Jump to content

Using set theory to define the number of solutions of polynomials?


VXXV

Recommended Posts

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 Qn(x) be the inverse of an nth-degree polynomial. Precisely, Qn(x)=|R(x)/Pn(x)|. Here, it is of my interest to define a number, J(n), that provides the maximum number of singular points of  Qn(x) or the maximum number of solutions of Pn(x)=0 since (Qn(x))-1 equals Pn(x). To that end, I have tried to create the following definition: 
J(n)=Sup{\gamma((Qn(x))-1=0):dQ \leq n} in which \gamma((Qn(x))-1=0) denotes the number of solutions of Pn(x)=0 and d indicates the degree of Q. Based on the above, I ask:
1. Is the above definition correct?
2. In set notation, may I use such definition "\gamma((Qn(x))-1=0)" to denote the number of solutions of Pn(x)=0?
3. Is there anything else to consider to define J(n)?

 

 

Link to comment
Share on other sites

30 minutes ago, VXXV 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.
Let Qn(x) be the inverse of an nth-degree polynomial. Precisely, Qn(x)=|R(x)/Pn(x)|. Here, it is of my interest to define a number, J(n), that provides the maximum number of singular points of  Qn(x) or the maximum number of solutions of Pn(x)=0 since (Qn(x))-1 equals Pn(x). To that end, I have tried to create the following definition: 
J(n)=Sup{\gamma((Qn(x))-1=0):dQ \leq n} in which \gamma((Qn(x))-1=0) denotes the number of solutions of Pn(x)=0 and d indicates the degree of Q. Based on the above, I ask:
1. Is the above definition correct?
2. In set notation, may I use such definition "\gamma((Qn(x))-1=0)" to denote the number of solutions of Pn(x)=0?
3. Is there anything else to consider to define J(n)?

 

 

I thought that the number of solutions to the general polynomial is always exactly equal to the degree of the polynomial, which in turn is given by MAX(n).

There may, of course be repeated or complex solutions in this reckoning.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.