# Phorgy Phynance

## Network Theory and Discrete Calculus – Differentiation Rules

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This post is part of a series

In the previous post, we introduced discrete calculus on a binary tree. In particular, we introduced two sets of basis 1-forms we’ll refer to as

1. Graph bases $\{du^+,du^-\}$ and
2. Coordinate bases $\{dt,dx\}$

related by

$du^\pm = \Delta u \mathbf{e}^\pm = \frac{\Delta u}{2\Delta t} dt \pm \frac{\Delta u}{2\Delta x} dx$

and saw that discrete 1-forms can be expressed in either left- or right-components forms

$\alpha = \overleftarrow{\alpha_t} dt + \overleftarrow{\alpha_x} dx = dt \overrightarrow{\alpha_t} + dx \overrightarrow{\alpha_x},$

where, in general, the left- and right-component 0-forms do not coincide, i.e.

$\overleftarrow{\alpha_t} \ne \overrightarrow{\alpha_t}$ and $\overleftarrow{\alpha_x} \ne \overrightarrow{\alpha_x}$

due to the noncommutativity of discrete 0-forms and discrete 1-forms.

### Product Rule

Recall the exterior derivative of a discrete 0-form $f$ may be expressed in left-component form as

$df = \overleftarrow{\partial_t f} dt + \overleftarrow{\partial_x f} dx$

where

\begin{aligned} \overleftarrow{\partial_t f} = \sum_{(i,j)} \left[\frac{f(i+1,j+1)+f(i-1,j+1) -2f(i,j)}{2\Delta t}\right] \mathbf{e}^{(i,j)} \end{aligned}

and

\begin{aligned} \overleftarrow{\partial_x f} = \sum_{(i,j)} \left[\frac{f(i+1,j+1) -f(i-1,j+1)}{2\Delta x}\right] \mathbf{e}^{(i,j)} \end{aligned}

Although the product rule is satisfied, i.e.

$d(fg) = (df)g + f(dg),$

note the discrete 0-form $g$ is on the right of the discrete 1-form $df$ and the discrete 0-form $f$ is on the left of the discrete 1-form $dg$. Attempting to express the product rule in left components, we find

\begin{aligned} d(fg) &= \overleftarrow{\partial_t (fg)} dt + \overleftarrow{\partial_x (fg)} dx \\ &= (\overleftarrow{\partial_t f} dt + \overleftarrow{\partial_x f} dx )g + f(\overleftarrow{\partial_t g} dt + \overleftarrow{\partial_x g} dx).\end{aligned}

In order to move $g$ to the left of the coordinate bases above, we need to know the commutation relations

$[dt,g]$ and $[dx,g].$

These commutation relations may be determined by noting that for any two discrete 0-forms $f$ and $g$, we have

$[df,g] = [dg,f].$

Therefore,

\begin{aligned} { [dt,g] }&= [dg,t] \\ &= \overleftarrow{\partial_t g} [dt,t] + \overleftarrow{\partial_x g} [dx,t] \\ &= \Delta t \overleftarrow{\partial_t g} dt + \Delta t \overleftarrow{\partial_x g} dx\end{aligned}

and

\begin{aligned} { [dx,g] }&= [dg,x] \\ &= \overleftarrow{\partial_t g} [dt,x] + \overleftarrow{\partial_x g} [dx,x] \\ &= \Delta t \overleftarrow{\partial_t g} dx + \frac{(\Delta x)^2}{\Delta t} \overleftarrow{\partial_x g} dt,\end{aligned}

where use has been made of the coordinate commutation relations in the previous post.

Putting everything together, we find the product rule above implies the left components satisfy

$\overleftarrow{\partial_t (fg)} = (\overleftarrow{\partial_t f}) g + f(\overleftarrow{\partial_t g})+ \Delta t (\overleftarrow{\partial_t f})(\overleftarrow{\partial_t g}) + \frac{(\Delta x)^2}{\Delta t} (\overleftarrow{\partial_x f})( \overleftarrow{\partial_x g})$

and

$\overleftarrow{\partial_x (fg)} = (\overleftarrow{\partial_x f}) g + f(\overleftarrow{\partial_x g})+ \Delta t (\overleftarrow{\partial_x f})(\overleftarrow{\partial_t g}) + \Delta t (\overleftarrow{\partial_t f})( \overleftarrow{\partial_x g})$

### Change of Variables

Change of variables is something straightforward, yet has many applications, so it is worth writing it down here for future reference.

Let $S$ be a discrete 0-form with

$dS = \overleftarrow{\partial_t S} dt + \overleftarrow{\partial_x S} dx.$

If $\overleftarrow{\partial_x S}$ is invertible, we can rewrite this as

$dx = \frac{1}{\overleftarrow{\partial_x S}} (dS - \overleftarrow{\partial_t S} dt).$

Given any other discrete 0-form $V$, we have

$dV = \left.\overleftarrow{\partial_t V}\right|_x dt + \overleftarrow{\partial_x V} dx = \left.(\overleftarrow{\partial_t V} \right|_x- \frac{\overleftarrow{\partial_x V}}{\overleftarrow{\partial_x S}} \left.\overleftarrow{\partial_t S}\right|_x) dt + \frac{\overleftarrow{\partial_x V}}{\overleftarrow{\partial_x S}} dS.$

From this, we can read off the discrete chain rules

$\overleftarrow{\partial_x V} = \overleftarrow{\partial_S V} \overleftarrow{\partial_x S}$

and

$\left.\overleftarrow{\partial_t V}\right|_S = \left.\overleftarrow{\partial_t V} \right|_x-\overleftarrow{\partial_S V} \left.\overleftarrow{\partial_t S}\right|_x.$

Written by Eric

January 8, 2012 at 11:20 pm