Talk:Degree of a polynomial

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[edit] Formal definition

I'm new to this Wikipedia thing, so I'll just post my opinion here before editing. Hope that someone with more experience could help.

The point is that there is no formal (mathematically correct) definition given. What you have is just an explanation/illustration.

Generally a polynomial is defind as a sequence of elements (called coefficients) from a ring indexed by natural numbers with the following characteristic: there is a natural number n (called the degree) so that the n-th coefficient is non-zero and all coefficients with index higher than are zero.

i.e. it is something like P={a0,a1,...,an,0,0,...} where a0,...an are ellements of R and 0 is the zero in R.

More formally: a polynomial is a function P from N (natural numbers) to R (where R is a ring) where there is n in N (the degree) so that P(n) is not zero and for all m>n P(m)=0.

The thing is that the definition of the degree is embedded in the definition of the polynomial. What would be the best way to include here the formal/mathematically correct definition of the degree - should we repeat the polynomial definition here? - AdamSmithee 22:25, 17 November 2005 (UTC)

[edit] Degrees 0 and 1

I've had a bit of a debate with someone regarding degrees 0 and 1 of a polynomial. My argument is this: the degree of 10x is 1, as the exponent on the variable is 1. But the degree of any constant term is 0. E.g: 5x0, as 0 is the exponent on x. To my surprise, they were arguing that 5 = 51, therefore the degree is 1. Surely the coefficients can't be used in determining the degree, otherwise the degree of 10x2 would be 3?! I'm not a mathematician so I'm a bit reluctant to change the article, but would it be reasonable to clarify this in the article? Or maybe I was wrong? --146.227.11.232 15:55, 13 January 2006 (UTC)

The degree = the highest power of X with a non-zero coeff. Indeed, the degree of the polynomial f = 5 = 5X0 is 0. The factorization of the coeff into primes (if available) doesn't change the degree of the polynomial AdamSmithee 09:07, 16 January 2006 (UTC)

[edit] fractional degree

degree of a function being "D", can |D|<1, while squreroots are 1/2 i'm not very sure, on this. but it isn't adressed in the artical

Thanks for asking. See Degree_of_a_polynomial#The_degree_computed_from_the_function_values. The degree of 3x1/2+5x is 1, while the degree of 3x1/2+17 is 1/2. I moved your question to the bottom of this discussion page where new contributions are expected to appear. You may sign your discussion entries automatically by four tildes '~'. Bo Jacoby 10:50, 9 August 2006 (UTC)

[edit] The degree of the zero polynomial is minus infinity

This is completely bogus. The degree of f(x) = 0 is 0. Ceroklis 20:17, 10 April 2007 (UTC)

The degree of f(x)=c is zero, but only if c is not 0. The degree of 0 is undefined (but many say that the degree is minus infinity so that it "sorts" below the constants). Here are a couple justifications for this statement:
  • In a Euclidean domain, the "norm" function (which is what "degree" is over the polynomial ring) is not defined at 0.
  • If f(x) has degree a and g(x) has degree b, we want the polynomial f(x)*g(x) to have degree a+b. Our polynomials should not "shrink" when we multiply them. If f(x)=0, then f(x)*g(x)=0, no matter what degree g(x) has. This will break many of our theorems. Therefore, we do not define the degree at 0. (This is a special case of the "norm" function for a euclidean domain which requires that deg(f*g)≥deg(f) and deg(f*g)≥deg(g).)
I'm not sure if this is enough reasoning for you... Does this make sense? - grubber 04:36, 11 April 2007 (UTC)


I does make sense, I have now figured it out. I guess the proper way to define this would go along those lines:
A polynomial over a ring R is a function P:\Bbb{N}\to{}R such that \{n\in\Bbb{N}|P(n)\neq0\} is finite.

  • Note that this definition, contrary to the one proposed by AdamSmithee above is correct in the sense that it doesn't exclude the zero polynomial.

The degree of a polynomial P is the highest n such that P(n)\neq{}0. Obviously this would then not be defined for the zero polynomial.
We should have these formal definitions somewhere, and then everything follows, that R[X] is a ring, that deg is a valuation, etc... Right now there are bits and pieces spread among the polynomial, degree of a polynomial and polynomial ring articles. Any idea on how to unify these articles ?
grubber: I have noted your new section on abstract algebra. This is a good idea but as said above it should be based on formal definitions and united with the other articles. In particular, the last sentence is misleading. deg(0) is not undefined because the norm is undefined at zero, it is undefined due to the way deg was defined. Deriving stuff in the proper order is important. Ceroklis 23:19, 11 April 2007 (UTC)

A ring is euclidean if there is some norm that satisfies the appropriate axioms. Finding such a norm is often very difficult, and there are often (perhaps always??) numerous norms whenever there is one (for example, if deg(f) is defined as we would expect and then we create a new function D, defined as D(f) = 42*deg(f), this is a valid norm). So, I say that degree is not defined at 0 because degree is a norm and norms are not defined at 0. But, it is something of a circular argument, so there probably is a better way to state it.
As far as your way of defining a polynomial, I'm not following. As I read your definition, if we let R be the real numbers, then how is exp(x) not a polynomial by your definition? - grubber 22:17, 12 April 2007 (UTC)
Your are confused by the distinction between a polynomial (an abstract object) and a polynomial function.
In my definition P(n) gives the coefficient of the nth power of the formal variable X.
To a polynomial P over R you associate a polynomial function f_P:R\to{}R defined by f_P(x)=\sum_{n\in\Bbb{N}|P(n)\neq{}0}{P(n)x^n}. We often identify P and fP because on \Bbb{R} or \Bbb{C}, the mapping P\mapsto{}f_P is injective, but it is important to understand that they are different objects. exp:\Bbb{R}\to{}\Bbb{R} is not a polynomial function, because there is no polynomial P over \Bbb{R} with fP = exp.
Another thing that may confuse you is that I define the polynomial as a function, instead of as a sequence, but that is equivalent. I find this way of doing things more elegant because you don't have bounds on the size of the sequence appearing anywhere. Ceroklis 00:51, 13 April 2007 (UTC)
Aha.. I thought the function P was evaluating the polynomial in the ring, and I didn't understand what N was (now I see it). OK, now that I understand the setup, notice that even in your definition, for the zero polynomial, "the highest n such that P(n)\neq{}0" is not defined, because \{n\in{}N|P(n)\neq0\} is an empty set. What would the maximum (or even supremum) of an empty set look like? You could make a theorem that if \{n\in{}N|P_1(n)\neq0\} \subseteq \{n\in{}N|P_2(n)\neq0\}, then deg(P_1)\in\{n\in{}N|P_2(n)\neq0\} and if P1 represents the zero polynomial, we can conclude that the degree of the zero polynomial is any arbitrary natural number (with the proper choice of P2). The joys of vacuously true statements on the empty set! I think the only way to resolve this conflict is to set the degree of the polynomial zero to be either "undefined" or explicitly defined as some negative number (since the empty set is contained in every one of these degree "sets", it must be strictly smaller than all of them). - grubber 16:39, 13 April 2007 (UTC)
Perhaps you should read more carefully; I wrote above: "Obviously this would then not be defined for the zero polynomial".Ceroklis 18:06, 13 April 2007 (UTC)
So you did, my mistake. So, I guess we really are in agreement on all this? - grubber 19:00, 13 April 2007 (UTC)
Yes. The question is now how to best integrate these formal definitions (or equivalent) in the various articles I mentioned. Ceroklis 21:49, 13 April 2007 (UTC)

[edit] Degree of i

i is the squart root of -1. So can someone explain what degree i and its factors are? --pizza1512 Talk Autograph 20:02, 13 April 2007 (UTC)

Over the complex, i is just a non-zero constant, like 2 for example. i.e. i = iz0 + 0z1 + 0z2 + .... So deg(i) = 0. Furthermore, it is already factored. Ceroklis 21:49, 13 April 2007 (UTC)