_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 2 }-{ x }_{ 3 } \right) } \times { y }_{ 2 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 3 }-{ x }_{ 0 } \right) \left( x_{ 3 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 2 } \right) } \times { y }_{ 3 }\)
\(=\frac { (x-6)(x-9)(x-11) }{ (5-6)(5-6)(5-11) } (12)+\frac { (x-5)(x-9)(x-11) }{ (6-5)(6-9)(6-9) } (13)+\frac { (x-5)(x-6)(x-11) }{ (9-5)(9-6)(9-11) } (14)+\frac { (x-5)(x-6)(x-9) }{ (11-5)(11-6)(11-9) } (16)\)
Put x = 10
\({ y }_{ (10) }=f\left( 10 \right) =\frac { 4(1)(-1) }{ (-1)(-4)(-6) } (12)+\frac { (5)(1)(-1) }{ (1)(-3)(-5) } (13)+\frac { 5(4)(-1) }{ 4(3)(-2) } (14)+\frac { (5)(4)(1) }{ 6(5)(2) } (16)\)}
= \(\frac { 1 }{ 6 } \left( 12 \right) -\frac { 13 }{ 3 } +\frac { 5\left( 14 \right) }{ 3\times 2 } +\frac { 4\times 16 }{ 12 } \)
= 14.6663

Let the missing entries by y₁ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero.
∴ Δ⁴y_k = 0 ⇒ (E - 1)⁴y_k = 0
(E⁴ - 4E³ + 6E²- 4E + 1) y_k = 0
Put k = 0 in (1) we get,
(E₄ - 4E₃ + 6E₂ - 4E + 1)y₀ = 0
⇒ y₄ - 4y₃ + 5y₂ - 4y₁ + y₀ = 0
⇒ y₄ - 4(15) + 6(8) - 4(y₁) + 0 = 0
⇒ y₄ - 60 + 48 - 4y₁ = 0
⇒ y₄ - 4y₁ - 12 = 0
⇒ y₄ - 4y₁ = 12  (2)
put k = 1 in (1) we get
(E⁴- 4E³ + 6E² - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ + y₁ = 0
⇒ 35 - 4y₄ + 6 (15) - 4 (8) +y₁ = 0
⇒ 35 - 4y₄ + 90 - 32 +y₁ = 0
⇒ 4y₄ +y₁ = -93  (3)

⇒ y₁ = \(\frac{-45}{-15}\) = 3
⇒ y₁ = 3.
Substituting y₁ = 3 in (2) we get,
y₄ - 4(3) = 12
y₄ - 12 = 12
⇒ y₄ = 12 + 12
⇒ y₄ = 24
Hence the missing entries are 3 and 24.

Since five values are given, the polynomial which fits the data is of degree four.
Hence Δ⁵y_k = 0 (i.e) (E−1)⁵y_k = 0
i.e., (E⁵ - 5E⁴ + 10E³ - 10E² + 5E - 1)y_k = 0
E⁵y_k - 5E⁴y_k+ 10E³y_k- 10E²y_k+ 5Ey_k - y_k = 0 (1)
Put k = 0 in (1)
E⁵y₀ - 5E⁴y₀+ 10E³y₀- 10E²y₀+ 5Ey₀ - y₀ = 0
y₅ − 5y₄ + 10y₃ − 10y₂ +5 y₁ - y₀ = 0
y₅ − 5(350) +10y₃−10(260)+5(220)− 200 = 0
y₅ + 10y₃ = 3450 (2)

Put k = 1 in (1)
E⁵y₁ - 5E⁴y₁+ 10E³y₁- 10E²y₁+ 5Ey₁- y₀ = 0
y₆ − 5y₅ + 10y₄ −1 0y₃ − y₁ = 0
430  −5y₅ +10(350) −10y₃ + 5(260)− 220 = 0
5y₅+10y₃ = 5010 (3)
(3) – (2) ⇒ 4y₅ = 1560
y₅ = 390
From (1)  390 +10y₃ = 3450
10y₃ = 3450 – 390
y₃ ≅ 306

Let x be the marks and y be the number of students
By converting the given series into cumulative frequency distribution, the difference table is as follows.

\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 45
\(\therefore\) x₀+nh = 45 , x0 = 40, h = 10 \(\Rightarrow n=\frac { 1 }{ 2 } \)
y_{(x = 45)} = \(31+\frac { 1 }{ 2 } \times 42+\frac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) }{ 2 } (9)+\cfrac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) \left( \frac { -3 }{ 2 } \right) }{ 6 } \times \left( -25 \right) +\frac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) \left( \frac { -3 }{ 2 } \right) \left( \frac { -5 }{ 2 } \right) }{ 24 } \times \left( -37 \right) \)
= \(31+21-\frac { 9 }{ 8 } -\frac { 25 }{ 16 } -\frac { 37\times 15 }{ 384 } \)
= 47.867 ≅ 48

(d) (E - 1)⁴y₃
Hint: (a), (b), (c) are different forms of ∆⁴y₃ but not (d).

(b) dependent variable
Hint: (a), (c), (d) are other names for x - variable but not (b).

(d) F(x + h)
Hint: (a), (b), (c) are values of E² F(x) but not (d).

(b) 3
Hint: The correct value is 5.

(c) linear polynomial

(b) Eⁿf(x) = f(x)
Hint: Eⁿf(x) = f(x + nh)