Q.Construct the forward difference table for the data $x=0,1,2,3,4$; $y=1,3,7,13,21$, and state $\Delta^3y_0$.

$\Delta(x^2)=(x+1)^2-x^2=2x+1$ when $h=1$.

The forward difference operator is defined by $\Delta f(x)=f(x+h)-f(x)$. With $h=1$ and $f(x)=x^2$:

$\Delta(x^2) = (x+1)^2 - x^2 = (x^2 + 2x + 1) - x^2 = 2x + 1.$

The backward difference operator is defined as $\nabla f(a)=f(a)-f(a-h)$.

In finite differences, $\nabla$ is the backward difference operator. For a step size $h$ it is defined by

$\nabla f(a) = f(a) - f(a-h).$

(Contrast with the forward difference $\Delta f(a)=f(a+h)-f(a)$.) So the correct expression subtracts the preceding value $f(a-h)$ from $f(a)$.

The backward difference $\nabla f(x)=f(x)-f(x-h)$; with $E^{-1}f(x)=f(x-h)$ this gives $\nabla\equiv 1-E^{-1}$, option (a).

Definitions. The shift operator satisfies $E^{-1}f(x)=f(x-h)$. The backward difference operator is

$\nabla f(x)=f(x)-f(x-h).$

Combine.

$\nabla f(x)=f(x)-E^{-1}f(x)=(1-E^{-1})f(x).$

Hence, as operators, $\nabla\equiv 1-E^{-1}$. (For contrast, the forward difference is $\Delta\equiv E-1$.)

$\Delta f(x)=f(x+1)-f(x)=2x+3$.

The forward difference with interval $h=1$ is $\Delta f(x)=f(x+1)-f(x)$.

$f(x+1)=(x+1)^2+2(x+1)+2=x^2+4x+5,$

$\Delta f(x)=(x^2+4x+5)-(x^2+2x+2)=2x+3.$

$(1+\Delta)(1-\nabla)=E\,E^{-1}=1$.

The standard operator relations are $E=1+\Delta$ and $\nabla=1-E^{-1}$, so $1-\nabla=E^{-1}$. Therefore

$(1+\Delta)(1-\nabla)=E\cdot E^{-1}=1.$

By definition $Ey_x=y_{x+h}$ and $\Delta y_x=y_{x+h}-y_x$; combining gives the standard operator relation $E=1+\Delta$.

The shifting (displacement) operator is defined by $Ey_x=y_{x+h}$, and the forward difference operator by

$\Delta y_x=y_{x+h}-y_x=Ey_x-y_x=(E-1)y_x.$

Therefore $\Delta=E-1$, which rearranges to

$E=1+\Delta.$

The operator $E$ advances the argument by one interval each time it acts, so $E(Ey_0)=E^2y_0=y_2$.

The shifting operator satisfies $E y_n=y_{n+1}$. Apply it in stages:

$E y_0=y_1,$

$E(E y_0)=E y_1=y_2.$

Equivalently $E^2 y_0=y_{0+2}=y_2$.

$\Delta f(x) = f(x+h) - f(x)$.

In the finite-differences part of the Tamil Nadu HSC Business Mathematics syllabus, the forward difference operator $\Delta$ is defined for a step size $h$ as the value one step ahead minus the current value:

$\Delta f(x) = f(x+h) - f(x).$

(For comparison, the backward difference is $\nabla f(x) = f(x) - f(x-h)$.)

$\nabla f(a) = f(a) - f(a-h)$.

The backward difference operator $\nabla$ with step size $h$ is defined as the current value minus the value one step behind:

$\nabla f(a) = f(a) - f(a-h).$

(For comparison, the forward difference is $\Delta f(a) = f(a+h) - f(a)$.)

Apply the forward-difference operator twice: $\Delta y_0 = y_1 - y_0$, then $\Delta^2 y_0 = \Delta y_1 - \Delta y_0 = y_2 - 2y_1 + y_0$.

Step 1 — First forward differences.

$\Delta y_0 = y_1 - y_0, \qquad \Delta y_1 = y_2 - y_1.$

Step 2 — Second forward difference.

$\Delta^2 y_0 = \Delta(\Delta y_0) = \Delta y_1 - \Delta y_0 = (y_2 - y_1) - (y_1 - y_0).$

Step 3 — Simplify.

$\Delta^2 y_0 = y_2 - y_1 - y_1 + y_0 = y_2 - 2y_1 + y_0.$