Set of all n-polynomials is Convex

24 Jul, 2024

The set of all $n$ th degree polynomials has the following structure:

{P (x) ∣ P (x) = \sum_{i = 0}^{n} a_{i} x^{i} \forall n \in ℕ_{0}}

This includes polynomials of every degree imaginable, provide your imagination only considers only the natural numbers, sans zero. Is this resulting infinite set of polynomials a convex set?

Yes.

Every convex set $C$ has a defining property: $θ x_{1} + (1 - θ) x_{2} \in C$ for two elements $x_{1}, x_{2} \in C$ and $0 \leq θ \leq 1$ . This requires that the sets to contain elements that can be found "in-between" the two generator elements $x_{1}$ and $x_{2}$ .

Keeping this in mind, we consider two polynomials of $n$ -th degree; $H (x)$ and $G (x)$ .

Respectively:

H (x) = h_{0} + h_{1} x + h_{2} x^{2} + \dots + h_{n} x^{n}

G (x) = g_{0} + g_{1} x + g_{2} x^{2} + \dots + g_{n} x^{n}

Adding both, we have:

\begin{matrix} θ G (x) + (1 - θ) H (x) = θ g_{0} + (1 - θ) h_{0} + θ g_{1} x + (1 - θ) h_{1} x + θ g_{2} x^{2} + (1 - θ) h_{2} x^{2} + \dots + θ g_{n} x^{n} + (1 - θ) h_{n} x^{n} \\ θ G (x) + (1 - θ) H (x) = (θ g_{0} + (1 - θ) h_{0}) + (θ g_{1} + (1 - θ) h_{1}) x + (θ g_{2} + (1 - θ) h_{2}) x^{2} + \dots + (θ g_{n} + (1 - θ) h_{n}) x^{n} \\ K (x) = θ G (x) + (1 - θ) H (x) = \underset{k_{0}}{\underset{⏟}{(θ g_{0} + (1 - θ) h_{0})}} + \underset{k_{1}}{\underset{⏟}{(θ g_{1} + (1 - θ) h_{1})}} x + \underset{k_{2}}{\underset{⏟}{(θ g_{2} + (1 - θ) h_{2})}} x^{2} + \dots + \underset{k_{n}}{\underset{⏟}{(θ g_{n} + (1 - θ) h_{n})}} x^{n} \end{matrix}

By restructuring the resultant polynomial, we have:

K (x) = k_{0} + k_{1} x + k_{2} x^{2} + \dots + k_{n} x^{n}

which is a new n-polynomial.

Until next time.