(i) \(\Delta \equiv E-1 \text { (or) } E \equiv 1+\Delta\)
(ii) \( \mathrm{E} \Delta \equiv \Delta \mathrm{E} \)
(iii) \( \nabla \equiv \frac{\mathrm{E}-1}{\mathrm{~F}}\)

\(\Delta \mathrm{f}(x)=\mathrm{f}(x+\mathrm{h})-\mathrm{f}(x) \text { and } \mathrm{E} \mathrm{f}(x)=\mathrm{f}(x+\mathrm{h})\)
Hence h is equal gap
\( \therefore \Delta =\Delta \mathrm{f}(x)=\mathrm{f}(x+\mathrm{h})-\mathrm{f}(x) \)
\(\Delta \mathrm{f}(x) =\mathrm{E} f(x)-\mathrm{f}(x) \)
\(\Delta \mathrm{f}(x) =(\mathrm{E}-1) \mathrm{f}(x) \)
\(\therefore \Delta =\mathrm{E}-1\)

f(0) = 1; f(1) = 3; f(3) = 55

Forward difference table:

\(\mathrm{y}=\mathrm{f}(x)=x^{2}+2 x+2\)

We know that \(f(3)-f(2)=\Delta f(2)\)
\( \because \mathrm{f}(2)-\mathrm{f}(1) =\Delta \mathrm{f}(1) \)
\(\Delta \mathrm{f}(2) =\Delta[\mathrm{f}(1)+\Delta \mathrm{f}(1)] \)
\( =\Delta \mathrm{f}(1)+\Delta^{2} \mathrm{f}(1) \)
\(\mathrm{f}(3) =\mathrm{f}(2)+\Delta \mathrm{f}(1)+\Delta^{2} \mathrm{f}(1)\)

\( \Delta\left(\frac{1}{x}\right) =\left(\frac{1}{x+1}-\frac{1}{x}\right) \)
\(=\frac{x-x-1}{(x+1)(x)}=\frac{-1}{(x)(x+1)} \)

\( \Delta \log a x =\log a(x+h)-\log a x \)
\( =\log \left(\frac{a x+a h}{a x}\right) \)
\( =\log \left(\frac{a x}{a x}+\frac{a h}{a x}\right) \)
\( =\log \left(1+\frac{h}{x}\right) \)

\( \Delta^{2} \mathrm{e}^{a x} =\Delta\left[\Delta \mathrm{e}^{a x}\right] \)
\(=\Delta\left[a^{a(x+h)}-e^{a x}\right] \)
\( =\Delta e^{a x}\left[e^{a h}-1\right] \)
\( =\left(e^{a h}-1\right) \Delta e^{a x} \)
\(=\left(e^{a h}-1\right)\left(e^{a(x+h)}-e^{a x}\right) \)
\( =\left(e^{a h}-1\right) e^{a x}\left(e^{a h}-1\right)=\left(e^{a h}-1\right)^{2} e^{a x}\)

\(\Delta \mathrm{e}^{x}=\mathrm{e}^{x+h}-e^{x}=e^{x}\left(e^{h}-1\right)\)

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