Finite Difference Operators

3.2. Finite Difference Operators#

3.2.1. Discrete Data Points and General Sequences#

For a function defined on a continuous domain

\[f=f(x), ~~ x\in \mathbb{R},\]

we can use a general sequence to represent it at discrete locations

\[\left\{f_j\right\}_{j=-\infty}^{+\infty},\]

where \(f_j = f(x_j)\).

Usually, though not necessarily, we let \(\left\{x_j\right\}_{j=-\infty}^{+\infty}\) be equally spaced, so

\[x_j= jh\]

where \(h=x_{j+1}-x_j\) is called the step size (or step length), \(h\in(0,+\infty)\).

Figure made with TikZ

Fig. 3.1 A general sequence \(\{f_j\}_{j=0}^{6}\).

Remark 3.1

In Section 2.2, a sequence is defined as a function of integers i.e.

\[y_j=y(j), \quad j \in \mathbb{Z}.\]

Here we generalise this concept to use a sequence to represent the function at discrete points,

\[y_j = y(x_j), \quad x_j = jh,~ j\in \mathbb{Z}\]

where \(x\) coordinates are not necessarily intergers.

3.2.2. Forward Difference Operator \(\Delta\)#

Definition 3.1 (Forward Difference)

Given a sequence \(\left\{f_j\right\}_{j=-\infty}^{+\infty}\), the first-order forward difference for an member \(f_j\) is the difference between the next member and itself:

(3.1)#\[\begin{equation} \Delta f_j = f_{j+1} - f_j, \end{equation}\]

and the \(k-\)th order forward difference is defined as:

(3.2)#\[\begin{equation} \Delta^k f_j= \Delta\left(\Delta^{k-1} f_j \right). \end{equation}\]

Example 3.1

  • \(\Delta f_0 = f_1 - f_0\)

  • \(\Delta f_{10}= f_{11} - f_{10}\)

Note

For function \(f(x)\), \(x\in\mathbb{R}\) with a step size \(h\),

\[\Delta f(x) = f(x+h)-f(x)\]

  • Second-order Forward Difference:

    \[\begin{align*} \Delta^2 f_j & = \Delta(\Delta f_j) \\ & = \Delta(f_{j+1} - f_j) \\ & = \Delta(f_{j+1}) - \Delta(f_j) \\ & = (f_{j+2} - f_{j+1}) - (f_{j+1} - f_j) \\ & = f_{j+2} - 2f_{j+1} + f_j \end{align*}\]

  • Third-order Forward Difference:

    \[\begin{align*} \Delta^3 f_j & = \Delta(\Delta^2 f_j)\\ & =\Delta(f_{j+2} - 2f_{j+1} + f_j) \\ & = (f_{j+3}-f_{j+2}) - 2(f_{j+2}-f_{j+1}) + (f_{j+1} - f_{j}) \\ & = f_{j+3} - 3f_{j+2} + 3f_{j+1} - f_j \end{align*}\]

  • Higher-order Forward Difference:

    \[ \Delta^n f_j = f_{j+n} - nf_{j+n-1} + \frac{n(n-1)}{2!}f_{j+n-2} - \frac{n(n-1)(n-2)}{3!}f_{j+n-3} + \dots \]

3.2.3. Backward difference operator \(\nabla\)#

Definition 3.2 (Backward Difference)

Given a sequence \(\left\{f_j\right\}_{j=-\infty}^{+\infty}\), the first-order backward difference for an member \(f_j\) is the difference between itself and the previous member:

(3.3)#\[\begin{equation} \nabla f_j = f_{j} - f_{j-1}, \end{equation}\]

and the \(k\)-th order backward difference is defined as:

(3.4)#\[\begin{equation} \nabla^k f_j= \nabla\left(\nabla^{k-1} f_j \right). \end{equation}\]

Example 3.2

  • \(\nabla f_1 = f_1 - f_0\)

  • \(\nabla f_0 = f_0 - f_{-1}\)

Note

For function \(f(x)\), \(x\in\mathbb{R}\) with a step size \(h\),

\[\nabla f(x) = f(x)-f(x-h)\]

  • Second-order Backward Difference:

    \[\begin{align*} \nabla^2 f_j & = \nabla(\nabla(f_j)) \\ & = \nabla(f_j - f_{j-1}) \\ & = \nabla(f_j) - \nabla(f_{j-1}) \\ & = (f_j - f_{j-1}) - (f_{j-1} - f_{j-2}) \\ & = f_j - 2f_{j-1} + f_{j-2} \end{align*}\]

  • Third-order Backward Difference:

    \[\begin{align*} \nabla^3 f_j & = \nabla(\nabla^2 f_j)\\ & =\nabla(f_{j} - 2f_{j-1} + f_{j-2}) \\ & = (f_{j}-f_{j-1}) - 2(f_{j-1}-f_{j-2}) + (f_{j-2} - f_{j-3}) \\ & = f_j - 3f_{j-1} + 3f_{j-2} - f_{j-3} \end{align*}\]

  • Higher-order Forward Difference:

    \[ \nabla^n f_j = f_j - nf_{j-1} + \frac{n(n-1)}{2!}f_{j-2} - \frac{n(n-1)(n-2)}{3!}f_{j-3} + \dots \]

3.2.4. Shift operators \(\E\) and \(\E ^{-1}\)#

Definition 3.3

Given a sequence \(\left\{f_j\right\}_{j=-\infty}^{+\infty}\), we define \(\E\) as an operator shifting a member in a sequence to the next member

(3.5)#\[\begin{equation} \E f_j = f_{j+1}, \end{equation}\]

and shifting a member forward \(k\) times gives

\[\E ^k f_j= \E (\E ^{k-1}f_j)= f_{j+k}.\]

Similarly, we define \(\E ^{-1}\) as an operator shifting a member in a sequence to the previous member

(3.6)#\[\begin{equation} \E ^{-1} f_j = f_{j-1} \end{equation}\]

and shifting a member backward \(k\) (\(k>0\)) times gives

\[\E ^{-k} f_j= \E ^{-1}(\E ^{-(k-1)}f_j)= f_{j-k}.\]

Note

The forward and backward shifting operations can also be applied to functions defined on continuous domains

  • \(\E f(x)= f(x+h)\),

  • \(\E ^{-1} f(x) = f(x-h)\).

If we do shifting for \(k\) times , then

  • \(\E ^k f(x)= f(x+kh)\)

  • \(\E ^{-k} f(x) = f(x-kh)\)

Questions

  • If we shift forward for \(k=0.9\) steps, what we will get?

\[\E ^{0.9} f_j= \E ^{0.9} f(x_j)= f(x_j+0.9h)=?\]

  • If we shift backward for \(k=0.9\) steps, what we will get?

\[\E ^{-0.9} f_j= \E ^{-0.9} f(x_j)= f(x_j-0.9h)=?\]

Theorem 3.1

The difference operators and shift operators have the following relations:

  • \(\E =1+\Delta\)

  • \(\E ^{-1}=1-\nabla\)

Proof (click to show)

Applying the operators on \(f(x)\)

  • \(\E =1+\Delta\)

\[\begin{split} \begin{aligned} (1+\Delta) f(x) & = f(x) + \Delta f(x) \\ & = f(x) + f(x+h)-f(x) \\ & = f(x+h) \\ & = \E f(x) \end{aligned} \end{split}\]

  • \(\E ^{-1}=1-\nabla\)

\[\begin{split} \begin{aligned} (1-\nabla) f(x) & = f(x) - \nabla f(x) \\ & = f(x) - \left[f(x)-f(x-h)\right]\\ & = f(x-h) \\ & = \E ^{-1} f(x) \end{aligned} \end{split}\]
Contents