My discovery about mathematics

Substitute the radius of the particle for the constant which means the radius
of the universe in Einstein equation. Robertson-Walker metric is a solution.
For example,the proton and the neutron in the deuteron are kept in a ball.
You can transform this ball into arbitrary forms by transformations in the space.
T^0,0 and the density don't change then. the mass and volume
don't change. General relativity(?) explains the invariance of the volumes of
the neucleons in the nuclear. The other theories cannot do it.
Maxwell equations in general relativity(?) explain the quadrupole moment
and the spin-orbit interaction. In general relativity(?), the gravity and the
nuclear force are the same force like the case of Newtonian mechanics.

You can explain results of general relativity and the quantum physics in
Newtonian mechanics.
Let (r,θ) be the circular coordinate system. Put u=r^-1.
The orbit of the particle by the central force is decided by
θ=∫(E-U(u))^2-1du+C.
You may assume that U(u) is an analytic function ∑_n∈Nc_nuⁿ.
When r is large enough,you can negrect the terms c_nuⁿ (n≥4).
u(θ) becomes an elliptic function whose period isn't 2π. When u is large,
you must treat U(u) as it is. This may be the nuclear force.
(u_θ)²=E-U(u)≥0
In the case of the electric force,E decreases by the radiation of the light.
(u_θ)²=E-U(u)=0, The orbit becomes a circul.
u_θθ=2^-1(E-U)^-2-1U_uu_θ=2^-1U_u.
When the orbit is a circle, 0=u_θθ=U_u
E=U(u) is a local maximum point of U(u). The E's are discrete values.

Naoto Meguro: Amateur. MSC2020: 03E30 . e-mail:

                                          1

ZF isn't consistent.

                                          2

Theorem 1. ZF isn't consistent.
Proof. Let A be the set of the axioms of ZF except the axiom of infinity.
You may set up that A doesn't include the individual symbols ω and S.
Let's write ∅=0, n∪{n}=n+1. Put A(ω)=∀n((n∈ω)→(n+1∈ω)).
Put B=A∪{A(ω)}. Let s(t) be a function symbol A doesn't include.
Put P(ω)=∀x((x∈ω)↔(s(x)={s(x+1)}))∧(S={x |∃n((n∈ω)∧(x=s(n)))}) and C=B∪{P(ω)}.
If A is consistent,C has a model M in which ω_M=0,S_M=0 and M |= ∀x(s_M(x)=0)
and is consistent. C |- ∀x((x∈S)→∃n((n∈ω)∧(x=s(n)={s(n+1)}∋s(n+1)∈S)).
By the axiom of regularity AR=∀x∃y((x≠0)→(y∈x)∧(x∩y=0)), C |- S=0. C |- ω=0.
C-{AR} |- AR→(ω=0). If ZF treating N as a set is consistent, it makes a model of C -{AR}
by putting ω={0}∪N,S={x|∃n((n∈{0}∪N)∧(x=s(n)))},
s(x)={s(x+1)} when x∈{0}∪N and s(x)={0,1} when x∉{0}∪N.
Let M" be this model. M" |= AR →({0}∪N=0). M" |= ¬AR.

s(t)=0 (t≠0) in the theorems of B. B |- P(ω)→((s(0)={0}={s(0+1)})→(0∈ω)).
B |- (({0}={0})→(0∈ω))→(0∈ω). B |- P(ω)→(0∈ω). B |- P(ω)→(ω≠0).
B |- P(ω)→(ω=0)∧(ω≠0). C |- (ω=0)∧(ω≠0). C isn't consistent. So A isn't consistent.
So ZF isn't consistent. ♦

                                           3

A isn't consistent. Even if you treat only finite sets, the set theory isn't consistent.
The mathematics which you know isn't consistent. The physics isn't consistent too.
The world isn't consistent and every phenomenon is possible. You can unite Ar and Ca²⁺.
As Poincaré

Naoto Meguro :Amateur. MSC2020:13B25. e-mail :

                               1

The theory of the polynomial ring isn't consistent if you assume the axioms of ZF.
The algebraic geometry treating the polynomial ring isn't consistent and proves every
conjecture with its denial if you assume the axioms of ZF.
The theory of the integral functions isn't consistent if you assume the axioms of ZF.
Riemann hypothesis(RH) is nonsense then. ZF is harmful.

                               2

Let's express the polynomial ring by the direct sum. (a,b)+(c,d)=(a+c,b+d). x(a,b)=(xa,xb).
Let R be a ring. R[x]=R⊕xR[x]={(r₁,x(r₂,x(r₃,…|r_i∈R, r_i=0_R except finite ones.}.
Theorem 1. The theory treating R[x] isn't consistent if you assume the axioms of ZF.
Proof. (a,b)={{a,a},{a,b}}∋{a,b}∋b. xⁿR[x]=xⁿR⊕xⁿ⁺¹R[x].
0_xⁿR[x]=(0_xⁿR,0_{xⁿ⁺¹R[x]})∋{0_xⁿR,0_{xⁿ⁺¹R[x]}}∋0_{xⁿ⁺¹R[x]}.
T={0_xⁿR[x]| n∈ω} and U={{0_xⁿR,0_{xⁿ⁺¹R[x]}}| n∈ω} are sets by the axiom of replacement.
T∋0_xⁿR[x]∋{0_xⁿR,0_{xⁿ⁺¹R[x]}}∈U. U∋{0_xⁿR,0_{xⁿ⁺¹R[x]}}∋0_{xⁿ⁺¹R[x]}∈T.
Put S=T∪U. (S≠∅)∧∀y((y∈S)→(y∩S≠∅)).
This is a contradiction by the axiom of regularity. ♦
The theory treating R[x₁]…[x_n] isn't consistent if you assume the axioms of ZF.
When the x_i's are indeterminates, the algebraic geometry treating V=R[x₁]…[x_n]/P"
treats ∪V=R[x₁]…[x_n] and isn't consistent if you assume the axioms of ZF.
Let's express the module of Maclaurin series of the integral functions by H=C⊕xH.
0_xⁿH∋{xⁿ0,0_xⁿ⁺¹H}∋0_xⁿ⁺¹H for ∀n∈N. This leads a contradiction by the axioms of ZF.
The complex function theory treating H isn't consistent if you assume the axioms of ZF.
(x-1)ζ(x) is an integral function. You can write (x-1)ζ(x)=∑_i∈Nb_ix^i-1 (b_i∈C) and can put
b=(b₁,x(b₂,x(b₃,…∈H. (x-1)ζ(x)=(x-c)∑_i∈Nc_ix^i-1 ⇒b=(0,x(c₁,x(c₂,x(c₃,… -c(c₁,x(c₂,x(c₃,….
RH=∀c((c∈C)∧∃h((h∈H)∧(b=(0,xh)-ch))→(Re(c)=2^-1)∨(Im(c)=0)) and ¬RH are proved
in the complex function theory if you assume the axioms of ZF. RH is nonsense then.
p(c,d)=0 ⇔ p(x,y)=(x-c)q+(y-d)r. Set up that R=Q[y] and p∈R[x] and
P=∀c∀d((c∈Q)∧(d∈Q)∧∃q∃r((q∈R[x])∧(r∈R)∧(p=(0_R,xq)-cq+((y-d)r,0_R[x])))→(cd=0)).
¬P is a denial of Fermat conjecture if p=(yⁿ⁺²-1,(0_R,(,…,(0_R,(xⁿ⁺²,0_R[x])…)=xⁿ⁺²+yⁿ⁺²-1.
P and ¬P are proved then if you assume the axioms of ZF by theorem 1.

                              3

If R[x₁]…[x_n] is a ghastly fake as Gauss said, the algebra after Abel is nonsense.
You couldn't admit it and must correct the set theory.
The algebraic geometry,the complex function theory and RH are nonsense until it.

                                1

The definition of the mapping by the graph has defect.
The theory of ⊗ isn't consistent if you define the mapping by the graph.

                                2

Let M be a module. Define f: M∋x→x+x∈M. Put F=Z/2Z and i=1_F.
Let G(g)={(x,g(x)) |(x∈A)∧(g(x)∈B)} be the graph of the mapping g: A∋x→g(x)∈B.
G(g⊗i)={(x⊗(1+2Z),y⊗(1+2Z)|(x,y)∈G(g)}.
Theorem 1. If you define the mapping by the graph, h: M∋x→x+x∈Im(f) ⇒ h=f.
Proof. G(f)={(x,x+x) |(x∈M)∧(x+x∈M)}={(x,x+x) | x∈M}={(x,x+x) |(x∈M)∧(x+x∈Im(f))}
=G(h). So h=f.♦
By the axiom of the equality, f=h ⇒ Im(f⊗i)=Im(h⊗i). (You will know that this is wrong.)
Theorem 2. If you define the mapping by the graph,the theory of ⊗ isn't consistent.
Proof. Put M=Z. M⊗F≅F. You may define M⊗F=F and x⊗(y+2Z)=xy+2Z in M⊗F=F.
f⊗i: M⊗F∋x⊗(1+2Z)→(x+x)⊗(1+2Z)∈M⊗F. f⊗i: F∋x+2Z→2x+2Z=0+2Z∈F. Im(f⊗i)≅{0}.
Im(f)⊗F=(2Z)⊗F≅F. You may define Im(f)⊗F=F and 2x⊗(y+2Z)=xy+2Z in Im(f)⊗F=F.
By theorem 1, f: M∋x→x+x∈Im(f). f⊗i:M⊗F∋x⊗(1+2Z)→(x+x)⊗(1+2Z)∈Im(f)⊗F.
f⊗i: F∋x+2Z→2x⊗(1+2Z)=x+2Z∈Im(f)⊗F=(2Z)⊗F=F. Im(f⊗i)=F. F≅{0}.
This is contradiction.♦
Theorem 3. The set theory which assumes the axioms of ZF and existence of the module Z
and defines the mapping by the graph isn't consistent.
Proof. Put M=Z. M, Im(f)=2Z={x+x |x∈M}, 1+2Z={x|(x∈M)∧(x∉2Z)}, F={2Z, 1+2Z}
and G(f)=G(h)={(x, x+x) |x∈M} are sets. So f=h.
The contradiction in the proof of theorem 2 is led when you define M⊗F=F and
x⊗(1+2Z)=x+2Z in M⊗F=F and (2Z)⊗F=F and (2x)⊗(1+2Z)=x+2Z in (2Z)⊗F=F. ♦
Theorem 4. The theory treating the free modules isn't consistent if you define the mapping
by the graph.
Proof. Z is a free module. M⊗N=Z^(M×N)/(0_M⊗0_N). Z^(M×N) and 0_M⊗0_N are free modules.♦
You may not be able to get the tensor products and the projective resolutions of the
modules in the consistent theory.

                                   3

You may have to reexamine the theorems proved by using ⊗.
Many conjectures may be back to the starting points or may become nonsense.

Naoto Meguro : Amateur. MSC 2020: 03B10.

                              1

The predicate logic using the quantification principle (QP) has defect.
It cannot treat the commutative algebra and the homological algebra.
It proves Riemann hypothesis(RH) plausibly.

                              2

Let a be a free variable. Let p(s) and h(s) be predicate symbols. Let t be a closed term.
Let U be a consistent axiom system. Define p(s)=¬(U |- q(s)) and h(s)=(U |- ¬q(s)).
q(s) doesn't include a.
Theorem 1. If U |- ¬q(t)), ¬(U |- q(s)) for every term s if you use QP.
Proof. ¬p(a)⇒ U |- ∀xq(x) ⇒ ¬p(t). p(t)→p(a) is true. p(t)→∀xp(x) is true. h(t) is true.
h(t)→p(t) is true. p(t) is true. So ∀xp(x) is true. ♦
Assume that U is the axiom system of the theory of the commutative ring A and t=1
and q(s)=¬(∃u((u∈A)∧(us=1))). ¬(U |- q(s)) for ∀s∈A if you use QP. But U |- q(0).
This is contradiction. U has a model in which A=Z/2Z and is consistent.
Let V be the set of the theorems of the metamathematics treating p(s),h(s),U and QP.
Theorem 2. The axiom system V isn't consistent if you use QP.
Proof. ¬p(a)→(¬p(1)) ∈V. p(1)→p(a) ∈V. h(1)→p(1) ∈V. h(1 )∈V. p(1) ∈V.
p(a) ∈V. V |- p(a). V |- ∀xp(x). V |- p(0). ¬p(0) ∈V. V |- ¬p(0). V |- p(0)∧(¬p(0)).
V isn't consistent.♦
The metamathematics about the commutative algebra isn't consistent and proves
that the logical formulas are provable if you use QP. For example,
V |- (U |- ∃x∃y∃z((x∈A)∧(y∈A)∧(z∈A)∧(xyz≠0)∧(xⁿ+yⁿ=zⁿ))) for ∀n∈N if you use QP.
Assume that U is the axiom system of the theory of ⊗ and t=Z and q(s)=Z⊗s≅{0}.
U |- ¬q( Z). So ¬(U |- q(s)) for every term s if you use QP. But U |- q({0}).
The metamathematics about ⊗ isn't consistent if you use QP and U is consistent.
Assume that U is the set of the theorems of the complex function theory and t=2 and
q(s)=(s∈C)∧(ζ(s)=0)∧(Re(s)≥0)∧(Re(s)≠2^-1). If you use QP, ¬(U |- q(s)) for ∀s∈C.
This means that you cannot find a counterexample of RH in the consistent complex
function theory if you use QP.

                              3

You must not use QP to evolve the consistent mathematics though it isn't natural.

Naoto Meguro : Amateur. MSC2020: 03B10.

                                        1

The predicate logic treating the equal sign has defect. You must correct it a little.

                                        2

Let s(t) be a function symbol.
Let X={∀x(x=x), ∀x(x≠s(x)), ∀x∀y∀z((y=z)→((x=y)↔(x=z)))} be an axiom system.
X has a model in which the object domain is {0,1} and 0=0,1=1,0≠1,1≠0,s(0)=1
and s(1)=0 and is consistent.
Theorem 1. The predicate logic treating the equal sign isn't consistent.
Proof. Let c be an individual symbol. Let a be a free variable. Let p(t) be a predicate symbol.
Define p(t)=(X |- c≠t). p(a)=( X |- c≠a)⇔(X |- ∀x(c≠x))⇒(X |- c≠c)=p(c).
p(a)→p(c) is true. ¬p(c)→(¬p(a)) is true. ¬p(c)→∀x(¬p(x)) should be true.
X |- c=c and X is consistent. So ¬p(c) is true. So ∀x(¬p(x)) should be true.
¬p(s(c)) should be true. But p(s(c)) is true. This is contradiction.♦
The axiom system X"={ p(a)→p(c), ¬p(c), p(s(c))} isn't consistent but you can set up that
the axioms in X" are true. Wrong one is the quantification principle. Let's correct it.
Let L be the set of the theorems of the predicate calculus. Set up that the elements of L
are the logical axioms. If q(a)∈L and a is a free variable,∀xq(x)∈L.
∀x(q(x)→r(x))→(∀yq(y)→(∀zr(z)) ∈L when ∀xq(x) and ∀xr(x) are logical formulas.
Set up that the rule of inference is only deduction principle.
The following theorems are proved then.
Theorem 2. Y∪{q} |- p ⇒ Y |- q→p. (q may include free variables. So X" is consistent.)
Theorem 3. Y∪{q} isn't consistent.⇒ Y |- ¬q.
Theorem 4. Y doesn't include the free variable a and Y |- q(a). ⇒ Y |- ∀xq(x).
Theorem 5. Y doesn't include the individual symbol c and Y |- q(c).⇒ Y |- ∀xq(x).
Theorem 6. Y doesn't include the individual symbol c and is consistent.⇒
Y∪{q(c)→∀q(x)} is consistent.
The completeness theorem is proved by these.

                                       3

You should correct the predicate logic to evolve the consistent mathematics.
Known theorems won't be lost by the above correction.

Naoto Meguro : Amateur. MSC 2020: 03B10.

                                     1

The predicate logic has defect. You must correct it a little.

                                     2

Theorem 1. The predicate logic or the set theory isn't consistent.
Proof. Assume that predicate logic and the set theory are consistent.
Let c and d be individual symbols. Let X={c∈d,d∉d} be an axiom system.
X has a model in which c=∅ and d={∅} and is consistent.
Let a be a free variable. Let p(t) be a predicate symbol. Define p(t)=(X |- t∉d).
p(a)=(X |- a∉d)⇔(X |- ∀x(x∉d))⇒(X |- c∉d)=p(c). p(a)→p(c)) is true.
¬p(c)→(¬p(a)) is true. ¬p(c)→∀x(¬p(x)) should be true. X |- c∈d and X is consistent.
So ¬p(c) is true.∀x(¬p(x)) should be true. ¬p(d) should be true. But p(d) is true.♦
Wrong one is quatification principle q→q(a). ⇒ q→∀xq(x).
Let's construct the predicate logic without using the free variables.
Set up that T(q₁,…,q_n) is a logical axiom when T(p₁,…,p_n) is a tautology and the q_i's
are closed logical formulas. (The p_i's are propositional variables.)
Set up that ∀xq(x)→q(t) is a logical axiom when t is a closed term and ∀xq(x) is a
closed logical formula.
Set up that ∀xq(x) is a logical axiom when q(c) is a logical axiom and c is an individual
symbol.
Set up that ∀x(q(x)→r(x))→(∀yq(y)→∀z r(z)) is a logical axiom when ∀xq(x) and
∀xr(x) are closed logical formulas. The rule of inference is only deduction principle.
Theorem 2. If the axiom system Y doesn't include the individual symbol c,
Y |- q(c) ⇒ Y |- ∀xq(x).
Proof. The case that q(c) is an axiom is trivial. If Y |- q(c) by Y |- r(c)→q(c) and Y |- r(c),
you may assume that Y |- ∀x(r(x)→q(x)) and Y |- ∀xr(x) by induction.
Y |- ∀x(r(x)→q(x))→(∀xr(x)→∀xq(x)). Y |- ∀xr(x)→∀xq(x). Y |- ∀xq(x).♦
The following theorems are formed in the above predicate logic treating only closed logical
formulas.
Theorem3. Y∪{q} |- r ⇒ Y |- q→r.
Theorem 4. If Y∪{q} isn't consistent, Y |- ¬q.
Theorem 5. If Y is consistent and doesn't include the individual symbol c,
Y∪{q(c)→∀xq(x)} is consistent.
You can get the completeness theorem without using the free variables.

                                       3

Known theorems won't be lost by the above correction though new theorems won't be
proved.

Naoto Meguro. Amateur. MSC2020 03C62.

                              1

You cannot prove Riemann hypothesis(RH) in the consistent mathematics.
It reveals defect of the set theory or the predicate logic.

                              2

Let T be an individual symbol. Let F be the set of the closed terms. Let G be the set of
the closed terms which don't include T.
Put S"={s|(s∈C)∧(ζ(-2s)=0)∧(Re(-2s)≠2^{- 1})}. RH means S"=N.
You are to treat S" and the elements of S" as individual symbols. S"∈G N⊂S"⊂G
Put U={s|(s∈F)∧(s∉S")} and Y={∃x(x∈T)}∪(∪_s∈U{s∉T}) and X=Y∪(∪_n∈N{n∉T}).
Let P be the proposition "There is a consistent set theory whose object domain is G.".
Theorem 1. X is a consistent axiom system if P is true.
Proof. Assume that P is true. By the definition T =S"-{1,2,…,n}, you get a consistent set
theory in which (s∈T⇔ s∈S"-{1,2,…,n}) and the object domain is F.
In this theory, s∉(S"-{1,2,…,n}) ⇒ s∉T. So 1∉T,…,n∉T.
s∈U ⇒ s∉S" ⇒ s∉S"-{1,2,…,n} ⇒ s∉T. n+1∈(S"-{1,2,…,n}). So n+1∈T. ∃x(x∈T).
The elements of X(n)=Y∪{1∉T, 2∉T,…, n∉T } are true and p∧(¬p) is false in this theory.
In this theory, the logical formulas proved by only logical axioms become true when you
substitute elements of F for the free variables. So ¬(X(n) |- p∧(¬p)) for ∀n∈N.
X is consistent.♦
Let p(t) be a predicate symbol. Let a be a free variable. Define p(t)=(X |- t∉T).
Let H be the set of the theorems of the standard theory whose object domain is F.
Put q=(X |- ∃x(x∈T))⇔ (X |- ¬∀x(x∉T)).
Theorem 2. RH is false if P is true.
Proof. Assume P. X is consistent. p(a)=(X |- a∉T)⇔(X |- ∀x(x∉T))⇒ ¬q. p(a)→(¬q) ∈H.
∀xp(x)→p(a) ∈H. ∀xp(x)→(¬q) ∈H. q→(¬∀xp(x)) ∈H. q∈H. ¬∀xp(x)∈H. ∃x(¬p(x))∈H.
If s∈F, s∉S"⇒ s∈U ⇒ p(s) and s∈N ⇒ p(s). ¬p(s)⇒ s∈S"-N then. ∃x(¬p(x))∈H⇒ S"-N≠∅.
RH is false.♦
The standard theory assumes that P is true. So RH is false in it if it is consistent.

                                3

If P is true, p(a)→(¬q). q→(¬p(a)). q→∀x(¬p(x)). ∀x(¬p(x)). ¬p(1). p(1).
The standard theory isn't consistent and proves every conjecture and its denial.
You may have to treat only finite sets as Gauss insisted. The above will be ghastly for him.
You couldn't define Γ(s) and ζ(s) then.

Naoto Meguro. Amateur. MSC 39A06.

                                  1

You can express the solution of the difference equation by the coefficients of it.
So there is an analytic formula giving a root of the algebraic equation.
These don't appear on the list of the formulas now.

                                  2

Set up that x_n=e_n (1≤n≤N) and x_n=∑_1≤k≤N c_kx_n-k (n≥N+1)
Theorem 1. n≥N+1 ⇒ x_n=∑_1≤p≤N e_p ∑_{k1+…+ks=n-p, 1≤ku≤N, ks≥N+1-p} ∏_1≤u≤s c_ku.
Proof. n≥N+1⇒ x_n=∑_1≤n-k1≤N e_n-k1c_k1 +∑_n-k1≥N+1 c_k1x_n-k1.
Put n-k₁-…-k_s=p. If s=1,k_s=k₁=n-p≥N+1-p.
By induction, you may assume that
n-k₁≥N+1 ⇒ x_n-k1=∑_1≤p≤Ne_p∑_{k2+…+ks=(n-k1)-p,1≤ku≤N, ks≥N+1-p} ∏_2≤u≤s c_ku.
Substituting these, you get theorem 1.♦
Theorem 2. n≥N+1 ⇒
x_n=∑_1≤p≤N e_p∑_1≤s≤n-p∫_0≤t≤2π(∑_1≤k≤N c_ke^ikt)^s-1(∑_{N+1-p≤k≤N} c_ke^ikt)e^it(p-n)dt/2π.
Proof. ∑_{k1+…+ks=n-p,1≤ku≤N, ks≥N+1-p} ∏_1≤u≤s c_ku is the coefficient of x^n-p in
(∑_1≤k≤N c_kx^k)^s-1(∑_{N+1-p≤k≤N} c_kx^k) etc..♦
(v∈C)∧(w∈C)∧(r∈R-{0})∧(r"∈R-{0})∧(v≠w)∧(|v|=|w|)∧(|v+r|=|w+r|)⇒
(Re(v)=Re(w))∧(Im(v)= - Im(w))∧(|v+r+r"i|≠|w+r+r"i|). ( i²=- 1.)
When f(x)=x^N+a₁x^N-1+…+a_N-1x+a_N=(x-w₁)…(x-w_N) and w_k≠w_h (k≠h) and a_N≠0,
you can set up that |w_k+p^{- 1}+q^{- 1}i|≠|w_h+p^{- 1}+q^{- 1}i|≠0 (k≠h) for p∈N and q∈N
for (|w_k|≠|w_h|)∨(|w_k+p^{- 1}|≠|w_h+p^{- 1}|)⇒(|w_k+p^{- 1}+q^{- 1}i|≠|w_h+p^{- 1}+q^{- 1}i|≠0)
if p and q are large enough.
When f(x-p^{- 1}-q^{- 1}i)=x^N-c₁x^N-1-…-c_N-1x-c_N and e₁=1 and e_d=0 (2≤d≤N),
x_n+1=∑_1≤k≤N d_k(w_k+p^{- 1}+q^{- 1}i)ⁿ⁺¹. (d₁,…,d_N)≠(0,…,0).
x_n+1/x_n→w+p^{- 1}+q^{- 1}i (n→∞). f(w)=0. w+p^{- 1}+q^{- 1}i → w+q^{- 1}i (p→ ∞)
w+q^{- 1}i→w (q→∞).
You get a formula giving a root of the algebraic equation written by the coefficients of it
and ∑, ∫ ,e^t, and lim.

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You can express the algebraic functions as explicit functions.