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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.