Polynomial-form - proper naming

26 Views Asked by At

In the mathematics literature, what are the names of all the possible different forms that a polynomial could take? That is, how would one, with words, differentiate between the following forms of an $n^{\textrm{th}}$-order polynomial $P_n(z)$?

$$P_n(z)=a_0\left(1-\frac{z}{z_1}\right)\left(1-\frac{z}{z_2}\right)\cdots \left(1-\frac{z}{z_n}\right) ~~~~ \textrm{\{}z_i\textrm{ are the roots\}} \tag{1}$$

$$P_n(z)=a_0+a_1z+a_2z^2+\cdots + a_nz^n \tag{2}$$

$$P_n(z)= a_0\left(1+b_1z(1+b_2z(1+\cdots ))\right) \tag{3}$$

$$\textrm{.... etc. .....}$$

2

There are 2 best solutions below

0
On BEST ANSWER

$Pn(z) = a_0 + a_1 z + a_2 z^2 \cdots a_n z^n$ is standard form

$Pn(z)=a_0(z−z_1)(z−z_2)⋯(z−z_n)$ I would call this factored form or a factorization or a product of irreducible polynomials.

$Pn(z)=a_0(1−\frac z{z_1})(1−\frac z{z_2})(1−\frac z{z_3})⋯((1−\frac z{z_n})$ Is a similar idea, but I don't have a name for this.

$Pn(z)=a_0 + a_1 (z - z_1) + a_2 (z - z_1)(z-z_2) + \cdots a_n \prod (z-z_i)$ Is a Newton polynomial. Which can be factored:

$Pn(z)=a_0 + (z - z_1)(a_1 + (z - z_2)(a_2+(z-z_3)\cdots)))$

0
On

A polynomial $f(z)\in R[z]$ is defined over, say, a ring $R$, and is written as in $(2)$, where the coefficients $a_i$ belong to $R$. So this has the name "polynomial of degree $n$ over $R$". $(2)$ is a special decomposition into "linear factors", assuming that all $z_i\neq 0$. For $R=\mathbb{C}$ the general decomposition is given by $$ P_n(z)=a_n(z-z_1)\cdots (z-z_n). $$ The name for this is the fundamental theorem of algebra. $(3)$ has no special name.

Sometimes different names are used for polynomial rings over various rings, e.g., polynomials $f\in \mathbb{Z}[x]$ are called "integer polynomials", and the equation $f(x)=0$ then is called a "Diophantine equation".