note-math

In flat space, linear and affine are often mixed. Similarly, in flat space, so are polynomials. Zero-order polynomials correspond to the use of affine

First deal with the case of one-dimensional real numbers

$\mathrm{\mathbb{N}}$ exponential power function $𝑘\in \mathrm{\mathbb{N}},\begin{array}{ccc}\mathrm{ℝ}& ⟶& \mathrm{ℝ}\\ 𝑣& ⟿& {𝑣}^{𝑘}\end{array}$

polynomial-function-1d_(tag) A polynomial function is a finite linear combination of power functions. (Affine) base point $𝑥$, (vector) offset $𝑣$

$$\begin{array}{ccc}𝑓\left(𝑥+𝑣\right)& =& {𝑎}_0+{𝑎}_1𝑣+\cdots +{𝑎}_{𝑛}{𝑣}^{𝑛}\\ & =& \sum _{𝑘=0..𝑛}{𝑎}_{𝑘}{𝑣}^{𝑘}\\ 𝑓\left(𝑥\right)& =& {𝑎}_0\end{array}$$

Polynomial function representation is not affine invariant, i.e. switching the base point $𝑥⇝𝑥+\mathrm{\Delta }=𝑦$ will result in a polynomial function representation of the same order but with different coefficients. Scaling $𝑣⇝𝜆𝑣$ is also the case

change-base-point-polynomial_(tag) Switch base point $𝑥⇝𝑥+\mathrm{\Delta }=𝑦$

$$\begin{array}{ccc}𝑓\left(𝑥+𝑣\right)& =& {𝑎}_0\left(𝑥\right)+{𝑎}_1\left(𝑥\right)𝑣+\cdots +{𝑎}_{𝑛}\left(𝑥\right){𝑣}^{𝑛}\\ 𝑓\left(𝑦+𝑤\right)& =& {𝑎}_0\left(𝑦\right)+{𝑎}_1\left(𝑦\right)𝑤+\cdots +{𝑎}_{𝑛}\left(𝑦\right){𝑤}^{𝑛}\end{array}$$

Represents the same affine function

$$𝑥+𝑣=𝑦+𝑤⟹𝑓\left(𝑥+𝑣\right)=𝑓\left(𝑦+𝑤\right)$$

then

$$\begin{array}{ccc}{𝑎}_{𝑝}\left(𝑦\right)& =& {𝑎}_{𝑝}\left(𝑥+\mathrm{\Delta }\right)\\ & =& \sum _{𝑘=𝑝..𝑛}{𝑎}_{𝑘}\left(𝑥\right)\left(\genfrac{}{}{0ex}{}{𝑘}{𝑝}\right){\mathrm{\Delta }}^{𝑘-𝑝}\end{array}$$

Proof $𝑓\left(𝑦+𝑣\right)=𝑓\left(𝑥+\left(\mathrm{\Delta }+𝑣\right)\right)$ expand the calculation, to compare coefficients, collect $𝑣$ power function terms, by the exchange of summation

$$\begin{array}{ccc}\sum _{𝑘=0..𝑛}{𝑎}_{𝑘}\left(𝑥\right){\left(𝑣+\mathrm{\Delta }\right)}^{𝑘}& =& \sum _{𝑘=0..𝑛}{𝑎}_{𝑘}\left(𝑥\right)\sum _{𝑝=0..𝑘}\left(\genfrac{}{}{0ex}{}{𝑘}{𝑝}\right){𝑣}^{𝑝}{\mathrm{\Delta }}^{𝑘-𝑝}\\ & =& \sum _{𝑝=0..𝑛}\left(\sum _{𝑘=𝑝..𝑛}{𝑎}_{𝑘}\left(𝑥\right)\left(\genfrac{}{}{0ex}{}{𝑘}{𝑝}\right){\mathrm{\Delta }}^{𝑘-𝑝}\right){𝑣}^{𝑘}\end{array}$$

If the base point is $0$ in the coordinates and the symbol $𝑣⇝𝑥$ is changed, then the polynomial function is expressed as $𝑓\left(𝑥\right)={𝑎}_0+{𝑎}_1𝑥+\cdots +{𝑎}_{𝑛}{𝑥}^{𝑛}$

Extending from polynomial as a finite linear combination to a countably infinite linear combination is called the $\mathrm{\mathbb{N}}$ exponential power series of a function

$$\begin{array}{ccc}𝑓\left(𝑥+𝑣\right)& \approx & {𝑎}_0+{𝑎}_1𝑣+\cdots +{𝑎}_{𝑛}{𝑣}^{𝑛}\\ 𝑓\left(𝑥+𝑣\right)& =& \underset{𝑛\to \infty }{\text{lim }}{𝑎}_0+{𝑎}_1𝑣+\cdots +{𝑎}_{𝑛}{𝑣}^{𝑛}\end{array}$$

The definitions of some functions do not come directly from the $\mathrm{\mathbb{N}}$ exponential power series, Example $\frac{1}{𝑥},\frac{1}{𝑧}$

In addition to $\mathrm{\mathbb{N}}$ as countably infinite data, $\mathrm{\mathbb{Z}},\mathrm{\mathbb{Q}}$ can also be used. The $\mathrm{\mathbb{N}}$ exponential power function ${𝑣}^{𝑘}$ is changed to the $\mathrm{\mathbb{Q}}$ exponential power function ${𝑣}^{\frac{𝑝}{𝑞}}$

• ${𝑣}^{-𝑘}=\frac{1}{{𝑣}^{𝑘}}$ requires multiplicative inverse

• ${𝑣}^{\frac{1}{𝑘}}=\sqrt[𝑘]{𝑣}$ requires solving the equation ${𝑤}^{𝑘}=𝑣$ and needs to deal with the issue of whether the number of solutions is unique

• ${𝑣}^{-𝑘}$ is unbounded at $𝑣=0$

• When $\frac{𝑝}{𝑞}\notin \mathrm{\mathbb{N}}$, the multiple derivatives will not be interrupted $\forall 𝑛,\left({𝑣}^{\frac{𝑝}{𝑞}}⇝\frac{𝑝}{𝑞}\cdots \left(\frac{𝑝}{𝑞}-𝑛+1\right){𝑣}^{\frac{𝑝}{𝑞}-𝑛}\ne 0\right)$

Here we only deal with $\mathrm{\mathbb{N}}$ power series for the time being, and refer to them as power series for short

Now deal with the high-dimensional case i.e. ${\mathrm{ℝ}}^{𝑑}\to {\mathrm{ℝ}}^{{𝑑}^{\prime }}$

If the range is $\mathrm{ℝ}$, we can also define the multiplication of functions $\left(𝑓𝑔\right)\left(𝑥\right)=𝑓\left(𝑥\right)𝑔\left(𝑥\right)$ and the multiplicative inverse $\left(\frac{1}{𝑓}\right)\left(𝑥\right)=\frac{1}{𝑓\left(𝑥\right)}$

First try to define polynomial functions and power series based on tensors i.e. multilinear

$$\underset{𝑘=0..𝑛}{⨂}{\mathrm{ℝ}}^{𝑑}$$

If not necessary, there is no need to take the linear direct sum of tensors of all orders ${⨁}_{𝑛=0..\infty }$ (called tensor algebra)

polynomial-function_(tag) Using the range ${\mathrm{ℝ}}^{{𝑑}^{\prime }}$ and multilinear function ${𝑎}_{𝑘}\in \text{Lin}\left({\left({\mathrm{ℝ}}^{𝑑}\right)}^{\otimes 𝑘}\to {\mathrm{ℝ}}^{{𝑑}^{\prime }}\right)$. Base point $𝑥$, offset $𝑣$, define polynomial function

$$\begin{array}{ccc}𝑓\left(𝑥+𝑣\right)& =& {𝑎}_0+{𝑎}_1𝑣+\cdots +{𝑎}_{𝑛}{𝑣}^{\otimes 𝑛}\\ & =& \sum _{0..𝑛}{𝑎}_{𝑘}{𝑣}^{\otimes 𝑘}\\ 𝑓\left(𝑥\right)& =& {𝑎}_0\end{array}$$

Affine transformations, i.e., changing the base point i.e. translation, or linear transformations i.e. $\text{GL}$ (including scaling), do not change the order of the polynomial

${𝑎}_1𝑣,{𝑎}_2{𝑣}^{\otimes 2}\in {\mathrm{ℝ}}^{{𝑑}^{\prime }}$ may not be collinear

Extending to $\mathrm{ℂ}$ is simple. For the cases of $\mathrm{ℍ},\mathrm{𝕆}$, due to non-commutativity and non-associativity, high-dimensional linear algebra and tensors require new processing methods

Different tensors may give the same polynomial, but the symmetric tensor is uniquely corresponding

Change notation

• ${𝑣}^{\odot 𝑘}⇝{𝑣}^{𝑘}$ power-tensor_(tag)
• $𝑣\odot 𝑤⇝𝑣𝑤$

The method to restore from a monomial of order $𝑛$ or a power tensor ${𝑣}^{𝑛}$ to a symmetric tensor of order $𝑛$ ${𝑣}_1\cdots {𝑣}_{𝑛}$ is difference

In the $𝑛$-th order monomial ${\left({𝑣}_1+\cdots +{𝑣}_{𝑛}\right)}^{𝑛}$ of ${𝑣}_1+\cdots +{𝑣}_{𝑛}$, there is a term ${𝑣}_1\cdots {𝑣}_{𝑛}$, but there are many other interfering terms

The whole problem is symmetric with respect to ${𝑣}_1,\dots ,{𝑣}_{𝑛}$, so a symmetric construction should be used

In the second order ${\left({𝑣}_1+{𝑣}_2\right)}^2-{𝑣}_1^2-{𝑣}_2^2=2{𝑣}_1{𝑣}_2$

difference-symmetric-tensor_(tag) Symmetric tensor $𝑛$-th order difference

$$\sum _{𝐼\subset \left\{1,\dots ,𝑛\right\}}{\left(-1\right)}^{\left|𝐼\right|-𝑛}{\left(\sum _{𝑖\in 𝐼}{𝑣}_{𝑖}\right)}^{𝑚}=\{\begin{array}{cc}0& \text{ if }𝑚<𝑛\\ 𝑛!\cdot {𝑣}_1\cdots {𝑣}_{𝑛}& \text{ if }𝑚=𝑛\\ \ne 0& \text{ else}\end{array}$$

Question Is there an intuitive understanding of the $𝑛$-th order difference?

successive-difference_(tag) The $𝑛$-th order difference can be written as $𝑛$ times the first-order difference

$$\sum _{{𝐼}_{𝑛}\subset \left\{𝑛\right\}}\cdots \sum _{{𝐼}_1\subset \left\{1\right\}}{\left(-1\right)}^{\left|{𝐼}_{𝑛}\right|-1}\cdots {\left(-1\right)}^{\left|{𝐼}_1\right|-1}{\left(\sum _{{𝑖}_1\in {𝐼}_1}{𝑣}_{{𝑖}_1}+\cdots +\sum _{{𝑖}_{𝑛}\in {𝐼}_{𝑛}}{𝑣}_{{𝑖}_{𝑛}}\right)}^{𝑛}$$

where ${𝐼}_{𝑘}\subset \left\{𝑘\right\}⟺{𝐼}_{𝑘}\in \text{Subset}\left\{𝑘\right\}=\left\{\varnothing ,\left\{𝑘\right\}\right\}$, ${\sum }_{{𝑖}_{𝑘}\in \varnothing }{𝑣}_{{𝑖}_{𝑘}}=0$, $𝐼≔{\bigcup }_{1..𝑛}{𝐼}_{𝑘}\subset \left\{1,\dots ,𝑛\right\}$

Due to the commutativity of summation, the order of successive differences does not affect the final result

Proof of #link(<difference-symmetric-tensor>)[]

The symmetry of the symmetric multilinear function ${𝑎}_{𝑚}$ allows the properties of #link(<difference-symmetric-tensor>)[difference] to be inherited

difference-polynomial_(tag) The $𝑛$-th order difference of $𝑓\left(𝑥+𝑣\right)={𝑎}_{𝑛}{𝑣}^{𝑛}$ is $𝑛!\cdot {𝑎}_{𝑛}\left({𝑣}_1\cdots {𝑣}_{𝑛}\right)$

The $𝑛$-th order difference of $𝑓\left(𝑥+𝑣\right)={𝑎}_{𝑛}{𝑣}^{𝑚},𝑚<𝑛$ is $0$

From this, we get that the polynomial function determines its multiple symmetric linear function representation Proof First, $𝑛$-difference gives the same ${𝑎}_{𝑛}$, after removing ${𝑎}_{𝑛}$ from both sides, it is still the same polynomial function, the order is $<𝑛$, continue $𝑛-1$ difference to get the same ${𝑎}_{𝑛-1}$ …

For power series, finite-order difference can never give zero

Formally, division and limits can be used to eliminate higher-order terms

$$\begin{array}{ccc}\frac{1}{{𝑡}^{𝑛}}\left({𝑎}_{𝑛}{\left(𝑡𝑣\right)}^{𝑛}+{𝑎}_{𝑛+1}{\left(𝑡𝑣\right)}^{𝑛+1}+\cdots \right)& =& {𝑎}_{𝑛}{𝑣}^{𝑛}+{𝑎}_{𝑛+1}{𝑣}^{𝑛+1}𝑡+\cdots \\ & =& {𝑎}_{𝑛}{𝑣}^{𝑛}+𝑜\left(1\right)\\ \underset{𝑡\to 0}{\text{ lim }}& =& {𝑎}_{𝑛}{𝑣}^{𝑛}\end{array}$$