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) @show eps(0.1) @show eps(0.01)

eps(1.0) = 2.220446049250313e-16 eps(0.1) = 1.3877787807814457e-17 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 6 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.

As an exercise to deepen your knowledge of issues related to finite precision arithmetic, you can play in Julia with examples such as

@show 1+eps(1.)≈1 @show 10 + eps(1.) - 10 @show 10 - 10 + eps(1.);

1 + eps(1.0) ≈ 1 = true (10 + eps(1.0)) - 10 = 0.0 (10 - 10) + eps(1.0) = 2.220446049250313e-16

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

\[ f'(x) = \lim_{\epsilon \rightarrow 0} \frac{f(x+\epsilon)-f(x)}{\epsilon} \]

Finite differencing directly follows from this definition by choosing a small $\epsilon$.

However, if $\epsilon$ is too large than there is error since this definition is asymptotic. If instead $\epsilon$ is too small, we get a roundoff error in the numerator.

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 so 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) @show sqrt(eps(Float64))

eps(Float64) = 2.220446049250313e-16 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:

\[ f(x+ih) = f(x) + f'(x)ih - \frac{1}{2}f''(x)h^2 + \mathcal{O}(h^3) \]

which we can re-arrange as:

\[ if'(x) = \frac{f(x+ih) - f(x)}{h} + \frac{1}{2}f''(x)h + \mathcal{O}(h^2) \]

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:

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

since $Im(f(x)) = 0$ and the next order term cancels for the same reason. With a sufficiently small choice of $h$, this is the **complex step differentiation** formula for calculating the derivative.

To understand the computational advantage, recall that $x$ is pure real, and thus $x+ih$ is a complex 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. 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.

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.

Setup (not necessary from the REPL):

using InteractiveUtils # only needed when using Weave

Each function requires two pieces of information and some particular "behavior", so we store these in a `struct`

. It's common to call this a dual number:

struct Dual{T} val::T # value der::T # derivative end

Each `Dual`

object represents a function. We define arithmetic operations to mirror performing those operations on the corresponding functions.

We must first import the operations from `Base`

:

Base.:+(f::Dual, g::Dual) = Dual(f.val + g.val, f.der + g.der) Base.:+(f::Dual, α::Number) = Dual(f.val + α, f.der) Base.:+(α::Number, f::Dual) = f + α Base.:-(f::Dual, g::Dual) = Dual(f.val - g.val, f.der - g.der) # Product Rule Base.:*(f::Dual, g::Dual) = Dual(f.val*g.val, f.der*g.val + f.val*g.der) Base.:*(α::Number, f::Dual) = Dual(f.val * α, f.der * α) Base.:*(f::Dual, α::Number) = α * f # Quotient Rule Base.:/(f::Dual, g::Dual) = Dual(f.val/g.val, (f.der*g.val - f.val*g.der)/(g.val^2)) Base.:/(α::Number, f::Dual) = Dual(α/f.val, -α*f.der/f.val^2) Base.:/(f::Dual, α::Number) = f * inv(α) Base.:^(f::Dual, n::Integer) = Base.power_by_squaring(f, n) # use repeated squaring for integer powers

We can now define `Dual`

s and manipulate them:

fd = Dual(3, 4) gd = Dual(5, 6) fd + gd

Main.##WeaveSandBox#291.Dual{Int64}(8, 10)

fd * gd

Main.##WeaveSandBox#291.Dual{Int64}(15, 38)

fd * (gd + gd)

Main.##WeaveSandBox#291.Dual{Int64}(30, 76)

It seems like we may have introduced significant computational overhead by creating a new data structure, and associated methods. Let's see how the performance is:

add(a1, a2, b1, b2) = (a1+b1, a2+b2)

add (generic function with 1 method)

add(1, 2, 3, 4) using BenchmarkTools a, b, c, d = 1, 2, 3, 4 @btime add($(Ref(a))[], $(Ref(b))[], $(Ref(c))[], $(Ref(d))[])

0.835 ns (0 allocations: 0 bytes) (4, 6)

a = Dual(1, 2) b = Dual(3, 4) add(j1, j2) = j1 + j2 add(a, b) @btime add($(Ref(a))[], $(Ref(b))[])

1.010 ns (0 allocations: 0 bytes) Main.##WeaveSandBox#291.Dual{Int64}(4, 6)

It seems like we have lost *no* performance.

@code_native add(1, 2, 3, 4)

.text ; ┌ @ audodiff.jmd:2 within `add` endbr64 movq %rdi, %rax ; │┌ @ int.jl:87 within `+` addq %rcx, %rsi addq %r8, %rdx ; │└ movq %rsi, (%rdi) movq %rdx, 8(%rdi) retq nopw %cs:(%rax,%rax) ; └

@code_native add(a, b)

.text ; ┌ @ audodiff.jmd:5 within `add` endbr64 movq %rdi, %rax ; │┌ @ audodiff.jmd:2 within `+` @ int.jl:87 vmovdqu (%rdx), %xmm0 vpaddq (%rsi), %xmm0, %xmm0 ; │└ vmovdqu %xmm0, (%rdi) retq nopw %cs:(%rax,%rax) ; └

We see that the data structure itself has disappeared, and we basically have a standard Julia tuple.

We can also define functions of `Dual`

objects, using the chain rule. To speed up our derivative function, we can directly hardcode the derivative of known functions which we call *primitives*. If `f`

is a `Dual`

representing the function $f$, then `exp(f)`

should be a `Dual`

representing the function $\exp \circ f$, i.e. with value $\exp(f(a))$ and derivative $(\exp \circ f)'(a) = \exp(f(a)) \, f'(a)$:

import Base: exp

exp(f::Dual) = Dual(exp(f.val), exp(f.val) * f.der)

exp (generic function with 15 methods)

```
fd
```

Main.##WeaveSandBox#291.Dual{Int64}(3, 4)

exp(fd)

Main.##WeaveSandBox#291.Dual{Float64}(20.085536923187668, 80.34214769275067 )

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**. To differentiate an arbitrary function, we define a generic function and then change the algebra.

hf(x) = x^2 + 2 a = 3 xx = Dual(a, 1)

Main.##WeaveSandBox#291.Dual{Int64}(3, 1)

Now we simply evaluate the function `h`

at the `Dual`

number `xx`

:

hf(xx)

Main.##WeaveSandBox#291.Dual{Int64}(11, 6)

The first component of the resulting `Dual`

is the value $h(a)$, and the second component is the derivative, $h'(a)$.

We can codify this into a function as follows:

derivative(f, x) = f(Dual(x, one(x))).der

derivative (generic function with 1 method)

Here, `one`

is the function that gives the value $1$ with the same type as that of `x`

.

Finally we can now calculate derivatives such as

derivative(x -> 3x^5 + 2, 2)

240

As a bigger example, we can take a pure Julia `sqrt`

function and differentiate it by changing the internal algebra:

function newtons(x) a = x for i in 1:300 a = 0.5 * (a + x/a) end a end @show newtons(2.0) @show (newtons(2.0+sqrt(eps())) - newtons(2.0))/ sqrt(eps()) @show newtons(Dual(2.0,1.0))

newtons(2.0) = 1.414213562373095 (newtons(2.0 + sqrt(eps())) - newtons(2.0)) / sqrt(eps()) = 0.3535533994436 264 newtons(Dual(2.0, 1.0)) = Main.##WeaveSandBox#291.Dual{Float64}(1.414213562 373095, 0.35355339059327373) Main.##WeaveSandBox#291.Dual{Float64}(1.414213562373095, 0.3535533905932737 3)

Notice that the computation with dual numbers gives a more precise estimate of the derivative, as we can check by comparing with the analytical solution:

1/2sqrt(2)

0.35355339059327373

How can we extend this to higher dimensional functions? For example, we wish to differentiate the following function $f: \mathbb{R}^2 \to \mathbb{R}$:

fquad(x, y) = x^2 + x*y

fquad (generic function with 1 method)

Recall that the **partial derivative** $\partial f/\partial x$ is defined by fixing $y$ and differentiating the resulting function of $x$:

a, b = 3.0, 4.0 fquad_1(x) = fquad(x, b) # single-variable function

fquad_1 (generic function with 1 method)

Since we now have a single-variable function, we can differentiate it:

derivative(fquad_1, a)

10.0

Under the hood this is doing

fquad(Dual(a, one(a)), b)

Main.##WeaveSandBox#291.Dual{Float64}(21.0, 10.0)

Similarly, we can differentiate with respect to $y$ by doing

fquad_2(y) = fquad(a, y) # single-variable function derivative(fquad_2, b)

3.0

Note that we must do **two separate calculations** to get the two partial derivatives; in general, calculating the gradient $\nabla$ of a function $f:\mathbb{R}^n \to \mathbb{R}$ requires $n$ separate calculations.

We can implement derivatives of functions $f: \mathbb{R}^n \to \mathbb{R}$ by adding several independent partial derivative components to our dual numbers.

We can think of these as $\epsilon$ perturbations in different directions, which satisfy $\epsilon_i^2 = \epsilon_i \epsilon_j = 0$, and we will call $\epsilon$ the vector of all perturbations. Then we have

\[ f(a + \epsilon) = f(a) + \nabla f(a) \cdot \epsilon + \mathcal{O}(\epsilon^2), \]

where $a \in \mathbb{R}^n$ and $\nabla f(a)$ is the **gradient** of $f$ at $a$, i.e. the vector of partial derivatives in each direction. $\nabla f(a) \cdot \epsilon$ is the **directional derivative** of $f$ in the direction $\epsilon$.

We now proceed similarly to the univariate case:

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

\[ \begin{align} (f \cdot g)(a + \epsilon) &= [f(a) + \nabla f(a) \cdot \epsilon ] \, [g(a) + \nabla g(a) \cdot \epsilon ] \\ &= f(a) g(a) + [f(a) \nabla g(a) + g(a) \nabla f(a)] \cdot \epsilon. \end{align} \]

We will use the `StaticArrays.jl`

package for efficient small vectors:

using StaticArrays struct MultiDual{N,T} val::T derivs::SVector{N,T} end function Base.:+(f::MultiDual{N,T}, g::MultiDual{N,T}) where {N,T} return MultiDual{N,T}(f.val + g.val, f.derivs + g.derivs) end function Base.:*(f::MultiDual{N,T}, g::MultiDual{N,T}) where {N,T} return MultiDual{N,T}(f.val * g.val, f.val .* g.derivs + g.val .* f.derivs) end

gcubic(x, y) = x*x*y + x + y (a, b) = (1.0, 2.0) xx = MultiDual(a, SVector(1.0, 0.0)) yy = MultiDual(b, SVector(0.0, 1.0)) gcubic(xx, yy)

Main.##WeaveSandBox#291.MultiDual{2, Float64}(5.0, [5.0, 2.0])

We can calculate the Jacobian of a function $\mathbb{R}^n \to \mathbb{R}^m$ by applying this to each component function:

fsvec(x, y) = SVector(x*x + y*y , x + y) fsvec(xx, yy)

2-element StaticArraysCore.SVector{2, Main.##WeaveSandBox#291.MultiDual{2, Float64}} with indices SOneTo(2): Main.##WeaveSandBox#291.MultiDual{2, Float64}(5.0, [2.0, 4.0]) Main.##WeaveSandBox#291.MultiDual{2, Float64}(3.0, [1.0, 1.0])

It would be possible (and better for performance in many cases) to store all of the partials in a matrix instead.

Forward-mode AD is implemented in a clean and efficient way in the `ForwardDiff.jl`

package:

using ForwardDiff, StaticArrays ForwardDiff.gradient( xx -> ( (x, y) = xx; x^2 * y + x*y ), [1, 2])

2-element Vector{Int64}: 6 2

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 $\mathbf{v}$. 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 derivative of $f$ at $x$. To see the equivalence of this to a directional derivative, write it out in the standard basis:

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

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

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

\[ f'(x)v \]

, also known as a *Jacobian-vector product*, or *jvp* for short, is the core object of forward-mode AD, since we can represent vector calculus with multidimensional dual numbers as follows. Let $d =[x,y]$ be the vector of dual numbers. We can represent it as:

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

where $d_0$ is the *primal* vector $[x_0,y_0]$, the $v_i$ are the vectors for the *dual* directions and the $\epsilon_i$s are the infinitesimal values such that $\epsilon_1 \epsilon_2 = 0$. If we work out the algebra, we 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. Because the information about $f(d_0)$ is shared between the calculations, this turns out to be more efficient than doing multiple applications of $f$. 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 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

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

where $e_i$ is the $i$-th canonical basis vector, then

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

computes all columns of the Jacobian simultaneously.

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

\[ D_0 = [d_1,d_2,d_3,\ldots,d_n] \]

\[ \Sigma = \begin{bmatrix} d_{11} & d_{12} & \cdots & d_{1n} \\ d_{21} & d_{22} & & \vdots \\ \vdots & & \ddots & \vdots \\ d_{m1} & \cdots & \cdots & d_{mn} \end{bmatrix} \]

\[ \epsilon=[\epsilon_1,\epsilon_2,\ldots,\epsilon_m] \]

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

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 $\Sigma$ simultaneously. Using multidimensional Taylor series expansions and doing the manipulations like before indeed implies that the arithmetic 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 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.

As an application, we will see how to solve nonlinear equations of the form $f(x) = 0$ for functions $f: \mathbb{R}^n \to \mathbb{R}^n$.

Since in general we cannot do anything with nonlinearity, we try to reduce it (approximate it) with something linear. Furthermore, in general we know that it is not possible to solve nonlinear equations in closed form (even for polynomials of degree $\ge 5$), so we will need some kind of iterative method.

We start from an initial guess $x_0$. The idea of the **Newton method** is to follow the tangent line to the function $f$ at the point $x_0$ and find where it intersects the $x$-axis; this will give the next iterate $x_1$.

Algebraically, we want to solve $f(x_1) = 0$. Suppose that $x_1 = x_0 + \delta$ for some $\delta$ that is currently unknown and which we wish to calculate.

Assuming $\delta$ is small, we can expand:

\[ f(x_1) = f(x_0 + \delta) = f(x_0) + Df(x_0) \cdot \delta + \mathcal{O}(\| \delta \|^2). \]

Since we wish to solve

\[ f(x_0 + \delta) \simeq 0, \]

we put

\[ f(x_0) + Df(x_0) \cdot \delta = 0, \]

so that *mathematically* we have

\[ \delta = -[Df(x_0)]^{-1} \cdot f(x_0). \]

Computationally we prefer to solve the matrix equation

\[ J \delta = -f(x_0), \]

where $J := Df(x_0)$ is the Jacobian of the function; Julia uses the syntax `\`

("backslash") for solving linear systems in an efficient way:

using ForwardDiff, StaticArrays function newton_step(f, x0) J = ForwardDiff.jacobian(f, x0) δ = J \ f(x0) return x0 - δ end function newton(f, x0) x = x0 for i in 1:10 x = newton_step(f, x) @show x end return x end fsvec2(xx) = ( (x, y) = xx; SVector(x^2 + y^2 - 1, x - y) ) x0 = SVector(3.0, 5.0) x = newton(fsvec2, x0)

x = [2.1875, 2.1875] x = [1.2080357142857143, 1.2080357142857143] x = [0.8109653811635519, 0.8109653811635519] x = [0.7137572554482892, 0.7137572554482892] x = [0.7071377642746832, 0.7071377642746832] x = [0.7071067818653062, 0.7071067818653062] x = [0.7071067811865475, 0.7071067811865475] x = [0.7071067811865476, 0.7071067811865476] x = [0.7071067811865475, 0.7071067811865475] x = [0.7071067811865476, 0.7071067811865476] 2-element StaticArraysCore.SVector{2, Float64} with indices SOneTo(2): 0.7071067811865476 0.7071067811865476

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 simultaneously, 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.