Can there be a polynomial with infinite solutions? If so, would it be a polynomial? Also, would it have an infinite solution set? Let's find out.
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\journal{Journal of Mathematical Stupidity}
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\title{Polynomials with infinite solutions}
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\author{Rahul Chhabra}
\address{St. Joseph's College, Prayagraj, Uttar Pradesh, India}
\begin{abstract}
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A solution to any polynomial $P(x)$ is a value of $x$ that satisfies $P(x) = 0$ . A polynomial of degree 1 forms an equation called a "linear" equation. A linear equation can be expressed as $ax + b = 0$. Its solution is then :
\[
x = \frac{-b}{a}
\]
For polynomials of degree 2, things go a little complicated :
\[
P(x) = ax^2 + bx + c
\]
Nonetheless, a quadratic equation can be solved rather easily with the help of the quadratic formula :
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Cubics and quartics each have 3 and 4 roots respectively.
Overall, any polynomial of degree $n$ has exactly $n$ solutions or roots.
But can there be a polynomial equation which has infinite solutions?
As it turns out - yes.
\end{abstract}
\begin{keyword}
Polynomial \sep solution \sep degree of a polynomial \sep linear
\sep quadratic \sep quadratic formula \sep roots \sep cubics \sep quartics
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\section{The null polynomial}
In general, any polynomial can be expressed as :
\[
P(x) = a_0 + a_1x + a_2x^2 + ... + a_ix^i
\]
The null polynomial is a polynomial that returns 0 for any value of $x$.
$P(x) = 0$
It can be understood better as a polynomial with every coefficient equal to 0:
\[
P(x) = 0 + 0x + 0x^2 + ... + 0x^i
\]
It will return 0 no matter what. Since it returns 0 for any given value of $x$, it has infinite solutions.\cite{1137202}
\label{S:1}
\section{A polynomial of degree infinity}
Most of the time when we consider polynomials, we consider the degree to be finite.
However, there are instances when a polynomial can be thought of as having an infinite degree. Like a power series.
One prime example is of a Taylor series.\cite{1137203}
\[
\sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{n!} (x-a)^n
\]
A Taylor series is defined to be infinite.
It is called a Maclaurin series when $a = 0$.
\label{S:2}
\section{Roots of a degree infinity polynomial}
By the fundamental theorem of algebra, we know that a degree $n$ polynomial's equation has $n$ roots.
Therefore, a polynomial of degree $\infty$ has $\infty$ solutions.
Given a polynomial $P(x)$, let us assume that it has a solution set $S$.
Let us also assume that the solution set is finite, i.e., there are a finite number of solutions for $P(x)$.
Now, let us take $n$ to be the largest root in $S$.
Since it is a root,
\[
P(n) = 0
\]
\begin{equation} \label{n_root}
a_0 + a_1n + a_2n^2 + ... = 0
\end{equation}
Let us also take any other arbitrary $m$.
Now, let us see if $m + n$ is a root of $P(x)$.
\begin{align*}
a_0 + a_1(n + m) + a_2(n + m)^2 + ... &= 0\\
a_0 + a_1n + a_1m + a_2(n^2 + 2mn + m^2) + ... &= 0\\
a_0 + a_1n + a_1m + a_2n^2 + a_2m^2 + 2a_2mn + ... &= 0
\end{align*}
We see that simplifying produces $a_in^i + a_im^i$ along with all the intermediate terms of a bionomial expansion of the form $(a \pm b)^n$.
Let the sum of all these intermediate terms be $T_{m + n}$.
Now,
\begin{align*}
a_1n + a_1m + a_2n^2 + a_2m^2 + ... + T_{m+n} &= 0\\
(a_1n + a_2n^2 + ...) + (a_1m + a_2m^2 + ...) + T_{m+n} &= 0
\end{align*}
But from \ref{n_root},
\begin{equation} \label{mn_root}
(a_1m + a_2m^2 + ...) + T_{m+n} = 0
\end{equation}
Since it is possible for the expression \ref{mn_root},
\[P(m+n)=0\]
But $n$ is by definition, the largest root, then, by contradiction, the solution set $S$ has to be infinite.
Therefore, in this case,there are infinite solutions to the polynomial $P(x)$.
But what if there is no value of $m$ that satisfies $T_{m+n}=0$? Then, in that case, the solution set can be thought of as not having as many elements as the degree of the polynomial.
Just like how
\[
x^2 + 1 - 2x = 0
\]
Has one solution $x=1$.
The solution set for this quadratic would be $\{1\}$.
Here, there are two roots - but they are both equal. The case when $m$ does not exist would also be similar.
In other words, the infinity of the solution set ($\infty_{S}$) would be smaller than or equal to the infinity of the degree of the polynomial ($\infty_{P}$), for any given infinite polynomial.
Or,
\begin{equation}
\infty_{S} \leqslant \infty_{P}
\end{equation}
\section{Such equations in action}
Here are a few examples :
\[
\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!} x^{2n+1}
\]
This is the Taylor series for the $sin(x)$ function. It is how your calculator get the value when you feed $x$ to it.
\section*{A note on other fields}
A field is a set on which the binary operations $+$, $-$, $\times$ and $\div$ are defined.
Besides integral fields, there are other fields where polynomials will behave differently and a polynomial with finite terms and of a finite degree can also have infinite solutions.
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