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Relativity of arithmetics as a fundamental symmetry of physics

Marek Czachor

Katedra Fizyki Teoretycznej i Informatyki Kwantowej, Politechnika Gda´nska, 80-233 Gda´nsk, Poland

Centrum Leo Apostel (CLEA), Vrije Universiteit Brussel, 1050 Brussels, Belgium

Arithmetic operations can be deﬁned in various ways, even if one assumes commutativity and

associativity of addition and multiplication, and distributivity of multiplication with respect to

addition. In consequence, whenever one encounters ‘plus’ or ‘times’ one has certain freedom of

interpreting this operation. This leads to some freedom in deﬁnitions of derivatives, integrals and,

thus, practically all equations occurring in natural sciences. A change of realization of arithmetics,

without altering the remaining structures of a given equation, plays the same role as a symmetry

transformation. An appropriate construction of arithmetics turns out to be particularly impor-

tant for dynamical systems in fractal space-times. Simple examples from classical and quantum,

relativistic and nonrelativistic physics are discussed.

PACS numbers: 05.45.Df, 02.30.-f

Symmetries of physical systems can be rather obvious

or very abstract. Lorentz transformations, discovered as

a formal symmetry of Maxwell’s equations, seemed ab-

stract until their physical meaning was understood by

Einstein. Theory of group representations, the corner-

stone of quantum mechanics and ﬁeld theory, had its

roots in Lie’s studies of abstract symmetries of diﬀer-

ential equations.

Einstein’s relativity, gauge invariance, Noether’s theo-

rems, or supersymmetry are prominent examples of sym-

metry principles in physics. Theory of group represen-

tations has taught us that diﬀerences in mathematical

realizations of a symmetry may directly reﬂect physi-

cal diﬀerences. Here we discuss a new type of symme-

try, occurring in any physical theory: The symmetry

of mathematical equations under modiﬁcations of arith-

metic operations, the induced modiﬁcations of deriva-

tives and integrals included. Similarly to other physical

symmetries, the symmetry maintains the form of relevant

equations, but may possess diﬀerent mathematical real-

izations. Fractal structures provide nontrivial examples.

A generalized arithmetics can lead to nontrivial continu-

ous dynamics in sets of measure zero, invisible from the

point of view of quantum mechanics. It opens a new

room for phenomena ‘coming out of nowhere’, such as

dark energy.

To begin with, let us consider a bijection f : X → Y ⊂

R, where X is some set. The map f allows us to deﬁne

addition, multiplication, subtraction, and division in X,

x ⊕ y = f

−1

f(x) + f(y)

, (1)

x y = f

−1

f(x) − f(y)

, (2)

x y = f

−1

f(x)f(y)

, (3)

x y = f

−1

f(x)/f(y)

. (4)

One easily veriﬁes the standard properties [1]: (1) asso-

ciativity (x⊕y)⊕z = x⊕(y ⊕z), (xy)z = x(yz),

(2) commutativity x ⊕ y = y ⊕ x, x y = y x, (3) dis-

tributivity (x ⊕ y) z = (x z) ⊕ (y z). Elements

0, 1 ∈ X are deﬁned by 0 ⊕ x = x, 1 x = x, which

implies f(0) = 0, f(1) = 1. One further ﬁnds x x = 0,

x x = 1, as expected [2]. In general, it is better to de-

ﬁne subtraction independently of addition since it may

happen that f(−x) is undeﬁned (think of the important

example [3–6] of the Cantor function f, where X is the

Cantor subset of [0, 1] and Y = [0, 1]). If 0 x exists,

one can denote it by x.

Practically the only diﬀerence between ⊕, , , and

+, −, ·, and / is that in general multiplication is not just

a repeated addition: x ⊕ x 6= 2 x. Multiplication and

addition are now truly independent.

Having all these arithmetic operations one can deﬁne

a derivative of a function A : X → X,

d

f

A(x)

d

f

x

= lim

h→0

A(x ⊕ h) A(x)

h, (5)

satisfying

d

f

A(x) B(x)

d

f

x

=

d

f

A(x)

d

f

x

B(x) ⊕ A(x)

d

f

B(x)

d

f

x

,

(6)

d

f

A(x) ⊕ B(x)

d

f

x

=

d

f

A(x)

d

f

x

⊕

d

f

B(x)

d

f

x

, (7)

d

f

A[B(x)]

d

f

x

=

d

f

A[B(x)]

d

f

B(x)

d

f

B(x)

d

f

x

. (8)

Now consider functions F : Y → Y and F

f

: X → X

related by

F

f

(x) = f

−1

F

f(x)

. (9)

Employing (5) and the fact that f(0) = 0 one ﬁnds

d

f

F

f

(x)

d

f

x

= f

−1

F

0

f(x)

, (10)

where F

0

(y) = dF/dy is the usual derivative in Y , deﬁned

in terms of +, −, ·, and /. It is extremely important to

note that (10) has been derived with no need of diﬀeren-

tiability of f. f(0) = 0 is enough to obtain a well deﬁned

arXiv:1412.8583v2 [math-ph] 2 Jan 2015

2

derivative. (10) is not the standard formula known for

composite functions since no derivatives of f occur. To

understand why functions of the form (9) are so essential

let us solve the diﬀerential equation

d

f

A(x)

d

f

x

= A(x), A(0) = 1 (11)

by assuming that A(x) = ⊕

∞

n=0

a

n

x

n

, where x

n

=

x · · · x (n times). Then, comparing term by term,

one ﬁnds the unique solution

A(x) = f

−1

e

f(x)

= exp

f

(x), (12)

fulﬁlling

exp

f

(x ⊕ y) = exp

f

x exp

f

y. (13)

Its inverse is

ln

f

x = f

−1

ln f(x)

, (14)

ln

f

(x y) = ln

f

x ⊕ ln

f

y. (15)

As our next example consider a classical harmonic oscil-

lator

d

2

f

x(t)

d

2

f

t

=

d

f

d

f

t

d

f

x(t)

d

f

t

= ω

2

x(t) (16)

where ω

2

= ω ω. The minus sign has to have a precise

meaning so here we assume that −f(x) = f (−x). Setting

x(t) = ⊕

∞

n=0

a

n

t

n

, one obtains

x(t) = C

1

sin

f

(ω t) ⊕ C

2

cos

f

(ω t) (17)

where

sin

f

x = f

−1

sin f(x)

, cos

f

x = f

−1

cos f(x)

, (18)

and C

1

, C

2

are constants. An instructive exercise is

to plot phase-space trajectories of the harmonic oscil-

lator corresponding to various choices of f. Fig. 1 shows

the trajectories for the Cantor function (extended to

all reals by f(−x) = −f(x)), and f (x) = x

n

, with

n = 1, 3, 5. All these trajectories represent a classical

harmonic oscillator that satisﬁes the usual law of ‘force

oppositely proportional to displacement’, with conserved

energy ‘ ˙x

2

+ ω

2

x

2

’, but with diﬀerent meanings of ‘plus’

and ‘times’.

One might still have the impression that what we do

is just standard physics in nonstandard coordinates. So,

consider the problem of a fractal Universe. Assume that

we live in a Universe of dimension 4 − , for example a

Cartesian product of fractals of the Cantor set variety.

Our physical equations have to be formulated in terms

of notions that are intrinsic to the Universe, but what

should be meant by a velocity, say? We have to subtract

positions and divide by time, but we have to do it in a

way that is intrinsic to the Universe we live in. Moreover,

from our perspective positions and ﬂow of time seem con-

tinuous even if they would appear discontinuous from an

exactly 4-dimensional perspective. We should not make

the usual step and turn to fractional derivatives [7], since

for inhabitants of the Cantorian (4 − )-dimensional Uni-

verse the velocity is just the ﬁrst derivative of position

with respect to time, and not some derivative of order

0 < α < 1.

Let us concentrate on the triadic Cantor-like set C

[0,1)

,

constructed as follows. Let us start with the right-open

interval [0, 1) ⊂ R, and let the (countable) set Y

2

⊂ [0, 1)

consist of those numbers that have two diﬀerent binary

representations. Let y = 0.b

1

b

2

· · · = 0.b

0

1

b

0

2

. . . be the

two representations of y ∈ Y

2

. Let us denote by 0.t

1

t

2

. . .

a ternary representation of some x ∈ [0, 1). Deﬁne

f

−1

: Y

2

→ R by f

−1

(y) = min{0.t

1

t

2

. . . , 0.t

0

1

t

0

2

. . . },

where t

j

= 2b

j

, t

0

j

= 2b

0

j

. Now, if y ∈ [0, 1) \ Y

2

then y

has a unique binary representation, say y = 0.b

1

b

2

. . . . So

let f

−1

: [0, 1)\Y

2

→ R be deﬁned by f

−1

(y) = 0.t

1

t

2

. . . ,

t

j

= 2b

j

. The triadic Cantor set is deﬁned as the image

C

[0,1)

= f

−1

[0, 1)

, and f : C

[0,1)

→ [0, 1), f = (f

−1

)

−1

is the required bijection between C

[0,1)

and the open in-

terval. For example, 1/2 ∈ Y

2

since 1/2 = 0.1

2

= 0.0(1)

2

.

We ﬁnd

f

−1

(1/2) = min{0.2

3

= 2/3, 0.0(2)

3

= 1/3} = 1/3. (19)

Accordingly, 1/3 ∈ C

[0,1)

while 2/3 /∈ C

[0,1)

. C

[0,1)

is

not exactly the standard Cantor set, but all irrational

elements of the Cantor set belong to C

[0,1)

(an irrational

number has a unique binary form), together with some

rational numbers such as 1/3. Note further that 0 ∈

C

[0,1)

, with f(0) = 0. We could proceed analogously with

1 /∈ [0, 1), since 1 = 1.(0)

2

= 0.(1)

2

possesses two binary

representations with min{2.(0)

3

, 0.(2)

3

} = min{2, 1} =

1. However, instead of including 1 in C

[0,1)

, let us shift

C

[0,1)

to the right by 1, thus obtaining C

[1,2)

on which we

employ the same f as before, but shifted up by 1. In this

way we can construct a fractal C =

P

k∈Z

C

[k,k+1)

, and

the bijection f : C → R. Explicitly, if x ∈ C

[0,1)

, then

x + k ∈ C

[k,k+1)

, f (x + k) = f(x) + k. Let us call the

resulting fractal the Cantor line, and f the Cantor-line

function.

An integral is deﬁned so that the fundamental law of

calculus,

Z

b

a

d

f

A(x)

d

f

x

d

f

x = A(b) A(a), (20)

holds true. Its explicit form reads

Z

b

a

F

f

(x) d

f

x = f

−1

Z

f(b)

f(a)

F (y)dy

!

, (21)

where

R

F (y)dy is the standard (say, Lebesgue) integral

in R.

The integral so deﬁned is not equivalent to the fractal

measure. Indeed, fractal measure of the Cantor set em-

bedded in an interval of length L is L

D

, where D = log

3

2.

Thus, for L = 1/3 one ﬁnds L

D

= 1/2. Since segments

[0, 1/3] and [2/3, 1] both have L = 1/3 they both have the

3

FIG. 1: [Color online] Phase-space trajectories of the har-

monic oscillator with ω = 1 and f (x) = x (black), f (x) = x

3

(red), f (x) = x

5

(green), and the Cantor-line function (blue).

Taking f (x) = x

n

with suﬃciently large n we would ﬁnd a

dynamics looking like a motion along a square.

FIG. 2: [Color online] Cantor-world oscillation in Cantorian

time. sin

f

(t) (black) and cos

f

(t) (red) for the Cantor function

f. Inhabitants of the Cantor-dust space-time would experi-

ence this as a continuous process.

same D-dimensional volume equal 1/2. Taking F

f

(x) = 1

we ﬁnd

Z

b

a

d

f

x =

Z

b

a

d

f

x

d

f

x

d

f

x = f

−1

f(b) − f (a)

, (22)

and

R

1/3

0

d

f

x = 1/3,

R

1

1/3

d

f

x = 1/3,

R

1

0

d

f

x = (1/3) ⊕

(1/3) = 1.

Now let us switch to higher dimensional examples.

First consider the plane, i.e. the Cartesian product

of two lines. One checks that sin

2

f

x ⊕ cos

2

f

x = 1,

cosh

2

f

x sinh

2

f

x = 1. Moreover, sin

f

, cos

f

, sinh

f

,

cosh

f

, functions satisfy the basic standard formulas such

as

sin

f

(a ⊕ b) = sin

f

a cos

f

b ⊕ cos

f

a sin

f

b (23)

and the like. Accordingly,

x

0

= x cos

f

α ⊕ y sin

f

α, (24)

y

0

= y cos

f

α x sin

f

α, (25)

deﬁnes a rotation. The rotation satisﬁes the usual group

composition rule, a fact immediately implying that one

FIG. 3: [Color online] Three proper-time hyperbolas from

Cantorian foliation of interior of the future light cone in 1+1

dimensional Cantorian Minkowski space. Generators of the

light-cone are given by Cantorian half-lines

can work with generalized-arithmetics matrix equations.

In an analogous way one arrives at Lorentz transforma-

tions in Cantorian Minkowski space, the Cartesian prod-

uct of four Cantor lines with the invariant form x

2

0

x

2

1

x

2

2

x

2

3

. Fig. 3 shows three proper-time hyperbo-

las in 1+1 dimensional Cantorian Minkowski space. This

is a Cantorian analogue of the empty-universe limit of the

Friedmann-Lemaitre-Robertson-Walker space-time. The

axes correspond to x

+

= x

0

⊕ x

1

and x

−

= x

0

x

1

light-

cone coordinates. In order to obtain the time coordinate

x

0

one ﬁrst computes

x

+

⊕ x

−

= x

0

⊕ x

0

= f

−1

2f(x

0

)

= f

−1

(2) x

0

(26)

and then -divides by f

−1

(2) ∈ C

+

.

Arithmetics of complex numbers requires some care.

One should not just take f : C → C due to the typical

multi-valuedness of f

−1

and the resulting ill-deﬁniteness

of ⊕ and . Deﬁnition of i as a π/2 rotation also does

not properly work since one cannot guarantee a correct

behaviour of i

n

if f is nonlinear. The correct solution

is the simplest one: One should treat complex numbers

as pairs of reals satisfying the following arithmetics

(x, y) ⊕ (x

0

, y

0

) = (x ⊕ x

0

, y ⊕ y

0

), (27)

(x, y) (x

0

, y

0

) = (x x

0

y y

0

, y x

0

⊕ x y

0

),(28)

i = (0, 1), (29)

supplemented by conjugation (x, y)

∗

= (x, −y). As

stressed in [8], the resulting complex structure is just

the standard one, but no mysterious ‘imaginary number’

is employed.

In this way we have arrived at quantum mechanics. As

our ﬁnal example let us solve the eigenvalue problem for

a 1-dimensional harmonic oscillator. Consider

ˆ

H

f

ψ

f

(x) = −α

2

d

2

f

ψ

f

(x)

d

2

f

x

⊕ β

2

x

2

ψ

f

(x)

= E

f

ψ

f

(x), (30)

4

where α, β are parameters. The normalized ground state

is

ψ

0f

(x) = f

−1

f(β)

πf(α)

1/4

e

−

f (β)f(x)

2

2f (α)

!

, (31)

with the eigenvalue E

0f

= α β. The excited states can

be derived in the usual way.

There are two peculiarities of the resulting quantum

mechanics one should be aware of. First of all, if f is a

Cantor-like function representing a fractal whose dimen-

sion is less that 1, then the real-line Lebesgue measure of

the fractal is zero. Keeping in mind that states in quan-

tum mechanics are represented by equivalence classes of

wave functions that are identical up to sets of measure

zero, we can remove the Cantor set, from R without alter-

ing standard quantum mechanics. Having removed the

Cantor line C from R we still can do ordinary quantum

mechanics on R \ C, whereas C itself can become a uni-

verse for its own, Cantorian theory. Removing C from R

does not mean that we impose some fractal-like bound-

ary conditions or that we consider a Schr¨odinger equation

with a delta-peaked potential of Cantor-set support [9].

We just use the freedom to modify wave functions on sets

of measure zero. So we can keep the standard Gaussian

f(x) = x ground state on R \ C, and employ the Canto-

rian ψ

0f

(x) on C. According to quantum mechanics the

resulting wave function belongs to the same equivalence

class as the usual Gaussian, and thus represents the same

state. However, now the energy is ~ω/2 + α β, with

α β ‘appearing from nowhere’. The analogy to dark

energy is evident. The additional energy is a real num-

ber so it can be added to ~ω/2, similarly to many other

energies that occur in physics and are additive in spite of

unrelated origins.

The second subtlety concerns physical dimensions of

various quantities occurring in f-generalized arithmetics.

Even the simple case of ω t may imply a necessity of

dimensionless ω and t if f is suﬃciently nontrivial. In

general we have to work with dimensionless variables x

in order to make f(x) meaningful. It is thus simplest

to begin with reformulating all the ‘standard’ theories

in dimensionless forms, similarly to c = 1 and ~ = 1

conventions often employed in relativity and quantum

theory.

Quantum mechanics has brought us to the issue of

probability. An appropriate normalization is ⊕

k

p

k

=

1 which, in virtue of f(1) = 1, implies

P

k

f(p

k

) =

P

k

P

k

= 1. We automatically obtain two coexisting

but inequivalent sets of probabilities, in close analogy

to probabilities P

k

and escort probabilities p

k

= P

q

k

oc-

curring in generalized statistics and multifractal theory

[10, 11]. Averages

hai

f

= ⊕

k

p

k

a

k

= f

−1

X

k

P

k

f(a

k

)

!

, (32)

have the form of Kolmogorov-Nagumo averages [11],

which implies the usual bounds a

min

≤ hai

f

≤ a

max

.

From the point of view of modiﬁed arithmetics the con-

straints one should impose on escort probabilities and

Kolmogorov-Nagumo averages are, though, completely

diﬀerent from those employed in nonextensive statistics

and R´enyi’s information theory [12], provided instead of

additivity one has ⊕-additivity in mind. The R´enyi’s lin-

ear or exponential f now can be replaced by a much wider

class of fs, and the analogue of CHSH-Bell inequality is

|hABi

f

⊕ hAB

0

i

f

⊕ hA

0

Bi

f

hA

0

B

0

i

f

| ≤ f

−1

(2), (33)

with f

−1

(2) = 2 for the Cantor-line function.

The modiﬁed calculus is as simple as the one one knows

from undergraduate education. What may be nontrivial

is to ﬁnd f if X is a suﬃciently ‘strange’ object. The

case of the Cantor set was quite obvious, but the choice

of f may be much less evident if X is a multifractal or a

higher-dimensional fractal.

In order to conclude, let us return to Fig. 1. All the

phase-space trajectories represent the same physical sys-

tem: A harmonic oscillator satisfying the Newton equa-

tion d

2

x/dt

2

= −ω

2

x, with the same physical parameters

for each of the trajectories. So how come the trajecto-

ries are diﬀerent? The answer is: Because the very form

of Newton’s equation does not tell us what should be

meant by ‘plus’ or ‘times’. This observation extends to

any theory that employs arithmetics of real numbers. It

would not be very surprising if some alternative arith-

metics proved essential for Planck-scale physics, where

fractal space-time is expected, or to biological modeling

where fractal structures are ubiquitous.

I am indebted to D. Aerts, J. Cie´sli´nski, M. Kuna and

J. Naudts for discussions and critical comments.

[1] All proofs are given in the Supplemental Material.

[2] Keeping the same symbols for 0, 1 ∈ X and 0, 1 ∈ R will

not lead to ambiguities.

[3] O. Dovgosheya, O. Martiob, V. Ryazanova, and M.

Vuorinenc, Expo. Math. 24, 1 (2006).

[4] G. Edgar, Measure, Topology, and Fractal Geometry, 2nd

edition, Springer, New York (2008).

[5] W. E. Thirring, A Course in Mathematical Physics, vol.

3, Springer, Berlin (1981).

[6] W. O. Amrein, Hilbert Space Methods in Quantum Me-

chanics, EPFL Press, Lausanne (2009).

[7] I. Podlubny, Fractional Diﬀerential Equations, Academic

Press, New York (1998).

[8] W. Rudin, Principles of Mathematical Analysis, 3rd edi-

tion, McGraw-Hill, New York (1976).

[9] C. P. Dettmann and N. E. Frankel, J. Phys. A 26, 1009

5

(1993).

[10] C. Tsallis, Introduction to Nonextensive Statistical Me-

chanics, Springer, London (2009).

[11] J. Naudts, Generalised Thermostatistics, Springer, Lon-

don (2011).

[12] A. R´enyi, MTA III. Oszt. K¨ozl. 10, 251 (1960). Reprinted

in Selected Papers of Alfred R´enyi, vol. 2, pp. 526–552,

Akad´emiai Kiad´o, Budapest (1976).

I. SUPPLEMENTAL MATERIAL

A. Arithmetic operations

Explicit cross-checks:

(x ⊕ y) ⊕ z = f

−1

[f(x ⊕ y) + f(z)]

= f

−1

h

f

h

f

−1

f(x) + f (y)

i

+ f (z)

i

= f

−1

[f(x) + f (y) + f(z)]

= f

−1

h

f(x) + f

h

f

−1

f(y) + f(z)

ii

= f

−1

[f(x) + f (y ⊕ z)]

= x ⊕ (y ⊕ z) (34)

(x ⊕ y) z = f

−1

[f(x ⊕ y) − f(z)]

= f

−1

h

f

h

f

−1

f(x) + f (y)

i

− f (z)

i

= f

−1

[f(x) + f (y) − f(z)]

= f

−1

h

f(x) + f

h

f

−1

f(y) − f(z)

ii

= f

−1

[f(x) + f (y z)]

= x ⊕ (y z) (35)

(x y) z = f

−1

[f(x y)f(z)]

= f

−1

h

f

h

f

−1

f(x)f(y)

i

f(z)

i

= f

−1

[f(x)f(y)f(z)]

= f

−1

h

f(x)f

h

f

−1

f(y)f(z)

ii

= f

−1

[f(x)f(y z)]

= x (y z) (36)

(x y) z = f

−1

[f(x y)/f(z)]

= f

−1

h

f

h

f

−1

f(x)f(y)

i

/f(z)

i

= f

−1

[f(x)f(y)/f(z)]

= f

−1

h

f(x)f

h

f

−1

f(y)/f(z)

ii

= f

−1

[f(x)f(y z)]

= x (y z)

= (x z) y (37)

Similarly one proves

(x y) z = (x z) y (38)

6

Distributivity

(x ⊕ y) z = f

−1

[f(x ⊕ y)f(z)]

= f

−1

h

f

h

f

−1

f(x) + f (y)

i

f(z)

i

= f

−1

[[f(x) + f (y)]f(z)]

= f

−1

[f(x)f(z) + f(y)f(z)]

= f

−1

f(f

−1

(f(x)f(z))) + f(f

−1

(f(y)f(z)))

= f

−1

[f(x z) + f(y z)]

= (x z) ⊕ (y z) (39)

(x ⊕ y) z = f

−1

[f(x ⊕ y)/f(z)]

= f

−1

h

f

h

f

−1

f(x) + f (y)

i

/f(z)

i

= f

−1

[[f(x) + f (y)]/f(z)]

= f

−1

[f(x)/f(z) + f(y)/f(z)]

= f

−1

f(f

−1

(f(x)/f(z))) + f(f

−1

(f(y)/f(z)))

= f

−1

[f(x z) + f(y z)]

= (x z) ⊕ (y z) (40)

B. Derivatives

Deﬁnition

F

◦

(x) = lim

h→0

F (x ⊕ h) F (x)

h (41)

Derivative of a sum

[F ⊕ G]

◦

(x) = lim

h→0

F (x ⊕ h) ⊕ G(x ⊕ h) F (x) G(x)

h

= lim

h→0

F (x ⊕ h) F (x) ⊕ G(x ⊕ h) G(x)

h

= lim

h→0

[F (x ⊕ h) F (x)] h ⊕ [G(x ⊕ h) G(x)] h

= F

◦

(x) ⊕ G

◦

(x) (42)

The Leibnitz rule

[F G]

◦

(x) = lim

h→0

F (x ⊕ h) G(x ⊕ h) F (x) G(x)

h

= lim

h→0

F (x ⊕ h) G(x ⊕ h) F (x) G(x ⊕ h) ⊕ F (x) G(x ⊕ h) F (x) G(x)

h

= lim

h→0

[F (x ⊕ h) F (x)] h G(x ⊕ h) ⊕ F (x) [G(x ⊕ h) G(x)] h

= F

◦

(x) G(x) ⊕ F (x) G

◦

(x) (43)

The chain rule

F [G]

◦

(x) = lim

h→0

F [G(x ⊕ h)] F [G(x)]

h (44)

= lim

h→0

F [G(x ⊕ h)] F [G(x)]

[G(x ⊕ h) G(x)] [G(x ⊕ h) G(x)] h (45)

Denote g = G(x ⊕ h) G(x), so that

G(x ⊕ h) = G(x) ⊕ G(x ⊕ h) G(x) = G(x) ⊕ g (46)

and

F [G(x)]

◦

= lim

g →0

F [G(x) ⊕ g] F [G(x)]

g lim

h→0

[G(x ⊕ h) G(x)] h (47)

= F

◦

[G(x)] G

◦

[x] (48)

7

C. Alternative proofs for derivatives

Consider

F

f

(x) = f

−1

F

f(x)

, (49)

G

f

(x) = f

−1

G

f(x)

, (50)

F

f

⊕ G

f

(x) = f

−1

f[F

f

(x)] + f[G

f

(x)]

(51)

= f

−1

F

f(x)

+ G

f(x)

(52)

= f

−1

(F + G)

f(x)

, (53)

F

f

G

f

(x) = f

−1

f[F

f

(x)]f[G

f

(x)]

(54)

= f

−1

F

f(x)

G

f(x)

(55)

= f

−1

F G

f(x)

(56)

The derivative

F

◦

f

(x) = lim

h→0

f

−1

F [f(x ⊕ h)]

f

−1

F [f(x)]

h (57)

= lim

h→0

f

−1

f

f

−1

F [f(x ⊕ h)]

− f

f

−1

F [f(x)]

h (58)

= lim

h→0

f

−1

F [f(x ⊕ h)] − F [f(x)]

h (59)

= lim

h→0

f

−1

F [f(x) + f(h)] − F [f(x)]

h (60)

= lim

h→0

f

−1

f

h

f

−1

F [f(x) + f(h)] − F [f(x)]

i

/f(h)

!

(61)

= lim

h→0

f

−1

F [f(x) + f(h)] − F [f(x)]

/f(h)

!

(62)

= f

−1

lim

h→0

F [f(x) + h] − F [f(x)]

/h

!

(63)

= f

−1

F

0

[f(x)]

(64)

Accordingly

[F

f

⊕ G

f

]

◦

(x) = f

−1

(F

0

+ G

0

)

f(x)

(65)

= f

−1

F

0

f(x)

+ G

0

f(x)

(66)

= f

−1

f[F

◦

f

(x)] + f[G

◦

f

(x)]

(67)

= F

◦

f

⊕ G

◦

f

(x) (68)

8

[F

f

G

f

]

◦

(x) = f

−1

(F G)

0

f(x)

(69)

= f

−1

(F

0

G + F G

0

)

f(x)

(70)

= f

−1

F

0

f(x)

G

f(x)

+ F

f(x)

G

0

f(x)

(71)

= f

−1

f

h

f

−1

F

0

f(x)

G

f(x)

i

+ f

h

f

−1

F

f(x)

G

0

f(x)

i

(72)

= f

−1

F

0

f(x)

G

f(x)

⊕ f

−1

F

f(x)

G

0

f(x)

(73)

= f

−1

"

f

h

f

−1

F

0

f(x)

i

f

h

f

−1

G

f(x)

i

#

⊕ f

−1

"

f

h

f

−1

F

f(x)

i

f

h

f

−1

G

0

f(x)

i

#

(74)

= f

−1

"

f

h

F

◦

f

(x)

i

f

h

G

f

(x)

i

#

⊕ f

−1

"

f

h

F

f

(x)

i

f

h

G

◦

f

(x)

i

#

(75)

= F

◦

f

G

f

(x) ⊕ F

f

G

◦

f

(x) (76)

Composition of functions

F

f

◦ G

f

(x) = f

−1

F

h

f

f

−1

G

f(x)

i

!

(77)

= f

−1

F

h

G

f(x)

i

!

(78)

= f

−1

F ◦ G

f(x)

(79)

and its derivative

[F

f

◦ G

f

]

◦

(x) = f

−1

(F ◦ G)

0

f(x)

(80)

= f

−1

F

0

G

f(x)

G

0

f(x)

(81)

= f

−1

f

h

f

−1

F

0

G

f(x)

i

f

h

f

−1

G

0

f(x)

i

!

(82)

= f

−1

f

h

f

−1

F

0

f

h

f

−1

G

f(x)

ii

f

h

f

−1

G

0

f(x)

i

!

(83)

= F

◦

f

G

f

(x)

G

◦

f

(x) (84)

D. Some explicit derivatives

Begin with F (x) = x. Then

F

◦

(x) = lim

h→0

F (x ⊕ h) F (x)

h, (85)

= lim

h→0

(x ⊕ h x) h, (86)

= lim

h→0

h h = 1 (87)

Now F (x) = x

n

. Using the Leibnitz rule

F

◦

(x) = ⊕

n

k=1

x

(n−1)

= f

−1

n

X

k=1

f(x

(n−1)

)

= f

−1

nf(x)

n−1

= f

−1

f[f

−1

(n)]f(x)

n−1

= f

−1

(n) x

(n−1)

(88)

9

E. Exponent

Now solve F

◦

= F . Let

F (x) = ⊕

∞

n=0

a

n

x

n

(89)

F

◦

(x) = ⊕

∞

n=1

a

n

⊕

n

k=1

x

(n−1)

(90)

= ⊕

∞

n=1

⊕

n

k=1

a

n

x

(n−1)

(91)

= a

1

⊕ a

2

x ⊕ a

2

x

| {z }

⊕ a

3

x

2

⊕ a

3

x

2

⊕ a

3

x

2

| {z }

⊕ . . . (92)

= a

0

⊕ a

1

x

| {z }

⊕ a

2

x

2

| {z }

⊕ . . . (93)

Therefore

a

0

= a

1

, (94)

a

1

x = a

2

x ⊕ a

2

x, (95)

a

2

x

2

= a

3

x

2

⊕ a

3

x

2

⊕ a

3

x

2

(96)

(97)

and so on. Explicitly,

f

−1

f(a

1

)f(x)

= f

−1

f(a

2

x) + f(a

2

x)

= f

−1

2f[f

−1

(f(a

2

)f(x))]

= f

−1

2f(a

2

)f(x)

(98)

f(a

1

)f(x) = 2f(a

2

)f(x) (99)

Let us start with a

0

= 1 = a

1

. Since f(1) = 1, we get

a

0

= a

1

= 2f(a

2

) = 1, (100)

a

2

= f

−1

(1/2) (101)

Next consider a

2

x

2

= a

3

x

2

⊕ a

3

x

2

⊕ a

3

x

2

f

−1

f(a

2

)f(x

2

)

= f

−1

3f(a

3

x

2

)

= f

−1

3f(a

3

)f(x

2

)

(102)

f(a

2

) = 3f(a

3

) = 1/2 (103)

a

3

= f

−1

(1/3!) (104)

Similarly a

n

= f

−1

(1/n!). So

F (x) = ⊕

∞

n=0

a

n

x

n

(105)

= ⊕

∞

n=0

f

−1

(1/n!) x

n

(106)

= ⊕

∞

n=0

f

−1

f(x

n

)/n!

(107)

= ⊕

∞

n=0

f

−1

f(x)

n

/n!

(108)

= f

−1

∞

X

n=0

f(x)

n

/n!

(109)

= f

−1

e

f(x)

(110)

F. Harmonic oscillator

Consider

F

◦◦

(x) = ω

2

F (x) (111)

10

with f (−x) = −f(x), so that x = −x. Recall

(x

n

)

◦

= ⊕

n

k=1

x

(n−1)

= f

−1

nf(x

(n−1)

)

= f

−1

nf(x)

n−1

= f

−1

(n) x

(n−1)

(x

n

)

◦◦

= f

−1

(n) f

−1

(n − 1) x

(n−2)

= f

−1

n(n − 1)

x

(n−2)

= f

−1

n(n − 1)f(x)

n−2

(112)

Let

F (x) = ⊕

∞

n=0

a

n

x

n

(113)

F

◦◦

(x) = ⊕

∞

n=2

a

n

f

−1

n(n − 1)f(x)

n−2

(114)

= ⊕

∞

n=2

f

−1

f(a

n

)n(n − 1)f(x)

n−2

(115)

= f

−1

∞

X

n=2

f(a

n

)n(n − 1)f(x)

n−2

(116)

= −ω

2

F (x) (117)

= f

−1

− f (ω)

2

f(F (x))

(118)

= f

−1

− f (ω)

2

∞

X

n=0

f(a

n

)f(x)

n

(119)

Then

− f (ω)

2

∞

X

n=0

f(a

n

)f(x)

n

=

∞

X

n=2

f(a

n

)n(n − 1)f(x)

n−2

(120)

=

∞

X

m=0

f(a

m+2

)(m + 2)(m + 1)f(x)

m

(121)

=

∞

X

n=0

f(a

n+2

)(n + 2)(n + 1)f(x)

n

(122)

and

− f (ω)

2

f(a

n

) = f(a

n+2

)(n + 2)(n + 1) (123)

−f(ω)

2

f(a

0

) = f(a

2

)2 (124)

−f(ω)

2

f(a

1

) = f(a

3

)3 × 2 (125)

−f(ω)

2

f(a

2

) = f(a

4

)4 × 3 (126)

−f(ω)

2

f(a

3

) = f(a

5

)5 × 4 (127)

−f(ω)

2

f(a

4

) = f(a

6

)6 × 5 (128)

−f(ω)

2

f(a

5

) = f(a

7

)7 × 6 (129)

(130)

Finally

f(a

7

) = −

f(ω)

2

7 × 6

f(a

5

) =

f(ω)

4

7 × 6 × 5 × 4

f(a

3

) = (−)

3

f(ω)

6

7!

f(a

1

) (131)

f(a

6

) = −

f(ω)

2

6 × 5

f(a

4

) =

f(ω)

4

6 × 5 × 4 × 3

f(a

2

) = (−)

3

f(ω)

6

6!

f(a

0

) (132)

f(a

5

) = −

f(ω)

2

×5 × 4

f(a

3

) = (−)

2

f(ω)

4

5!

f(a

1

) (133)

f(a

4

) = −

f(ω)

2

4 × 3

f(a

2

) = (−)

2

f(ω)

4

4!

f(a

0

) (134)

f(a

3

) = (−)

1

f(ω)

2

3!

f(a

1

) (135)

f(a

2

) = (−)

1

f(ω)

2

2!

f(a

0

) (136)

11

and

F (x) = f

−1

∞

X

n=0

f(a

n

)f(x)

n

(137)

= f

−1

f(a

0

) −

f(ω)

2

2!

f(a

0

)f(x)

2

+

f(ω)

4

4!

f(a

0

)f(x)

4

+ . . .

+ f (a

1

)f(x) −

f(ω)

2

3!

f(a

1

)f(x)

3

+

f(ω)

4

5!

f(a

1

)f(x)

5

+ . . .

!

(138)

= f

−1

f(a

0

)

1 −

f(ω)

2

f(x)

2

2!

+

f(ω)

4

f(x)

4

4!

+ . . .

+

f(a

1

)

f(ω)

f(ω)f(x) −

f(ω)

3

f(x)

3

3!

+

f(ω)

5

f(x)

5

5!

+ . . .

!

(139)

= f

−1

f(a

0

) cos[f(ω)f(x)] +

f(a

1

)

f(ω)

sin[f(ω)f(x)]

!

(140)

= f

−1

f(a

0

) cos f

f

−1

[f(ω)f(x)]

+

f(a

1

)

f(ω)

sin f

f

−1

[f(ω)f(x)]

!

(141)

= f

−1

f(a

0

) cos f(ω x) +

f(a

1

)

f(ω)

sin f(ω x)

!

(142)

= f

−1

f

f

−1

h

f(a

0

) cos f(ω x)

i

+ f

f

−1

h

f(a

1

)

f(ω)

sin f(ω x)

i

!

(143)

= f

−1

f

f

−1

h

f(a

0

)f

f

−1

[cos f(ω x)]

| {z }

cos

f

(ωx )

i

+ f

f

−1

h

f(a

1

)

f(ω)

f

f

−1

[sin f(ω x)]

| {z }

sin

f

(ω x)

i

!

(144)

= a

0

cos

f

(ω x) ⊕ a

1

ω sin

f

(ω x) (145)

G. Properties of trigonometric and hyperbolic functions

Values at 0:

sin

f

(0) = f

−1

sin f(0)

= f

−1

sin 0

= f

−1

(0) = 0, (146)

cos

f

(0) = f

−1

cos f(0)

= f

−1

cos 0

= f

−1

(1) = 1, (147)

sinh

f

(0) = f

−1

sinh f(0)

= f

−1

sinh 0

= f

−1

(0) = 0, (148)

cosh

f

(0) = f

−1

cosh f(0)

= f

−1

cosh 0

= f

−1

(1) = 1. (149)

Pitagorean identity

sin

2

f

x ⊕ cos

2

f

x = f

−1

f(sin

2

f

x) + f(cos

2

f

x)

(150)

= f

−1

f(sin

f

x)

2

+ f (cos

f

x)

2

(151)

= f

−1

sin

2

f(x) + cos

2

f(x)

= f

−1

(1) = 1 (152)

Hyperbolic identity

cosh

2

f

x sinh

2

f

x = f

−1

f(cosh

2

f

x) − f(sinh

2

f

x)

(153)

= f

−1

f(cosh

f

x)

2

− f (sinh

f

x)

2

(154)

= f

−1

cosh

2

f(x) − sinh

2

f(x)

= f

−1

(1) = 1 (155)

12

Formulas for sums of arguments

sin

f

(a ⊕ b) = f

−1

sin f

f

−1

f(a) + f (b)

!

(156)

= f

−1

sin

f(a) + f (b)

!

(157)

= f

−1

sin f(a) cos f(b) + cos f(a) sin f(b)

!

(158)

= f

−1

f

f

−1

sin f(a) cos f(b)

+ f

f

−1

cos f(a) sin f(b)

!

(159)

= f

−1

sin f(a) cos f(b)

⊕ f

−1

cos f(a) sin f(b)

(160)

= f

−1

f

f

−1

sin f(a)

f

f

−1

cos f(b)

!

⊕ f

−1

f

f

−1

cos f(a)

f

f

−1

sin f(b)

!

(161)

= f

−1

f

sin

f

a

f

cos

f

b

⊕ f

−1

f

cos

f

a

f

sin

f

b

(162)

= sin

f

a cos

f

b ⊕ cos

f

a sin

f

b (163)

sin

f

(a b) = f

−1

sin f

f

−1

f(a) − f (b)

!

(164)

= f

−1

sin

f(a) − f (b)

!

(165)

= f

−1

sin f(a) cos f(b) − cos f(a) sin f(b)

!

(166)

= f

−1

f

f

−1

sin f(a) cos f(b)

− f

f

−1

cos f(a) sin f(b)

!

(167)

= f

−1

sin f(a) cos f(b)

f

−1

cos f(a) sin f(b)

(168)

= f

−1

f

f

−1

sin f(a)

f

f

−1

cos f(b)

!

f

−1

f

f

−1

cos f(a)

f

f

−1

sin f(b)

!

(169)

= f

−1

f

sin

f

a

f

cos

f

b

f

−1

f

cos

f

a

f

sin

f

b

(170)

= sin

f

a cos

f

b cos

f

a sin

f

b (171)

13

cos

f

(a ⊕ b) = f

−1

cos f

f

−1

f(a) + f (b)

!

(172)

= f

−1

cos

f(a) + f (b)

!

(173)

= f

−1

cos f(a) cos f(b) − sin f(a) sin f(b)

!

(174)

= f

−1

f

f

−1

cos f(a) cos f(b)

− f

f

−1

sin f(a) sin f(b)

!

(175)

= f

−1

cos f(a) cos f(b)

f

−1

sin f(a) sin f(b)

(176)

= f

−1

f

f

−1

cos f(a)

f

f

−1

cos f(b)

!

f

−1

f

f

−1

sin f(a)

f

f

−1

sin f(b)

!

(177)

= f

−1

f

cos

f

a

f

cos

f

b

f

−1

f

sin

f

a

f

sin

f

b

(178)

= cos

f

a cos

f

b sin

f

a sin

f

b (179)

and so on.

H. Integral

Assume integrals are deﬁned as inverses of the derivative

Z

F

◦

f

(x) dx = F

f

(x) ⊕ const (180)

Z

b

a

F

◦

f

(x) dx = F

f

(b) F

f

(a) (181)

For

F

◦

f

(x) = f

−1

F

0

f(x)

(182)

we get

Z

b

a

F

◦

f

(x) dx = f

−1

F

f(b)

f

−1

F

f(a)

(183)

= f

−1

F

f(b)

− F

f(a)

(184)

= f

−1

Z

f(b)

f(a)

F

0

(x)dx

(185)

So, for a general

F

f

(x) = f

−1

F

f(x)

(186)

we deﬁne

Z

b

a

F

f

(x) dx = f

−1

Z

f(b)

f(a)

F (x)dx

!

(187)

14

Let us cross-check

Z

x

a

F

f

(y) dy

!

◦

=

"

f

−1

Z

f(x)

f(a)

F (y)dy

!#

◦

(188)

= lim

h→0

"

f

−1

Z

f(x⊕h)

f(a)

F (y)dy

!

f

−1

Z

f(x)

f(a)

F (y)dy

!#

h (189)

= lim

h→0

"

f

−1

Z

f(x)+f(h)

f(a)

F (y)dy −

Z

f(x)

f(a)

F (y)dy

!#

h (190)

= f

−1

"

lim

h→0

Z

f(x)+f(h)

f(a)

F (y)dy −

Z

f(x)

f(a)

F (y)dy

!

/f(h)

#

(191)

= f

−1

F

f(x)

= F

f

(x) (192)

In particular

Z

b

a

(x)

◦

dx =

Z

b

a

1 dx = b a, (193)

Z

b

a

F

◦

f

(x) dx ⊕

Z

c

b

F

◦

f

(x) dx = F

f

(b) F

f

(a) ⊕ F

f

(c) F

f

(b) = F

f

(c) F

f

(a) (194)

On the Cantor set

Z

1/3

0

1 dx = (1/3) 0 = f

−1

f[0.0(2)

3

] − f(0)

= f

−1

0.0(1)

2

= 0.0(2)

3

= 1/3, (195)

Z

1

1/3

1 dx = 1 (1/3) = f

−1

f(1) − f [0.0(2)

3

]

= f

−1

0.1

2

= 1/3 (196)

Z

1

0

1 dx = 1 0 = f

−1

f(1) − f (0)

= 1, (197)

(1/3) ⊕ (1/3) = f

−1

2f(1/3)

= f

−1

(1) = 1 (198)

I. Bell inequality

For any Kolmogorov-Nagumo average

hai

f

= f

−1

X

k

P

k

f(a

k

)

!

, (199)

one ﬁnds the standard bounds

a

min

≤ hai

f

≤ a

max

(200)

since f is strictly monotonic.

Let

p

j

= N

j

⊕

K

k=1

N

k

= f

−1

f(N

j

)

P

K

k=1

f(N

k

)

!

, (201)

⊕

K

j=1

p

j

= ⊕

K

j=1

N

j

⊕

K

k=1

N

k

= 1 (202)

15

If all p

j

are equal then

p

j

= f

−1

f(N

j

)

P

K

k=1

f(N

k

)

!

= f

−1

1

K

= f

−1

f(1)/f[f

−1

(K)]

= 1 f

−1

(K)

(203)

Recall that f(1) = 1. Assume that f(−x) = −f(x), so that f(−1) = −1. Consider

hABi

f

= ⊕

kl

p

kl

a

k

b

l

(204)

− 1 = min{a

k

b

l

} ≤ hABi

f

≤ max{a

k

b

l

} = 1 (205)

hCi

f

= hABi

f

⊕ hAB

0

i

f

⊕ hA

0

Bi

f

hA

0

B

0

i

f

= ⊕

kl

p

kl

a

k

b

l

⊕ a

k

b

0

l

⊕ a

0

k

b

l

a

0

k

b

0

l

= ⊕

kl

p

kl

a

k

(b

l

⊕ b

0

l

) ⊕ a

0

k

(b

l

b

0

l

)

= ⊕

kl

p

kl

c

kl

(206)

So

min{c

kl

} ≤ hCi

f

≤ max{c

kl

} (207)

c

kl

= a

k

(b

l

⊕ b

0

l

) ⊕ a

0

k

(b

l

b

0

l

)

= f

−1

f

a

k

(b

l

⊕ b

0

l

)

+ f

a

0

k

(b

l

b

0

l

)

!

= f

−1

f

f

−1

h

f(a

k

)f(b

l

⊕ b

0

l

)

i

+ f

f

−1

h

f(a

0

k

)f(b

l

b

0

l

)

i

!

= f

−1

f(a

k

)f(b

l

⊕ b

0

l

) + f(a

0

k

)f(b

l

b

0

l

)

!

= f

−1

f(a

k

)f

f

−1

h

f(b

l

) + f(b

0

l

)

i

+ f (a

0

k

)f

f

−1

h

f(b

l

) − f(b

0

l

)

i

!

= f

−1

f(a

k

)

f(b

l

) + f(b

0

l

)

+ f (a

0

k

)

f(b

l

) − f(b

0

l

)

= f

−1

a

k

(b

l

+ b

0

l

) + a

0

k

(b

l

− b

0

l

)

(208)

− f

−1

(2) ≤ c

kl

≤ f

−1

(2) (209)

hCi = ⊕

kl

p

kl

f

−1

a

k

(b

l

+ b

0

l

) + a

0

k

(b

l

− b

0

l

)

= f

−1

X

kl

f(p

kl

)

a

k

(b

l

+ b

0

l

) + a

0

k

(b

l

− b

0

l

)

= f

−1

X

kl

P

kl

a

k

(b

l

+ b

0

l

) + a

0

k

(b

l

− b

0

l

)

16

Finally

− f

−1

(2) ≤ hCi ≤ f

−1

(2) (210)

For the Cantor-line function f(2) = f(0) + 2 = 2, so f

−1

(2) = 2.