Fourier is a mathematical poem.

Lord Kelvin

A great many physical systems are governed by (up to) second-order linear partial differential equations. One of the neater features of linear partial differential equations is that if you add two solutions to the equation, or multiply a solution by some constant scaling, you get another solution; linear combinations of solutions are themselves solutions. This makes it practical, and often desirable, to identify some family of simple solutions for which every solution is simply a linear combination of members of the chosen family. Sometimes, decomposing a function into such a linear combination can make it easier to see how that function will interact with the differential equation; this can even be helpful when the simple functions in the family aren't solutions – e.g. if their interaction with the differential equation is suitably straightforward. Where rotations are prominent in a theory, the spherical harmonics may be used; where translations are more prominent, however, sinusoids prove more apt.

The Fourier transform decomposes any function from a Euclidean space to scalars into a linear combination of sinusoids: for ({complex}: f |V), (| (: exp(i.k·x) ←x :) ←k :dual(V)) provides the easy form of these (with i as a square root of −1); when it is important to only work with real scalars, it may still make sense to use these (and not mind the complex intermediate values) but otherwise one can use the union of {(: sin(k·x) ←x |V): k in dual(V)} and {(: cos(k·x) ←x |V): k in dual(V)}. These functions have the nice property that the second derivative is a negative multiple of the function itself; f'' = −k×k.f. This tends to simplify their relationships with partial differential equations.

Since any function from one Euclidean space to another can be reduced (via a basis of the destination) to a family of functions from a Euclidean space to scalars, one can apply the Fourier transform to any function between Euclidean spaces: however, discussing the case with Euclidean destination gains me little but complicates the discussion, so I shall work with scalar destination here.

For a function ({complex}: f |V) from some linear space to the complex scalars, we have ({complex}: Fourier(f) |dual(V)) characterized by

- f = (: integral(: exp(i.k·x).Fourier(f,k) ←k |dual(V)) ←x |V)

from which we need to be able to infer Fourier(f) from f. Fourier determined that the correct answer was of form

- Fourier(f) = (: integral(: exp(−i.k·x).f(x) ← 2.π.x |V) ←k |dual(V))

(It'll become evident, below, that the scaling by 2.π could be split between Fourier and its inverse; the above is just one definition one can use for the Fourier transform, and I don't actually know what Fourier originally did about the scaling.) To check this answer, we need to evaluate, for any x in V (expecting f(x) as answer):

- integral(: exp(i.k·x).integral(: exp(−i.k·y).f(y) ←2.π.y |V) ←k |dual(V))
- = integral(: integral(: exp(i.k·(x−y)) ←k |dual(V)).f(y) ←2.π.y |V)

The inner integral, over dual(V), reduces to integral(: 1←k
|dual(V)) when x and y are equal; this is clearly infinite. When x and y
differ, the integrand is a sinusoid in k; strictly, formally, the integral thus
cannot converge. We shall, in due course, see that this
inner integral yields a very strange function

of x−y (known as
Dirac's delta function). In the face of this oddity, it proves constructive to
approach the problem from another direction.

In order to justify the definition of the Fourier transform and, in particular, the scaling it involves, let us first study its inverse; and restrict attention to a particularly well-behaved family of functions.

Since we're doing integration, we also need to specify a metric (it defines
the unit of length and notion of orthogonality that we need in order to define
integration); this is, formally, a linear iso (dual(V):g|V) which is
symmetric

, i.e. it gives g(u,v) = g(v,u) for all v, u in V; and it'll
usually also be positive definite

, i.e. g(v,v) >0 for every non-zero v
in V. (Strictly, we don't need to specify a metric; we only need to specify the
square root of its determinant, which maps volume elements in V to their
(scalar) volumes; but in practice it is usual to specify a metric, and this
simplifies the discussion.) Our integration over V is tacitly defined in terms
of the metric, g, via √det(g), where det maps linear maps (on
finite-dimensional spaces) to their determinants. If we shift co-ordinates in
V, we'll need a factor of the determinant of a linear map that describes the
co-ordinate change (the Jacobean of the change of co-ordinates).

For any metric G on our linear space V, we can define the gaussian ({positives}: exp(−G(x,x)/2) ←x |V). We can translate this through any vector in V and multiply it pointwise by an arbitrary sinusoid to get a ({scalars}: |V) version not centred on zero (as the given one is). Thus we can define a mapping W for which

- W(G,m,h) = (: exp(i.h·x −G(x−m,x−m)/2) ←x |V)

for each metric G on V, each (mid-point) m in V and each (wave co-vector) h in dual(V). That may seem like a hairy mess to be defining, but let's look at the effect of applying our unscaled transform to one of these with G positive-definite, taking dim to be the dimension of V;

- (: integral(: exp(i.(k+h)·x −G(x−m,x−m)/2) ←x |V) ←k |dual(V))
this is insensitive to us re-naming k+h to k and x−m to x, so we can re-write it as

- = (: integral(: exp(i.k·(x+m) −G(x,x)/2) ←x+m |V) ←k−h |dual(V))
integration over all of V is insensitive to translation of the origin, so (pulling the obvious k·m factor out as a constant in the integration) this is just

- = (: exp(i.k·m).integral(: exp(i.k·x −G(x,x)/2) ←x |V) ←k−h |dual(V))
Now, G is positive definite and symmetric; there's a theorem of linear algebra which then lets us select a basis (dual(V):q|dim) which simultaneously diagonalises both g and G; i.e. for which g = sum(: N(e).q(e)×q(e) ←e |dim) and G = sum(: q(e)×q(e) ←e |dim) for some ({reals}: N |dim). Dual to this basis will be a basis (V:b|dim) of V for which q(e)·b(j) = 0 unless e is j, in which case q(e)·b(e) = 1. This will express k·x as sum(: k·b(e).q(e)·x ←e |dim) and G(x,x) as sum(: q(e)·x.q(e)·x ←e |dim). Thus we obtain

- = (: exp(i.k·m).integral(: exp(sum(: (i.k·b(e) −q(e)·x/2).q(e)·x ←e |dim)) ←x |V) ←k−h |dual(V))
- = (: exp(i.k·m).integral(: product(: exp((i.k·b(e) −q(e)·x/2).q(e)·x) ←e |dim) ←x |V) ←k−h |dual(V))
but integrating x over V allows q(e)·x to take every real value, independently for each e in dim, enabling us to separate integration out among these co-ordinates; and q diagonalises g so g's integration over V is equivalent to integrating over these co-ordinates for x and multiplying by the square root of g's determinant in these co-ordinates — i.e. by √product(N) = √det(g/G) – giving us

- = (: exp(i.k·m).product(: integral(: exp((i.k·b(e) −t/2).t) ←t |{reals}) ←e |dim).√det(g/G) ←k−h |dual(V))
- = (: exp(i.k·m).product(: integral(: exp(−(t.t −2.i.t.k·b(e))/2) ←t |{reals}) ←e |dim).√det(g/G) ←k−h |dual(V))
using n = k·b(e), we have t.t −2.i.t.n = (t−i.n).(t−i.n) +n.n so can shift each t-origin by i.k·b(e), and then exploit some helpful properties of analytic functions on the complex plane (the shift moved integration from {reals} to {i.n + reals}, but the function integrated has no poles at finite input, and dies away good and fast towards infinity, which lets us shift the integration back), to turn this into

- = (: exp(i.k·m).product(: exp(−k·b(e).k·b(e)/2).integral(: exp(−t.t/2) ←t |{reals}) ←e |dim).√det(g/G) ←k−h |dual(V))
and the t-integral, which is the only remaining integration, is inside the product over e in dim, but independent of e; so we just get power(dim) of it; while the remaining product over e is multiplying together outputs of exp, so could equally sum the inputs to exp over e and feed the result to exp:

- = (: exp(i.k·m).exp(−sum(: k·b(e).k·b(e) ←e |dim)/2).power(dim, integral(: exp(−t.t/2) ←t |{reals})).√det(g/G) ←k−h |dual(V))
meanwhile, the inverse of G is just sum(: b(e)×b(e) ←e |dim); call this H for now and observe that sum(: k·b(e).k·b(e) ←e |dim) is just H(k,k) = k/G\k, so we have

- = (: exp(i.k·m).exp(−H(k,k)/2).power(dim, integral(: exp(−t.t/2) ←t |{reals})).√det(g·H) ←k−h |dual(V))
and we can evaluate the t-integral (by exploiting the two-dimensional case); use u = t/√2 to give integral(: exp(−t.t/2) ←t |{reals}) = integral(: exp(−u.u) ←u :).√2 = √(2.π), to get

- = (: exp(i.k·m −H(k,k)/2).power(dim, √2.π).√det(g·H) ←k−h |dual(V))
- = (: exp(i.k·m −H(k,k)/2).√(power(dim, 2.π).det(g·H)) ←k−h |dual(V))
- = (: exp(i.(k+h)·m −H(k+h,k+h)/2) ←k |dual(V)).√det(2.π.g·H)
- = exp(i.h·m).W(H,−h,m).√det(2.π.g·H)

(phew). Now that we know how W transforms, we can define a re-scaled version of it which splits the dependence on G-or-H, h and m between the side being transformed and the side that comes out as the answer;

- w(G,m,h)
- = root(4, det(G/g)).exp(−i.h·m/2).W(G,m,h)
- = (: exp(i.h·(x+m/2) −G(x,x)/2).root(4, det(G/g)) ←x+m |V)

(wherein root(4) is just the reverse of power(4), i.e. root(4, t) is just √√t). Note that g's inverse is the metric on dual(V) we would use to define integral(: |dual(V)), so that where w(G) uses G/g, w(H) must use H·g; and det(H·g).det(G/g) = 1, so we obtain

- (: integral(: exp(i.k·x).w(G,m,h,x) ←x |V) ←k |dual(V)) = w(H,−h,m).power(dim, √(2.π))

and we can sink the power of 2.π in the integration by scaling x; so define an operator

- F = (: (: integral(: exp(i.k·x).f(x) ←x.√(2.π) |V) ←k |dual(V)) ← (:f|V) :)

to obtain

- F(w(G,m,h)) = w(H,−h,m)

with H and G mutually inverse; we then obtain

- F(F(w(G,m,h))) = w(G,−m,−h)
- F(F(F(w(G,m,h)))) = w(H,h,−m)
- F(F(F(F(w(G,m,h))))) = w(G,m,h)

so that F&on;F&on;F&on;F acts as the identity on pointwise products of gaussians and sinusoids. Functions which die away to zero reasonably fast outside some finite region of some linear space can necessarily be expressed as integrals of w over the available range of values for metric, centre-point and wave co-vector (i.e. G, m and h); F is linear, so transforms each w-function separately and maps such a superposition to the matching superposition of transformed w-functions; thus F&on;F&on;F&on;F also acts as the identity on such functions. It should also be noted that

- w(G, −m, −h, x)
- = root(4, det(G/g)).exp(−i.(−h)·(−m)/2).exp(−i.h.·x −G(x+m,x+m)/2)
but G(x+m,x+m) = G(−x−m,−x−m), so this is

- = root(4, det(G/g)).exp(−i.h·m/2).exp(i.h.·(−x) −G((−x)−m,(−x)−m)/2)
- = w(G, m, h, −x)

so that F&on;F is actually just (: (: f(−x) ←x :) ←f :), at least on the span of sinusoidal gaussians (which is everything we're justified in supposing we can play with). We can then introduce negate = (: −x ←x :) to write this as F(F(f)) = f&on;negate. This, in turn, means that

- F(F(F(: f |V)))
- = F(f&on;negate)
- = (: integral(: exp(i.k·x).f(−x) ←x.√(2.π) |V) ←k |dual(V))
- = (: integral(: exp(−i.k·x).f(x) ← −x.√(2.π) |V) ←k |dual(V))
- = (: integral(: exp(i.k·x).f(x) ← −x.√(2.π) |V) ← −k |dual(V))
- = F(f)&on;negate

Now Fourier's inverse is defined to be the unscaled form of F; and F's inverse is just F&on;F&on;F; so, on any given linear space of dimension dim, Fourier's inverse is F.power(dim, √(2.π)) and Fourier = F&on;F&on;F/power(dim, √(2.π)), which (again sinking the power of 2.π in the integration variable, where it joins up with the one already there) is what I gave earlier.

Having used the above to justify the definitions, I'll mainly work with F, only sometimes with Fourier and its inverse, hereafter; g and W shall drop out of the discussion, while w will only be relevant for the following digression.

Note that w(G,m,h) is centred on

m but spreads out around it to an
extent characterized by G's unit sphere, {x in V: G(x,x) = 1}, which gets
proportionately smaller as G gets bigger. Likewise, w(H,−h,m) is centred
on −h, but spreads out to an extent characterized by H's unit sphere, {k
in dual(V): H(k,k) = 1}, which gets proportionately bigger as G gets bigger
(because that makes H get smaller). Thus the less blurred a function is, the
more blurred its transform is bound to be.

Specifically, for each basis member b(e), with e in dim, w(G,m,h)'s spread in the b(e) direction is proportional to √N(e) while w(H,−h,m)'s spread in the corresponding q(e) direction is proportional to 1/√N(e). If we combine two w-functions, with spreads a and A in some direction, their combined spread will be (the separation of their centres plus) at least the maximum of a and A; their transforms will have spreads 1/a and 1/A yielding a combined spread of (the separation of the two wave co-vectors plus) the maximum of these; when we multiply these two combined spreads we thus get at least (and generally more than) 1. As for a simple sum, so for an arbitrary linear combination of w's, hence for any function to which we have any business applying the Fourier transform.

This imposes a lower bound on the product of the uncertainties in any two functions which are one another's Fourier transforms, classically the position and wave co-vector of any function of which those are characteristic properties. In quantum mechanics, the momentum of a particle is just Planck's constant times the particle's wave covector (passed through the metric's inverse to turn it into a vector), so the uncertainties in the particle's position and momentum must, when multiplied, yield Planck's constant times the lower bound. Thus Heisenberg's famous uncertainty is a natural property of the Fourier transform.

The w-functions also serve as archetype for the notion of a
wave packet

; they're localized, like a particle, but also have
wave-nature. It's thus natural to interpret a photon (for example) as a
w-function at any given moment of time. Suppose we have a wave-packet that's
w(G,m,k) at time t=0. We know w(G,m,k) is, with H as G's inverse,
F(w(H,k,−m)). This expresses our wave-packet as a superposition of
spatial sinusoids at that moment; and sinusoids are solutions to the
electromagnetic field equations. A sinusoid with spatial wave co-vector q is
the state, at t=0, of a solution which is a sinusoid whose space-time co-vector
has q for its spatial co-ordinates, along with a time-like co-ordinate equal to
the magnitude of q. Taking g as the inverse of our 3-space metric, we have
√g(q,q) as the time-like component, so the phase exp(i.q·x) changes
to exp(i.(q·x +c.t.√g(q,q))) at time t. Thus our wave packet
varies with time as integral(: (: exp(i.q·x) ←x
:).w(H,k,−m,q).exp(i.c.t.√g(q,q)) ←q :), which ceases being a
simple wave-packet once t isn't 0. Thus, although very nice as an archetype for
a wave-packet, w's sinusoidal Gaussians aren't quite satisfactory as an actual
description of photons ...

Now consider a transformation which maps a given (: f |V) to

- (: integral(: exp(i.Q.k·x).f(x) ←x.√(2.π) |V) ←k |dual(V))

This is trivially just (: F(f, Q.k) ←k |dual(V)). Applying it to its own output, we get

- (: integral(: exp(i.Q.k·x).F(f, Q.k) ←k.√(2.π) |V) ←x |dual(V))

which is inescapably just

- (: integral(: exp(i.k·x).F(f, k) ←k.√(2.π)/Q |V) ←x |dual(V))

which is just F(F(f)).power(dim, Q) = power(dim, Q).f&on;negate. Now F is just Fourier's inverse with a scaling of power(dim, √(2.π)), so F&on;F is just Fourier's inverse squared with a scaling of power(dim, 2.π). If we applied a Q-scaling to Fourier's inverse's exp(...)'s input, the result would compose with itself to yield power(dim, Q/(2.π)).f&on;negate ←f, so choosing Q = 2.π gives us an operator

- E = (: (: integral(: exp(2.π.i.k·x).f(x) ←x |V) ←k |dual(V)) ←(:f|V) :)

for which E&on;E = (: f&on;negate ←f :) = F&on;F. Thus E, like F,
is a fourth root of the identity; and the only place it involves a scaling is in
the input to exp. Now, ({phases}: exp(2.π.i.t) ←t |{reals}) is a
periodic function with period 1, and our transformation is using this as its
canonical sinusoid – rather than cos(t) + i.sin(t) = exp(i.t) ←t
– with which to decompose any function as a superposition of sinusoids.
In effect, this amounts to choosing the whole period

(or turn) as unit of
angle, in preference to the radian.

So I hope I've satisfied you that F, which is just Fourier's inverse
judiciously scaled, is simply a fourth root

of the identity on (a broad
class of) functions between linear spaces; hence that Fourier is sensibly
defined (at least on the relevant broad class of functions). There are some
serious analytic questions which deserve to be addressed (as should be evident
from my first, naïve, attack at justifying my definition of Fourier) if
Fourier is to be applied to arbitrary functions between linear spaces; the
integrals are all too prone to turning out divergent when examined in detail.
Rather than go into those, I'm going to flagrantly ignore the problems, derive
some orthodox results (which should perhaps be thought of as slogans

rather than theorems) and then produce a justification for them. Because it
happens to avoid lots of scrappy little powers of 2.π, I'm going to use E
rather than F or Fourier.

I showed earlier that, at least on those inputs where E is well-defined,
E&on;E = (: (: f(−x) ←x :) ←f :) = (: f&on;negate ←f :).
Thus E is a square root of compose after negate

, which (in its turn) is
manifestly a square root of the identity (since negate is self-inverse). So
let's look at E&on;E again: given (: f |V) for some linear space V of dimension
dim,

- f&on;negate
- = E(E(f))
- = (: integral(: exp(2.π.i.x·k).E(f,k) ←k |dual(V)) ←x |V)
- = (: integral(: integral(: exp(2.π.i.k·(x+y)).f(y) ←y |V) ←k |dual(V)) ←x |V)
- = (: integral(: integral(: exp(2.π.i.k·(x+y)) ←k |dual(V)).f(y) ←y |V) ←x |V)

so define

- (:δ|V) = (: integral(: exp(2.π.i.k·z) ←k |dual(V)) ←z |V)

polymorphically for each linear space V; we then have

- (: f(−x) ←x |V) = (: integral(: δ(x+y).f(y) ←y |V) ←x |V)

whence, selecting a particular input −x,

- f(x) = integral(: δ(y−x).f(y) ←y |V) = integral(: δ(y).f(y+x) ←y |V)

so that integrating the pointwise product of δ with some other function selects that other function's value where δ's input is zero. That's quite a clever trick for a function to do. In particular, it says that δ is 0 everywhere but at zero, where I already noted that it is infinite (since it's integral(: 1←k |dual(V)), and dual(V) is a linear space over the reals, so unbounded).

Of course, if we apply either E or Fourier's inverse to δ we trivially obtain (: 1 ←k |dual(V)), so (:δ|V) = Fourier(: 1←k |dual(V)). This also gives us Fourier(:δ|V) = (: 1/power(dim, 2.π) ←k |dual(V)) and F(:δ|V) = (: 1/power(dim, √(2.π)) ←k |dual(V)). This is constant, so applying F to it involves integrating a simple sinusoid over a vector space, which is divergent; but that's OK, we already know E&on;E = F&on;F is just (: f&on;negate ←f :) so F(F(δ)) = δ&on;negate = E(E(δ)). Likewise, E(E(E(δ))) = E(δ)&on;negate, and if you can't distinguish that from E(δ) it's because you'd have had trouble distinguishing δ&on;negate from δ.

Now, reasonable sane folk might plausibly be upset at these games, so perhaps it's time I told you about distributions. A distribution on some domain V is formally a linear map ({scalars}: h :{map ({scalars}: :V)}). It naturally induces a natural extension to (W: :{map (W: :V)}), for arbitrary linear space W, by applying h to components in W; and h is equally a member of dual({map ({scalars}: :V)}). Now, when the V in question is a finite collection, we can define a particular distribution on it called sum which just adds together all the outputs of the mapping from V that it's given. In such a case, any other distribution h on V can be written as (: sum(: H(v).f(v) ←v :V) ← (:f|V) :) for some ({scalars}: H |V) defined by H(v) = h({1}::{v}), h's output when its input maps v to 1 (and ignores all other members of V, effectively mapping them to 0).

In light of this last, I define a binary operator @ for which k@h is defined whenever h is a distribution on some V and k is a mapping ({scalars}:k:V); in that case, k@h = (: h(: k(v).f(v) ←v :V) ←(:f:V) :). In these terms, the above expressed h as H@sum. It is worth noting that, when k@h is defined, h is a distribution on V, hence a member of dual({map ({scalars}::V)}), and k is a member of {map ({scalars}::V)}; so it's worth bearing in mind that I may eventually work out how to generalize @ to combine k@h whenever k and h are members of mutually dual linear spaces.

So, now that I've told you what distributions are, you might make sense of
the fact that: integral is a distribution; most distributions we deal with on
continua (e.g. linear spaces over {reals}) are of form f@integral for some
scalar function f; δ isn't really a scalar function, but
δ@integral

is a distribution. Thus, when orthodoxy says

- integral(: δ(y).f(y+x) ←y :) = f(x)

what it really means is to talk about a distribution Δ, to be thought of as δ@integral, for which

- Δ(: f(y+x) ←y :) = f(x)

which is a perfectly well-defined distribution.

It is manifest from the structure of its definition that F is linear. This implies that F(: f(x) +h×e(x) ←x :) = F(f) +h×F(e) where meaningful. In this section, I'll look into how F interacts with some other transformations one can apply to a function. It'll also turn out to be worth looking into F's action on powers and polynomials, for all that they don't strictly meet the criterion for eligibility as inputs to F.

Given a function (W:f|V), for linear spaces V and W, it should be immediately apparent from the form of F that scaling by a sinusoid or shifting origin will each transform to the other:

- F(: f(x−m) ←x |V) = (: exp(i.k·m).F(f,k) ←k |dual(V))
- F(: exp(i.q·x).f(x) ←x |V) = (: F(f,k+q) ←k |dual(V))

We can also ask for f's derivative, ({linear (W:|V)}: f' |V), whose transform will give, for any k in dual(V),

- F(f', k)
- = integral(: exp(i.k·x).f'(x) ←x.√(2.π) |V)
to which we can apply

integration by parts

; the integrand is one of the two terms in the integral of the x-derivative of exp(i.k·x).f(x), so it's - = change(: exp(i.k·x).f(x).√(2.π) ←x.√(2.π) |V)
−integral(: i.exp(i.k.·x).f(x)×k ←x.√(2.π) |V)
but, for f to be amenable to F, it must die off fast enough, as x grows, to make the former zero, leaving the latter, which is just

- = −i.F(f, k)×k

so that differentiation just scales a function's transform by −i times its input (necessarily tensored on the right so as to make a linear (W:|V) as the output),

- F(f') = (: i.F(f,k)×k ←k :)
- E(f') = (: 2.π.i.E(f,k)×k ←k :)

In particular, with N = power(dim, √(2.π)), N.F(δ) = (: 1←k :) so N.F(δ') = (: i.k ←k :), N.F(δ'') = (: −k×k ←k :), N.F(δ''') = (: −i.k×k×k ←k :), N.F(δ'''') = (: k×k×k×k ←k :) and so on; when V = {reals}, dim = 1 this tells us that F maps any linear combination of derivatives of δ to a corresponding polynomial. Likewise,

- F(: f(x)×x ←x |V)
- = (: integral(: exp(i.k·x).f(x)×x ←x |V) ←k |dual(V))
- = (: −i.d(F(f,k))/dk ←k |dual(V))
- = −i.F(f)'

In particular, F(: 1←x :) is just N.δ so F(: x←x :) is −i.N.δ', F(: x×x ←x :) is −N.δ'', F(: x×x×x ←x :) is i.N.δ''', F(: x×x×x×x ←x :) is N.δ'''' and so on; with V = {reals}, dim = 1 we thus find that F maps any polynomial to a suitable linear combination of the derivatives of δ.

Earlier I noted that it's possible to handle real functions using just sin and cos, rather than the complex phase sinusoid (: exp(i.t) ←t :), so I guess I'd better show how that's done, too. I'll demonstrate this for E, but similar can be done for F or Fourier.

Suppose we have a ({reals}: f |V); then we have

- E(f)
- = (: integral(: exp(2.π.i.k·x).f(x) ←x |V) ←k |dual(V))
- = (: integral(: cos(2.π.k·x).f(x) ←x |V) ←k |dual(V))

+ i.(: integral(: sin(2.π.k·x).f(x) ←x |V) ←k |dual(V))

so introduce

- C = (: ({reals}: integral(: cos(2.π.k·x).f(x) ←x |V) ←k |dual(V)) ←({reals}:f|V) :)
- S = (: ({reals}: integral(: sin(2.π.k·x).f(x) ←x |V) ←k |dual(V)) ←({reals}:f|V) :)

so that we have E(f) = C(f) +i.S(f). (Note, in passing, that the functions (: cos(2.π.t) ←t :) and (: sin(2.π.t) ←t :) are just (: Cos(t.turn) ←t :) and (: Sin(t.turn) ←t :), in terms of the trigonometric functions of angle, rather than their dimensionless cousins implicitly using radian as unit of angle.) Now, E(E(f)&on;negate) is f, so our inverse transformation will be

- f
- = C(E(f)&on;negate) +i.S(E(f)&on;negate)
- = C(C(f)&on;negate +i.S(f)&on;negate) +i.S(C(f)&on;negate +i.S(f)&on;negate)
- = C(C(f)&on;negate) +i.C(S(f)&on;negate) +i.S(C(f)&on;negate) −S(S(f)&on;negate)
and we know f is real, while C, S and negate preserve real-ness; so the two imaginary terms must necessarily cancel one another, leaving us with

- = C(C(f)&on;negate) −S(S(f)&on;negate)

Now, composing S(f) after negate just replaces k with −k in sin(2.π.k·x) in the integral; and sin(−2.π.k·x) = −sin(2.π.k·x) so S(f)&on;negate is simply −S(f). Likewise, composing C(f) after negate gives us cos(−2.π.k·x) in place of cos(2.π.k·x); but (: cos(−t) ←t :) = cos, so this is no change and we have C(f)&on;negate = C(f). Thus the inverse transform is simply:

- f = C(C(f)) +S(S(f))

The equivalent for F, in place of E, would define C and S with k·x, without any factor of 2.π, as input to sin and cos; while multiplying the input to the integrand by √(2.π) or (equivalently) dividing the integral by a factor of power(dim, √(2.π)); this would give C&on;C +S&on;S as the identity, again.

In discarding the two complex terms, I reasoned that they must cancel; just for the record, aside from their factor of i, they're just

- C(S(f)&on;negate) +S(C(f)&on;negate)
- = −C(S(f)) +S(C(f))
- = (: integral(: integral(: f(y).(Sin(k·x.turn).Cos(k·y.turn)
−Cos(k·x.turn).Sin(k·y.turn)) ←y |V) ←k |dual(V)) ←x |V)
in which the sine of a difference appears, yielding

- = (: integral(: integral(: f(y).Sin(k·(x−y).turn) ←y |V) ←k |dual(V)) ←x |V)
- = (: integral(: integral(: −f(y+x).Sin(k·y.turn) ←y+x |V) ←k |dual(V)) ←x |V)
- = (: integral(: −f(y+x).integral(: Sin(k·y.turn) ←k |dual(V)) ←y |V) ←x |V)
and integral(: Sin(k·y.turn) ←k |dual(V)) is zero when y is zero; otherwise, it's the integral of a sinusoid, so strictly divergent but, by analogy with δ being zero except at zero, it's persuasively zero (this can alternatively be justified by cutting dual(V) in half, with {k: k·y = 0} as the cut, and expressing one half as {−k: k in H} with H as the other half; then our integral over dual(V) becomes an integral over H of two terms, one of which is Sin(k·y.turn), the other is the same for −k, which cancels it exactly).

So I'd be on slightly dodgy ground claiming the two complex terms cancel, but for the fact that we know in advance that our answer is real. Note, though, that their cancellation tells us that S and C commute: C&on;S = S&on;C, at least when acting on ({real}: |{real}) functions. Since they are also linear, that should suffice to ensure they still commute on complex functions.

Written by Eddy.