Suggest edit — polynomial

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title: "polynomial"
source: https://www.jemoka.com/posts/kbhpolynomial/
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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&rsquo;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