polynomial — Jemoka Knowledge Base

A polynomial is a polynomial constituents a function p: \mathbb{F} \to \mathbb{F} coefficient a_0, \dots, a_{m} \in \mathbb{F} requirements A polynomial is defined by:

\begin{equation} p(z)=a_0+a_1z+a_2z^{2}+\dots +a_{m}z^{m} \end{equation}

for all z \in \mathbb{F} additional information degree of a polynomial \deg p A polynomial’s degree is the value of the highest non-zero exponent. That is, for a polynomial:

\begin{equation} p(z) = a_0+a_1z+\dots +a_{m}z^{m} \end{equation}

with a_{m} \neq 0, the degree of it is m. We write \deg p = m. A polynomial =0 is defined to have degree -\infty Of course, a polynomial with degree n, times a polynomial of degree m, has degree mn. We see that:

\begin{equation} x^{n}x^{m} = x^{n+m} \end{equation}

\mathcal{P}(\mathbb{F}) \mathcal{P}(\mathbb{F}) is the set of all polynomials with coefficients in \mathbb{F}. \mathcal{P}(\mathbb{F}) is a vector space over \mathbb{F} We first see that polynomials are functions from \mathbb{F}\to \mathbb{F}. We have shown previously that F^s is a Vector Space Over F. Therefore, we can first say that \mathcal{P}(\mathbb{F}) \subset \mathbb{F}^{\mathbb{F}}. Lastly, we simply have to show that \mathcal{P}(\mathbb{F}) is a subspace. zero exists by taking all a_{m} = 0 addition is closed by inheriting commutativity and distributivity in \mathbb{F} scalar multiplication is closed by distributivity Having satisfied the conditions of subspace, \mathcal{P}(\mathbb{F}) is a vector space. \blacksquare \mathcal{P}_{m}(\mathbb{F}) For m\geq 0, \mathcal{P}_{m}(\mathbb{F}) denotes the set of all polynomials with coefficients \mathbb{F} and degree at most m. product of polynomials see product of polynomials polynomial of operator see polynomial of operator