In flat spaces, linear and affine are often mixed. Similarly, in flat spaces, polynomials are also like this. Zero-order polynomials correspond to the base point in affine.
First, handle the one-dimensional real case.
exponential power function
[polynomial-function-1d] Polynomial functions are finite linear combinations of power functions. (Affine) base point , (vector) offset
The representation of a polynomial function is not affine invariant, i.e., switching the base point yields a representation of the same degree but with different coefficients. Scaling is also like this.
[change-base-point-polynomial] Switching base point
Then
Proof Expand and compute, compare the coefficients of to get the expression for . Use the commutativity of summation to collect power function terms.
If using as the base point in coordinates and changing the symbol , then the polynomial function is expressed as
Generalizing from polynomials as finite linear combinations to countably infinite linear combinations, called exponential power series of functions.
Some functions are not defined directly from exponential power series. Example can be defined directly through division, without needing to use power series for definition.
Besides as countable infinite data, can also be used.
Changing the exponential power function to the exponential power function introduces new aspects that require attention.
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requires a multiplicative inverse.
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requires solving the equation and addressing the issue of whether the solution is unique.
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is unbounded at .
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When , repeated differentiation does not terminate .
For now, we only deal with power series, abbreviated as power series.
Now handle the higher-dimensional case i.e. .
If the codomain is , we can additionally define function multiplication and multiplicative inverse .
First, attempt to base the definitions of polynomial functions and power series on tensors i.e. multilinearity.
If not necessary, there is no need to take the linear direct sum for tensors of all orders (known as the tensor algebra) for now.
[polynomial-function] Using codomain and multilinear functions . Base point , offset , define polynomial function
Affine transformations, i.e., changing the base point i.e. translation, or linear transformations i.e. (including scaling), do not change the degree of the polynomial.
may not be collinear.
Generalization to is straightforward. For the cases of , due to non-commutativity and non-associativity, it becomes very troublesome.
There may be different tensors that give the same polynomial, but a polynomial corresponds to a unique symmetric tensor .
Change notation.
- [power-tensor]
Using difference techniques, one can recover the -th order symmetric tensor from the -th order monomial or power tensor .
In the -th order monomial of , there is a term , but there are many other interfering terms.
The entire problem is symmetric in , so a symmetric construction should be used.
In second order
[difference-symmetric-tensor] Symmetric tensor th order difference
Question Is there an intuitive understanding of th order difference?
[successive-difference] th order difference can be written as successive first-order differences
where , ,
Due to the commutativity of summation, the order of successive differences does not affect the final result
Proof of difference-symmetric-tensor
Fully expand
A function can be viewed as a function . Thus
Define weight
Prop For any non-empty finite set ,
Proof
All can be classified according to , each corresponds to choices (combination). Thus
for
define
define
There is a bijection
Weight
Final condition
- When , it always holds for all
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When , it fails only when is a bijection i.e. is all th order permutations, then
- and
Symmetric tensors make
Remaining
There are permutations of order
So the result is
The symmetry of a symmetric multilinear function allows the property of difference to be inherited
[difference-polynomial] The -th order difference of is
The -th order difference of is
From this we obtain
Prop A polynomial function determines its symmetric multilinear function representation
Proof Suppose polynomial has two symmetric multilinear function representations
First, the -th order difference gives the same . Simultaneously removing the -th order term yields a polynomial of degree and its two symmetric multilinear representations . Induction on yields the conclusion
For power series, finite-order differences can never yield zero
Formally, division and limits can be used to eliminate higher-order terms