I May I use set theory to define the number of solutions of polynomials?

[V9999]

TL;DR Summary

Here, I present a few silly doubts on how to define the maximum number of solutions of a polynomial using set notation and theory.

Let $Q_{n}(x)$ be the inverse of an nth-degree polynomial. Precisely,

$$Q_{n}(x)=\displaystyle\frac{1}{P_{n}(x)}$$,

It is of my interest to use the set notation to formally define a number, $J_{n}$ that provides the maximum number of solutions of $Q_{n}(x)^{-1}=0$. Despite not knowing how to proceed, below you may find my attempt.

Let the maximum number of solutions of $Q^{-1}_{n}(x)=0$ be

$$J_{n}=\text{Sup}\{\pi(Q^{-1}_{n}(x)=0):\partial Q_{n}^{-1}(x) \leq n\}$$,

in which $\partial$ denotes "the degree of" and $\pi(Q^{-1}_{n}(x)=0)$ is the number of solutions of $Q^{-1}_{n}(x)=0$.

Based on the above, I ask:

1. Is the above definition correct?
2. How may improve and formally define $J_{n}$ using proper notation of set theory and mathematics? That is to say, is there anything else that I should consider to define $J_{n}$ in the way stated above?
3. In order to define the degree of a polynomial, should I consider $n \in \mathbb{N}$ or $n \in \mathbb{Z}$?

Thanks in advance.

 

[Office_Shredder]

Why are we even talking about Q instead of P?

 

[martinbn]

Why not just write that $J_n=n$?

 

[WWGD]

$\frac {1}{P_n(x)}$ here, will never be $0$ on the Reals, if I understood you correctly, except possibly in the limit, when $P_n$ grows without bound. Notice you first wrote $Q_n(x)^{-1}$ and then you used $Q_n^{-1}(x)$.
For your second question, the degree of a polynomial may be 0 or a Natural number; never a negative Integer.

 

[martinbn]

$\frac {1}{P_n(x)}$ here, will never be $0$ on the Reals, if I understood you correctly, except possibly in the limit, when $P_n$ grows without bound. Notice you first wrote $Q_n(x)^{-1}$ and then you used $Q_n^{-1}(x)$.
For your second question, the degree of a polynomial may be 0 or a Natural number; never a negative Integer.

In his notations $Q^{-1}$ is just $P$, uless he means inverse function, which i doubt because it wouldnt exist.

 

[V9999]

Why are we even talking about Q instead of P?

Hi, Office_Shredder. I hope you are doing well.
First, thanks for commenting. Second, I could be P since $(Q_{n}(x))^{-1}$ is $P_{n}$. In light of the foregoing, the definition would be $Sup\{\pi(P_{n}(x)=0):\partial P \leq n\}$. Based on the above, is there anything else that I should consider to define $J_{n}$? Thanks in advance.

 

[V9999]

$\frac {1}{P_n(x)}$ here, will never be $0$ on the Reals, if I understood you correctly, except possibly in the limit, when $P_n$ grows without bound. Notice you first wrote $Q_n(x)^{-1}$ and then you used $Q_n^{-1}(x)$.
For your second question, the degree of a polynomial may be 0 or a Natural number; never a negative Integer.

Hi, WWGD. I hope you are doing well.
Thanks for commenting. In my definition stated above, I am interested in the singular points of $Q_{n}(x)$, which are obtained by the zeros or solutions of $P_{n}=0$. In as much as $(Q_{n}(x))^{-1}=P_{n}$, then I have considered $(Q_{n}(x))^{-1}=P_{n}=0$. Is that incorrect? Thanks again.

 

[V9999]

Why not just write that $J_n=n$?

Hi, martinbn, I hope you are doing well. Thanks for commenting.
It could be $n$. However, I would prefer $J_{n}$ rather than simply $n$.

 

[martinbn]

Hi, martinbn, I hope you are doing well. Thanks for commenting.
It could be $n$. However, I would prefer $J_{n}$ rather than simply $n$.

Why!? It is the maximal number of roots a polynomial of degree $n$ can have, which is $n$.

 

[Mark44]

Let
$Q_{n}(x)$ be the inverse of an nth-degree polynomial. Precisely,

$$Q_{n}(x)=\displaystyle\frac{1}{P_{n}(x)}$$,

Your definition for $Q_n(x)$ would be the reciprocal of $P_n(x)$, not the inverse.

For example, if $f(x) = 2x + 3$, then $f^{-1}(x) = \frac 1 2(x - 3)$. Note that the reciprocal of f would be $\frac 1 {2x + 3} \ne \frac 1 2 (x - 3)$.

The operation involved in a function and its inverse is function composition. The operation involved in a function and its reciprocal is multiplication.

 

[WWGD]

Your definition for $Q_n(x)$ would be the reciprocal of $P_n(x)$, not the inverse.

For example, if $f(x) = 2x + 3$, then $f^{-1}(x) = \frac 1 2(x - 3)$. Note that the reciprocal of f would be $\frac 1 {2x + 3} \ne \frac 1 2 (x - 3)$.

The operation involved in a function and its inverse is function composition. The operation involved in a function and its reciprocal is multiplication.

And 0s would be singularities. Op's framing seemed a bit confusing to me.

 

[V9999]

Your definition for $Q_n(x)$ would be the reciprocal of $P_n(x)$, not the inverse.

For example, if $f(x) = 2x + 3$, then $f^{-1}(x) = \frac 1 2(x - 3)$. Note that the reciprocal of f would be $\frac 1 {2x + 3} \ne \frac 1 2 (x - 3)$.

The operation involved in a function and its inverse is function composition. The operation involved in a function and its reciprocal is multiplication.

Hello, Mark44. I hope you are doing well!
Thanks a lot for your insightful comment. That is exactly what I was thinking. However, how may I mathematically define the "reciprocal" of $Q_n(x)$?
That is to say, is there a specific notation to define the reciprocal of $Q_n(x)$? Thanks in advance, and my apologies for the delay.

 

[Mark44]

Hello, Mark44. I hope you are doing well!
Thanks a lot for your insightful comment. That is exactly what I was thinking. However, how may I mathematically define the "reciprocal" of $Q_n(x)$?
That is to say, is there a specific notation to define the reciprocal of $Q_n(x)$?

How about $\frac 1 {Q_n(x)}$?

 

[Mark44]

It's not clear to me what you are trying to do.
In post #1 you gave this definition:

$J_{n}=\text{Sup}\{\pi(Q^{-1}_{n}(x)=0):\partial Q_{n}^{-1}(x) \leq n\}$

But since, after some time, we learned that $Q_n(x)$ is just the reciprocal (or multiplicative inverse) of $P_n(x)$, the above could be rewritten as $J_{n}=\text{Sup}\{\pi(P_{n}(x)=0):\partial (P_{n}(x) \leq n\}$

Some comments:
1. It's good that you defined what you mean by $\partial(P_n(x))$, because I haven't seen that symbol used to mean "number of solutions of". In any case, as pointed out by @martinbn in post #9, a polynomial of degree n has n solutions.
2. What does the notation $\pi(P_n(x))$ mean?
3. What are you trying to do? In your summary, you said

Here, I present a few silly doubts on how to define the maximum number of solutions of a polynomial using set notation and theory.

As noted above, a polynomial of degree n has n roots, some of which might be complex. It seems to me that you are trying to obsccure a very simple idea with complicated set-theoretic notation.

Please clarify what you are trying to do, and why you want to do this.