0

I'm taking an intro to numerics class and stumbled across the following practice problem:

Let $\Pi_n$ be the linear space of polynomials of degree less than or equal to $n\in \mathbb{N}_0$.
Show that every polynomial in $\Pi_n$ which has more than n complex roots (including their multiplicity) vanishes identically.

HINT: Make use of the fundamental theorem of algebra and use induction.

I have three questions here:

(A) Is this question asking me to show that such a polynomial is equal to $0$?

(B) If the answer to (A) is yes, why would $0$ be the polynomial with this property? This doesn't make sense to me.

(C) Also, I'm confused as to how such a polynomial could even exist at all/in $\Pi_n$ in the first place.

I would greatly appreciate any support here.

Prem
  • 14,696
l337n00b
  • 593
  • What doesn't make sense? That $0$ is a polynomial? That it has degree less than $n$? That it has more than $n$ roots? That it belongs to $\Pi_n$ in addition to belonging to some larger class of polynomials? – David K Oct 14 '24 at 12:05
  • There is a subtle difference between what you wrote in the title and the actual problem statement. It is the words "less than or." Did you miss (or fail to fully appreciate) those words when reading the problem statement, and is that the source of your confusion? – David K Oct 14 '24 at 12:09
  • This works over $\mathbb C$ but not $\mathbb Z/8\mathbb Z$: consider $x^2-1$ – J. W. Tanner Oct 14 '24 at 12:34
  • @DavidK My confusion stems from the question of how a polynomial with infinite roots can be in $\Pi_n$ if all polynomials in $\Pi_n$ can have at most n roots by the fundamental theorem of algebra. – l337n00b Oct 14 '24 at 12:47
  • 2
    If you're confused how to reconcile this exercise with the Fundamental Theorem of Algebra, try writing the theorem in the question exactly as you learned it and say how the theorem appears to be contradicted. – David K Oct 14 '24 at 13:50
  • The statement of the fundamental theorem of algebra excludes the zero polynomial – J. W. Tanner Oct 14 '24 at 14:00
  • I have updated my answer to match your confusion in your reply to @DavidK : Do let me know whether that works out. – Prem Oct 14 '24 at 14:10
  • 1
    @Prem thanks, i got it now! – l337n00b Oct 14 '24 at 16:51

1 Answers1

1

(C) How such a polynomial could even exist at all in $\Pi_n$ in the first place.

Definition we might generally use is $P(x)=\Sigma_{i=0}^{i=n} a_i x^x$ is Degree $n$ provided $a_n \not = 0$
Of course , we also include $P(x)=a_0$ to have Degree $0$ when $a_0 \not = 0$
There are certain authors who might additionally include $P(x)=0$ to have Degree $0$ or sometimes Degree $-\infty$

This then indicates that $\Pi_n$ (which is the Set of Polynomials with maximum Degree $n$) does include $P(x)=0$

(A) Is this question asking me to show that such a polynomial is equal to $0$ ?
YES

(B) If the answer to (A) is yes, why would $0$ be the polynomial with this property?
It is the only Polynomial with more than $n$ roots , it has infinite number of roots.

ADDENDUM 1 :

OP says via comment to question in response to user David K :
"My confusion stems from the question of how a polynomial with infinite roots can be in $\Pi_n$ if all polynomials in $\Pi_n$ can have at most $n$ roots by the fundamental theorem of algebra."

Confusion is resolved by this :
Definition of $\Pi_n$ is not "Set of Polynomials having at most $n$ roots".
Definition of $\Pi_n$ is "Set of Polynomials having at most $n$ Degree".
$P(x)=0$ is Degree at most $n$
$P(x)=0$ has more than $n$ roots , having infinite number of roots.

ADDENDUM 2 :

It is essential that $\Pi_n$ contain $P(x)=0$ , since it is the linear Space where we can add two Polynomials to generate a new Polynomial.
We can consider $\Pi_9$ where we could easily add $x$ to $-x$ , add $x^2+x+1$ to $-x^2-x-1$ , add $8x^8$ to $-8x^8$ , add $6x^6+2x+2$ to $-6x^6-2x-2$ , all of which will generate a new Polynomial $P(x)=0$ , which must occur in $\Pi_9$

Prem
  • 14,696