Research Publication (math): An Old Math Logic Idea I Had a Few Months Ago

Note from my "Notes app" on my MacBook Air are posted below.  I haven't reviewed this carefully, I don't think I included anything embarrassing in this file.  Basically, it's an idea for a formal theory that can formalize some things about calculus that does not have the ability to prove Godel's Incompleteness Theorems within it.  There's also a cool number theory algorithm buried in here somewhere, I forgot what it does or how it works though.  Note, as mentioned below, Google AI assisted in my selection of this idea...though all of the subsequent math/calculations/ideas were 100% all my own.  The cool algorithm is, I think, for mapping circuits to polynomials.  I mostly forgot, though.

 Clean Calculus Summary - Current

12-7-2025, 11:45 a.m.

Formal Theory:  Taylor Calculus (TC).

by Philip J. White

philipjwhite2025@gmail.com

(Citation:  Google AI provided me with the idea of devising a formal theory that axiomatizes calculus, in response to a question about hobbies I should pursue.  All of the research, aside from cited sources, is my own.)

Our domain is:

all infinite rational-component valued vectors that can be generated from quantified boolean expressions including predicate symbols, variables, and the constant symbols

Note:  The constant symbols alone do not make up everything in the domain.

Our constant symbols are:

<0, 0, 0, 0, … , 0 , …>

and

<-1, -1, -1, … , -1 , … >

and

<0,1,2,3,4,…,>

and

<1, 0, 0, 0, … 0 , …> , <0, 1, 0, 0, … 0 , …> , <0, 0, 1, 0, … 0 , …> , …

and

<2, 0, 0, 0, … , 0 , …> , <0, 2, 0, 0, … , 0 , …> …

and

<3, 0, 0, 0, …> ….

…

(Infinitely many integers, from 1 to 2 to 3 to infinity)

and

<1, 1, 1, 1, … ,>

<2, 2, 2, … ,>

<3, 3, 3, …,>

…

We can construct some integer valued vectors (i.e., < c1, c2, c3, … , >, where c1, … , are all integer constants) using the constant symbols and multiplication, e.g., <1, 1, 1, …> * <0,2, 0, …> * <0,0,5, …>.  Note, we can’t express *all vectors* that way, because they’re uncountably infinite.  We can also construct many rational-valued vectors…see the 5 predicate symbols below.

Note:  There are countably infinitely many constant symbols.  We can index them and write c_1, c_2, … , by beginning with the two finite sets of constant symbols above, and then “dovetailing” over the two countably infinite sets of constant symbols given above.

Our predicate symbols are:

shift(x,y)

We can multiply and divide any Taylor series by x with this predicate symbol and its inverse.

taylorDotProduct(x,y,z)

The domain of this function is restricted; y cannot have any 0 components.

taylorReciprocal(x,y)

Multiplicatively invert all of the coefficients, and it equals y; this function and its inverse are not defined for vectors that have 0 as any of their components—restricted domain—and we don’t require the ability to multiply an inverse as the LHS of an expression, e.g., we do not ever have to do x * y = z and then x^(-1) * x * y = x^(-1) * z; we do have to use the equation, x*y*y^(-1) = z*y^(-1) .

equals(x,y)

This relation “returns true” if the two infinite vectors are equal.

compose(s,y,z)

In Taylor series y written as a vector, let x= the Taylor series s, so we’ll take y(s), and this new Taylor series will be represented as z.

Note:  We can write every Taylor vector as a summation series statement, i.e., something like sum(s_a*x^(a-r),a,1,infinity), and then define s_a = … .

Axioms:

(can be enumerated to capture precisely the predicate symbols listed above)

Provable Theorems:

- Taylor Calculus (TC) can express, compose, and prove solutions to certain circuit decision and function problems.

——————————————————————————————————————————————————————————————————

Proof Idea:  We can write any circuit as a function that maps finitely many nonnegative integers to finitely many nonnegative integers.  We can capture that with a finite polynomial, and we can construct any finite integer-coefficient-valued polynomial out of a Taylor series that has a “final non-zero component,” where all components after the “final” component are equal to 0.  We can do this using the predicate symbols and the constant symbols.  Next, we can use the rational root theorem and techniques that are similar to the Chinese remainder theorem to construct the coefficients of a polynomial, one by one, that will yield a polynomial that precisely captures all of the ordered pairs specified by the finite domain and finite codomain function described above.

Here are some more notes about this proof idea:

Let P(x) be a polynomial of finite degree such that it does not have a non-zero constant term that is the coefficient of x^0.

P(x0) + c  =  y0

Here, x0 is the x coordinate of one of the points, and y0 is the coordinate of one of the other points.

We can re-write this as:

P(x0) + c - y0  =  0

Thus, by the rational root theorem, since all of our solutions are integers, we have that given the solution p/1, p, the numerator, is a divisor of the constant term c-y0, and 1 is a divisor of the leftmost coefficient (the coefficient of x^k, where k is the degree of the polynomial).  The second fact is trivial, but the first fact is very important.

So we have:

c - y0 (mod d0)  =  0

The unknown we want to solve for is d.

The good news is, we have all the constraints we need.  Consider all k of the points, (x0,y0) , … , (x_k, y_k).

We can write:

c - y0 (mod d0) = 0

c - y1 (mod d1)  =  0

c - y2 (mod d2)  =  0

…

c - y_j (mod d_j)  =  0

Quite easily, we can re-write this as:

c (mod d0) = y0

c (mod d1)  =  y1

c (mod d2) = y2

…

c (mod d_j)  = y_j

The key is, we already have the values for d0, d1, d2.  The value of d0 is the value of x0, etc.

So it’s really:

c (mod x0) = y0

c (mod x1) = y1

…

c (mod x_j)  =  y_j

We may not have that x0, x1, … , x_j are pairwise coprime, so using the Chinese Remainder Theorem might be tricky.  According to this resource, however, we can still solve the system of modular equations:  https://math.stackexchange.com/questions/2294043/how-to-solve-congruence-systems-with-noncoprime-moduli

Thus, assuming that resource is correct and that we can still solve systems of modular equations with non-pairwise-coprime moduli, we can completely solve for a good, even minimal value of c.

So we can now in a sense absorb c into the y_i values.  The new equation is:

P’(x) + d*x^1 =y0 - c

The idea is, P’ is a polynomial that has 0 for its coefficients at x^0 and x^1.  Now we want to solve for d.

Now, if we substitute in the value of x0 for x, we get:

P’(x0) + d*(x0) =y0 -c

We substituted in x0 for x, that is why it is now d*x0.

Since all of these values are known constants, we can easily re-write this equation (and the equation for x1, x2, …) as:

P(x0) +(d*x0+c-y0)  =  0

Now, we can apply the rational root theorem again.  We don’t have to treat P(x0) as having a constant subbed in for it; we can revert to: P(x), and leave d*x0+c-y0 the same.

We have:

P’(x) + d*x + c-y  =  0

Now, we need to access d as separate from the x0  value.  Recall that we have coordinate pairs (x0,y0),(x1,y1),(x2,y2), … .  We are currently working with (x0,y0), and we will generalize the argument to address the other coordinate pairs soon.

Let’s do this:

P’(x) +d*x =y-c

We will use the information that we got from the previous use of the techniques that are similar to Chinese Remainder Theorem.  We got the value of c.

P’(x_i) +d*x^1 + c  =  y_i

We have the c value that satisfies all (x_i,y_i) pairs.

P’(x0) + d*(x0)^1 + c-y0  =  0

If we have a constant value for x0, we can use that.

The modulus is x0.

Divide the whole non-zero LHS of the equation by x, i.e., x0, and divide the RHS (0) by that as well:

Now, let P’# stand for:  P’(x)/x.  The degree will be one less.  Each coefficient applies to a slightly different value, i.e., each coefficient applies to the polynomial term with the of power of that term x^s reduced so that we have it applying to x^(s-1) .  The final value, the coefficient of x^2, becomes the coefficient value for x^1.

P’#(x0) + d + (c-y0)/x0  =  0

This time, the rational root theorem yields:

d+(c-y0)/x0 (mod x0)  =  0

Remember, we already had:  c-y0 (mod x0) = 0, for all of the different x_i values even.  So we can *DIVIDE* arithmetically, since x0 divides c-y0, neglecting number theory, (c-y0)/x0, and get an integer.  Call that integer J(y0).

Thus we have:  d + J(y0) (mod x0) = 0

Recall that all values for x0 and y0 and x_i and y_i in general are the same—they’re derived from the coordinate pairs.

So overall, extending this argument to all of the coordinate pairs, x0, x1, x2, … , x_k, we have:

d (mod x0) = -I(y0)

d (mod x1) = -I(y1)

…

We can solve this for d, the next coefficient in the polynomial, with the techniques mentioned in the link above that are similar to the Chinese Remainder Theorem, too.

We continue this process, moving to the left in the polynomial as far as we need to, until we have solved for all of the coefficients.  Remember that the leftmost term in the polynomial has coefficient equal to 1.

——————————————————————————————————————————————————————————————————

 - TC can express, compose, and prove solutions to certain decision and function problems about infinitely differentiable everywhere differentiable functions.

Proof Idea:  We can construct the derivative of any Taylor series.  We do this using the shift predicate function and the taylorDotProduct function.  To begin with, we take the constant symbol vector <0, 1, 2, 3, … > and Taylor dot product it with the vector that represents the Taylor series in question.  Then, we apply shift, to move the coefficients one to the left.  This yields a vector that is precisely the derivative of the initial Taylor series.  We can define integration using wfs and this derivative function that we can build from the constant symbols and predicate symbols.  Note, we seem to be unable to construct:  addition, modulus, limits, or Turing machines.  To prove this is difficult; if TC is provably consistent and complete, then it is provably true that addition, modulus, and Turing machines cannot be defined.  It is not clear how to prove that limits are not able to expressed in TC, but the epsilon-delta definition would appear to be difficult to formalize without a notion of subtraction.

Key Conjecture:

• “The formal theory TC is consistent and complete”…is a theorem of ZFC, though probably not of TC.

If this conjecture is true:

We can construct a calculus formal theory, Taylor Calculus (TC), such that TC is complete and consistent.  The theory is able to handle scalar multiplication if we treated the domain objects that are infinite vectors as scalars when their components are all equal to the same natural number.  Addition is not defined within the theory.  We can express and prove facts about scalar multiplication, or, if we interpret the vectors in a different way, about derivatives and integrals of functions.

Additional Theorems That Would Follow From This Conjecture If Proved:

 - Addition cannot be constructed in TC, and thus neither can Turing machines be.

• Of critical importance:  There are no undecidable integrals that use continuous infinitely and everywhere differentiable functions, i.e., every integral either has a solution to it, or, the solution is *provably* undefined, including in ZFC.  “Integration problems are decidable.”