y = f(x) = x³+2x+1 for x = 1,2,3,4,5

The Forward difference table is given below

Let the diameter be x and area be y.
To find y when x = 82, use Newton's forward interpolation form
∴ x0 + nh = 82 ⇒ 80 + n(5) ⇒ 82 - 80 = 2
⇒ n = \(\frac{2}{5}\) = 0.4
The difference table is

∴ y(82) = y0 + \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ \(\frac { n(n+1)(n+2)(n-3) }{ 4! } ​​{ \triangle }^{ 4 }{ y }_{ 0 }​​​​​​​​​​​​​​\)

+ \(\frac{(0.4)(0.4-1)(0.4-2)}{6}\)(-2)

= 5026 + 259.2 + (0.4) (-0.6) (20) + \(\frac{(0.4)(-0.6)(-1.6)(-1)}{3}\) + \(\frac{(0.4)(-0.6)(-1.6)(-2.6)}{3}\)
= 5026 + 259.2 - 4.8 - 0.128 - 0.1664
= 5280.10
∴ When the diameter is 82, area of circle is 5280.1 (≅ 5281)
To find y when x = 91, use Newton's backward interpolation formula.
∴ x_n + nh = 91 ⇒ 100 + n(5) = 91
⇒ 5n = 91 - 100
⇒ 5n = -9 ⇒ n = \(\frac{-9}{5}\) = 1.8
Newton's backward interpolation formula is
y_(x=xn+nh) = yn + \(\frac { n }{ 1! } \triangledown { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }​​​​​​​​​​​​​​\)
y(x = 91) = 7854 - 1.8 (766) +

\(+\frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{3!}\)(4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + (-1.8) (-0.8) (20) + (-1.8) (-0.8) (0.2) + \(\frac{(-1.8)(-0.8)(0.2)(1.2)}{6}\) (4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + 28.8 + 0.288 + 0.2304
⇒ y_{(x = 91)} = 6504.5
Hence when the diameter is 91, area is 6504.5 ≅ 6504

Since x = 43 lies in the beginning of the table, use.
Newton's forward interpolation formula.
∴ x₀+ nh = 43 ⇒ 40 + n(10) = 43 ⇒ 10n
=43 - 40 = 3
⇒ n = \(\frac{3}{10}\) = 0.3
Newton's forward interpolation formula is
y_(x=x0+nh) = y₀\(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }​​​​​​​​​​​​​​\)
The difference table is

= 184 + 6 + (0.3)(-0.7)
= 184 + 6 - 0.21
= 189.79
∴ when x = 43, y = 189.79.
To find y when x = 84, use Newton's backward interpolation formula
∴ x_n + nh = 84
90 + n(10) = 84 ⇒ 10n = 84 - 90 = -6
⇒ n = \(\frac{-6}{10}\) = -6
Newton's backward interpolation formula is
∴ y_(x=xn+nh)= y_n + \(\frac { n }{ 1! } \triangledown { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }​​​​​​​​​​​​​​\) 

= 304 - 16.8 + (-0.6) (0.4)
= 304 - 16.8 - 0.24
= 304 - 17.04 = 286.96
∴ when x = 84, y = 286.96

Given
The difference table is

To findy when x = x ⇒ x₀+ nh = x ⇒ 1 + n (1) = x ⇒ n = x-1
Newton's forward interpolation formula is
y_(x=x) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ .....
y_(x=x) = 1 + (x-1)(-2) + \(\frac { (x-1)(x-2) }{ 2 } (4)+\frac { (x-1)(x-2)(x-3) }{ 6 } (-8)+\frac { (x-1)(x-2)(x-3)(x-4) }{ 24 } (16)\)
⇒ y = 1 - 2x + 2(x² - 3x + 2) - \(\frac{4}{3}\) (x - 1) (x - 2) (x - 3) + \(\frac{2}{3}\) (x - 1) (x - 2) (x - 3) (x -4)
⇒ y = 3 - 2x + 2x² - 6x + 4 - \(\frac{4}{3}\) [(x² - 3x + 2) (x- 3)] + \(\frac{2}{3}\) [(x² - 3x + 2)(x² - 7x + 12)]
⇒ y = 2x² - 8x+ 7- \(\frac{4}{3}\) [x³- 3x²+ 2x- 3x² + 9x- 6] + \(\frac{2}{3}\) [x⁴ - 3x³ + 2x² - 7x³ + 21x² - 14x + 12x² - 36x + 24]
⇒ y = 2x²-8x+7- \(\frac{4}{3}\) x³ + 4x² - \(\frac{8}{3}\) x + 4x² - 12x + 8 + \(\frac { { 12x }^{ 4 } }{ 3 } -\frac { 20 }{ 3 } { x }^{ 3 }+\frac { 70 }{ 3 } { x }^{ 2 }-\frac { 100x }{ 3 } +\frac { 48 }{ 3 } \)
⇒ y = \(\frac{2}{3}\) x⁴ + x³ \(\left( \frac { -4 }{ 3 } \frac { -20 }{ 3 } \right) \) + x²\(\left( 2+4+4+\frac { 70 }{ 3 } \right) \) + x \(\left( -8-\frac { 8 }{ 3 } -12-\frac { 100 }{ 3 } \right) \) + 31
⇒ y =  \(\frac{2}{3}\) x⁴ - 8x³ + \(\frac{100}{3}\) x² - 56x + 31 which is the required polynomial.

Given

Here the intervals are unequal
∴ By Lagranges interpolation formula, we have
x₀ = 0, x₁ = 1, x₂ = 3, x₃ = 4
y₀ = -12, y₁ = 0, y₂ = 6, y₃ = 12 and x = x.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (x-1)(x-3)(x-4) }{ (0-1)(0-3)(0-4) } (-12)+\frac { (x-0)(x-3)(x-4) }{ (1-0)(1-3)(1-4) } (0)+\frac { (x-0)(x-1)(x-4) }{ (3-0)(3-1)(3-4) } (6)+\frac { (x-1)(x-3)(x-4) }{ (4-0)(4-1)(4-3) } (12)\)
= \(\frac { (x-1)(x-3)(x-4) }{ (-1)(-3)(-4) } (-12)+0+\frac { x(x-1)(x-4) }{ (3)(2)(-1) } (6)+\frac { x(x-1)(x-3) }{ (4)(3)(1) } (12)\)
= +[(x - 1)(x - 3)(x - 4)] - x (x - 1)(x -  4) + x(x - 1)(x - 3)
= +[(x³-4x+3)(x-4)] -x(x²-5x+4) + x(x²-4x + 3)
= - (x³ - 8x²+ 19x - 12) - 4x² + 3x
= (x - 4)(x² - 4x + 3) - x (x² - 5x + 4) + x(x² - 4x + 3)

= x³ - 7x² + 19x - 12.

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

Let the missing entries be y₂ 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
⇒ (E - 1)⁴y_k = 0
⇒ (E₄- 4E₃ + 6E₂ - 4E + 1) y_k = 0  (1)
Put k = 0 in (1) we get,
y₄ - 4y₃ + 6y₂ - 4y₁ + y₀ = 0
y₄ - 4(18) + 6y₂ - 4(11) + 7 = 0
⇒ y₄ - 72 + 6y₂ - 44 + 7 = 0
⇒ y₄ + 6y₂ = 109    (2)
Put k = 1 in (1) we get,
(E₄- 4E₃ + 6E₂ - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ +y₁ = 0
⇒ 32 - 4 (y₄) + 6 (18) - 4y₂ + 11 = 0
32 - 4y₄ + 108 - 4y₂ + 11 = 0
⇒ -4y₄ - 4y₂ + 151 = 0

Adding,           
⇒ y₂ = 14.25
Substituting y₂ = 14.25 in (2) we get,
y₄ + 6 (14.25) = 109
⇒ y₄ + 85.5 = 109
⇒ y₄ = 109 - 58.5
⇒ y₄ = 23.5

Using interpolation, find the value of f(x) when x =1
Here the intervals are unequal. By Lagrange's interpolation formula, we have
x₀=3, x₁ = 7, x₂ = 11, x₃ = 19
y₀ = 42, y₁ = 43, y₂ = 47, y₃ = 60 and x = 15
∴ Y =f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (15-7)(15-11)(15-19) }{ (3-7)(3-11)(3-19) } \times 42+\frac { (15-3)(15-7)(15-19) }{ (11-3)(11-7)(11-19) } \times 43+\frac { (15-3)(15-7)(15-19) }{ (11-3)(11-7)(11-19) } \times 47+\frac { (15-3)(15-7)(15-11) }{ (19-3)(19-7)(19-11) } \times 60\)

= \(\frac{21}{2}-43+70.5+15\)
= 10.5 - 43 + 70.5 + 15
y= 53
Hence when x = 15, f(x) = 53.

Given:

Here, the intervals are unequal:
∴ By Lagrange's interpolation formula, we have
x₀ = 1982, x₁ = 1983, x₂ = 1984, x₃ = 1986
y₀ = 150, y₁ = 235, y₂ = 365, y₃ = 525 and x = 1985
=\(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)  \(\frac { (1985-1983)(1985-1984)(1985-1986) }{ (1982-1983)(1982-1984)(1982-1986) } \times 150+\frac { (1985-1982)(1985-1984)(1985-1986) }{ (1983-1982)(1983-1984)(1983-1986) } \times 235+\frac { (1985-1982)(1985-1983)(1985-1986) }{ (1984-1982)(1984-1983)(1984-1986) } \times 365+\frac { (198 5-1982)(1985-1983)(1985-1984) }{ (1986-1982)(1986-1983)(1986-1984) } \times 525\)

\(=\frac { (2)(1)(-1) }{ (-1)(-2)(-4) } \times 150+\frac { (3)(1)(-1) }{ (1)(-1)(-3) } \times \) 
= 37.5 + 235 + 547.5 + 131.25
= 481.25

We know  f(x), ∇f(x + h), ∇f(x + 2h),... are the first order differences of f(x).
= f(x) - 2f(x - h) + f(x - 2h)
Consider ∇²f(x) = ∇[∇f(x))]
= ∇[f(x) - f(x - h)]
= ∇ f(X) - ∇ f(x - h)
= [f(x) - f(x - h)] - [f(x - h) - f(x - 2h)]
= f(x) - f(x - h) - f(x - h) - f(x - 2h)
∴ ∇² f(x), ∇² f(x + h), ∇² f(x + 2h) ... are the second order differences of f(x).

We know Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1

Since three values of f(x) are given, we assume that the polynomial is of degree two.
⇒ Δ³(f(x₀)) = 0
⇒ ∆³(y_o) = 0
⇒ (E - 1)³ y_o= 0
⇒ (E³ - 3E² + 3E - 1) y_o= 0
⇒ y₃- 3y₂+ 3y₁ - y_o= 0
⇒ 157 - 3 (126) + 3y₁ - 100 = 0
⇒ y₁ = 107
∴ The missing term is 107.

By data we have,
x₀ = 0, x₁ = 1, x₂ = 3
y₀ = 5, y₁ = 6, y₂ = 50 and x = 2
Using Lagrange's formula, we get
\(y=\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 0 }-{ x }_{ 1 }) } { y }_{ 2 }\)
= \(5\times \frac { (2-0)(2-3) }{ (0-1)(0-3) } +6\times \frac { (2-0)(2-3) }{ (1-0)(1-3) } +50\times \frac { (2-0)(2-1) }{ (3-0)(3-1) } \)

= \(\frac { 10 }{ 3 } -6+\frac { 50 }{ 3 } =\frac { 60 }{ 3 } -6=20-6=14\)
∴ f(2) = 14

Since only two values of yare given, the polynomial which fits the data is of degree 1.
Hence 2nd differences are zeros
∴ Δ²(y₀) = 0
⇒ (E-1)²y_o=0
⇒(E² - 2E + 1) y_o = 0
⇒y₂ - 2y₁ +y_o = 0
⇒34 - 2y₁ + 51 = 0
​​​​​​​⇒85 - 2y₁ =0
​​​​​​​⇒2y₁ + 85 ​​​​​​​⇒ y₁ = \(\frac{85}{2}\)
​​​​​​​⇒y₁ = 42.5