[Math] Is $f(x)=0$ a polynomial function

functionspolynomials

Is $f(x)=0$ a polynomial function?

we know that constant functions are polynomials of degree zero

But, does $f(x)=0$ follow this definition?

Best Answer

$f(x)=0$ is a polynomial function, with degree $-\infty$ (by convention).

In this way, $\deg(fg) = \deg(f) + \deg(g)$ and $\deg(f+g) \leq \max(\deg (f), \deg (g))$ are true for any polynomials $f$ and $g$.

Related Solutions

[Math] Why isn’t $\frac{1}{x}$ a polynomial

If you want a more formal "proof", you can suppose for contradiction that $1/x$ is in fact equal to some expression $a_k x^n$ of the form $ \sum_{k=0}^n a_k x^n $:

$$ \frac{1}{x} = a_0 + a_1 x + \cdots + a_n x^n$$

multiplying through by $x$ gives:

$$ 1 = a_0 x + a_1 x^2 + \cdots + a_n x^{n+1}$$

Setting $x=0$ gives: $$ 1 = a_0 \cdot 0 + \cdots + a_n \cdot 0^{n+1} = 0$$

a contradiction, so we must have that $1/x$ is not a polynomial.

Map from polynomial to polynomial function

It may help to think of the polynomial ring $R[X]$ differently, at least for a short moment, as the group $\bigoplus_{n\in\Bbb{N}}R$ with multiplication defined by $$(c_k)_{k\in\Bbb{N}}\cdot(d_k)_{k\in\Bbb{N}}=\left(\sum_{j=0}^kc_jd_{k-j}\right)_{k\in\Bbb{N}}.$$ Here of course the sequences $(c_k)_{k\in\Bbb{N}}$ and $(d_k)_{k\in\Bbb{N}}$ correspond to the polynomials $$\sum_{k\in\Bbb{N}}c_kX^k \qquad\text{ and }\qquad \sum_{k\in\Bbb{N}}d_kX^k,$$ where the sums are finite by definition of the direct product. The newly defined product of the sequences above then indeed corresponds to the product of these polynomials, so this ring is isomorphic to $R[X]$. In this ring the powers of the indeterminate $X$ correspond to the standard basis elements of the direct sum. They are in no way functions from $R$ to $R$.

Now every such sequence does define a function $R\ \longrightarrow\ R$ by substitution, i.e. by plugging in the elements of $R$. In this way the sequence $(c_k)_{k\in\Bbb{N}}\in R[X]$ defines the function $$R\ \longrightarrow\ R:\ r\ \longmapsto\ \sum_{k\in\Bbb{N}}c_kr^k.$$

Of course the same ideas work for the polynomial ring in $n$ indeterminates, by repeating this construction $n$ times.

Best Answer

Related Solutions

[Math] Why isn’t $\frac{1}{x}$ a polynomial

Map from polynomial to polynomial function

Related Question