Floating point arithmetic is relatively scaled, which means that the precision that you get from calculations is relative to the size of the floating point numbers. Generally, you have 16 digits of accuracy in (64-bit) floating point operations. To measure this, we define *machine epsilon* as the value by which `1 + E = 1`

. For floating point numbers, this is:

eps(Float64)

2.220446049250313e-16

However, since it's relative, this value changes as we change our reference value:

@show eps(1.0)

eps(1.0) = 2.220446049250313e-16

@show eps(0.1)

eps(0.1) = 1.3877787807814457e-17

@show eps(0.01)

eps(0.01) = 1.734723475976807e-18 1.734723475976807e-18

Thus issues with *roundoff error* come when one subtracts out the higher digits. For example, $(x + \epsilon) - x$ should just be $\epsilon$ if there was no roundoff error, but if $\epsilon$ is small then this kicks in. If $x = 1$ and $\epsilon$ is of size around $10^{-10}$, then $x+ \epsilon$ is correct for 10 digits, dropping off the smallest 10 due to error in the addition to $1$. But when you subtract off $x$, you don't get those digits back, and thus you only have 6 digits of $\epsilon$ correct.

Let's see this in action:

ϵ = 1e-10rand() @show ϵ

ϵ = 5.7848124379111953e-11

@show (1+ϵ)

1 + ϵ = 1.0000000000578482

ϵ2 = (1+ϵ) - 1 (ϵ - ϵ2)

-4.631898182815095e-17

See how $\epsilon$ is only rebuilt at accuracy around $10^{-16}$ and thus we only keep around 6 digits of accuracy when it's generated at the size of around $10^{-10}$!

To start understanding how to compute derivatives on a computer, we start with *finite differencing*. For finite differencing, recall that the definition of the derivative is:

\[ f'(x) = \lim_{\epsilon \rightarrow \infty} \frac{f(x+\epsilon)-f(x)}{\epsilon} \]

Finite differencing directly follows from this definition by choosing a small $\epsilon$. However, choosing a good $\epsilon$ is very difficult. If $\epsilon$ is too large than there is error since this definition is asymtopic. However, if $\epsilon$ is too small, you receive roundoff error. To understand why you would get roundoff error, recall that floating point error is relative, and can essentially store 16 digits of accuracy. So let's say we choose $\epsilon = 10^{-6}$. Then $f(x+\epsilon) - f(x)$ is roughly the same in the first 6 digits, meaning that after the subtraction there is only 10 digits of accuracy, and then dividing by $10^{-6}$ simply brings those 10 digits back up to the correct relative size.

This means that we want to choose $\epsilon$ small enough that the $\mathcal{O}(\epsilon^2)$ error of the truncation is balanced by the $O(1/\epsilon)$ roundoff error. Under some minor assumptions, one can argue that the average best point is $\sqrt(E)$, where E is machine epsilon

@show eps(Float64)

eps(Float64) = 2.220446049250313e-16

@show sqrt(eps(Float64))

sqrt(eps(Float64)) = 1.4901161193847656e-8 1.4901161193847656e-8

This means we should not expect better than 8 digits of accuracy, even when things are good with finite differencing.

The centered difference formula is a little bit better, but this picture suggests something much better...

The problem with finite differencing is that we are mixing our really small number with the really large number, and so when we do the subtract we lose accuracy. Instead, we want to keep the small perturbation completely separate.

To see how to do this, assume that $x \in \mathbb{R}$ and assume that $f$ is complex analytic. You want to calculate a real derivative, but your function just happens to also be complex analytic when extended to the complex plane. Thus it has a Taylor series, and let's see what happens when we expand out this Taylor series purely in the complex direction:

\[ f(x+ih) = f(x) + f'(x)ih + \mathcal{O}(h^2) \]

which we can re-arrange as:

\[ if'(x) = \frac{f(x+ih) - f(x)}{h} + \mathcal{O}(h) \]

Since $x$ is real and $f$ is real-valued on the reals, $if'$ is purely imaginary. So let's take the imaginary parts of both sides:

\[ f'(x) = \frac{Im(f(x+ih))}{h} + \mathcal{O}(h) \]

since $Im(f(x)) = 0$ (since it's real valued!). Thus with a sufficiently small choice of $h$, this is the *complex step differentiation* formula for calculating the derivative.

But to understand the computational advantage, recal that $x$ is pure real, and thus $x+ih$ is an imaginary number where **the $h$ never directly interacts with $x$** since a complex number is a two dimensional number where you keep the two pieces separate. Thus there is no numerical cancellation by using a small value of $h$, and thus, due to the relative precision of floating point numbers, both the real and imaginary parts will be computed to (approximately) 16 digits of accuracy for any choice of $h$.

The derivative measures the **sensitivity** of a function, i.e. how much the function output changes when the input changes by a small amount $\epsilon$:

\[ f(a + \epsilon) = f(a) + f'(a) \epsilon + o(\epsilon). \]

In the following we will ignore higher-order terms; formally we set $\epsilon^2 = 0$. This form of analysis can be made rigorous through a form of non-standard analysis called *Smooth Infinitesimal Analysis* [1], though note that nilpotent infinitesimal requires *constructive logic*, and thus proof by contradiction is not allowed in this logic due to a lack of the *law of the excluded middle*.

A function $f$ will be represented by its value $f(a)$ and derivative $f'(a)$, encoded as the coefficients of a degree-1 (Taylor) polynomial in $\epsilon$:

\[ f \rightsquigarrow f(a) + \epsilon f'(a) \]

Conversely, if we have such an expansion in $\epsilon$ for a given function $f$, then we can identify the coefficient of $\epsilon$ as the derivative of $f$.

Thus, to extend the idea of complex step differentiation beyond complex analytic functions, we define a new number type, the *dual number*. A dual number is a multidimensional number where the sensitivity of the function is propagated along the dual portion.

Here we will now start to use $\epsilon$ as a dimensional signifier, like $i$, $j$, or $k$ for quaternion numbers. In order for this to work out, we need to derive an appropriate algebra for our numbers. To do this, we will look at Taylor series to make our algebra reconstruct differentiation.

Note that the chain rule has been explicitly encoded in the derivative part.

\[ f(a + \epsilon) = f(a) + \epsilon f'(a) \]

to first order. If we have two functions

\[ f \rightsquigarrow f(a) + \epsilon f'(a) \]

\[ g \rightsquigarrow g(a) + \epsilon g'(a) \]

then we can manipulate these Taylor expansions to calculate combinations of these functions as follows. Using the nilpotent algebra, we have that:

\[ (f + g) = [f(a) + g(a)] + \epsilon[f'(a) + g'(a)] \]

\[ (f \cdot g) = [f(a) \cdot g(a)] + \epsilon[f(a) \cdot g'(a) + g(a) \cdot f'(a) ] \]

From these we can *infer* the derivatives by taking the component of $\epsilon$. These also tell us the way to implement these in the computer.

To encode this information in Julia, we introduce a new number type, called `Dual`

, containing a value and derivative at some point $a$ (which is not usually explicitly recorded):

struct Dual val::Float64 partial::Float64 end val(x::Dual) = x.val partial(x::Dual) = x.partial

partial (generic function with 1 method)

Now we define the rules of dual number arithmetic:

import Base: +, * +(f::Dual, g::Dual) = Dual(val(f) + val(g), partial(f) + partial(g)) *(f::Dual, g::Dual) = Dual(val(f) * val(g), val(f)*partial(g) + val(g)*partial(f))

* (generic function with 351 methods)

To speed up our derivative function, we can directly hardcode the derivative of known functions which we call *primitives*. For example, for `sin`

we can add a rule like:

\[ \sin(a + \epsilon b) = \sin(a) + \epsilon b \cos(a). \]

Base.sin(f::Dual) = Dual(sin(val(f)), cos(val(f)) * partial(f))

For functions where we don't have a rule, we can recursively do dual number arithmetic within the function until we hit primitives where we know the derivative, and then use the chain rule to propagate the information back up.

Under this algebra, we can represent a + \epsilon $ as `Dual(a, 1)`

. Thus, applying `f`

to `Dual(a, 1)`

should give `Dual(f(a), f'(a))`

. This is thus a 2-dimensional number for calculating the derivative without floating point error, **using the compiler to transform our equations into dual number arithmetic**.

For a function $f: \mathbb{R}^n \to \mathbb{R}$ the basic operation is the **directional derivative**:

\[ \lim_{\epsilon \to 0} \frac{f(\mathbf{x} + \epsilon \mathbf{v}) - f(\mathbf{x})}{\epsilon} = [\nabla f(\mathbf{x})] \cdot \mathbf{v}, \]

where $\epsilon$ is still a single dimension and $\nabla f(\mathbf{x})$ is the direction in which we calculate.

We can directly do this using the same simple `Dual`

numbers as above, using the *same* $\epsilon$, e.g.

\[ f(x, y) = x^2 \sin(y) \]

\[ \begin{align} f(x_0 + a\epsilon, y_0 + b\epsilon) &= (x_0 + a\epsilon)^2 \sin(y_0 + b\epsilon) \\ &= x_0^2 \sin(y_0) + \epsilon[2ax_0 \sin(y_0) + x_0^2 b \cos(y_0)] + o(\epsilon) \end{align} \]

so we have indeed calculated $\nabla f(x_0, y_0) \cdot \mathbf{v},$ where $\mathbf{v} = (a, b)$ are the components that we put into the derivative component of the `Dual`

numbers.

If we wish to calculate the directional derivative in another direction, we could repeat the calculation with a different $\mathbf{v}$. A better solution is to use another independent epsilon $\epsilon$, expanding $x = x_0 + a_1 \epsilon_1 + a_2 \epsilon_2$ and putting $\epsilon_1 \epsilon_2 = 0$.

In particular, if we wish to calculate the gradient itself, $\nabla f(x_0, y_0)$, we need to calculate both partial derivatives, which corresponds to two directional derivatives, in the directions $(1, 0)$ and $(0, 1)$, respectively.

Note that another representation of the directional derivative is $f'(x)v$, where $f'(x)$ is the Jacobian or total of $f$ at $x$. To see the equivalence of this to a directional derivative, write it out in the standard basis:

Written out in the standard basis, we have that:

\[ w_i = \sum_{j}^{m} J_{ij} v_{j} \]

Now write out what $J$ means and we see that:

\[ w_i = \sum_j^{m} \frac{df_i}{dx_j} v_j = \nabla f_i(x) \cdot v \]

**The primitive action of forward-mode AD is f'(x)v!**

This is also known as a *Jacobian-vector product*, or *jvp* for short.

We can thus represent vector calculus with multidimensional dual numbers as follows. Let $d =[x,y]$, the vector of dual numbers. We can instead represent this as:

\[ d = d_0 + v_1 \epsilon_1 + v_2 \epsilon_2 \]

where $d_0$ is the *primal* vector $[x_0,y_0]$ and the $v_i$ are the vectors for the *dual* directions. If you work out this algebra, then note that a single application of $f$ to a multidimensional dual number calculates:

\[ f(d) = f(d_0) + f'(d_0)v_1 \epsilon_1 + f'(d_0)v_2 \epsilon_2 \]

i.e. it calculates the result of $f(x,y)$ and two separate directional derivatives. Note that because the information about $f(d_0)$ is shared between the calculations, this is more efficient than doing multiple applications of $f$. And of course, this is then generalized to $m$ many directional derivatives at once by:

\[ d = d_0 + v_1 \epsilon_1 + v_2 \epsilon_2 + \ldots + v_m \epsilon_m \]

For a function $f: \mathbb{R}^n \to \mathbb{R}^m$, we reduce (conceptually, although not necessarily in code) to its component functions $f_i: \mathbb{R}^n \to \mathbb{R}$, where $f(x) = (f_1(x), f_2(x), \ldots, f_m(x))$.

Then

\[ \begin{align} f(x + \epsilon v) &= (f_1(x + \epsilon v), \ldots, f_m(x + \epsilon v)) \\ &= (f_1(x) + \epsilon[\nabla f_1(x) \cdot v], \dots, f_m(x) + \epsilon[\nabla f_m(x) \cdot v] \\ &= f(x) + [f'(x) \cdot v] \epsilon, \end{align} \]

To calculate the complete Jacobian, we calculate these directional derivatives in the $n$ different directions of the basis vectors, i.e. if

\[ d = d_0 + e_1 \epsilon_1 + \ldots + e_n \epsilon_n \]

for $e_i$ the $i$th basis vector, then

\[ f(d) = f(d_0) + Je_1 \epsilon_1 + \ldots + Je_n \epsilon_n \]

computes all columns of the Jacobian simultaniously.

Instead of thinking about a vector of dual numbers, thus we can instead think of dual numbers with vectors for the components. But if there are vectors for the components, then we can think of the grouping of dual components as a matrix. Thus define our multidimensional multi-partial dual number as:

\[ D = D_0 + \Sigma \epsilon \]

where $D_0$ is a vector in $\mathbb{R}^n$, $\epsilon$ is a vector of dimensional signifiers and $\Sigma$ is a matrix in $\mathbb{R}^{n \times m}$ where $m$ is the number of concurrent differentiation dimensions. Each row of this is a dual number, but now we can use this to easily define higher dimensional primitives.

For example, let $f(x) = Ax$, matrix multiplication. Then, we can show with our dual number arithmetic that:

\[ f(D) = A*D_0 + A*\Sigma*\epsilon \]

is how one would compute the value of $f(D_0)$ and the derivative $f'(D_0)$ in all directions signified by the columns of $\Sigma$ simultaniously. Using multidimensional Taylor series expansions and doing the manipulations like before indeed implies that the arithematic on this object should follow:

\[ f(D) = f(D_0) + f'(D_0)\Sigma \epsilon \]

where $f'$ is the total derivative or the Jacobian of $f$. This then allows our system to be highly efficient by allowing the definition of multidimensional functions, like linear algebra, to be primitives of multi-directional derivatives.

The above techniques can be extended to higher derivatives by *adding more terms to the Taylor polynomial*, e.g.

\[ f(a + \epsilon) = f(a) + \epsilon f'(a) + \frac{1}{2} \epsilon^2 f''(a) + o(\epsilon^2). \]

We treat this as a degree-2 (or degree-$n$, in general) polynomial and do polynomial arithmetic to calculate the new polynomials. The coefficients of powers of $\epsilon$ then give the higher-order derivatives.

For example, for a function $f: \mathbb{R}^n \to \mathbb{R}$ we have

\[ f(x + \epsilon v) = f(x) + \epsilon \left[ \sum_i (\partial_i f)(x) v_i \right] + \frac{1}{2}\epsilon^2 \left[ \sum_i \sum_j (\partial_{i,j} f) v_i v_j \right] \]

using `Dual`

numbers with a single $\epsilon$ component. In this way we can compute coefficients of the (symmetric) Hessian matrix.

To make derivative calculations efficient and correct, we can move to higher dimensional numbers. In multiple dimensions, these then allow for multiple directional derivatives to be computed simultaniously, giving a method for computing the Jacobian of a function $f$ on a single input. This is a direct application of using the compiler as part of a mathematical framework.

John L. Bell,

*An Invitation to Smooth Infinitesimal Analysis*, http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdfBell, John L.

*A Primer of Infinitesimal Analysis*Nocedal & Wright,

*Numerical Optimization*, Chapter 8Griewank & Walther,

*Evaluating Derivatives*

Many thanks to David Sanders for helping make these lecture notes.